tag:blogger.com,1999:blog-32146486079968395292019-11-16T19:01:10.996+00:00The Winding NumberInteresting insights on math.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.comBlogger68125tag:blogger.com,1999:blog-3214648607996839529.post-15615412938537077922019-11-07T02:02:00.000+00:002019-11-07T16:25:11.202+00:00Financial derivatives, payoff functions and portfolios: motivationWhen I first saw the definitions of several financial assets, I found them completely arbitrary -- it's not that I didn't get the reason one would have them, but rather that I saw no way to immediately understand them or a starting point for reasoning about them mathematically. Other than what was perhaps the most basic asset -- stocks (and also bonds, etc., which are really the "same" idea -- I prefer to think of stocks and bonds as being on a continuum based on riskiness) and their baskets -- all the derivatives (and things that aren't called derivatives) based on them seemed really artificial in their construction.<br /><br />But this isn't exactly unfamiliar territory, is it? You've seen unmotivated definitions in mathematics, and you've seen that you need to put in quite a bit of effort to really motivate them and understand why they make perfect sense -- you've seen that, e.g. <a href="https://thewindingnumber.blogspot.com/">here</a>.<br /><br />So let's do the same thing with finance.<br /><br /><hr /><br />Let's start with a simple one: <b>shorting</b>. <br /><br />There is a certain asymmetry in the definitions of longing and shorting, isn't there? It's the "borrowing a stock" part of the definition of shorting that introduces this asymmetry.<br /><br />But if you've spent any time thinking about economics, the idea of borrowing something you don't have should be familiar -- it's what you do when you don't have any investment capital to start with, but you think you can grow the value of what you've borrowed by e.g. investing it in a stock. Let's phrase this in a slightly different (and by "slightly different", I mean "take the buying-selling dual of") way:<br /><br /><b>How to invest in a stock without money at hand:</b>&nbsp;Borrow some money, immediately "sell" the money for some stocks -- after some time has passed, "buy" back the money by returning the stocks. If the value of the stocks have increased, you'll get more money in return and be able to repay the loan.<br /><br />This is <i>precisely</i>&nbsp;symmetric to the situation of shorting -- <b>longing an asset</b>&nbsp;just means <i>shorting money</i>&nbsp;-- or more precisely, <b>shorting the rest of the market</b>.<br /><br />The apparent asymmetry between longing and shorting comes back from the fact that you are much more likely to already own some of "the rest of the market" than to own a particular stock -- for example, the unbounded losses of shorting arise from the fact that it's much easier for a single stock's value to skyrocket than for money's -- so in longing, there may still be ways for you to earn the money to repay it even if the value of the stock drops, i.e. the value of your other assets (e.g. your labour or property) relative to money would not have dropped.<br /><br />One advantage of this approach is that it is conceptually interesting -- and will hopefully allow us to transfer insights and ideas between stocks and shorts (except when certain approximations may be involved) -- another is that it immediately nullifies "moral" criticism of shorting, from e.g. Elon Musk, as it is really just the same as investing in the "rest of the market".<br /><br /><div class="twn-pitfall">Wait a minute -- but what if you actually just invested in "the rest of the market"? That would clearly have a much lower return than shorting the stock directly, right? Except you're thinking about investing in the rest of the market by paying money, not by paying the stock you're betting against -- that's a bet for the rest of the market against money, not against said stock.</div><br /><hr /><br />Well, shorting was an example where we wanted to bet that the price of an asset goes down. But in general, we may have any sort of weird prediction on the price of an asset -- maybe that it will "fluctuate a lot", or that it "won't exceed a certain level", or that it "will go up but only to a point", or that it "will reach a certain range". You may have any sort of elaborate <i>probability distribution</i>&nbsp;$\rho(x)$ on the value $x$ of the asset after a period of time. Given such a distribution, what you'd want to do (<b>ignoring risk</b>) is to maximise your expected return (minus the cost of the transaction and of buying the contract, of course):<br /><br />$$\chi=\int {\rho (x)f(x)dx}$$<br />Where $f(x)$ is the payoff you get if the asset reaches the price $x$ -- this is called the <b>payoff function</b>.<br /><br />Well, why not just take $f(x)$ to be arbitrarily high? Because the contract will be really expensive, of course. How expensive? Predicting that would require:<br /><ul><li>not only the $\rho$ distribution on this asset as believed by each seller and buyer in the market</li><li>but also the amount of capital they have and their beliefs about the future behavior of other assets in the market contracts on which they could buy instead</li></ul><div>And that is still not to mention the fact that people do not maximise the expected value of profit per say, but have varying levels of risk aversion.<br /><br />But that's alright -- we don't need to predict that. That price is crunched for us by the market and is the market price of the contract -- it is the <b>market price</b>. What's more important is to estimate $\chi = E_{\rho}[f(x)]$. Well, in fact, if we're concerned with <b>risk</b>, then we'd also be interested in the variance of the distribution -- and in general, an individual may also have a skewness or kurtosis preference (an example of a kurtosis preference would be among gamblers, who want heavy tails for the "big win").<br /><br />In fact, $\chi$ can depend on multiple underlying assets:<br /><br />$$\chi=E_\rho[f(\mathbf{x})]$$<br />Where $\mathbf{x}$ is the vector of prices of each underlying asset. In fact, this multivariate $f$ can represent your entire <b>portfolio</b> of derivatives on assets. If $f(\mathbf{x})$ can be written as a sum of functions of each component, this can be considered as some number of separate univariate derivatives -- the reason such a portfolio is still useful is that of risk management, if we use a $\rho$ that has some correlations.<br /><br /><div class="twn-pitfall">There is an alternative definition of the payoff function, where it is $f(x)$ minus the cost of purchasing it, i.e. a profit/loss function. This is actually a better way of thinking about things (to see why, try to find the payoff function for a short) -- the only problem is that not every function can be a profit/loss function.</div><br /><hr /><br />Get some practice constructing various financial derivatives, i.e. constructing derivatives that have a given payoff function (using the first definition). I'll do one:<br /><br />$$f(x)=(a-x)I(x&lt;a)$$<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-3s_6o7haZyo/XcN1Ut7NCDI/AAAAAAAAF1w/QZaoPDdnOjEEs1A8fhGATkNot3K8gC9AQCLcBGAsYHQ/s1600/put.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="896" data-original-width="720" height="320" src="https://1.bp.blogspot.com/-3s_6o7haZyo/XcN1Ut7NCDI/AAAAAAAAF1w/QZaoPDdnOjEEs1A8fhGATkNot3K8gC9AQCLcBGAsYHQ/s320/put.png" width="257" /></a></div>Such a function would be a useful alternative to shorting, as it doesn't allow arbitrary losses.<br /><br />The whole discontinuity of the function really suggests to me a fundamental change in behaviour at the point $x=a$ -- like you just don't make the trade if $x\ge a$. This decision can only be made once the final price is discovered, so you must have bought a contract that gave you the <i>option</i>&nbsp;to make a transaction: that transaction must be <i>selling</i>, it must be executed after the price is realised, but it must be at price $a$, which is initially fixed.<br /><br />This is called a <b>put&nbsp;option</b>&nbsp;-- you buy the <i>option</i> to sell a stock (that you buy at the time of exercising the option) at a pre-decided price.</div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-35216312825555588562019-10-17T01:57:00.000+01:002019-10-17T01:57:45.759+01:00Sigma fields are Venn diagramsThe starting point for probability theory will be to note the difference between <b>outcomes</b>&nbsp;and <b>events</b>.<br /><br />An <b>outcome</b>&nbsp;of an experiment is a fundamentally non-empirical notion, about our theoretical understanding of what states a system may be in -- it is, in a sense, analogous to the "microstates" of statistical physics. The set of all <i>outcomes</i>&nbsp;$x$ is called the <b>sample space</b>&nbsp;$X$, and is the fundamental space to which we will give a probabilistic structure (we will see what this means).<br /><br />Our actual observations, the events, need not be so precise -- for example, our measurement device may not actually measure the exact sequence of heads and tails as the result of an experiment, but only the total number of heads, or something -- analogous to a "macrostate". But these measurements <i>are</i>&nbsp;statements about what microstates we know are possible for our system to be in -- i.e. they correspond to sets of outcomes. These sets of outcomes that we can "talk about" are called <b>events</b>&nbsp;$E$, and the set of all possible events is called a <b>field</b>&nbsp;$\mathcal{F}\subseteq 2^X$.<br /><br />For instance: if our sample space is $\{1,2,3,4,5,6\}$ and our measurement apparatus is a guy who looks at the reading and tells us if it's even or odd, then the field is $\{\varnothing, \{1,3,5\},\{2,4,6\},X\}$. We simply <i>cannot</i>&nbsp;talk about sets like $\{1,3\}$ or $\{1\}$. Our information just doesn't tell us anything about sets like that -- when we're told "odd", we're never hinted if the outcome was 1 or 3 or 5, so we can't even have prior probabilities -- we can't even give probabilities to whether a measurement was a 1 or a 3.<br /><br />Well, what kind of properties characterise a field? There's actually a bit of ambiguity in this -- it's clear that a field should be closed under <b>negation and <i>finite</i>&nbsp;unions</b> (and finite intersections follow via de Morgan) -- if you can talk about whether $P_1$ and $P_2$ are true, you can check each of them to decide if $P_1\lor P_2$ is true (and since a proposition $P$ corresponds to a set $S$ in the sense that $P$ says "one of the outcomes in $S$ is true", $\lor$ translates to $\cup$). But if you have an infinite number of $P_i$'s, can you really check each one of them so that you can say without a doubt that a field is closed under arbitrary union?<br /><br />Well, this is (at this point) really a matter of convention, but we tend to choose the convention where the field is closed under <b>negation and <i>countable</i>&nbsp;unions</b>. Such a field is called a <b>sigma-field</b>. We will actually see where this convention comes from (and why it is actually important) when we define probability -- in fact, it is required for the idea that one may have a uniform probability distribution on a compact set in $\mathbb{R}^n$.<br /><br /><hr /><br />A beautiful way to understand fields and sigma fields is in terms of venn diagrams -- in fact, as you will see, <b>fields are precisely a formalisation of Venn diagrams</b>. I was pretty amazed when I discovered this (rather simple) connection for myself, and you should be too.<br /><br />Suppose your experiment is to toss three coins, and make "partial measurements" on the results through three "measurement devices":<br /><ul><li><b>A:</b>&nbsp;Lights up iff the number of heads was at least 2.</li><li><b>B:</b>&nbsp;Lights up iff the first two coins landed heads.</li><li><b>C:</b>&nbsp;Lights up iff the third coin landed heads.</li></ul>What this means is that $A$ gives you the set $\{HHT, HTH, THH, HHH\}$, $B$ gives you the set $\{HHH, HHT\}$, $C$ gives you the set $\{HHH, HTH, THH, TTH\}$. Based on precisely which devices light up, you can decide the truth values of $\lnot$'s and $\lor$'s of these statements, i.e. complements and unions of these sets -- this is the point of fields, of course.<br /><br />Or we could visualise things.<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-_GSgU8wZr3o/XaXXGandEiI/AAAAAAAAFxI/DgbNQDmdnMgiiFFaxcM9_x8h37s0F5nVQCEwYBhgL/s1600/venn%2Bsigma.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="588" data-original-width="948" height="247" src="https://1.bp.blogspot.com/-_GSgU8wZr3o/XaXXGandEiI/AAAAAAAAFxI/DgbNQDmdnMgiiFFaxcM9_x8h37s0F5nVQCEwYBhgL/s400/venn%2Bsigma.png" width="400" /></a></div>Well, the Venn diagram produces a partition of $X$ corresponding to the equivalence relation of "indistinguishability", i.e. "every event containing one outcome contains the other"? The <i>field</i>&nbsp;consists precisely of any set one can "mark" on the Venn diagram -- i.e. unions of the elements of the partition.<br /><br />A consequence of this becomes immediately obvious:<br /><br /><b>Given a field $\mathcal{F}$ corresponding to the partition $\sim$, the following bijection holds: $\mathcal{F}\leftrightarrow 2^{X/\sim}$.</b><br /><br />Consequences of this include: the cardinalities of finite sigma fields are precisely the powers of two; there is no countably infinite finite field.<br /><br /><hr /><br />Often, one may want to some raw data from an experiment to obtain some processed data. For example, let $X=\{HH,HT,TH,TT\}$ and the initial measurement is of the number of heads:<br /><br />\begin{align}<br />\mathcal{F}=&amp;\{\varnothing, \{TT\}, \{HT, TH\}, \{HH\},\\<br />&amp; \{TT, HT, TH\}, \{TT, HH\}, \{HT, TH, HH\}, X \}<br />\end{align}<br />What kind of properties of the outcome can we talk about with certainty given the number of heads? For example, we can talk about the question "was there at least one heads?"<br /><br />$$\mathcal{G}=\{\varnothing, \{TT\}, \{HT, TH, HH\}, X\}$$<br />There are two ways to understand this "processing" or "re-measuring". One is as a function $f:\frac{X}{\sim_\mathcal{F}}\to \frac{X}{\sim_\mathcal{G}}$. Recall that:<br /><br />\begin{align}<br />\frac{X}{\sim_\mathcal{F}}&amp;=\{\{TT\},\{HT,TH\},\{HH\}\}\\<br />\frac{X}{\sim_\mathcal{G}}&amp;=\{\{TT\},\{HT,TH,HH\}\}<br />\end{align}<br />Any such $f$ is a permissible "<b>measurable function</b>", as long as $\sim_\mathcal{G}$ is at least as coarse a partition as $\sim_\mathcal{F}$. In other words, a function from $X/\sim_1$ to $(X/\sim_1)/\sim_2$ is always measurable.<br /><br />But there's another, more "natural", less weird and mathematical way to think about a re-measurement -- as a function $f:X\to Y$, where in this case $Y=\{0,1\}$ where an outcome maps to 1 if it has at least one heads, and 0 if it does not.<br /><br />But there's a catch: knowing that an event $E_Y$ in $Y$ occurred is equivalent to knowing that <i>an</i> outcome in $X$ mapping to $E_Y$ occurred -- i.e. that the event $\{x\in X\mid f(x)\in Y\}$ occurred. Such an event must be in the field on $X$, i.e.<br /><br />$$\forall y\in\mathcal{F}_Y,f^{-1}(y)\in\mathcal{F}_X$$<br />This is the condition for a <b>measurable function</b>, also known as a <b>random variable</b>.<br /><br /><hr /><br />One may observe certain analogies between the measurable spaces outlined above, and topology -- in the case of countable sample spaces, there actually is a correspondence. The similarity between a Venn diagram and casual drawings of a topological space is not completely superficial.<br /><br />The key idea behind fields is mathematically a notion of "distinguishability" -- if all we can measure is the number of heads, $HHTTH$ and $TTHHH$ are identical to us. For all practical purposes, we can view the sample space as the partition by this equivalence relation. They are basically the "same point".<br /><br />It's this notion that a <b>measurable function</b> seeks to encapsulate -- it is, in a sense, a&nbsp;<b>generalisation of a function</b> from set theory. A function <b>cannot distinguish indistinguishable points</b>&nbsp;-- in set theory, "indistinguishability" is just equality, the discrete partition; a measurable function&nbsp;<b>cannot distinguish indistinguishable points</b> -- but in measurable spaces, "indistinguishability" is given by some equivalence relation.<br /><br />Let's see this more precisely.<br /><br />Given sets with equivalence relations $(X,\sim)$, $(Y,\sim)$, we want to ensure that some function $f:X\to Y$ "lifts" to a function $f:\frac{X}{\sim}\to\frac{Y}{\sim}$ such that $f([x])=[f(y)]$. <br /><br /><b>(Exercise:</b>&nbsp;Show that this (i.e. this "definition" being well-defined) is equivalent to the condition $\forall E\in\mathcal{F}_Y, f^{-1}(E)\in \mathcal{F}_X$. It may help to draw out some examples.)<br /><br />Well, this expression of the condition -- as $f([x])=[f(y)]$ -- even if technically misleading (the two $f$'s aren't really the same thing) give us the interpretation that a measurable function is one that <i>commutes with the partition</i>&nbsp;or&nbsp;<i>preserves the partition</i>.<br /><br />While homomorphisms in other settings than measurable spaces do not precisely follow the "cannot distinguish related points" notion, they do follow a generalisation where equivalence relations are replaced with other relations, operations, etc. -- in topology, a continuous function preserves limits; in group theory, a group homomorphism preserves the group operation; in linear algebra, a linear transformation preserves linear combinations; in order theory, an increasing function preserves order, etc. In any case, a homomorphism is a function that does not "break" relationships by creating a "finer" relationship on the target space.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-73594883711501591652019-10-02T00:05:00.001+01:002019-10-02T00:05:47.153+01:00The Killing form; factorising non-Abelian Lie groupsIt could be fun to try and define a "dot product" on a Lie algebra.<br /><br />You know, you might've already realised that the cross product is a Lie bracket of sorts -- you know, given its antisymmetry and the whole $a^\mu b^\nu - a^\nu b^\mu$ representation of the wedge product and all that. It's a short exercise to verify that the Lie algebra $\mathfrak{so}(3)$ of $SO(3)$ is the algebra of skew-symmetric matrices, and with the Lie bracket $XY-YX$ is isomorphic to $\mathbb{R}^3$ with the cross product.<br /><br />Well, the dot product on $\mathbb{R}^3$ has an interesting connection to $SO(3)$ -- it is precisely the form that is invariant under the action of $SO(3)$. Well, but that's $SO(3)$ acting on $\mathbb{R}^3$ -- what is that action in the notation of $\mathfrak{so}(3)$? As it turns out (and you can work this out), it is <b>precisely the adjoint map</b> $\mathrm{Ad}_gX:=gXg^{-1}$ which corresponds to this "rotating $X$ by $g$". It's not really that unexpected, if you ask me -- conjugation is always the natural way to transform matrices in linear algebra when vectors are multiplied on the left.<br /><br />So the "dot product" is an $\mathrm{Ad}$-invariant bilinear form. In fact, adding a symmetricity requirement allows us to just bother with norms (as a symmetric inner product can be determined from the norm, through the cosine rule). Conjugation basically allows you to determine the "<b>contours</b>" of this norm or inner product. The question is: can we determine the bilinear form -- up to scaling -- just from "<b>$\mathrm{Ad}$-invariant symmetric bilinear form</b>"&nbsp;alone?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-5hjfT6Pw29w/XYyu1CJnKtI/AAAAAAAAFvQ/Fq0LC1Bgqv0QBcW8I3_5DabJkyLOX254QCLcBGAsYHQ/s1600/conjugation%2Bcontour.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="461" data-original-width="601" height="244" src="https://1.bp.blogspot.com/-5hjfT6Pw29w/XYyu1CJnKtI/AAAAAAAAFvQ/Fq0LC1Bgqv0QBcW8I3_5DabJkyLOX254QCLcBGAsYHQ/s320/conjugation%2Bcontour.png" width="320" /></a></div><br />This is equivalent to asking "<i>is the orbit of some non-zero $X$ under conjugation by $G$ equal to $\mathfrak{g}$?</i>" (so that the norm of that $X$ would suffice to determine all norms -- do you see why?) Well, this is equivalent to asking "<i>is $X$ contained in some non-trivial ideal?</i>" (prove that these are equivalent!), and this is equivalent to asking "<i>does $\mathfrak{g}$ have any non-trivial ideals?</i>" (do you see why?)<br /><br />A Lie algebra without nontrivial ideals is called a&nbsp;<b>simple Lie algebra</b>. Our demonstration above shows that a simple Lie algebra has a unique $\mathrm{Ad}$-invariant symmetric bilinear form, determined by the value of $\langle X, X\rangle$ for some non-zero $X$.<br /><br /><div class="twn-furtherinsight">Even before we actually derive what this form must look like, we can derive one important consequence of automorphism invariance: $\langle X, [X, Y]\rangle = 0$ (prove it!), i.e. the tangent to an automorphism curve is perpendicular to the position vector at every point. The understanding of the group as acting as a "rotation group" on its Lie algebra in the adjoint representation really makes sense!</div><br /><div class="twn-beg">Someone tell me if they know how one may "derive" the trace-form formula from this characterisation rather than pulling it out of the blue and <em>then</em> proving it is the unique $\mathrm{Ad}$-invariant symmetric bilinear form. Here's something I started to write:<br /><br />Here's an idea for the base length (i.e. to define the scaling): $X$ has length 1 iff the length of $[X,Y]$ equals the length of $Y$ for all $Y$ perpendicular to $X$ -- equivalently: $\forall V\in\mathfrak{g}, |[X,[X,V]]|=|[X,V]|$. We need to check that this condition is well-defined, i.e. that:<br /><ol><li>Given an $X$, $|[X,[X,U]]|=|[X,U]|$ for some $U$ not a multiple of $X$ implies that $|[X,[X,V]]|=|[X,V]|$ for all $V$.</li><li>$X$ satisfying $|[X,[X,V]]|=|[X,V]|$ implies that all conjugates $gXg^{-1}$ of it satisfy it too. This is trivial from considering $V=gV'g^{-1}$ (since the identity is true for all $V$).</li></ol>Is the first one even true outside $\mathfrak{so}(3)$ -- for all simple Lie algebras?</div><br />One may come up with the idea of defining a form $\langle X, Y\rangle = \mathrm{tr}[X,[Y,\cdot]]$ (example of some weak motivation -- the vector triple product $x\times(x\times v)$ has as eigenvectors the vectors $v$ perpendicular to $x$ and the eigenvalues depend on the length of $x$) and check that this is indeed an $\mathrm{Ad}$-invariant symmetric bilinear form, and is thus unique up to scaling for simple Lie algebras. This form is called the <b>Killing form</b>.<br /><br /><hr /><br /><b>Factorisation of Lie groups</b><br /><br />We have seen the classification of connected Abelian Lie groups: they are products of circles and lines. We wonder if such a classification is possible for more general Lie groups.<br /><br />The natural way to "factorise" groups by taking quotients over normal subgroups -- we wonder if this means that all Lie groups can be written as direct products of <b>simple Lie groups</b> (groups that don't have a nontrivial connected normal subgroup -- can you see why "connected" matters?). Well, not really -- the quotients need not be subgroups at all, after all. Instead, the "factorisation" takes the form of what is known as a <b>group extension</b>. A group for which it <i>is</i>&nbsp;a direct product is called a <b>reductive Lie group</b>&nbsp;-- and its Lie algebra is the direct sum of simple Lie algebras, or a <b>reductive Lie algebra</b>.<br /><br /><div class="twn-pitfall">It is more conventional in the literature to define a simple Lie algebra excluding the one-dimensional/abelian case. In this definition, direct sums of simple Lie algebras are <b>semisimple Lie algebras</b>, and reductive Lie algebras are direct sums of semisimple and abelian Lie algebras.</div><br />TBC: Cartan's criterion, solvability, nilpotency<br />Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-75397706828241147852019-09-23T16:55:00.000+01:002019-09-23T16:55:06.334+01:00Lie group topologyI'll assume you have a basic understanding of general topology -- if not, consult the <a href="https://thewindingnumber.blogspot.com/p/2204.html">topology articles here</a>. Most of the abstract stuff and "weird" cases are not really important, because it is easy to see that Lie groups are manifolds.<br /><br />We need to be careful while studying the topology of Lie groups, because we already have an intuitive picture of a Lie group, and we need to be careful to prove all the things we just "believe" to be true.<br /><br />The main point of the topology of a Lie group is that the group elements define the "flows" on the manifold. What this means is that <b>left-multiplication is a homeomorphism</b>, and it's not absurd to say that <b>inversion is a homeomorphism</b>, because it represents a "reflection" of the manifold. That these conditions make sense is confirmed by looking at the proofs of the following "obvious" facts.<br /><br /><b>(1) In a connected group, a neighbourhood of the identity generates the entire group,</b>&nbsp;i.e. $H\le G\land H\in N(1)\implies H=G$ for connected $G$.<br /><br />Let's think about why this is true. Why does $H$ need to be a neighbourhood -- why must it contain an open set containing the identity? Suppose instead we just knew it contained a set $Q$ that looked like this:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-GKetHN8XvRw/XYZe0gFDizI/AAAAAAAAFuo/TNWiIWG-kno_P6VXRJvktYb9_SlRGZjtgCLcBGAsYHQ/s1600/not%2Ba%2Bneighbourhood.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="378" data-original-width="346" height="320" src="https://1.bp.blogspot.com/-GKetHN8XvRw/XYZe0gFDizI/AAAAAAAAFuo/TNWiIWG-kno_P6VXRJvktYb9_SlRGZjtgCLcBGAsYHQ/s320/not%2Ba%2Bneighbourhood.png" width="291" /></a></div><br />Well, $H$ still contains the orange point, but we cannot say it contains the purple point, because it's perfectly happy not containing it -- it's not like we have some vertical element in the Lie group that if you multiplied to some point in $Q$, you'd get the purple point. But instead if $Q$ was an open neighbourhood of the identity:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-MnEQeC8NSoQ/XYZkr1agIEI/AAAAAAAAFu0/v0TxF35QuT4Ctb1FVdxM4y76Rw4-nzWqgCLcBGAsYHQ/s1600/neighbourhood%2Ba%2Byes.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="393" data-original-width="370" height="320" src="https://1.bp.blogspot.com/-MnEQeC8NSoQ/XYZkr1agIEI/AAAAAAAAFu0/v0TxF35QuT4Ctb1FVdxM4y76Rw4-nzWqgCLcBGAsYHQ/s320/neighbourhood%2Ba%2Byes.png" width="301" /></a></div>Then the purple point has to be in $H$, because $Q$ contains flows in "all directions" on the group. To actually prove that every point will be contained in $H$ -- well, we know that the point is (will eventually be) that $H$ is the connected component of $G$ (and since $G$ is connected, $H=G$) -- let's just show that $H$ is both open and closed, i.e. nothing in $H$ touches its exterior, and nothing in its exterior touches $H$. Here's the proof:<br /><ul><li><b>Nothing in $H$ touches anything -- </b>Suppose $\exists x\in H, x\in\mathrm{cl}(H')$. Then $xQ$ contains a point in $H'$.</li><li><b>Nothing outside $H$ touches it</b>&nbsp;-- Suppose $\exists x\in H', x\in\mathrm{cl}(H)$. Then $xQ$ contains a point in $H$, so $x$ must be in $H$.</li></ul>We're really just formalising the notion of "translating $Q$ to its edges to extend $H$ further and further". The key fact we've used here is, of course, that left-multiplication is a homeomorphism, so $xQ$ is still an open set.<br /><b><br /></b><b>(2) The connected component of the identity is a subgroup.</b><br /><b><br /></b> The idea is that taking two elements $g,h$ of the connected component, their product should remain in the connected component. Once again, this follows from the <b>continuity of left-multiplication</b>&nbsp;-- considering the action of left-multiplication by $g$ on the connected component, its continuity implies that the image must remain connected.<br /><b><br /></b><b>(3) If a subgroup contains a neighbourhood of the identity, it contains the connected component of the identity.</b><br /><div><b><br /></b> Corollary to (1) and (2).<br /><b><br /></b></div><div><b>(4) The connected component of the identity is a&nbsp;<i>normal</i>&nbsp;subgroup.</b><br /><b><br /></b> Conjugation is a continuous map.<br /><b><br /></b><b>(5) Open subgroups are closed.</b><br /><b><br /></b> Corollary to (3). Alternate proof: the complement is the union of some cosets, which are open sets too. A weaker theorem can be made of closed sets -- closed subgroups with finite index are open.<br /><br />What this means: any open subgroup is a union of connected components.<br /><br /><b>(6) Intuition for compact subgroups</b><br /><b><br /></b>How can a Lie group possibly "close in on itself"? Surely we keep "extending" an open neighbourhood $W$ of the identity by observing that $xW$ must be in the subgroup? The idea is that these translations of $W$ form an <b>open cover of the group, if it has a finite subcover</b>, then it makes sense for the group to close in on itself. By playing around with different open neighbourhoods $W$ and taking some suitable unions, one can see that this is equivalent to the condition that every open cover has a finite subcover, i.e. the group is compact.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-cHRhxfTndyo/XYcEqmYku_I/AAAAAAAAFvA/S4UjQafMaX81u0p1emQk8-6kgy8yEWbXgCLcBGAsYHQ/s1600/open%2Bcover%2B-%2Bselect.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="393" data-original-width="408" height="308" src="https://1.bp.blogspot.com/-cHRhxfTndyo/XYcEqmYku_I/AAAAAAAAFvA/S4UjQafMaX81u0p1emQk8-6kgy8yEWbXgCLcBGAsYHQ/s320/open%2Bcover%2B-%2Bselect.png" width="320" /></a></div><br /></div><div><b>(7) A compact, connected Abelian Lie group is a torus.</b></div><div><b><br /></b> This is a generalisation of "a finite Abelian group is the direct product of cyclic groups".<br /><br />The idea behind the proof is that in the Abelian case, the exponential map is a homomorphism from the Lie algebra to the Lie group, but the Lie algebra cannot detect compactness in the Lie group -- the kernel of the exponential map can. We know from our study of the exponential map that it has a discrete kernel, and in the Abelian case is surjective -- thus the Lie group is homeomorphic to $\mathbb{R}^n/\mathbb{Z}^n$, which is an $n$-torus.<br /><b><br /></b></div><div><b>(8) A connected Abelian Lie group is a cylinder (direct product of a torus and an affine space)</b><br /><b><br /></b>Analogous to above, except $\mathbb{R}^m/\mathbb{Z}^n$ where $m\ge n$.</div><div></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-4155573325156580872019-09-16T19:47:00.001+01:002019-09-16T19:47:30.522+01:00Mixed states II: decoherence; important measures of purity and entropy<b>Decoherence</b><br /><br />At the end of this section, you should be able to:<br /><ul><li>appreciate why the density matrix is really a great way of expressing states, even for pure states (they uniquely determine the dynamics of the system, without any "overall phase", etc.)</li><li>develop an intuition for measurement, even "inadvertent" measurement</li><li>understand on a somewhat high level how classical physics arises as a limit of quantum physics</li><li>hang out with Wigner's friend</li><li>admit that complex phases matter in quantum mechanics and link them to interference</li></ul><br />Let's talk about <b>measurement</b>.<br /><br />Suppose we have a system that we wish to measure it under an operator whose eigenvectors are $|0\rangle_A$ and $|1\rangle_B$. The idea is that we have some measurement apparatus, and their original combined state evolves from something like:<br /><br />$$|\psi\rangle_{AB}=(\lambda|0\rangle_A+\mu|1\rangle_B)\otimes|0\rangle_B$$<br />To the entangled state:<br /><br />$$|\psi\rangle_{AB} = \lambda|0\rangle_A\otimes|0\rangle_B+\mu|1\rangle_A\otimes|1\rangle_B$$<br />Then observing the apparatus is sufficient to observe the system. The idea is that ultimately, the observer himself (or his "knowledge") are the apparatus, and the he entangles with the system to measure it.<br /><br />Well, we know that often, we end up seeing things we didn't really want to. After all, physics does not care about your wants and preferences. In fact, in pretty much any situation, information about the system <i>will</i>&nbsp;<b>leak</b> out into the surroundings in some specific way. For example, Schrodinger's cat leaks information about the life of the cat by making the environment smelly, i.e. the state evolves from:<br /><br />$$|\psi\rangle_{AB}=(\lambda|\mathrm{alive}\rangle+\mu|\mathrm{dead}\rangle)\otimes|\mathrm{clean}\rangle$$<br />To the entangled state:<br /><br />$$|\psi\rangle_{AB}=\lambda|\mathrm{alive}\rangle\otimes|\mathrm{clean}\rangle+\mu|\mathrm{dead}\rangle\otimes|\mathrm{smelly}\rangle$$<br />What this means is that the density matrix of the cat evolves as:<br /><br />$$\left[ {\begin{array}{*{20}{c}}{{{\left| \lambda&nbsp; \right|}^2}}&amp;{\lambda \bar \mu }\\{\mu \bar \lambda }&amp;{{{\left| \mu&nbsp; \right|}^2}}\end{array}} \right] \mapsto \left[ {\begin{array}{*{20}{c}}{{{\left| \lambda&nbsp; \right|}^2}}&amp;0\\0&amp;{{{\left| \mu&nbsp; \right|}^2}}\end{array}} \right]$$<br />(Check that I got the right transpose.) OK, what happened here?<br /><br />Recall that the probabilities of collapsing to $|0\rangle$ and $|1\rangle$ are determined purely by the elements on the diagonal -- the off-diagonal elements, or the <b>coherences</b>, are only relevant for collapsing on to some combination of $|0\rangle$ and $|1\rangle$. What's going on here is that when the environment entangles with the system, it has "kinda" already observed it -- like your Wigner's friend. It "knows" that the system isn't in $|0\rangle+|1\rangle$, and even though you haven't observed the environment yet (you haven't smelled it), you know how the combined state has evolved, and the probability has become a <b>classical probability</b>, because the quantum stuff has already been observed -- by the environment.<br /><br /><b>The idea behind decoherence is the same idea that ensures that the Wigner's friend scenario is consistent.</b><br /><b><br /></b> "Eventually", "all" the information about the system will leak into the environment -- i.e. in principle, we should be able to determine anything about the system from measuring the environment, and our uncertainty about the system arises entirely from our <b>completely classical uncertainty</b> about the environment -- so the density matrix becomes a classical one, i.e. a <b>diagonal one</b>&nbsp;(the off-diagonal terms go to zero).<br /><br />What basis is it diagonal in? In the basis corresponding to the states of the environment -- i.e. if the environment can be in states $|0\rangle_B$ and $|1\rangle_B$, then the states of the system that precisely induce these states of the environment form the preferred basis. These are often called the "<b>environmentally selected basis</b>".<br /><br />This process is called <b>decoherence</b>. You may also hear the terms <b>pointer states</b> (for the preferred basis), <b>einselection</b> (<i>environmentally induced selection</i>&nbsp;of the preferred basis), or <b>Quantum Darwinism</b> (what the heck?) -- but they're really synonymous. We'll just use the fancy words when they're grammatically useful.<br /><br />Well, the following may not be completely clear, but you should at least be able to appreciate that it is true: the off-diagonal terms <i>approach</i> zero, rather than hit it. Why? Although the system leaks information into the surroundings, we aren't really certain about what we're inferring about the system from the environment -- a live cat may be smelly too, etc. So the pointer states are not exactly orthogonal, either.<br /><br />The precise behavior of decoherence depends on the Hamiltonian of the system -- e.g. predicting the generation of the smelliness of the air from the state of the cat based on what's going on microscopically is something that could be done in principle by solving a really complicated Schrodinger equation. You can, given a Hamiltonian, at least make order-of-magnitude estimates of at how much time and at how macroscopic a scale (i.e. with how many degrees of freedom) does the system begin to behave in a way that can be described as classical.<br /><br /><div class="twn-pitfall">Decoherence does <em>not</em> remove the need for wavefunction collapse -- one still needs the observer to note an observation, collapsing the system.</div><br />TBC: purity, entropy, correlation functionsAbhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-29756933463475304592019-09-13T20:36:00.000+01:002019-09-25T13:23:46.173+01:00Lie group homomorphismsBecause a Lie group is fundamentally a group that is also a manifold, we'd like to define a Lie group homomorphism as one that is both a <i>group homomorphism</i>, and <i>smooth</i>. For this, though, we need to define what it means to differentiate a group homomorphism.<br /><br />Recall that the general notion of a derivative is the idea of "how does the map work locally"? Letting a general function $f:G\to H$ map a curve $\gamma(t)$, it should be easy to see that $\gamma'(t)$ transforms as $(f\circ\gamma)'(t)$ (make sure that this makes sense -- think in terms of the chain rule, or write it out in limit form, or just in terms of the image of the curve).<br /><br />Consequently this leads to the <b>differential&nbsp;</b>$df:dG\to dH$ (where $dG$ is the Lie Algebra of $G$) defined as $df(\gamma'(0))=(f\circ\gamma)'(0)$. Some short exercises:<br /><ul><li>Confirm that this is equivalent to saying that $df(X)$ is the <b>directional derivative</b> of $f$ in the $X$ direction.</li><li>Differentiate $f(xyx^{-1}y^{-1})$ with respect to $x$ in the $X$ direction at $x=1$ (hint: this is a direct application of the definition of the differential in reverse).</li><li>Convince yourself that any derivative operator commutes with $df$, i.e. $D(df(X))=df(D(X))$.</li></ul>It should be intuitively clear that if $f$ is a homomorphism, its local effect should be to act as a homomorphism of the Lie algebra as it should preserve all local structure. We can easily show that:<br /><ol><li>Since $df$ is a derivative of $f$, its value must be a <b>linear map</b> (like the Jacobian). This applies to the derivative as an operator on the tangent space of any manifold -- $f$ doesn't need to be a group homomorphism at all.</li><li>It <b>preserves the Lie bracket</b>. Take $f(xyx^{-1}y^{-1})=f(x)f(y)f(x)^{-1}f(y)^{-1}$ and differentiate it once with respect to $x$ in the $X$ direction at $x=1$, obtaining: $df(X-yXy^{-1})=df(X)-f(y)df(X)f(y)^{-1}$, simplify and differentiate it with respect to $y$ in the $Y$ direction at $y=1$ to get: $df([Y,X])=[df(Y),df(X)]$.</li></ol><hr /><br /><b>The adjoint map</b><br /><br />The Lie Bracket $[Y,X]$ is <em>not</em> the derivative of conjugation $gxg^{-1}$, so you don't have to worry -- the Lie Bracket is not a Lie algebra homomorphism (it doesn't preserve Lie Brackets), the derivative of conjugation at the identity is zero. That's unfortunate -- our <a href="https://thewindingnumber.blogspot.com/2019/06/derivations-and-jacobi-identity.html">explanation of the Jacobi identity</a>&nbsp;("a derivation acts through the Lie Bracket as a derivation on the space of derivations where multiplication is given by the Lie Bracket") really indicated that it has something to do with it.<br /><br />The Lie Bracket <em>is</em> the derivative of conjugation $xgx^{-1}$. OK, so?<br /><br />Here's the idea: $\mathrm{Ad}(x)(y)=xyx^{-1}$ defines a homomorphism $\mathrm{Ad}:G\to\mathrm{Aut}(G)$. Its differential $\mathrm{ad}:dG\to d\mathrm{Aut}(G)$ can be confirmed to be the Lie Bracket $\mathrm{ad}(X)(Y)=[X,Y]$. So preservation of the Lie Bracket means:<br /><br />$$\mathrm{ad}([X,Y])=[\mathrm{ad}(X),\mathrm{ad}(Y)]$$<br />This is <b>precisely the Jacobi identity</b>! So the Lie bracket <i>is</i> a Lie algebra homomorphism, from a Lie algebra to the Lie algebra of half-filled Lie brackets.<br /><br />There is indeed a relationship between this "homomorphism" understanding of the Jacobi identity and the "derivation" understanding. In general, given a curve $\phi:\mathbb{R}\to\mathrm{Aut}(G)$, differentiating $\phi(t)(gh)=\phi(t)(g)\phi(t)(h)$ at $t=0$ we see that its derivative $d\phi$ satisfies the product rule, i.e. is a derivation (in fact this is true even when $G$ is not a group -- often a Lie group arises this way, as the automorphism group of some object and these derivations then form its Lie algebra). This implies<br /><br />$$d\mathrm{Aut}(G)\subseteq\mathrm{Der}(dG)$$<br />So $[X,\cdot]$ is a derivation, and the map from $X$ to $[X,\cdot]$ is a Lie algebra homomorphism $dG\to\mathrm{Der}(dG)$. This really does give us a much more general way to look at everything we talked about in the <a href="https://thewindingnumber.blogspot.com/2019/06/derivations-and-jacobi-identity.html">last article</a>.<br /><br /><div class="twn-pitfall">Wait -- shouldn't it be an equality? <a href="https://thewindingnumber.blogspot.com/2019/06/derivations-and-jacobi-identity.html">I thought all derivations were part of the Lie Algebra</a>? Ah, but there the derivations on $M$ formed the Lie Algebra of $\mathrm{Aut}(M)$, i.e. $d\mathrm{Aut}(M)=\mathrm{Der}(M)$. So indeed $d\mathrm{Aut}(dG)=\mathrm{Der}(dG)$. This makes sense, indeed $\mathrm{Aut}(G)\subseteq \mathrm{Aut}(dG)$. It's interesting to think about when it is that the Lie algebra has "more" automorphisms than the Lie group.</div><br /><div class="twn-furtherinsight">One may wonder if all automorphisms of a group are a conjugation by something -- or equivalently, if all automorphisms of a Lie algebra are a derivation of some kind. We will later see a special classification of Lie group for which this is true -- in general, the conjugation automorphisms are called the <b>innner automorphisms</b> of the group and are denoted as $\mathrm{Inn}(G)$. The group of <em>all</em> endomorphisms (invertible linear transformations $dG\to dG$) of a Lie algebra, meanwhile are denoted as $\mathrm{End}(dG)$, and it's easy to see that this occurs iff the Lie algebra is Abelian.</div><br /><b>Exercise:</b> Show that the map $\mathrm{Ad}:G\to \mathrm{Aut}(G)$ is injective iff $G$ has a trivial center.<br /><br />So if $G$ has trivial center and all its automorphisms are inner, it is isomorphic to $\mathrm{Aut}(G)$ and is called <b>complete</b>.<br /><br /><hr /><br /><b>The determinant map</b><br /><b><br /></b> The determinant is a homomorphism $\det:GL_F(n)\to F$ from any matrix group. The first thing we'd like to do with this is find its differential $\det'$ (which will be an $F$-valued function on $M_F(n)$). By definition of the differential:<br /><br />$$\det' A = \lim_{\varepsilon\to 0}\frac{\det (I+\varepsilon A)-1}{\varepsilon}$$<br />It's easy to prove by writing out the entries of the matrix as $\delta_{ij}+\lambda_{ij}\varepsilon$ and performing induction on the dimension of the matrix that this is equivalent to:<br /><br />$$\det'A=\mathrm{tr} A$$<br /><hr /><br /><b>Lie algebra homomorphisms in detail: ideals</b><br /><b><br /></b>Well, Lie algebra homomorphisms are a specific category of vector space homomorphisms, aren't they? It's not enough that they preserve the linear structure, they must preserve the Lie bracket too. Well, let's study them in more detail -- like a crash course through linear algebra, but with Lie algebra instead.<br /><br />What does the kernel of a Lie algebra homomorphism $A$ look like? Well, because the homomorphism preserves linear combinations, the kernel must be a linear subspace -- similarly because the homomorphism preserves the Lie bracket, we must have that $Av=0\implies \forall w\in\mathfrak{g}, A[v,w]=0$, i.e. the kernel must be closed under derivations from $\mathfrak{g}$: $[\mathfrak{g},\mathfrak{i}]\subseteq\mathfrak{i}$. Such a subalgebra is called an <b>ideal</b>.<br /><br /><b>Exercise:</b>&nbsp;Show that the Lie algebra of a normal subgroup is an ideal (careful -- it's not as obvious as you might think -- but still pretty obvious).Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-60780241954584002052019-09-10T19:40:00.000+01:002019-09-16T19:59:02.859+01:00Time evolution, Schrodinger and Heisenberg pictures, Noether's theoremSo far, we have discussed quantum mechanics without any reference to changes across time. You might think we could just upgrade $\psi(x)$ to $\psi(x,t)$ and e.g. an observable $Q_t$ measuring a value $q$ at time $t$ would have eigenvectors whose cross-section at $t$ are of the form $\delta(x-a)$. But this would mean the entire $\psi(x,t)$ is the state of the object, rather than there being a state at each value of $t$, and time would be an observable. This is clearly not what we want (right now -- although to be consistent with special relativity we will need to treat space and time on an equal footing later in this series).<br /><br />Instead, a more appropriate approach is to say that the state is a function of time $|\psi(t)\rangle$ and the evolution of the state is given by some operation $|\psi(t)\rangle=U[|\psi(0)\rangle]$. <br /><br />How do we know that $U$ is a <b>linear operator</b>? What does it mean for $U$ to be a linear operator anyway? The only sense in which such a linearity can be tested is by looking at a state in a superposition. So suppose $|\psi(0)\rangle=|\psi_1(0)\rangle+|\psi_2(0)\rangle$. Now $|\psi(t)\rangle=U[|\psi_1(0)\rangle+|\psi_2(0)\rangle]$ -- this is from the perspective of some observer Alice.<br /><br />But if another observer Bob had previously observed and collapsed the system to $|\psi_1(0)\rangle$ at time 0, then according to him, the state should evolve to $U[|\psi_1(0)\rangle]$, and if he had observed the system in $|\psi_2(0)\rangle$, his knowledge of the system would evolve to $U[|\psi_2(0)\rangle]$.<br /><br />So according to Alice, who doesn't know what Bob has observed (she has not observed him), her knowledge of the system can also be written as $U[|\psi_1(0)\rangle]+U[|\psi_2(0)\rangle]$. Thus<br /><br />$$U[|\psi_1(0)\rangle+|\psi_2(0)\rangle]=U[|\psi_1(0)\rangle]+U[|\psi_2(0)\rangle]$$<br />I.e. $U$ is linear, so we can write it as a linear operator as in $U|\psi(0)\rangle$. (The above scenario is called <b>Wigner's friend</b>)<br /><br />$U$ is also clearly a <b>unitary&nbsp;operator</b>, as it must preserve all lengths.<br /><br />We can consider infinitesimal time evolutions $U_t(dt)$ representing evolution of the state from $t$ to $t+dt$. Then:<br /><br />$$|\psi(t)\rangle=U_0(dt)\dots U_{t-dt}(dt)|\psi(0)\rangle$$<br />This <a href="https://thewindingnumber.blogspot.com/p/2202.html">product integral</a>&nbsp;can be written alternatively as:<br /><br />$$|\psi(t)\rangle=\mathcal{T}\left\{e^{\int \ln U_t(dt)}|\psi(0)\rangle\right\}$$<br /><div class="twn-furtherinsight">$\mathcal{T}$ is the <b>time-ordering operator</b> which orders a product like $H(t_1)H(t_2)$ in order of ascending $t$ in an expansion. Can you see why this is necessary (hint: $e^{AB}\ne e^{A+B}$ for noncommuting $A,B$).</div><br /><div class="twn-pitfall">Oh, and it's not actually an operator -- not even in the math sense, it's a "formal operation", one that takes a <em>form</em> or sentence (rather than its value) -- in this case the $\exp$ Taylor expansion -- and changes it some way.</div><br />$\ln U_t(dt)$ is an infinitesimal, and it's easy to see that it is equal to $U_{t}'(0)dt$ -- a member of the Lie algebra. We know, of course, that the Lie Algebra of the unitary group is comprised of <b>anti-Hermitian operators</b>&nbsp;(this can be checked without Lie Algebra, of course), and so $iU_{t}'(0)$ is Hermitian. From Lie Algebra, we can tell that this represents a <b>generator of time translations</b> -- and from a little experience of classical mechanics, we want this to represent energy. So for dimensional consistency with energy, we write:<br /><br />$$H(t)=i\hbar U_{t}'(0)$$<br />(Why $\hbar$ and not $h$? Because $U_{t}'(0)$ is basically already in "radians per second".) This is called the&nbsp;<b>Hamiltonian operator</b>.&nbsp;It determines $U_t$, and thus describes the time evolution of a state. How exactly? Since:<br /><br />$$|\psi(t+dt)\rangle=U_t(dt)|\psi(t)\rangle$$<br />We can write re-arranging:<br /><br />$$\frac{\partial |\psi(t)\rangle}{\partial t}=-\frac{i}{\hbar}H(t)|\psi(t)\rangle$$<br />This is the most general form of the <b>Schrodinger equation</b>. Note that the earlier exponential equation is the "general solution" to this equation -- obviously not very useful, rewritten as what is known as the&nbsp;<b>Dyson series</b>:<br /><br />$$|\psi(t)\rangle=\mathcal{T}\left\{e^{-i/\hbar\int H(t) dt}|\psi(0)\rangle\right\}$$<br />It's easy to show from this that the evolution of a density matrix $\rho(t)$ is similarly:<br /><br />$$\frac{\partial\rho(t)}{\partial t}=-\frac{i}\hbar [H, \rho]$$<br />Which is the <b>von Neumann equation</b>, whose solution is given by:<br /><br />$$\rho(t)=\mathcal{T}\left\{e^{-i/\hbar\int H(t) dt}\rho(0)e^{i/\hbar \int H(t) dt}\right\}$$<br />These should all appear as obvious special cases of Lie theoretic results.<br /><br /><hr /><br /><div class="twn-pitfall">$H(t)$ is <em>not</em> the same as $i\hbar\partial/\partial t$. $H(t)$ is a Hermitian operator, i.e. an observable, while $\partial/\partial t$ does not act on the Hilbert space at all. One could also see what could wrong by equating the two in the "solution to the Schrodinger equation" above. The Schrodinger equation does not say that $H$ and the time-derivative are equal in general -- rather, it says that they are the same <i>on a valid state vector</i>&nbsp;$|\psi(t)\rangle$ -- you cannot just "factor this out".</div><br />So the Hamiltonian is fundamentally what determines the dynamics of a quantum system. Give me a Hamiltonian, and you've given me a theory. The Schrodinger equation (or equivalently the von Neumann equation) above is just an axiom of quantum mechanics/of any quantum mechanical theory.<br /><br /><hr /><br />Can we talk about the velocity and acceleration observables for a moment? Actually, we can't, because they fundamentally have to do with time evolution, and we can't have observables that depend on the time-evolution of the state -- observables must act on the Hilbert space. But we can define observables that predict how the state will evolve (like the Hamiltonian with the Schrodinger equation).<br /><br />Doing this systematically is where the Heisenberg picture comes in.<br /><br />What does this mean? Everything we've discussed so far is the <b>Schrodinger picture</b>, where the state evolves on a fixed background basis created by the observables' eigenvectors -- so observables represent <b>active transformations</b>. Instead, we can have a completely different picture of reality, the <b>Heisenberg picture</b>, where we view time-evolution as simply viewing the state in a different basis -- then the observables represent <b>passive transformations</b>.<br /><br />OK, so how do we do this? Remember how every question in quantum mechanics can fundamentally be asked in terms of expectation values (specifically those of Hermitian projections). The expected value of an observable at time $t$ of course evolves as:<br /><br />$$\langle A\rangle(t) = \langle\psi|U^*(t)AU(t)|\psi\rangle$$<br />In the Schrodinger picture, we attach the $U(t)$ to $|\psi(0)\rangle$ to make $\langle A\rangle(t)=\langle\psi(t)|A|\psi(t)\rangle$. In the Heisenberg picture instead, we attach the $U(t)$ to the $A(0)$, writing $\langle A\rangle(t)=\langle\psi|A(t)|\psi\rangle$.<br />From differentiating conjugation in $A(t)=U^*(t)A(0)U(t)$, we get:<br /><br />$$\frac{dA}{dt}=\frac{i}\hbar [H, A]$$<br />This is the <b>Heisenberg equation</b>. Immediately, it yields:<br /><br />$$\begin{array}{l}\frac{{dX}}{{dt}} = \frac{i}{\hbar }\left[ {H,X} \right]\\\frac{{{d^2}X}}{{d{t^2}}} =&nbsp; - \frac{1}{{{\hbar ^2}}}\left[ {H,\left[ {H,X} \right]} \right]\end{array}$$<br />Thinking of the evolution of $X$ as a translation of the co-ordinate system, etc., what this does is give us two conditions on what the Hamiltonian should look like for a "Euclidean" system:<br /><br />$$\begin{array}{l}\left[ {H,X} \right] =&nbsp; - \frac{{i\hbar }}{m}P\\\left[ {H,\left[ {H,X} \right]} \right] = \frac{{{\hbar ^2}}}{m}U'(x)\end{array}$$<br />This gives us yet another strong reason (besides the fact that the Hamiltonian generates time-translations, that the "eigenvectors" of $\partial/\partial t$ are the energy states by the de Broglie theorem (but not really), etc.) to suspect that the Hamiltonian represents the <i>energy</i>&nbsp;of the system. Indeed if we use:<br /><br />$$H=\frac1{2m}P^2+U(x)$$<br />We can confirm those conditions above. Well, this is certainly not the only Hamiltonian compatible with classical mechanics, so at this point, I'll just say that this is confirmed by experiment, and is an axiom of the quantum theory of Euclidean systems.<br /><br /><b>Exercise:</b>&nbsp;By taking expectation values in the Heisenberg equation, show that $m\frac{d}{dt}\langle x\rangle =\langle p\rangle$ and $\frac{d}{dt}\langle p\rangle = -\langle U'(x)\rangle$ under the Euclidean Hamiltonian. This is called the <b>Ehrenfest theorem</b>.<br /><br /><hr /><br />I'll discuss one final application of the Heisenberg formalism: it makes <b>Noether's theorem</b>&nbsp;completely trivial.<br /><br />Indeed, $dA/dt=0$ iff $[H,A]=0$ iff $\forall\tau, H=e^{-i/\hbar A\tau}He^{i/\hbar A \tau}$. Then $A$ is a conserved quantity and conjugation with it as an infinitesimal generator represents a symmetry of the Hamiltonian.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-44775225261506995002019-09-06T10:56:00.001+01:002019-09-06T10:57:22.841+01:00Supplementary definitions: levels of abstractionAlthough we have been able to generalise a lot of our theorems about the topological properties of metric spaces to topological spaces, several others remain out of our reach, and we need to find the right level of abstraction that makes them true -- and we can usually do this by trying to prove the theorem and seeing what it "needs", then if this theorem actually characterises the space, proving the need is equivalent to the theorem.<br /><br /><b>Separation axioms</b><br /><i><br /></i><a href="https://thewindingnumber.blogspot.com/2019/08/topology-iii-example-topologies.html">We've seen the first one.</a>&nbsp;TFAE:<br /><ul><li>$X$ is a T0/Kolmogorov/distinguishable space.</li><li>All points in $X$ are distinguishable.</li><li>For any pair of points, there exists an open set containing exactly one of them.</li></ul>Now,&nbsp;in metric spaces, statements like "there is a point of a set (or sequence) $S$ in every open neighbourhood of $x$" are equivalent to "there are an infinite number of points of $S$ in every open neighbourhood of $x$", i.e. <b>every open neighbourhood of a limit point of $S$ has infinite points in $S$</b>. The key here is that at any step in the process, an open neighbourhood of $x$ can be constructed that does not contain any of the previous points in the sequence, i.e. there exists an open neighbourhood of $x$ not containing some finite number of points.<br /><br />This is possible, e.g. if we decide that <b>finite sets are closed</b>&nbsp;(or equivalently <b>singletons are closed</b>). This is an iff -- if a finite set $S$ isn't closed, there are points in $\mathrm{cl}(S)$ that aren't in $S$, but it's impossible for there to be an infinite number of points in $S$ in a neighbourhood of such a point, as $S$ is finite.<br /><br /><div class="twn-pitfall">Wait, what about for points in $S$ itself -- aren't these all also limit points of $S$? We sneakily changed the definition of limit points from $x\in\mathrm{cl}(S)$ to $x\in\mathrm{cl}(S-\{x\})$, i.e. to exclude sequences that include the point itself. So since excluding a set from a finite set keeps it finite, it is still closed, and thus $x$ is not re-included when you close it (so it is not a limit point at all).</div><br /><div class="twn-pitfall">Wait, what about the discrete topology on a finite set? All finite sets are closed, but how could we have an infinite number of anything? The key is that there are no "limit points" of any set. So the only T1 finite spaces are the discrete ones.</div><br />Note how this implies a stronger form of T0 -- T0 said that for every pair of points, at least one of them has an open neighbourhood excluding the other. Well, in a T1 space, we can always just remove any finite number of points from an open neighbourhood and it will remain an open neighbourhood -- so in particular, this means that for every pair of points, <b>each has an open neighbourhood excluding the other</b>, or <b>any two points are separated</b>. In fact, this is clearly an iff (take some finite intersections).<br /><br />The closedness of finite sets has plenty of other implications that match our intuition. You will see some in the list of exercises.<br /><br />So for now, TFAE:<br /><ul><li>$X$ is a T1/Frechet space.</li><li>All points in $X$ are separated.</li><li>For any pair of points, each has an open neighbourhood excluding the other.</li><li>Each open neighbourhood of a limit point of $S\subseteq X$ contains an infinite number of members of $S$.</li><li>Finite sets are closed/singletons are closed.</li></ul>We've seen spaces in which limits are not unique. You might think that topological distinguishability should suffice to have unique limits -- is this right?<br /><br />Let's go through the proof of the uniqueness of limits on metric spaces. Suppose $f$ has two limits $L_1$ and $L_2$ at $c$. Now does it really suffice that there is a neighbourhood of $L_1$ not containing $L_2$? Not necessarily -- it may be the case that this neighbourhood of $L_1$ intersects every neighbourhood of $L_2$ because there's just that amount of indiscretion around $L_2$. What's important is that there exist disjoint neighbourhoods of $L_1$ and $L_2$.<br /><br />So the <b>uniqueness of limits</b>&nbsp;is equivalent to the existence of&nbsp;<b>disjoint neighbourhoods</b>&nbsp;for distinct points. This is variously called a Hausdorff space&nbsp;or&nbsp;a T2 space. So TFAE:<br /><ul><li>$X$ is a T2/Hausdorff space.</li><li>All points in $X$ are separated by neighbourhoods.</li><li>Any pair of points has a pair of respective neighbourhoods disjoint from each other.</li><li>Limits of nets/filters are unique.</li></ul>Here's another obvious fact about the topology of metric spaces: any continuous function $f:S\to\mathbb{R}$ on a closed (not just compact) set $S$ may be <b>extended to a continuous function</b> $F:X\to\mathbb{R}$. The proof of this relies on the ability to "connect" points between the pieces of $S$:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-4AjUE-rMf7w/XW6U-Y-8NpI/AAAAAAAAFsc/lUlqnRsFsTozNnVVnjSpLmYZOr0vB4xnQCLcBGAs/s1600/tietze.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="583" data-original-width="887" height="210" src="https://1.bp.blogspot.com/-4AjUE-rMf7w/XW6U-Y-8NpI/AAAAAAAAFsc/lUlqnRsFsTozNnVVnjSpLmYZOr0vB4xnQCLcBGAs/s320/tietze.png" width="320" /></a></div><br />You might think that this just means "having a continuous function between two points", but this is only because of the choice of a one-dimensional space -- in general, the "boundary" of each connected component is not just a finite number of points. What does suffice is that for disjoint closed sets $S$ and $T$ there <b>exists a continuous function $h:X\to\mathbb{R}$ "separating" them</b>, i.e. such that $h(S)=\{0\}$, $h(T)=\{1\}$. Then a bunch of such functions can just be added to $f$ to obtain $F$. This is trivally an iff.<br /><br />One may try further to "prove" the existence of such a continuous function $h$. Well, on $\mathbb{R}$, we have a notion of a "midpoint" between the endpoints of the closed sets -- we could have the value of $h$ at this point be $1/2$, and then continue the construction for all "Dyadic rational"-points.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-I7rwx7Ij0Rk/XW6jNqxfszI/AAAAAAAAFso/ntO_QJ1rhHoW4p1KmyIjvSrIO3fq47nQgCLcBGAs/s1600/tietze-urysohn.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="583" data-original-width="887" height="210" src="https://1.bp.blogspot.com/-I7rwx7Ij0Rk/XW6jNqxfszI/AAAAAAAAFso/ntO_QJ1rhHoW4p1KmyIjvSrIO3fq47nQgCLcBGAs/s320/tietze-urysohn.png" width="320" /></a></div><br /><br />In general, we don't really need the notion of a midpoint, but we do need some point with "space around it", i.e. open neighbourhoods of each closed set that don't contain the point. We only need the case where $S$ and $T$ are in a connected component, so this is equivalent to the <b>existence of open neighbourhoods of disjoint closed sets</b>. This is also clearly an iff, as one may consider preimages of open sets in $[0,1]$. Thus TFAE:<br /><ul><li>$X$ is a T4/normal space.</li><li>Every pair of disjoint closed sets has respective disjoint open neighbourhoods.&nbsp;</li><li>Every pair of disjoint closed sets can be separated by a continuous function (<i>Urysohn's lemma</i>).</li><li>Every continuous map on a closed subset can be extended to a continuous map on $X$ (<i>Tietze's extension theorem</i>).</li></ul><b><br /></b> <b>Countability</b><br /><b><br /></b> <a href="https://thewindingnumber.blogspot.com/2019/08/topology-ii-kuratowski-closure-topology.html">We've already seen the first one.</a>&nbsp;TFAE:<br /><ul><li>$X$ is a first-countable space.</li><li>Every point in $X$ has a countable neighbourhood basis.</li><li>$\lim_{x\to a}f(x)=L\iff\forall (x_{n\in\mathbb{N}})\to a, f(x_n)\to f(a)$.&nbsp;</li><li>Every filter has a corresponding sequence.</li></ul>Several other cardinality functions can be&nbsp;defined and other axioms of countability can be introduced, e.g. sequantial spaces, separable spaces and second-countable spaces, but we cannot really motivate them at this point. (Second-countability in particular important when dealing with metrisability, integration and related issues, as it gives rise to the notion of "paracompactness".)<br /><br /><div class="twn-exercises">For the following statements, find the "least restrictive" level of abstraction necessary, and decide if the property characterises the level (i.e. is it an iff?):<br /><ol><li>Every subset $S$ is the union of all the closed sets contained in $S$.</li><li>Every subset $S$ is the intersection of all the open sets containing $S$.</li><li>Every nonempty open set contains a nonempty closed set.</li><li>Every point $x$ admits a nested neighbourhood basis, i.e. a neighbourhood basis totally ordered by $\subseteq$.</li><li>Closed subset $T$ of a compact space $S$ is compact.</li><li>Compact space is closed.</li><li>There is no smallest neighbourhood of any point (i.e. $\forall x\in X,\forall N\in N(x), \exists M\in N(x), N\not\subseteq M$).</li></ol><b>Solutions:</b><br /><ol><li>T1. This is equivalent to asking that every point in $S$ is contained in a closed subset of $S$, which is possible if the singleton is closed. For equivalence, consider a singleton $S$.</li><li>T1. Dual to Exercise 1.</li><li>T1 suffices, but is not equivalent (e.g. the indiscrete topology or any other topology full of clopen sets). Maybe if you change to proper containment.</li><li>First-countability. (both sides of equivalence are clear -- think directed set, etc.)</li><li>Any topological space suffices. We need to be careful with the proof here because the notions of closedness and compactness are intuitively very similar. Here it is: any net on $T$ is a net on $S$ and thus has a convergent subnet with limit is in $S$ -- but its limit must also be in $T$ because $T$ is closed.</li><li>Hausdorff suffices (probably not equivalent, but who cares). The thing is that in general, even on a compact space, although a convergent sequence may have a convergent subsequence in $S$, it may have other limits outside $S$. But in a Hausdorff space, limits are unique.</li><li>"Not Alexandrov". Let's characterise the topologies that <i>do</i>&nbsp;have smallest neighbourhoods -- these are necessarily the ones for which arbitrary intersections of open sets are open, and are called <i>finitely-generated topologies</i>&nbsp;or <i>Alexandrov topologies</i>.&nbsp;</li></ol></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-16050679316255550342019-08-29T10:36:00.001+01:002019-10-07T01:59:37.662+01:00Supplementary definitions: compactness, compactification, hemicompactnessLet's think about the <b>Bolzano-Weierstrass theorem</b> in real analysis (every sequence in a compact set has a convergent subsequence). What would it take to generalise this theorem to a topological space $X$?<br /><br />First, let's write down the statement: we know that <i>sequences</i>&nbsp;aren't very fundamental to arbitrary topological spaces, so the statement we're looking for would be something like: <b>every net in a compact set has a convergent subnet.</b>&nbsp;The question is what the definition of "compact" is a general topological space generalising the "closed and bounded" definition for metric spaces (and how to prove the generalised statement from this definition).<br /><br />Let's try to formulate some equivalent ways of stating that. It is trivial to see that the filter associated with a subnet is a filter refinement -- so we can state it as: <b>every filter in a compact set has a convergent refinement.</b>&nbsp;This also implies that <b>every ultrafilter is convergent</b>&nbsp;(as an ultrafilter has no proper refinement), and by the ultrafilter lemma, the implication is an iff. Of course equivalently, <b>every ultranet is convergent</b>&nbsp;(where an ultranet is obviously a net that for all $S$ is either eventually in $S$ or in $S'$).<br /><br /><div class="twn-pitfall">You might think of ultranets (nets corresponding to ultrafilters) as all sorts of stuff, like "all convergent nets are ultra", or as generalisations of monotonic sequences -- all these are wrong, and it's a useful exercise to write down counter-examples to this on the real line (hint for the monotonic sequences thing: take the union of a bunch of sets such that a tail set of your sequence is in neither this union nor its complement). You might also think that the "ultrafilter" formulation of the statement is related to the "halving intervals" proof of the Bolzano-Weierstrass theorem, but the filter base generated by this mechanism does not generate an ultrafilter (construct a set that isn't in the filter and its complement isn't either).</div><br />It should be clear that if we only wanted <i>sequences</i>&nbsp;having convergent subsequences, requiring the convergent refinement for <i>every</i>&nbsp;filter is only necessary needed if every filter has a corresponding sequence, and having a convergent refinement only suffices if every filter has a corresponding sequence -- <a href="https://thewindingnumber.blogspot.com/2019/08/topology-ii-kuratowski-closure-topology.html">recall that this occurs precisely in a <b>first-countable space</b></a>. The statement here is that <b>in and only in (inn?) a first-countable space, compactness implies sequential compactness</b>.<br /><br />OK -- we still haven't discovered what this generalised "compactness" is. What is the hypothesis of the generalised Bolzano-Weierstrass theorem?<br /><br />Let's consider two example spaces where the Bolzano-Weierstrass theorem doesn't apply:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-2m8xVdBtDwU/XWdaD3OuzoI/AAAAAAAAFsI/0CdUa43sB4UgMabz1O2egvYKS3HAVgpxQCLcBGAs/s1600/bw%2Bcounter%2B-%2B1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="562" data-original-width="722" height="249" src="https://1.bp.blogspot.com/-2m8xVdBtDwU/XWdaD3OuzoI/AAAAAAAAFsI/0CdUa43sB4UgMabz1O2egvYKS3HAVgpxQCLcBGAs/s320/bw%2Bcounter%2B-%2B1.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-EbWrNvAAR3A/XWdanTEToTI/AAAAAAAAFsQ/5JqzxZFkClITXhfe4JfuKLwZix_SYuVVgCLcBGAs/s1600/bw%2B-%2Bcounter%2B-%2B2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="414" data-original-width="351" height="320" src="https://1.bp.blogspot.com/-EbWrNvAAR3A/XWdanTEToTI/AAAAAAAAFsQ/5JqzxZFkClITXhfe4JfuKLwZix_SYuVVgCLcBGAs/s320/bw%2B-%2Bcounter%2B-%2B2.png" width="271" /></a></div>These two are "basically the same" kind of set -- a set with some <b>limit points removed</b>. Except in the second case, we can actually say "it's a set with some limit points removed", because these limit points exist in the space the set is in, while in the first case, the limit points have been removed from the containing space <i>itself</i>. You know, let's actually just talk in terms of the subspace topology on the set from now on to avoid having to make a distinction between the two "cases".<br /><br />Well, can we recover some notion of the limit point after it's been removed? A trick you may be familiar with to find limits is to take intersections -- we could just say that if this is true for all proper filters $\phi$, we have a compact set:<br /><br />$$\bigcap_{N\in\phi}N\ne\varnothing$$<br />Right?<br /><br />Not right. This intersection will actually every often be empty -- it only measures if the net has a <b>constant&nbsp;subnet</b> (check that this is true). What we need to do is to consider the <b>closures</b>&nbsp;of each set in the filter, so that the limit points, if any, are actually included -- and since the closures are members of the filter too (being supersets), we can just consider the <b>intersection of all closed members of the filter</b>.<br /><br /><div class="twn-pitfall">Note how we cannot just say something about the closure differing from the sequence -- do you see why?</div><br />So we can write e.g. <br /><br />$$\forall \phi\ne \pi(S), \bigcap_{\mathrm{closed}\ N\in\phi}N\ne\varnothing$$<br />Now while this is a perfectly good condition on its own, we see that it contains some redundant sets -- we're taking an intersection, so this kind of naturally "subsumes" the filter's defining properties of closure under finite intersections and supersets. We can just get rid of these sets and consider a bunch of sets that generate the filter -- this is called a <b>filter sub-basis</b>. Any set of sets generates a filter, but we specifically want to generate a proper filter -- this requires that every finite intersection of sets in the filter sub-basis is non-empty -- this is called the <b>finite intersection property</b>&nbsp;(FIP).<br /><br />So we have our definition of a compact set: <b>any family of closed sets satisfying the finite intersection property has a nonempty intersection</b>.<br /><br />Or we could state in terms of the contrapositive: <b>any family of closed sets with empty intersection has a finite subfamily with empty intersection</b>.<br /><br />And then we could take the open-closed dual of the statement: <b>any open cover of $S$ has a finite subcover</b>.<br /><br />The last one is the most common formulation of compactness.<br /><ul></ul>What the Bolzano-Weierstrass theorem really gives us is an interpretation of compact sets as "kinda" like finite sets -- not in terms of the number of points, but in terms of the amount of "space". You have only a finite amount of "space" to move around in, you can't just <b>escape to infinity</b>, where <b><i>infinity</i> is a point not in the set</b>. The Bolzano-Weirstrass theorem is really a "generalisation of the infinite pigenhole principle" (which is precisely the statement of having "limited space"), and in fact yields the latter for the discrete topology where the only convergent sequences are the constant ones.<br /><br />So in contexts where we want to "generalise" facts about finite sets, like in Lie theory, it's natural that compact sets are of importance. To really drive the point home, consider the following reformulation of the open cover definition:<br /><br /><b>For a family of sets, a compact union of some of these sets has any property that every finite union of them does.</b><br /><b><br /></b> Or further, defining an <b>extensive property</b>&nbsp;as a property of open sets such that $P(S)\land P(T)\implies P(S\cup T)$, a <b>locally true</b>&nbsp;property as a property satisfied by some neighbourhood of every point in a space and a <b>globally true</b>&nbsp;property as a property satisfied by the universal set:<br /><br /><b>Every locally true extensive property is globally true iff the space is compact.</b><br /><b><br /></b> This formulation of the compactness leads to several well-known results about compact sets, including: <i>A continuous real-valued function on&nbsp;a compact set is bounded</i>.<br /><br />More on the local/global correspondence at <a href="https://www.math.ucla.edu/~tao/preprints/compactness.pdf">Terry Tao, "Compactness and Compactification"</a>.<br /><br />So here's a list of things we've seen are equivalent to compactness:<br /><ul><li>$S$ is a compact set.</li><li>Every locally true extensive property on $S$ is globally true.</li><li>If $S$ is a union of sets, it has any property that all finite unions of them have.</li><li>All open covers of $S$ have a finite subcover.</li><li>All families of closed sets with the FIP in $S$ have a nonempty intersection.</li><li>All ultrafilters on $S$ converge.</li><li>All filters on $S$ have a convergent refinement.</li><li>All convergent nets in $S$ have a convergent subnet.</li></ul><div><br /></div><hr /><br /><b>Compactification</b><br /><div><b><br /></b></div><div>The obvious question is how we can make a space compact by adding in some points, like we can in the "subset without its entire boundary" example (just take the closure). In this sense, compactification is a "generalisation" of closure, where instead of just adding points that already exist in the space, we add completely new points: points we call <b>infinity</b>. Making the correspondence precise, we are asking for an embedding of the space such that the closure of the embedded set is the "compactification" (the minimum such embedding <i>is</i>&nbsp;the compactification).</div><div><br /></div><div>Obviously, we have some freedom as to how to make this compactification -- as to how we can "glue the loose ends" of the space. For example, we could assign a single point $\infty$ to both sequences increasing without bound and those decreasing without bound (compactifying $\mathbb{R}$ into a circle), or we could define $+\infty$ and $-\infty$ differently, compactifying it into a interval.&nbsp;</div><div><br /></div><div>There are still questions on how we ensure, in general, that e.g. $(n)$ and $(n^2)$ both "converge" to the same point.</div><div><br /></div><div>Are they necessarily the same point?</div><div><br /></div><div>Suppose we add a point, call it $\infty$ to represent the limit of the sequence $(n)$. What does it mean to say that $(n)$ converges to $\infty$? It means that $(n)$ is eventually in every neighbourhood of $\infty$ -- every net corresponds to a filter, and in this case we have the neighbourhood filter of $\infty$ -- the set of all sets containing a cofinite number of positive integers. Similarly, $(n^2)$ generates the filter of all sets containing a cofinite number of squares.</div><div><br /></div><div><i>These are different points.</i>&nbsp;Could we <i>make</i>&nbsp;them the same point? For this, we'd need both $(a_n)$ and $(b_n)$ to be eventually in every neighbourhood of $\infty$ -- this corresponds to the filter of sets containing all but a finite number of points in each sequence. So if we wanted two infinities, $+\infty$ and $-\infty$, the neighbourhoods of $+\infty$ would be the filter of sets containing all but a finite number of points of any sequence diverging to $+\infty$ -- equivalently:<br /><br />$$\begin{array}{l}N( + \infty ) = \{ S\mid \exists r,\forall x &gt; r,x \in S\} \\N( - \infty ) = \{ S\mid \exists r,\forall x &lt; r,x \in S\} \end{array}$$<br />But if we just wanted a single point at infinity,<br /><br />$$N(\infty)=\{S\mid\exists r,\forall |x|&gt;r, x\in S\}$$<br />How would this <b>one-point compactification</b> work in general? We'd like to assign $\infty$ as the limit in $X\cup\{\infty\}$ of every net that does not have a limit point in $X$ (this will automatically also create a limit point at $\infty$ for nets that diverge but also have a limit point in $X$, like $n\sin(n)$) -- i.e. we want a filter that every divergent net is eventually in. Well, the point of the Bolzano-Weierstrass theorem is that a divergent net <b>eventually escapes every compact set</b>, i.e. it is eventually in every cocompact set. So in general, we have a neighbourhood <b>basis for the point at infinity</b>:<br /><br />$$B(\infty)=\{S\cup\{\infty\}\mid S\subseteq\Phi_X\land \mathrm{compact}\ S'\}$$<br />Or in terms of open sets,<br /><br />$$\Phi_{X\cup\{\infty\}}=\Phi_X\cup\{S\cup\{\infty\}\mid\mathrm{compact}\ S'\}$$<br />This is called the <b>one-point compactification</b>&nbsp;or the <b>Alexandroff compactification</b>&nbsp;of $X$.<br /><br />But the one with two infinities was cool too. The idea was that the two infinities at the far ends of $\mathbb{R}$ are somehow "disconnected" -- this is in contrast to e.g. $\mathbb{R}^{n&gt;1}$, where the "limits" of any divergent sequence <i>must</i>&nbsp;be&nbsp;connected by a giant loop at infinity. If we want each infinity to be a connected component of its own, what we need is a filter base of connected sets converging to a point at infinity.<br /><br />What do I mean? In general, all compactifications are obtained by considering some&nbsp;<b>partitions by non-compact sets of co-compact sets</b>&nbsp;-- in the case of the one-point compactification, these subsets are the trivial ones. In the case of the "plus infinity, minus infinity" compactification, the partition is the set of&nbsp;<b>connected components</b>.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-zcc5chPEu_0/XXPONNzxg3I/AAAAAAAAFtQ/NU71X-grBVgTWw0dRAXWQ9pGpy7sFFPfwCLcBGAs/s1600/compactification%2B-%2Bend.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="484" data-original-width="728" height="265" src="https://1.bp.blogspot.com/-zcc5chPEu_0/XXPONNzxg3I/AAAAAAAAFtQ/NU71X-grBVgTWw0dRAXWQ9pGpy7sFFPfwCLcBGAs/s400/compactification%2B-%2Bend.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Illustration of why we need a partition of the co-compact sets -- the above is <i>not</i>&nbsp;a partition, and there are sequences escaping every one of the four filters drawn. This is actually an illustration of why the end-compactification of the plane has only one infinity.</td></tr></tbody></table>OK, so how do we actually make this construction? The thing is that we have co-compact sets like $(-\infty,-2)\cup(-1,1)\cup(1,\infty)$ -- but $(-1, 1)$ is not really a connected component we care about. We care about the <b>connected components of a co-compact set that are a part of a chain that every co-compact set has connected components in</b>. One way to do this is as follows:<br /><ol><li>Construct an infinite nested basis $U_1\supseteq U_2\supseteq \dots$ of co-compact sets -- i.e. so that every co-compact set contains one of these sets.</li><li><i>Each chain</i>&nbsp;$N_1\supseteq N_2\supseteq\dots$ of connected components is a neighbourhood basis for a point at infinity, called an <b>end</b>.</li></ol>This is the <b>end compactification</b>.<br /><br />There are just three questions:<br /><ol><li>Is this independent of the co-compact set basis we use?</li><li>Does such a basis always exist?</li><li>Is the result always compact (this includes questions you may have about the existence of connected component chains, etc.)?</li></ol>To prove the first, it suffices to show that for any two co-compact bases $U$ and $V$, if a set contains some $N_i$, a set in some connected component chain of $U$, it contains an $M_j$, a set in some connected component chain of $U$ (this creates a natural bijection between the ends arising from $U$ and those from $V$). Because connected components are a partition, it suffices to show that for any two co-compact bases $U$ and $V$, if a set contains some $U_i$, it must contain a $V_j$ -- which is clearly true. So:&nbsp;<b>TRUE</b>.<br /><br />For the second -- this is equivalent to asking the dual question: is there a chain of compact sets such that every compact set is contained in a set in the chain? A simpler, equivalent way to phrase the question is: is there a chain of compact sets whose interiors cover $X$? This is called <b>exhaustion by compact sets</b>, or <b>hemicompactness</b>&nbsp;-- and as you may have guessed, <i>not</i>&nbsp;all spaces have it. So:&nbsp;<b>FALSE</b>.<br /><br />TBC: why 3 is false, general version, Gromov boundary, Stone-Cech, why Hausdorff compactifications are preferred, add labels<br /><br />A stupid joke -- the seven Cs: closed (x2), compact, connected, convergent, continuous, covers (x2). Maybe eight: co-.</div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-75405237479567066302019-08-28T05:35:00.000+01:002019-09-20T12:55:17.920+01:00Supplementary definitions: connectedness, clopen setsWhat does it mean for a space to have $n$ "separate" parts?<br /><br />Well, let's start from the basics -- how do we count the $n$ parts? We label them $1,...n$, i.e. we have a surjection $\ell:X\to [n]$. Now, for any point in $X$ mapping to some $k\le n$, we should be able to jiggle it around a bit so that it still maps to $k$, otherwise the other "part" mapping to something else will be connected to this one. In other words, $\forall k, \ell^{-1}(\{k\})\in\Phi_X$.<br /><br />Thus, we say that $X$ has $n$ <b>separate components</b> if there exists a <b>continuous surjective labeling function</b> $\ell:X\to [n]$, where $[n]$ is equipped with the discrete topology -- equivalently, we say that $X$ is the <b>union of $n$ nonempty disjoint open sets</b>&nbsp;(the "nonempty" part comes from the "surjective" part) (we'll call this an <i>open partition</i>). We can then define a <b>connected space</b>&nbsp;as one that cannot be written as an open partition.<br /><br />What about a set being connected? If we similarly define maps from the set to a labeling topology, we can see that what we're looking for here is precisely the notion of a subspace topology -- a set is a <b>connected set</b>&nbsp;if taken as a subspace, it is a connected space. This is clearly equivalent to saying that the set cannot be <i>covered</i>&nbsp;by two non-empty disjoint open sets, i.e. does not have a <i>open disjoint cover</i>.<br /><br />Let's define three parallel concepts to define the notion of a <b>connected component</b>:<br /><ul><li>A <i>maximal&nbsp;component partition (i)</i>&nbsp;of a space $X$ is an open partition of $X$ with the largest possible value of $n$ -- a <i>maximal&nbsp;component (i)</i>&nbsp;is a set in this partition.&nbsp;</li><li>A <i>open connected partition (ii)</i>&nbsp;of a space $X$ is an open partition of $X$ such that each component is connected -- a <i>open connected component (ii)</i>&nbsp;is a set in the partition.</li><li>A <i>connected component (iii)</i>&nbsp;of a space $X$ is a connected subset $S$ all of whose proper supersets are disconnected. A <i>connected component list (iii)</i>&nbsp;is the list of all such distinct subsets.</li></ul>(You can tell (i) is going to be ugly -- maximums and all don't play well with infinite cardinalities)<br /><br />So many questions.<br /><ol><li><b>Does a maximal component partition (i) always exist?</b>&nbsp;If $n$ must be an integer, then FALSE (e.g. $\mathbb{Q}$); If $n$ is any cardinal number, then TRUE (as the cardinality of the space is an upper bound). Note that with cardinal numbers, just because there is a maximum $n$ doesn't mean there is a maximal such partition, i.e. one that cannot be partitioned any further. That would be a open connected partition (ii).</li><li><b>Does an open connected partition (ii) always exist?</b>&nbsp;FALSE. E.g. $\mathbb{Q}$ with the standard topology -- any open set contains an infinite number of rational numbers, which is clearly disconnected.</li><li><b>Is a connected component list (iii) a partition of $X$?</b>&nbsp;TRUE.</li><ol type="a"><li><b>(Disjoint)</b> The union of two distinct connected components is disconnected by assumption, so can be written as an open partition, which create an open disjoint cover of each connected component -- this can only be true of an open set if the sets in the cover are precisely the connected components -- since they are disjoint, the conclusion follows.&nbsp;</li><li><b>(Union)</b>&nbsp;There is a unique connected component around each point, and it is the union of all connected sets containing the point (Proof: suffices to show this union is connected. An open disjoint cover of the union would be an open disjoint cover of each of these sets, and since none of these sets are disjoint, this would imply some of them are disconnected, contradiction.), and is clearly nonempty. The union of the connected components around each point contains each point.</li></ol><li><b>Sameness between partitions (notation: <i>is (i) implies is (ii)</i>)</b></li><ol type="a"><li><b>(i) imp (ii)?</b>&nbsp;FALSE in general, e.g. $\mathbb{Q}$.</li><li><b>(i) imp (iii)?</b>&nbsp;FALSE in general, e.g. $\mathbb{Q}$.&nbsp;</li><li><b>(ii) imp (iii)? </b>TRUE.</li><li><b>(ii) imp (i)? </b>TRUE.&nbsp;Suffices to show every open partition has at most as many sets as an open connected partition (ii). From 9, there exists a (ii) with at least as many sets than the given partition. From 6, this (ii) is unique.</li><li><b>(iii) imp (i)?</b>&nbsp;FALSE in general, e.g. $\mathbb{Q}$.</li><li><b>(iii) imp (ii)?</b>&nbsp;FALSE in general, as above, e.g. $\mathbb{Q}$. But if (ii) exists, then TRUE, by 4c and 7.</li></ol><li><b>Is a maximal component partition (i) unique up to permutation?</b>&nbsp;FALSE, as we saw with $\mathbb{Q}$.</li><li><b>Is an open connected partition (ii) unique up to permutation?</b>&nbsp;TRUE, by 4c and 7.</li><li><b>Is a connected component list (iii) unique up to permutation?</b>&nbsp;TRUE, from the "uniqueness" lemma in 3b.</li><li><b>Is every set $S$ in an open partition a union of maximal components (i)?</b>&nbsp;Ambiguous since the maximal components are not unique. If we want this to be true for any maximal component partition, this clearly FALSE, e.g. $\mathbb{Q}$. If we're saying there exists a maximal component partition such that this is true, then TRUE -- just consider the maximal component partitions of each such set and note that together they form a maximal component partition of the space.</li><li><b>Is every set $S$ in an open partition a union of open connected components (ii)?</b>&nbsp;TRUE by 4c and 10.</li><li><b>Is every set $S$ in an open partition a union of connected components (iii)?</b>&nbsp;By 3, it suffices to show that every connected component is either disjoint from $S$ or contained in it. This is true as otherwise their union would be a connected superset of the connected component.</li></ol>So here's our conclusion from all this: (iii) should be our definition of a <b>connected component</b>&nbsp;-- a connected subset whose supersets are all disconnected, or by construction, the union of all connected sets containing a given point. When these components are open, they are the same as in (ii) (which is the only time (ii) exists) -- they are open sets that partition your space. And when the number of connected components is finite, they are the same as (i).<br /><br />The reason we initially thought the definitions would all be equivalent is that we were only visualising cases with a <b>finite</b> number of connected components. In this case, it's easy to see that the connected components are open -- just keep partitioning with open sets until you have as many as the number of connected components. If you can partition any further, you have a open disjoint cover of a connected component, which is a contradiction. This construction cannot be reproduced with $\mathbb{Q}$.<br /><br />By the way, we've used (and proven) the following obvious fact throughout the above series of exercises: <b>the union of two non-disjoint connected sets is connected</b>.<br /><br />Here are some more facts:<br /><ol><li><b>Closure of a connected set is connected.</b>&nbsp;(i.e. something touching a connected set is "connected to it")&nbsp;Suppose not. Then the disjoint open cover of $\mathrm{cl}(S)$ is a disjoint open cover of $S$ unless $\mathrm{cl}(S)-S$ is open, which it isn't (as every open set containing a member of $\mathrm{cl}(S)$ intersects $S$).</li><ol type="a"><li>(Corollary) <b>A connected component is a closed set.</b></li><li>(Corollary) <b>The union of two touching connected sets is connected.</b></li></ol><li>The topology on $X$ is the disjoint union topology of those of the disjoint open sets partitioning it. (exercise left to the reader)</li><ol type="a"><li>(Corollary) If a space $X$ can be partitioned into $n$ disjoint open sets each homeomorphic to $A$, its topology is the product topology $A\times [n]$ where $[n]$ has the discrete topology.</li></ol></ol>We also have the following interpretation of connected components: they are the <i>equivalence classes</i>&nbsp;of the relation "there exists a connected set containing both points".<br /><br />Make sure you understand why the following are true, and why the "finite" qualifiers are needed where we've used it.<br /><ul><li>A connected component, and thus a finite union of connected components, is always closed.</li><li>If there are only a finite number of connected components, a connected component, and thus a (finite) union of connected components is always open.</li></ul><div>Also: <b>continuous functions preserve connectedness</b>&nbsp;(consider the preimage of a connected component). This is a generalisation of the <b>intermediate value theorem</b> (do you see why?).</div><hr /><b><br /></b> <b>Clopen sets</b><br /><br />You may have noticed that when we have a finite number of connected components, each component is both closed and open. Closed as always, because nothing touches a connected component. Open, because the complement is a finite union of closed sets and thus closed. When we have infinite components, sometimes some infinite unions act this way. Sets that are both open and closed are called <i>clopen</i>.<br /><br />In fact, every <b>clopen set $S$ is necessarily a union of (possibly infinite) connected components</b>. This is an instructive proof to work through -- the proof you'd see is: suppose for contradiction that $\exists x\in X$ touching both $S$ and $S'$. Then $x\in S$ implies $S'$ not closed and $x\in S'$ implies $S$ not closed, contradiction.<br /><br />The intuition behind the proof is this: we can understand a closed set as a set where all convergent nets lying inside converge inside, and an open set as a set where all nets converging inside lie inside. So if we stuck a set "touching" $S$ (i.e. with some point touching them both), then for any point on their border, we could construct sequences on both sides converging to it. Then both $x\in S$ and $x\in S'$ lead to contradictions.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-cws3z90Q3lg/XXJ53dtCDmI/AAAAAAAAFtE/7ksvQTi_CBorGu6PnlPgDScr8OztXU9WQCLcBGAs/s1600/clopen.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="454" data-original-width="489" height="297" src="https://1.bp.blogspot.com/-cws3z90Q3lg/XXJ53dtCDmI/AAAAAAAAFtE/7ksvQTi_CBorGu6PnlPgDScr8OztXU9WQCLcBGAs/s320/clopen.png" width="320" /></a></div>The open set/closed set formalism is a way to simplify this whole reasoning without having to worry about nets and their limits.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-5481427728427210122019-08-25T21:23:00.000+01:002019-09-09T09:11:35.338+01:00Topology III: example topologies, inherited topologies, distinguishabilityLet's think of some example topologies one may introduce on a set -- this is easiest in the open set formalism. Prove the statements we make.<br /><ul><li>The <b>discrete topology</b>, where $\Phi=\wp(X)$, which are also the closed sets. The neighbourhoods are simply the principal filters, i.e. $N(x)=\{S\subseteq X\mid x\in S\}$. The closure operator is just the identity and $\sim$ is just $\in$. The limit of a function is simply its value at the point. All functions <i>from</i> a discrete space are continuous everywhere. This basically adds no structure at all to a set, as each point is essentially "separated" from each other point. Clearly, any bijection between sets acts as a homeomorphism between the corresponding discrete spaces.</li><li>The <b>indiscrete topology</b>,&nbsp;where $\Phi=\{\varnothing, X\}$, which are also the closed sets. The neighbourhoods are $N(a)=\{X\}$. Every point touches every non-empty set, and the closure of a non-empty set is just $X$. Every point is the limit of every function. All functions <i>to</i> an indiscrete space are continuous. In other words, every point is "basically the same", cannot be distinguished.</li><li>The <b>kinda-sorta topology</b>, where $\Phi=\wp(S)\cup \{X\}$ for some $S\subset X$. You should be able to see by now that this topology has a bunch of discrete points and a bunch of indsistinguishable points. A neighbourhood of a point in $S$ touches only the sets containing it, but a neighbourhood of a point outside $S$ touches every non-empty set -- so the closure of a set is just the set united with $S'$. Not very interesting, so I'll stop.</li><li>The <b>cofinite topology</b>&nbsp;on an infinite set, where $\Phi = \{S\mid \mathrm{finite}\ S'\}\cup\{\varnothing\}$ -- the closed sets are the finite sets and $X$ -- a neighbourhood of a point is a cofinite set containing that point. A point touches a set if that set is infinite or contains that point -- so the closure of a finite set is the identity and the closure of an infinite set is $X$. For a function $f:X\to Y$ where $X$ is cofinite, its limit at $a$ is a point such that for each of its neighbourhoods, almost all values of $x$, including $a$, map into it. The function is continuous at $a$ if for each neighbourhood of $f(a)$, almost all $f(x)$ are in it -- it is continuous everywhere if every open set in $Y$ contains almost all $f(x)$. In particular, if $Y$ is also cofinite, the limit can only be equal to $f(a)$, which it is iff the function is finite-to-one at every point besides $a$ -- and a function that is continuous everywhere is a finite-to-one function.</li><li>The <b>cocountable topology</b>&nbsp;on the real numbers is basically the same as above except with "countable" replacing finite everywhere.</li><li>The <b>cobounded topology</b>&nbsp;on a metric space -- again kinda the same idea.</li><li>The <b>co(finite volume) topology</b>&nbsp;on a measurable space -- same idea, I guess? Check and find out.</li><li>The <b>first-n topology</b>&nbsp;on the naturals where a set is open if it is of the form $\{x&lt; n\}$ for some (natural or infinite) $n$. The neighbourhoods of a point $x$ are all the sets containing $\{1,2\dots x\}$, a point touches a set if the set contains a point less than or equal to it, the closure is $\mathrm{cl}(S)=\{x\ge\min S\}$. The limit of a function $f:X\to Y$ ($X$ equipped with the first-n topology) at a point $a$ is $L$ if $f(x)$ is topologically indistinguishable from $L$ for all $x\le a$. If $Y$ is also first-n, then the function's limit is any upper bound on its value for $x\le a$ -- the function is continuous at $a$ if its value at $a$ is higher than that at any lower value, and is continuous everywhere if it is (not strictly) increasing.</li></ul><div>You should get some kind of feel for open sets now -- if you can find an open set containing one point but not another, they can be distinguished in some sense. This is the definition of <b>topological&nbsp;distinguishability</b>.<br /><br />One may then define a partial order on the topologies on a set based on <b>fineness</b>&nbsp;and <b>coarseness</b>&nbsp;(this is called the <b>comparison of topologies</b>), where $\Phi_1\le\Phi_2\iff \Phi_1\subseteq\Phi_2$. The discrete topology is the finest topology and the indiscrete/trivial topology is the coarsest topology. A continuous function $f:X\to Y$ remains continuous if $X$ becomes finer or $Y$ becomes coarser, but a homeomorphism does not remain continuous in either situation (but can if they both change in a specific way)</div><div><br /></div><div>For fun, let's try to visualise the cofinite topology (keep the set $\mathbb{N}$ in mind) -- the only sets with any sort of interiors are the cofinite sets. A <i>finite</i>&nbsp;set is closed, has no interior (it's its own boundary, i.e. it's a bunch of discrete points) and doesn't touch anything outside (because it's closed). An <i>infinite </i>set touches every single point, making them all part of their boundary -- it is closed at the points it contains and open at the points it doesn't contain. This explains the nature of the open and closed sets to us: if a point is contained in a finite set, it is located on its boundary. No matter how large you make the set, as long as it is not cofinite, it cannot contain the point on its interior, as its complement is infinite and thus touches the point.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/--WNBTBBh0c0/XWD4mlvGaOI/AAAAAAAAFrM/89M7Pfd3Tzc2PtWNAc475M0X16Kij857gCLcBGAs/s1600/cofinite.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="760" data-original-width="764" height="397" src="https://1.bp.blogspot.com/--WNBTBBh0c0/XWD4mlvGaOI/AAAAAAAAFrM/89M7Pfd3Tzc2PtWNAc475M0X16Kij857gCLcBGAs/s400/cofinite.png" width="400" /></a></div><div>You can see that the description above -- while basically a very intuitive explanation of how the topological features of the space interact -- is basically equivalent to rigorous topological arguments. So our axioms of topology really do describe spaces with some very similar properties to ones we're used to dealing with.</div><div><br /><b>More examples: inherited topologies</b></div><div><br /></div><div>We've already seen one kind of inherited topology: that inherited from a metric space -- where the open sets are just sets with some "wiggle-room" for each element.<br /><br />Here are some obvious "inherited topologies" -- topologies determined by some other structure:</div><div><ul><li><b>Subspace topology -- </b>The open sets define a generalised "wiggle room" in a set $X$ -- for a function to be continuous, it must be similar to its value for some "wiggle room" in its domain. In a subspace (subset)&nbsp;$S$&nbsp;of that space, we don't really know or care how a function on $S$ behaves outside $S$ -- for it to be continuous <i>in</i>&nbsp;$S$ at a point, the only wiggle room we care about are is wiggle room inside $S$ (e.g. the Heavside step function is continuous on the nonnegative reals) -- and since the open sets on $X$ define what wiggle room is in this topological space, $\Phi_{S}=\{O\cap S: O\in\Phi_X\}$. Check that the axioms apply.</li><li><b>Disjoint union topology -- </b>You can't really invert the subspace topologies on a partition to necessarily recover the original topology -- you necessarily lose information about how the subspaces stick to each other when you cut it up. But you can still just "stitch together" some topological spaces, i.e. take a disjoint union of some sets -- it's important that you don't cut anything while doing so, so the "stitching" (i.e. the canonical injections) need to be continuous. There are multiple ways to define such a stitching, but the canonical disjoint union topology is the finest among them -- we let <i>all</i>&nbsp;sets $U$ whose pre-images $p_i^{-1}(U)$ in each $X_i$ are open be open. I.e. $\Phi_{\sum X_i}=\{O\mid \forall i, O\cap X_i\in\Phi_{X_i}\}$.</li><li><b>Product topology -- </b>Continuity on $X\times Y$ means being able to wiggle a little bit in any direction on $X\times Y$ and keep the function's value similar. The only way to generally define this notion of a general wiggle space is as an arbitrary union of little "squares" (called <b>open cylinders</b>), like in calculus, i.e. with some abuse of notation: $\Phi_{\prod X_i}=\{\bigcup \prod O_i : O_i \in \Phi_{X_i}\}$.</li><li><b>Quotient topology -- </b>Quite often, we want to "collapse" a topology tying together some nearby or otherwise related points. Such a relation is an equivalence relation $\sim$, and what we need is a topology on the equivalence classes. The idea behind the topology is that the equivalence classes of nearby points are still nearby, i.e. so that the function $q(x)=[x]$ is continuous while not becoming too indiscrete: i.e. $\Phi_{X/\sim}=\{S\subseteq X/\sim\mid q^{-1}(S)\in\Phi_X\}$.</li><li><b>Initial topology -- </b>The subspace topology can be understood as asking that the embedding of the subspace into the parent space is continuous, while not adding too many open sets. This notion can be generalised -- we can ask for the coarsest topology on a set $X$ that makes a family of functions from $X$ continuous. This is a generalisation of the subspace and product topologies</li><li><b>Final topology -- </b>Dually, we may ask for the finest topology on a set $X$ that makes a family of functions to $X$ continuous. This is a generalisation of the disjoint union and quotient topologies.</li></ul></div><div><b>Elementary theorems (exercises)</b></div><div><ol><li>Figure out if these statements are True/False:</li><ol type="a"><li>A set is closed iff it is closed under limits of nets.</li><li>The closure of a set equals the intersection of all closed sets containing the set.</li><li>The interior of a set equals the union of all open sets contained in the set.</li><li>A set equals the union of all closed sets contained in the set.</li><li>A set equals the intersection of all open sets containing the set.</li><li>Having an empty interior implies being closed.</li><li>Having an empty interior is equivalent to being a subset of its boundary.</li><li>$\forall N\in N(x), y\in N(x)\implies x=y$.&nbsp;</li><li>Limits are unique.</li><li>If $S$ and $T$ are disjoint from each other's closures, their closures are disjoint.</li><li>The product topology is the initial topology with respect to the projection maps.</li><li>$\sum_I X$ with the disjoint union topology is homeomorphic to&nbsp;$X\times I$ (with $I$ endowed with a discrete topology) with the product topology.</li></ol><li>Explain (not prove!) why the interval $[0,1)$ is not homeomorphic to $S^1$.</li></ol><b>Answers</b><br /><ol><li>True/False:</li><ol type="a"><li><b>TRUE.</b>&nbsp;(forward) Suppose a net in $S$ converges to a point $a$ in $S'$. Since $S'$ is open, there is an open set around $a$ contained in $S'$, and this open set must have points of $a$. (backward) Suppose $S$ is not closed, so there is a point in $S'$ whose neighbourhoods all intersect $S$. We can use these neighbourhoods to construct a net converging to it.</li><li><b>TRUE.</b>&nbsp;Suffices to show that the closure of $S$ is closed, and is contained in every closed set containing $S$. (1) The set of all points around which there exists an open set not intersecting $S$ is open is an open set, as it is the union of all these "existing" open sets. (2) Take a point $p$ outside a closed set $C\supseteq S$. Then $C'$ is an open set containing $p$ that does not intersect $S$, thus $p$ does not touch $S$. So every point that does touch $S$ is in $C$.</li><li><b>TRUE.</b>&nbsp;Dual to the above.</li><li><b>FALSE.</b>&nbsp;See e.g. the trivial topology. True for a metric space, though.</li><li><b>FALSE.</b>&nbsp;Dual to the above.</li><li><b>FALSE.</b>&nbsp;Remember the infinite non-cofinite sets in the cofinite topology? This isn't even true on a metric space (poke a hole in a curve).</li><li><b>TRUE.</b>&nbsp;Obviously.</li><li><b>FALSE.</b>&nbsp;If you look through the proof for metric spaces, you'll see that it relies on the ability to create an open set that separates the two non-equal points. This is topological distinguishability -- the kind of space for which this statement is true is a <i>T1 space</i>; if we had the symmetric condition, it would be true in a <i>T0 space</i>.</li><li><b>FALSE. </b>Remember the first-n topology? The proof requires having disjoint neighbourhoods of distinct points -- this requires being a <i>T2 space</i>, also known as a "Hausdorff space".</li><li><b>FALSE.</b>&nbsp;Consider $\mathbb{R}^-$ and $\mathbb{R}^+$.</li><li><b>TRUE.</b>&nbsp;Requiring the projection maps to be continuous just means that each $p_i^{-1}(U)$ is open. The topology generated by these is the topology described -- so what is it? Intersecting these sets across $i$ as $\bigcap_i p_i^{-1}(U)$ leads to the open cylinders, which generate the product topology.</li><li><b>TRUE.</b>&nbsp;$p_i^{-1}(U)\cap X_i = U$.</li></ol><li>Because the preimage of the open set $[0,1/2)$ is not open in the circle. This goes back to the fact that because we're cutting the circle, the neighbourhood of 0 becomes "easier". Again, this is <i>not</i>&nbsp;a proof, just a proof that cutting doesn't work.</li></ol></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-18479511458813798142019-08-22T19:12:00.001+01:002019-09-08T13:38:30.136+01:00Topology II: Kuratowski closure topology, nets, neighbourhood basis<a href="https://thewindingnumber.blogspot.com/2019/08/topology-limits-continuity-and.html">So far</a>, we've described two axiomatisations of toplogy: in terms of <b>neighbourhoods</b>&nbsp;and in terms of <b>open sets</b>. While the neighbourhoods definition was a natural extension of our understanding of topological structure being in terms of limits, the open sets definition is kind of hard to wrap your head around, as far as I can see. A continuous function doesn't even preserve open sets (the definition of a continuous function isn't "preserves neighbourhoods" in the neighbourhood formulation either, but at least we know where it comes from). It's not openness in particular that's important, we could formalise topology in terms of their complements the closed sets, too.<br /><br />In the following set of exercises, we will build an alternative axiomatisation of topology based on the notion of <b>touching</b>, which will perhaps give us some explanation of why the notion of openness and closedness are important to topology.<br /><ol><li>Consider the relation of "touching" between a point and a set (it can't be between a point and a point, but it can be with a set -- kind of for the same reason that a sequence tends to a point but there is no point in the sequence that equals that point). How would you write this in terms of the open set topology? (<b>Ans:</b>&nbsp;every open set containing $x$ intersects $S$)</li><li>Now formulate an open set in terms of touching. (<b>Hint:</b>&nbsp;formulate a closed set first, the open set is its complement) (<b>Ans:</b>&nbsp;no point in $S$ touches $S'$)</li><li>Great. Find out what axioms we need on the touching operation $\sim$ to prove the three axioms of open sets.</li><ol type="a"><li>$X$ is open requires -- (<b>Ans:</b>&nbsp;$\not\exists x \sim \varnothing$)</li><li>$O_1\cap O_2\in\Phi$ requires -- (<b>Ans:</b>&nbsp;$x\sim S\cup T\Rightarrow x\sim S \lor x\sim T$)</li><li>$\bigcup_\lambda O_\lambda \in\Phi$ requires -- (<b>Ans:</b>&nbsp;$x\sim S\subseteq T \Rightarrow&nbsp;x\sim T$)</li></ol><li>Given a touching relation $\sim$, we can produce the set of open sets $\{S\mid \forall x\in S, x\not\sim S'\}$. From this, we can produce the relation $\bar\sim$, given by: $\forall T\, \mathrm{st.} (x\in T \land \forall y\in T, y\not\sim T'), S\cap T\ne\varnothing$. Try proving this is equivalent to $x\sim S$, and see what axioms you need.</li><ol type="a"><li>$\Leftarrow$ requires -- (<b>Ans:</b>&nbsp;None. Suppose $S\cap T$ were empty. Then $S\subseteq T'$, so by 3c-ans, $x\sim T'$, contradiction.)</li><li>$\Rightarrow$ requires -- (<b>Hint:</b>&nbsp;&gt;given such a $T$, construct a smaller $T$ whose intersection with $S$ is empty if $x\not\sim S$) (<b>Ans:</b>&nbsp;$S\subseteq\mathrm{cl}(S)$ and $\mathrm{cl}(\mathrm{cl}(S))\subseteq\mathrm{cl}(S)$. Suppose $x\not\sim S$. Then for any $T$ satisfying the LHS (e.g. the universe by 3a-ans), consider $T\cap\mathrm{cl}(S)'$ where $\mathrm{cl}(S)$ is the set of all points touching $S$ -- $x\in T\cap \mathrm{cl}(S)'$ by assumption; now given $y\in T\cap \mathrm{cl}(S)'$ we want to show $y\not\sim T'\cup \mathrm{cl}(S)$ (for this to contradict the claim that $S\cap(T\cap\mathrm{cl}(S)')\ne\varnothing$, we need that $S\subseteq\mathrm{cl}(S)$ -- <i>this is new!</i>). Suppose that $y\sim T'\cup\mathrm{cl}(S)$ and apply 3b-ans: $y\sim T'$ contradicts $y\in T$ as $T$ is open; now we just want $y\sim\mathrm{cl}(S)$ to imply $y\in\mathrm{cl}(S)$ -- <i>this is new!</i>)</li></ol><li>So rewrite our axioms in terms of the <b>closure operator</b> $\mathrm{cl}$ as follows -- and this is completely equivalent to our earlier "touching" description of the closure operator as $x\sim S\iff x\in\mathrm{cl}(S)$:</li><ol type="a"><li>$\mathrm{cl}(\varnothing)=\varnothing$</li><li>$\mathrm{cl}(S\cup T)\subseteq\mathrm{cl}(S)\cup\mathrm{cl}(T)$</li><li>$S\subseteq T\Rightarrow \mathrm{cl}(S)\subseteq\mathrm{cl}(T)$</li><li>$S\subseteq\mathrm{cl}(S)$</li><li>$\mathrm{cl}(\mathrm{cl}(S))\subseteq\mathrm{cl}(S)$</li></ol><li>To get yourself comfortable with this closure operator, check if the following statements are true or false:</li><ol type="a"><li>$\mathrm{cl}(S)\cap\mathrm{cl}(T)\subseteq\mathrm{cl}(S\cap T)$ (<b>Ans: </b>False -- consider two disjoint sets sharing a boundary)</li><li>$\mathrm{cl}(S\cup T)=\mathrm{cl}(S)\cup\mathrm{cl}(T)$ (<b>Ans:</b>&nbsp;True -- in fact can replace 5b and 5c)</li><li>$\mathrm{cl}(\mathrm{cl}(S))=\mathrm{cl}(S)$ (<b>Ans:</b>&nbsp;True -- it's also equivalent to the statement that $\mathrm{cl}(S)$ is closed, i.e. that no point in $\mathrm{cl}(S)'$ touches $\mathrm{cl}(S)$ -- can you see why?)</li><li>$S\subseteq\mathrm{cl}(T)\Rightarrow \mathrm{cl}(S)\subseteq\mathrm{cl}(T)$ (<b>Ans:</b>&nbsp;True -- in fact can replace 5c and 5e)</li><li>$\mathrm{cl}(S\cap T)\subseteq\mathrm{cl}(S)\cap\mathrm{cl}(T)$ (<b>Ans:</b>&nbsp;True -- in fact the arbitrary-intersection version of this is the reason we have the equivalent 5c, from 3c-ans, and can replace it)</li></ol><li>From 5c and the answer to 6a, you might see an analogy with <b>limits of sequences</b> -- indeed, a "set" is kind of a sequence, and its closure is to add the <b>limit points</b> of the sequence to the set. Indeed, the definition of a <b>continuous function</b>, as we will see, is $f(\mathrm{cl}(S))\subseteq\mathrm{cl}(f(S))$, i.e. all <b>limit points remain limit points</b> (and for a homeomorphism, the reverse inclusion is also true). This definition should be fairly obvious as it just says "if $x$ touches $S$, $f(x)$ touches $f(S)$", i.e. nothing gets ripped apart in the process. So $\mathrm{cl}$ is <b>a generalisation of a limit</b>&nbsp;to any subset of $X$.</li><li>Well, actually, this is not clearly a stronger condition than "convergent sequence limits are preserved" -- that actually requires that convergent sequences remain convergent, that you don't just add a new limit point like you can for non-convergent sequences.</li></ol><div>But this does get us thinking -- we saw above that "preserves the limit points of every set" fully describes a continuous function. But does "preserves the limits of convergent sequences" -- which we used to motivate the idea of topology in the first place -- still characterise a continuous function? Is $x_n\to a\Rightarrow f(x_n)\to f(a)$ (for all sequences $(x_n)$) equivalent to $x\to a \Rightarrow f(x)\to f(a)$ like it does for standard metric spaces? Certainly the backward implication is correct.</div><div><br /></div><div>Let's go through the proof of the forward implication on metric spaces.</div><div><br /></div><div>Suppose $f$ is not continuous at $a$. So $\exists\varepsilon&gt;0$ such that $\forall\delta&gt;0,\exists x\,\mathrm{st.} |x-a|&lt;\delta, |f(x)-f(a)|\ge\varepsilon$. We want to construct a sequence $(x_n)$ converging to $a$ so that $\forall N, \exists k \ge N, |f(x_k)-f(a)|\ge\varepsilon$. We construct the $n$th element of the sequence from the choice function for $\delta = 1/n$, which is an $x$ within $1/n$ of $a$ such that $|f(x)-f(a)|\ge\varepsilon$.&nbsp;</div><div><br /></div><div>OK -- how can we generalise this to an arbitrary topological space?&nbsp;</div><div><br /></div><div>Suppose $f$ is not continuous at $a$. So $\exists N\in N(f(a))$ such that $f^{-1}(N)\notin N(a)$, i.e. $\forall M\in N(a), \exists x\in M, f(x)\notin N$ (make sure you can tell that these are indeed equivalent). We want to construct a sequence $(x_n)$ converging to $a$ so that $\forall K, \exists k\ge K, f(x_k)\notin N$. Now here's the deal: if we can construct a sequence of $M$s in $N(a)$ that ultimately converge to the point $a$, like we could in the metric case, we are done.</div><div><br /></div><div>What does it mean to "ultimately converge to the point $a$"? We need that the choices are eventually contained in every neighbourhood of $a$ -- or to write it down concisely: we want a sequence of sets $M_i\in N(a)$ such that $\forall N\in N(a), \exists i, M_i\subseteq N$. This is called a <b>neighbourhood basis</b>&nbsp;-- and particularly since the domain of $i$ is the natural numbers, a <b>countable neighbourhood basis</b>.</div><div><br /></div><div>Well, so does every topology admit a countable neighbourhood basis to every point? As it turns out, there are counter-examples.</div><div><br /></div><div>So we've generalised a bit beyond the notion of preserving limits of sequences, and this is OK -- the natural numbers aren't <i>that</i>&nbsp;fundamental, are they? I would say that preserving the limit points of sets as in pt. 7 is a more fundamental notion than preserving the limits of convergent sequences. In any case, different terms exist for different levels of specialisation, such as:</div><div><ul><li>Pre-topological space (where you can have "multiple layers of boundaries" -- this is what happens if you leave out idempotence of closure, the arbitrary union axiom of open sets, or the wiggle-room axiom of neighbourhoods)</li><li>Topological space</li><li>T0-space (distinguishability)</li><li>T1-space</li><li>T2-space or Hausdorff space (uniqueness of limits of nets)</li><li>Alexandrov topology (with arbitrary intersection <i>and</i>&nbsp;union)</li><li>First-countable space (for which the "limits of sequences" thing suffices)</li><li>Second-countable space</li><li>Separable space</li><li>Uniform space</li><li>Metrisable space</li></ul><div>In any case -- although we do not always have a countable neighbourhood basis, we <i>do</i> always have a neighbourhood basis for a point -- the entire neighbourhood filter itself. And while the neighbourhood filter isn't a countable set, it is a <b>directed set</b>&nbsp;(a poset where every two elements have a shared superior). The generalisation of a sequence to a directed domain is called a <b>net</b>.&nbsp;</div><div><br /></div><div>We can define the limit of a net from a directed set $D$ in the same way as usual with filters: $\forall N\in N(a), f^{-1}(N)\in N(+\infty)$, or $\forall N\in N(a), \exists K\in D, \forall k \ge K, x_k\in N$. Then our generalisation of the "limits of a sequence" motivation for topology is:</div></div><div><br /></div><div><b>A map is continuous if it preserves the limits of all convergent nets.</b></div><br />Of course, the convergence of nets is equivalent to the convergence of filters.<br /><br /><hr /><br /><b>Dense sets</b><br /><b><br /></b>Let's discuss a quick application of closure -- recall how $\mathbb{Q}$ is&nbsp;<b>dense</b>&nbsp;in $\mathbb{R}$. This can be formulated in numerous ways, but the simplest way is probably "every open set in $\mathbb{R}$ intersects $\mathbb{Q}$. Does this remind us of something? Yes, of course -- it's the definition of closure in terms of open sets, i.e. $\mathrm{cl}(\mathbb{Q})=\mathbb{R}$.<br /><br />And this is obviously true -- it's the definition of $\mathbb{R}$, isn't it? So there's a very natural explanation for the rationals being dense in the reals.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-66451297101226641012019-08-16T05:39:00.000+01:002019-09-06T12:49:31.841+01:00Topology I: limits, continuity and homeomorphism, the neighbourhood filter, open setsThe idea of a topology, although often linked to geometry* in some classifications, is perhaps better understood as something that has fundamentally more to do with groups, vector spaces and such. You know, when talking about groups, what we really care about is the <i>structure</i>&nbsp;-- we don't really care if we're dealing with $U(1)$ or $SO(2)$ or $\mathbb{Z}/2\pi\mathbb{Z}$ -- if two groups are isomorphic, they have the same "structure" and it's this structure we care about. We're interested in the properties of the group that are invariant under isomorphism -- these are the "group theoretic" properties. <br /><br />Anyway -- a group homomorphism $f:G\to H$ is a function that <b>preserves the group structure</b>, or alternatively <b>commutes with multiplication</b>, i.e. $f(x\cdot y) = f(x)\cdot f(y)$ (you might see the "commuting" analogy better if you write $x\cdot y$ as $\mathrm{mult}(x,y)$. Similarly, a vector space homomorphism or "linear transformation" $f:V\to W$ is a function that <b>preserves the linear structure</b>&nbsp;or alternatively <b>commutes with linear combination</b>/commutes with addition and scalar multiplication, i.e. $f(ax+y)=af(x)+f(y)$.<br /><br /><div class="twn-furtherinsight">Here's another example: the homomorphisms of order theory are increasing functions. Do you see why?</div><br /><div class="twn-beg">*: It's not completely unrelated to geometry, though, which as we discussed <a href="https://thewindingnumber.blogspot.com/2019/04/geometry-positive-definiteness-and.html">here</a>&nbsp;is the theory of the invariants of a manifold under some symmetries. Well, symmetries are analogous to isomorphisms, but different -- symmetries form a group, in that the transformation from one state of the manifold to another can be judged to be the "same transformation" as that between two other states, i.e. $gx=y$ and $gx'=y'$. But there is no natural way to "parameterise" group isomorphisms so that you can compose them -- is this what one calls a "groupoid"? I don't know.</div><br />In general,&nbsp;commuting&nbsp;means some sort of preservation or non-interference policy. Here's another example: a <b>continuous function</b>&nbsp;$f:X\to Y$ is a function that <b>commutes with the limit</b>, i.e.<br /><br />$$f\left(\lim_{x\to a}x\right)=\lim_{x\to a}f(x)$$<br />The properties of a space that are invariant under continuous functions are precisely what encompass the theory of <b>topology</b>&nbsp;-- so a&nbsp;<b>topological homomorphism</b>&nbsp;is a "<b>continuous function</b>", and a&nbsp;<b>topological isomorphism</b>&nbsp;is a continuous function with a continuous inverse or a "<b>homeomorphism</b>". The&nbsp;<b><i>structure</i>&nbsp;of a topological space is precisely the limit operation</b>. It's not a binary operation like group multiplication $(\cdot):G^2\to G$ or vector addition $(+):V^2\to V$ or a unary function family like scalar multiplication $(\cdot):K\times V\to V$, but an operator family $(\lim):(\mathcal{I}\to X) \to (\mathcal{I}\to X)$ where $\mathcal{I}$ is fixed to be the domain set.<br /><br /><div class="twn-furtherinsight">We could also phrase this in an equivalent more natural manner -- a continuous function is one that preserves convergent sequences and their limits ("Cauchy" is incorrect on incomplete spaces, e.g. $1/x$ is continuous on $\mathbb{R}^+$ but maps the Cauchy sequence $(1/n)$ to $(n)$).</div><br />Well, our equation above doesn't even make sense for <b>incomplete spaces</b> like $\mathbb{R}^+$ or $\mathbb{Q}$ ($\lim_{x\to a}f(x)$ is not necessarily defined), but we still want to talk about the "topology" of such spaces. In any case, we've talked about spaces without even describing what the objects of the space are, or how the "space" arises from the set (endowing with a metric is too much information that isn't invariant under continuous functions).<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/7/79/Neighborhood_illust1.svg/777px-Neighborhood_illust1.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="753" data-original-width="777" height="310" src="https://upload.wikimedia.org/wikipedia/commons/thumb/7/79/Neighborhood_illust1.svg/777px-Neighborhood_illust1.svg.png" width="320" /></a></div><br />How would we generalise the notion of a limit to an arbitrary space? The key is something called a "<b>filter</b>" specifically a "<b>neighbourhood filter</b>". Filters are interesting mathematical objects with wide mathematical applications, but in this context, the idea behind a neighbourhood filter is that we want a sort of "convergence" of a poset of sets to a point. You would recall from basic real analysis that expression $\min(\delta_1,\delta_2)$ and $\max(N_1,N_2)$ are of importance to many theorems -- this is taking the intersection of two neighbourhoods and forming a new one. This leads to the following axioms:<br /><br /><b>Axioms for the neighbourhood topology</b><br /><ol><li>$\forall N\in N(x), x\in N$ (to relate the neighbourhoods to the points themselves)</li><li>$\forall N\in N(x), \exists M\in N(x), M\subseteq N\land \forall y\in M, N\in N(y)$ (to ensure the sets are only neighbourhoods to points on their "interior")</li><li>$\forall N\in N(x), \forall N' \supseteq N, N' \in N(x)$ (to express the notion of convergence -- we really want a "smallest neighbourhood", but since there is no such thing, we want a sequence of neighbourhoods that get arbitrarily small)</li><li>$\forall N_1, N_2\in N(x), N_1\cap N_2\in N(x)$ (the intersection thing)</li><li>$\forall x\in X, N(x)\ne\varnothing$ (this is not strictly neceessary, but there is a point without neighbourhoods, you could just remove it from the set and not lose anything topological -- do you see why? -- as it is not topologically related to any other point -- do you see why? -- think about what continuous functions do with them)</li></ol><div>The last three are the axioms of a filter, and the first two specify the <i>neighbourhood</i>&nbsp;filter in particular. Then the definition of a limit is:<br /><br />$$\lim_{x\to a}f(x)=L\iff \forall U\in N(L),f^{-1}(U)\in N(a)$$<br />And the definition of continuity is:<br /><br />$$\forall U\in N(f(a)),f^{-1}(U)\in N(a)$$<br />It's a useful exercise to confirm that with the standard neighbourhoods on $\mathbb{R}$, this reduces to the standard definition of the limit. It's also useful to generalise the standard properties of the limit with this definition -- start with the uniqueness of the limit. You'll see that this axiomatisation really does capture the basic "idea" of the limit.<br /><br />(If the second axiom is a bit unclear, the point is that the set $[0,1)$ is not the neighbourhood of the point 0 even though it contains it -- a neighbourhood must contain an epsilon-ball around the point. Convince yourself that this is necessary by constructing a situation where $f^{-1}(U)$ is $[0,1)$ but $f$ is discontinuous.)<br /><br />These axioms are the <i>axioms for a topology defined in terms of neighbourhoods</i>, or a <b>neighbourhood topology</b><b>&nbsp;</b>-- any choice of neighbourhoods satisfying these axioms define a topology on the underlying set. <br /><br />The second axiom is appalling with its number of quantifiers, but it gives some solid idea of what a neighbourhood really is -- one can show it's equivalent to $\forall x\in\mathrm{Int}A, \mathrm{Int} A\in N(x)$ (where $\mathrm{Int}A$ is the interior of $A$, the set of points of which $A$ is a neighbourhood), i.e. that $\mathrm{Int}A$ is an <b>open set</b> and thus equivalent to $\forall N\in N(x), \mathrm{Int} N\in N(x)$. It is then easy to see that this implies the notion in our head that "<i>a neighbourhood of a point is a set that contains an open set containing that point</i>".<br /><br />OK, so we have these (circular!) definitions of neighbourhoods and open sets in terms of each other:<br /><br /><b>Relationship between neighbourhood topology and "associated" open set topology</b><br /><ol><li>An open set is a set that is the neighbourhood of each of its points: $O\in\Phi\iff \forall x \in O, O\in N(x)$</li><li>A neighbourhood of a point is a set containing an open set containing the point: $N\in N(x)\iff \exists O\in\Phi, x\in O\subseteq N$</li></ol>Now, so far we've been starting with neighbourhoods and defining open sets from them with (1) -- and our discussion above shows that this respects (2), but instead, we could start with a definition of topology in terms of open sets and define neighbourhoods from them with (2), then check that it respects (1). So we need to find out what <i>axioms</i>&nbsp;an <b>open set topology</b> must satisfy such that the associated neighbourhoods satisfy the axioms for a neighbourhood topology, and that respects (1) (i.e. such that the open sets arising from these neighbourhoods are the same as the original open sets).<br /><br />To find these axioms, we'll simply try and "prove" each of the neighbourhood topology axioms and see what axioms we could use on the open sets to do so.<br /><br /><b>Figuring out the right open set axioms: Part I</b><br /><ol><li>We want $x\in N(x)$. But this follows from the definition of the neighbourhood in (2), so no axioms on the open sets are imposed here.</li><li>Same as above -- you can trivially confirm that the open set contained in the neighbourhood is the $M$ we want. That these two axioms follow from the definition was the point of the definition (2).</li><li>Same as above.</li><li>What's the open set around $x$ contained by $N_1\cap N_2$? The simplest answer is $O_1\cap O_2$. So this produces an axiom: <b>open sets are closed under finite intersection</b>.&nbsp;</li><li>Every point having a neighbourhood is equivalent to&nbsp;<b>every point being contained in an open set</b>.&nbsp;</li></ol>Now to ensure the $\text{open sets}\to \text{neighbourhoods} \to \text{open sets}$ chain ends up where it started -- what this is basically doing is ensuring the converse $\text{neighbourhood topology}\Rightarrow\text{open set topology}$, because we will be able to show that these "open set axioms" will be true for the&nbsp; the open sets arising from a neighbourhood topology, and the point is to find out what axioms are needed to ensure these are the same as the original open set topology and thus the axioms apply to the original open set topology. Do you see why this is kind of a tricky game? Let's hope it works.<br /><br /><b>Figuring out the right open set axioms: Part II</b><br /><br />The neighbourhoods arising from the open set topology $\Phi$ are $N(x)=\{S\subseteq X\mid\exists O\in\Phi,x\in O\subseteq S\}$. The open sets arising from these are $\Phi' = \{S\subseteq X\mid \forall x\in S, S\in N(x)\}$. Or:<br /><br />$$\Phi'=\{S\subseteq X\mid \forall x\in S, \exists O\in\Phi, x\in O\subseteq S\}$$<br />It is clear that $\Phi\subseteq\Phi'$. We want to show the converse, that $\Phi'\subseteq\Phi$, i.e.<br /><br />$$\forall x\in S, \exists O\in\Phi, x\in O\subseteq S\Rightarrow S\in \Phi$$<br />Or in English: "if every point in $S$ has an open set around it contained in $S$, then $S$ is an open set". Recognising that $S$ is the union of all the $O$'s, it is easy to show that this is equivalent to: if $O_\lambda$ is an arbitrary family of sets in $\Phi$, then $\bigcup_\lambda O_\lambda \in \Phi$, i.e. <b>open sets are closed under arbitrary union</b>.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-p30Ty7KDm1I/XVJ-PNIlyBI/AAAAAAAAFqk/FkRJ6MhfhW4T6wsZvSVr62R1yNIe0vsAQCLcBGAs/s1600/neighbourhoods%2Bunion.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="619" data-original-width="443" height="320" src="https://1.bp.blogspot.com/-p30Ty7KDm1I/XVJ-PNIlyBI/AAAAAAAAFqk/FkRJ6MhfhW4T6wsZvSVr62R1yNIe0vsAQCLcBGAs/s320/neighbourhoods%2Bunion.png" width="229" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The union of every such blue circle you can create is equal to $S$.</td></tr></tbody></table>So we now have an axiomatisation of topology in terms of open sets, the open sets topology: we have a topology on a set $X$ given by a set of its subsets $\Phi$ satisfying:<br /><br /><b>Axioms for the open set topology: 1st ed.</b><br /><ol><li>Every point is contained in an open set: $\forall x\in X, \exists O\in\Phi, x\in O$.</li><li>Closure under finite intersection: $\forall O_1, O_2\in \Phi, O_1\cap O_2\in\Phi$.</li><li>Closure under arbitrary union: $\forall (O_\lambda)\in\Phi^\mathcal{I}, \bigcup_\lambda O_\lambda\in\Phi$.&nbsp;</li></ol>(Exercise: show that these axioms are implied on the open sets by the neighbourhood topology axioms. This is important, because at least with the finite intersection axiom, we just chose the "simplest axiom" available to us, which was stronger than the actual axiom we required, which was $\forall x\in X, \forall (O_1, O_2 \in \Phi)\ni x, \exists O\in\Phi, x\in O\subseteq O_1\cap O_2$. Well, along with the arbitrary union axiom this implies finite intersection, anyway.)<br /><br />There's actually a further re-formulation one can make: <i>every point is in an open set</i> is, due to the arbitrary union axiom, equivalent to <b>the universe is an open set</b>. So our <b>final formulation of topology</b> is:<br /><br /><b>Axioms for the open set topology: 2nd ed.</b><br /><ol><li>$X\in\Phi$.</li><li>Closure under finite intersection: $\forall O_1, O_2\in \Phi, O_1\cap O_2\in\Phi$.</li><li>Closure under arbitrary union: $\forall (O_{\lambda\in\mathcal{I}})\in\Phi^\mathcal{I}, \bigcup_\lambda O_\lambda\in\Phi$.&nbsp;</li></ol>Recall that as the fifth axiom of the neighbourhood topology was <b>optional</b> -- $X\in\Phi$ is similarly optional -- without it, the topology would basically just be a topology over the union of all open sets. But you'd then have a topology with "invisible points" (or "topologically invisible points"), which is kinda stupid.<br /><br />Sometimes, $\varnothing\in\Phi$ is also added as an axiom, but this can be proven from the arbitrary union axiom, as the empty set is simply the empty union. You can see why this must be true, as indeed each point of the empty set vacuously has it as a neighbourhood.</div><br /><div class="twn-beg">We were able to show that a topology can be determined completely by its open sets -- but the sense in which the open sets form the structure of the topological space is quite different from seeing the limit operator as the structure of the topological space. I wonder if some nice analogy to groups or linear spaces exists. <b>Can a group be determined by its subgroups?</b> Open sets are basically <b>sub-(topological spaces)</b>&nbsp;-- aren't they? All limits within an open subspace are the same as in the original (you can see this rigorously when we define subspace topologies). See <a href="https://math.stackexchange.com/questions/3323761/is-a-finite-group-determined-by-its-subgroups"><b>my math stackexchange question</b></a>.</div><br /><div class="twn-furtherinsight">If you want to really appreciate the motivation for the axioms of a neighbourhood, you need to understand a neighbourhood as a notion of "wiggling", e.g. when we say "for all neighbourhoods", we mean "for even the slightest wiggle", and the reason this translates this way is that supersets are part of the filter.</div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-90343427949562538012019-08-03T15:27:00.000+01:002019-09-23T16:58:35.153+01:00Orthogonal group, indefinite orthogonal group, orthochronous stuffThis post appears in the <a href="https://thewindingnumber.blogspot.com/p/1103.html">Linear Algebra</a>&nbsp;and <a href="https://thewindingnumber.blogspot.com/p/2101.html">Special Relativity</a>&nbsp;courses.<br /><br />There are several ways to see that the matrices satisfying $A^*A=I$ are related to rotations in some way, other than just expanding out the components like a dumb pygmy chimp -- no, we are the normal chimp:<br /><ol><li>Write it as $A^TIA=I$ -- i.e. the set of matrices that <b>preserve the identity quadratic form</b>. The identity quadratic form corresponds to the $n$-sphere (e.g. a circle), so we're looking for transformations that preserve the $n$-sphere. A clearer way to see this is that preserving the quadratic form $I$ is equivalent to preserving the valuation $x^TIy$ for all $x, y$, i.e. $(Ax)^TI(Ay)=x^Ty$, so it preserves the value of each contour.</li><li>With the same logic as above, $(Ax)^T(Ay)=x^Ty$, i.e. the <b>preservation of the Euclidean dot product</b>&nbsp;means that all lengths and angles are preserved. These are called "rigid rotations", and are basically the kind of stuff we can do to a sheet of paper without compressing or stretching it in any way -- i.e. if we nudge a vector by a certain angle, every other vector should also be nudged by the same angle.</li></ol><div>What kind of transformations preserve the unit sphere?&nbsp;</div><br /><div class="twn-furtherinsight">The reason this is a good way of understanding things is that there are plenty of other such "dot products" you can define in mathematics, corresponding to <b>different geometries</b>&nbsp;-- each can be based on the bilinear form it preserves, see this <a href="https://thewindingnumber.blogspot.com/2019/04/geometry-positive-definiteness-and.html">later linear algebra article</a> for more details, relating to isomorphisms of such geometries etc.</div><br />As for discriminating between rotations and reflections, suppose we define rotations in a completely geometric way -- for a matrix to be a rotation, all its eigenvalues are either 1 or in pairs of unit complex conjugates.<br /><br />What do the eigenvalues of orthogonal matrices look like? For each eigenvalue, you need $\overline{\lambda}\lambda=1$, i.e. all the eigenvalues are unit complex numbers. If a complex eigenvalue isn't <b>paired with a corresponding conjugate</b>, you will not get a real-valued transformation on $\mathbb{R}^n$. Meanwhile if an eigenvalue of -1 isn't paired with another -1 -- i.e. if there are an <b>odd number of reflections</b> -- you get a reflection. In this sense, the "<b>conjugate eigenvalues</b>" property of rotations can be seen as a <b>generalisation of the "$s_1s_2=r$" property</b> which you may have learned from plane geometry or dihedral groups. The orthogonal (or rather unitary) transformations that do not behave this way are precisely the rotations.<br /><br />The similarity between unpaired unit complex eigenvalues and unpaired -1's is interesting, by the way -- when thinking about reflections, you might have gotten the idea that reflections are $\pi$-angle rotations in a higher-dimensional space -- like the vector was rotated through a higher-dimensional space and then landed on its reflection -- like it was a discrete snapshot of a process as smooth as any rotation.<br /><br />Well, <b>now you know what this higher-dimensional space is</b> -- precisely $\mathbb{C}^n$. And the determinant of a unitary matrix also takes a continuous spectrum -- the entire unit circle. In this sense (among other senses) complex linear algebra is <b>more "complete" than real linear algebra</b>. In fact, you will see in Lie theory that the group $SO(n)$ is connected but $O(n)$ is not, while $SU(n)$ and $U(n)$ are both connected. Can you see why?<br /><br />(<a href="https://math.stackexchange.com/questions/68119/why-does-ata-i-det-a-1-mean-a-is-a-rotation-matrix/3177807#3177807">original version of above originally posted to math stackexchange</a>)<br /><br />Well, here, we benefited from the fact that the product of two reflections is a rotation -- so we could just enforce the "even number of flips", i.e. that $\det A=1$, to specify rotations. But what if we're dealing with one of the "generalised geometries" we discussed? What if instead of preserving $I$, we wanted the group $O(m\mid n)$, i.e. that preserves some $\mathrm{diag}(m\mid n)$ with $m$ 1's and $n$ -1's along the diagonal?<br /><br />Well, then we don't have rotations between the "1"-labeled (spatial) axes and the "-1"-labeled (temporal) axes, only <b>boosts</b>. But compositions between such reflections form <i style="font-weight: bold;">rotations</i>! So simply restricting that $\det A = 1$ will -- while still forming a group $SO(m \mid n)$ -- retain all these rotations which can only be understood as compositions of reflections.<br /><br />So how do we extract the transformations we want? (What transformations <i>do</i>&nbsp;we want? The ones that correspond to changes of reference frame, in special relativity language -- well, in the sense of Lie theory, this means we're looking for the "<b>component connected to the identity</b>" -- do you see why?)<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/9/91/HyperboloidOfTwoSheets.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="456" data-original-width="406" height="320" src="https://upload.wikimedia.org/wikipedia/commons/9/91/HyperboloidOfTwoSheets.png" width="284" /></a></div><br />Let's think about this more clearly. Start by noting that not all reflections in spacetime preserve the Minkowski metric $\mathrm{diag}(m\mid n)$ -- only those that preserve the invariant hyperboloids. In the case of 3+1-spacetime, this means infinite spatial reflections and one time-reversal -- in the case of a general $m+n$-spacetime, this means infinite <b>spatial reflections</b> and infinite <b>temporal reflections</b>&nbsp;(in any $m+n-1$-plane whose normal vector is temporal, not to be confused with time-like). When you multiply an odd temporal reflection with an odd spatial reflection, you get an even time-space rotation, which is in $SO(3\mid 1)$. <br /><br />$$A = \left[ {\begin{array}{*{20}{c}}{{A_t}}&amp;B^T\\C&amp;{{A_s}}\end{array}} \right]$$<br />(Note on notation: we'll use ${A_T} = \left[ {\begin{array}{*{20}{c}}{{A_t}}&amp;0\\0&amp;I\end{array}} \right]$ and analogously ${A_S} = \left[ {\begin{array}{*{20}{c}}I&amp;0\\0&amp;{{A_s}}\end{array}} \right]$, where $A_t$ and $A_T$ are "basically the same thing", and analogously for $A_s$ and $A_S$ -- in particular $\det A_t=\det A_T$ and $\det A_s=\det A_S$.)<br /><br />We see the problem: instead of just mandating $\det A=1$, we must mandate that the <b>temporal minor</b>&nbsp;and the <b>spatial minor</b>&nbsp;of the matrix both have determinant 1, $\det A_t=\det A_s = 1$. But this isn't right -- if you have a boost, i.e. some mixing between the space and time co-ordinates, then $A\ne A_TA_S$ and the component determinants are multiplied by a Lorentz factor (even though still $\det A = 1$). So we mandate instead that $\det A_t&gt;0$, $\det A_s&gt;0$ (equivalently $\ge 1$). Such transformations are called the <b>proper orthochronous Lorentz transformations</b>, because in the context of special relativity they are proper Lorentz transformations that do not flip time:<br /><br />$$SO^{+}(3 \mid 1)=\{A\in O(3 \mid 1) \mid \det A_t &gt;0, \det A_s &gt;0\}$$<br />OK, how do we show $SO^{+}(m\mid n)$&nbsp;is a subgroup? You might get the notion that because of the "<b>two sheets hyperbola</b>" topology of the group, the sheet connected to the identity must be a <b>subgroup</b> (and the other sheet a <b>coset</b>) because moving about on the sheet keeps you on the sheet (and that's what group multiplication is -- moving about on the sheet). The formal way to say this is to say that the map $A\mapsto \mathrm{sgn}(\det A_t )$ is a group <b>homomorphism</b> to the cyclic group $\{1,-1\}$, so its kernel is necessarily a <b>normal subgroup</b> (do you see how these are the same thing?).<br /><br />So the key is to prove that for two matrices satisfying $\mathrm{sgn}(\det A_t )&gt;0$, their product does too. A proof of the $SO^+(m\mid 1)$ case (relevant for relativity) can be found <a href="https://physics.stackexchange.com/questions/36384/how-to-prove-that-orthochronous-lorentz-transformations-o1-3-form-a-group/36425#comment1111428_36425">here</a>&nbsp;-- I'm not sure how that proof can be appropriately generalised to $SO^+(m\mid n)$. I've written out the first few steps here:<br /><ol><li>Multiply the two matrices $A$ and $\tilde{A}$ to show $(A\tilde{A})_t=A_t\tilde{A}_t+B^T\tilde{C}$. We want to show the determinant of this is positive.</li><li>From multiplying out $A^T\eta A=\eta$ and $A\eta A^T=\eta$, we see that $A_t^2-C^TC=A_t^2-B^TB=I$ and analogous for $\tilde{A}$.</li><li>So $\det((A\tilde{A})_t-A_t\tilde{A}_t)=\det(B^T\tilde{C})=\sqrt{\det(A_t^2-I)\det(\tilde{A}_t^2-I)}$</li><li>Well, I'm not sure how to proceed at this point. Does $\det(X-PQ)=\det((P^2-I)(Q^2-I))^{1/2}$ imply that $\det P\ge1\land\det Q\ge1\Rightarrow \det X&gt;0$?</li></ol>Well, I can't think of a way to continue -- and certainly one can think of a much wider category of problems like this, where we have a much simpler topological picture in our heads than rubbish algebra like the above would betray. So we need a <i>topological way of looking at Lie groups</i>.<br /><br />You might think of just considering something like the orbit of a vector -- e.g. the unit time vector -- under the group for the topology, but this <b>does not fully describe the topology of the group</b>. As an illustration, in the above example, for $n&gt;1$, the orbit of the time vector under $O^+(m\mid n)$ is actually connected (prove this -- you need to count the number of sheets a general hyperbola has), while the entire topology of the group is actually disconnected, as we will see. A simple way to see that these are two different topologies is that spatial rotations/reflections leave the unit time vector unchanged and therefore all correspond to a single point on the orbit.<br /><br />This will be our starting point to motivate the study of the topology of a Lie group in the&nbsp;<a href="https://thewindingnumber.blogspot.com/p/1203.html">Lie theory</a>&nbsp;articles.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-69286950634292416892019-07-08T15:26:00.000+01:002019-07-14T09:34:31.596+01:00Mixed states I: density matrix, partial trace, the most general Born ruleIn the&nbsp;<a href="https://thewindingnumber.blogspot.com/2019/07/systems-and-sub-systems-tensor-product.html">last article</a>, we saw that sub-systems entangled with other sub-systems did&nbsp;<b>not have well-defined pure states</b>&nbsp;themselves -- just like correlated random variables don't have their own probability distributions. Since pretty much everything you see in the real world is entangled with&nbsp;<i>something</i>&nbsp;-- has correlations with some other thing -- this is a problem. One can't just consider the "state of the entire universe" when you just want to study a single electron or&nbsp;<i>something</i>.<br /><br />Wait -- why can't we just consider the&nbsp;<b>marginal distributions</b>, like we do in statistics? OK, suppose we start with the system -- with $|\phi\rangle$ and $|\varphi\rangle$ an orthonormal basis:<br /><br />$$|\psi\rangle = \frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$$<br />Naively, you may think that the state of the first sub-system $|\psi_1\rangle$ may be given by $|\psi'_1\rangle=\frac1{\sqrt2}|\phi\rangle+\frac1{\sqrt2}|\varphi\rangle$. Certainly, if we're measuring the subsystem with an operator with eigenvalues $|\phi\rangle$ and $|\varphi\rangle$, you have 50% probabilities of each. But to say that two things are in the same state requires that they produce the same outcome for&nbsp;<b>any</b>&nbsp;measurement, not just that one. Does our sub-system behave exactly like $|\psi'_1\rangle$ <b>for all observables</b>? Recall that in the last article, we showed that collapsing the first sub-system onto $|\chi\rangle$ collapses the entire system into the state:<br /><br />$$|\chi\rangle\otimes\left(<br />\langle\chi|\varphi\rangle|\phi\rangle+\langle\chi|\phi\rangle|\varphi\rangle\right)$$<br />To calculate the probability amplitude of this collapse, we may take the inner product of this with the original state -- you can compute this, and see the answer comes down to $1/\sqrt2$, i.e. there's a <b>probability of $1/2$ of the first subsystem collapsing to <i>any</i>&nbsp;such eigenstate $|\chi\rangle$</b>. You can use <i>any</i> observable in this two-dimensional state space, and the sub-system would collapse into <b>either eigenstate with probability exactly $1/2$</b>.<br /><br />This is a <i>completely different situation</i> from if the state of the first subsystem were simply a <b>pure state</b> like $|\psi'_1\rangle$.<br /><br />The situation we're dealing with is called a <b>mixed state</b> -- an example of a mixed state, in line with the motivating examples we had at the beginning of the course -- is <b>unpolarised light</b>. In fact, the state we described above models precisely <b>unpolarised light involving two photons</b>&nbsp;(is it obvious why?).<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/Circular_Polarizer_Creating_Clockwise_circularly_polarized_light.svg/1086px-Circular_Polarizer_Creating_Clockwise_circularly_polarized_light.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="393" data-original-width="800" height="157" src="https://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/Circular_Polarizer_Creating_Clockwise_circularly_polarized_light.svg/1086px-Circular_Polarizer_Creating_Clockwise_circularly_polarized_light.svg.png" width="320" /></a></div><br />The basic idea behind mixed states is that we have some uncertainty as to <i>what the state of a particle is</i>&nbsp;-- we don't know if the particle has state $|\phi\rangle$ or $|\varphi\rangle$ -- it has a 50% chance of either. This is a classical probability, rather than a quantum one, and different from the state being a superposition of these states, as we just saw above.<br /><br /><div class="twn-furtherinsight">Does this sort of thing occur with multivariate distributions in statistics? Suppose we have a multivariate distribution $\psi(x,y)$ and extract the marginal $x$-distribution $\phi(x) = \int_y \psi(x,y) dy$. Certainly this $\phi(x)$ gives us the right probability densities of each $x$-value. But the analog of considering general states like $|\chi\rangle$ is to make a <b>transformation of the domain</b> -- like a Fourier transform -- and consider probability densities in the transformed domain.<br /><br />As an exercise, write down a multivariate Fourier transform expression for $\hat{\psi}(\omega_1,\omega_2)$ and use it to compute the $\omega_1$-marginal probabilities $\hat{\phi}(\omega_1)$ -- compare this to what you would get if you were to Fourier-transform the $x$-marginal $\phi(x)$ directly.</div><br /><hr /><br />But saying "it has 1/2 probability of being in $|\phi\rangle$ and 1/2 probability of being in $|\varphi\rangle$" is clearly an <b>overdetermination</b>. As we saw above, this resulting state has a 1/2 probability of collapsing onto any state -- this is a statement that doesn't depend on $|\phi\rangle$ and $|\varphi\rangle$, the behaviour of the state is the same if you describe it instead as having "1/2 probability of being in $\frac1{\sqrt2}(|\phi\rangle+|\varphi\rangle)$ and 1/2 probability of being in $\frac1{\sqrt2}(|\phi\rangle-|\varphi\rangle)$". These two are <b>different statistical ensembles</b>&nbsp;but in the <b>same mixed state</b>.<br /><br />You can see similarly that "50% left-polarised + 50% right-polarised" is the same mixed state as "50% left-circular + 50% right-circular" -- they're both just <i>unpolarised light</i>.<br /><br /><b>What's the general condition for two statistical ensembles to produce the same observations?</b><br /><b><br /></b> Given statistical ensemble $\left(\left(p_i,|\psi_i\rangle\right)\right)$ and $\left(\left(p_i,|\psi'_i\rangle\right)\right)$, they are the same mixed state if for all $|\chi\rangle$, the probabilities of collapsing onto $|\chi\rangle$ is the same, i.e.<br /><br />$$\sum_i p_i|\langle\psi_i|\chi\rangle|^2=\sum_i p_i|\langle\psi'_i|\chi\rangle|^2$$<br />Well, each side of this equation is just the evaluation of a quadratic form for the vector $|\chi\rangle$ -- and <b>two quadratic forms are identically equal on all vectors if and only if their matrix representations are the same</b>. Well, what's the matrix representation? In the basis of $|\psi_i\rangle$, it's just the matrix of probabilities $p_i$. The way to write this in Bra-ket notation is to factor out the $|\chi\rangle$s:<br /><br />$$\left\langle \chi&nbsp; \right|\left( {\sum\limits_i {{p_i}\left| {{\psi _i}} \right\rangle \left\langle {{\psi _i}} \right|} } \right)\left| \chi&nbsp; \right\rangle&nbsp; = \left\langle \chi&nbsp; \right|\left( {\sum\limits_i {{p_i}\left| {{{\psi '}_i}} \right\rangle \left\langle {{{\psi '}_i}} \right|} } \right)\left| \chi&nbsp; \right\rangle<br />$$<br />This quadratic form in between, representing a mixed state, is called the <b>density matrix</b>&nbsp;and can be used to completely specify mixed states. In this sense, it is a <b>generalisation of the state vector</b>, which can only be used to represent pure states.<br /><br />$$\rho={\sum\limits_i {{p_i}\left| {{\psi _i}} \right\rangle \left\langle {{\psi _i}} \right|} }<br />$$<br />You may confirm that indeed:<br /><br />$$\frac12|\phi\rangle\langle\phi|+\frac12|\varphi\rangle\langle\varphi|=\frac12\left(\frac{|\phi\rangle+|\varphi\rangle}{\sqrt2}\frac{\langle\phi|+\langle\varphi|}{\sqrt2}\right)+\frac12\left(\frac{|\phi\rangle-|\varphi\rangle}{\sqrt2}\frac{\langle\phi|-\langle\varphi|}{\sqrt2}\right)$$<br />In fact, there is a simpler way to see that those two ensembles are the same: the density matrix is simply the <b>Gram matrix of the ensemble</b>&nbsp;-- you take the states in the ensemble, weighted by $\sqrt{p_i}$ in a matrix $Y$, and $\rho=Y^*Y$. Well, $Y^*Y=Y'^*Y' \iff Y'=UY$&nbsp; for some unitary $Y$, i.e. the ensembles are rotations of each other.<br /><br /><hr /><br /><b>Properties of the density matrix, generalised Born's rule, etc.</b><br /><br />Here's something that's obvious: the density matrix is <b>nonnegative-definite ("positive-semidefinite") Hermitian and unit-trace</b>&nbsp;-- and all such matrices can represent density matrices.<br /><br />Well, so it's a Hermitian operator -- does it represent any interesting observable? Not really. It's an observable, sure, but not an interesting one (you might say it measures something's being in one of the ensemble states -- written in an orthonormal basis -- and whose eigenvalues are the mixing ratios, etc. -- but what if two mixing ratios are the same? Its behaviour is just bizarre and useless, really).<br /><br />We saw earlier that the probability of a density matrix collapsing into a state $|\chi\rangle$ is given by $\langle\chi|\rho|\chi\rangle$.<br /><br /><div class="twn-pitfall">This is <b>completely different</b> from the generalised Born's rule we saw earlier which took the form $\langle\psi|L|\psi\rangle$! There, the <em>state</em> was the vector and the information on the projection space was the quadratic form in between. Here, the state is the quadratic form in between while the state being projected onto is the vector. This is just a generalisation of the simple Born rule $\langle\chi|\psi\rangle\langle\psi|\chi\rangle$, as far as I can see. If anyone comes up with a connection between it and the generalised Born rule for pure states, tell me.</div><br />This brings the question, though -- what's the <i>most generalised Born's rule</i>&nbsp;we can come up with? What is the probability of a <b>mixed state collapsing into some eigenspace of a Hermitian projection operator</b>?<br /><br />Well, given the ensemble $((p_i,|\psi_i\rangle))$ (you can start writing ensembles with their density matrices now if you like, like $\sum p_i|\psi_i\rangle\langle\psi_i|$ -- but I just want to reaffirm that our result will indeed be in terms of the density matrix), the probability is:<br /><br />$$\sum_i p_i\langle\psi_i|L|\psi_i\rangle$$<br />This is hardly useful -- it's not in terms of the density matrix at all. But look at each term -- what's $p_i\langle\psi_i|L|\psi_i\rangle$? $L|\psi_i\rangle$ is the $i$th column of $L$ in the $(|\psi_i\rangle)$-basis -- the inner product $\langle\psi_i|L|\psi_i\rangle$ is the $i$th entry of this column. Multiplying this by $p_i$ gives us the <i>dot product of the $i$th row of $\rho$ with the $i$th column of $L$</i>. The <i>sum</i>&nbsp;of these for all $i$ gives us the trace of $\rho L$:<br /><br />$$\ldots = \mathrm{tr}(L\rho)$$<br />This is the <b>most general form of Born's rule</b>. Note that our derivation could have also applied to finding the <b>expectation value</b> of a general operator $A$ under the density matrix $\rho$ (recall that Hermitian projection operators are basically "indicator variables" whose expectation values represent probabilities), indeed generally:<br /><br />$$\langle A\rangle_\rho=\mathrm{tr}(L\rho)$$<br />(note that $\mathrm{tr}(V)=\sum_{i} \langle i | V|i \rangle$ for any basis $(|i\rangle)$, which you should show.)<br /><br />It is also trivial to show that upon collapse given by Hermitian projection operator $L$, the density matrix <b>collapses</b> to:<br /><br />$$\rho'=\frac1{\mathrm{tr}\left(L\rho L\right)}{L\rho L}=\frac1{\mathrm{tr}\left(L\rho\right)}{L\rho L}$$<br />Generalising the pure state collapse to $|\psi'\rangle=\frac{1}{\langle \psi | L | \psi\rangle}L|\psi\rangle$. One may check that the above expression reduces to $|\chi\rangle\langle\chi|$ in the case where $L=|\chi\rangle\langle \chi|$.<br /><br /><hr /><br /><b>Partial trace, trace</b><br /><b><br /></b> We started our discussion considering the pure state $\frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$ and asking for the mixed state of the first sub-system. We computed the inner product of this state with its projection under the operator $|\chi\rangle\langle\chi|\otimes1$ -- this tells us the evaluation of the quadratic form $\langle\chi|\rho_A|\chi\rangle$ at all vectors $|\chi\rangle$, which determines the quadratic form $\rho_A$ of the first state.<br /><br />So what exactly did we do -- in general? Starting with a density matrix $\rho$ on $H_1\otimes H_2$, we compute the probability of the first sub-system appearing in state $|\chi\rangle$: it's $\mathrm{tr}((|\chi\rangle\langle\chi|\otimes 1)\rho)$. So we try to find a density matrix $\rho_1$ satisfying, for all states $|\chi\rangle$:<br /><br />$$\mathrm{tr}((|\chi\rangle\langle\chi|\otimes 1)\rho)=\mathrm{tr}(|\chi\rangle\langle\chi|\rho_1)$$<br /><br /><b>Exercise: </b>Let $V$ be an operator on $H_1\otimes H_2$. We define its&nbsp;<b>partial trace</b> on $H_2$ as $\mathrm{tr}_2(V)=\sum_{j}\langle j|V|j\rangle$ for basis $(|j\rangle)$ of $H_2$ (where the inner product is done by extending operators by tensoring them with the identity). Show that the density matrix $\rho_1$ is given by:<br /><br />$$\rho_1=\mathrm{tr}_2(\rho)$$<br />I.e. show that for operators of the form $A\otimes I$: $\mathrm{tr}[(A\otimes I)\rho]=\mathrm{tr}_1(A\,\mathrm{tr}_2\rho)$.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-75142887905617048262019-07-06T08:52:00.000+01:002019-07-15T12:34:26.503+01:00Systems and sub-systems, the tensor product, quantum entanglementWhat if we want to describe probabilities relating to multiple objects -- a system of objects?<br /><br />You might think it's sufficient to just write down the state vector of each individual object -- but this doesn't really tell us the entire picture. Suppose for example we're considering the state of Schrodinger's cat, less popularly known as&nbsp;<b>Schrodinger's cuckoo</b>, where the box contains TNT and the cuckoo bird. The state of the TNT is $|\mathrm{unexploded}\rangle+|\mathrm{exploded}\rangle$ and the state of the cuckoo is $|\mathrm{alive}\rangle+|\mathrm{dead}\rangle$ -- right?<br /><br /><div class="twn-pitfall">Never trust a cuckoo.</div><br /><div class="twn-furtherinsight">If I'm in charge of the box, the state of the cuckoo will definitely be $|\mathrm{dead}\rangle$ even if the state of the TNT is $|\mathrm{unexploded}\rangle$</div><br />Not exactly. There is a <b>correlation</b> between the state of the TNT and the state of the cuckoo that goes missing here -- the cuckoo is dead if and only if the TNT has exploded, and the cuckoo is alive if and only if the TNT is unexploded. In fact, defining the state vectors separately doesn't really make any sense -- we're just assigning the coefficients based on overall probabilities, and as we will see, this is really mixing up quantum and classical probabilities in a certain way whereas the state vector is supposed to only show quantum probabilities.<br /><br />Well, this is really the same question as what we do when we have multiple correlated random variables in statistics -- we define a <b>"joint" probability function</b> on a <b>"joint" phase space</b> that is the <b>Cartesian product of the original phase spaces</b>.<br /><br />You may be inclined to claim that similarly, our new Hilbert space should be the Cartesian product of the original Hilbert spaces. But Hilbert spaces are different in a fundamental sense from these phase spaces -- every point on the classical phase space is an independent state in the Hilbert space -- and general <b>vectors are <i>distributions</i></b>&nbsp;on the classical phase space. So like how the <em>cardinalities</em> of the classical phases are multiplied, the <b><em>dimensions</em> of the Hilbert spaces are multiplied</b>.<br /><br /><div class="twn-furtherinsight">The reason that we sometimes draw an analogy between the classical state space and the quantum state space is that the state vectors are really the "real objects" in quantum mechanics, and the Hilbert space shows the possible configurations of the state in this sense.</div><br />The product we want of Hilbert spaces -- which is <i>not</i>&nbsp;the Cartesian product -- is called a <b>tensor product of linear spaces</b>&nbsp;-- given an orthogonal basis $(|\phi_1\rangle,|\phi_2\rangle,\ldots)$ for the first Hilbert space and $(|\psi_1\rangle,|\psi_2\rangle,\ldots)$ for the second, their tensor product is spanned by new vectors which we denote as<br /><br />$$\left( {\begin{array}{*{20}{c}}{|{\phi_1}\rangle&nbsp; \boxtimes |{\psi_1}\rangle ,|{\phi_1}\rangle&nbsp; \boxtimes&nbsp;|{\psi_2}\rangle ,...,}\\{|{\phi_2}\rangle&nbsp; \boxtimes&nbsp;|{\psi_1}\rangle ,|{\phi_2}\rangle&nbsp; \boxtimes&nbsp;|{\psi_2}\rangle ,...,}\\ \vdots \end{array}} \right)$$<br /><div class="twn-pitfall">We're using $\boxtimes$ instead of $\otimes$ in the above enumeration of the basis, because we haven't yet defined the tensor product of states. The idea is that $|\phi_i\rangle\boxtimes|\psi_j\rangle$ are just placeholders, and we will shortly state that they are/can be the tensor product $|\phi_i\rangle\otimes|\psi_j\rangle$, which we will define now.</div><br />Certainly, this can represent any possible state in which the combined system of two objects can be in. What we need is a way to express the state of a combined system of two independent things in this "larger" Hilbert space -- i.e. a <b>map</b>&nbsp;from $H_1\times H_2\to H_1\otimes H_2$ that takes the (pure) states of two independent objects in $H_1$ and $H_2$ and outputs their state as a combined system in $H_1\otimes H_2$ -- we will call this product the <b>tensor product of vectors</b>, and denote it by the same symbol $\otimes$.<br /><br />OK, so what's the map? Certainly, $|\phi_i\rangle\otimes|\psi_j\rangle$ must form an <b>orthogonal basis</b> for $H_1\otimes H_2$ (why? think about this for a while -- they're clearly <b>orthogonal, as they are mutually exclusive</b> -- you can't be in "$|\phi_i\rangle$ and $|\psi_j\rangle$" and "$|\phi_{i'}\rangle$ and $|\psi_{j'}\rangle$" unless $(i,j)=(i',j')$; <b>spanning is proven similarly</b>, as considering the $|\phi_i\rangle$s and $|\psi_j\rangle$s as eigenstates of some operators $X$ and $Y$ on $H_1$ and $H_2$, then if one performs the operation of "observing $X$ and $Y$" -- and we can do this because the objects are independent -- then because the objects must be found in one of $|\phi_i\rangle$ and one of $|\psi_j\rangle$, the system must be found in one of $|\phi_i\rangle\otimes|\psi_j\rangle$ -- thus its original state was a linear combination of such states).<br /><br />OK, so<br /><br /><br />(p_1|\phi_1\rangle+p_2|\phi_2\rangle+\ldots)\otimes(q_1|\psi_1\rangle+q_2|\psi_2\rangle+\ldots)\\<br />\begin{align}<br />=\ &amp; r_{11}|\phi_1\rangle\otimes|\psi_1\rangle + r_{12}|\phi_1\rangle\otimes|\psi_2\rangle+\ldots+\\<br />&amp;r_{21}|\phi_2\rangle\otimes|\psi_1\rangle + r_{22}|\phi_2\rangle\otimes|\psi_2\rangle+\ldots+\\<br />&amp; \vdots<br />\end{align}<br />What are the coefficients $r_{ij}$?<br /><br />Well, it's fairly obvious that $|r_{ij}|^2=|p_{i}|^2|q_{j}|^2$ -- that the <b><i>probabilities</i> are multiplicative</b>, this is tautological given what we want our product to represent -- the probability that the system is found in the state $|\phi_i\rangle\otimes|\psi_j\rangle$ is the probability that the objects are found in states $|\phi_i\rangle$ and $|\psi_j\rangle$, which is the product of the respective probabilities, as they are independent objects.<br /><br />Is it also true that the <b>probability amplitudes are multiplicative</b>, i.e. $r_{ij}=p_iq_j$?<br /><br />This may seem hard to prove, but the idea is quite simple: suppose we observe the state with the observables $X$ and $Y$, and find it in the state $|\phi_i\rangle\otimes|\psi_j\rangle$. Well, then if $r_{ij}=u_{ij}|r_{ij}|$ for some unit complex number $u_{ij}$, then from the right-hand-side, we must have collapsed to $u_{ij}|\phi_i\rangle\otimes|\psi_j\rangle$. So we must have $u_{ij}=1$.<br /><br />So indeed the product we're looking for is exactly the <b>tensor product from tensor algebra</b>.<br /><br /><div class="twn-furtherinsight">Here's a thing worth noting -- we've been referring to "systems" and "objects" as if they are somehow completely distinct things. But are they? The cat's state is itself a <b>tensor product</b> of a massive number of different states belonging to each <b>elementary particle</b> in its body, and lives already in a massive Hilbert space, because the "object" is itself a system. We will use the term <b>subsystem</b>&nbsp;instead of <b>object</b>&nbsp;from now.</div><br /><hr /><br />Alright: so we now know that elements of the tensored Hilbert space are all states, and only the ones that are <b>factorable</b> into an element of $H_1$ and $H_2$ represent subsystems that are independent. This is precisely how only <b>factorable probability mass/density functions</b> represent independent variables in statistics. Otherwise the variables are correlated -- not necessarily linearly correlated, but correlated.<br /><br />Such correlations can, of course, exist in our quantum mechanical theory, too -- like the cuckoo-TNT system we mentioned earlier. These are called <b>quantum correlations</b>&nbsp;or <b>quantum entanglement</b>.<br /><br />Why the fancy name? Because its consequences may seem superficially kinda "surprising". It's also a demonstration of quantum mechanics being different from classical mechanics, because without entangled states, the dimension of $H_1\otimes H_2$ would indeed be the sum of those of $H_1$ and $H_2$ rather than their product, like with phase spaces in classical mechanics.<br /><br />OK, what kind of surprising consequences?<br /><br />They're basically all of the following nature: suppose we have a state given by:<br /><br />$$\frac1{\sqrt2} (|\phi\rangle\otimes|\psi\rangle+|\psi\rangle\otimes|\phi\rangle)$$<br />I.e. two entangled particles where we know that they are in <b>two distinct states, but we don't know which is which</b>. Such a state can certainly be produced -- how? Just put two identical independent particles in a box then do a <b>"partial" measurement</b> -- a "<b>peek</b>" -- (which can be achieved, e.g. by some logic gates) that checks if they're in the same state or not, and uncovers no other information.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-RdQ446MRCZ0/XR-4FMLPe8I/AAAAAAAAFnI/Wdq0C5sq0csOQ7oa3N_WDTclzTPPFqMNACLcBGAs/s1600/alibobeshwar-1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="472" data-original-width="1144" height="132" src="https://1.bp.blogspot.com/-RdQ446MRCZ0/XR-4FMLPe8I/AAAAAAAAFnI/Wdq0C5sq0csOQ7oa3N_WDTclzTPPFqMNACLcBGAs/s320/alibobeshwar-1.png" width="320" /></a></div><br />Now separate the particles spatially -- there's nothing wrong with this, they're still a system, which still has a state -- and give one to Alice and the other to Bob. Now if Bob looks at his particle and sees it in $|\psi\rangle$, he immediately <i><b>knows</b></i>&nbsp;that Alice could only observe her particle to be in state $|\phi\rangle$ -- <b>there's nothing Alice can do to change this outcome</b>.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-IgUTjPAGjsk/XR-4JqV0FJI/AAAAAAAAFnM/K5TDzVtWn_Qco4PtrHfMEkryqkq5iVJMQCLcBGAs/s1600/alibobeshwar-2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="540" data-original-width="1144" height="151" src="https://1.bp.blogspot.com/-IgUTjPAGjsk/XR-4JqV0FJI/AAAAAAAAFnM/K5TDzVtWn_Qco4PtrHfMEkryqkq5iVJMQCLcBGAs/s320/alibobeshwar-2.png" width="320" /></a></div><br />(you may worry that spatially separating the particles alters the state in some important way -- but it doesn't: the states $|\phi\rangle$ and $|\psi\rangle$ are both transformed individually that doesn't change the entangled structure of the combined state -- make sure this makes sense to you. But if it makes you happy, you could imagine the particles were already spatially separated when they were first entangled.)<br /><br />OK, perhaps you don't find this particularly surprising or unintuitive -- I don't either. But perhaps you do -- perhaps you think there's a violation of locality -- and the reason you do is because you haven't yet fully accepted <a href="https://thewindingnumber.blogspot.com/2017/08/three-domains-of-knowledge.html">logical positivism</a>. Let's consider what locality entails for each observer in the set-up, and see if it's violated:<br /><ul><li><b>Alice:</b>&nbsp;From Alice's perspective, Bob opening his box is just another way to observe her particle -- or rather, she can <b>observe Bob's brain</b> that contains the information, which collapses the state from her perspective. But this is <b>perfectly local</b> -- it takes time for information to propagate from Bob to her. Alternatively, if she doesn't observe Bob's brain and <b>just observes her own box</b> later, that's when her state collapses to $|\phi\rangle$, and she then learns that Bob had collapsed his state into $|\psi\rangle$ -- but as Bob <b>cannot choose what his state vector&nbsp; collapses</b>&nbsp;to, so he can't send her any information through entanglement. Even if there were a large number of entangled systems this way, the distribution of the states Alice can observe is the same whether or not Bob has collapsed his states (you can confirm this -- this is an idea called the <b>no-communication theorem</b>&nbsp;which we will discuss later in more mathematical detail).</li><li><b>Bob: </b>Certainly, Bob acquires knowledge of something far away, but no information actually propagated from Alice to him -- he just observed his own box.</li><li><b>another observer: </b>Charlie, who stands somewhere between Alice and Bob, too takes time to observe Bob's brain.</li></ul><div>So there really isn't a violation of locality. This isn't surprising at all -- certainly one could have <b>classical correlations</b>&nbsp;too. You could just juggle two distinct particles in a box and give them to each person, and Bob discovering his particle allows him to determine Alice's particle.&nbsp;</div><div><br /></div><div>The difference between the classical case and the quantum case is that in the classical case you could <i>pretend</i>&nbsp;that there's some <b>hidden truth</b> that is just not known to the observers. Quantum mechanics forbids any such hidden truth (as confirmed by commutator relations), and forces you to accept logical positivism, and there cannot be a "universal observer" as such a notion is inherently non-local. But the fact that correlation isn't non-local doesn't depend on whether you have&nbsp;<b>metaphysical notions of hidden truths</b>&nbsp;in classical physics -- it is a physical question, and is the <b>same in the classical and quantum cases</b>.<br /><br /><hr /><br />Are we done writing down our algebra of tensor products? We still haven't discussed how&nbsp;<b>inner products</b>&nbsp;and <b>projections&nbsp;</b>of tensor products behave. The basic question is "how do we <b>upgrade/combine operators</b> from $H_1$ and $H_2$ to $H_1\otimes H_2$? Let's start with the simple case of a factorable state in the form $|\phi\rangle\otimes|\varphi\rangle$. Suppose we apply a projection operator $X$ on the first particle. Have we made any observation on the second state? No -- just an identity projection. Or we could make an observation, a projection $Y$. So we can say that for the combined observation $X\otimes Y$,<br /><br />$$(X\otimes Y)(|\phi\rangle\otimes|\varphi\rangle)=(X|\phi\rangle)\otimes(Y|\varphi\rangle)$$<br />And an upgrade from $H_1$ to $H_1\otimes H_2$ is just tensoring with the identity $X\otimes 1$.<br /><br />But the full range of operators on $H_1\otimes H_2$ is a lot more complicated. We could consider <b>entangled states</b>. We could consider operators that are entangled (<b>"partial measurement" operators</b> like we described -- think about what these are). How would measurements on linear combinations of states look like (we know they <i>should</i>&nbsp;apply linearly, but let's show that)?<br /><br />Suppose we have a state in the form $\frac1{\sqrt2}|\phi\rangle\otimes|\varphi\rangle+\frac1{\sqrt2}|\varphi\rangle\otimes|\phi\rangle$. What exactly is this? We had two independent subsystems each in state $\frac1{\sqrt2}|\phi\rangle+\frac1{\sqrt2}|\varphi\rangle$, then we made an observation that showed they were in two distinct states -- we don't know which is in which. Now we make an observation and collapse the first subsystem to $|\chi\rangle$.<br /><br /><b>How does this alter the state of sub-system 2?</b><br /><br />OK, so "was" (quotation marks! quotation marks!) the system in $|\phi\rangle\otimes|\varphi\rangle$ or $|\varphi\rangle\otimes|\phi\rangle$? This is a question for <b>Bayes' theorem</b>.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-pf_gBpB1fu4/XSGtBgjdpoI/AAAAAAAAFns/zgumWWJQ2k8MIPnK3NiRd72_5fCIZDSGwCLcBGAs/s1600/bayes.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="396" data-original-width="560" height="226" src="https://1.bp.blogspot.com/-pf_gBpB1fu4/XSGtBgjdpoI/AAAAAAAAFns/zgumWWJQ2k8MIPnK3NiRd72_5fCIZDSGwCLcBGAs/s320/bayes.png" width="320" /></a></div><br /><div class="twn-pitfall">The above diagram is for <em>illustration only</em>. There is no real hidden truth (as we will see a few articles from now) of whether the state was initially $|\phi\rangle\otimes|\varphi\rangle\otimes|\phi\rangle$ or otherwise. But the probabilities still obey all the standard laws, such as Bayes's theorem, so tree diagrams make sense to illustrate this.</div><br />So the "probability that the system was in $|\phi\rangle\otimes|\varphi\rangle$" (quotation marks! quotation marks!) is:<br /><br />$$\frac{\frac12|\langle\chi|\phi\rangle|^2}{\frac12|\langle\chi|\phi\rangle|^2 + \frac12|\langle\chi|\varphi\rangle|^2}$$<br />(which if $\langle\phi|\varphi\rangle=0$ is just $|\langle\chi|\phi\rangle|^2$) And analogously for the other possibility. So the collapse of sub-system 1 to $|\chi\rangle$ collapses the entire state to<br /><br />$$|\chi\rangle\otimes\left(\frac1{\sqrt2}\langle\chi|\phi\rangle\cdot|\varphi\rangle+\frac1{\sqrt2}\langle\chi|\varphi\rangle\cdot|\phi\rangle\right)$$<br />Or some normalisation thereof if $\langle\phi|\varphi\rangle\ne0$. You can confirm that if $|\chi\rangle=|\phi\rangle$ or $|\chi\rangle=|\varphi\rangle$, this reduces to $|\phi\rangle\otimes|\varphi\rangle$ or $|\varphi\rangle \otimes |\phi\rangle$ respectively as we expect.<br /><br />You can check that this is <b>precisely what you get from applying the projection operator</b> $|\chi\rangle\langle\chi|\otimes 1$ as a linear operator to the original state. The above argument can be repeated for a general vector in the tensored space, yielding the linearity of the tensored operator.<br /><br /><div class="twn-exercises">What about <b>inner products of tensored states</b>? Convince yourself that the inner product of $|\phi_1\rangle\otimes|\phi_2\rangle$ and $|\chi_1\rangle \otimes |\chi_2\rangle$ is $\langle\phi_1|\chi_1\rangle\langle\phi_2|\chi_2\rangle$.</div><br />There was the other case we mentioned -- we may have operators that are themselves entangled. What does this mean? Suppose we start with the factorable state:<br /><br />$$\frac12|\phi\rangle\otimes|\phi\rangle+\frac12|\phi\rangle\otimes|\varphi\rangle+\frac12|\varphi\rangle\otimes|\phi\rangle+\frac12|\varphi\rangle\otimes|\varphi\rangle$$<br />Then perform the observation corresponding to "are the states different from each other?" This is a projection onto the plane spanned by $(|\phi\rangle\otimes|\varphi\rangle,|\varphi\rangle\otimes|\phi\rangle)$, perpendicular to the plane spanned by $(|\phi\rangle\otimes|\phi\rangle,|\varphi\rangle\otimes|\varphi\rangle)$ (confirm that this is true based on our discussion above of applying inner products to tensored states) -- we can write this as:<br /><br />$$\left(|\phi\rangle\otimes|\varphi\rangle\right)\left(\langle\phi|\otimes\langle\varphi|\right) +<br />\left(|\varphi\rangle\otimes|\phi\rangle\right)\left(\langle\varphi|\otimes\langle\phi|\right)<br />$$<br />(check that this, and the system of "distributing bras" makes sense). Or alternatively:<br /><br />$$|\phi\rangle\langle\phi|\otimes|\varphi\rangle\langle\varphi|+|\varphi\rangle\langle\varphi|\otimes|\phi\rangle\langle\phi|$$<br />One can check that applying this operator to the factorable state indeed results in the entangled state (up to normalisation). For better insight, perform the operation on the factored form of the state.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-_grXpKdhg2s/XSI6bApIi8I/AAAAAAAAFoE/ohtG-MZb51I1g34o5G5fiXQlZ5ch3BQ6wCLcBGAs/s1600/factorable%2Bstates%2Band%2Boperators.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1400" data-original-width="808" height="400" src="https://1.bp.blogspot.com/-_grXpKdhg2s/XSI6bApIi8I/AAAAAAAAFoE/ohtG-MZb51I1g34o5G5fiXQlZ5ch3BQ6wCLcBGAs/s400/factorable%2Bstates%2Band%2Boperators.png" width="230" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">In the representation shown in the above diagram, a <b>factorable state</b>&nbsp;is one given by a <b>single&nbsp;square region</b>, while a <b>factorable projection operator</b>&nbsp;is one that selects a square region out of the maximally distributed state. With this notion, it becomes clear that <b>an entangled operator (one that cannot be factored into operators on each Hilbert space) is the only&nbsp;kind of operator that projects a factorable state into an entangled state</b>.&nbsp;</div><br /><div class="twn-pitfall">We're not saying that entangled operators <em>always</em> project factorable states onto entangled ones -- you can just measure an irrelevant property of the system (e.g. you know each particle is either in the UK or France, and you check "are they both in India?"). But they are the only operators that <em>can</em>, and for any such operator, there exist factorable states that it entangles (almost by definition).</div><br /><div class="twn-pitfall">Note that we're only talking about <em>projection operators</em> above. We could certainly have factorable observables that enforce a partial measurement -- e.g. $X_1\otimes X_2$, which measures the product of the positions of the two particles -- but the projection operators onto each of the eigenstates of this operator are not factorable (check this).</div><br /><div class="separator" style="clear: both; text-align: left;">The following rules then determine the action of our operations on the tensored Hilbert space.</div><div class="separator" style="clear: both; text-align: left;"></div><ol><li>$(A\otimes B)(|\phi\rangle\otimes|\varphi\rangle)=(A|\phi\rangle)\otimes(B|\varphi\rangle)$ -- the tensored operators <b>associate</b>&nbsp;with the corresponding states.</li><li>The tensor product of two linear operators is <b>linear</b>.</li><li>The image of a <b>linear</b> combination of operators is the linear combination of the images, i.e. $(A+B)|\psi\rangle=A|\psi\rangle+B|\psi\rangle$.</li></ol></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-73839945827608305552019-06-17T02:38:00.001+01:002019-07-16T21:01:20.735+01:00Derivations and the Jacobi IdentityLet's consider a new way to think of the Lie algebra to a group -- instead of just considering the tangent vector to be <i>at</i>&nbsp;the identity, we could smear it across the group to form a <b>vector field</b>, resolving questions of whether our tangent space "really needs to be" at the identity (the exponential map in matrix representation only exists in the traditional form if we're talking about tangent vectors at the identity, but we're free to write down the Lie algebra in this way).<br /><br />But not every vector field is a valid element of the Lie algebra. We need the vector field to be "<b>constant</b>" across the manifold in some sense so that that constant vector it equals is the tangent-space-at-the-identity element it corresponds to. But what exactly do we mean by "constant" on a Lie Group?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Wv4qP6_NVOY/XQJ0FqkQ5HI/AAAAAAAAFmE/sImh5ae4NY82cWm_nUlQmc4bWe9zQEqlACLcBGAs/s1600/leftinvariantvectorfield.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="633" data-original-width="633" height="320" src="https://1.bp.blogspot.com/-Wv4qP6_NVOY/XQJ0FqkQ5HI/AAAAAAAAFmE/sImh5ae4NY82cWm_nUlQmc4bWe9zQEqlACLcBGAs/s320/leftinvariantvectorfield.png" width="320" /></a></div>In the case of the unit circle in the complex plane, we have an idea of what we want -- the vector field $T(M)$ is constant over the group if it is determined by the value at the identity as $T(M)=MT(0)$.<br /><br />Is this preserved in the matrix representation of the group? Well, yes, because the correspondence between complex numbers and spiral matrices is a homomorphism. We can use this as a motivation to define the condition for a vector field to be a Lie algebra on a matrix Lie group -- it needs to be a <b>left-invariant vector field</b>, i.e. we need that the value of the vector field determined as $T(M)=MT(0)$.<br /><br /><div class="twn-furtherinsight">Why left-invariant? Why not right-invariant? Why matrix multiplication at all? The choices made here are certainly arbitrary to some extent. When we study <b>abstract lie algebra</b>, we'll just have "left-multiplication by $M$" being replaced by a group action and the usage of matrix multiplication is a <b>choice of representation</b>. In the context of abstract Lie algebra, the "left-multiplication by $M$ we're interested in is really the <i>derivative</i>&nbsp;of the group homomorphism $M:G\to G$, which is a linear map between the tangent spaces at $I$ and $M$. You can show that this map is represented by matrix left-multiplication given a matrix representation (i.e. letting the group be $GL(n,\mathbb{C})$).</div><br /><hr /><br />Ok, why did we just do that? Why did we upgrade our tangent vectors to vector fields? If it wasn't obvious already, the <b>noncommutativity</b> of a Lie group is "the" feature of importance in a Lie group, at least in some neighbourhood of the identity (we will later find out exactly the kind of features that aren't determined by just the Lie bracket -- the important keywords here are <b>connected</b>&nbsp;and <b>compact</b>) -- if the Lie group is commutative, then the Lie algebra is just a vector space with no additional structure, and the Lie group is a "basically unique" choice.<br /><br />In our discussions of noncommutativity in the <a href="https://thewindingnumber.blogspot.com/2019/05/an-easy-way-to-see-closure-under-lie.html">last article</a>, we repeatedly referred to <i>flowing along a vector</i>&nbsp;-- the nature of noncommutativity is inherently "dynamical" in this sense. So we need to talk about <i>differentiating along the corresponding vector field to a tangent vector</i>.<br /><br />So let's upgrade our vector fields to derivative operators, or <b>derivations</b>&nbsp;$D$. These are operators on functions $f:G\to \mathbb{R}$ that tell you the derivative of $f$ in the direction of the vector field -- the left-invariant ones are a certain generalisation of the directional derivative operators.<br /><br />Well, what exactly is a derivation? On Euclidean space, directional derivatives can be imagined as stuff of the form $f\mapsto\vec{v}\cdot\nabla f$ -- but this requires the concept of a dot product which is quite weird within the context of matrix groups. But if you try to work this out on the unit circle (do it!), you might get an idea: we can define a <b>curve</b>&nbsp;$\gamma:\mathbb{R}\to G$ passing through a point and consider:<br /><br />$$f\mapsto(f\circ \gamma)'(t)$$<br />At the point and you get precisely the directional derivative in the direction $\gamma'(t)$ (show that this is right in Euclidean space, and make sure you understand why it is right/makes sense -- it's the chain rule, and a certain analogy exists to projecting matrices onto subspaces in linear algebra). And if we just want tangent vectors at the identity, we can just consider the operation $f\mapsto(f\circ \gamma)'(0)$.<br /><br />OK. Let's try to "<b>abstract out</b>" the properties of a derivation $D$, i.e. something that just allows us to define what a derivation is, abstractly, that is equivalent to being an operator of the above form.<br /><br />What makes an operator a directional derivative? Certainly it must be a linear operator -- but not every linear operator is a directional derivative. The key idea behind a directional derivative is that $D(f(x))$ is <b>determined in a specific way by $D(x)$, the rate at which $x$ changes</b> in the specified direction.<br /><br />How do we use this? Well, if you think about it a little bit, we can restrict $f$ to be analytic -- so we need:<br /><ol><li>$D(x)$ predicts $D(x^n)$ in the right way -- this is ensured by the <b>product rule</b>&nbsp;-- $D(fg)=f\ Dg + g\ Df$.</li><li>$D(x^n)$ for all $n$ predicts $D(a_0+a_1x+a_2x^2+\ldots)$ in the right way -- this is ensured by <b>linearity</b>.</li></ol><div class="twn-beg">If anyone can motivate the definition of a derivation without restricting to analytic functions, tell me.</div><br />An operator that satisfies these two properties is called a <b>derivation</b>&nbsp;-- one can prove additional properties from these axioms fairly easily, e.g. $D(c)=0$ for constant $c$, etc.<br /><br /><hr /><br />Let's think about why this whole construction above makes sense.<br /><br />Let $G$ be the group of translations of $\mathbb{R}$ -- one can parameterise them by the translated distance as $\Delta(p)$ with composition given by $\Delta(p)\Delta(q)=\Delta(p+q)$. Well, this is isomorphic to the additive group on the reals, and in turn to the multiplicative positive real numbers. We can consider the group to be acting on real analytic functions by translations of the domain: $\Delta_pf(x):=f(x+p)$ The Lie algebra is just spanned by the derivative of $\Delta(p)$ at the identity, that is:<br /><br />$$\Delta '(0) = \lim\limits_{h \to 0} \frac{{\Delta (h) - 1}}{h} = \frac{d}{{dx}}$$<br />And our Lie algebra members are all real multiples of $d/dx$ -- these are precisely the directional derivatives on $\mathbb{R}$. Similar constructions can be made on $\mathbb{R}^n$, or a general automorphism group.<br /><br />So we see that the "derivations" construction of the Lie algebra actually are <b>the tangent vectors on the Lie group identified as the automorphism group of some object</b>. If you've ever done some differential geometry, this gives you the motivation for treating partial derivatives as basis vectors.<br /><br /><div class="twn-pitfall">Our discussion of derivations so far works both for derivations (general vector fields on the manifold) and&nbsp;<b>point-derivations</b>&nbsp;(basically tangent vectors at a specific point). Under the first interpretations, though, we're&nbsp;<b>not actually interested in all derivations</b>, only the left-invariant ones. For example, in the example above, an operation of the form of $p(x)\frac{d}{dx}$ is linear and satisfies the product rule:<br /><br />$$p\frac{d(f\cdot g)}{dx}=g\cdot p\frac{df}{dx}+f\cdot p\frac{dg}{dx}$$<br />And why shouldn't it? It corresponds to a vector field all right -- $xe_x$. But this is not a&nbsp;<i>left-invariant vector field</i>.</div><br /><div class="twn-furtherinsight">Interpret the <b>Taylor series as the exponential map</b> from the Lie algebra to the Lie group! Make the "similar construction" in the multivariate case ($\mathbb{R}^n$) and interpret the <b>multivariate taylor series</b> as an exponential map -- i.e. that $\Delta=\exp\nabla$</div><div><div><br /></div><div><hr /></div><div><br /></div>The first thing that we can do with our formalism of point-derivations is give another proof of closure under the Lie Bracket:&nbsp;</div><div><br /></div><div>$$[D_1,D_2](fg)=f[D_1,D_2]g+g[D_2,D_1]f$$</div><div>I.e. that the Lie Bracket of two derivations is also a derivation. Check that the above is correct by expanding stuff out and using the product rule for $D_1$ and $D_2$.</div><div><br /></div><div>There's another way that derivations can be used to show closure under the Lie Bracket, which shows more closely the connection to the product rule for the second derivative discussed in the <a href="https://thewindingnumber.blogspot.com/2019/05/an-easy-way-to-see-closure-under-lie.html">previous article</a>.<br /><br />One might wonder if, like the directional derivative at the identity in the $c'(0)$ direction is given by $(f\circ c)'(0)$, the directional derivative at the identity in the $c''(0)$ direction may be given as $(f\circ c)''(0)$. Well, in general:<br /><br />$$(f\circ c)''(t)=c''(t)\cdot\nabla f(t)+c'(t)\frac{d}{dt}\nabla f(t)$$<br />Which since $c'(0)=0$, at $t=0$ is simply equal to the first term, the directional derivative in the $c''(0)$&nbsp; direction. So we just need to show that $f\mapsto (f\circ c)''(0)$ is a derivation. This follows from the <b>Leibniz rule for the second derivative</b>, and the fact that the first derivative of $c$ is zero.</div><div><br /></div><div><hr /><br /><div>OK, one more thing before we actually do something useful -- something we haven't done before in other ways.<br /><br />This is an <b>extended pitfall prevention</b>, because I fell into this pit myself. When thinking about left-invariance of a vector field $D$, I formulated the idea in my head this way: the idea is that under $D$, we should get the same result if we differentiate (derivate?) $f$ at 0 or if we translate it forward by $x$ and derivate it at $x$. i.e. where $\phi^h$ represents the translation $f(x)\mapsto f(x-h)$, we want:<br /><br />$$D=\phi^{h}D\phi^{-h}$$</div></div><div>(<i>THIS IS WRONG!</i>&nbsp;This is a pitfall prevention, not an actual result!) And I looked at some simple Abelian cases, like the additive real group and the circle group and thought this was clearly true.<br /><br />But it's wrong. How do we know that? Well, let's consider the group action $\phi^{-h}D\phi^h$ -- certainly at $h=0$, it's the identity, so let's differentiate it (against $h$) at 0. We get, where $d\phi_0$ is the derivative of $\phi$ at 0:<br /><br />$$[d\phi_0, D]$$<br />Which isn't zero. So my argument must be wrong -- I must have assumed abelian-ness somehow.<br /><br />Here's the problem: the final left-multiplication by $\phi^h$ is fine -- it just brings the derived function back to the origin, but "translating the function forward and then differentiating it" messes things up when the direction you're differentiating in doesn't commute with the direction of translation. Draw some pictures of curved surfaces to convince yourselves of this.<br /><br />So left-multiplication determines a sort of "parallel transport" on the Lie Group, while right-multiplication is an "alternative" way to compare vectors in different tangent spaces, and its disagreement with left-multiplication determines the non-commutativity of the group. Well, this choice of left-multiplication vs right-multiplication is really a convention, arising from the choice of representation.<br /><br /><hr /></div><div><br />OK, the useful thing: Suppose we're interested in "nested Lie brackets" $[X,[Y,Z]]$. We're talking about conjugating $[Y,Z]$ as $\phi^p[Y,Z]\phi^{-p}$ where $d\phi_0=X$ so that to first-order in $p$:<br /><br />$$\phi^p[Y,Z]\phi^{-p}=[Y,Z]+p[X,[Y,Z]]$$<br />Since conjugation is a homomorphism, we can also write:<br />\begin{align}<br />\phi^p[Y,Z]\phi^{-p} &amp;= [\phi^pY\phi^{-p},\phi^pZ\phi^{-p}] \\<br />&nbsp;&amp;= [Y+p[X,Y],Z+p[X,Z]] \\<br />&nbsp;&amp;= [Y,Z] + p([Y,[X,Z]]+[[X,Y],Z])\\<br />\Rightarrow [X,[Y,Z]]&amp;=[Y,[X,Z]]+[[X,Y],Z]<br />\end{align}<br />Now, couldn't we have just have proven this by expanding everything out as commutators? Sure, but this provides more insight as to what's going on -- you might notice the resemblance to the product rule. Indeed, this identity -- <b>the Jacobi identity</b>&nbsp;-- is perhaps best stated as:<br /><br />"<b>A derivation $X$ acts through the Lie Bracket as a derivation on the space of derivations where "multiplication" is given by the Lie Bracket.</b>"<br /><br />In this sense, it's actually quite expected -- it results from the fact that the Lie Bracket is a bilinear operator obtained from <b>differentiating a group symmetry, conjugation</b> -- this mandates that it is a derivation.<br /><br />As it turns out, the Jacobi identity, along with the antisymmetry and the bilinearity, determines the Lie Algebra -- it is enough to "abstract out" the properties of a Lie Algebra. Why? This is something we will see over several articles, which will then allow us to motivate abstract Lie algebra.</div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-13410626595773071622019-06-03T01:38:00.000+01:002019-07-08T14:39:24.152+01:00Dealing with eigenspaces; noncommuting variables and another postulateAt the end of the&nbsp;<a href="https://thewindingnumber.blogspot.com/2019/06/projection-operators-generalised-borns.html">last article</a>, you might've wondered how one might talk about 3-dimensional position -- so far, we've only considered an operator representing a one-dimensional position, e.g. to find the $x$-co-ordinate of something. This is obviously insufficient. What we need to measure three-dimensional position is&nbsp;<b>three separate measurements</b>&nbsp;of the different spatial dimensions.<br /><br />But there's a problem here -- we know that upon an observation, the state vector is modified, in that it is replaced by some eigenstate of the observable in question. So after measuring the $x$-co-ordinate, if we measure the $x$-co-ordinate afterwards, are we "really measuring" the $y$-co-ordinate of the particle as it was in its initial state, or have we shaken it around a bit?<br /><br />Let's try to think very precisely about what's going on here. The first question to ask is -- how do the eigenstates of the $X$ operator look like?<br /><br />Well, because it's an observable, it must have eigenstates that produce a full eigenbasis. But if each eigenvalue <b>corresponded to just one eigenstate</b>, then we would only have information about the $x$-positions of particles, which is clearly insufficient to represent the entire state of a particle. So we must have each eigenvalue -- each $x$-position -- correspond to <b>an infinitude of states, an eigenspace,</b> corresponding to each position with the same $x$-position (which, remember, is their eigenvalue), <b>and their superpositions thereof</b>. And this makes a lot of sense -- each position is a state, but these positions give us the same values for the $x$-position.<br /><br />What this means is that each function of the form $g(y,z)\delta(x-x_0)$ is an eigenstate of the $X$ operator, with eigenvalue $x_0$. So we can have something like $g(y,z)=\delta(y-y_0)\delta(z-z_0)$, which would also be an eigenstate of the $Y$ and $Z$ operators (with eigenvalues $y_0$ and $z_0$ respectively), or some other linear combination, which would no longer be an eigenstate of $Y$ and $Z$.<br /><br />OK. So what happens when we observe $X$ taking the value $x$? You might think that the state just turns into some <b>randomly chosen eigenstate </b>with the observed eigenvalue $x$. But if you think about it, this would be quite <b>unphysical</b>, as this would mean our $X$-observation would <i><b>magically change</b></i> <b>our knowledge about the $y$ and $z$ positions</b> too (for example, if the state collapsed into a state that is also an eigenstate of $Y$, we would have accidentally completely measured the $y$ position) -- but we can certainly design experiments in which an observation of an $x$ position does not so radically rattle the particle in the $y$ and $z$ directions.<br /><br /><div class="twn-furtherinsight">Another way to think about this is that the eigenvalues of an operator are the measurements we're getting out. If a state is already in an eigenspace corresponding to the eigenvalue $\lambda$, and we "measure" the observable again (i.e. do nothing), the state shouldn't change.</div><br />So we don't want the observation to change the $y$ and $z$ probability information in any way -- so what we're looking for is a <b>projection&nbsp;of the state into the eigenspace</b> with eigenvalue $x$. This is in line with our discussion of the generalised Born's rule in the <a href="https://thewindingnumber.blogspot.com/2019/06/projection-operators-generalised-borns.html">last article</a>&nbsp;-- but it is an <b>additional postulate</b> of quantum mechanics, or rather generalises the existing postulate about states projecting randomly into eigenstates.<br /><br /><div class="twn-furtherinsight">Weren't the eigenvalues completely irrelevant? You ask. You can just make the eigenvalues whatever you want, they're just labels to the eigenstates, right? Not really -- the eigenvalues <em>are</em> exactly what you measure. You can choose to measure any function of them, but if you use a function that isn't injective, you are measuring <em>less information</em> about the system, and you're collapsing the state "less" in this sense.</div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Et48Tk7LJKk/XPGTZ8d7geI/AAAAAAAAFlY/hfBffc8ndk0iGTaYB3E1f1YlgaoIj1VYwCPcBGAYYCw/s1600/projection.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="394" data-original-width="296" height="320" src="https://1.bp.blogspot.com/-Et48Tk7LJKk/XPGTZ8d7geI/AAAAAAAAFlY/hfBffc8ndk0iGTaYB3E1f1YlgaoIj1VYwCPcBGAYYCw/s320/projection.png" width="240" /></a></div><br />(<i>Unlike in the projection above, the subspace being projected onto upon measurement of X is itself an infinite-dimensional space, spanned by the different positions Y and Z an take. Oh, and we have to normalise the projected state.</i>)<br /><br />Something that we've seen in this discussion above is that there is a <b>common eigenbasis</b> for $X$, $Y$ and $Z$ -- specifically, the <b>position basis</b>,&nbsp;the basis of Dirac delta distributions centered at the different points in three-dimensional space. From linear algebra, this is equivalent to saying that $X$, $Y$ and $Z$ commute.<br /><br />What this means is that when you then go on to measure $Y$, and then $Z$, you end up in a state that is a common eigenstate of $X$, $Y$ and $Z$ -- so that you have precise values for each co-ordinate of the positions. And as the $X$ information is only altered in the $X$ observation, etc. so the probability distributions for each variable is the same <b>regardless of the order</b> you measure it in -- so <b>three-dimensional position is indeed well-defined</b> in quantum mechanics.<br /><br /><hr /><br /><div class="twn-pitfall">Just to be clear, the fact that $X$, $Y$ and $Z$ commute is a <b>postulate</b> -- equivalent to the physical claim that each position in space is in fact an eigenstate, that we can in fact pinpoint the position of a particle exactly. We <em>cannot</em> do this e.g. for position and momentum -- $(x,p)$ pairs cannot be considered eigenstates, as there is no simultaneous eigenstate of $X$ and $P$. So for example, you <b>can't just construct a spacefilling curve</b> in $(x,p)$ space to measure position and momentum simultaneously, because the parameters of the curve would simply not have any corresponding eigenstates. The $(x,p)$ space <b>does not exist in the Hilbert space</b>, there are no states that precisely put down the values of position and momentum. It is possible to construct quantum mechanical theories -- called <b>non-commutative quantum theories</b> -- in which the $(x,y,z)$ space isn't in the Hilbert space either, so that our perception of three-dimensional positions must necessarily be approximate.<br /><br />We're assuming here that this is not so, that three-dimensional space does form an eigenbasis for the $X$, $Y$ and $Z$ operators, that the representation of the $Y$ operator in the $X$ basis is indeed $\psi(x,y,z)\mapsto y\psi(x,y,z)$, not something weird and fancy.</div><br /><hr /><br />A very different picture arises when you have noncommuting variables. Suppose two operators $X$ and $P$ don't commute, i.e. there is no common eigenbasis for them. So once you observe $X$ and put it in some eigenspace of $X$, there is a non-zero probability that the state will have to be projected out of this $X$-eigenspace when $P$ is measured.<br /><br />So this means that the observables $X$ and $P$ <b>cannot be measured simultaneously</b>. Some specific bounds on the uncertainties will be discussed in the <a href="https://thewindingnumber.blogspot.com/2019/06/position-momentum-bases-fourier.html">next article</a>. For now, let's demonstrate an example of two noncommuting variables: <b>position</b>&nbsp;and <b>momentum</b>&nbsp;(in the same direction).<br /><br /><b>NOTE: We will show in the next article the given results about momentum being $-i\hbar \frac{\partial}{\partial x}$, etc. Just intuit them out here from its eigenvectors.</b><br /><br />As we've shown before, the position and momentum operators can be given in the position basis as $x$ and $-i\hbar\partial/\partial x$ respectively. What this means is that given a wavefunction $\psi(x)$, it transforms under these operators as $x\psi(x)$ and $-i\hbar\psi'(x)$ respectively (check that this makes sense -- especially for the position case -- and also that one can go the other direction and show that the <b>corresponding eigenvectors</b> of the position operator must be <b>Dirac delta functions</b>).<br /><br />So do these operators commute? Clearly not -- the eigenbasis of one is Dirac delta functions in $x$, the other's is sinusoids in $x$. But we can also verify this computationally:<br /><br />\begin{align}XP &amp;= -i\hbar x \frac{\partial}{\partial x}\\<br />PX\{\psi(x)\}&amp;=-i\hbar\frac{\partial}{\partial x}(x\psi(x))\\<br />&amp;= -i\hbar\left[\psi(x)+x\psi'(x)\right]\\<br />\Rightarrow PX &amp;= -i\hbar x\frac{\partial}{\partial x} -i\hbar\end{align}<br /><br />So we have the commutator $i[X,P]=-\hbar$ (why do we talk about $i[A,B]$? Because as it is easy to see, for any Hermitian $A$ and $B$, this is Hermitian, while $[A,B]$ is simply anti-Hermitian). This is the "purest" commutator -- a (scaled) Identity operator. Since we didn't use any other properties of position and momentum, <b>this is a property of all observables that are Fourier transforms of each other/canonically conjugate observables&nbsp;</b>(more on this in the next article).<br /><br /><hr /><br /><b>Exercise:</b> Write down the most generalised form of Born's rule accounting for generalised eigenspaces (the answer is identical to what we've already written, but make sure you understand it). Show, as in <a href="https://thewindingnumber.blogspot.com/2019/06/projection-operators-generalised-borns.html">the last article</a>, that the probability density of finding a particle somewhere in three-dimensional space is $|\Psi(x,y,z)|^2$ -- make sure you define $\Psi(x,y,z)$ clearly!<br /><br />Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-35716964878934175752019-06-02T19:17:00.001+01:002019-06-04T10:48:31.516+01:00Position, momentum bases and operators, Fourier transform, uncertaintyIn this article, we'll assume the de Broglie relation for all particles -- i.e. that their momentum is given by $p=hf$. This is actually quite an incredible assumption, even if not surprising -- we've accepted that a particle is a wave in the sense of probability (the wave describes the probability amplitude densities of finding it at some point), but why at all should the spatial frequency of the probability wave relate to its momentum?<br /><br />Well, it's natural for you to find this assumption unsatisfactory. We've been quite liberal in assuming the de Broglie relation earlier when <a href="https://thewindingnumber.blogspot.com/2019/05/from-polarisation-to-quantum-mechanics.html">motivating quantum theory</a>, too -- we'll later produce some motivation for the de Broglie relation for photons, and discuss derivations <i>from</i>&nbsp;quantum mechanics, axiomatising our theory clearly to eliminate circularities. But for now, let's not.<br /><br />The key point of $p=hf$ is that for a sinusoidal wave $e^{i \cdot 2\pi f \cdot x}$ (so the probability density is uniform, and the standard deviation in the observation of the particle's position is infinite), the momentum takes a specific <i>definite</i>&nbsp;value, $hf$, with zero standard deviation.<br /><br />Well, what if the wavefunction isn't a simple sinusoid, but some other distribution $\Psi(x)$? If you did all the assigned exercises in the <a href="https://thewindingnumber.blogspot.com/2019/05/from-polarisation-to-quantum-mechanics.html">first article</a>, you should know the answer (if not, work it out before reading on). Classically, if you could write that wavefunction as a sum of sinusoids (i.e. use a Fourier transform), then each sinusoid would have its own momentum and there would be some chunk of your matter in each of those momenta, forming a momentum distribution. In quantum mechanics, you can't have chunks of a single quantum, so you this distribution is a probability distribution (still a probability <i>amplitude</i>&nbsp;distribution, because we want superposition). We'll use the notation $\Psi(p)$ to represent this "<b>momentum-space wavefunction</b>", and we'll see why soon.<br /><br />So it's not too hard to see that the frequency distribution is simply the Fourier transform of $\Psi(x)$, while the momentum-space wavefunction is given by:<br /><br />$$\Psi(p)=\frac1h \mathcal{F}_x^{p/h}(\Psi(x))$$<br />Where $\mathcal{F}_x^{p/h}(\Psi(x))$ is the Fourier transform of $\Psi(x)$ (which is a function of $f$) written with the variable substitution $f=p/h$. Note that we're considering the non-normalised Fourier transform, in terms of ordinary frequencies.<br /><br />Well, $\Psi(x)\, dx$ and $\Psi(p)\, dp$ are just the representations of the state vector in the position and momentum bases respectively. So the <b>inverse&nbsp;Fourier transform acts as a <i>change-of-basis matrix</i></b>&nbsp;from the position basis to the momentum basis. I.e.<br /><br />$$|\psi\rangle_P=F|\psi\rangle_X$$<br />This change-of-basis matrix $F^{-1}$ precisely represents the <b>eigenstates of the momentum operator</b> written in the position basis, and the corresponding eigenvalues are the actual values of the momenta. So we have eigenstates $\frac1h e^{ix \cdot 2\pi p / h} dp$ with corresponding eigenvalues $p$.<br /><br />Before going any further, let's make sure we know exactly what this means: our change-of-basis matrix $F^{-1}$ is an uncountably infinite-dimensional "matrix" whose "indices" are denoted as $(x,p)$ in the rows-by-columns format. Its general entry is $\frac1h e^{ix \cdot 2\pi p / h} dp$, and each column -- here's the important bit -- each column holds <i>p</i>&nbsp;constant and varies <i>x</i>, i.e. each column, i.e. each eigenstate of $P$ is a function of $x$.<br /><br />Anyway, so we're looking for a linear operator $P$ solving the eigenvalue problem (and we're just ignoring the scalar multiples):<br /><br />$$P e^{ix \cdot 2\pi p / h} = pe^{ix \cdot 2\pi p / h}$$<br />It should be quite clear that the operator we're looking for is:<br /><br />\begin{align}P &amp;= \frac{h}{2\pi i}\frac{\partial}{\partial x} \\<br />&amp;= -i\hbar \frac{\partial}{\partial x} \end{align}<br />We need to be clear that this is the representation of the momentum operator <i>in the position basis</i>&nbsp;-- in the momentum basis, its representation is simply "$p$" (i.e. its action on each eigenstate $|p\rangle$ is to multiply it by $p$). Similarly, it should be easy to show that in the momentum basis,<br /><br />$$X=i\hbar\frac{\partial}{\partial p}$$<br /><b>Exercise:</b>&nbsp;make sure you clearly know and understand what the eigenvectors and eigenvalues of $X$ and $P$ are, in both the position and momentum bases. Hint: something about the Dirac delta function.<br /><br /><hr /><br /><b>Derivation of Heisenberg and Robertson-Schrodinger uncertainty principles</b><br /><b><br /></b>We can derive a variety of "uncertainty principles" -- inequalities showing trade-off between the certainties of two observables -- with some basic algebraic manipulation. It is important to note that none of these individual uncertainty principles is really much more fundamental than any of the others (or at least I don't see in what way they can be) -- one can always make stronger bounds for the uncertainty, and many stronger bonds exist than the ones we're showing here -- but the <i>concept</i>&nbsp;of an uncertainty principle is crucial, in that it demonstrates the rigorously difference between <b>quantum mechanics and statistical physics</b>. In general, the noncommutativity of observables (having no shared eigenstates) is something that has no analog in classical physics.<br /><br />OK. So we'll show two statements about the product of uncertainties of two observables, $(\langle A^2\rangle - \langle A\rangle^2)^{1/2}(\langle B^2 \rangle - \langle A \rangle^2)^{1/2}&nbsp;$. Once again, there is nothing special about the specific relations we will show -- we can consider other combinations than products, like $\Delta a^2 + \Delta b^2$, and indeed, there exist uncertainty relations for such terms.<br /><br />Defining $A'=A-\langle A\rangle$ and $B'=B-\langle B\rangle$ for <b>Hermitian</b>&nbsp;(this is important!) $A$ and $B$, we see that:<br /><br />\begin{align}<br />\langle A'^2\rangle \langle B'^2 \rangle &amp;= \langle \psi | A'^2 | \psi \rangle \langle \psi | B'^2 | \psi \rangle \\<br />&amp;= \langle A' \psi | A' \psi \rangle \langle B' \psi | B' \psi \rangle \\<br />&amp;\ge |\langle \psi | A' B' | \psi \rangle| ^ 2 \\<br />&amp;= \left|\frac12 \langle\psi|A'B'+B'A'|\psi\rangle + \frac12\langle\psi|A'B'-B'A'|\psi\rangle\right|^2 \\<br />&amp;= \frac14 |\langle\psi|A'B'+B'A'|\psi\rangle|^2 + \frac14|\langle\psi|A'B'-B'A'|\psi\rangle|^2 \\<br />&amp;= \frac14 |\langle \{A-\langle A\rangle, B-\langle B\rangle\} \rangle| ^2 + \frac14 |\langle [A,B]\rangle|^2\\<br />&amp;= \frac14 |\langle\{A,B\} \rangle - 2\langle A\rangle \langle B\rangle |^2 + \frac14|\langle[A,B]\rangle|^2\\<br />\Rightarrow \Delta a\,\Delta b &amp;\ge \frac12 \sqrt{|\langle\{A,B\} \rangle - 2\langle A\rangle \langle B\rangle |^2&nbsp;&nbsp;+ |\langle [A,B]\rangle|^2}<br />\end{align}<br />This is the&nbsp;<b>Robertson-Schrodinger relation</b>.<br /><br />(Guide in case you get stuck somewhere -- <i>line 3</i>, Cauchy-Schwarz inequality; <i>line 4</i>, splitting into Hermitian and anti-Hermitian parts; <i>line 5</i>, magnitude of a complex number -- I'm not sure if I can give any better motivation for specifically considering the product of the standard deviations -- like I said, these specific relations are not really that fundamental. I guess we just want to illustrate the <b>point</b> of "the" uncertainty principle, regardless of the specific ways in which it is treated, and would like to get a simple form for it, regardless of how weak or strong it may be.)<br /><br />One may weaken the inequality further, writing (and this is equivalent to having ignored the real part in line 4, saying the magnitude of a complex number is at least that of the imaginary part):<br /><br />$$\Delta a\,\Delta b \ge \frac12 |\langle [A,B]\rangle|$$<br />This is the <b>Heisenberg uncertainty relation</b>. In particular, in the <a href="https://thewindingnumber.blogspot.com/2019/06/dealing-with-eigenspaces-noncommuting.html">last article</a>, we showed that for the position and momentum operators, $[X,P]=i\hbar$. So in this case, we get the celebrated identity:<br /><br />$$\Delta x\, \Delta p \ge \frac{\hbar}{2}$$<br />For canonically conjugate $X$ and $P$.<br /><br />As mentioned before, other stronger uncertainty relations exist for general observables. Some examples can be found on the Wikipedia page <a href="https://en.wikipedia.org/wiki/Stronger_uncertainty_relations">Stronger uncertainty relations</a>&nbsp;(<a href="https://en.wikipedia.org/w/index.php?title=Stronger_uncertainty_relations&amp;oldid=874251670">permalink</a>).Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-77330539512350933022019-06-01T13:13:00.000+01:002019-06-03T16:12:45.229+01:00Projection operators, generalised Born's rule, position basis, wavefunctionAt the end of the&nbsp;<a href="https://thewindingnumber.blogspot.com/2019/05/from-polarisation-to-quantum-mechanics.html">last article</a>, I asked you to investigate Born's rule for continuous variables like position and momentum.<br /><br />Well, the problem is that if $x$ is continuously distributed (i.e. we have an operator $X$ whose eigenvalues form a continuous spectrum $\Sigma_X$), typically $P(x=\lambda)=0$ -- and this gives us very little information about the actual probability distribution. What we're really interested in is $P(x\in B)$ for $B$ some subset of $\Sigma_X$.<br /><br /><div class="twn-furtherinsight">Technically, we need $B$ to be a "Borel subset", or "measurable subset". We will be omitting several such technicalities in the article, such as the need for the spectral theorem to define a "projection-valued measure" or "spectral measure" on an operator with a continuous spectrum -- this is something that will be covered in the <a href="https://thewindingnumber.blogspot.com/p/1103.html">MAO1103: Linear Algebra course</a>.</div><br />First, let's think about $P(x\in B)$ in the countable case. One can write $B=\{\lambda_1,\ldots\lambda_n\}$, and then simply say that<br /><br />$$P(x\in B)=\sum |\langle\psi|\phi_k\rangle|^2$$<br />But the term on the right is a Pythagorean sum -- specifically, it is the length-squared of the vector formed by summing all the projections of $|\psi\rangle$ onto the eigenstates $|\phi_1\rangle\ldots|\phi_k\rangle$. But this is the same as the length of the projection of $|\psi\rangle$ onto the span of these eigenstates.<br /><br />(<b>Note on notations:</b>&nbsp;From here onwards, we will use the notation $|\lambda\rangle$ to refer to the eigenvector corresponding to the eigenvalue $\lambda$ (if the eigenspace has dimension more than 1, we'll figure something out). We will use the notation $\{|B\rangle\}$ to refer to the span of the eigenvectors corresponding to the eigenvalues in $B$.)<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Et48Tk7LJKk/XPGTZ8d7geI/AAAAAAAAFlQ/_RK3VpoUgvoRp1dfUpK_zzE98L-8mqwfQCLcBGAs/s1600/projection.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="394" data-original-width="296" height="320" src="https://1.bp.blogspot.com/-Et48Tk7LJKk/XPGTZ8d7geI/AAAAAAAAFlQ/_RK3VpoUgvoRp1dfUpK_zzE98L-8mqwfQCLcBGAs/s320/projection.png" width="240" /></a></div>So we could just define a Hermitian <b>projection operator</b> $L_X(B)$ for any subset $B$ of the spectrum of $X$ -- it is an easy exercise to write down an explicit form for $L_X(B)$ in terms of the eigenvectors of $X$.<br /><br />Then the probability $P(x\in B)$ is simply $|L_X(B)|\psi\rangle|^2$. Recalling that a Hermitian projection operator satisfies $L^*=L=L^2$, we can write the <b><i>generalised Born's rule</i></b>&nbsp;as:<br /><br />\begin{align}P(x\in B) &amp;= |L_X(B)|\psi\rangle|^2\\ &amp;= \langle\psi|L_X(B)|\psi\rangle\end{align}<br />Well, this is interesting! In the last article, you proved that the <b>expected value</b> of an observable $X$ given a state $|\psi\rangle$ is given by $\langle\psi|X|\psi\rangle$. But here we have a <i><b>probability</b></i>&nbsp;given by the same expression. So we want to interpret our projection operators as some sort of "observable" -- we can omit the "Hermitian", since all observables are Hermitian.<br /><br />There's another place you might've seen something like this, and that is with <i>indicator variables</i> in probability and statistics -- <i>the expected value of an indicator variable for an event is the probability that the event occurs</i>.<br /><br />Try to interpret these projection operators as observables that are analogous to "indicators" in some sense. If you think a little about it, you might see exactly what these observables represent: the eigenvalues of $L_X(B)$ are all 1 and 0 -- if the value "1" is realised, the state has been projected into the $\{|B\rangle\}$ -- and if the value "0" is realised, it hasn't.<br /><br />So projection operators are a special type of observable, measuring the answer to "<b>Yes/No questions</b>" -- if the answer to "is the system in one of the states $\{|B\rangle\}$?" is <b>yes</b>, the observable $L_X(B)$ takes the value 1 -- if the answer is <b>no</b>, then it takes the value 0. So it is precisely an "<b>indicator variable</b>" for $\{|B\rangle\}$.<br /><br />We have seen such projection operators, of course, in the context of polarisation -- where the operator represented whether or not the photon has passed through. Indeed, one may formulate quantum mechanics entirely in terms of projection operators, as any question can be formulated with some number of Yes/No questions (the key reason why this can be done, as we will see -- is that these "yes/no questions" all commute, i.e. the corresponding projection operators share an eigenbasis). Let's not.<br /><br /><hr /><br />Well, this can be generalised in the straightforward way to an operator with a continuous spectrum, resulting in the same expression. We can also calculate probability <i>densities</i>&nbsp;using this result. Let $X$ be an operator with continuous spectrum $\Sigma_X$ -- then we can write the state $|\psi\rangle$ in the eigenbasis of $X$:<br /><br />$$|\psi\rangle = \int_{\Sigma_X} |x\rangle\, \Psi(x)\, dx$$<br />Where $\Psi(x)\, dx=\langle\psi|x\rangle$ are the coefficients of the state in the eigenbasis, i.e. the probability amplitudes -- we call $\Psi(x)$ the <b>wavefunction</b>, and it represents <i>probability amplitude densities</i>. Then for some set $M\subseteq \Sigma_X$ of eigenvalues $L_X(B)|\psi\rangle$ is the projection:<br /><br />$$L(M)|\psi\rangle = \int_B |x\rangle\, \Psi(x)\, dx$$<br />And one may calculate the dot product, noting that complex dot products require taking the complex conjugate:<br /><br />$$\langle \psi | L_X(B) | \psi \rangle = \int_B \Psi^*(x)\, \Psi(x)\, dx$$<br />Which gives us an expression for the <b>probability density function</b>&nbsp;on $\Sigma_X$ as:<br /><br />\begin{align}\rho(x) &amp;=\Psi^*(x)\,\Psi(x) \\<br />&amp;=|\Psi(x)|^2\end{align}<br />And this applies to any operator with a continuous spectrum, like position and momentum.<br /><br /><hr /><br /><div class="twn-pitfall">Some texts define the eigenvectors $|x\rangle$ of a continuous-spectrum observable differently from us -- it is often conventional to let $|x\rangle$ be infinitely large so that $\langle x_1|x_2\rangle = \delta(x_1-x_2)$. This is so that the amplitudes $\langle\psi|x\rangle$ are not infinitesimal, but instead $\langle\psi|x\rangle=\Psi(x)$ (without multiplication by $dx$). For consistency with discrete spectra, we do not use this convention.</div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-2410056145919502942019-05-29T23:11:00.000+01:002019-06-01T13:18:04.420+01:00From polarisation to quantum mechanics: states, observables, Born's lawLike most texts on the theory, I will motivate the mathematics of quantum mechanics from the example of polarisation -- mostly because it's a very accessible example of stuff being wavelike. From this example, we will be able to motivate: the <b>state vector</b>&nbsp;(generalising the polarisation), <b>state vector collapse</b>&nbsp;(the event of polarisation),&nbsp;<b>observables and their eigenvalues</b>&nbsp;(stuff like energy, number of photons, etc.), <b>eigenstates and their orthogonality&nbsp;</b>(polarisation basis), <b>noncommuting operators and uncertainty</b>&nbsp;(the noncommuting of lenses).<br /><br />The key feature of quantum mechanics -- the fundamentally probabilistic nature -- comes from the following two facts, confirmed by experiments (the famous experiments here are the <b>double-slit experiment</b> and <b>photoelectric effect</b> respectively):<br /><br /><ul><li>Everything is a <b>wave</b>&nbsp;-- objects behave as waves, following the superposition principle and the waves represent densities of observations at large scales.</li><li>Everything is a <b>particle </b>-- which manifests itself in the form of some stuff, like energy and momentum, coming in little quanta.</li></ul><div><br /></div><div>This is the principle of <b>wave-particle duality</b>. You may realise how this implies a probabilistic description, but the following example should make it quite clear: consider a wave of light, with energy $hf$ (so it's a single photon) polarised at angle $\theta$ to the horizontal -- and it passes through a horizontal polarising filter. Well, then the wave that passes through would be a horizontally polarised wave with energy $hf\cos^2\theta$, right?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-g0BSGI23_Tw/XO5d5XpI5SI/AAAAAAAAFks/y51gFRd5w1MYIRBI54_95upCVVU-hiKQACLcBGAs/s1600/4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="545" data-original-width="615" height="283" src="https://1.bp.blogspot.com/-g0BSGI23_Tw/XO5d5XpI5SI/AAAAAAAAFks/y51gFRd5w1MYIRBI54_95upCVVU-hiKQACLcBGAs/s320/4.jpg" width="320" /></a></div><br />But this is <i>impossible</i>, since energy levels in quantum mechanics are quantised -- you can't <i>have</i>&nbsp;$\cos^2\theta$ of a photon, you can only have integer multiples of a photon. But the fact that energy drops as $E\cos^2\theta$ is something that you can verify at your home, using sunglasses -- what the heck?<br /><br />The key point is that the empirical verification of the $\cos^2\theta$ business that you can do at home is on a <b><i>macroscopic</i>&nbsp;level</b>, when you have a large number of photons $E=Nhf$. So something occurs with the photons on a <b>microscopic level</b> such that when you try it with a <b>large number of photons</b>, $\cos^2\theta$ of the photons pass through.<br /><br />Well, this is <b>precisely the (frequentist) definition of probability</b>! A single photon passes through the filter with a <i>probability</i>&nbsp;of $\cos^2\theta$ so that for a large number of photons, $\cos^2\theta$ of the photons pass through. This is a non-trivial result -- wave-particle duality makes no mentions of probability as such, it just tells us that stuff is both a particle and a wave, but this simple condition in itself implies&nbsp;a probabilistic, non-deterministic reality.<br /><br /><hr /><br />Similar thought experiments can illustrate the probabilistic nature of other things (the "things" in question here will soon be called "eigenstates"): <b>position</b>&nbsp;is easy -- consider a standing wave photon in a box (this can easily be constructed). This is uniformly distributed throughout the box -- so how much of the energy is in some chunk of the box?<br /><br /><b>Momentum</b>&nbsp;is trickier, but shouldn't be too hard if you're familiar with Fourier transforms -- what's the analog of a "box" in momentum-space? Well, consider a concentrated pulse of light -- this can be written, via a Fourier transform, as the sum of several light waves of different momenta (i.e. frequencies), each wave with some lower energy. Taking "some chunk" of this "box" amounts to filtering some specific frequencies of the light. This can be done easily, e.g. with a colour filter -- so how much of the energy is contained in the waves with these specific momenta?<br /><br />In both cases, the key point is that you can't have a fraction of the energy of the photon at these positions/momenta, so you must have a <i>probability</i>&nbsp;of measuring the photon to be in a specific range of positions or a specific range of positions -- to be in a specific region or in a specific region of momentum-space.<br /><br /><hr /></div><div><br />The fundamental point here can be made for any quantity $X$: if you can filter out the "part" of a collection of particles that has $X$ in a certain subset of its range, then on a microscopic level, is <i>probabilistic</i>. The act of "filtering out the parts with a certain $X$", applied to a single particle, is just the act of checking if a particle is in a certain $X$-interval, and is called <b>measurement</b>. Any quantity that you can measure is called an <b>observable</b>.&nbsp;</div><div><br /></div><div class="twn-furtherinsight">Something like polarisation is really a form of measurement -- you're <i>finding out</i>&nbsp;whether or not the photon is in a certain polarisation $|\phi_{\parallel}\rangle$. You may have another observable, corresponding to a different polarisation -- even one that is orthogonal to the first polarisation -- $|\phi_{\perp}\rangle$ and still get that the photon is in $|\phi_\perp\rangle$. There is nothing wrong with this, as we just know beforehand that the photon is in $|\phi_\parallel\rangle$ <em>or</em> $|\phi_\perp\rangle$. If you perform the polarisation with $|\phi_\perp\rangle$ <em>after</em> the polarisation with $|\phi_\parallel\rangle$, you will find that the photon doesn't pass through, as you know for sure that the photon is not in both $|\phi_\parallel\rangle$ <em>and</em> $|\phi_{\perp}\rangle$.<br /><br />Now, you may have certain psychological issues with this, as have many in history -- however, you might want to note that the aim of quantum mechanics is not to fix your psychological, psychiatric etc. problems but to explain nature. You need to accept logical positivism and learn to <em>shut up and calculate</em> to be comfortable with quantum mechanics -- I recommend reading <a href="https://thewindingnumber.blogspot.com/2017/08/three-domains-of-knowledge.html">Three Domains of Knowledge</a>.</div><div><br /></div><div>So whatever calculus we invent to describe these probabilistic phenomena, it is going to apply to all <i>observables</i>.</div><div><br /></div><div>In our first example, the polarisation of the photon can be represented by a unit vector which we will denote as $|\psi\rangle$. The polarising filter has two special axes, represented by unit vectors $|\phi_{\parallel}\rangle$ and $|\phi_\perp\rangle$ -- these are special in the sense that an incoming photon polarised as $|\phi_{\parallel}\rangle$ or $|\phi_\perp\rangle$ will simply be scaled, by factors of 1 and 0 respectively -- so these form an <b>eigenbasis</b> for a certain operator.</div><div><br /></div><div>Well, we said that the photon passes through (with polarisation $|\phi_{\parallel}\rangle$) with probability $\cos^2\theta$ -- this arises simply from considering the <i>amplitude of $|\psi\rangle$ in the direction of $|\phi_\parallel\rangle$</i>. So we can write the <b>probability that the photon ends up in a state</b> $|\phi\rangle$ as $|\langle\psi|\phi\rangle|^2$ where $\langle\psi|\phi\rangle$ is called the corresponding "<b>probability amplitude</b>".<br /><br /><b>This expression, $P(x=\lambda)=|\langle\psi|\phi_\lambda\rangle|^2$ is called Born's rule.</b></div><div><br /></div><div>Let's get back to the eigenbasis -- what exactly is this an eigenbasis of? We said that the corresponding eigenvalues are 1 and 0, so this gives us a complete description of the operator. Note that this operator depends <i>only</i>&nbsp;on the observable (namely "number of photons in the $|\phi_{\parallel}\rangle$ direction), not on the state or any other feature of the observation. So we decide to <b><i>call</i>&nbsp;this operator/matrix the "observable"</b>, and its eigenvalues are the values of the observable that can be measured.<br /><br />To find properties of these observables, the natural way is to note that the only feature we've really required of them is Born's rule, i.e. the probabilistic interpretation -- so we can apply the axioms of probability and see what they apply in the context of these observables.<br /><br /><ul><li><b>$P(E)\ge 0$ --</b>&nbsp;imply that the observables are over either the reals or complexes, so that $|\langle\psi|\phi\rangle|^2\in \mathbb{R}$ in the first place. The nonnegativity then follows.</li><li><b>$P(\Omega)=1$ and </b><b>$P\left(\bigcup_i E_i\right) = \sum_i P(E_i)$ for disjoint $E_i$ -- </b>this, along with the second axiom, implies that $\sum |\langle\phi|\psi\rangle|^2 = 1=|\langle\psi|\psi\rangle|^2$ where the sum is taken over all eigenstates $|\phi\rangle$ of the operator. As this must be true for all states $|\psi\rangle$, the thing on the left must be a Pythagorean sum, so the $|\phi\rangle$s must form an orthogonal basis. This implies that all <b>observables are normal operators</b>.</li></ul><div><br /></div><div>The latter fact is very important, and can also be seen in the following way -- if you a system is in one eigenstate, it cannot possibly collapse onto another eigenstate (the probabilistic interpretation is: if you know for sure the value of the symbol is a thing, it's that thing) -- so we must have $|\langle \phi_1|\phi_2\rangle|^2=0$ for all eigenstates $|\phi_1\rangle$ and $|\phi_2\rangle$.</div><div><br /></div><div>Another restriction we add is that the observables be not only normal, but <b>Hermitian operators</b>&nbsp;in particular, so they have real eigenvalues. This may seem an odd choice, but it makes sense, as any normal operator may be uniquely written as $X_H+iX_{AH}$ where $X_H$ and $X_{AH}$ are Hermitian, and $X_H$ and $X_{AH}$ commute, so any complex observation can be done unambiguously as two real observations. So we stick to real eigenvalues.</div><div><br /></div><div>This also makes it essential that we allow complex operators rather than just real ones (the two choices were given to us from the first probability axiom), so that this decomposition is possible. Later, we will see concrete examples of this with commutators $[X,Y]$, which must be multiplied by $i$ to turn Hermitian. We will also see more fundamental reasons to choose complex numbers in QM.</div><div><br /></div><div><div><hr /></div><div><br /><b>Exercise:</b>&nbsp;Show that the expected value of an observable $X$ given a state $\psi$ can be given as $\langle \psi|X|\psi \rangle$ (i.e. $\psi^*X\psi$ in conventional notation).<br /><br /><b>Exercise:</b>&nbsp;Explain Born's rule with other observables, like position and momentum. Explain why it holds in general.</div></div></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-88310947406843342162019-05-22T00:57:00.000+01:002019-05-30T13:59:14.989+01:00What's with e^(-1/x)? On smooth non-analytic functions: part IWhen you first learned about the Taylor series, your intuition probably went something like this: you have $f(x)$, the derivative at this point tells you how $f$ changes from $x$ to $x+dx$ (which tells you $f(x+dx)$), the second derivative tells you how $f'$ changes from $x$ to $x+dx$, which recursively tells you $f(x+2\ dx)$, the third derivative tells you $f(x+3\ dx)$, and so on -- so if you have an <i>infinite </i>number of derivatives, you know how <i>each</i>&nbsp;derivative changes, so you should be able to predict the <i>full global behaviour of the function</i>, assuming it is infinitely differentiable (smooth) throughout.<br /><br />Everything is nice and dandy in this picture. But then you come across two disastrous, life-changing facts that make you cry for those good old days:<br /><ol><li><b>Taylor series have <i>radii of convergence</i> -- </b>If I can predict the behaviour of a function up until a certain point, why can't I predict it a bit afterwards? It makes sense if the function becomes rough at that point, like if it jumps to infinity, but even functions like $1/(1+x^2)$ have this problem. Sure, we've heard the explanation involving complex numbers, but why should we care about the complex singularities (here's a question: do we care about quaternion singularities?)? Specifically, a Taylor series may have a zero radius of convergence. Points around which a Taylor series has a zero radius of convergence are called <b>Pringsheim points</b>.</li><li><b>Weird crap --</b> Like&nbsp;$e^{-1/x}$. Here, the Taylor series <i>does</i>&nbsp;converge, but it converges to the wrong thing -- in this case, to zero. Points at which the Taylor series doesn't equal a function on any neighbourhood, despite converging, are called <b>Cauchy points</b>.</li></ol><div>In this article, we'll address the <b>weird crap -- </b>$e^{-1/x}$ (or "$e^{-1/x}$ for $x&gt;0$, 0 for $x= 0$" if you want to be annoyingly formal about it) will be the example we'll use throughout, so if you haven't already seen this, go plot it on Desmos and get a feel for how it looks near the origin.<br /><br /><i>Terminology:</i>&nbsp;We'll refer to <b>smooth non-analytic functions</b>&nbsp;as <b>defective functions</b>.<br /><br /></div><hr /><br /><div>The thing to realise about $e^{-1/x}$ is that the Taylor series -- $0 + 0x + 0x^2 + ...$ -- <i>isn't wrong</i>. The truncated Taylor series of degree $n$ is the <i>best polynomial approximation </i>for the function near zero, and none of the logic here fails for $e^{-1/x}$. There is honestly no other polynomial that better approximates the shape of the function as $x\to 0$.<br /><div><br /></div><div>If you think about it this way, it isn't too surprising that such a function exists -- what we have is a function that <b>goes to zero</b> as $x\to 0$ <b>faster than any polynomial</b> does. I.e. a function $g(x)$ such that</div><div>$$\forall n, \lim\limits_{x\to0}\frac{g(x)}{x^n}=0$$</div><div>This is not fundamentally any weirder than a function that escapes to infinity faster than all polynomials. In fact, such functions are quite directly connected. Given a function $f(x)$ satisfying:</div><div>$$\forall n, \lim\limits_{x\to\infty} \frac{x^n}{f(x)} = 0$$</div><div>We can make the substitution $x\leftrightarrow 1/x$ to get</div><div>$$\forall n, \lim\limits_{x\to0} \frac{1}{x^n f(1/x)} = 0$$</div><div>So $\frac1{f(1/x)}$ is a valid $g(x)$. Indeed, we can generate plenty of the standard smooth non-analytic functions this way: $f(x)=e^x$ gives $g(x)=e^{-1/x}$, $f(x)=x^x$ gives $g(x)=x^{1/x}$, $f(x)=x!$ gives $g(x)=\frac1{(1/x)!}$ etc.<br /><br /></div><div><hr /><br /><div></div></div><div>To better study what exactly is going on here, consider Taylor expanding $e^{-1/x}$ around some point other than 0, or equivalently, expanding $e^{-1/(x+\varepsilon)}$ around 0. One can see that:</div></div><div>$$\begin{array}{*{20}{c}}{f(0) = {e^{ - 1/\varepsilon }}}\\{f'(0) = \frac{1}{{{\varepsilon ^2}}}{e^{ - 1/\varepsilon }}}\\{f''(0) = \frac{{ - 2\varepsilon&nbsp; + 1}}{{{\varepsilon ^4}}}{e^{ - 1/\varepsilon }}}\\{f'''(0) = \frac{{6{\varepsilon ^2} - 6\varepsilon&nbsp; + 1}}{{{\varepsilon ^6}}}{e^{ - 1/\varepsilon }}}\\ \vdots \end{array}$$</div><div>Or ignoring higher-order terms for our purposes,</div><div>$$f^{(N)}(0)\approx(1/\varepsilon)^{2N}e^{-1/\varepsilon}$$</div><div>Each derivative $\frac{e^{-1/\varepsilon}}{\varepsilon^{2N}}\to0$ as $\varepsilon\to0$, but they each approach zero <i>slower</i>&nbsp;than the previous derivative, and somehow that is enough to give the sequence of derivatives the "kick" that they need in the domino effect that follows -- from somewhere at $N=\infty$ (putting it non-rigorously) -- to make the function grow as $x$ leaves zero, even though all the derivatives were zero at $x=0$.</div><div><br /></div><div><hr /><br /></div><div><i>But</i>&nbsp;we can still make it work -- by letting $N$, the upper limit of the summation approach $\infty$ <i>first</i>, before $\varepsilon\to 0$. In other words, instead of directly computing the derivatives $f^{(n)}(0)$, we consider the terms</div><div>$$\begin{array}{*{20}{c}}{f_\varepsilon^{(0)}&nbsp;= f(0)}\\{{{f}_\varepsilon^{(1)} }(0) = \frac{{f(\varepsilon ) - f(0)}}{\varepsilon }}\\{{{f}_\varepsilon^{(2)} }(0) = \frac{{f(2\varepsilon ) - 2f(\varepsilon ) + f(0)}}{{{\varepsilon ^2}}}}\\{{{f}_\varepsilon^{(3)} }(0) = \frac{{f(3\varepsilon ) - 3f(2\varepsilon ) + 3f(\varepsilon ) - f(0)}}{{{\varepsilon ^3}}}}\\ \vdots \end{array}$$</div><div>And write the generalised <b>Hille-Taylor series</b>&nbsp;as:</div><div>$$f(x) = \mathop {\lim }\limits_{\varepsilon&nbsp; \to 0} \sum\limits_{n = 0}^\infty&nbsp; {\frac{{{x^n}}}{{n!}}f_\varepsilon ^{(n)}(0)}$$</div><div>Then $N\to\infty$ before $\varepsilon\to0$ so you "reach" $N\to\infty$ first (or rather, you get large $n$th derivatives for increasing $n$) before $\varepsilon$ gets to 0.</div><div><br /></div><div>Another way of thinking about it is that the "local determines global" stuff makes sense to predict the value of the function at $N\varepsilon$, countable $N$, but it's a stretch to talk about uncountably many $\varepsilon$s away, which is what a finite neighbourhood is. But with these difference operators in the Hille-Taylor series, one ensures that each neighbourhood is a finite multiple of $h$ away at any point, so the differences determine $f$.<br /><br /><hr /></div><div><b>Very simple (but fun to plot on Desmos) exercise: </b>use $e^{-1/x}$ or another defective function to construct a "<b>bump function</b>", i.e. a smooth function that is 0 outside $(0, 1)$, but takes non-zero values everywhere in that range.<br /><br />Similarly, construct a "<b>transition function</b>", i.e. a smooth function that is 0 for $x\le0$, 1 for $x\ge1$. (hint: think of a transition as going from a state with "none of the fraction" to "all of the fraction")<br /><br />If you're done, play around with this (but no peeking):&nbsp;<a href="https://www.desmos.com/calculator/ccf2goi9bj"><b>desmos.com/calculator/ccf2goi9bj</b></a></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-37345142881056927582019-05-13T00:35:00.000+01:002019-05-13T00:38:26.043+01:00The Cauchy Riemann Equations: what do they really mean?<b>Question: <a href="https://math.stackexchange.com/a/3197879/78451">Geometrical Interpretation of Cauchy Riemann equations?</a></b><br /><br />One might think that being differentiable on $\mathbb{R}^2$ is sufficient for differentiability on $\mathbb{C}$. But the Jacobian of an arbitrary such function doesn't have a natural complex number representation.<br /><br />$$<br />\left[ {\begin{array}{*{20}{c}}<br />{\partial u/\partial x} &amp; {\partial u/\partial y} \\<br />{\partial v/\partial x} &amp; {\partial v/\partial y}<br />\end{array}} \right]<br />$$<br />Another way of putting this is that no complex-valued derivative (see below for an example) you can define for an arbitrary function fully captures the local behaviour of the function that is represented by the Jacobian.<br /><br />$$<br />\frac{df}{dz} = \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right) + i\left(\frac{\partial v}{\partial x}-\frac{\partial v}{\partial y}\right)<br />$$<br />The idea is that we should be able to define a complex-valued derivative "purely" for the value $z$, without considering directions, i.e. we want to consider $\mathbb{C}$ one-dimensional in some sense (the sense being "as a vector space"). More precisely, the derivative in some direction in $\mathbb{C}$ should determine the derivative in all other directions in a natural manner -- whereas on $\mathbb{R}^2$, the derivatives in *two* directions (i.e. the gradient) determines the directional derivatives in all directions. <br /><br />If you think about it, this is quite a reasonable idea -- it's analogous to how not every linear transformation on $\mathbb{R}^2$ is a linear transformation on $\mathbb{C}$ -- only spiral transformations are.<br /><br />$$<br />\left[ {\begin{array}{*{20}{c}}<br />{a} &amp; {-b} \\<br />{b} &amp; {a}<br />\end{array}} \right]<br />$$<br />How would we generalise differentiability to an arbitrary manifold? Here's an idea: <b>a function is differentiable if it is locally a linear transformation</b>. So on $\mathbb{R}^2$, any Jacobian matrix is a linear transformation. But on $\mathbb{C}$, only Jacobians of the above form are linear transformations -- i.e. the only linear transformation on $\mathbb{C}$ is <b>multiplication by a complex number</b>, i.e. a spiral/amplitwist. So a complex differentiable function is one that is locally an amplitwist (geometrically), which can be stated in terms of the components of the Jacobian as:<br /><br /><br />\begin{align}<br />\frac{\partial u}{\partial x} &amp; = \frac{\partial v}{\partial y} \\<br />\frac{\partial u}{\partial y} &amp; = - \frac{\partial v}{\partial x} \\<br />\end{align}<br /><br />This is precisely why you shouldn't (and can't) view complex differentiability as some basic first-degree smoothness -- there is a much richer structure to these functions, and it's better to think of them via the transformations they have on grids.Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-20494292960509634462019-05-06T22:21:00.001+01:002019-08-13T09:56:23.623+01:00Lie Bracket, closure under the Lie Bracket(If you're just here for the easy way to see closure, skip ahead to <a href="https://www.blogger.com/blogger.g?blogID=3214648607996839529#closure">Closure under the Lie Bracket</a>)<br /><br />In the <a href="https://thewindingnumber.blogspot.com/2019/04/introduction-to-lie-groups.html">previous article</a>, I introduced Lie Groups and Lie Algebras by talking about Lie Algebras as a parameterisation for the Lie Group -- we said that the elements of the Lie Group could be written as exponentials of these parameters (not uniquely, sure, but they can be written in this way). Some things to note here:<br /><ul><li>What we've called "Lie Groups" refers only to <b><i>connected</i>&nbsp;Lie Groups</b>, as motivation. In general, the theory of Lie groups considers <b>any group that is also a manifold </b>-- for instance, the non-zero real numbers are also a Lie Group (even though their Lie Algebra is identical to that of the positive real numbers -- can you see why?). We will hereby use this more general definition.</li><li>It's not really true that any Lie group can be parameterised in this fashion by writing each element as an exponential of a Lie Algebra element -- even for connected groups. This shouldn't be surprising -- given a term of the form $\exp X$ and a term $\exp Y$, their product $\exp X\exp Y$ is in the group by closure, but it isn't necessarily equivalent to $\exp(X+Y)$ on a non-Abelian group (could it be the exponential of something else? We'll find out later).</li><li>A <i>parameterisation</i> of this form is not the same as a <i>co-ordinate system</i>.</li></ul>The last point is what we will concentrate on in this article, because not being described fully by the Lie algebra is what makes things interesting, right?<br /><br />What is a co-ordinate system on a manifold? Well, they key point is that any element of the manifold can be decomposed in terms of its components along the co-ordinates. On a Lie Group, this means that there should exist a "basis" for the Lie Group $\exp(X_1),\ldots\exp(X_n)$ corresponding to the basis $X_1,\ldots X_n$ for the Lie Algebra vector space such that every element of the Lie Group can be written as products of powers of these elements, and any rearrangement of the terms in the product should leave it invariant (i.e. the elements should commute with each other).<br /><br /><div class="twn-pitfall">Note that it <em>is</em> possible to decompose elements of a connected Lie Group as a product of <em>some</em> exponentials, but this is different from there being specifically $n$ elements that one can write any Lie group element as products of.</div><br />But clearly, this can only be possible if the group is <i>Abelian</i>, commutative. This is a special case of the more general fact that only a <b>holonomic basis</b> gives rise to a co-ordinate system on a manifold. The idea is -- a closed loop should produce no overall group action. If you <b>flow</b> $\varepsilon$ in the $X$ direction, then flow $\varepsilon$ in the $Y$ direction, then flow $\varepsilon$ back in the $X$ direction and flow $\varepsilon$ back in the $Y$ direction, you should end up back where you started. If you don't, then the resulting difference is the infinitesimal "<b>group commutator</b>" of the Lie Group:<br /><br />$$e^{\varepsilon X}e^{\varepsilon Y}e^{-\varepsilon X}e^{\varepsilon Y}$$<br />One can check via a Taylor expansion that this is equal, to second order, to:<br /><br />$$1+\varepsilon^2(XY-YX)$$<br />The first thing to note about this is that the $\varepsilon^1$ term is zero -- this may seem like a surprising coincidence, but perhaps it isn't that surprising (I mean, there's nothing else it <i>could</i>&nbsp;be, right? If the commutator was to first-order $1+\varepsilon z$, $\exp z$ would be equal to 1, and so it would give no characterisation at all of the amount of non-commutativity of the flows $X$ and $Y$) -- it's analogous to vector calculus, where the <b>curl</b> of a vector field is proportional to $\varepsilon^2$ (i.e. a line integral along the curve is proportional to its area, so you divide it by this area in the definition of curl, etc.).<br /><br />The second-order term, $XY-YX$, is more interesting. This may seem weird because so far, we've been considering the Lie algebra purely as a <b>vector space</b>, with addition and scalar multiplication being the only things going on. But clearly, this cannot be the entire picture, or a connected Lie group would be characterised entirely by the dimension of its Lie algebra. This operation -- the <b>Lie Bracket</b>&nbsp;or <b>Lie Algebra commutator</b>&nbsp;represented by $[X,Y]$ -- as we will see, gives some additional structure to the Lie Algebra, and in fact characterises it (we'll see what this means).<br /><br />So far, we've obtained no motivation for why this operation $XY-YX$ is actually of any significance. Sure, it appeared in our second-order approximation for the group commutator, but is the group commutator we defined really so great? Surely there could be other ways one could measure the non-commutativity of a group. And the $\varepsilon^2$ business is <i>weird</i>. Things that arise proportional to $\varepsilon$ live in the tangent space, in the Lie Algebra. Where does $[X,Y]$ even live?<br /><br />Two facts will convince us that the Lie Bracket is indeed the "right" measure of non-commutativity of a Lie Algebra:<br /><br /><ul><li><b>The Lie Algebra is closed under the Lie Bracket -- </b>we will see that in fact, $[X,Y]$ lives <i>in the lie algebra</i>, so it is in fact a binary operation on the Lie Algebra, and really does add structure to the Lie Algebra.</li><li><b>It characterises the entire Lie Algebra -- </b>not only is it <i>part</i>&nbsp;of the structure of the Lie Algebra, it characterises the entire structure of the Lie Algebra. What this means is that defining the Lie Bracket on the vector space allows a full characterisation of the part of the group connected to the identity (the "connected part" of the group), so we can say that any Lie Algebras with the same dimension and Lie Bracket are isomorphic.</li></ul><div><br /></div><hr /><br /><a href="https://www.blogger.com/null" id="closure" name="closure"><b>Closure under the Lie Bracket</b></a><br /><br />If you're like me, you might've thought of several analogous situations to our $1+\varepsilon^2(XY-YX)$ expression -- e.g. in (complex) analysis, at a point where the derivative of a function is zero, the function is characterised by its <i>second</i>&nbsp;derivative (consult Needham's <i>Complex Analysis</i>, p. 205-207 for an explanation). Another example is -- if the first derivative of a function is zero, the second derivative satisfies the product rule (this is actually directly related, in a way we won't go into now).<br /><br />Here's an idea you <i>might</i>&nbsp;think of: as we discussed earlier, the infinitesimal group commutator is $e^{\varepsilon X}e^{\varepsilon Y}e^{-\varepsilon X}e^{-\varepsilon Y}= 1+\varepsilon^2 (XY - YX) + O(\varepsilon^3)\in G$. But for a moment let $\varepsilon$ not be infinitesimal. So $\varepsilon (XY - YX) + O(\varepsilon^2)\in \mathfrak{g}$, the Lie Algebra corresponding to Lie Group $G$, so by scaling $XY-YX+O(\varepsilon)\in\mathfrak{g}$ and by connectedness of the vector space $XY-YX\in\mathfrak{g}$.<br /><br />But this argument is <b>incorrect</b> -- this becomes obvious if you try to formally write it down -- In general, $1+\varepsilon T\in G$ does <b>not</b>&nbsp;imply $T\in\mathfrak{g}$ for non-infinitesimal $\varepsilon$. It's close to an element in $\mathfrak{g}$ (for small $\varepsilon$), but how close? You might get the feeling that it is "sufficiently close", in that the limit $\varepsilon\to0$ of the sequence $\left(c_\varepsilon(X,Y)-1\right)/\varepsilon^2$ (where $c_\varepsilon(X,Y)$ is the group commutator) indeed ends up in the Lie Algebra.<br /><br />To make this feeling formal, consider instead the curve parameterised differently as $\gamma(\varepsilon)=e^{\sqrt\varepsilon X}e^{\sqrt\varepsilon Y}e^{-\sqrt\varepsilon X}e^{-\sqrt\varepsilon Y}$. Then $\gamma'(0)=XY-YX$, and we're done.<br /><br /><div class="twn-furtherinsight">think about the Taylor expansion here of this new curve for a while</div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0tag:blogger.com,1999:blog-3214648607996839529.post-66478014476020575272019-04-22T12:57:00.000+01:002019-04-22T13:00:07.702+01:00Trace, Laplacian, the Heat equation, divergence theoremThe aim of this article is to help build an intuition for the trace of a matrix, "the sum of the elements on the diagonal" -- the basic idea is that the trace is an "average" of some sort, an average of the action of an operator or a quadratic form. We'll make this idea clearer with an example from classical physics: the heat equation.<br /><br /><hr /><br />Consider an $n$-dimensional space with some temperature distribution $T(\vec{x},t)$. We wish to set up a differential equation for this function.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-WDr_mgo-qEg/XL2NKs0J8iI/AAAAAAAAFfc/RQlvQLKSZDklWwkAOd7jkVaq-XXcwCzcACLcBGAs/s1600/lawofcooling.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="177" data-original-width="361" height="156" src="https://4.bp.blogspot.com/-WDr_mgo-qEg/XL2NKs0J8iI/AAAAAAAAFfc/RQlvQLKSZDklWwkAOd7jkVaq-XXcwCzcACLcBGAs/s320/lawofcooling.png" width="320" /></a></div>In the case that $n = 1$, this differential equation is exceedingly easy to write down, considering the difference $(T(x+dx)-T(x))-(T(x)-T(x-dx))$ as the double-derivative upon division by $dx^2$. More rigorously, what we're doing here is applying a <b>localised version of the fundamental theorem of calculus</b>. I.e. we're writing down:<br /><br />\begin{align}<br />\lim_{\Delta x \to 0} \frac{1}{\Delta x}(T'(x + \Delta x) - T'(x)) &amp;= \lim_{\Delta x \to 0} \frac{1}{{\Delta x}}\int_x^{\Delta x} {T''(x)dx}&nbsp; \\<br />&amp; = T''(x)<br />\end{align}<br /><br />More generally, we may consider the $n$-dimensional case.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-HZE4Or8E8tU/XL2YeooEtuI/AAAAAAAAFfw/akx8XXDjp5clCqNjy4LzSmkFV3Fk0-KzACLcBGAs/s1600/laplace.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="522" data-original-width="605" height="276" src="https://1.bp.blogspot.com/-HZE4Or8E8tU/XL2YeooEtuI/AAAAAAAAFfw/akx8XXDjp5clCqNjy4LzSmkFV3Fk0-KzACLcBGAs/s320/laplace.png" width="320" /></a></div>Analogously to before, one may try to look at temperature flows in each direction -- here, we have an <i>integral</i>, done on the boundary of an infinitesimal region $V$ (this symbol will also represent the volume of the region):<br /><br />$$\frac{{\partial T}}{{\partial t}} = \lim_{V \to 0} \frac{\alpha }{V}\int_{\partial V} {\hat u\,dS \cdot \vec \nabla T}$$<br />At this point, one may apply the divergence theorem, converting this to:<br /><br />$$\frac{{\partial T}}{{\partial t}} = \mathop {\lim }\limits_{V \to 0} \frac{\alpha }{V}\int\limits_V {\vec \nabla&nbsp; \cdot \vec \nabla T\;dV}&nbsp; = \alpha{\left| {\vec \nabla } \right|^2}T$$<br />In this sense, the divergence theorem is analogous to the fundamental theorem of calculus for manifolds with boundaries that are more than one-dimensional (see the bottom of the page for a link to a formalisation/an abstraction based on this analogy). But there are more ways to intuitively understand this. Note how the Laplacian is the trace of the Hessian matrix (note: we use $\vec{\nabla}^2$ to refer to the Hessian and $\left|\vec\nabla\right|^2$ to refer to the Laplacian):<br /><br />$${\left| {\vec \nabla } \right|^2}T = {\mathop{\rm tr}} \left({\vec{\nabla} ^2}T\right)$$<br />The trace of a matrix is fundamentally linked to some notion of <i>averaging</i>&nbsp;-- the simplest interpretation of this is that it is the mean of the eigenvalues. But more relevant to our situation, it can be shown that the trace of a matrix is the expected value of the quadratic form defined by the matrix on the unit sphere -- or on a general sphere $S$:<br /><br />$${\mathop{\rm tr}} A = \frac{1}{S}\int_S {\frac{{\Delta {x^T}A\,\Delta x}}{{\Delta {x^T}\Delta x}}\,dS}$$<br />One may check that taking the limit as $\Delta x \to 0$, substituting $\nabla^2$ for the operator and writing&nbsp;${\overrightarrow \nabla ^2}f\,d\vec x = \overrightarrow \nabla&nbsp; f$, one gets the original "average of directional derivatives" expression.<br /><br /><div class = "twn-furtherinsight">Can you interpret the other coefficients of the characteristic polynomial in terms of statistical ideas?</div><br /><hr /><br /><div><b>Further reading:</b></div><div><ul><li>Using the "infinitesimal region" idea to define divergence, curl and Laplacian rigorously: <a href="https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/formal-definitions-of-divergence-and-curl/a/formal-definition-of-divergence-in-two-dimensions">Khan Academy</a></li><li>An abstraction based on the "analogy" between FTC, Divergence Theorem, Navier-Stokes Theorem, etc. <a href="https://en.wikipedia.org/wiki/Stokes%27_theorem">Stokes' theorem (Wikipedia)</a></li></ul></div>Abhimanyu Pallavi Sudhirhttp://www.blogger.com/profile/17891661511466934614noreply@blogger.com0