In 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?
Well, it's natural for you to find this assumption unsatisfactory. We've been quite liberal in assuming the de Broglie relation earlier when motivating quantum theory, too -- we'll later produce some motivation for the de Broglie relation for photons, and discuss derivations from quantum mechanics, axiomatising our theory clearly to eliminate circularities. But for now, let's not.
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 definite value, $hf$, with zero standard deviation.
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 first article, 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 amplitude distribution, because we want superposition). We'll use the notation $\Psi(p)$ to represent this "momentum-space wavefunction", and we'll see why soon.
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:
$$\Psi(p)=\frac1h \mathcal{F}_x^{p/h}(\Psi(x))$$
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.
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 inverse Fourier transform acts as a change-of-basis matrix from the position basis to the momentum basis. I.e.
$$|\psi\rangle_P=F|\psi\rangle_X$$
This change-of-basis matrix $F^{-1}$ precisely represents the eigenstates of the momentum operator 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 $p$.
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 p constant and varies x, i.e. each column, i.e. each eigenstate of $P$ is a function of $x$.
Anyway, so we're looking for a linear operator $P$ solving the eigenvalue problem (and we're just ignoring the scalar multiples):
$$P e^{ix \cdot 2\pi p / h} = pe^{ix \cdot 2\pi p / h}$$
It should be quite clear that the operator we're looking for is:
$$\begin{align}P &= \frac{h}{2\pi i}\frac{\partial}{\partial x} \\
&= -i\hbar \frac{\partial}{\partial x} \end{align}$$
We need to be clear that this is the representation of the momentum operator in the position basis -- 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,
$$X=i\hbar\frac{\partial}{\partial p}$$
Exercise: 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.
Derivation of Heisenberg and Robertson-Schrodinger uncertainty principles
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 concept of an uncertainty principle is crucial, in that it demonstrates the rigorously difference between quantum mechanics and statistical physics. In general, the noncommutativity of observables (having no shared eigenstates) is something that has no analog in classical physics.
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} $. 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.
Defining $A'=A-\langle A\rangle$ and $B'=B-\langle B\rangle $ for Hermitian (this is important!) $A$ and $B$, we see that:
$$\begin{align}
\langle A'^2\rangle \langle B'^2 \rangle &= \langle \psi | A'^2 | \psi \rangle \langle \psi | B'^2 | \psi \rangle \\
&= \langle A' \psi | A' \psi \rangle \langle B' \psi | B' \psi \rangle \\
&\ge |\langle \psi | A' B' | \psi \rangle| ^ 2 \\
&= \left|\frac12 \langle\psi|A'B'+B'A'|\psi\rangle + \frac12\langle\psi|A'B'-B'A'|\psi\rangle\right|^2 \\
&= \frac14 |\langle\psi|A'B'+B'A'|\psi\rangle|^2 + \frac14|\langle\psi|A'B'-B'A'|\psi\rangle|^2 \\
&= \frac14 |\langle \{A-\langle A\rangle, B-\langle B\rangle\} \rangle| ^2 + \frac14 |\langle [A,B]\rangle|^2\\
&= \frac14 |\langle\{A,B\} \rangle - 2\langle A\rangle \langle B\rangle |^2 + \frac14|\langle[A,B]\rangle|^2\\
\Rightarrow \Delta a\,\Delta b &\ge \frac12 \sqrt{|\langle\{A,B\} \rangle - 2\langle A\rangle \langle B\rangle |^2 + |\langle [A,B]\rangle|^2}
\end{align}$$
This is the Robertson-Schrodinger relation.
(Guide in case you get stuck somewhere -- line 3, Cauchy-Schwarz inequality; line 4, splitting into Hermitian and anti-Hermitian parts; line 5, 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 point 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.)
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):
$$\Delta a\,\Delta b \ge \frac12 |\langle [A,B]\rangle|$$
This is the Heisenberg uncertainty relation. In particular, in the last article, we showed that for the position and momentum operators, $[X,P]=i\hbar$. So in this case, we get the celebrated identity:
$$\Delta x\, \Delta p \ge \frac{\hbar}{2}$$
For canonically conjugate $X$ and $P$.
As mentioned before, other stronger uncertainty relations exist for general observables. Some examples can be found on the Wikipedia page Stronger uncertainty relations (permalink).
Insights on interesting things.
Showing posts with label wavefunction. Show all posts
Showing posts with label wavefunction. Show all posts
Projection operators, generalised Born's rule, position basis, wavefunction
At the end of the last article, I asked you to investigate Born's rule for continuous variables like position and momentum.
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 $\Sigma_X$.
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
$$P(x\in B)=\sum |\langle\psi|\phi_k\rangle|^2$$
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.
(Note on notations: 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 $B$.)
So we could just define a Hermitian projection operator $L_X(B)$ 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 $X$.
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 generalised Born's rule as:
$$\begin{align}P(x\in B) &= |L_X(B)|\psi\rangle|^2\\ &= \langle\psi|L_X(B)|\psi\rangle\end{align}$$
Well, this is interesting! In the last article, you proved that the expected value of an observable $X$ given a state $|\psi\rangle$ is given by $\langle\psi|X|\psi\rangle$. But here we have a probability 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.
There's another place you might've seen something like this, and that is with indicator variables in probability and statistics -- the expected value of an indicator variable for an event is the probability that the event occurs.
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.
So projection operators are a special type of observable, measuring the answer to "Yes/No questions" -- if the answer to "is the system in one of the states $\{|B\rangle\}$?" is yes, the observable $L_X(B)$ takes the value 1 -- if the answer is no, then it takes the value 0. So it is precisely an "indicator variable" for $\{|B\rangle\}$.
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.
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 densities 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 $X$:
$$ |\psi\rangle = \int_{\Sigma_X} |x\rangle\, \Psi(x)\, dx $$
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 wavefunction, and it represents probability amplitude densities. Then for some set $M\subseteq \Sigma_X$ of eigenvalues $L_X(B)|\psi\rangle$ is the projection:
$$ L(M)|\psi\rangle = \int_B |x\rangle\, \Psi(x)\, dx $$
And one may calculate the dot product, noting that complex dot products require taking the complex conjugate:
$$ \langle \psi | L_X(B) | \psi \rangle = \int_B \Psi^*(x)\, \Psi(x)\, dx $$
Which gives us an expression for the probability density function on $\Sigma_X$ as:
$$\begin{align}\rho(x) &=\Psi^*(x)\,\Psi(x) \\
&=|\Psi(x)|^2\end{align}$$
And this applies to any operator with a continuous spectrum, like position and momentum.
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 $\Sigma_X$.
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 MAO1103: Linear Algebra course.
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
$$P(x\in B)=\sum |\langle\psi|\phi_k\rangle|^2$$
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.
(Note on notations: 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 $B$.)
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 generalised Born's rule as:
$$\begin{align}P(x\in B) &= |L_X(B)|\psi\rangle|^2\\ &= \langle\psi|L_X(B)|\psi\rangle\end{align}$$
Well, this is interesting! In the last article, you proved that the expected value of an observable $X$ given a state $|\psi\rangle$ is given by $\langle\psi|X|\psi\rangle$. But here we have a probability 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.
There's another place you might've seen something like this, and that is with indicator variables in probability and statistics -- the expected value of an indicator variable for an event is the probability that the event occurs.
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.
So projection operators are a special type of observable, measuring the answer to "Yes/No questions" -- if the answer to "is the system in one of the states $\{|B\rangle\}$?" is yes, the observable $L_X(B)$ takes the value 1 -- if the answer is no, then it takes the value 0. So it is precisely an "indicator variable" for $\{|B\rangle\}$.
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.
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 densities 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 $X$:
$$ |\psi\rangle = \int_{\Sigma_X} |x\rangle\, \Psi(x)\, dx $$
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 wavefunction, and it represents probability amplitude densities. Then for some set $M\subseteq \Sigma_X$ of eigenvalues $L_X(B)|\psi\rangle$ is the projection:
$$ L(M)|\psi\rangle = \int_B |x\rangle\, \Psi(x)\, dx $$
And one may calculate the dot product, noting that complex dot products require taking the complex conjugate:
$$ \langle \psi | L_X(B) | \psi \rangle = \int_B \Psi^*(x)\, \Psi(x)\, dx $$
Which gives us an expression for the probability density function on $\Sigma_X$ as:
$$\begin{align}\rho(x) &=\Psi^*(x)\,\Psi(x) \\
&=|\Psi(x)|^2\end{align}$$
And this applies to any operator with a continuous spectrum, like position and momentum.
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.
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