### Dealing with eigenspaces; noncommuting variables and another postulate

At the end of the last article, 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 three separate measurements of the different spatial dimensions.

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?

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?

Well, because it's an observable, it must have eigenstates that produce a full eigenbasis. But if each eigenvalue corresponded to just one eigenstate, 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 an infinitude of states, an eigenspace, corresponding to each position with the same $x$-position (which, remember, is their eigenvalue), and their superpositions thereof. And this makes a lot of sense -- each position is a state, but these positions give us the same values for the $x$-position.

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 $Z$.

OK. So what happens when we observe $X$ taking the value $x$? You might think that the state just turns into some randomly chosen eigenstate with the observed eigenvalue $x$. But if you think about it, this would be quite unphysical, as this would mean our $X$-observation would magically change our knowledge about the $y$ and $z$ positions 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.

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.

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 projection of the state into the eigenspace with eigenvalue $x$. This is in line with our discussion of the generalised Born's rule in the last article -- but it is an additional postulate of quantum mechanics, or rather generalises the existing postulate about states projecting randomly into eigenstates.

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 are 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 less information about the system, and you're collapsing the state "less" in this sense.

(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.)

Something that we've seen in this discussion above is that there is a common eigenbasis for $X$, $Y$ and $Z$ -- specifically, the position basis, 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.

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 regardless of the order you measure it in -- so three-dimensional position is indeed well-defined in quantum mechanics.

Just to be clear, the fact that $X$, $Y$ and $Z$ commute is a postulate -- 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 cannot 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 can't just construct a spacefilling curve 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 does not exist in the Hilbert space, there are no states that precisely put down the values of position and momentum. It is possible to construct quantum mechanical theories -- called non-commutative quantum theories -- 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.

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.

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.

So this means that the observables $X$ and $P$ cannot be measured simultaneously. Some specific bounds on the uncertainties will be discussed in the next article. For now, let's demonstrate an example of two noncommuting variables: position and momentum (in the same direction).

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.

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 corresponding eigenvectors of the position operator must be Dirac delta functions).

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:

\begin{align}XP &= -i\hbar x \frac{\partial}{\partial x}\\ PX\{\psi(x)\}&=-i\hbar\frac{\partial}{\partial x}(x\psi(x))\\ &= -i\hbar\left[\psi(x)+x\psi'(x)\right]\\ \Rightarrow PX &= -i\hbar x\frac{\partial}{\partial x} -i\hbar\end{align}

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, this is a property of all observables that are Fourier transforms of each other/canonically conjugate observables (more on this in the next article).

Exercise: 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 the last article, 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!