Today in class (Intro to QM) we went over a couple of commutators. Among them was $[x, V]$, where $V=V(x)$ is a potential. What the teacher said to prove this is zero was: let's assume $V$ is analytic and can be expanded in a power series. Then we can take the commutator of $x$ and each term in the series, and since $[x, x^n] = 0$ everything is zero.

I immediately thought of something much simpler: In coordinate space, both $x$ and $V$ are simply multiplication. For any function $\psi$, $xV\psi = Vx\psi$ trivially, because we're just taking products. Therefore, they commute.

Is there anything wrong with my reasoning?

  • $\begingroup$ It seems to me that your solution assumes $[x,\,V]=xV-Vx=0$ a priori and does not actually prove the case. $\endgroup$ – Kyle Kanos Oct 25 '13 at 18:13
  • $\begingroup$ There is nothing wrong, but let be more precise. Consider a function $\psi$ of $x$ (and optionnaly of $t$). In this case, you have, by definition : $X \psi(x) = x\psi(x)$ and $V(X) \psi(x) = V(x)\psi(x)$. The commutation is obvious. For a function of $p$ (and optionnaly of $t$), you will have $X \psi(p) = -i \hbar\frac{\partial}{\partial p}\psi(p)$, and $V(X) \psi(p) = V(-i \hbar\frac{\partial}{\partial p})\psi(p)$, and the commutation is obvious too. $\endgroup$ – Trimok Oct 25 '13 at 18:32

There is nothing wrong with your reasoning; the issue is that your starting assumption that the potential can be treated as a multiplication operator needs to be justified. Here's why:

Let's consider the potential operator $V$ acting on a state $|\psi\rangle$. Let $X$ denote the position operator. The definition of the position space representation of $V$ acting on the position space wavefunction $\psi$ is as follows: \begin{align} V\psi(x) = \langle x| V|\psi\rangle \end{align} As you note, if we could somehow show that there exists some function $\tilde V$ such that \begin{align} V\psi(x) = \tilde V(x)\psi(x) \tag{$\star$} \end{align} then we would have \begin{align} XV\psi(x) &= X(\tilde V(x)\psi)(x) = \tilde V(x)X\psi(x) = \tilde V(x) x\psi(x) \end{align} while we would also have \begin{align} VX\psi(x) = V(x\psi)(x) = xV\psi(x) = x\tilde V(x)\psi(x) = \tilde V(x)x\psi(x) \end{align} which would give $XV = VX$ as desired. The issue is that that property $(\star)$ does not come for free from the definition above it. If, however, $V$ is, for example, defined to be some analytic function $\tilde V$ of the position operator, then that property (modulo some mathy subtleties) does hold because for any positive integer power $n$ of $X$, one has \begin{align} X^n\psi(x)=\langle x|X^n|\psi\rangle = x^n\langle x|\psi\rangle = x^n\psi(x) \end{align} by acting $X$ on the left repeatedly. One then uses this for each term in the power series expansion of $V$ in $X$ to obtain the property $(\star)$.

Note. I am abusing notation here slightly and using the same symbol $V$ for the position operator and its position space representation.

  • $\begingroup$ Is there any case in which the potential operator isn't simply a multiplication? $\endgroup$ – Javier Oct 25 '13 at 18:46
  • $\begingroup$ @JavierBadia Well, I'm not sure about cases that are physically well-motivated, but mathematically it's really easy to concoct such things. The easiest way I can think of would be to assume that the potential is depends on $P$ as well. You might then ask if there is any way to make it depend only on the position operator and still have it not be multiplication. Well, in that case, you would need to somehow define what it means for an operator to be a non-analytic function of another operator, and I'm not sure how such a thing would work. $\endgroup$ – joshphysics Oct 25 '13 at 18:50
  • $\begingroup$ @JavierBadia Actually I'm not even sure non-analytic functions of $X$ (whatever that would mean), would be good enough. In any case, that's kind of beside the point. $\endgroup$ – joshphysics Oct 25 '13 at 19:13
  • $\begingroup$ @JavierBadia: in a momentum basis for your state, it is definitely not multiplication. $\endgroup$ – Jerry Schirmer Oct 25 '13 at 21:07

I agree with joshphysic's answer that your reasoning is right, but you need to justify $V(x)$'s being a multiplication operator (in position space, that is).

Maybe the way I tend to look at this may help: I find in matters like these physical motivations tend to be the most intellectually fulfilling.

You are presumably dealing with a first quantized wave equation with a "semiclassical" model of interaction with the outside world - say the Schrödinger or Dirac equation with a conservative potential impressed on the quantum particle's Hamiltonian by a central, approximately unshifting charged nucleus. The quantum nature of the origins of this potential are ignored for simplicity - otherwise we would be dealing with a quantum many body problem.

So once the eigenvalue $x$ of the position observable $X$ has been measured by applying the observable to the quantum particle's state, the standard quantum postulates say that the particle must be in the position eigenstate corresponding to $x$. So at the instant just after the measurement, the particle's position is certain, and so the only sensible value we can postulate for the "potential energy" measurement is $v(x)$, where $v(x)$ is the classical potential function, and the value of the classical potential is certain for the instant after the measurement, once we know the position is $x$.

Likewise for any other position measurement with position eigenstate.

Therefore we see that every position eigenstate is an eigenstate of the observable we need to build for the semiclassical potential, and since the position eigenstates are complete, i.e. any quantum state is a superposition of these states, then we see that we have just fully defined the diagonalisation of the potential energy observable. Namely, it is diagonal in position co-ordinates, i.e. it is the multiplication operator $V \psi(x) = v(x) \psi(x)$.

Your main result also follows from considering the fact that position eigenstates make the semiclassical potential certain, i.e. they are all eigenstates and the only eigenstates for the potential. Given suitable reasonable assumptions about the operators and quantum state space concerned, operators commute if and only if they have the same eigenstates.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.