Momentum operator acting on a bound state doesn't return an eigenvalue although kinetic energy operator does. Why?

Question

We know that $[\hat{H}, \hat{P}]=i\hbar\frac{\mathrm{d}V}{\mathrm{d}x}$, therefore $\hat{H}$ and $\hat{P}$ commute if $\frac{\mathrm{d}V}{\mathrm{d}x}=0$, which is true for $V=0$. In that case the operators share the same eigenstates. As a matter of fact, both of the operators should return an eigenvalue if they operate on an eigenstate.

Now if we take a plane wave state $|\psi\rangle = Ae^{ikx}$, we see the above argument holds: \begin{align} \hat{P}\text{ operator gives:}& & -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}|\psi\rangle &=\hbar k|\psi\rangle \\ \hat{H}\text{ operator gives:}& & -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}|\psi\rangle &=\frac{\hbar^2k^2}{2m}|\psi\rangle \end{align}

But if we go to the particle in a box problem, then $|\psi\rangle =\sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x)$, where $k=\frac{n\pi}{a}$, so \begin{align} \hat{P}\text{ operator gives:}& & -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}|\psi\rangle &= -i\hbar\frac{n\pi}{a}\sqrt{\frac{2}{a}}\cos\left(\frac{n\pi}{a}x\right) \neq\hbar\frac{n\pi}{a}|\psi\rangle \\ \hat{H}\text{ operator gives:}& & -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}|\psi\rangle &= -\frac{\hbar^2}{2m}\left(\frac{n\pi}{a}\right)\left(-\frac{n\pi}{a}\right) \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a}x\right) =\frac{\hbar^2k^2}{2m}|\psi\rangle \end{align}

So in a bound state, $\hat{H}$ returns an eigenvalue but $\hat{P}$ doesn't, although the commutation condition is satisfied. (For the harmonic oscillator potential, I am not surprised by a similar result as the commutation criteria are not satisfied in the first place). Why is this?

Answer 1

For the specific case of the particle-in-a-box problem, the momentum operator is much trickier to handle than what you're allowing for, but that's not the issue at the heart of your current confusion.

If you know that two operators $A$ and $B$ commute, i.e. $[A,B]=0$, then there's two common ways to understand the consequences, one of which is right and one of which is wrong:

• You are guaranteed the existence of at least one shared eigenbasis of both operators; but

• You are not guaranteed that all eigenbases of $A$ will be eigenbases of $B$, or vice versa.

(For other occurrences of the misconception that the second point does happen, see e.g. this answer or this one.)

Thus, the 'contradiction' you observe is also present for the case of the free particle, without the box: the wavefunction $\psi(x)=\cos(kx)$ is an eigenfunction of $\hat H=\frac{1}{2m}\hat P^2$, but it is not an eigenfunction of $\hat P$. That is, the fact that the hamiltonian's eigenfunctions over a compact interval are not momentum eigenfunctions is not surprising and it is not specific to that configuration.

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But, that said, you can still ask something like

well, OK, so some $\hat H$ eigenfunctions in the particle-in-a-box problem are not $\hat P$ eigenfunctions and that's not a problem, but $[\hat H,\hat P]=0$ is still true, so shouldn't I be guaranteed a shared eigenbasis, even if it isn't the one I started with?

and it's a reasonable question. Here what happens is that the subtleties with the compact-interval momentum operator kick in, at a much deeper level than just the commutation: the result, in full, reads

If two self-adjoint operators $A$ and $B$ commute, then there exists at least one shared eigenbasis,

and it breaks because $\hat P$-in-a-box is not a self-adjoint operator: it is symmetric, but it has domain problems that prevent it from being self-adjoint. The consequences of this are as deep as they get: there simply isn't a momentum eigenbasis in this Hilbert space. You can extend the momentum operator to make it self-adjoint; this extension is not unique, but there is a reasonable choice (setting $\alpha=0$ in V. Moretti's answer) that comes close to making $\hat H$ the square of the extended $\hat P_\alpha$, but ultimately this can't work, as they have different domains. (More specifically, the domain of $\hat H$ is contained in the domain of $\hat P_\alpha$, but the eigenfunctions of $\hat P_\alpha$ do not fall into that subspace.)

Answer 2

To add some context on the answer by Emilio Pisanty, your question stems from your insufficient understanding of the crucial rôle of functional analysis in Quantum Mechanics. This rôle is definitely not spelled out in the standard university education and it's missing from the majority of books which serve as course textbooks out there.

When you say "We know that $[H,P] = i\hbar \frac{d V(x)}{dx}$", you MUST understand that there is a whole library on functional analysis (I can give you six books off the top of my head) which could serve one to prove this statement by spelling out precise conditions under which this is true.

So you may probably not understand the answer to this question you've been kindly provided with, but I am sure you do understand that it's not your fault that the books you were reading did not contain elements of functional analysis in (rigged) Hilbert spaces with application in Quantum Mechanics.