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Checking a thought I had... Given two potentials, $V(\mathbf{r})$ and its mirror potential $V(\mathbf{-r})$, the momentum probability distributions would necessarily be equivalent, right?

I'm thinking yes, because these potentials would respectively give rise to energy eigenfunctions $\psi_n(\mathbf{r})$ and $\psi_n(\mathbf{-r})$. Then from

$$\phi_n(\mathbf{k})=\frac{1}{(\sqrt{2 \pi})^{3}} \int_{\mathbf{D}} \psi_n(\mathbf{r})e^{-i \mathbf{k} \cdot \mathbf{r}}(\mathbf{k}) \mathrm{d}^{3} r$$

we see that the momentum wavefunctions would be $\phi_n(\mathbf{k})$ and $\phi_n(\mathbf{-k})$ respectively. And

$$\rho(\mathbf{p}) = |\phi_n(\mathbf{k})|^2 = |\phi_n(\mathbf{-k})|^2$$

What are the fundamental properties that lead us here? That is, I'm looking to better understand this result in terms of abstract properties. It seems basic, but I wonder if there's more there. Thanks.

edit: I just realized I assumed $\psi_n$ was real in making the statement about $\rho(\mathbf{p})$, since I had been thinking about 1-D asymmetric infinite square well potentials when these thoughts arose. I need to consider the case otherwise.

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  • $\begingroup$ @MaxStammer The entire system is mirrored and momentum is conserved under that symmetry, which is why this occurs, yes? It's been a while since I've used this language, so I was hoping someone could help formalize my thoughts. $\endgroup$ – suneater Nov 25 '19 at 18:52
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    $\begingroup$ There is no reason for the last equality you posted should hold, unless $\phi_{n}(\mathbf{k})$ had spherical symmetry (up to a phase factor) for some reason. You are making a transformation $V(\mathbf{r})\to V^{\prime}(\mathbf{r})=V(-\mathbf{r})$. The only things you can say for sure is that $\psi^{\prime}(\mathbf{r})=\psi(-\mathbf{r})$ and $\phi^{\prime}(\mathbf{k})=\phi(-\mathbf{k})$, where the last one follows from the properties of the Fourier transform, but note that these are different wavefunctions because they are solutions to a different Schrödinger's equation. $\endgroup$ – user137661 Nov 25 '19 at 19:06
  • $\begingroup$ Momentum conservation on the other hand is a different thing. First it comes from translation symmetry, not reflection symmetry. And it can be verified if it works for your particular Hamiltonian by evaluating $[\hat H, \hat{\mathbf{p}}]$, in general any (hermitian) operator that commutes with the Hamiltonian represents an observable that is a constant of the motion (i.e., it is conserved). $\endgroup$ – user137661 Nov 25 '19 at 19:11
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The correct statement is the following:

Consider two systems with different potentials. The first system has a potential $V(\vec{r})$, and the second one a potential $\tilde{V}(\vec{r}) = V(-\vec{r})$ with corresponding energy eigenstates $\psi_n(\vec{r}), \tilde{\psi}_n(\vec r)$. First of all, it will be true that the phase factors of the energy eigenstates can be chosen so that $\tilde{\psi}_n(\vec{r}) = \psi_n(-\vec{r})$.

Then you can easily show by a change of variables $\vec{r}\to\vec{r}' = -\vec{r}$ in the Fourier transform that $\mathcal{F}[\tilde{\psi}_n](\vec{k}) = \mathcal{F}[\psi_n](-\vec{k})$. However, the two will generally not have the same momentum distribution but only one connected by inversion. They would only have the same momentum distribution if the potential $V(\vec{r})$ was spherically symmetric about some point. (I will leave it to the dear reader as an exercise to show the potentials need not be spherically symmetric about the origin of coordinates for this to hold.)

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