Momentum probability distribution for $V(\mathbf{r})$ vs $V(\mathbf{-r})$ is the same? 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.
 A: 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.) 
