Born's interpretation for momentum operator Hi I have a basic QM question: Given a state vector $|\psi(t) \rangle$, at some time $t$, we can project this onto the position basis, $\langle \vec{r}| \psi(t) \rangle = \psi(\vec{r},t)$. Then from Born's probabilistic interpretation, the square of the norm $$|\psi(\vec{r},t)|^2$$ represents the probability density at a time $t$ of finding the particle at in a volume $d^3r$ located between $\vec{r}$ and $\vec{r} + d \vec{r}$. 
Consider the the momentum eigenvector $|p\rangle$, by projecting this onto the position basis we get $\langle \vec{r}| \vec{p} \rangle = Ae^{\frac{i \cdot \vec{p}\cdot \vec{r}}{\hbar}} = \psi(\vec{r})$. Does it follow that $|\psi(\vec{r})|^2d^3r$ gives the probability of finding the particle with momentum $\vec{p}$ in the volume element $d^3r$? If so, is this a separate postulate or does it follow from Borns postulate or something else?  
Thanks for any help.
 A: 
Does it follow that $|\psi(\vec{r})|^2 d^3r$ gives the probability of
  finding the particle with momentum $\vec{p}$ in the volume element $d^3r$?

Yes.

If so, is this a separate postulate or does it follow from Borns
  postulate or something else?

No, it's a special case of one the principles of quantum mechanics. 
By the way, there is a problem with eigenstates of momentum operator, and it's the problem of non-normalizability of momentum's eigenstates for infinite size space. But it's not a serious physical question as there is not any particle in the nature that has an exact momentum.
A: 
Does it follow that $|\psi(\vec{r})|^2d^3r$ gives the probability of finding the particle with momentum $\vec{p}$ in the volume element $d^3r$?

The Born interpretation is not valid for non-normalizable functions such as $e^{ipx/\hbar}$. It is believed to be valid for any normalized function $\psi$; $|\psi(\vec{r})|^2d^3r$ is assumed to be probability that the particle is in the region $d^3r$ around  $r$, while nothing is assumed about its momentum.
This does not mean the particle does not have momentum, it only means the Born rule in above form gives no information on it.

If so, is this a separate postulate or does it follow from Borns postulate or something else? 

There is another variant of the Born rule, which says $|\tilde{\psi}(\vec{p})|^2d^3p$ gives probability that the particle has momentum that belong to the neighbourhood $d^3 p$ of vector $\vec{p}$. Here $\psi(\vec{p})$ is the Fourier transform of $\psi(\vec{r})$. This rule is independent assumption, not implied by the first one.
