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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.

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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.

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    $\begingroup$ @Alex: It is better to denote your $\psi(\vec{r})$ as $\psi_{\vec{p}}(\vec{r})$. $\endgroup$ – Vladimir Kalitvianski Jul 19 '16 at 18:32
  • $\begingroup$ @Hosein Which principle of quantum mechanics? $\endgroup$ – Alex Jul 19 '16 at 21:10
  • $\begingroup$ @Alex One of the principles says that if we measure a physical quantity, the outcome of measurement will be one the eigenvalues of the linear operator which corresponds to that physical quantity(which we call $\hat{A}$) and state of the system after measurement will be the corresponding eigenstate. And the probability to obtain one specific eigenvalue (which I call $A_n$) equals to $|\langle A_n|\psi\rangle |^2$ where $|\psi \rangle$ is the state of system before measurement and $|A_n\rangle$ is the corresponding eigenstate. $\endgroup$ – Hosein Jul 20 '16 at 6:45
  • $\begingroup$ @Hosein Okay I understand that, but how does that imply that $|\psi_{p}(\vec{r})|^2d^3r$ is the probability density of finding the particle with momentum $p$ in the volume $d^3r$ where $\psi_{p}(r) = \langle \vec{r}|\vec{p} \rangle$ is one of the eigenfunctions of the momentum operator projected onto the position basis? $\endgroup$ – Alex Jul 20 '16 at 21:31
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    $\begingroup$ @Alex When a particle has momentum $\vec{p}$ then it's state is $|\vec{p}\rangle$. $\endgroup$ – Hosein Jul 21 '16 at 7:45
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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.

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  • $\begingroup$ Yes that is clear, but I am referring to $|\psi_{p}(\vec{r})|^2d^3r$, where $\psi_{p}(\vec{r}) = \langle \vec{r}| \vec{p} \rangle$, representing the probability density of finding the particle with momentum $p$ in volume $d^3r$ rather than your $|\Phi(p)|d^3p$, where $\Phi(p) = \langle \vec{p}| \psi \rangle$ for some state vector $| \psi \rangle$, which is the probability density of finding particle in momentum range $d^3p$. $\endgroup$ – Alex Jul 20 '16 at 21:37

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