- #1

jmcelve

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## Homework Statement

Consider the matrix elements of [itex] \hat{x} [/itex] in momentum space. That is, evaluate [itex] \langle p | \hat{x} | \psi(t) \rangle [/itex] in terms of the momentum space wave equation [itex] \langle p | \psi(t) \rangle [/itex].

## Homework Equations

[tex] \langle x | p \rangle = \frac{1}{\sqrt{2 \pi \hbar}} e^{ipx/ \hbar} [/tex]

[tex] I = \int | x \rangle \langle x | dx [/tex] [tex] I = \int | p \rangle \langle p | dp [/tex]

## The Attempt at a Solution

So when we insert the identity, we obtain our integral: [tex] \int_{-\infty}^{+\infty} dx \langle p | \hat{x} | x \rangle \langle x | \psi(t) \rangle[/tex]

The thing is, I know we have to insert the identity again to obtain:

[tex] \int_{-\infty}^{+\infty} dx dp' \langle p | \hat{x} | x \rangle \langle x | p' \rangle \langle p' | \psi(t) \rangle [/tex]

(Assume it's a double integral.)

This was necessary so that we could get our ket psi in terms of the momentum basis. We then act [itex]\hat{x}[/itex] on [itex] |x \rangle [/itex] and get an x in our integral. Then we insert the Fourier transforms and obtain:

[tex] \int_{-\infty}^{+\infty} dx dp' x e^{ipx/ \hbar} e^{-ip'x/ \hbar} \psi(p', t) = \int_{-\infty}^{+\infty} dx dp' x e^{i(p-p')x/ \hbar} \psi(p', t) [/tex]

I'm a little lost here. I see that the exponential will become a Dirac delta, which will bring p into the psi (making it the form it should be). I can also see where the differential form of x in the p basis comes out of the exponential. But I don't see how the integral with respect to x will disappear.

**EDIT** I've solved it. No need to respond.

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