Quantum Mechanics, Momentum Space

[NeoDevin]

1. The problem statement, all variables and given/known data
Show that

<x> = \int \Phi^* \left(-\frac{\hbar}{i}\frac{\partial}{\partial p} \right) \Phi dp


2. Relevant equations

\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} \Psi(x,t)dx

\Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \Phi(p,t)dp

3. The attempt at a solution

I started out with

<x> = \int^{\infty}_{-\infty} \Psi^* x \Psi dx

Using the above equation for \Psi(x,t) (and it's conjugate) gives:

\Psi^* (x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} \Phi(p,t)dp

and

x\Psi(x,t) = \frac{x}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \Phi(p,t)dp

= \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} x e^{\frac{ipx}{\hbar}} \Phi(p,t)dp

= \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} \frac{\partial}{\partial p} e^{\frac{ipx}{\hbar}} \Phi(p,t)dp

= \frac{1}{\sqrt{2\pi\hbar}} \left[ \left( e^{\frac{ipx}{\hbar}} \Phi(p,t) \right) \bigg|^{\infty}_{-\infty} -\int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \frac{\partial}{\partial p}\Phi(p,t)dp \right]

= -\frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \frac{\partial}{\partial p} \Phi(p,t) dp

Substituting into the original equation for <x> then gives

<x> = \int^{\infty}_{-\infty}\left( \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty}e^{\frac{-ipx}{\hbar}} \Phi^* (p,t) dp \right) x\Psi(x,t) dx

= \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} e^\frac{-ipx}{\hbar} (x\Psi(x,t))dx dp

= -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \frac{\partial}{\partial p} \Phi(p,t) dp dx dp

= -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} \frac{\partial}{\partial p} \Phi(p,t) \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} e^{\frac{ipx}{\hbar}} dx dp dp

= -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} \frac{\partial}{\partial p} \Phi(p,t) \int^{\infty}_{-\infty} dx dp dp

I'm pretty sure I messed up somewhere, since that integral is infinite...

Any help would be appreciated.

[NeoDevin]

I also tried the reverse, starting with the expression you're supposed to get for <x>, and working back from there using similar methods... but it gives me the same problem.

[jpr0]

Your calculation seems fine up until the point where you substitute your expression for x\Psi(x,t). You should be integrating over two dummy variables in your final expression, say p and p^{\prime}, but you have written both dummy variables as the same variable p. The expression you derived for x\Psi(x,t) in terms of an integral over p, change p to p^{\prime} and everything should work out.

[NeoDevin]

So then for the second last line we end up with
<x> = -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} \frac{\partial}{\partial p'} \Phi(p',t) \int^{\infty}_{-\infty} e^{\frac{-i(p-p')x}{\hbar}} dx \ dp' \ dp

Is that integral doable?

[Dick]

You should recognize the x integral as a delta function.

[NeoDevin]

Oh, right... Duh. :blushing: Got it now