# Infinite square well, momentum space

• NeoDevin
In summary, the momentum-space wave function for the nth stationary state of the infinite square well is given by Phi_n(p,t) = \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \int^{\infty}_{-\infty} e^{-ipx/\hbar} \sin(\frac{n\pi}{a}x) dx.f

#### NeoDevin

Problem:
Find the momentum-space wave function $\Phi_n(p,t)$ for the $n$th stationary state of the infinite square well.

Equations:

$$\Psi_n(x,t) = \psi_n(x) \phi_n(t)$$

$$\psi_n(x) = \sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x)$$

$$\phi_n(t) = e^{-iE_n t/\hbar}$$

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

Attempt:

$$\Phi_n(p,t) = \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \int^{\infty}_{-\infty} e^{-ipx/\hbar} \sin(\frac{n\pi}{a}x) dx$$

$$= \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \frac{1}{2i} \int^{\infty}_{-\infty}\Bigg(e^{i(\frac{n\pi}{a} - \frac{p}{\hbar})x} - e^{i(\frac{-p}{\hbar} - \frac{n\pi}{a})x}\Bigg) dx$$

$$= \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \frac{1}{2i} 2\pi \Bigg(\delta(\frac{n\pi}{a} - \frac{p}{\hbar}) - \delta(\frac{p}{\hbar} + \frac{n\pi}{a})\Bigg)$$

This doesn't seem right to me. Do I have this right, or am I missing something somewhere?

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OOOOPS. Except you DON'T want to integrate x from -infinity to +infinity. The wavefunction only lives in the box.

Oh, right... thanks

I corrected my previous flip answer.

I know, I saw. Thanks.

OOOOPS. Except you DON'T want to integrate x from -infinity to +infinity. The wavefunction only lives in the box.

He SHOULD integrate from - to + infinity, it's just that outside the box the wavefunction is strictly 0...

What happens when you only integrate from 0 to a? When I try I get a seemingly uselessly messy answer, is there some trick to use delta functions or something I am missing?

Hi, See Schbert

http://www.ecse.rpi.edu/~schubert/Course-ECSE-6968%20Quantum%20mechanics/Ch03%20Position&momentum%20space.pdf [Broken]

Regards.

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