# Finding the momentum-space wave function for the infinite square well

1. Feb 6, 2012

### epsilonjon

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

Find the momentum-space wave function for the nth stationary state of the infinite square well.

2. Relevant equations

Nth state position-space wavefunction:

$\Psi_n(x,t) = \sqrt(\frac{2}{a})sin(\frac{n\pi}{a}x)e^{-iE_nt/\hbar}$.

Momentum operator in position space:

$\hat{p} = -i\hbar\frac{d}{dx}$.

3. The attempt at a solution

As a result of an earlier question I know that you can write

$\Psi_n(x,t) = \sqrt(\frac{2}{a})sin(\frac{n\pi}{a}x)e^{-iE_nt/\hbar}$
$= \sqrt(\frac{2}{a})[\frac{e^{i n\pi x/a}-e^{-i n\pi x/a}}{2i}]e^{-iE_nt/\hbar}$
$= \frac{-i}{\sqrt(2a)}[\sqrt(a)f_{n\pi\hbar/a}(x) - \sqrt(a)f_{-n\pi\hbar/a}(x)]e^{-iE_nt/\hbar}$
$= \frac{-i}{\sqrt(2)}[f_{n\pi\hbar/a}(x) - f_{-n\pi\hbar/a}(x)]e^{-iE_nt/\hbar}$(*)

where $f_{p}(x) = \frac{1}{\sqrt(a)}e^{ipx/\hbar}$ are the momentum eigenfunctions (when we're in position space) normalized over 0≤x≤a.

Now when you write $\Psi_n(x,t)$ as a sum (or integral) of momentum eigenfunctions then $\Phi_n(p,t)$ is the coefficient of these eigenfunctions right? So combining this with (*) do we not get that

$\Phi(p,t) = \frac{-i}{\sqrt(2)}e^{-iE_nt/\hbar}$ when $p=n\pi\hbar/a$;
$\Phi(p,t) = \frac{i}{\sqrt(2)}e^{-iE_nt/\hbar}$ when $p=-n\pi\hbar/a$;
$\Phi(p,t) = 0$ otherwise .

And note that the sum of the squares of these adds up to 1, as it should.

But now if I just did it without thinking so much, and did

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

then I get the result

$\Phi_n(p,t) = \frac{\sqrt(a\pi)}{\sqrt(\hbar)}\frac{ne^{-iE_nt/\hbar}}{(n\pi)^2 - (ap/\hbar)^2}[1-(-1)^ne^{-ipa/\hbar}]$.

Now this function does peak at those "special" momentum values $p=\pm n\pi\hbar/a$, but it is non-zero for the other values.

So how come the two methods do not match up? The first says that I am certain to measure one of the two momentum values $p=\pm n\pi\hbar/a$, but the second says I can also get others. Which is correct?

Thanks for any help!

Last edited: Feb 6, 2012
2. Feb 6, 2012

### vela

Staff Emeritus
It's because the $\psi(x) = 0$ outside the well, but the linear combination of fpn(x) and f-pn(x) doesn't vanish. In other words, you can't express the spatial wave function as a linear combination of just the two f's.