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

[epsilonjon]

Homework Statement

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

Homework 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}.

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!

 

[vela]

It's because the $\psi(x) = 0$ outside the well, but the linear combination of f_pn(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.