(adsbygoogle = window.adsbygoogle || []).push({}); 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:

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

Momentum operator in position space:

[itex]\hat{p} = -i\hbar\frac{d}{dx}[/itex].

3. The attempt at a solution

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

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

[itex]= \sqrt(\frac{2}{a})[\frac{e^{i n\pi x/a}-e^{-i n\pi x/a}}{2i}]e^{-iE_nt/\hbar}[/itex]

[itex]= \frac{-i}{\sqrt(2a)}[\sqrt(a)f_{n\pi\hbar/a}(x) - \sqrt(a)f_{-n\pi\hbar/a}(x)]e^{-iE_nt/\hbar}[/itex]

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

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

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

[itex]\Phi(p,t) = \frac{-i}{\sqrt(2)}e^{-iE_nt/\hbar}[/itex] when [itex]p=n\pi\hbar/a[/itex];

[itex]\Phi(p,t) = \frac{i}{\sqrt(2)}e^{-iE_nt/\hbar}[/itex] when [itex]p=-n\pi\hbar/a[/itex];

[itex]\Phi(p,t) = 0[/itex] 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

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

then I get the result

[itex]\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}][/itex].

Now this function does peak at those "special" momentum values [itex]p=\pm n\pi\hbar/a[/itex], 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 [itex]p=\pm n\pi\hbar/a[/itex], but the second says I can also get others. Which is correct?

Thanks for any help!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**