Probability for a momentum value for particle in a box

[pedroobv]

[SOLVED] Probability for a momentum value for particle in a box

Homework Statement

This is the problem 7.37 from Levine's Quantum Chemistry Textbook.

Show that, for a particle in a one-dimensional box of length l, the probability of observing a value of p_{x} between p and p+dp is:

\frac{4|N|^{2}s^{2}}{l(s^{2}-b_{2})}\left(1-(-1)^{n}cos(bl)\right)dp​

Where s\equiv n\pi l^{-1} and b\equiv p\hbar^{-1}. The constant N is to be chosen so that the integral from minus infinity to infinity of the previous equation is unity.

Homework Equations

Since p has a continue range of eigenvalues, the probability of finding a value of p beetween p and p+dp for a system in the state \Psi is

|\left\langle g_{p}(x)|\Psi (x,t) \right\rangle |^{2}dp​

where g_{p} are the eigenfunctions of \hat{p}.

For a particle in a box:

\Psi = \left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)​

and the eigenfunctions of the momentum operator are:

g=Ne^{ipx/\hbar}​

The Attempt at a Solution

I made the corresponding substitutions to obtain the integral:

\int ^{0}_{l}Ne^{-ipx/\hbar}\left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)dx​

Which after computing and using some identities gives:

\int ^{0}_{l}Ne^{-ipx/\hbar}\left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)dx = \frac{\sqrt{2}Ns\left[1-(-1)^{n}e^{-ibl}\right] }{\sqrt{l}\left(s^{2}-b^{2}\right)}​

Which looks a lot like the square root of the expression that I seek, however I'm stuck at this point. Any help will be appreciated, if more informations is needed please tell me.

Thanks in advance

 

[pedroobv]

Sorry, I made a mistake in the first equation. Instead of

\frac{4|N|^{2}s^{2}}{l(s^{2}-b_{2})}\left(1-(-1)^{n}cos(bl)\right)dp​

Must be:

\frac{4|N|^{2}s^{2}}{l(s^{2}-b^{2})^{2}}\left(1-(-1)^{n}cos(bl)\right)dp​

 

[pedroobv]

Ok thanks anyway, I already solved it. I forgot to take the absolute value.