# Probability for a momentum value for particle in a box

[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.

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