# Commpute probability in momentum space of particle in box (after walls are removed)

1. Apr 28, 2012

### Don'tKnowMuch

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

A particle is initially in the nth eigenstate of a box of length 2a. Suddenly the walls of the box are completely removed. Calculate the probability to find that the particle has momentum between p and p + dp. Is energy conserved?

2. Relevant equations

solution for particle in a box of length L ---->
ψ(x)(sub n) = (2/L)^1/2*sin(n∏x/L)

Fourier Transform of momentum space ----->

ζ(p) = [ 1/(2∏h-bar)^(1/2) ]*∫ e^(-ipx/h-bar)*ψ(x)

3. The attempt at a solution

I need the probability that the particle is between p & (p+dp). So my plan was to put the ψ(x)(sub n) into the Fourier transform, and then square that to get the probability. My trouble lies in the integral. I am not sure how to compute it. Would the bounds of the integral be p to (p+dp)?

2. Apr 28, 2012

### sgd37

Re: commpute probability in momentum space of particle in box (after walls are remove

you dont have to do the integral because the P[p+dp,p] = ζ^2(p)dp since ζ^2(p) is the probability density function

3. Apr 28, 2012

### Don'tKnowMuch

Re: commpute probability in momentum space of particle in box (after walls are remove

This is my understanding, there is a wave function in position space defined for a box of length 2a. This state of definite position is being projected onto a state which has deinfite momentum

ζ(p)^2 = 1/(2∏h-bar) *∫ e^(-ipx/h-bar)^2*( 1/(2a)*sin^2(n∏x/2a))dx

There's still an integral hanging around though. I feel like it needs to be integrated. If that were the case, would i change the sin(stuff) to an exponential and see what comes out?

4. Apr 28, 2012

### francesco85

Re: commpute probability in momentum space of particle in box (after walls are remove

Hi, in my opinion the computations you have proposed are right. It may be easier to write the exponential as cos+i sin and then use the formulae (from wikipedia :) to compute integrals like sin( ax) cos( bx) or sin( ax) sin( bx) .
Best,
Francesco

5. Apr 28, 2012

### Don'tKnowMuch

Re: commpute probability in momentum space of particle in box (after walls are remove

Looking at the sine term i can simplify the argument a little by setting......

k' = (n∏/2a)

making the sine term------> sin(k'x)

Looking at the exponential, i can say that.......

p/h-bar = k

Putting these in the integral looks something like.....

ζ(p) = 1/(2∏h-bar)^(1/2)*∫ e^(-ikx)*(1/2i)(e^ik'x-e^-ik'x) dx......

which simplifies to....

ζ(p) = (1/2i)*1/(2∏h-bar)^(1/2)*∫ e^(-ix(k' - k))-(e^-ix(k' + k)) dx......

I have been using every resource at my disposal to crack this one, and to no avail. I saw that the solution to this integral is a combination of delta functions, but i just can't put it together.