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

[Don'tKnowMuch]

Homework Statement

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?

Homework 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)

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)?

 

[sgd37]

you don't have to do the integral because the P[p+dp,p] = ζ^2(p)dp since ζ^2(p) is the probability density function

 

[Don'tKnowMuch]

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?

 

[francesco85]

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) .

Francesco

 

[Don'tKnowMuch]

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.