Solving for Psi(x,t) in an Infinite Square Well Potential

[stunner5000pt]

Consider a particle of mass m in the normal ground sate of an infinite square well potential of width a/2. Its normalized wave function at time t=0 is given by

\Psi(x,0) = \frac{2}{\sqrt{a}} \sin \frac{2 \pi x}{a} for 0 <x <a/2
0 elsewhere

At this time the well suddenly changes to an infinite square well with width a without affectring the wave function.

By writing Psi(x,t) as a linear superposition of the energy eigenfunctions of the new potentaail find the probability taht a subsequency measurement of the enrgy will yield th result
E_{1} = \frac{\hbar^2 \pi^2}{2ma^2}
{Hint: A linear superposition of square well eigenfunctions is a Fourier sine series and teh coefficients of teh series are given by simple integrals)

Now we know that
\Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x)

and the wave function didnt change
from teh hint
if i find the coefficients all i do is find an approximation to teh wave function. the new potentail would be

V(x) = 0 for 0<x<a
infinity elsewhere

and we know the wave function for that potential it is
\Psi_{2}(x,0) = \sqrt{\frac{2}{a}} \sin\frac{\pi x}{a}
for the first exceited state

so am i to find Psi 1 in terms of Psi 2??

 

[Galileo]

You have to find the probability a measurement of the energy yields E_1. That probability is equal to |c_1|^2.

You can get the term c1 from
\Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x)
by taking the inner product with \psi_1 on both sides, since the eigenstates are orthonormal.

 

[stunner5000pt]

You have to find the probability a measurement of the energy yields E_1. That probability is equal to |c_1|^2.

You can get the term c1 from
\Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x)
by taking the inner product with \psi_1 on both sides, since the eigenstates are orthonormal.

so with the change in dimensions of the well, the fact taht the wave funcion didnt change allows us to compute Cn using the wave function given to us??

so
c_{1} = \frac{2}{a/2} \int_{0}^{a/2} \sin \frac{\pi}{a} x \Psi(x,0) dx

wher \Psi(x,0) = \frac{2}{\sqrt{a}} \sin \frac{2 \pi x}{a}

right?

 

[Galileo]

so with the change in dimensions of the well, the fact taht the wave funcion didnt change allows us to compute Cn using the wave function given to us??

You can ALWAYS do that. If you expand your wave function on an orthonormal basis:

|\Psi\rangle = \sum_n c_n|\phi_n\rangle
then we can find the coefficients by taking the inner product:
c_n = \langle \phi_n|\Psi\rangle

(Maybe the notation is new to you, but I hope you get what it means).

so
c_{1} = \frac{2}{a/2} \int_{0}^{a/2} \sin \frac{\pi}{a} x \Psi(x,0) dx

right?

It could possibly be a factor of \frac{2\sqrt{2}}{a} in front of there, but otherwise it looks good.