Solving for Wave Function in an Infinite Square Well

[silverthorne]

Hi guys, this assignment is driving me nuts! Thank you very much for the help!

Homework Statement

Consider the infinite square well described by V=0, -a/2<x<a/x, and V=infinity otherwise. At t=0, the system is given by the equation

\Psi(x,0) = C_{1} \Psi_{1}(x) + C_{2} \Psi_{2}(x)
\Psi(x,0) = \frac{1}{\sqrt{2}} \sqrt{\frac{2}{a}}cos\left \frac{\pi x}{a} \right + \frac{1}{\sqrt{2}} \sqrt{\frac{2}{a}}sin\left \frac{2 \pi x}{a} \right

(a) Obtain \Psi (x,t)
(b) Use this \Psi (x,t) to calculate <H>, delta H, <x> and <p>.
(c) What can you say about the result you obtained from part (b). Explain.

Homework Equations

\psi(x,0)=\sum_{n=1}^{\infty}c_n\psi_n(x)

\psi_{n}(x)= \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right

E_{n}=\frac{n^2\pi^2\hbar^2}{2ma^2}

c_{n}=\int_{0}^{a} \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right \psi(x,0)dx

The Attempt at a Solution

Um...this problem is kind of similar to the infinite well problem posted below earlier...I want to know if the formulae up there are the right one to use first...before I blindly apply it and do the integrals...

the second term in the wave function looks like an eigenfunction for the energy...but the first one is a cosine so I am not sure what to do there...do I need to...split them up?

The equations above are for the infinite well from o to a...but this question is from -a/2 to a/2...so I am not sure if the eigenfunctions \Psi(x) change ...

I also know \Psi (x,t)can be obtained from multiplying \Psi (x,0)by the appropriate phase factor once the \Psi_{n} (x,0) is written as a linear combination of the energy eigenfunctions...but then there's the cosine in the first term...

 

[v0id]

In this case, both cosine and sine forms are eigenfunctions of the energy operator. This is because we no longer have the boundary conditions \Psi(0,0) = \Psi(a,0) = 0.

I would think the complete form \Psi(x,t) is easily obtained from the initial condition by simply appending the time-dependence factor to \Psi(x,0) to yield \Psi(x,t) = \Psi(x,0) \exp{\left( \frac{iEt}{\hbar} \right)}. But then again, I'm one of your fellow students who also needs to get this done by Monday 12:00h.

Moreover, once you have \Psi(x,t), the rest of the question is very easy. You know what the Hamiltonian operator simplifies to in this case (I'm assuming "delta H" refers to the variance \sigma and not the standard deviation).

 

[silverthorne]

Ah...thanks! Yes, I forgot the fact that the reason why only the sines are the eigenfunctions is because of the initial conditions! So yeah, the cosine here works just fine. So I guess it's just \Psi(x,t) = \frac{1}{\sqrt{2}} \sqrt{\frac{2}{a}}cos\left \frac{\pi x}{a} \right \exp{\left( \frac{iE_{0} t}{\hbar} \right)}+ \frac{1}{\sqrt{2}} \sqrt{\frac{2}{a}}sin\left \frac{2 \pi x}{a} \right \exp{\left( \frac{iE_{1} t}{\hbar} \right)}, with E_{0}=\frac{\pi^2\hbar^2}{2ma^2} and E_{1}=\frac{2^2\pi^2\hbar^2}{2ma^2}?

 

[v0id]

This is very strange. If we run with the assumption that the time-dependence factor can be tacked on so primitively, then the expected value of the Hamiltonian is a function of time.

I see no other way to derive the time-dependent form of \Psi(x,t) from only \Psi(x,0).

 

[silverthorne]

Um...well, I am stuck then if we can't just tack on the exponential factor. How should I go about in doing this problem?

 

[v0id]

Sorry, maybe it does work. I just tried solving for \langle H \rangle in both Maple and Maxima, and the two CASs give different results. Maple gives a single value, but I think Maxima is unable to simplify the final expression to that value.

EDIT: My mistake. Maxima is able to simplify the expression perfectly. The input was wrong.

 

[silverthorne]

It's ok! So I guess the answer to part a is as above...

Anyway, I very much appreciate your help!