Another inifinite square well (1 Viewer)

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

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

[tex]\Psi(x,0) = C_{1} \Psi_{1}(x) + C_{2} \Psi_{2}(x)[/tex]
[tex]\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[/tex]

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

2. Relevant equations

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


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

3. 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 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/ I am not sure if the eigenfunctions [tex]\Psi(x)[/tex] change ...:cry:

I also know [tex]\Psi (x,t)[/tex]can be obtained from multiplying [tex]\Psi (x,0)[/tex]by the appropriate phase factor once the [tex]\Psi_{n} (x,0)[/tex] is written as a linear combination of the energy eigenfunctions...but then there's the cosine in the first term...:mad: :confused:
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In this case, both cosine and sine forms are eigenfunctions of the energy operator. This is because we no longer have the boundary conditions [tex]\Psi(0,0) = \Psi(a,0) = 0[/tex].

I would think the complete form [tex]\Psi(x,t)[/tex] is easily obtained from the initial condition by simply appending the time-dependence factor to [tex]\Psi(x,0)[/tex] to yield [tex]\Psi(x,t) = \Psi(x,0) \exp{\left( \frac{iEt}{\hbar} \right)}[/tex]. 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 [tex]\Psi(x,t)[/tex], 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 [tex]\sigma[/tex] and not the standard deviation).
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 [tex]\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)}[/tex], with [tex]E_{0}=\frac{\pi^2\hbar^2}{2ma^2}[/tex] and [tex]E_{1}=\frac{2^2\pi^2\hbar^2}{2ma^2}[/tex]?
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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 [tex]\Psi(x,t)[/tex] from only [tex]\Psi(x,0)[/tex].
Um...well, I am stuck then if we can't just tack on the exponential factor. :confused: How should I go about in doing this problem?
Sorry, maybe it does work. I just tried solving for [tex]\langle H \rangle[/tex] 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.
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It's ok! So I guess the answer to part a is as above...

Anyway, I very much appreciate your help!!

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