How Does Time Evolution Affect a Particle's Wave Function in a Quantum System?

[Bashyboy]

This is Problem comes from Griffiths Quantum Mechanics textbook; specifically, it is problem 2.5 (b).

A particle in an infinite square well has its initial wave function an even mixture of the first two stationary states:

$\displaystyle \Psi(x,0) = A[\psi_1(x) + \psi_2(x)]$

Here is the part of the problem that I am having a little trouble with:

(b) Find $\displaystyle \Psi(x,t)$ and $\displaystyle |\Psi(x,t)|^2$|. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let $\omega \equiv \frac{\pi^2 \hbar}{2ma^2}$

According to the answer key, even after t=0, the wave function continues to be a mixture of the first two stationary states. Why is that? I am having a little difficulty understanding this. Why can't it be a new 'mixture?'

As far as I understand, to calculate the wavefunction for all future times, we use the equations

$\displaystyle \Psi (x,t) = \sum_{n=1}^{\infty} c_n \sqrt{\frac{2}{a}} \sin \left( \frac{n \pi}{a} x \right) e^{-i(n^2 \pi^2 \hbar/2ma^2)t}$

and

$\displaystyle c_n = \sqrt{\frac{2}{a}} \int_0^a \sin \left( \frac{n \pi}{a} x \right) \Psi (x,0) dx$

 

[vela]

I'm not sure I understand your question. $\Psi$ will evolve with time, but it will remain a superposition of $\psi_1$ and $\psi_2$.

 

[Bashyboy]

That's precisely my question. Why will it remain as a superposition of the two states, even after t=0?

 

[vela]

At t=0, $c_n = 0$ for $n>2$, right? So the coefficient of $\psi_n$ for $n>2$ will be 0 as well for $t>0$.

 

[Bashyboy]

Oh, so the coefficients do not change with time?

 

[vela]

The coefficients of $\psi_1$ and $\psi_2$ change with time as they pick up a phase factor. If you look at the formula for $\Psi(x,t)$, you should be able to see that the coefficient of $\psi_1$, for instance, is $c_1 e^{-i\omega t}$. But the $c_n$'s are defined in terms of $\Psi$ when $t=0$, so they're just constants.

 

[Fredrik]

Are you familiar with the time evolution operator in the form $U(t)=e^{-iHt/\hbar}$? See if you can find it in your book. Once you understand it, you should find it easy to use it to answer your question.