# Relative phase in infinite square well

1. Feb 18, 2005

### painfive

This is a problem from my introductory quantum mechanics class. It's Griffifth's problem 2.6, if anyone has that book. The problem says to investigate the effect of adding two steady state solutions with a relative phase. Namely:

$$\Psi(x,0) = A [ \psi_1(x) + e^{i \phi} \psi_2 (x) ]$$

where:

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

He says to compare this to the case with no relative phase, and to study the special cases of phi=pi/2 and pi. But I get that the only difference is that it effectively shifts the time origin. For example, I get:

$$|\Psi(x,t)|^2 = \frac{1}{a}\left[\sin^2\left(\frac{\pi x}{a}\right) + \sin^2\left(\frac{2\pi x}{a}\right) + 2\sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{2\pi x}{a}\right) \cos\left(\frac{3 \pi^2 \hbar t}{2 m a^2} - \phi\right) \right]$$

This isn't anything that needs to be investigated, and the special cases look meaningless, which makes me think I've done something wrong. Have I?

2. Feb 18, 2005

### dextercioby

What do you mean they look meaningless...?I think not.You added two wavefunctions corresponding to 2 stationary states and now judge what it came out of this addition.

Daniel.

3. Feb 18, 2005

### painfive

I mean adding pi or pi/2 to that fraction with h's and a's isn't going to give anything special.

4. Feb 18, 2005

### dextercioby

I'd say that u've gotten yourself a nostationary state...I don't see other effect...

Daniel.