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Relative phase in infinite square well

  1. Feb 18, 2005 #1
    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:

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

    where:

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

    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:

    [tex] |\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] [/tex]

    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. jcsd
  3. Feb 18, 2005 #2

    dextercioby

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    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.
     
  4. Feb 18, 2005 #3
    I mean adding pi or pi/2 to that fraction with h's and a's isn't going to give anything special.
     
  5. Feb 18, 2005 #4

    dextercioby

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

    Daniel.
     
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