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The Time Independence of Normalization

  1. Jun 8, 2014 #1
    On page 13 of Griffith's "Introduction to Quantum Mechanics 2nd ed" David goes into a long (relatively speaking) proof of why a normalized pair of quantum state vectors will not at some time later become "un-normalized". It seems like just putting the Psi's in a braket the e^(-it) "time dependence" term would just cancel out -- showing that normalization is not time dependent. Could anyone take a shot at why this is not the case -- other than the fact he has not developed braket notation or shown that the time dependence a separate exponent factor? Maybe I just answered my own question :D Is there anything wrong with showing it this way?

    Chris Maness
  2. jcsd
  3. Jun 8, 2014 #2
    The general wave function is built of a linear combination of the eigenfunctions. When you do this, the exponential time-dependent phases do not cancel out. When he shows that the normalization remains constant in time, he shows it for a general wave function that is a solution of just the Schrodinger equation, and not the eigenvalue equation in general.
  4. Jun 8, 2014 #3


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  5. Jun 9, 2014 #4
  6. Jun 9, 2014 #5


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    Well, it's enough to know that

    [itex]\int \psi^* H \psi dx[/itex]

    is real. That's because Schrodinger's equation implies that:

    [itex]\dfrac{d}{dt} \int \psi^* \psi dx = \dfrac{2}{\hbar} Im(\int \psi^* H \psi dx)[/itex]

    where [itex]Im[/itex] means the imaginary part.
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