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Quantum mechanics: Expectation values

  1. Aug 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Let's say that i have a wave function
    [tex]\Psi (x,t) = A \cdot \exp ( - \lambda \cdot \left| x \right|) \cdot \exp ( - i\omega t)[/tex]

    I want to find the expectation value for x. For this I use
    [tex]\left\langle x \right\rangle = \int_{ - \infty }^\infty x \left| \Psi \right|^2 dx[/tex].

    Why must I set t = 0 when doing this?
     
  2. jcsd
  3. Aug 8, 2008 #2

    nrqed

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    You don't.

    Just leave t arbitrary and you will see that it will drop out by itself.
     
  4. Aug 9, 2008 #3
    Yes, you are right. But that is when I have normalized the wave function first.

    In my book, when he normalizes this function he uses t=0, so A = sqrt(lambda). This way he uses t=0 when finding <x>.

    But I can't do both (both use t=t and t=0). So what is more correct?
     
  5. Aug 9, 2008 #4

    Dick

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    ??? The normalization is independent of t. exp(i*w*t) times it's complex conjugate is 1. For all t.
     
  6. Aug 9, 2008 #5
    My mistake. I just squared the whole expression, when I should have multiplied with the conjugate (we are dealing with complex numbers, of course).

    Thanks to both of you.
     
  7. Aug 10, 2008 #6

    nrqed

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    You are welcome.

    By the way, it is a general property of wavefunctions: if you normalize them at any time, they will remain normalized at all times (this assumes that the wavefunction is square integrable, that the potential is real, etc). So one can in principle fix the time to be some value (any value!) before normalizaing but this is never necessary because the time dependence *always* drops out when we normalize.
     
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