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Homework Help: Vanishing Wavefunction

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    If <x> and <p> are the expectation values of x and p formed with the wave-function of a one-dimensional system, show that the expectation value of x and p formed with the wave-function vanishes. The wavefunction is:

    [tex]\phi(x)=exp(-\frac{i}{h}\langle p\rangle x)\psi(x+\langle x\rangle)[/tex]


    Basically I don't know how to start this problem. Do I plug in the integral forms of the expectation values into the exponential and distribute the psi over x and [tex]\langle x\rangle[/tex]?
     
  2. jcsd
  3. Sep 28, 2010 #2

    fzero

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    You left out some important text, but I found the problem set with a google search. [tex]\langle x\rangle, \langle p\rangle[/tex] are computed with the wavefunction [tex]\psi(x)[/tex], i.e.

    [tex] \langle x\rangle = \int x \psi^*(x) \psi(x) \, dx, \ldots[/tex]

    You are asked to compute expectation values with the wavefunction [tex]\phi(x)[/tex]:

    [tex] \langle x\rangle_\phi = \int x \phi^*(x) \phi(x) \, dx, \ldots[/tex]

    These expressions can be reduced to expectation values for certain quantities in the wavefunction [tex]\psi(x)[/tex]. You don't have to do any integrals explicity, just some algebra and derivatives.
     
  4. Sep 29, 2010 #3
    I've set it up the way you suggested, and the exponential terms computing [tex]\langle x \rangle[/tex] cancel. However there is a [tex]\langle x \rangle[/tex] term in [tex]\psi(x+\langle x \rangle)[/tex]. Is this not a discrepancy to have the value you are trying to calculate inside the equation for calculating its value?

    Specifically,

    [tex]\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}[/tex]
     
  5. Sep 29, 2010 #4

    fzero

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    Well [tex]\langle x \rangle[/tex] is just a number (it came out of another integral over x, so it doesn't depend on x). There's still something you have to do to this integral though.
     
  6. Sep 29, 2010 #5
    If [tex]\psi(x+\langle x \rangle)[/tex] is distributable (meaning [tex]f(x+y)=f(x)+f(y)[/tex]) then I come up with the following:

    [tex]\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}[/tex]
    [tex]\int{ x[ (\psi^{*}x+\psi^{*}\langle x \rangle)( \psi x+\psi\langle x \rangle)\,dx}[/tex]
    [tex]\int{ (\psi^{*}\psi x^{3}+2\psi^{*}\psi x^{2}\langle x \rangle+\psi^{*}\psi x\langle x \rangle^{2})\,dx}[/tex]
    [tex]\langle x^{3}\rangle +2\langle x^{2}\langle x \rangle\rangle + \langle x\langle x\rangle^{2}\rangle[/tex]

    ...which doesn't make any sense to me. Also the "distributable" assumption I made does not work for exponentials, since [tex]f(x+y)=f(x)f(y)[/tex] for exponential functions.
     
  7. Sep 29, 2010 #6

    fzero

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    Yeah wavefunctions will almost never have that property since they're always related to exponential functions. Think about how you can convert this to an expectation value in [tex]\psi(x)[/tex] by making a substitution in the integration variable.
     
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