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What is the expected momentum value for a real wavefunction?

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that a real valued wavefunction ##\psi(x)## must have ##\langle \hat{p}\rangle = 0##: Show that if we modify such a wavefunction
    by multiplying it by a position dependent phase ##e^{iax}## then ##\langle \hat{p}\rangle = a##

    2. Relevant equations
    ##\hat{p} = -i \hbar \frac{\partial}{\partial x}##

    3. The attempt at a solution
    The problem is to work out that the expected value is 0, but it is easy to do with that fact in mind. You can say that the imaginary part must equal 0 and so the LHS must also equal 0.

    However, if I try to work it through I get:

    ##\langle \hat{p}\rangle = -i\hbar \int_{-\infty}^{\infty} \psi(x) \frac{\partial \psi(x)}{\partial x} \mathrm{d}x##

    Even if I convert the partial derivative into a complete derivative I still don't know how to proceed. Does anyone have any better ideas? Any help would be much appreciated.
     
  2. jcsd
  3. Apr 3, 2015 #2
    I've solved it myself!

    Here's how:

    ##\langle \hat{p}\rangle = -i\hbar \int_{-\infty}^{\infty} \psi(x) \frac{\partial \psi(x)}{\partial x} \mathrm{d}x##

    ##\implies \langle \hat{p}\rangle = -i\hbar \int_{-\infty}^{\infty} \psi(x) \frac{\mathrm{d} \psi(x)}{\mathrm{d} x} \mathrm{d}x##

    Because ##\psi## is only a function of ##x##.

    ##\implies \langle \hat{p}\rangle = -i\hbar \int_{0}^{0} \psi(x) \mathrm{d} \psi(x)##

    This works because ##\psi(\infty) = \psi(-\infty) = 0##
     
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