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Expectation Value of Momentum

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a wave function [tex]\psi (x,t) = R(x,t) exp(i S(x,t))[/tex] what is the expectation value of momentum?

    2. Relevant equations
    [tex] <f(x)> = \int^{\infty}_{-\infty} \psi^* f(x) \psi dx[/tex]

    [tex] \hat{p} = -i \hbar \frac{\partial}{\partial x} [/tex]

    3. The attempt at a solution
    This is for an intro to modern class so I don't really have a formal background with eigenvalues/vectors yet so this is a bit confusing.

    Can I just say that [tex] <p> = \int^{\infty}_{-\infty} \psi^{*} \hat{p} \psi dx = \int^{\infty}_{-\infty} \hat{p} \psi^{*} \psi dx[/tex] ?

    If so by the normalization condition <p> = -i hbar which I don't think can be the case.


    [tex]<p> = \int^{\infty}_{-\infty} R exp(-i S) \hat{p} R exp(i S) dx[/tex]

    [tex]= -i\hbar \int^{\infty}_{-\infty} R exp(-i S) * [R' exp(i S) + i R S' exp(i S)]dx [/tex]

    [tex]= -i\hbar \int^{\infty}_{-\infty} R R' + i R^2 S' dx [/tex]

    = ???

    I don't really see anything from there...

    I'm tempted to just say 0, but I'm not sure that the function being evaluated is odd.
    Last edited: Feb 13, 2009
  2. jcsd
  3. Feb 13, 2009 #2


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    No, you cannot.
    The first equality is true, but the second is not. You cannot just pull an operator through a function, because
    [tex]\psi^* \frac{\partial}{\partial x} \psi \neq \frac{\partial}{\partial x} \psi^* \psi[/tex]
    which is ambiguous too, as it could mean either
    [tex]\frac{\partial}{\partial x} \left( \psi^* \psi \right) = \frac{\partial \psi^*}{\partial x} \psi + \psi^* \frac{\partial\psi}{\partial x} [/tex]
    [tex]\left( \frac{\partial}{\partial x} \psi^* \right) \psi[/tex]

    Instead you just plug in the wave-function. Start by writing down what is
    [tex]\hat p \psi(x, t) [/tex]
  4. Feb 13, 2009 #3
    I all ready did apply the momentum operator like that in my derivation:

    [tex]i\hbar [R' exp(i S) + i R S' exp(i S)][/tex]

    I just assumed that the second expression was wrong and did it the (semi)correct way.
  5. Feb 13, 2009 #4


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    Ah, I see that now.
    Yes you did it right.
    The result isn't something very beautiful, but I suppose one would not expect that... after all the wave function depends on R and S, both of which are arbitrary functions on spacetime, so one cannot "predict" the expectation value on physical grounds.
  6. Feb 13, 2009 #5
    Ah, all right I just had in my mind that it would come out to be something the come be simplified a bit more. Thanks.
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