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Derivation of momentum expectancy

  1. Nov 19, 2011 #1

    A_B

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    I'm trying to figure out why

    [tex] \lim_{x \to +\infty} \left(x \Psi^* \frac{\partial \Psi}{\partial x} \right) = 0 [/tex]

    This is what i've done so far:


    Since [itex]\Psi[/itex] must go to zero faster than [itex]x^{-1/2}[/itex] as [itex]x \to +\infty[/itex] we have

    [tex]
    \begin{align*}
    \frac{\partial \Psi}{\partial x} &< \frac{d}{dx} (x^{-1/2}) \\
    &= -\frac{1}{2} x^{-3/2}
    \end{align*}
    [/tex]

    So

    [tex]
    \lim_{x \to +\infty} \left(x \Psi^* \frac{\partial \Psi}{\partial x} \right) < \lim_{x \to +\infty} \left(-\frac{1}{2} \Psi^* x^{-1/2} \right) = 0
    [/tex]

    Since [itex]\Psi^* \to 0[/itex] as [itex]x \to \infty[/itex].


    I think this proves that the limit must be smaller than zero, but here I'm stuck.


    Help is much appreciated,
    A_B
     
  2. jcsd
  3. Nov 20, 2011 #2

    diazona

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    Homework Helper

    I think you're misunderstanding the meaning of a limit. The limit as [itex]x\to\infty[/itex] of a function isn't an actual value that the function takes on. It only means that the value of the function gets closer and closer to the limiting value as [itex]x[/itex] gets larger than larger. And when you have an expression involving a limit, it's "code" for thinking about what happens to the value of the expression as the limit is approached. You don't literally evaluate the limit and use it as a number.

    In a sense,
    [tex]\lim_{x\to\infty}x\Psi^* \frac{\partial\Psi}{\partial x} < \lim_{x\to\infty}-\frac{1}{2}\Psi^* x^{-1/2}[/tex]
    should be treated as if it read
    [tex]\lim_{x\to\infty}\biggl[x\Psi^* \frac{\partial\Psi}{\partial x} < -\frac{1}{2}\Psi^* x^{-1/2}\biggr][/tex]
    You can then use something akin to the limit comparison test.
     
  4. Nov 21, 2011 #3

    A_B

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    I thought the limit comparison test was for checking convergence infinite series?


    I'll write out my original question in it's context.

    I'm studying from Griffiths Introduction to Quantum Mechanics. The problem arises in the derivation of the expectation value of momentum:

    (All integrals are from -infinity to +infinity)
    Starting from the expectation value of position
    [tex]
    \left<x\right> = \int x\left|\Psi\right|^2 dx
    [/tex]

    The expectation value for velocity is the time derivative of this
    [tex]
    \begin{align*}
    \left<v\right>=\frac{d\left<x\right>}{dt} &= \int x \frac{\partial}{\partial}\left|\Psi\right|^2 dx \\
    &= \frac{i\hbar}{2m}\int x \frac{\partial}{\partial x}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) dx
    \end{align*}
    [/tex]

    Partial integration then gives
    [tex]
    =-\frac{i \hbar}{2m} \left[ \int \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) dx - \left.\left(x\Psi^* \frac{\partial \Psi}{\partial x} - x\frac{\partial \Psi^*}{\partial x} \Psi \right)\right|_{-\infty}^{+\infty} \right]
    [/tex]


    Griffiths says the boundary term equals zero so
    [tex]
    =-\frac{i\hbar}{2m} \int \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) dx
    [/tex]


    My question is why that boundary term is zero.


    A_B
     
  5. Nov 21, 2011 #4
    The boundary term is zero because the assumption is that the wavefunction vanishes at x=∞. Otherwise it wouldn't be normalized correctly. I know that Griffiths mentions this in the beginning chapters.
     
  6. Nov 21, 2011 #5

    dextercioby

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    Psi(x) is a test function, a Schwartz function so those limits are 0.
     
  7. Nov 21, 2011 #6

    A_B

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    Hi,

    I understand why ψ(x) vanishes as x→∞, the problem is that the boundary term is not only ψ(x), but it involves terms of the form xψ*(∂ψ/∂x). So, to me at least, it's not entirely obvious why these terms must vanish. I'm looking for a mathematical argument for why this is so.

    dextercioby: Griffiths mentions that the wavefunction must go to zero faster than 1/√|x|. Which I understand follows from the requirement that the wavefunction is normalizable. Why should be a Schwartz function, which goes to zero faster than any inverse power of x?


    Thanks
    A_B
     
  8. Nov 21, 2011 #7

    dextercioby

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    Because the momentum operator and all its powers can be rendered essentially self-adjoint on the real line, iff the domain of this operator is the Schwartz space.
     
  9. Nov 27, 2011 #8

    diazona

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    Yes, it is. That's why I mentioned doing something akin to the limit comparison test: you're working with a function, not a series. The limit comparison test tells you to compute the ratio of [itex]a_n/b_n[/itex] as [itex]n\to\infty[/itex], so here you compute the limit of the ratio between the two functions,
    [tex]\lim_{x\to\infty}\frac{x\Psi^* \frac{\partial\Psi}{\partial x}}{-\frac{1}{2}\Psi^* x^{-1/2}}[/tex]
    The point I'm trying to make is that your calculations in your original post were correct, but they did not show that the limit of the former function must be strictly less than zero. The limit could be equal to zero.

    Sorry about the delayed response, by the way; I was on vacation for the week and there were some problems with internet access.
     
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