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Show the expectation value is non negative

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    The kinetic energy is given by [itex]\left\langle E_{kin} \right\rangle = \frac{\left\langle \widehat{p}^2 \right\rangle}{2m}[/itex]

    In Dirac notation we have

    [itex]\left\langle E_{kin} \right\rangle = \frac{1}{2m} \left\langle \widehat{p}\Psi | \widehat{p}\Psi \right\rangle[/itex]

    2. Relevant equations
    We are asked to give this in integral form and involving [itex]\left| \frac{\partial \Psi}{\partial x} \right|[/itex]

    [itex] \left\langle E_{kin}\right\rangle =\dfrac{1}{2m} \displaystyle\int_{-\infty}^{\infty}\widehat{\textrm{p}}_{x}^{*}\Psi^{*}(x,t)\widehat{ \textrm{p}}_{x} \Psi(x,t) dx [/itex]

    Which then becomes

    [itex] \dfrac{\hbar^{2}}{2m} \displaystyle\int_{-\infty}^{\infty}\left|\dfrac{\partial\Psi}{\partial x}\right|^{2} dx [/itex]

    3. The attempt at a solution
    This is not a problem but we are also asked to confirm explicitly that this cannot be negative in value

    [itex] \dfrac{\hbar^{2}}{2m} \displaystyle\int_{-\infty}^{\infty} \frac{\partial^2}{\partial x^2} \left| \Psi \right|^{2} dx [/itex]

    Is this valid? As the constants on the left are positive and the square of the modulus of a complex function is non negative.

    Thanks
     
  2. jcsd
  3. May 14, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The square of a first derivative of a function is not the second derivative of that function squared.


    Isn't the answer obtained directly from:
     
  4. May 14, 2013 #3
    Yes, thanks.
     
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