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Testing if the momentum operator is Hermitian

  1. Sep 8, 2008 #1
    Hi. I'm not too good at maths and I'm having some trouble figuring out the basics of what to do with complex conjugates of functions.

    Our lecturer has set a couple problems requiring us to test if a few operators are Hermitian. Before I can get to those I thought I'd test the basic momentum operator: [tex](\frac{\hbar}{i} \frac{\partial}{\partial x})[/tex].

    Using integration by parts:

    [tex]\int_{-\infty}^\infty \Psi^\ast(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi(x,t) dx = \frac{\hbar}{i} [\Psi^\ast(x,t)\Psi(x,t)]^\infty_{-\infty} - \int_{-\infty}^\infty \Psi(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi^\ast(x,t) dx [/tex]

    Now how do I deal with the first term? Does it reduce to 0 somehow? I recongnise that the second term should be equal to the LHS for the momentum operator to be shown as Hermitian.

    Cheers. Kaan

    EDIT: sorry, I found the answer here
    Last edited: Sep 8, 2008
  2. jcsd
  3. Sep 8, 2008 #2


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    Staff: Mentor

    [itex]\Psi[/itex] must be normalizable, therefore [itex]\Psi^* \Psi[/itex] must go to zero as x goes to infinity in either direction.
  4. Sep 8, 2008 #3
    Yes first term on RHS goes to zero because of the normalization condition on the wavefunction [tex]\Psi(x,t) ->0 [/tex] as [tex] x-> \infty [/tex] ( or [tex] -\infty [/tex] ).

    So you're left with:

    [tex]\int_{-\infty}^\infty \Psi^\ast(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi(x,t) dx = \int_{-\infty}^\infty \Psi(x,t) (-\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi^\ast(x,t) dx [/tex]

    Thus the Hermitian condition, [tex]\int_{-\infty}^\infty \Psi^\ast(x,t) \hat{P} \Psi(x,t) dx = \int_{-\infty}^\infty \Psi(x,t) \hat{P^\ast} \Psi^\ast(x,t) dx [/tex]

    is satisified since the conjugate of the momentum operator is just minus the momentum operator.
  5. Nov 5, 2009 #4
    Is this a valid proof?

    Given an arbitrary ket [tex]|a\rangle[/tex]:

    [tex]\langle a|\hat{\textbf{p}}a\rangle[/tex]

    [tex]=\langle \hat{\textbf{p}}^{\dagger}a|a\rangle[/tex]

    If [itex]\hat{\textbf{p}}[/itex] is Hermitian, then



    [tex]\langle \hat{\textbf{p}}^{\dagger}a|a\rangle = \langle \hat{\textbf{p}}a|a\rangle[/tex]

    [tex]=\left(\langle a|\hat{\textbf{p}}a\rangle\right)^{*}[/tex]
  6. Nov 5, 2009 #5


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    Homework Helper

    You can't assume P is hermitean, you have to prove such thing ! You can't prove it in the abstract Dirac language. You must choose a Hilbert space. Define the operators and only then check for properties such as continuity/boundedness, hermiticity, unitarity, etc.
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