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Expected value of momentum

  1. Mar 9, 2010 #1
    [itex]
    \langle p\rangle=-i\hbar\int \Psi^*\frac{\partial}{\partial x}\Psi dx
    [/itex]

    Apparently, in general, we get a complex number for momentum. What does it mean, since p is an observable?
     
  2. jcsd
  3. Mar 9, 2010 #2

    Matterwave

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    The values are real, and can only be real, the imaginary numbers will cancel out. The momentum operator, p, being hermitian, guarantees this.
     
  4. Mar 9, 2010 #3
    Ok, thank you.
     
  5. Mar 9, 2010 #4

    SpectraCat

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    In a little more detail, if the wavefunction is purely real, then you will always find that <p>=0 (try this out if you like, for example on the Harmonic oscillator wavefunctions).

    If the wavefunction is complex, then you can separate the real and imaginary parts ... the imaginary part will of course be multiplied by i, and so you will end up with i2 in your final answer, which will be completely real. (The real part will yield a result of zero, as covered above).
     
  6. Mar 12, 2010 #5
    Hi.

    By partial integration, as ψ vanishes at infinity x,
    ∫ψ*∂x ψ dx = [ψ*ψ] - ∫(∂x ψ*)ψ dx = - ∫ψ∂xψ* dx.
    By adding the both and dividing by two,
    -ih'∫ψ*∂x ψ dx = -ih'{ ∫ψ*∂x ψ dx - ∫ψ∂xψ* dx}/2
    = -ih'/2 ∫ψ*∂x ψ dx + ih'/2 ∫ψ∂xψ* dx
    = -ih'/2 ∫ψ*∂x ψ dx + its conjugate complex
    = real number

    Regards.
     
    Last edited: Mar 12, 2010
  7. Mar 12, 2010 #6
    Another way is to take into consideration that the derivative of a real function has opposite parity. So the integrand is odd, resulting in a null integral in the symmetric limits. I think.
     
  8. Mar 12, 2010 #7
    Interesting proof, thank you.
     
  9. Mar 12, 2010 #8
    Since everyone seem to agree this momentum is "real", can someone please explain what kind of momentum is that - spatial, angular, spin? Does it apply to any particle or object or just photons or just bound electrons, or what? How do we measure it experimentally?
     
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