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The 'i' in the momentum operator

  1. May 27, 2010 #1
    The momentum operator is -i*h bar * derivative wrt x. But won't this lead sometimes to complex expected values of momentum? What does this mean physically, since complex values can't be measured.
     
  2. jcsd
  3. May 27, 2010 #2

    Doc Al

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    Staff: Mentor

    All operators in QM, including this one, are Hermitian. The eigenvalues of a Hermitian operator are real, not complex. (Which is a good thing, as you point out. Complex values would be a problem.)
     
  4. May 27, 2010 #3
    The i is actually needed to make the momentum operator Hermitian!
     
  5. May 27, 2010 #4

    Doc Al

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    Exactly! :smile:
     
  6. May 27, 2010 #5
    The operator d/dx is not Hermitian, so you add a -i to it and the i's cancel out to -1 and then the - cancels out to +1
     
  7. May 27, 2010 #6

    jtbell

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    Others have given the reasons why the expectation value of p must always be real.

    From a practical point of view, when you calculate such an expectation value, you usually get an expression that contains i, to start with. But you can always eliminate i from these expectation values by using identities such as

    [tex]\sin x = \frac{1}{2i} \left( e^{ix} - e^{-ix} \right)[/tex]

    If you can't, then you've made a mistake in your algebra somewhere. :wink:
     
    Last edited: May 28, 2010
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