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Expectation value of non-physical observable

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data
    This may be incredibly obvious, but I just need to check. Of course we all know that physical observables must yield real expectation values. What if you tried to calculate, say, <xV(d/dx)>, where x is the position, d/dx is a first derivative, and V is the potential? This isn't an observable...so do we automatically know the answer is zero(or non-real)?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 6, 2013 #2

    vela

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    Nope.
     
  4. Mar 6, 2013 #3
    So after searching around I agree with your answer, but I am not sure why this is so. For example, <xp> is not zero but has a complex component. xp isn't really considered a physical observable though...is it? Is the only restriction on for a real expectation value that the quantity be a "physically meaningful" observable such as momentum, energy, etc..?
     
  5. Mar 6, 2013 #4

    dextercioby

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    The basic requirement is for the linear operators which describe quantum observables to be self-adjoint. xp isn't. Self-adjoint operators have a spectral decomposition with a purely real spectrum, so that every expectation value on a state in their domain is real.
     
  6. Mar 6, 2013 #5
    However, the combination

    [tex]D = \frac{XP+PX}{2},[/tex]

    which looks the same classically, is self adjoint. One can often construct valid observables by getting the operator ordering right.
     
  7. Mar 7, 2013 #6
    So since this is self adjoint...then how to we interpret this physically?
     
  8. Mar 8, 2013 #7
    Simply as the product of position and momentum, I would think. We could in principle design a clever experiment to measure D.
     
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