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Shouldn't the operator be applied to both the wf anD its modulus?

  1. Mar 29, 2013 #1

    jshrager

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    Gold Member

    Why is it:

    $$ \langle A \rangle = \int dV ~ \psi^* (\hat{A} \psi) $$

    As opposed to:

    $$ \langle A \rangle = \int dV ~ (\hat{A} \psi^*) (\hat{A} \psi) $$

    The op can substantially change the wf, so it would seem to make more mathematical sense (at least from a linear algebra pov) to operate on both the wf and its mod, otherwise, as we do it now (first expr) you're getting a very odd product. Is this is a property of the way that wfs are formulated? Are they a special case where this isn't required? Or maybe it's a magical property of hermetian operators? Or maybe it just works and I shouldn't think too hard about it.
     
  2. jcsd
  3. Mar 29, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    1) it works
    2) the second formula is not linear
    3) There is probably some deeper reason why this works
     
  4. Mar 29, 2013 #3

    kith

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    Science Advisor

    If you start with the standard definition of the expectation value of a random variable <A> = Ʃi piai -where the ais are the possible outcomes and pi is the probability to obtain the i-th- your first formula follows from the QM postulates about these quantities. I don't know if this is satifying to you. ;-)
     
    Last edited: Mar 29, 2013
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