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The expectation of an expection (relating to Wick's Theorem)

  1. Feb 25, 2010 #1
    Hi:

    If we want to work out the expectation of:

    <0|T(φ1φ2)|0>

    ie. <0|<0|T(φ1φ2)|0>|0>

    apparently it is acceptable to pull out the <0|T(φ1φ2)|0>:

    So <0|<0|T(φ1φ2)|0>|0>=<0|T(φ1φ2)|0><0|I|0>

    I do realise this is a really stupid question, but I want to be 100% sure. Is this simply because an expectation is always a constant, not an operator which acts on a state? Can you always pull out an expectation in this way?

    Thanks.
     
  2. jcsd
  3. Feb 25, 2010 #2

    Matterwave

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    Gold Member

    As far as I know, you can always pull out the expectation that way. I know of no cases where the expectation value is not simply just a number (or value). I don't see any point in finding the "expectation of the expectation value" tho...It's always just the same as the expectation value itself.
     
  4. Feb 25, 2010 #3
    thanks matterwave.

    Yes indeed, the expectation of an expectation is the expectation itself. A better example is the following:

    Apparently this expression equals the one below it:

    <0|(<0|T(φ1φ2)|0>)φ3|0>

    =<0|T(φ1φ2)|0><0|φ3|0>

    So you're saying one can always pull out the <0|T(φ1φ2)|0> in this way?
     
  5. Feb 25, 2010 #4

    SpectraCat

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    Yes ... it is essentially a constant, and can be treated like any other constant. You are right to be cautious though, since with expressions Dirac notation you need to be conscientious about re-ordering bra's and ket's, since it can change the meaning of the expression in general. However in this case you are fine.
     
  6. Feb 26, 2010 #5
    Thanks SpectraCat.
     
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