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Vacuum expectation values of combinations of ##a^\dagger## and ##a##

  1. May 23, 2013 #1
    I am slightly confused on how do we calculate vacuum expectation values of product of creation and annihilation operators for bosons, e.g. ##\langle 0| a_{k_1} a^\dagger_{k_2} a_{k_3} a^\dagger_{k_4} |0 \rangle##

    If i commute ##k_3## and ##k_4##:

    $$\langle 0| a_{k_1} a^\dagger_{k_2} a_{k_3} a^\dagger_{k_4} |0 \rangle = \langle 0| a_{k_1} a^\dagger_{k_2} |0 \rangle \delta(k_3-k_4) + \langle 0| a_{k_1} a^\dagger_{k_2} a^\dagger_{k_4} a_{k_3} |0 \rangle $$
    Wouldn't the second term give zero automatically because ## a |0 \rangle = 0## ?
     
  2. jcsd
  3. May 23, 2013 #2

    strangerep

    User Avatar
    Science Advisor

    Yes.
     
  4. May 23, 2013 #3
    thanks.
     
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