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Normal ordering of bosonic commuators

  1. Oct 29, 2014 #1
    I am trying to understand wick's theorem and normal ordering mostly from Peskin and Schroeder. Now I have this problem with how normal ordering is defined. It seems to me that if you take the normal ordering of a commutator it should always be zero.

    Here is what I understand normal ordering to be. If there is some operator [itex]\hat{A}[/itex] which is a product of bosonic operators [itex]a_p[/itex] and [itex] a_p^{\dagger}[/itex] , then normal ordering of of [itex] \hat{A}[/itex] is [itex] N(\hat{A})[/itex] where the creation operators are all moved to the left and the annihilation operators are all moved to the right. This essentially would mean that inside a normal ordering all operators can be commuted. For example,
    [tex] N(a_pa_k^{\dagger}) = N(a_k^{\dagger}a_p) = a_k^{\dagger}a_p [/tex]

    Now taking the normal ordering of a commutator,
    [tex] N([a_p,a_k^{\dagger}]) = N(a_pa_k^{\dagger} - a_k^{\dagger}a_p) = N(a_pa_k^{\dagger}) -N(a_k^{\dagger}a_p) = 0 [/tex]

    But if I had used the fact that the commutator of [itex] [a_p,a_k^{\dagger}] = \delta^{(3)}(p-k) [/itex] which is a c-number, then

    [tex] N([a_p,a_k^{\dagger}]) = N(\delta^{(3)}(p-k)) = \delta^{(3)}(p-k) [/tex]

    So there is a contradiction. Can someone explain to me what is the right way to think about normal ordering?
  2. jcsd
  3. Oct 29, 2014 #2


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    Science Advisor

    Normal ordering is not linear:
    [tex]N(a a^\dagger) = N(1 + a^\dagger a) = N(1) + N(a^\dagger a) = 1 + a^\dagger a \neq a^\dagger a[/tex]
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