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Time ordering in QFT

  1. Dec 18, 2013 #1
    I have asked this question once, but noone seemed to notice it, so I'll try again. In my book the time ordering operator is used to rewrite an operator product:

    U(β,τ)A(τ)U(τ,τ')B(τ')U(τ',0) = T_τ(U(β,0)A(τ)B(τ'))

    To refresh your memories the time ordering operator T_τ orders operators according to time such that:
    T_τ(A(τ)B(τ')) = A(τ)B(τ') for τ>τ' and B(τ')A(τ) for τ'>τ
    And the operator U(t,t') is a unitary operator that propagates a state from t' to t and has the property that:

    I am still unsure how the rewriting is done though. One key ingredient is to use the property of the unitary above to write:
    This way we have:
    And i think the idea is then to insert in the expression and use time-order but I am not sure how to.
  2. jcsd
  3. Dec 18, 2013 #2


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    There is an implicit assumption here that ##\beta>\tau>\tau'>0##.

    Start with
    Then substitute in
    to get
    Now rearrange the operators so that time labels decrease as you go left to right:
    The labels are now in time-order, so the time-ordering symbol can be dropped:
  4. Dec 19, 2013 #3
    hmm okay I thought it was something like that, but I am still unsure though. Which time do you assing to the operator U(t1,t2)? It propagates a state from t1 to t2, so it is not really a function of one time - or am I missing something?
  5. Dec 19, 2013 #4
    Last edited by a moderator: May 6, 2017
  6. Dec 19, 2013 #5


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    You can think of it as a product of many operators at closely spaced times, and break it up as needed; this is what I did above.
  7. Dec 19, 2013 #6
    So I should basically assign to U(t1,t2) a value of time between t1 and t2?
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