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Proof of Wick's Theorem for 3 fields

  1. Jul 4, 2017 #1
    1. The problem statement, all variables and given/known data

    Question attached:

    wick3question.png
    2. Relevant equations

    Using the result from two fields that

    ## T(\phi(x) \phi(y))= : \phi(x) \phi(y) : + G(x-y)##

    Where ##G(x-y) = [\phi(x)^+,\phi(y)^-] ##

    ## : ## denotes normal ordered

    and ##\phi(x)^+ ## is the annihilation operator part , and ## \phi(x)^- ## is the creation operator part.


    3. The attempt at a solution

    Assume non-trivially that ## z^0 > x^0 > y^0 ##

    Then ## T(\phi(z),\phi(x),\phi(y)) = \phi(z) T(\phi(x) \phi(y)) ##

    ##=(\phi(z)^+ + \phi(z)^-) T (\phi(x),\phi(y)) ##

    Since ##\phi(z)^-## is already normal ordered, look at the term multiplied by ##\phi(z)^+##:

    ##=\phi(z)G(x-y) + \phi(z)^+:\phi(x)\phi(y): ## (1)

    The term to be concerned with from

    ##\phi(z)^+:\phi(x)\phi(y):## is ##\phi(z)^+\phi(x)^-\phi(y)^-=\phi(x)^-\phi(z)^+\phi(y)^- +[\phi(z)^+,\phi(x)^-]\phi(y)^-= \phi(x)^-\phi(y)^-\phi(z)^+ +\phi(x)^-[\phi(z)^+,\phi(y)^-] + [ \phi(z)^+,\phi(x)^-]\phi(y)^-##

    So putting this with (1) I have

    ## T(\phi(z),\phi(x),\phi(y)) = : \phi(z) (\phi(x) \phi(y)): + [ \phi(z)^+,\phi(x)^-]\phi(y)^- +\phi(x)^-[\phi(z)^+,\phi(y)^-] +\phi(z)(G(x-y)) ##

    So looking at the solution the last term is right, but the other propagator terms , should have a factor of both the creation and annihilation parts of the field, ##\phi(y)^- + \phi(y)^+ ## and ## \phi(x)^+ + \phi(x)^- ## , multiplying the propagator? and should be multiplying the RHS of the propagator rather than the LHS ? I'm not sure what I have done wrong...

    Many thanks in advance.
     
  2. jcsd
  3. Jul 9, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Jul 10, 2017 #3
    got empty pockets again hun
     
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