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Wick's Theorem for free fields only

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  1. Jan 16, 2015 #1
    I'm not quite following why Wick's Theorem only applies to free fields. What part of the argument depends on a free field assumption?
     
  2. jcsd
  3. Jan 16, 2015 #2

    atyy

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  4. Jan 16, 2015 #3
    It can be used for that application, but the actual statement of the theorem just relates time-ordered operator products to normal-ordered operator products and contractions. I'm not seeing why the operators in question have to be free-field.
     
  5. Jan 16, 2015 #4

    vanhees71

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    You cannot introduce even a definition of normal ordering without a mode decomposition of a field in terms of annihilation and creation operators with respect to some single-particle basis. Such a decomposition only exists for free fields and that's why one uses the interaction picture.

    In addition, it's important to note that Wick's theorem is valid if and only if the initial state is given by a statistical operator of the form ##\hat{\rho}=\exp(-\hat{A})##, where ##\hat{A}## is an appropriate single-particle operator. The special case of vacuum QFT, i.e., when ##\hat{\rho}=|\Omega \rangle \langle \Omega |## can be taken as the zero-temperature limit and zero-chemical potential(s) limit of the grand-canonical ensemble, ##\hat{\rho}=\exp(-\beta \hat{H})/Z## with ##Z=\mathrm{Tr} \exp(-\beta{\hat{H}})##, ##\beta=1/(k_B T)##.

    For a very good introduction on these issues, see

    Danielewicz, P.: Quantum Theory of Nonequilibrium Processes II. Application to Nuclear Collisions, Ann. Phys. 152, 305–326, 1984
    http://dx.doi.org/10.1016/0003-4916(84)90093-9 [Broken]
     
    Last edited by a moderator: May 7, 2017
  6. Jan 18, 2015 #5
    Thank you, that puts me on the trail.
     
  7. Jan 19, 2015 #6

    vanhees71

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    Yes, this is the best source to start learning about the Schwinger-Keldysh real-time formalism (NB: If Keldysh is expected to be in the audience listening to your talk, better call it the "Keldysh real-time formalism" :-)).
     
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