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  Questions about the equilvalent form of Wick's theorem ?

+ 1 like - 0 dislike

I have met Wick's theorem first in this book fundamentals of many-body physics (by Wolfgang Nolting) when talk about the perturbation expansion of zero temperature Green's function. Later in the perturbation expansion of finite temperature Green's function I wolud meet this theorem again. In both theories the Wick's theorem is only considered as a operator identity;by which you can transform the time-order product into sums of contraction and normal product.You can see the theorem has nothing to do with the system Hamiltonian that you are caring,no matter the Hamiltonian is quartic (interacting) or quadratic (noninteracting).

Then in this paper Expansion of nonequilibrium Green's functions (by Mathias Wagner ) the author told a general Wick's theorem proved by Danielewicz (Danielewicz,Ann,Phys.152,239(1984)). The general Wick's theorem guarantee the equivalence of the two statements:

  1. Wick's theorem holds exactly (in the form of operator identity);
  2. The operators to be averaged are noninteracting and the initial density matrix is a one-particle density matrix.

I have struggled with the Danielewicz's paper but I still cannot figure out the deep connection between these two statements,can anyboy help me to work out the proof of Danielewicz?Any supporting materials to his proof will also be appreciated.

asked Sep 8, 2017 in Theoretical Physics by Kohn (15 points) [ no revision ]

I believe the point about Wick's theorem qua operator identity among creation and annihilation operators versus Wick's theorem qua n-point correlation functions of Gaussian distributions is that the latter come from always quadratic Hamiltonians, which can be quantized using creation and annihilation operators, whereby the correlation function identities follow from the operator identities.

The perturbative expansion is of the interacting theory in terms of the free theory; thus Wick's theorem is always applied to a product of free field operators only - even when using it for the interacting case. 

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