How to compute the absolute simplest case of operator product expansion

Question

The operator product expansion seems to be a quite useful tool. In an attempt to find a full concise complete computation, involving deriving the coefficients, and introducing the taxonomy associated with using this tool, I have hit a wall. The gentle starting point  of  A(x)B(y) =sum C * (new Operator) is usually stated and the rest assumed to be common knowledge.  If we consider the case of a simple free field (composite field), how does one  find the coefficients and what is this new operator and what is its physical significance.  I think I tend to fair well when things are laid out fully and explicitly, not abstractly.   I think may be the definition and example  with energy momentum tensors shown in most conformal field theory textbooks is not what I am looking for.  I am looking for hopefully an explicit very  simple free quantum field theory showing fully how this is done.

Comments

Hi fake-student-silicon, we do have LaTex on PO, just in case you did not know ...

I once had a similar question

http://www.physicsoverflow.org/6878/systematic-approach-calculate-individual-operator-expansion?show=6878#q6878

but on that thread there is not yet an explicit calculation of a specifi example in the free field case.

Maybe you can make it a bit more clear already in the title, that you want to see the explicit calculation of a specific example?

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For free fields one can simply apply Wick's theorem to products of normally ordered composite operators like $:\phi^2(x)::\phi^2(y):=4[\phi(x)\phi(y)]:\phi(x)\phi(y):+2[\phi(x)\phi(y)]^2$, where I use "[ ]" to denote contractions.

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Hey guys thanks for the comments. I have seen  how it is done in in CFT, it seems they compute the variation of phi, and then compare it with [Q, A], this gives another variation. Then you should be able to read off OPE. Can someone actually show me all the steps, and can someone please help me with how this is done for like a simple scalar field theory.  @Dilaton  @Jia Yiyang