Problem with Wick's theorem at first order

by GQuinta
Tags: order, theorem, wick
GQuinta is offline
Jan29-14, 11:28 AM
P: 1
Hey there, first post here!

I've been struggling with a detail in Second Quantization which I really need to clear out of my head. If I expand the S-matrix of a theory with an interaction Hamiltonian [itex] H_I(x) [/itex] then I have

[itex] S - 1= \int^{+\infty}_{-\infty} d^4 x H_I(x) + \int^{+\infty}_{-\infty} \int^{+\infty}_{-\infty} d^4 x d^4 y T[ H_I(x) H_I(y) ] + ... [/itex]

where the T operator is unnecessary in the first term. Now, if I choose a [itex] \overline{\psi}(x)\psi(x) [/itex] theory for example, the first term gives some contributions which I can calculate most easily by doing the expansion [itex] \psi(x) = \psi^+(x) + \psi^-(x) [/itex], which is the essence of Wick's theorem. I know the contributions will be trivial, but the point is Wick's theorem is not defined for the same spacetime points, so I can't understand why everywhere I look people assume implicitly that

[itex] \overline{\psi}(x)\psi(x) = T[ \overline{\psi}(x)\psi(x) ] [/itex]

in the first term of [itex] S-1[/itex] if the time ordering operators aren't even the same, since this one has a minus sign in its definition. As far as I can see, what everyone says is that since the T operator in the first term can be there, then when substituting [itex] H_I(x) [/itex] we simply retain the operator and use Wick's theorem like the spacetime points were different and impose x=y at the end, but this doesn't make any sense since the operators T are different.

Sorry for the long text... Thanks!
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andrien is offline
Jan30-14, 07:12 AM
P: 987
Well,the answer is quite subtle.You cannot take the product of field evaluated at same space time point,it leads to very short distance ultraviolet problem.It is however possible to evaluate it in free field theory case,by defining the time ordered product in a special way taking out the divergence part(subtracting it).It is not possible to do it in interacting field case,in which you have to absorb this short distance information into some constant which multiply with the finite(ultraviolet finite) part.You actually have an expansion of this sort called Operator product expansion.However if you are bothered with the first term here,then good news is that for a physically possible process where it arises for example in QED,it vanishes between the initial and final possible states.

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