Problem with Wick's theorem at first order

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SUMMARY

The discussion centers on the application of Wick's theorem in the context of second quantization, specifically regarding the S-matrix expansion with an interaction Hamiltonian \( H_I(x) \). The user raises a critical point about the assumption that \( \overline{\psi}(x)\psi(x) = T[ \overline{\psi}(x)\psi(x) ] \) when evaluating the first term of \( S - 1 \). It is established that Wick's theorem is not defined for the same spacetime points, leading to potential misunderstandings in calculations. The resolution involves recognizing that the product of fields at the same point leads to ultraviolet divergences, which can be managed in free field theory but not in interacting cases.

PREREQUISITES
  • Understanding of second quantization principles
  • Familiarity with the S-matrix formalism
  • Knowledge of Wick's theorem and time-ordering operators
  • Basic concepts of quantum field theory, particularly QED
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  • Study the implications of ultraviolet divergences in quantum field theory
  • Learn about Operator Product Expansion (OPE) and its applications
  • Explore the differences between free and interacting field theories
  • Investigate the role of time-ordering in quantum electrodynamics (QED)
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Quantum field theorists, physicists working with quantum electrodynamics, and students seeking to deepen their understanding of Wick's theorem and its implications in particle physics.

GQuinta
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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 H_I(x) then I have

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) ] + ...

where the T operator is unnecessary in the first term. Now, if I choose a \overline{\psi}(x)\psi(x) theory for example, the first term gives some contributions which I can calculate most easily by doing the expansion \psi(x) = \psi^+(x) + \psi^-(x), 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

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

in the first term of S-1 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 H_I(x) 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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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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