Hey there, first post here!(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums - The Fusion of Science and Community**

# Problem with Wick's theorem at first order

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Problem with Wick's theorem at first order

Loading...

**Physics Forums - The Fusion of Science and Community**