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
$$ S – 1= \int^{+\infty}_{-\infty} d^4x\, H_I(x) + \int^{+\infty}_{-\infty} \int^{+\infty}_{-\infty} d^4 x\, d^4 y\, T[ H_I(x) \, H_I(y) ] + … $$
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 in this example will be trivial, but the point is Wick's theorem is not defined for the same spacetime points and everywhere I 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. Basically it is assumed implicitly that
$$ \overline{\psi}(x)\,\psi(x) = T[ \overline{\psi}(x)\,\psi(x) ] $$
in the first term of $ S-1 $, but the time ordering operators aren't even the same, since this one has a minus sign in its definition. The point is, the first term will always be
$$ \int^{+\infty}_{-\infty} d^4x\, H_I(x) $$
independently if I chose to put there the T operator (without the minus sign) or not, so for fermions I should have $\overline{\psi}(x)\,\psi(x)$ there and not $T[ \overline{\psi}(x)\,\psi(x) ]$, since they are not the same as far as I can see.
Best Answer
The main point is that in the context of the Dyson series it is by definition implicitly assumed that the time ordering operator $T$ does not act/operate/re-order operators inside the interaction Hamiltonian $H_I(t)$. In plain English: It does not 'see' what operators that $H_I(t)$ consists of. It only re-orders between different appearances of the interaction Hamiltonian, i.e.
$$\tag{1}T [H_I(t_1)]~=~H_I(t_1),$$ $$\tag{2}T [H_I(t_1)H_I(t_2)]~=~\Theta(t_1-t_2) H_I(t_1)H_I(t_2)~+~ \Theta(t_2-t_1) H_I(t_2)H_I(t_1),$$ and so forth.