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.

  • $\begingroup$ Why are you talking about the Wick theorem, since it seems that your problem is with the Time ordering operator. And what different time ordering operator are you talking about ? What minus sign ? $\endgroup$
    – Adam
    Commented Jan 31, 2014 at 23:14
  • $\begingroup$ because my problem is about using Wick's theorem with two fermionic fields at the same spacetime point in the first order term of S-1. In particular, why can we consider the time ordering operator to be there so we can use Wick's theorem. And the time ordering operator for fermionic fields has a minus sign in its definition if you use dirac spinors... but just in case I wasn't clear enough, it's this minus sign $$ T[\overline{\psi}(x) \psi(x) ] = \theta(x-y) \overline{\psi}(x)\psi(y) - \theta(y-x) \psi(y) \overline{\psi}(x)$$ I was talking about. $\endgroup$
    – user38680
    Commented Jan 31, 2014 at 23:25

1 Answer 1


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.

  • 4
    $\begingroup$ Exactly, that's why I think that in the case $H_I(t) = \overline{\psi}(x)\psi(x)$ the first term cannot be calculated using Wick's theorem. Even in $\phi^4$ theory the first term has the product $\phi(x)\phi(x)\phi(x)\phi(x)$, and people use Wick's theorem nontheless, using the result where the spacetime points are all different and then put all spacetime points equal, giving terms like $D_F(x-x)$ that would not appear if you simply made the substitution $\phi(x) = \phi(x)^+ + \phi(x)^-$ and simplified by hand. My question is exactly why does everyone use Wick's theorem in the first term. $\endgroup$
    – user38680
    Commented Feb 1, 2014 at 1:49
  • $\begingroup$ The interaction Hamiltonian $H_I(t)$ could in principle be re-arranged in normal-ordered form. $\endgroup$
    – Qmechanic
    Commented Feb 1, 2014 at 2:02
  • $\begingroup$ It is best to work with the normal ordered form of the interaction hamiltonian. Then, you need not include any self-contractions (contractions involving the same vertex) in the Wick expansion. $\endgroup$
    – suresh
    Commented Feb 1, 2014 at 2:05

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