Please see this question for a clear background of the notation I use. My issue is that I want to use Wick's theorem to calculate the amplitude of meson $\psi(p_1)\psi(p_2)\rightarrow\psi(p_1')\psi(p_2')$ scattering. I can quickly get to the point where I need to evaluate:

$\langle p_1',p_2'|:\psi_1^\dagger\psi_1\psi_2^\dagger\psi_2:|p_1,p_2\rangle$

In my notes, the next step is

$\langle p_1',p_2'|:\psi_1^\dagger\psi_1\psi_2^\dagger\psi_2:|p_1,p_2\rangle=\langle p_1',p_2'|\psi_1^\dagger\psi_2^\dagger|0\rangle\langle0|\psi_1\psi_2|p_1,p_2\rangle$

where he seems to have changed orders of the fields and inserted $|0\rangle\langle0|$, $|0\rangle$ being the vacuum of the free theory. Can anyone explain what this particular step is? From thereafter the derivation is pretty clear to me


2 Answers 2


You have to put the first term (which in not in normal order, hence the dots) in normal order. The vacuum mapping is the identity matrix so you simple CAN put it.

  • 4
    $\begingroup$ Hi John, can you expand on your answer a bit? $\endgroup$ Dec 15, 2013 at 23:54
  • $\begingroup$ You insert a complete set of states (it includes all multi-particle states) -- however, only the vacuum state contributes. You need to verify this. $\endgroup$
    – suresh
    May 16, 2014 at 23:38
  • 1
    $\begingroup$ Could you provide a hint on how to verify? $\endgroup$
    – user37222
    Mar 31, 2015 at 10:46

Expand the normal order: $$ :\psi^\dagger_1\psi_1\psi^\dagger_2\psi_2:=\psi^\dagger_1\psi^\dagger_2\psi_1\psi_2. $$ (creator operator before annihilation operator). Then insert unit operator $I=|0\rangle\langle0|$ between them. Meaning of this expression is that, the total amplitude include two amplitudes. One is for annihilating initial particles of momentums $p_1, p_2$ to vacuum state, and one is for creating final particle of momentum $p'_1,p'_2$ from vacuum state. Due to the rule of probability multiplication, this must be a product.


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