I'm currently following David Tong's notes on QFT. In the section on calculating transition amplitudes using Wick's theorem, he gives an example using a scalar Yukawa theory with real scalar field $\phi$ and complex scalar field $\psi$. At a specific point on p.59 he finds a matrix element of a normal ordered string of operators using the following procedure:

$$<p_1',p_2'|:\psi^\dagger(x_1)\psi(x_1)\psi^\dagger(x_2)\psi(x_2):|p_1,p_2> \\\\=<p_1',p_2'|\psi^\dagger(x_1)\psi^\dagger(x_2)|0><0|\psi(x_1)\psi(x_2)|p_1,p_2>\tag{3.48}$$

I'm having trouble understanding how he went from the first to the second line. I'm sure I could brute force my way through the calculation and get the same thing by writing all the fields in terms of creation and annihilation operators, but I'd like to know if there's an easier/intuitive way he changes the order of the operators and adds a vacuum projection operator to get the result.


1 Answer 1



  1. First recall that schematically $$\psi~\sim~ b+c^{\dagger}, \qquad \psi^{\dagger}~\sim~ b^{\dagger}+c \qquad \text{and} \qquad |p_1,p_2\rangle~\sim~ b^{\dagger}b^{\dagger}|0\rangle,$$ cf. bullet points on p. 54 and eq. (3.45) in Tong's notes.

  2. Next normal-order the creation and annihilation operators in eq. (3.48).

  3. An operator term schematically of the form $\sim b^{\dagger}b^{\dagger}bb$ in the sandwich (3.48) is the only surviving contribution.

  4. Therefore we may assume that the operator ordering in the sandwich (3.48) is $\psi^{\dagger}\psi^{\dagger}\psi\psi$.

  5. We can conveniently make point 3 manifest by inserting the projector$^1$ $|0\rangle\langle0|$ in the middle of $\psi^{\dagger}\psi^{\dagger}\psi\psi$.


$^1$ Here we are assuming that the normalized vacuum state $|0\rangle$ with no particles and no anti-particles is unique up to a phase factor.

  • $\begingroup$ The ordering makes sense now, thanks! Although we're not allowed to just treat the vacuum projector as a completeness relation, are we? I was under the impression it had to be the integral over momentum of a projection of momentum eigenstates. $\endgroup$ Nov 1, 2023 at 14:19
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 1, 2023 at 14:32

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