# Use of projection operator $|0\rangle\langle 0|$ in a specific example in QFT

I am still a beginner in QFT and I am reading the notes by David Tong. On Page 59 of the notes, in equation (3.48), the author writes $$\langle p'_1,p'_2|\colon\psi^\dagger(x_1)\psi(x_1)\psi^\dagger(x_2)\psi(x_2)\colon|p_1, p_2\rangle\overline{\phi(x_1)\phi(x_2)} \\ = \langle p'_1,p'_2|\psi^\dagger(x_1)\psi^\dagger(x_2)|0\rangle\langle0|\psi(x_1)\psi(x_2)|p_1, p_2\rangle \overline{\phi(x_1)\phi(x_2)}$$

Is the operator $|0\rangle\langle0|$ the identity in this case ? If not, I do not understand how it can be wedged in between the operators in the second line.

• What Tong does makes no sense at all. If I were you I'd give up those notes and read Weigand's, which are way better. – AccidentalFourierTransform Dec 14 '16 at 19:26
• I was about to ask the same question. Appreciate a good answer too! – user56963 Dec 14 '16 at 19:32

I have faced the same problem, and I reached the conclusion that he inserted a completeness of the Fock space, something like $$1 = \sum_{q_1, q_2, \ldots} | q_1 , q_2, \ldots \rangle \langle q_1 , q_2, \ldots | = |0 \rangle \langle 0 | + \sum_{q} | q \rangle \langle q | + \sum_{q,k} | q, k \rangle \langle q ,k | + \cdots$$ between the two couples of operators (normal ordered). Now I think that we can discard all terms but the first because all others result in a non-matching between the number of particles after the action of the creation / annihilation operators.
• $\psi \sim b + c ^\dagger$ and $\psi \sim c + b^\dagger$; the normal ordering pushes the $c$'s and the $b$'s at the right; the $c$'s applied to the two particle state give $0$, whereas the $b$'s survive and that is all you need. (I am not 100% sure) – user139175 Dec 14 '16 at 22:49