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.

  • $\begingroup$ 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. $\endgroup$ Dec 14, 2016 at 19:26
  • $\begingroup$ I was about to ask the same question. Appreciate a good answer too! $\endgroup$
    – user56963
    Dec 14, 2016 at 19:32

1 Answer 1


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.

  • $\begingroup$ how did the normal ordering disappear? and how are you so sure that the projector over two particles vanishes? (this would be obvious if it werent for the normal ordering) $\endgroup$ Dec 14, 2016 at 20:21
  • $\begingroup$ $\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) $\endgroup$
    – user139175
    Dec 14, 2016 at 22:49
  • $\begingroup$ For the records, I am now quite confident in explaining that relation by expanding psi and psi dagger in terms of operator b and c, applying the normal ordering and then inserting the completeness relation. If you expand the right hand side, the other terms cancel as well. $\endgroup$
    – user139175
    Dec 17, 2016 at 0:20

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