2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – AccidentalFourierTransform Dec 14 '16 at 19:26
  • $\begingroup$ I was about to ask the same question. Appreciate a good answer too! $\endgroup$ – user56963 Dec 14 '16 at 19:32
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – AccidentalFourierTransform Dec 14 '16 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) $\endgroup$ – yoric Dec 14 '16 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. $\endgroup$ – yoric Dec 17 '16 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.