2
$\begingroup$

In Matthew D.Schwartz book, 'Quantum Field Theory and the Standard Model' page 227/8 he talks about Feynman diagrams in QED, specifically $e^{-}e^{-} \rightarrow e^{-}e^{-}$ u and t channel tree level scattering amplitudes.

He then proceeds to say that these two diagrams represent contriubtions to the fourier transform of the greens function $\langle\Omega| T \psi(x_1)\bar\psi(x_3)\psi(x_2)\bar\psi(x_4)|\Omega\rangle $ where $x_1,x_2,x_3,x_4$ correspond to the two incoming electrons and two outgoing electrons respectively.

My confusion comes from the fact that $\psi(x)$ annihilates an electron/ creates a positron. With $t_1<t_3$ and $t_2 <t_4$ then surely the bars are on the wrong spinors; to me this looks like positron- positron scattering.

Edit: I have found a source that has the LSZ reduction formula for fermions (http://fma.if.usp.br/~burdman/QFT1/lecture_14.pdf). Page 4 it derives the LSZ reduction formula for $e^{-}e^{-} \rightarrow e^{-}e^{-}$. Ignoring the pre-factors and post-factors (which, to my understanding, forces the contribution of the greens function not relevant to the scattering problem to be zero?) it shows the relevant correlation function for the problem does indeed have the bars on the opposite spinors (formula 14.10). So this is a mistake in the Schwartz book?

$\endgroup$
1
  • $\begingroup$ I have the exact same question coming from Schwartz. FYI, I checked the printing corrections and couldn't find any that relates to this issue. $\endgroup$
    – dor00012
    Feb 3, 2023 at 17:28

1 Answer 1

1
$\begingroup$

The LSZ reduction formula is a method to calculate S-matrix elements (the scattering amplitudes) from the time-ordered correlation functions of a quantum field theory. The time-ordering operator $T$ manhandle the fields so that fields at a later time are at the left of fields at an earlier time.

As the LSZ formula performs an exhaustive integration over the space-time variables $x^\mu$, all the combinations are covered.

$\endgroup$
1
  • $\begingroup$ I don't think this answers the question. You are right of course, but the integration you've mentioned doesn't change the fact that created (bar) spinors are for the initial state and annihilated (no bar) spinors are for the final state. In Schwartz there seems to be a contradiction. In other words, the position of the states (which is integrated) is not related to whether or not they are initial or final. $\endgroup$
    – dor00012
    Feb 3, 2023 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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