# Operator Product Expansion in Massless 2D QED

In Peskin & Schroeder chapter 19 page 656, where the axial current anomaly of massless 2D QED is discussed, the authors go from: $$\bar\psi(x+\varepsilon/2)\Gamma(x)\psi(x-\varepsilon/2)\tag{19.25}$$ (where $\Gamma(x)$ is some operator) to: $$\bar\psi(x+\varepsilon/2)\Gamma(x)\psi(x-\varepsilon/2) \tag{19.27}$$ (where now the two Fermionic fields are contracted) to: $$Tr\left[\Gamma(x)S_F(\varepsilon)\right] \tag{19.27}$$ (where $S_F(x)$ is the Fermion propagator between a spacetime interval $x$)

I really don't understand these transitions and would appreciate any help with how to do them. In particular:

1) Is it implicitly understood (from the very beginning of this derivation) that the axial current is in fact time-ordered (so that we can employ Wick's theorem) and always assume it operates on the vacuum (so that normal ordered terms vanish)?

2) Why does the contraction of the two Fermion fields over the $\Gamma$ operator lead to a trace, as if we had a loop?

• Observe that the trace is invariant under cyclic permutation, this should give you the answer to your second question. – ACuriousMind Jul 13 '14 at 13:11
• Comments to the question (v1): The first and second eqs. look the same. The second and third eqs. numbers are the same. Is that on purpose? – Qmechanic Jul 14 '14 at 14:43
• @Qmechanic, thanks for your comment. The first and second equation look the same because I didn't know how to denote contraction in MathJax, but I wrote in words that the Fermion fields are (suddenly) contracted in the second equation. The third equation indeed has the same number as the second equation, in P&S, because that equation has two equal signs, but there is a transition happening inside that same-numbered equation. – PPR Jul 14 '14 at 16:51
• about the first question: there are 3 equivalent answers. a) the current operator is a local operator defined at one single point of spacetime only, so it doesn't really matter whether you look at time ordered points, irrespectively of your regularization procedure. b) the time ordering does indeed appear in the operator product expansion that you mention in the title. It is thus ok to select the most divergent contribution taking time ordered contractions. c) all green-functions, time ordered or not, share the same divergent behavior (since the difference is in the $i\epsilon$ prescription) – TwoBs Jul 18 '14 at 12:14

$$(S_F)_{\alpha\beta}(x-y) = \langle T\{\psi_{\alpha}(x)\bar\psi_{\beta}(y)\} \rangle$$
$$\langle T\{\bar\psi(x)\Gamma\psi(y)\} \rangle = \Gamma_{\alpha\beta} \langle T\{\bar\psi_{\alpha}(x)\psi_{\beta}(y)\} \rangle = -\Gamma_{\alpha\beta} (S_F)_{\beta\alpha}(x-y) = -\mathrm{tr} [\Gamma S_F(x-y)]$$