2

On page 212 footnote 18 says:

Remember that, in a theory with complex or Grassmann fields, only contractions $\sim \langle \bar{\psi}\psi\rangle_0$ exist, i.e., there is a total of $n!$ distinct contributions to a contraction $\langle \bar{\psi} \psi \cdots \psi \rangle_0$ of $2n$ field operators.

My question is why only these contractions $\sim \langle \bar{\psi}\psi\rangle_0$ exist. I was thinking it is related to the fact that if $c$ is a Grassmann number, then $c^2 = 0$, but the footnote says this applies both with complex or Grassmann fields.

For reference: $\langle \cdots \rangle_0 \equiv \frac{\int D \phi e^{-S_0 \, [\phi]} \; \; \; ( \cdots )}{\int D \phi e^{-S_0 \, [\phi]}}$,

the functional average over the Gaussian action $S_0 \equiv S |_{g = 0} \; $.

Qmechanic
  • 201,751
Physics Enthusiast
  • 217

1 Answers1

4

This is because of a global $U(1)$ phase symmetry $$\phi^{\prime}~=~e^{i\theta}\phi, \qquad \bar{\phi}^{\prime}~=~e^{-i\theta}\bar{\phi}, $$ for the complex fields. Only correlators $$\langle F[\phi,\bar{\phi}] \rangle_0~=~\langle F[\phi^{\prime},\bar{\phi}^{\prime}] \rangle_0~=~e^{iq\theta}\langle F[\phi,\bar{\phi}] \rangle_0$$ with no net $U(1)$ charge $q$ can be non-zero, cf. e.g. my related Phys.SE answer here.

Qmechanic
  • 201,751