A short question, when I am studying QFT-P&S's book, try to use completeness relation (7.2) to expand the two-point correlation function: $$\langle\Omega|\hat T{\phi(x)\phi(y)}|\Omega\rangle\tag{7.3}$$ in to (7.3), P&S say:

the term $\langle\Omega|\phi(x)|\Omega\rangle$ & $\langle\Omega| \phi(y)|\Omega\rangle$(or for spin half field: $\langle\Omega|\psi(x)|\Omega\rangle$ & $\langle\Omega|\bar \psi(y)|\Omega\rangle$) are usually zero by symmetry; for higher-spin field it's zero by Lorentz invariance.

My question is, know how to prove this claim? (both for $\phi$ and $\psi$)

This is the original P&S book: (P. 212)

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2 Answers 2


In the cases that Peskin & Schroeder are referring to, both the action and the path-integral measure have the $\mathbb{Z}_2$ global symmetry $ \phi(x) \mapsto -\phi(x).$

This means that the full quantum theory has that symmetry, and hence so do correlation functions. Therefore $$\left<\Omega\middle\vert \phi(x) \middle\vert\Omega\right> = 0,$$ since it is odd under that $\mathbb{Z}_2$, assuming that the vacuum does not break the symmetry.

  • $\begingroup$ then for $\langle\Omega|\psi(x)|\Omega\rangle$ & $\langle\Omega|\bar \psi(y)|\Omega\rangle$ is the same reason? $\endgroup$ Nov 3, 2022 at 15:23
  • $\begingroup$ @aFishinDiracSea yes $\endgroup$ Nov 3, 2022 at 16:38
  • $\begingroup$ But looks like Peskin & Schroeder say “for higher-spin field it's zero by Lorentz invariance.” It looks like implie us use Lorentz invariance to derive $=0$ for non-zero spin field. $\endgroup$ Nov 4, 2022 at 14:00
  • 1
    $\begingroup$ @aFishinDiracSea this phys.SE answer shows the claim by Lorentz invariance. This is the formally correct way to do it, if you assume that $\bar{\psi}(x)$ is independent of $\psi(x)$. But if you relate them by complex conjugation (decorated appropriately with $\gamma$ matrices and transposition) my answer holds. $\endgroup$ Nov 4, 2022 at 14:17
  • $\begingroup$ thanks a lot! is very helpfull! $\endgroup$ Nov 4, 2022 at 14:41

Let us mention for completeness that even if there is no symmetry to ensure the vanishing of the 1-point functions $\langle\phi\rangle$, the vanishing $\langle\phi\rangle=0$ is often imposed as a renormalization condition, cf. e.g. Ref. 1 and my Phys.SE answer here.


  1. M. Srednicki, QFT, 2007; eq. (9.2). A prepublication draft PDF file is available here.

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