$$\langle0\lvert T\hat\phi(x_1)\hat\phi(x_2)\hat\phi(x_3)\rvert0\rangle$$

I'm looking for an intuition to it, if not an actual interpretation. Otherwise, I know how to get the result 0 using Wick's theorem. The same question applies to any n-point function with odd n.

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    $\begingroup$ Are you assuming that the cubic interaction vanishes? Are you assuming a $\phi\to -\phi$ symmetry? $\endgroup$
    – Qmechanic
    Commented Jan 30, 2021 at 11:30
  • $\begingroup$ I'm only thinking that since we only have singly contracted terms, there'll aways be a non-contracted field whose expectation value vanishes. $\endgroup$ Commented Jan 30, 2021 at 11:37

2 Answers 2


You have not specified what theory you are working with, but I am assuming you are working with free scalar field theory (possibly including a $\phi^4$ interaction term).

Intuition is a bit hard to define, but perhaps the easiest way to see this is that your Lagrangian is invariant under $\phi\rightarrow -\phi$ (a $\mathbb{Z}_2$ symmetry), so the field $\phi$ and the field $-\phi$ are physically indistinguishable.

Thus, any product which includes an odd number of fields must vanish.

The three-point function can be interpreted as "starting from vacuum, what's the probability of creating three particles and ending up back in the vacuum?". If your particles have charge, then you cannot create an odd number of them whilst conserving charge. Here the charge is a $\mathbb{Z}_2$ quantity rather than a U(1) electric charge (as it would be in e.g. complex scalar field theory), which may be a bit harder for intuition but the math works the same way.

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    $\begingroup$ Nice answer! Just a small suggestion: rather than say that $\phi$ and $-\phi$ are the same (which I am not sure about), perhaps it is better to say that the $Z_2$ symmetry means that physical observables cannot depend on the sign of $\phi$. $\endgroup$
    – Andrew
    Commented Jan 30, 2021 at 14:31
  • $\begingroup$ Yes, it's indeed free scalar field theory. Thank you, that helps. $\endgroup$ Commented Jan 30, 2021 at 14:40
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    $\begingroup$ Thanks for the suggestion @Andrew, I have edited the answer accordingly $\endgroup$
    – G.Lang
    Commented Jan 30, 2021 at 15:28

Because you average over a state with a definite number of particles, while the field operator changes the number of particles by 1. This is of course specific to how youd efine the averaging (i.e., your ground state in this case) and the field operators. Superconductivity and superfluidity are the examples where the ground state does not have a finite number of particles, so odd averages like $\langle\phi(x)\rangle$ do not necessarily vanish.


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