A question about a consequence of symmetry in $\phi^4$ theory

Why does the symmetry $\phi→-\phi$ mean that an amplitude can be written as $\alpha+\beta p^2+\gamma p^4+...$ without the odd terms in $p$? I understand that, due to this symmetry, any diagram in $\phi^4$ has an even number of external legs, because otherwise the correlation function of the external fields is zero. So any diagram can be written in the form $$V(p^2) \left( \frac{i}{p^2−m^2} \right)^n$$ where n is even and $V(p^2)$ is the expression for the amplitude without the external legs. Expanding $V(p^2)$ in $p$ will, of course, give only even powers of p, as will the expansion of $i/(p^2−m^2)^n$, but that is true also for odd $n$, corresponding to an odd number of external legs. So where does this symmetry play a role here?

• Note that $p^4$ really means $(p^2)^2$. What you would mean by $p^3$ would be $(p^2)^{3/2}$. The latter is a fractional power of $p^2$. Such terms could never appear from Feynman diagrams. – Prahar Jun 24 '15 at 9:48
• @Prahar why does $p^4$ mean $(p^2)^2$? – DanielSank Jun 24 '15 at 10:08
• Because there is no way of contracting 4 $p^\mu$ other than $p^\mu p_\mu p^\nu p_\nu = (p^2)^2$. – romanovzky Jun 24 '15 at 10:12
• @Prahar I understand that such terms can never appear when the propagator is a function of $p^2$, but, in a book I'm reading, "Quantum Field Theory for the Gifted Amateur", on page 289, the author writes the integral for the Saturn (sunset) diagram as a series with terms involving only even powers of p and gives the $\phi\rightarrow-\phi$ symmetry as the justification for that. I'm trying to understand it from that direction. – Dave Reikher Jun 24 '15 at 17:14
• Hm, I don't think that's the correct justification. Your reasoning seems completely correct. – David Vercauteren Jun 29 '15 at 8:01