Why the symmetry of $\phi^4$ excludes the odd diagams?

Question

I have a follow-up question from this post: Suppose $$ L\supset \lambda\phi^4 $$

This term is invariant under $\phi\rightarrow-\phi$, Peskin and Schroeder (p.323) said this implies that all amplitudes with an odd number of external legs will vanish. Therefore, the superficial divergent constant could be found as $\omega = 4-b$, where $b = 0,2,4$.

However, I don't quite understand why those amplitudes vanish in the first place. In my last example (linked above), we have
$$ L\supset g\phi^3 $$ (where $L$ is in 6 dimensions), we don't have this symmetry, so there are both odd and even diagrams.

Answer 1

This can be viewed through the lens of symmetry or simple combinatorics.

1. States with a definite number of particles can be divided into orthogonal sectors which are respectively even and odd under $\phi\rightarrow -\phi$. These sectors are the eigenspaces of the (idempotent) operator $\hat O : \phi \mapsto -\phi$, with respective eigenvalues $1$ and $-1$. If there are an odd number of external legs, then the incoming and outgoing states belong to different sectors. But since $H_{int} \sim \phi^4$ is invariant under that operation (i.e. it commutes with $\hat O$), then $\hat O H_{int} \hat O = H_{int}$ and so $$\langle f | H_{int}|i\rangle = \langle f|\hat O^2 H_{int} \hat O^2 |i\rangle = (-1) \langle f|\hat O H_{int} \hat O|i\rangle = -\langle f|H_{int} |i\rangle$$ $$\implies \langle f|H_{int} |i\rangle = 0$$ More generally, one can say that if $\hat O$ is a symmetry of the Hamiltonian, then time evolution cannot mix its eigenspaces together (i.e. a state in one eigenspace cannot evolve under $ H_{int}$ into a state in a different eigenspace).

2. The vertices in $\phi^4$ theory are all of degree $4$, meaning that each vertex is connected to four legs. Each internal leg is connected to two vertices, and each external leg is connected to one; convince yourself that this implies that $$4V = 2I + E$$ where $V,I,$ and $E$ are the numbers of vertices, internal legs, and external legs, respectively. As a trivial result, $E = 4V-2I$ is necessarily even.

Neither one of these explanations works for $\phi^3$, because (1) $[\phi^3, \hat O]\neq 0$ and (2) $E=3V-2I$ does not imply that $E$ must be even.

Answer 2

Your typical diagram is the fourier transform of some quantity like $$\langle \Omega|T\phi(x_1)...\phi(x_n)|\Omega\rangle.$$ Under a parity transform $U$ defined via $$U\phi \ U^{-1}=-\phi,$$ the ground state $|\Omega\rangle$ transforms trivially since the Hamiltonian is invariant.

Thus $$\begin{aligned}\langle \Omega|T\phi(x_1)...\phi(x_n)|\Omega\rangle&=\langle \Omega|U^{-1}UT\phi(x_1)U^{-1}U...\phi(x_n)U^{-1}U|\Omega\rangle \\&=(-1)^n\langle \Omega|T\phi(x_1)...\phi(x_n)|\Omega\rangle\end{aligned},$$ which means that all diagrams with odd $n$ vanish.

Answer 3

The point is that there is actually at least two different $\phi^4$ theories:

1. one $\phi^4$ theory with $\mathbb{Z}_2$-symmetry excluding odd terms.
2. another $\phi^4$ theory without $\mathbb{Z}_2$-symmetry containing all possible terms.