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

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.

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.

• Thanks so much for the answer! Is 'sector' the same as 'vector'?
– IGY
Commented Dec 21, 2022 at 19:48
• @IGY No, by orthogonal sectors I mean orthogonal subspaces $U$ and $V$ such that the Fock space can be written $F = U\oplus V$. Commented Dec 21, 2022 at 20:00

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.

• Thanks so much for the answer!! Why the $\phi^3$ term doesn't exclude odd $n$ in this scheme?
– IGY
Commented Dec 21, 2022 at 22:50
• Because the groundstate for $\phi^3$ theory is not invariant under $U$ so the above result doesn't apply. Commented Dec 21, 2022 at 23:01
• Thanks, is it correct if I write $U|\Omega\rangle = -|\Omega\rangle$, and $\langle\Omega|U^{-1} = -\langle\Omega|$ for $\phi^3$? (and that factor becomes $(-1)^{n+2}$?)
– IGY
Commented Dec 21, 2022 at 23:09
• No the groundstate does not transform that way, actually the groundstate doesn't even exist for a purely $\phi^3$ state since the Hamiltonian is not bounded from below. Commented Dec 22, 2022 at 0:56

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.
• Thanks so much for the answer! 1. What do you mean by 'monomial terms'? 2. Are we usually using the first type of $\phi^4$ theory?
– IGY
Commented Dec 21, 2022 at 22:35
• 1. I updated the answer. Commented Dec 21, 2022 at 23:32
• Just to be extra clear: the difference between these two theories is whether we include a $\phi^3$ term or not (and $\phi$, for tadpoles, if you wish...) Commented Dec 21, 2022 at 23:36
• @AccidentalFourierTransform Thank you!!
– IGY
Commented Dec 22, 2022 at 0:06