No. With only $\lambda_0\phi^6$, the free energy has an exact <Z2 symmetry> $\phi\mapsto-\phi$. Integrating out fast modes and rescaling preserve that symmetry, whereas $\phi^5$ is odd. Therefore no five-point vertex and no correction to $\gamma(\zeta)$ can be generated at any order in $\lambda_0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 2 / e / ii / Solution

No. Even when the two masses differ, the free energy retains an independent <Z2 symmetry> for each component: $\phi_1\mapsto-\phi_1$ and $\phi_2\mapsto-\phi_2$. The bilinear $\phi_1\phi_2$ is odd under either transformation, so integrating out modes cannot generate it.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 3 / b / iii / Solution

{c}
{title2=$\mathbb Z_2$ symmetry}

A Z2 symmetry is generated by an operation whose square is the identity. For a real scalar field, the sign symmetry $\phi\mapsto-\phi$ forbids every odd power of $\phi$.


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 Incoming links: Z2 symmetry