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.

Solution

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

 Solution

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