# Why do degenerate ground states mix in perturbation theory?

I'm interested in the Ising model and I'm reading ahead a little about it. I picked up this article that states that the Ising model has degenerate ground states (all up and all down), which is good so far, but then he says that these degenerate states mix in the perturbative expansion of the state (page 468), which leaves me wondering why. He also uses this to justify the thermodynamic limit of $$N\to\infty$$.

• Am I understanding it right, that these states mix in the $$N$$th order of the expansion, so taking the limit means we don't have finite mixing?

• If I am right, why do they mix? And why at $$N$$?

Unfortunately I only know basic perturbation theory, but I can look at references and things if you have a suggestion. Thank you!

• There's nothing deep here. It's just the observation that, analogous to (14.33), the perturbation term $\hat{V}$ flips one spin, so to connect the all-up to the all-down state, you need a product of $N_s$ terms. This is just to argue that, starting out in one symmetry-broken state, one can formally evolve to the other state by this Hamiltonian, but not in practice because $N_s$ is larger than any reasonable time scale. Mar 8 '21 at 5:25
• @AronBeekman Oh, ok, so its just due to the spin flipping that happens in the expansion? Mar 8 '21 at 5:53
• Yes. You should convince yourself there's a non-zero matrix element $\langle \text{all down} | \hat{V}^{N_s} | \text{all up} \rangle$. That's what Fradkin means by mixing. Mar 8 '21 at 7:18