# Spontaneous symmetry breaking confusion

I was reading the spontaneous symmetry breaking in the case of double well . And I come to the following statement

When there is a degeneracy, the ground state no longer has to obey the symmetry of the Hamiltonian.

But I don't understand what is meant by "ground state no longer symmetric under Hamiltonian". Could someone explain it in a lighter manner (I mean without using too much QFT and advanced condenser matter stuffs). The example I read was from Sakurai

Let $$\{g\}$$ be the set of symmetries of the Hamiltonian. That is, there is a unitary operator $$U(g)$$ such that $$U^\dagger(g)HU(g)=H$$ for all $$g$$. The result we want comes from representation theory and can be stated as follows:

Each degenerate eigenspace of the Hamiltonian transforms within itself under the symmetries of the Hamiltonian. (Assuming no accidental degeneracy)

What This Means:

If a set of states $$\{|\psi_i\rangle\}_{i=1}^n$$ are degenerate, then for any $$g$$ and $$|\psi_i\rangle$$, we have $$U(g)|\psi_i\rangle = \sum_{j=1}^n a_j |\psi_j\rangle$$ Notice that the resulting superposition is of course still a state with the same energy since all the states which compose it have the same energy. So when an energy eigenstate $$|\psi\rangle$$ is non-degenerate, this means that up to a global phase, we have $$U(g)|\psi\rangle =|\psi\rangle$$ for all $$g$$. Clearly then, a non-degenerate state is left invariant (transformed into itself) by any symmetry operation. It has the full symmetry of the Hamiltonian.

When two states $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ are degenerate, it means that the symmetry operations $$\{g\}$$ can mix these two states. That is, $$U(g)|\psi_1\rangle = \alpha |\psi_1\rangle + \beta|\psi_2\rangle$$ for some $$\alpha,\beta\in\mathbb{C}$$ and similarly for $$|\psi_2\rangle$$. So each of the two states are not left invariant by themselves. They do not independently possess the full symmetry of the Hamiltonian, but are only symmetric as a doublet. This similarly applies for high degenerate spaces.

If the Hamiltonian possesses some symmetry - parity, in the case of the double well - its natural to ask if the eigenvectors of the Hamiltonian have the same symmetry. If $$PHP^\dagger = H$$ and $$H|\psi\rangle = E|\psi\rangle$$, can we conclude that $$P|\psi\rangle = \mu|\psi\rangle$$ for some $$\mu$$?

In your case, this is asking whether eigenstates of a parity-invariant Hamiltonian have definite parity themselves, and the answer is generically no. It's not hard to show that $$P|\psi\rangle$$ is an eigenstate of $$H$$ with eigenvalue $$E$$: $$HP|\psi\rangle = (PHP^\dagger)P|\psi\rangle = PH|\psi\rangle = E(P|\psi\rangle)$$ If the spectrum of $$H$$ is non-degenerate, then there is only one linearly independent eigenstate for each eigenvalue, which would indicate that $$P|\psi\rangle = \mu |\psi\rangle$$ for some constant $$\mu$$. However, the same is not true if there is degeneracy. In that case, all we can conclude is that $$|\psi\rangle$$ and $$P|\psi\rangle$$ belong to the same eigenspace of $$H$$.

That being said, $$[H,P]=0$$ means that it's possible to construct a basis of simultaneous eigenstates of $$H$$ and $$P$$. That means we can make an energy eigenbasis out of states of definite parity. But it's important to remember that a general energy eigenstate does not have definite parity unless the spectrum of $$H$$ is non-degenerate.