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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

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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.

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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.

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