My understanding
Consider a quantum theory with Hilbert space and Hamiltonian $(\mathcal{H}, H)$. The dynamical symmetry group of $H$ is a $d$-dimensional Lie group $G$ and a projective unitary representation $\pi: G \to U(\mathcal{H})$ such that the induced Lie algebra representation $\pi: \mathfrak{g} \to \mathfrak{gl}(\mathcal{H})$ outputs represented generators $\{\pi(X_i)\}$ that all commute with the Hamiltonian, i.e., $[H, \pi(X_i)] = 0$ for all $i = 1, 2, ..., d$. Hereafter, a dynamical symmetry will be referred to as a symmetry.
Spontaneous symmetry breaking refers to the phenomenon in which there exists a stable state $\lvert \psi \rangle \in \mathcal{H}$ such that \begin{equation} \pi(g)\lvert \psi \rangle \not\sim \lvert \psi \rangle \end{equation} where "$\not\sim$" means not equal modulo a global phase. This state splits $G$ into a stabilizer subgroup \begin{equation} H_{\lvert \psi \rangle} := \{g \in G : \pi(g) \lvert \psi \rangle \sim \lvert \psi \rangle \} \end{equation} and the spontaneously broken symmetries $G/H_{\lvert \psi \rangle}$, which does not generally form a group. The stabilizer is dependent on the stable state under consideration. Recall that group theory tells us \begin{equation} \lvert \psi' \rangle = \pi(g) \lvert \psi \rangle \implies H_{\lvert \psi' \rangle} = \pi(g)^{-1} H_{\lvert \psi \rangle} \pi(g). \end{equation} Hence, the stabilizers of stable states in the same orbit of $\pi(G)$ are related by conjugation.
My questions
- What is a (mathematically precise if possible) definition of a stable state?
- Why does spontaneous symmetry breaking solely focus on stable states, as opposed to non-stable states which still spontaneously break symmetries?
