Isn't there a unique vacuum of the Yang-Mills quantum theory?

The theta vacua$$^1$$ of a Yang-Mills quantum theory are given by $$|\theta\rangle=\sum\limits_{n=-\infty}^{\infty}e^{in\theta}|n\rangle.$$ In Srednicki's Quantum Field Theory, he claims that the energy of a theta vacuum is proportional to $$-\cos\theta$$ so that the theta vacuum labelled by $$\theta=0$$ has the minimum energy. For other values of $$\theta\neq 0$$, the corresponding theta vacua $$\{|\theta\rangle\}$$ will have higher energies. Why are then the states $$\{|\theta\rangle\}$$ with $$\theta\neq 0$$ are referred to as vacua?

$$^1$$ As far as I understand, due to the possibility of tunnelling mediated by instantons the states $$|n\rangle$$ are not the true ground states of the theory. Gauge field configurations labelled by $$n$$ are minimum energy configurations only in the classical theory.

• 1. Are $|n\rangle$ true vacua of the theory? 2. For a fixed $\theta$, one has eigenstates $|\theta\rangle$ of $H$. I have no problem with that. But for any $\theta$, energy is not minimum. Only for $\theta=0$, the energy is minimum. So shouldn't the state $|\theta=0\rangle$ be called the theta vacuum and the rest of the states $|\theta\rangle$ with $\theta\neq 0$ simply as eigenstates of $H$? @ACuriousMind
– SRS
Oct 31, 2018 at 19:43

There is no tunnelling between the $$\theta$$-vacua, so each of them forms a ground state for its own superselection sector of the space of states, and it is wholly irrelevant how their absolute energy compares with the other $$\theta$$-vacua (without tunnelling, why would the energy matter?). Such a ground state is usually called a vacuum.
• Did I say there is tunnelling between theta vacua? I said there is tunnelling between $|n\rangle$ states. @ACuriousMind
• So two different $\theta$ vacua belong to two different Hilbert spaces? But you agree that two different $\theta$ vacua are not degenerate unless we add different constants to make them degenerate? @ACuriousMind