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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.

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  • $\begingroup$ 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 $\endgroup$ – SRS Oct 31 '18 at 19:43
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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.

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  • $\begingroup$ Did I say there is tunnelling between theta vacua? I said there is tunnelling between $|n\rangle$ states. @ACuriousMind $\endgroup$ – SRS Oct 31 '18 at 19:45
  • $\begingroup$ @SRS No. I just reiterated that there is none. $\endgroup$ – ACuriousMind♦ Oct 31 '18 at 19:46
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    $\begingroup$ 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 $\endgroup$ – SRS Oct 31 '18 at 19:53

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