Motivation for the shape of the theta vacua

Question

I understand that the reason why we construct the theta vacua is because instantons allow tunnelling between different vacuum states, $\left|n\right>$. This means that we have to consider a real vacuum state $$ \left|\Omega\right> = \sum_n a_n \left|n\right>. $$ Now, my question is why do we have $a_n = \frac{1}{\sqrt{2\pi}}\,e^{i\theta n}$, so that $$ \left|\theta\right> = \dfrac{1}{\sqrt{2\pi}}\,\sum_n e^{i\theta n} \left|n\right>. $$

I see that this allows some neat properties like orthogonality, $$ \left<\theta\right| \mathcal{O} \left|\theta'\right> = 0 \quad \text{if} \quad \theta \neq \theta', $$ or it being an eigenstate of the Hamiltonian, $$ H \left|\theta\right> = \lambda \left|\theta\right>, $$ but is the theta vacuum the only vacuum structure that allows this for QCD? Why is this the correct QCD vacuum (at least when there are no fermions)?

Answer 1

As boring as this might sound, the simple reason why $\left|\theta \right>$ is the correct vacuum for QCD is because it is an eigenstate of any Hamiltonian which verifies $$ \left<n \right| H \left|m \right> = f(n-m),\tag{1} $$ which is easily proven: \begin{align} H \left|\theta \right> &= \dfrac{1}{\sqrt{2\pi}}\,\sum_{n} e^{i n \theta} H\left|n \right> = \dfrac{1}{\sqrt{2\pi}}\,\sum_{n,m} e^{i n \theta} \left|m \right> \left<m \right| H \left|n \right> \\ &= \dfrac{1}{\sqrt{2\pi}}\,\sum_{n,m} e^{i n \theta} \left|m \right> f(m-n) = \dfrac{1}{\sqrt{2\pi}}\,\sum_{n,m} f(m-n) e^{i (n-m) \theta} e^{i m \theta} \left|m \right> \\ &= \dfrac{1}{\sqrt{2\pi}}\,\sum_{q,m} f(q) e^{i q \theta} e^{i m \theta} \left|m \right> = \sum_{q} f(q) e^{i q \theta} \dfrac{1}{\sqrt{2\pi}}\,\sum_{m} e^{i m \theta} \left|m \right>\\ &= \left( \sum_{q} f(q) e^{i q \theta} \right) \left|\theta \right> \equiv \lambda \left|\theta \right>. \end{align}

See chapter 93 of Sredniki for a proof that QCD satisfies (1), or this SE answer.