Is there an argument for using the $\theta$-vacuum for a Yang-Mills theory that works regardless of the presence of fermions?

Question

Consider a Yang-Mills theory, possibly including fermions. It has many possible vacua $\{|n\rangle\}$ labelled by integer winding number $n$, defined as the Maurer-Cartan topological invariant: for the gauge element $g_{(n)}$ and corresponding unitary large gauge transformation $U(g_{(n)})$ we have $$ |n\rangle = U(g_{(n)})|0\rangle, \quad n = \frac{i}{24\pi^{2}}\int \limits_{S^{3}} d^{3}\theta \epsilon^{ijk}\text{tr}\big[ g_{(n)}\partial_{i}g_{(n)}^{-1}g_{(n)}\partial_{k}g_{(n)}^{-1}g\partial_{k}g_{(n)}^{-1}\big] $$ What is the most theory-independent argument which shows that the vacuum of the non-abelian gauge theory must correspond to the $\theta$-vacuum state $$ |\theta\rangle = \sum_{n = -\infty}^{\infty}e^{in\theta}|n\rangle? $$

Examples of arguments which are not complete for me

1. Consider a pure YM theory without fermions. In order to argue why we have to use the $\theta$-vacuum as the ground state, people show that the Hamiltonian $H$ is non-diagonal in the basis $\{|n\rangle\}$: $$ \langle n|H|m\rangle \simeq e^{-\frac{8\pi^{2}}{g^{2}}|n - m|} $$ and therefore, vacuum tunnelling is possible. This requires us to diagonalize the Hamiltonian, and the $\theta$-vacuum basis is the diagonal basis.

But this argument works fine only when semiclassical approximation is valid, and also only if massless fermions are not included.

2. The first argument, however, is valid only for pure Yang-Mills theory and breaks down when massless fermions are included, since massless fermions suppress the tunnelling. People then use an argument based on the cluster decomposition principle (or CDP). The detailed argument is shown e.g. in "The structure of the gauge theory vacuum" by Callan Jr. One introduces a conserved operator $$ \tilde{Q}_{5} =\int d^{3}\mathbf r (J_{0,5} - 2K_{0}) , $$ where $K_{0}$ is defined as $$ G_{\mu\nu,a}\tilde{G}^{\mu\nu}_{a} = 2\partial_{\mu}K^{\mu}, $$ and by using this charge shows that the VEV of non-zero 2c chirality operator $B(\mathbf x)$ (i.e., $[\tilde{Q}_{5}, \mathbf B(\mathbf x)] = 2c \mathbf B(\mathbf x)$) show that the VEV $$ \langle n| B(\mathbf x)B^{\dagger}(0)|n\rangle $$ doesn't satisfy the CDP $$ \lim_{|\mathbf x| \to \infty}\langle n| B(\mathbf x)B^{\dagger}(0)|n\rangle = \lim_{|\mathbf x| \to \infty}\langle n|B(\mathbf x)|n\rangle \langle n|B(0)|n\rangle $$ The $\theta$-vacuum is the solution of this problem.

But this argument (its detailed part) depends on presence of fermions. Precisely, we introduce the chirality and operates with chirality operator $\tilde{Q}_{5}$.

What do I want?

I want some argument (possibly purely mathematical) which shows that we must choose the $\theta$-vacuum as the ground state of the YM theory (if it exists) independently from the precise field content, in particular independent of the presence of massless fermions. Is there such an argument?

Answer 1

Consider euclidean Yang Mills on $R^3\times S_1$. The euclidean partition function is a sum over different topological sectors labeled by $$ q = \frac{1}{32\pi^2}\int d^4 x \, G_{\mu\nu}^a\tilde{G}^{\mu\nu\, a} $$ Now perform a Fourier transform with respect to $q$, that is determine the partition function $Z(\theta)=\sum_q e^{iq\theta}Z_q$. If we wish to study the ground state ($T=0$) we let the size of the $S_1$ go to infinity, The partition function will now sample the ground state wave function. We can define the averages of any operator as $\langle\theta |O|\theta\rangle\equiv \frac{1}{Z(\theta)}\sum_q e^{iq\theta}O_q$, which effectively defines $|\theta\rangle$.

Answer 2

The first argument you give, which you consider not complete as it only works in the semi-classical approach can be shown for more general cases. This is explained in Section 93 of Sredniki, as stated by @tparker.

In general, $\left|\theta \right>$ diagonalises 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>. \tag{2} \end{align} Now, the delicate part, which is treated in Sredniki is proving that (1) is satisfied. I'm just going to copy the argument (I hope this doesn't go against the rules of the comunity!).

To see this, consider making a gauge transformation by $U_k(x)$, where $U_k(x)$ has winding number $k$. The product of two maps with winding numbers $n$ and $k$ is a map with winding number $n + k$. Thus, making a gauge transformation by $U_k(x)$ converts a field configuration with winding number n to one with winding number $n+k$. In the quantum theory, the gauge transformation is implemented by a unitary operator $U_k$, and we should have $$ U_k\left|n\right> = \left|n+k \right>,\tag{3} $$ On the other hand, the hamiltonian, which is built out of field strengths, must be invariant under time-independent gauge transformations: $$ U_k H U_k^\dagger = H ,\tag{4} $$ Inserting factors of $I = U_k^\dagger U_k$ on either side of $H$ in $\left<n \right| H \left|m \right> $, and using eqs. (3) and (4), we find $$ \left<n \right| H \left|m \right> = \left<n +k\right| H \left|m +k\right>. \tag{5} $$ We conclude that $\left<n \right| H \left|m \right>$ depends only on $|n-m|$.

Thus completing the proof of (2).