# Why doesn't the $\theta$ Angle Renormalize?

The $$\theta$$ term for Yang-Mills takes the form $$L_{\theta}=\frac{\theta}{64\pi^2}\varepsilon^{\mu\nu\rho\sigma}F^a_{{\mu\nu}}F^a_{\rho\sigma}$$

A fact that I have heard is that $$\theta$$ does not run under renormalization. I understand that this term is topological. I have been told by peers that this explains why it does not run, however I would like a much better explanation for this fact. Is there a detailed proof of this statement? Perhaps a reference would help!

• As a comment, an analogous argument would say that the Chern-Simons level should also not run since the quantity is topological, but we know that it indeed receives finite renormalization from an Maxwell term in 2+1D. – Aaron Apr 13 at 0:06
• If being a topological term meant that $\theta$ couldn't run, there would be no strong CP problem. This is pure Yang-Mills not the standard model, but you need to be careful with arguments like this. – octonion Apr 13 at 4:15

The reason is that it is a total derivative. When people say that that's because it is topological they mean total derivative.$${}^1$$

A reference that explains is is, for example, Marcos Mariño's notes on large $$N$$. See around $$(5.29)$$.

The argument is roughly as follows: $$q(x) \equiv \frac{1}{64\pi^2}\epsilon^{\mu\nu\rho\sigma}F^a_{{\mu\nu}}F^a_{\rho\sigma} = \partial_\mu K^\mu\,,$$ where $$K^\mu$$ is actually identical to the Chern-Simons term in $$d=3$$, times $$\epsilon_{\mu\nu\rho\lambda}$$. In Fourier transform this means $$\tilde{q}(p) = p^\mu \tilde{K}_\mu(p)\,\underset{p\to0}{\longrightarrow} 0\,.$$ Now the $$\theta$$ dependent partition function would be $$Z[\theta] = \frac1{\mathcal{N}}\int \mathcal{D}[A,\psi,\ldots]\,\exp\left({\mathcal{L}_{\mathrm{QCD}} + \theta\int \mathrm{d}x\,q(x)}\right) \,.$$ The perturbative expansion looks like $$Z[\theta] = Z + \theta Z^{(1)} + \frac{\theta^2}{2!} Z^{(2)} + \cdots\,,$$ and each term is of the form $$Z^{(n)} = \int \prod_{i=1}^{n-1} \frac{\mathrm{d}^4k_i}{(2\pi)^4}\langle \tilde{q}(k_1)\cdots \tilde{q}(k_{n-1})\tilde{q}(k_1+\cdots + k_{n})\rangle\,.$$ The last argument shows up when Fourier transforming $$\langle q(x_1) \cdots q(x_n)\rangle$$ as a consequence of translational invariance, because one could send $$x_i \to x_i - x_n$$. Due to momentum conservation the sum of the momenta vanishes and thus $$\tilde{q}(k_1+\cdots + k_n) = \tilde{q}(0) = 0\,.$$ This means that $$Z[\theta]$$ does not depend on $$\theta$$ at any order in perturbation theory and thus one can ignore it in all perturbative computations. In particular, there cannot be perturbative corrections to $$\theta$$ itself.

$$\qquad{}^1$$ As a note on the comment under the original post: the Chern-Simons action is famously topological but not a total derivative. So the seeming inconsistency with the statement that topological terms do not give perturbative effects is due to a terminology issue.

Edit:

Even though it is very clearly written in the linked notes and it has been discussed in the comments, I still want to emphasize that the true partition function $$Z[\theta]$$ does depend on $$\theta$$ through non perturbative effects. For instance by summing infinitely many diagrams (large $$N$$ limit) or via lattice computations. A complete review about the subject is Ettore Vicari, Haralambos Panagopoulos.

• Honestly I don't know whether $\theta$ runs in pure Yang-Mills or not, but I am skeptical about your argument that it can't run since it is a total derivative. The 2D CP(1) sigma model has a theta term that can be written as a total derivative and as far as I understand it runs. See for instance Fig 1 in this paper (arxiv.org/abs/1706.06598) – octonion Apr 13 at 4:25
• I am not saying that it can't run at all. The statement is that it can't run perturbatively. That is, you'll never get any renormalization by computing finitely many diagrams. There are however non perturbative effects that involve resumming infinitely many diagrams and it does run under those. In the link I provided indeed he does large $N$ planar limit and computes $Z^{(2)} \neq 0$. – MannyC Apr 13 at 4:32
• By the way, there is a lot of literature for $4d$ too. If you are interested, it is all reviewed in Vicari-Panagopoulos. – MannyC Apr 13 at 4:39
• Thanks for the link. I just want to stress that your answer could be a little misleading since non-perturbative effects are after all what the $\theta$ term is all about, and someone could easily read your answer and come to the mistaken conclusion that $\theta$ can't run since it is a total derivative (which is more or less what OP wanted to confirm). – octonion Apr 13 at 4:50
• I completely agree, this is why while you were typing the comment I was atually making an edit. If you have any other suggestions I'll implement them :) – MannyC Apr 13 at 4:53