I've been learning about the theta parameter of QCD and I'm confused about the fact that it's supposed to be very small but at the same time some sources say that the Yang-Mills theory should be invariant to 2$\pi$ shifts in $\theta$

Sources that says that $\theta$ is small:

(1)The CP Puzzle in the Strong Interactions (See page 6)

(2) TASI Lectures on The Strong CP Problem (See page 19)

Sources that say that Yang-Mills theory is invariant under $2\pi$:

(3) Notes on Supersymmetry (See page 6)

(4) Non-Perturbative Dynamics Of Four-Dimensional Supersymmetric Field Theories (See page 4)

Now source (1) (amongst others) tells us that there exists a term of the form:

$\theta n = \theta (1/32 \pi^2) \int d^4 x \ \epsilon^{\mu\nu\rho\sigma} \ F^a_{\mu\nu} F^a_{\rho\sigma}$

in the Lagrangian of our QCD, where the $\int d^4 x \ \epsilon^{\mu\nu\rho\sigma} \ F^a_{\mu\nu} F^a_{\rho\sigma}$ part gives CP violation. Thus we require $\theta = 0$ if want no CP violation. However to me this implies that the theory is not invariant under a shift in $2\pi$ of $\theta$ as $\theta = 2\pi$ would not make the above term disappear (you'd end up with $2\pi n$ rather than $0 \times n = 0$).

Similarly how can we say that $\theta$ has to be small if $\theta$ can be shifted by any multiple of $2\pi$ to give the same theory? Is it the case that when we say '$\theta$ is small' we actually mean '$\theta$ modulo $2\pi$ is small'. Also is it the case that the above equation some how disappears for all values of $\theta$ that are multiples of $2\pi$, and not only $\theta = 0$?

Theta vacuum effects on QCD phase diagram - page 2 seems to imply that the CP violating term disappears for all value of $\theta$ that are multiples of $\pi$.


1 Answer 1


Remember that the theta term appears in an exponential $e^{i\theta n}$ inside the path integral. If $\theta n$ shifts by $2\pi N$, for any integer $N$, the exponential is unchanged, and all path integrals have the same value.

The integral $n = \frac{1}{32\pi^2} \int F \wedge F$ is not arbitrary either. It's a topological invariant, and it's normalized so that it will be an integer. Indeed, on flat $\mathbb{R}^4$ with your normalization, it's equal to $2 \nu$, where $\nu \in \mathbb{Z}$ is the instanton number. (You can find the argument in Weinberg, Vol II, p 450-2.) But for now let's just take it as granted that that $n$ is always integer-valued. It follows then that shifting $\theta \to \theta + 2\pi M$ for any integer $M$ sends $e^{i\theta n}$ to $e^{i\theta n + i 2\pi M n} = e^{i\theta n}$.

Which means that only the value of $\theta$ mod $2\pi$ affects physics. So when someone says that $\theta$ is small, they mean that $\theta$ mod $2\pi$ is small, as you guessed.

  • 1
    $\begingroup$ Is it correct that the effective term $\delta \mathcal{L}_\textrm{eff} \propto \theta Tr(\tilde{F}F)$, which isn't periodic, appears from a small $\theta$ approximation $e^{i\theta} \approx 1 + i\theta$? i.e. the effective Lagrangian is periodic, but not in our approximation... $\endgroup$
    – innisfree
    May 15, 2013 at 15:54
  • 1
    $\begingroup$ @innisfree, you have $\exp(iS)$ in functional integral, so shifts in action $S$ that are multiples of $2\pi$ are irrelevant. $\endgroup$ May 15, 2013 at 16:33
  • $\begingroup$ Thank you very much, a very detailed and helpful answer. $\endgroup$ May 15, 2013 at 16:51
  • $\begingroup$ @PeterKravchuk Yes, that's true. But written in full, the Lagrangian itself would be invariant under those shifts? I mean, suppose we kept a $\theta^2$ term in the expansion, then not even the path integral would be invariant... $\endgroup$
    – innisfree
    May 15, 2013 at 17:17
  • 1
    $\begingroup$ @innisfree, I dont get what expansion you are talking about. The lagrangian is not invariant, but it is $\exp(i\int\mathcal{L}d^dx)$ which matters. It is invariant. $\endgroup$ May 15, 2013 at 17:21

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