# Understanding the prefactor $\frac{\theta g^2}{32\pi^2}$ of the $F\tilde{F}$ term in Yang-Mills theories

The most general Yang-Mills (YM) action consistent with Lorentz invariance, gauge invariance and renormalizability should contain a term $$\kappa F_{\mu\nu a}\tilde{F}^{\mu\nu a}\tag{1}$$ where $$\kappa$$ is a proportionality constant. Such a term is relevant even if the theory is completely classical. The reason is as is as follows. Though it can be shown that $$(1)$$ can be reduced to a total divergence term of the form $$\partial_\mu K^\mu$$, the four-vector $$K^\mu$$ need not vanish at the boundary making its effects admissible. Therefore, even the classical equation of motion will be modified though $$\kappa$$ remains undetermined classically. The exact term is, however, $$\frac{\theta g^2}{32\pi^2}F_{\mu\nu a}\tilde{F}^{\mu\nu a}\tag{2}$$ where $$\theta$$ is a parameter that characterizes the nontrivial vacuum state of YM theory. Apart from the famous factor $$32\pi^2$$ which is a topological effect, presence of $$\theta$$ in the proportionality constant reminds us that it is a quantum effect and have to be fixed in the quantum version of the theory.

Question $$1$$ Can we show that the quantum theory fixes the value of $$\kappa$$ to be $$\frac{\theta g^2}{32\pi^2}$$ starting with a classical YM Lagrangian which is already augmented by the term $$(1)$$?

I was going through Srednicki's book (Pages $$598-599$$). It doesn't exactly show this. It starts with a classical action $$S$$ that does not include $$\kappa F\tilde{F}$$ term. It considers transitions from one $$\theta$$ vacuum to another (See Eqs. 93.40-93.42) and ultimately shows that this transition amplitude is an integral over all field configurations $$A$$ weighted by $$e^{iS^\prime}$$ instead of $$e^{iS}$$ where $$S^\prime=S-\frac{\theta g^2}{32\pi^2}F\tilde{F}.\tag{3}$$

Question $$2$$ Why doesn't this calculation start with $$S+\kappa F\tilde{F}$$ instead of $$S$$ in which case we would have ended up with $$e^{iS^\prime}$$ with a different $$S^\prime$$ than $$(3)$$.

Srednicki says that he neglected such terms in the classical Lagrangian because it doesn't affect equation of motion which I cannot agree.

• This question is a little unclear. The title and first 90% make it sound like you're asking "why is $\kappa = \theta g^2 / 32 \pi^2$", but later it sounds like you already know, and your real question is what measurable consequences the parameter $\kappa$ (or equivalently $\theta$) has. Is that a fair summary? – knzhou Nov 21 '19 at 19:11
• If so, one consequence of $\theta$ is the neutron electric dipole moment, which may be computed in chiral perturbation theory. – knzhou Nov 21 '19 at 19:12
• @knzhou I hope the edited question is clearer. I have another related question which I will ask separately. – SRS Nov 21 '19 at 19:58
• It would be very useful if you clarified what exactly you think the current answers are missing. They seem good to me, so I wouldn't know what else to say. If you explained what is wrong with them maybe somebody else could try to fill in the gaps. – AccidentalFourierTransform Dec 3 '19 at 0:50
• I will try. To be honest, there is nothing wrong with the existing answers. Srednicki's approach is this: if the classical Lagrangian have $\theta=0$, quantum corrections will induce $\theta\neq 0$ term in the Lagrangian. This solely arises as a result of the nontrivial vacuum structure. However, it is not really possible to start with a QCD Lagrangian with $\theta=0$ even classically as the boundary term doesn't vanish. So a better way to start the classical theory is with $\theta\neq0$ from the beginning. Wouldn't that be a fair and more general thing to do? @AccidentalFourierTransform – SRS Dec 3 '19 at 10:34

The normalization is of course arbitrary, because we can always redefine what we mean by $$\theta$$. However, given that topological charge is quantized, we know that $$\theta$$ is periodic. This means it makes sense to define the topological part of the action so that $$\theta$$ has period $$2\pi$$, that is $$S = \theta Q_{top}$$ where $$Q_{top}\in Z$$ is an integer. Standard QFT textbooks show that $$Q_{top} =\frac{g^2}{32\pi}F_{\mu\nu}\tilde{F}^{\mu\nu}$$ is an integer.

Regarding question 2: Obviously, Srednicki could have started from the action that includes the $$\theta$$ term (so this may in part be a pedagogical question). However, it is indeed true that you can determine the tunneling path (the instanton) without including the $$\theta$$ term. All the $$\theta$$ term does is give a phase $$\exp(\pm i\theta)$$ to the tunneling amplitude. More generally, you can compute the QCD partition function at non-zero $$\theta$$ by computing the partition function at $$\theta=0$$, and then assigning a factor $$\exp(i\theta n)$$ to the sector with topological charge $$n$$.

• Any insight about question 2? @Thomas – SRS Nov 23 '19 at 15:35
• @srs Added a short comment. – Thomas Nov 23 '19 at 19:49
• Thanks! This is exactly my concern. "More generally, you can compute the QCD partition function at non-zero $\theta$ by computing the partition function at $\theta=0$, and then assigning a factor $\exp(i\theta n)$ to the sector with topological charge $n$." Do you have a reference for this? @Thomas – SRS Nov 27 '19 at 15:21
• @srs This is kind of clear from the definition of the partition function. It is used in lattice QCD, see, for example, equ.(8) in arxiv.org/abs/1306.2919 – Thomas Nov 28 '19 at 1:09

For your first question, fixing $$\kappa$$ depends on what our convention is for $$\theta$$. Our convention is we want $$\theta$$ to be a parameter such that $$\theta$$ and $$\theta+2\pi$$ lead to the same physics. Given that is the case, since $$\theta$$ appears in the action like $$\exp(i\theta \kappa\int d^4x F\bar{F})$$ we clearly want to choose $$\kappa$$ to be such that $$\kappa \int d^4x F\bar{F}$$ is always an integer. This is a purely classical condition on field configurations in the path integral. Quantum mechanics is only entering through the notion that what matters is $$\exp(iS)$$ and not the action itself.

For your second question, Srednicki is using the action without a theta term because this is the action we would use to find a transition amplitude between states with definite $$n$$.

Imagine instead we are considering the ordinary quantum mechanics problem of a particle in a sinusoidal potential. The actual energy eigenstates will be non-localized and have some pseudomomentum corresponding to $$\theta$$. But it is perfectly well-defined to ask what is the transition amplitude for the particle to go from one particular minimum of the sinusoid to another particular minimum in some finite time. And the action you will use is just the ordinary action in a sinusoidal potential. The analogue of this is what Srednicki is doing with the states of definite $$n$$.

• I'm not at ease with the second part of the answer. In particular, you wrote "this is the action we would use to find a transition amplitude between states with definite n" In considering this transition amplitude, how would you justify throwing away the $F\tilde{F}$ term in the action? @octonion – SRS Nov 23 '19 at 14:59
• @SRS, If you were to calculate the transition amplitude going from one well to another in the quantum mechanics problem in a sinusoidal potential, would you include a theta term in the action of the path integral? No, theta doesn't appear here yet in this problem. The theta term is selecting a particular state to take expectations around in the path integral. Different thetas are different states within the same theory. Now perhaps there is another valid interpretation where it is a genuine parameter modifying the Hamiltonian, but that's not what Srednicki is doing – octonion Nov 24 '19 at 12:35
• I am not sure why or how a theta term will appear in quantum mechanics? @octonion – SRS Nov 26 '19 at 7:02
• @SRS, Is that meant to be a dismissal or are you trying to understand the answer? Do you see the connection between a periodic potential in QM and the multiple classical energy minimum Srednicki calls $|n\rangle$? – octonion Nov 26 '19 at 9:41
• "Now perhaps there is another valid interpretation where it is a genuine parameter modifying the Hamiltonian, but that's not what Srednicki is doing" That's exactly my point. Our philosophy of writing QFT Lagrangians are Lorentz invariance, gauge invariance and perhaps also renormalizability. If you agree on this, I see no reason to discard the $F\tilde{F}$ term in the starting Lagrangian. In quantum mechanics, we are not driven by any such symmetry requirement to write the Hamiltonian. The Hamiltonian for a particle moving in a potential is always $H=\textbf{p}^2/2m+V(\textbf{r})$.@octonion – SRS Nov 27 '19 at 14:49