# Are theta vacua topologically protected?

In discussions of Yang-Mills instantons it is often stated that one should sum in the path integral over all contributions of fluctuations around all the topologically distinct vacua labelled by winding number $$n$$.

Usually there follows a discussion on $$\theta$$-vacua, which are basically a linear combination of $$n$$-vacua, in the sense that $$\lvert \theta \rangle = \sum e^{in\theta} \lvert n \rangle$$ and that the procedure of summing over all fluctuations around instanton contributions in the path integral could be replaced by applying the usual rules to a Lagrangian where the term $$\sim \theta (*F, F)$$ is added, where $$(.,.)$$ denotes the Cartan inner product.

I don't understand how this can be. Doing this, one seems to replace the path integral as a sum over contributions from different $$n$$-vacua by just the path integral expanded in a unique $$\theta$$-vacuum. Indeed, this seems to come down to summing over a particular linear combination of $$n$$-vacua. But WHY should you only take into account only a particular $$\theta$$-vacuum, even though the $$\theta$$-vacua are inequivalent (they have different energy density)?

Why should e.g. QCD not just be studied in the true vacuum (the one with the lowest energy density) which is just $$\theta=0$$? Is this because, the $$\theta$$-vacua, too are topologically protected?

the procedure of summing over all fluctuations around instanton contributions in the path integral could be replaced by applying the usual rules to a Lagrangian where the term $$\sim \theta (*F, F)$$ is added, where $$(.,.)$$ denotes the Cartan inner product.

I don't think you are thinking about this correctly. If you do not add the $$\theta$$ term that is the same as $$\theta=0$$. Even if $$\theta=0$$, you still need to sum over field configurations in the path integral with arbitrary number of instantons. The $$\theta$$ term just weights these different field configurations differently in such a way that you are in a particular non-zero $$\theta$$ vacuum.

You are certainly free to study QCD with $$\theta=0$$ if you want to, the point is there is an extra parameter which tells us about different sectors of QCD (built upon different states $$|\theta\rangle$$), and we can study these sectors just by adding a $$\theta$$ term to the path integral.

Different values of $$\theta$$ correspond to different sectors in the sense that a state from one sector can't evolve to that from another sector. This is less like a topological protection, and more like an ordinary superselection rule (like between states of different charge). There is a trivial shift symmetry $$|n\rangle\rightarrow |n+1\rangle$$ of the theory, and this corresponds to $$\theta$$ being conserved (it is a bit like a conserved momenta in field space)

• Ok, I agree, but could you add an explanation on why should a particular sector $\theta$ be considered? Why would this sector be stable even though the energy density of the $\theta=0$ sector, corresponding to equal contributions from all $n$ configurations, is minimal? Is this vacuum itself topologically protected? Apr 23, 2019 at 15:59
• @MichaelAngelo, I added another section. Apr 23, 2019 at 16:20
• I don't understand. $\theta$ is a parameter of the theory. There doesn't have to be a superselection sector for every value of every parameter. Apr 23, 2019 at 19:45
• @RyanThorngren, It's not just a parameter. It is more like a value of momentum like I said in my answer. Take a look at OP's description of states $|\theta\rangle$ which can be defined within the same theory Apr 23, 2019 at 20:02
• @RyanThorngren, Okay fair enough, but the difference is these ground states differing by $\theta$ have no overlap, and they really are superselection sectors. The shift transformation I mentioned gives different phases to the different sectors, but we think of the shift as not 'real' just a large gauge transformation. So a superposition of states from different sectors doesn't make sense. Apr 23, 2019 at 20:33