Are Instantons Massless?

Question

That is, are the only field configurations which give a non-zero winding number ones in which the Fourier transform includes a factor like $\theta(k^0)\hat{D}\delta(k^2)$, where $\hat{D}$ is some differential operator acting on the delta function?

Below I will explain why I think this may be the case, and hopefully we can have a discussion about it:

Say we consider the theory of a massless fermion $\psi$ coupled to some gauge field $A^a_{\mu}$ in $4$ spacetime dimensions. The Lagrangian I would like to consider is $$L=i\bar{\psi}\gamma^{\mu}\nabla_{\mu}\psi-\frac{1}{4g^2}f^a_{\mu\nu}f^{a\mu\nu}$$We know that this theory has a chiral anomaly of the form $$\partial_{\mu}J^{\mu}_5=-\frac{C}{32\pi^2}\varepsilon^{\mu\nu\rho\sigma}f^a_{\mu\nu}f^a_{\rho\sigma}$$ Where $\text{tr}(t_at_b)=C\delta_{ab}$ and $t_a$ are the generators of the gauged Lie group, and the representation $\psi$ transforms under. Note first of all that if we integrate both sides we get

$$Q_5(t=\infty)-Q_5(t=-\infty)=-2C\nu$$ Where $\nu$ is the instanton winding number.

At this point, I would now like to consider the Wilsonian Effective Lagrangian $L_{\Lambda}$, for which all of the modes of mass $m>\Lambda$ are integrated out. Under a chiral transformation $\psi\to e^{i\alpha\gamma_5}\psi$, we know that $L_{\Lambda}\to L_{\Lambda}-\alpha\frac{C}{32\pi^2}\varepsilon^{\mu\nu\rho\sigma}f^a_{\mu\nu}f^a_{\rho\sigma}$.

For this to be the case, it would be essential that none of the massive modes of mass $m>\Lambda$ have non-zero winding number. This must be case, or else instantons would affect the transformation properties of $S_{\Lambda}=\int d^4xL_{\Lambda}$ under a chiral transformation.

This must also be the case for all values of $\Lambda$, since this chiral transformation does not depend on $\Lambda$ either. Therefore the only room we have for instantons is them to be massless, and have the factor mentioned above in their Fourier transform.

The statement I left in bold is what this argument hinges on, and is also a statement that I am not certain of it's validity. I know for sure that if we only integrated out the high mass modes of $\psi$, that would be the way $L_{\Lambda}$ transforms. However, the fact that we also integrate out the high mass modes of $A^a_{\mu}$ potentially screws this nice transformation up.

I would really appreciate getting both feedback on the argument, and whether or not the statement in bold is true!

Answer 1

Introductory pedantry

Asking whether "an instanton" is massless is a meaningless question: A field configuration does not correspond to a quantum state, quantum states are rather (in one representations) functionals in the fields. Things that are not a quantum state do not have a mass since you can't apply the mass operator $P_\mu P^\mu$ to them.

Nevertheless, an instanton - as an extremum of the classical action - is frequently taken as the starting point of perturbation theory, and more or less corresponds to some sort of "perturbative vacuum state" in this sense. It would be meaningful to ask whether or not we have to assign a "mass" to this vacuum, if it were not additionally complicated that the "true" vacua of an instantonic theory are the $\theta$-vacua which are superpositions of the vacua with definite instanton number, see also this answer of mine.

The modes of the quantum field do not intrinsically carry any "winding number", as your question implies in passing. They are operators. A winding number is a characteristic of a field configuration, or of a state built by creation/annihilation operators above a particular perturbative vacuum, not of an operator.

In any case, instantons are not excited states that would be related to the naive modes of the quantum field and therefore the effect off the Wilsonian cutoff on them needs to be considered more carefully.

Wilsonian flow in gauge theories

A detailed reasoning about how Wilsonian effective actions work for non-Abelian gauge theories would far exceed the scope of this answer. A good starting point is Pawlowski's "Wilsonian Flows in Non-Abelian Gauge Theories" (PDF link), where in particular chapter VI and appendix B are relevant to the question at hand.

The upshot is this: The instantonic effects persist, in a suitably renormalized manner, along the entire Wilsonian flow, i.e. they are never cut off. But the chiral transformation you write down does not hold for the Wilsonian action.

What one has to establish for the persistence of the instantonic effects is that the zero modes of the regularized Dirac operator don't vanish along the flow, since the difference in chirality of these zero modes is why the instantons appear as the anomalous terms in the first place (cf. its derivation via the Fujikawa method, see also this answer of mine).

Then one has to actually compute the zero-mode contribution in terms of the fields with the cutoff in place to see how the term violating the chiral symmetry looks in practice.