I'm confused about fermion zero modes in relation to instantons. I understand that instantons can create fermion zero modes, but it's not clear to me when a fermion has a zero mode.

For example, consider QCD with nonzero fermion masses: i.e. there are explicit mass terms in the Lagrangian. Do these fermions have zero modes? Are they created by the QCD instanton?

Is it generically possible to have a zero mode when the field has an explicit mass term, and if so, do those zero modes behave differently in an instanton background than the zero modes of massless fermions?


  • $\begingroup$ According to 0802.1862, top of page 30, massive spinors can have no zero mode. But then I'm confused: why do we talk about writing 't Hooft operators for non-perturbative effects in the Standard Model when none of the Standard Model fermions are massless? Is there a difference when the masses come from spontaneous symmetry breaking? $\endgroup$ – Pengpeng Xu Jun 21 '13 at 13:55

Fermionic zero modes on an instanton background are in one-to-one correspondence with solutions to the Dirac equation $$(i D_\mu \gamma^\mu - m)\Psi = 0$$ where the partial derivative $D_\mu$ contains the gauge field term with the gauge field defining the instanton solution substituted into it.

We want to study these solutions in the Euclidean spacetime. The zero mode should be "concentrated" near the region where the instanton is nontrivial, much like the bosonic zero modes.

Effectively, the momentum of the zero mode in the spacetime is zero almost everywhere: there's no plane wave left. Much like the Dirac equation with a zero mass has a solution in terms of a spacetime-constant spinor, the Dirac equation on the instanton background has a zero mode. But when the mass is nonzero, there's no way to find a solution to the Dirac equation. Far away from the instanton, the gauge terms are negligible and we're supposed to get a plane wave but there's no such configuration because the momentum of such a planewave must be timelike and there are no timelike directions in the Euclideanized spacetime!

So for a nonzero mass $m$, there will be no fermionic zero modes on the instanton.

One may study what happens when $m$ is adjusted from $m=0$ to a nonzero $m$, from a value which admits zero modes to a value that doesn't. The evolution is continuous. In the $m=0$ case, one finds zero modes for the left-handed Weyl spinor and one for the right-handed Weyl spinor. These two zero modes "team up" to create a nonzero mode that is allowed to "escape" to nonzero (timelike) momentum.

Quite generally, the zero modes may be stuck there if one has Weyl spinors but the nonzero modes contain twice as many degrees of freedom and they may move freely in the energy. This is the basic idea behind the index theorems: the number of fermionic zero modes (minus the number of fermionic zero modes with some "chirality" reversed) in various SUSY and similar theories is an integer that is invariant under all continuous deformations of the Hamiltonian. When there is one left-handed and one right-handed solution, they may pair up and disappear from the list of the zero modes but that doesn't change the difference that defines the index.

Concerning the "puzzling" paper, note that the masses of most SM fermions are much lighter than the electroweak scale where $SU(2)\times U(1)_Y$ operates and that determines the typical length scale of the electroweak instantons; and they're even smaller than the QCD scale. So there must exist a sense in which the masses may be neglected for most of these instantons. One may derive these zero modes by assuming $m=0$ and the correction coming from the small yet nonzero mass implies that these modes are "almost zero" nonzero modes rather than strict zero modes.

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  • $\begingroup$ Thanks Lubos, that cleared up a lot. From an "intuitive" perspective I am happy with the idea that we can treat very light fermions as "almost massless," but I'm having trouble seeing where we pay for this approximation. $\endgroup$ – Pengpeng Xu Jun 21 '13 at 14:20
  • $\begingroup$ For example, the 't Hooft operator is there because the instanton vanishes unless it can create zero modes. When the fermions have small masses, does that mean that one can have instanton effects without zero mode creation? (Suppressed by m(fermion)/v ?) Alternately, how is the 't Hooft operator modified to account for the fact that the fermionic fields aren't actually massless? How does the 't Hooft operator interpolate between zero and "small" fermion masses? $\endgroup$ – Pengpeng Xu Jun 21 '13 at 14:23
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    $\begingroup$ Dear Pengpeng, good questions. The instanton-induced interaction has the huge $\exp(-S)$ suppression, which is what really makes it small, and a prefactor. For tiny masses and "almost zero modes", the prefactor must be composed of $1$ (which doesn't include the factors from the modes because they are not zero modes) PLUS terms that are proportional to the near zero modes, too. For instantons much smaller (size in $x$ space) than the mass scale, the second term proportional to the fermion near-zero-mode-having fields is more important than the term $1$. $\endgroup$ – Luboš Motl Jun 22 '13 at 6:09
  • $\begingroup$ Note that in an accurate treatment where you don't neglect the masses, the leading term in the 't Hooft interaction has prefactor "1" - i.e. denies the existence of the zero modes. But all instantons produce higher-order interactions, too, and the extra interaction with the extra factors of the fermions is there, too, even if no zero modes exist (like a loop diagram on top of the instanton). So there's no contradiction between the approximation and the exact result. $\endgroup$ – Luboš Motl Jun 22 '13 at 6:26

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