A fermion zero mode is a zero eigenfunction, $$i\gamma^\mu(\partial_\mu-iA_\mu)\psi=0$$ The number of zero modes is apparently related to the instantons of the gauge field.

But now my question is about 'ordinary' solutions to the Dirac equation. Even if there is no gauge field and even in Euclidean space with a mass term the Dirac equation has solutions $$(i\gamma^\mu\partial_\mu + m)\psi=0$$ For instance, a possible basis choice in 2D is, $\gamma^0=\sigma^1, \gamma^1=\sigma^2$. Then there is the simple solution with no $x^1$ dependence, $$(\psi_L,\psi_R)=(e^{i m x^0},e^{i m x^0}).$$ Why are these ordinary solutions not considered when zero modes are considered? In Luboš Motl's answer here, he goes so far as to say solutions with non-zero mass don't exist in Euclidean space, but I don't see why not, I just explicitly found an obvious one. Is there some extra condition that goes into the definition of the zero mode that I am missing?

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    $\begingroup$ zero modes are normalisable. $\endgroup$ – AccidentalFourierTransform Jan 21 at 20:31
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    $\begingroup$ @AccidentalFourierTransform, why is that important? Plane wave solutions are important in bosonic path integrals $\endgroup$ – octonion Jan 21 at 20:34
  • $\begingroup$ @AccidentalFourierTransform, Also if we work on the torus, this is normalizable $\endgroup$ – octonion Jan 21 at 20:37

The eigenvalues of the Euclidean Dirac opertor are of the form $i\lambda+m$, $\lambda,m,\in {\mathbb R}$. Instanton backgrounds can allow solutions with $\lambda=0$, but if $m\ne 0$ you cannot get $i\lambda+m=0$. Your confusion comes from the fact that the Euclidean Dirac operator with hermitian $\gamma^\mu$ obeying $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu= 2\delta^{\mu\nu}$ is $$ \gamma^\mu \partial_\mu+m. $$ There is no "$i$" before the gammas. This absence of "$i$" is essential precisely because there must be no zero modes in the Euclidean theory so that the Euclidean propagator $(\gamma^\mu \partial_\mu+m)^{-1}$ always exists. It's only when we go back to Minkowski signature that we can go "on shell."
Your zero mode solution has $p_0=m$ which is a Minkowski energy momentum relation.

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  • $\begingroup$ Thanks for the answer, that's simple enough. So just to be clear the zero mode equation written in Motl's answer that I linked to is mistakenly for Minkowski space, even though he is talking about Euclidean? $\endgroup$ – octonion Jan 22 at 0:56
  • $\begingroup$ There are different sign and gamma conventions so I can't say what Lubos had in mind. Even in Minkowski space people write Dirac differently between the mostly plus metric and the mostly minus metric. One has "$i$" before the gamma and the other does not as changing the sign of the metric multiplies the gammas by "$i$." What is true that the plane wave Euclidean eigenvalues $\Lambda$ obey $|\Lambda|^2=p_0^2+{\bf p}^2+m^2$ and cannot be zero, while in Minkowski they are real and obey $\Lambda^2=-p_0^2+{\bf p}^2+m^2$ and can can be zero when $p_0^2={\bf p}^2 +m^2$ $\endgroup$ – mike stone Jan 22 at 12:26
  • $\begingroup$ Thanks. By the way, there are formal solutions involving an real exponential to this Dirac equation as well, but I suppose to define the notion of eigenvalue of the Dirac operator and make sense of the path integral we need to restrict the function space somehow. I was hoping to get a better understanding of this, but that is a different question I guess! $\endgroup$ – octonion Jan 22 at 21:36
  • $\begingroup$ The Euclidean Dirac opertor $D\!\! /=\gamma^\mu D_\mu$, with $D_\mu$ the covariant derivative on spinors, is skew-self-adjoint on a compact closed Riemanian manifold, so has a well defined eigenvlaue problem with complete sets of $L^2(M)$ eigenfunctions. In this it is little different from the laplacian on $M$ . $\endgroup$ – mike stone Jan 23 at 15:06

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