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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?

octonion
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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.

mike stone
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