The fermionic zero modes to the "standard" $SU(2)$ 't Hooft instantons are known as follows (see page 42 in this article):

For a 't Hooft instanton (self-dual field strength tensor) in the regular gauge, centered at the origin, the massless Dirac equation admits a unique right-handed normalizable fundamental-representation Weyl spinor solution (up to an overall constant):

$$\psi_{\alpha,a}(x)=\begin{pmatrix}0 \\ \lambda_{\alpha,a}(x) \end{pmatrix}\,\,\, \textrm{(Right-handed)}$$

$$\lambda_{\alpha,a}(x)=\frac{\epsilon_{\alpha a}}{(x^2+\rho^2)^{3/2}}$$

where $\alpha=1,2$ is the spinor index, $a=1,2$ is the color/gauge index, and $\epsilon_{\alpha a}$ is the standard Levi-Civita tensor which surprisingly mixes spinor and color indices. By definition, this solution satisfies the equation:

$$(\sigma^{\mu})^{\alpha \alpha'}(\partial_\mu \lambda_{\alpha', a}-iA^{m}_{\mu}(T_m)_{ab}\lambda_{\alpha', b})=0$$

where $A^m_{\mu}$ is the 't Hooft instanton gauge field, $T_m$ are the 3 $SU(2)$ infinitesimal generators (here in the fundamental representation, so they're just Pauli matrices), and $\sigma^\mu=(\sigma^i,i)$.

In the singular gauge, the corresponding solution is:

$$\lambda_{\alpha,a}(x)=\frac{1}{x (x^2+\rho^2)^{3/2}}x_{\mu}(\sigma^{\dagger \mu})_{\alpha,a}$$

[My Question] What about non-zero modes? In other words, what is the spectrum of the Dirac operator in the background of an $SU(2)$ self-dual instanton? ($\gamma^{\mu}D_{\mu}\psi=\lambda \psi$)

Do the corresponding fermionic non-zero modes have analytic solutions? What about orthogonality relations between these eigenfunctions?

I'm having a hard time finding answers to this online...


I am not aware of any attempts to write down the non-zero mode solutions. However, we are (typically) not really interested in individual non-zero mode solutions, but in the non-zero mode propagator $$ S(x,y)=\sum_{\lambda\neq 0} \frac{\psi_\lambda(x)\psi_\lambda^\dagger(y)}{\lambda} $$ and this object has been determined, see here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.