Modified Dirac Equation

With the additional gauge potential ${\bf A}$, our Dirac eigenvalue problem differs from HCM. From (11) we have

$$[{\boldsymbol \alpha}\cdot ({\bf R}-\gamma_5\, {\bf A}) + g \beta (\varphi^R -i\, \gamma_5\, \varphi^I) \Psi = E\, \psi$$ (18)

Observe that $C\equiv \alpha^3$ anti-commutes with all the matrices on the left side of (16). Therefore if $\psi_E$ is an eigenfunction with eigenvalue $E$, $M\psi_E$ belongs to eigenvalue $- E$, and zero modes can be chosen as eigenstates of $M$.

Next we show that the gauge interaction in (16) does not affect the zero modes found by HCN at ${\bf A}= 0$. To this end, we adopt the Coulomb gauge and present ${\bf A}$ as $A^i = \varepsilon^{ij}\, \partial_j \, A$. Also it is true $\alpha^i\, \gamma_5 = -i\, \varepsilon^{ij}\, \alpha^i\, M$. Thus the kinetic term in (16) also is $e^{-A C}\, {\boldsymbol \alpha}\cdot {\bf R}\, e^{-A C}$ and (16) becomes

$$({\boldsymbol \alpha}\cdot {\bf R}+ g\, \beta\, [\varphi^R - i \, \gamma_5\, \varphi^I]) \quad (e^{-AM}\, \Psi) \quad =\quad E\ (e^{AM} \, \Psi)$$ (19)

and $e^{-AM}\, \Psi$ satisfies the HCM equation at $E=0$. Comparison with (15) shows that $A (r) = n \, a\, (r) / r$, so that the infinity $A^\prime \, (r)$ lends to $-\frac{n}{2}\ l n \, r$, and the zero modes with the gauge interaction differ from the HCM modes by factors $r^{\pm} \, n/2$. This does not affect nomalizability because the zero modes are exponential damped by the interaction with $\varphi$. Finally, since the HCN mode as well as ours has the form for $$n =-1, \Psi = \left( \begin{array}{c} 0\\ v\\ v\\ 0 \end{array}\right)$$, we see that indeed it is an $M$ eigenstate, with eigenvalue $-1$. Fermion number fractionalization in the gauge theory is now established by the same reasoning as in HCM.