Calculate the determinant of the Dirac operator Suppose the Dirac operator
$$
D = a_{\mu}\gamma^{\mu} + b_{\mu}\gamma^{\mu}\gamma_{5} - m
$$
How to calculate the logarithm of its determinant
$$
\text{ln}\left[\text{det}D\right]?
$$
I think that for $m = 0$, for example, it is possible somehow to separate the contributions from the different chiralities sectors, but completely don't understand how.
 A: It is pretty obvious that $\ln(\det D)=\frac{1}{2}\ln(\det D^2)$ (if the logarithm in the problem is well-defined).
Let us introduce notation $A=a_\mu\gamma^\mu$, $B=b_\mu\gamma^\mu$, then $$D^2=A^2-B^2+m^2-2 m A-2 m B \gamma^5+A B\gamma^5-B A\gamma^5= a_\mu a^\mu-b_\mu b^\mu-2 m D-m^2+(A B-B A)\gamma^5.$$ Therefore, if we calculate eigenvalues of $(A B-B A)\gamma^5$, we can calculate the eigenvalues of D. On the other hand, $(A B-B A)\gamma^5$ is block-diagonal (say, in the chiral representation of gamma-matrices), so all you need is to calculate eigenvalues of 2x2 matrices (and the contributions from different chiralities are indeed separated, if I understand you correctly). Maybe you can simplify that task even more, but I haven't looked that far. And I cannot guarantee that my answer does not contain errors.
EDIT(11/27/2016): Additionally, I suspect that the eigenvalues of $(A B-B A)\gamma^5$ can be best expressed via the invariants of the "electromagnetic field" ($F_{\mu\nu}=a_\nu b_\mu-b_\nu a_\mu$): $F_{\mu\nu}F^{\mu\nu}$ and $\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}$.
EDIT(11/29/2016) My guess of the previous edit was correct. We need to calculate the eigenvalues of $(A B-B A)\gamma^5$. However, matrix $$S=A B-B A=a_\mu\gamma^\mu b_\nu\gamma^\nu-b_\mu\gamma^\mu a_\nu\gamma^\nu=$$ $$=F_{\mu\nu}\gamma^\mu\gamma^\nu,$$ where $F_{\mu\nu}$ is an antisymmetric tensor, commutes with $\gamma^5$, so there are eigenstates of both $S$ and $\gamma^5$. It is more convenient for me to calculate the eigenvalues of $S$, rather than $S \gamma^5$. If we consider the invariants of tensor $F_{\mu\nu}$ $$I_1=F_{\mu\nu}F^{\mu\nu}=2(-(F_{01})^2-(F_{02})^2-(F_{03})^2+(F_{12})^2+(F_{13})^2+(F_{23})^2)$$ and $$I_2=\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}=4(F_{03}F_{12}-F_{02}F_{13}+F_{01}F_{23}),$$ then one can directly check that the eigenvalues of $S$ are $\pm\sqrt{-2(I_1+i I_2)}$ and $\pm\sqrt{-2(I_1-i I_2)}$, unless I messed something up. I used the chiral representation of the gamma-matrices and the notation of the book by Itzykson and Zuber. I guess $S \gamma^5$ has the same eigenvalues.
