Hermiticity of Dirac Operator $\gamma^{\mu}D_{\mu}$ and Expansion in eigenmodes

Question

I'm interested to know under what conditions $\gamma^{\mu}D_{\mu}$ is a hermitian operator.

I am studying the Fujikawa method of anomalies and I see that many sources have different answers for this. Some claim that $\gamma^{\mu}D_{\mu}$ is hermitian or possibly anti hermitian in Euclidean space, or that $i\gamma^{\mu}D_{\mu}$ is hermitian in Minkowski space. I need the hermiticiy condition so that I can expand the Dirac spinors $\Psi(x)$ and $\bar{\Psi}(x)$ in a basis of orthonormal eigenvectors of the Dirac operator to perform the path integral.

Another question I have is this expansion. I would like something of the form

$$\Psi(x) = \sum_n \psi_n(x)a_n$$

$$\bar{\Psi}(x) = \sum_n \bar{\psi_n}(x)\bar{b}_n$$

where $a_n$ and $\bar{b}_n$ are elements of a Grassmann algebra. But how is $\bar{\psi}_n$ defined? Some references define $\bar{\psi}_n(x)$ = ${\psi}(x)^{\dagger}$, but I guess it depends on how you are defining your inner product.

Thanks, I've attached some references below.

Fujikawa 1) https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.42.1195

Fujikawa 2) https://journals.aps.org/prd/abstract/10.1103/PhysRevD.21.2848

TASI Anomalies https://arxiv.org/abs/hep-th/0509097

Answer 1

The (operator-valued) matrix $\not D=\gamma\cdot D$ is anti-hermitian with respect to the inner product $$ \langle v,u\rangle:=\int_{\mathbb R^d} \bar v(x)u(x)\ \mathrm dx\tag1 $$ where $$ \bar v(x):= v^\dagger(x)\gamma^0\tag2 $$

Indeed, write $$ \not D=\not\!\partial-i\not A\tag3 $$

It is clear that the second factor satisfies \begin{equation} \begin{aligned} \overline{i\not A}&=-i\overline{\not A}\\ &=-i\not A \end{aligned}\tag4 \end{equation} where we have used the fact that $\gamma^\mu$ satisfies $$ \gamma^0\gamma^{\mu\dagger}\gamma^0=\gamma^\mu\tag5 $$ and that $A^\mu$ is hermitian.

On the other hand, the first factor in $(3)$ satisfies $$ \begin{aligned} \langle v,\not\!\partial u\rangle&=\int_{\mathbb R^d} \bar v(x)\not\!\partial u(x)\ \mathrm dx\\ &=-\int_{\mathbb R^d} \bar v(x)\overset{\leftarrow}{\not\!\partial}u(x)\ \mathrm dx\\ &=-\int_{\mathbb R^d} \overline{(\bar{\not\!\partial }v)}(x)u(x)\ \mathrm dx\\ &=-\langle \not\!\partial v,u\rangle \end{aligned}\tag6 $$ where in the first equality we integrated by parts and in the last one we used $(5)$.

From this we learn that $\not D$ satisfies $\overline{\not D}=-\not D$, which is equivalent to the statement that $\not D$ is anti-hermitian with respect to $\langle\cdot,\cdot\rangle$, as claimed.

Note that this notion of anti-hermiticity guarantees that $i\not D$ is diagonalisable, with real eigenvalues and orthogonal eigenvectors, by the usual reasons (spectral theorem, etc.): $$\tag7 i\not Dv_i=\lambda_i v_i\qquad\Rightarrow\qquad\begin{cases}\lambda_i\in\mathbb R\\\langle v_i,v_j\rangle\propto\delta_{ij}\end{cases} $$