Prove the hermiticity of $\sigma_{\mu\nu} F^{\mu\nu}$ In exercise 4.2, "Relativistic Quantum Mechanics" by Bjorken-Drell, an additional term is added  to the Dirac Hamiltonian such that new equation of motion is
$$\left(i\gamma_\mu{\nabla^\mu}-e_{i}\gamma_\mu{A^\mu}+\frac{\kappa_{i}e}{4M_{i}}\sigma_{\mu\nu}F^{\mu\nu}-M_{i}\right)\psi(x)=0.$$
The first two and fourth terms generate the Dirac equation. The third term has to be hermitian where $\kappa_{i}$ is a real number and $e$ is the electron charge. How do I prove it?
My attempt:
$$\left(\sigma_{\mu\nu}F^{\mu\nu}\right)^{\dagger}$$
$$=(-i)\gamma_{\nu}^{\dagger}\gamma_{\mu}^{\dagger}F^{\mu\nu}$$
$$=-i\begin{cases}
\begin{array}{c}
-\gamma_{0}\gamma_{i}F^{i0}\\
\gamma_{i}\gamma_{j}F^{ji}
\end{array} & \begin{array}{c}
\nu=0,\mu=i\\
\nu=i,\mu=j
\end{array}\end{cases}$$
$$=\begin{cases}
\begin{array}{c}
\sigma_{0i}F^{i0}\\
-\sigma_{ij}F^{ji}
\end{array} & \begin{array}{c}
\nu=0,\mu=i\\
\nu=i,\mu=j
\end{array}\end{cases}$$
$$=\begin{cases}
\begin{array}{c}
-\sigma_{0i}F^{i0}\\
+\sigma_{ij}F^{ij}
\end{array} & \begin{array}{c}
\nu=0,\mu=i\\
\nu=i,\mu=j
\end{array}\end{cases}$$
In this way, I can't find it to be hermitian.
 A: It is wrong trying to prove that
$$
\sigma_{\mu\nu}F^{\mu\nu}
$$
is hermitian. Actually it's not!
The right way is to prove that the action
$$
S= \int dx^4\bar{\psi}(x)\left(i\gamma_\mu{\nabla^\mu}-e_{i}\gamma_\mu{A^\mu}+\frac{\kappa_{i}e}{4M_{i}}\sigma_{\mu\nu}F^{\mu\nu}-M_{i}\right)\psi(x)
$$
is hermitian or equivalently to prove that the counterpart of the equation of motion
$$
\left(i\gamma_\mu{\nabla^\mu}-e_{i}\gamma_\mu{A^\mu}+\frac{\kappa_{i}e}{4M_{i}}\sigma_{\mu\nu}F^{\mu\nu}-M_{i}\right)\psi(x)=0
$$
for $\bar{\psi}(x)$ is
$$
\bar{\psi}(x)\left(i\gamma_\mu{\nabla^\mu}-e_{i}\gamma_\mu{A^\mu}+\frac{\kappa_{i}e}{4M_{i}}\sigma_{\mu\nu}F^{\mu\nu}-M_{i}\right)=0.
$$
Therefore, it can be easily seen that for the extra non-Dirac term, the proof of hermiticity boils down to proving
$$
\gamma_0\left(\sigma_{\mu\nu}F^{\mu\nu}\right)^{\dagger}\gamma_0= \sigma_{\mu\nu}F^{\mu\nu}
$$
or
$$
\gamma_0\left(\sigma_{\mu\nu}\right)^{\dagger}\gamma_0= \sigma_{\mu\nu}.
$$
The proof is pretty straight forward, and I will leave it yourself. A hint:
$$
\left(\gamma_\mu\right)^{\dagger}= \gamma_0\gamma_\mu\gamma_0.
$$

Added note:
The extra term:
$$
\bar{\psi}(x)\left(\frac{\kappa_{i}e}{4M_{i}}\sigma_{\mu\nu}F^{\mu\nu}\right)\psi(x)
$$
is of dimension 5, thus not renormalizable and excluded from the Standard Model.
However, from an effective field theory point of view, it should be part of the Lagrangian, since it observes all the required symmetries: the Lorentz symmetry and the $U_{EM}(1)$ gauge symmetry. The catch is that the term is extremely small
($\kappa_{i}$ is small) under low energy circumstances. That said, this sort of terms would be unavoidably important right after the big bang, where the high energy Planck scale effect is front and center.
