In 3+1 dimensions, define the Dirac operator as
$$\tag{1} \not D = \gamma^\mu (\partial_\mu -i A_\mu )$$
where $A_\mu$ is a $U(1)$ gauge field. Define the following inner product between spinor fields:
$$\tag{2} \langle u ,v \rangle = \int \bar{u}v .$$
It's easy to show that $i\not D$ is hermitian with respect to this inner product — e.g. see this answer and this slightly clearer answer. Usually, I'd immediately conclude that $i\not D$ has only real eigenvalues, but here I'm not so sure. The inner product (2) is not positive definite, since $\bar{u}v = u^\dagger \gamma^0 v$ and $\gamma^0$ has negative eigenvalues. So the usual proof that the eigenvalues of a hermitian operator are real does not work.
As a reminder, here's the usual proof. Assume $H$ is hermitian. Let $u$ be a nonzero eigenvector with eigenvalue $\lambda$. By hermiticity, $\langle Hu, u\rangle = \langle u, Hu \rangle$. Hence $\lambda^* \langle u, u\rangle = \lambda \langle u,u\rangle$. Assuming $\langle \cdot,\cdot\rangle$ is positive definite, we can conclude $\lambda^* = \lambda$.
But this last step relies crucially on positive definiteness: otherwise, we could have a nonzero eigenvector $u$ with $\langle u,u \rangle =0$. So, in principle, it seems $i\not D$ could have imaginary eigenvalues.
Does $i\not D$ actually have imaginary eigenvalues, or does something prevent this from happening?
Best Answer
No, the eigenvalues of the Dirac operator are in general not real. In fact, the theory of Dirac operators in Lorentzian signature is somewhat exotic - most mathematical texts consider only the Riemannian case, where the Dirac operator is just straightforwardly self-adjoint under the canonical positive-definite inner product induced by the Riemannian metric.
Additionally, it depends on whether you mean "eigenvalues" in the strict sense of mathematicians as elements of the point spectrum or the non-strict sense of physicists as just any element of the spectrum (as in: do you consider the position operator $x$ on $L^2(\mathbb{R}^3,\mathrm{d}x)$ to have no eigenvalues or all of $\mathbb{R}$ as eigenvalues?).
"Spectra of the Dirac Operator of Pseudo-Riemannian Spin Manifolds" by Reincke proves for instance that on a Lorentzian Torus the spectrum of the Dirac operator is always $\mathbb{C}$, but the point spectrum is indeed real. On Minkowksi space, the spectrum is $\mathbb{C}$ and the point spectrum is empty, so indeed it is the "standard" case that the Dirac operator has non-real spectrum!
I could not find any other known Dirac spectra for the Lorentzian case, so there appears to be no example for non-real point spectrum of a Lorentzian Dirac operator.