The massless Dirac Hamiltonian is given by $H = -i \gamma^0 \gamma^i \partial_i \equiv -i \alpha^i \partial_i $. If I define an inner product of spinors as
$$ ( \psi , \phi ) = \int d^n x \psi^\dagger \phi$$
then we have
$$ (\psi , H \phi ) = \int d^n x \psi^\dagger ( - i \alpha^i \partial_i \phi) = \int d^n x (i \partial_i \psi^\dagger \alpha^i ) \phi = \int d^nx (-i\alpha^i \partial_i \psi)^\dagger \phi =(H\psi, \phi) $$
where I have used the fact that $(\alpha^i)^\dagger = \alpha^i$ and integrated by parts, so I would conclude that $H$ is Hermitian. However, we are frequently told that, for spinors, we must really use the Lorentz invariant inner product
$$ \langle \psi , \phi \rangle = \int d^n x \bar{\psi} \phi =\int d^n x \psi^\dagger \gamma^0 \psi$$
which is used, for example, when writing down the Dirac action. In this case, the Dirac Hamiltonian is not Hermitian w.r.t. this inner product:
$$ \langle \psi , H \phi \rangle = \int d^n x \psi^\dagger \gamma^0 (-i \alpha^i \partial_i \phi) = \int d^n x (i \partial_i \psi^\dagger \gamma^0 \alpha^i) \phi = \int d^n x (-i \alpha^i \gamma^0 \partial_i \psi)^\dagger \phi \neq \langle H \psi , \phi \rangle$$
where I have also used $(\gamma^0)^\dagger = \gamma^0$.
My questions
The Dirac equation $i \gamma^\mu \partial_\mu \Psi = 0$ can be moulded into the Schrodinger form as $$ i \partial_t \Psi = H \Psi$$
by splitting up the space and time parts, where $H$ is defined as above. If I make the usual phase ansatz $\Psi(t,\mathbf{x}) = \psi(\mathbf{x})e^{-iEt}$, then we have the time-independent Schrodinger equation
$$ H \psi = E \psi$$
However, from above, the Hamiltonian $H$ is not Hermitiain w.r.t. the inner product $\langle \cdot , \cdot \rangle$, which seems strange. My questions are the following:
- The Hermiticity of the Hamiltonian seems to depend upon which inner product we use. Does it matter that the Hamiltonain is not Hermitian w.r.t. the inner product $\langle \cdot , \cdot \rangle$?
- Hermitian operators have orthogonal eigenstates, but this seems to depend heavily on the choice of inner product. Am I right to conclude that the solutions to the Dirac equation are not orthogonal w.r.t. the inner product $\langle \cdot , \cdot \rangle$?