# Are the solutions to the Dirac equation orthogonal?

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:

1. 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 $$\langle \cdot , \cdot \rangle$$?
2. 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 $$\langle \cdot , \cdot \rangle$$?

You need to use the $$\langle \psi|\chi\rangle=\int \psi^\dagger \chi\,d^3x$$ to do Dirac single particle quantum mechanics. There is no conflict with Lorentz invariance because rewriting the Dirac equation as $$i\partial_t \psi =H_{\rm Dirac} \psi$$ has already broken explicit Lorentz invariance. Of course $$\psi^\dagger \psi$$ is not Lorentz invariant, but then neither would any form of probablity density as Lorentz transformations alter the volume.