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.

Once you use this inner product the energy eigenfunctions are orthonormal in the usual manner of self-adjoint operators.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.