Relativistic Von Neumann equation?

In non-relativistic quantum theory, Schrödinger's equation can be re-expressed using the density matrix $\rho=|\psi\rangle \langle\psi|$ as the Von Neumann equation: $$i\partial_t \rho = \frac{1}{\hbar}[H,\rho]$$

Now for spin-$\frac{1}{2}$ fermions relativistic theory introduces the Dirac equation, to which no exposition on the relativistic counterpart to the equation for the density matrix is given (or, at least, I have never seen one). Out of curiosity I derived what I think is the analogue: $$i\gamma^\mu\partial_\mu\rho=\frac{1}{\hbar c}[H_0,\rho]$$

Where $H_0=mc^2+V$ is the "rest hamiltonian." The issue is that the $mc^2$ term is not an operator, so it should be canceled in the commutator, but that would lead to the nonsensical conclusion that mass has no effect on the matrix. My first thought is to replace that term with an energy operator as $H_0\to i\hbar\partial_t$ or $H_0\to\hat H$, the "standard" hamiltonian used in the schrodinger equation, but I do not know how to make that notion rigorous, nor if that substitution can reproduce the Dirac equation.

Is it valid to make such a substitution or otherwise express the Dirac equation in a similar way?

• In the Hamiltonian the mass is an operator. It's $m \gamma^0$. – Ryan Thorngren Aug 14 '18 at 7:31
• @RyanThorngren when you write $m\gamma^0$ is your $m$ the rest mass? – user195162 Aug 14 '18 at 15:39
• I'm just saying that if you want to write the Dirac equation as a time-dependent Schrodinger equation, then you have to move the $\gamma^0$ which is in front of $\partial/\partial t$ over to the right hand side. $m$ is whatever mass appears in the Dirac equation (the bare rest mass I suppose). – Ryan Thorngren Aug 14 '18 at 17:43

The Dirac equation is a Schrödinger-like equation, with the Hamiltonian being: \begin{align} H_0&=γ^{0}mc^{2}+γ^{0}γ^{k}cp^{k}\\&=γ^{0}mc^{2}-iℏγ^{0}γ^{k}c\frac{∂}{∂x_{k}} \end{align} Hence, the von Neumann equation remains the same in the relativistic case.