Is total mass conserved for free Dirac fermions? I am studying quantum field theory and stumbled across the following problem:
Is the total mass conserved for free Dirac fermions? I.e., does the total mass operator commute with the Dirac Hamiltonian (with field operators $\psi(\bf{x})$):
$$H_D = \int \text{d}{\bf{x}} \hspace{0.2cm}\psi^\dagger({\bf{x}}) \gamma^0 [i \gamma^\mu \partial_\mu +m] \psi({\bf{x}}) $$
$$M_\text{tot} = \int \text{d}{\bf{x}} \hspace{0.2cm} m \psi^\dagger({\bf{x}}) \gamma^0 \psi({\bf{x}}) $$
Is it then true that $[H_D, M_\text{tot}]=0$?
My intuition tells me this should hold true, because in the free theory (without interaction via e.g. gauge fields) there should be no particle-antiparticle creation that would change the total mass, but I have so far been unable to prove it. Thank you for your help (:
 A: The expression that you denoted $M_\text{tot}$ is not the total mass operator. Loosely speaking, it ignores the fact that the motion of a system's constituents contributes to the system's total mass. This is true even for free Dirac fermions.
The correct total mass-squared operator is $H^2-(\vec P)^2$, where $H$ is the Hamiltonian (the generator of translations in time) and $\vec P$ are the total momentum operators (the generators of translations in space). This operator commutes with $H$ because $\vec P$ commutes with $H$, which in turn follows from the fact that $H$ is invariant under translations.
This expression for the total mass-squared operator assumes that the constant terms in $H$ and $\vec P$ have been adjusted so that the vacuum state has zero energy and zero momentum. This is allowed because the fact that these operators generate translations is not affected by a change in the constant term, and it is appropriate because "particle" is defined with respect to the vacuum (lowest-energy) state.
