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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 (:

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    $\begingroup$ As a comment: If you remove the $m$ then considering whether the probability is conserved? $\endgroup$ – jim Jul 12 at 15:10
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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.

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  • $\begingroup$ Thank you for the insightful answer! I understand now that I need to define $M_{tot}= H^2 - (P)^2$. What would the form of the operator $P$ be in terms of the fields? And do you have an intuition of what the $M_{tot}$ I defined above then would correspond to? Is it the bare total mass of system of particles with no relative motion with respect to each other (i.e. the sum of their rest masses)? $\endgroup$ – B. Croydon Jul 15 at 11:40
  • $\begingroup$ @B.Croydon Remember the square-root: $M_\text{tot} = \sqrt{H^2-(P)^2}$. The momentum operators (generators of translations in space) have the form $P_k\sim\int d^3x\ \psi^\dagger\partial_k\psi$. By the way, the Hamiltonian (energy operator, generator of translations in time) has the form $H\sim\int d^3x\ \psi^\dagger\partial_0\psi\sim \int d^3x\ \psi^\dagger \gamma^0 (i\gamma^k\nabla_k+m)\psi$. To check these, you can verify that the commutators of $\psi$ with $\vec P$ and $H$ are $\sim\vec\partial\psi$ and $\sim\dot\psi$, respectively. I've omitted factors of $\pm i$, but that's the idea. $\endgroup$ – Chiral Anomaly Jul 15 at 13:45
  • $\begingroup$ Thank you, I got it now and it worked! $\endgroup$ – B. Croydon Jul 15 at 18:15

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