# Lagrangian for a free antimatter particle

Context

The Lagrangian, $$L$$, of a free particle is derived many places including in Section 8 of Landau's Theory of Classical Fields. The well-known result for a free material particle of mass $$m$$ and speed $$v$$ is $$L = -mc^2\sqrt{1-\frac{v^2}{c^2}},$$ where $$c$$ is the speed of light. It is known that in many respects a quantity assigned to a particle (say an electron) is equal and opposite to that of its antiparticle (say a positron).

Question

My question is this: Is the Lagrangian of a free antiparticle $$L = +mc^2\sqrt{1-\frac{v^2}{c^2}}?$$

• Is this question motivated from Dirac's equation derivation? Mar 4 at 21:52

My question is this: Is the Lagrangian of an antiparticle $$L = +mc^2\sqrt{1-\frac{v^2}{c^2}}?$$

No. An antiparticle has the same mass as its particle partner.

The antiparticle has the opposite charge from its partner, not the opposite mass.

So, in this sense the term "antimatter" is a bit of a misnomer. What is "anti" is the charge not the mass.

The Lagrangian of a classical charged particle $$\mathrm Q$$ in an EM field is given by

$$L_{\mathrm Q}(V,\mathbf A)=-mc^2\sqrt{1-\frac{v^2}{c^2}}+q\mathbf v \cdot \mathbf A -q V,$$ where $$q$$ is the charge, $$\mathbf A$$ is the vector potential and $$V$$ the electric potential.

For its anti-particle $$\overline{\mathrm Q}$$, we have

$$L_{\overline{\mathrm Q}}(V,\mathbf A)=-mc^2\sqrt{1-\frac{v^2}{c^2}}-q\mathbf v \cdot \mathbf A +q V.$$

It's the charge signs that change, not the mass. When we set $$V=0$$ and $$\mathbf A=0$$, we get $$L_{\rm Q}(0,0)=L_{\overline{\mathrm Q}}(0,0)$$ thus the same free particle Lagrangians.

So, the kind of Lagrangian you are using from Landau describes a free particle as they understood within the context of the special theory of relativity. Now, particles like electrons and their anti-particles, positrons, are best understood as quantum particles, thus the purely relativistic description does not apply. However, if you wanted to ignore quantum effects, and just treat them as relativistic particles,(which is done quite often and is very useful) then the Lagrangian as you have from Landau applies equally well to both particles and anti-particles. For example, anti-particles are characterized by an opposite charge, i.e. negative for electrons and positive for positrons, however, both have energy that is positive.

Requested extra information (see comments below)

The Lagrangian formulation can be used to derive the "equation of motion", as it were, for a particle/anti-particle in ordinary Quantum mechanics, however, the only correct description for electrons/positrons that takes into account both relativistic and quantum effects is that of quantum field theory, where the specific field theory is known as the Dirac field. However, for the sake of argument and to answer your question, the Lagrangian which treats the electron or positron as a non-relativistic quantum free particle is given as: $$\mathcal L={\hbar^2\over 2m}\nabla\psi\cdot\nabla\psi^*+{\hbar\over 2i}(\psi^*\dot\psi-\psi\dot\psi^*).$$ This Lagrangian density (this theory is a continuum theory) leads to the Schrodinger equation for a free-particle: $$-{\hbar\over 2m}\nabla^2\psi=i\hbar{\partial\psi\over\partial t}.$$ Where of course, $$|\psi|^2$$ is to be interpreted as a probability distribution for finding the particle in some region of space.

However, do not underestimate the utility of Landau's treatment of the electron or positron, as it were, in terms of special relativity. This kind of description is quite useful, even if it is not the full story. Relativistic kinematical treatments are quite useful in particle experiments for example.

• I'd be curious to note what the Lagrangian of a free (anti)particle is in quantum mechanics. Can you include this in your answer? Can you give a citation? Thanks Mar 4 at 21:18
• @MichaelLevy Sure, I have added some notes on Lagrangians and free particles in quantum mechanics. Mar 4 at 21:37
• @MichaelLevy See Goldstein's Classical Mechanics chapter 13 for facts on Lagrangian densities for continuous systems and fields (the problems at the end of the chapter give a Lagrangian for Schrodinger's equation). Mar 4 at 21:42

No, what you call "particle" and what you call "antiparticle" is a mere convention. In both cases, the relativistic Lagrangian is given by $$L= -mc^2(1-\vec{v}^2/c^2)^{1/2}$$.