In the scalar field Lagrangian the mass term is given by

$$m^2 \phi^2.$$

But the equivalent term in Maxwell's Lagrangian for electromagnetism is


But I don't know why the latter is correct. Could anyone help me getting to it?

  • $\begingroup$ Maxwell electrodynamics is massless, you're asking about the Proca action. $\endgroup$ – Ryan Unger Jun 9 '16 at 11:45
  • $\begingroup$ The coupling a scalar field $\phi$ with Maxwell field $A_{\mu}$ is a consequence of gauge invariance. For $U(1)$ gauge theory this corresponds to invariance of the Lagrangian under $\partial_{\mu}\rightarrow \partial_{\mu}+igA_{\mu}$. If you add a term in the Lagrangian like $A^{\mu}A_{\mu}$, which is no longer invariant under the gauge transformations. That is why respecting the gauge invariance to mass term is allowed for Maxwell's field. $\endgroup$ – AMS Jun 9 '16 at 12:38

The fact that $m^2A_\mu A^\mu$ is both Lorentz invariant tells that this term is an allowed term in the Lagrangian. But it does not explain why $m^2 A_\mu A^\mu$ represent the mass term.

The reason why $m^2\phi^2$ is called the mass term of the scalar field is that after quantization, each single particle state $|p\rangle$ in the Fock space is characterized by definite eigenvalue $m^2$ for the operator $P_\mu P^\mu$ i.e., $$P_\mu P^\mu|p\rangle=p_\mu p^\mu|p\rangle=m^2|p\rangle$$ which amounts to the relativistic dispersion relation $E^2-\textbf{p}^2=m^2$. Therefore, one can identify $m$ as the mass of the quanta of the theory.

The reason is same for a massive Proca field ("massive photon"). If you start with a term $m^2A_\mu A^\mu$ in the Lagrangian, after quantization, the operator $P_\mu P^\mu$ yields the eigenvalue $m^2$ on the one-particle states. It is the same $m^2$ that appear in the relativistic dispersion relation for that particle.

  • 1
    $\begingroup$ Simple good answer. $\endgroup$ – gented Jun 4 '19 at 13:43

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