# [Physics] How are Lagrangians in QFT constructed

actionfield-theorylagrangian-formalismvariational-principle

Various particle equations (like the K-G equation, the Dirac equation, the Proca equation etc.) in QFT are derived by applying the Euler-Lagrange equations to the Lagrangian density. But how are these Lagrangian densities constructed without reference to the particle equations?

Symmetry, stability and dimension analysis. You can consider a scalar field theory, for instance. A dynamical action for such a theory must be

$S = \int d^4 x \, \, \partial_\mu \phi \partial^\mu \phi$

because

i. Lorentz symmetry indicates that all the indices must be properly contracted

ii. The field equations must not exceed second order in derivative

iii. You can add a potential term $V(\phi) = m^{4-n} \phi^n$ just by dimension analysis. Note that this term satisfies the first two conditions as well.

You can also consider a massless vector field $A_\mu$. In this case, you must satisfy two symmetries

a. Lorentz (means all indices must be contracted)

b. $U(1)$ (Means that the action must be invariant under the variation $\delta A_\mu = \partial_\mu \Lambda$. Thus, you must use an invariant object, which is $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$.

Thus, the kinetic action that satisfies that symmetry principle as well as the stability (second order field equation) is

$S = \int d^4 x \, \, F_{\mu\nu} F^{\mu\nu}$

You can add a potential term to this Lagrangian, i.e. $m^{4-n} (A_\mu A^\mu)^n$, as in the scalar case. In this case, you can no longer satisfy the $U(1)$ symmetry, so you have to give up on that. Note that choosing $n=2$ would lead to Proca.