# How are Lagrangians in QFT constructed?

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?

• Modern day theorists are all about Effective field theory. Here, the first step is to identify all the symmetries of a system. Then, we write down the most general Lagrangian that respects all of these symmetries, including all higher dimensional operators. Terms that are quadratic in this Lagrangian give rise to the free field equations, while cubic and higher order terms lead to interactions. – Prahar May 12 '16 at 20:48
• Educated guessing, like with all other theories and models in physics. – CuriousOne May 12 '16 at 20:50
• @Qmechanic I can see what you mean. But I still feel uncomfortable with it. – descheleschilder May 12 '16 at 23:49

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.

• We could also argue that for terms with too many fields or operators, these become irrelevant under RG flow. – Aurey May 12 '16 at 22:36
• The answer begs the question why a principle of least action should hold. Does stating that U(1) symmetry means that the action is invariant under the variation of A is equivalent to stating thatU(1) symmetry means that a Lagrangian is invariant under local gauge transformations? – descheleschilder May 13 '16 at 0:04
• @descheleschilder Least action is the classical approximation. The real principle is stationary action, and stationary points in the action dominate the contribution to the path integral (e.g.). That is, all paths contribute to quantum mechanics, the stationary action path is just the dominant contribution. – Sean E. Lake Aug 26 '17 at 15:31
• Why must the field equations not contain derivatives or order higher than 2? Is this a symmetry requirement? – Geoffrey Nov 14 '17 at 0:25