# Lagrangian of a photon?

I can't find this in any textbook, so I must not understand something about it. What is the Lagrangian of a photon? Would it be just h*nu?

Photons have Spin 1. The general Lagrangian for Spin 1 particles is called the Proca Lagrangian and if put into the Euler Lagrange euquation yields the Proca equation. In addition, photons are massless. Therefore putting $m=0$ in the Proco yields the correct Lagrangian for photons. If you put this Lagrangian (i.e. the Proca with $m=0$ ) into the Euler Lagrange equation you get the inhomogeneous Maxwell equation.

You can find the Lagrangian, for example, here

Orodruin
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There is no such thing as the lagrangian of a photon. Photons are quantum excitations of the electromagnetic field, which has a Lagrangian, essentially the lagrangian quoted by unknown1111.

bhobba
massless spin-1 = photon, carry only a kinitic term in L ;

L=-\frac{1}{4}F^{2}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=-\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^{2}

OK, then how does one calculate the action (S) for the amplitude of a photon?

phi = (const) exp[(i/h_bar)S]

Orodruin
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OK, then how does one calculate the action (S) for the amplitude of a photon?
Which part of
There is no such thing as the lagrangian of a photon. Photons are quantum excitations of the electromagnetic field,
was unclear? You need to specify exactly what it is you are trying to do.

bhobba
Feynman & Hibbs, p. 29, eqn 2.15:

I can't seem to get the eqn editor to work.

Feynman & Hibbs, p.29, eqn 2.15: phi[x(t)] = const e^(I/h-bar)S[x(t)]

p.26: S = integral[L(x-dot,x,t) dt]

So if a photon has no Lagrangian, how does one calculate the action, amplitude, probability, etc. for a photon?

Orodruin
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So if a photon has no Lagrangian, how does one calculate the action, amplitude, probability, etc. for a photon?
You don't. You compute the action of the electromagnetic field and correlation functions (essentially amplitudes) between different excitations of the field.

Orodruin
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