# Doubts on representations of poincare group and QFT

I am studying Poincare group and encountered the term massless representations of the Poincare group. I know Poincare group is studied by the studying the little group of various momenta, massless and massive. Does massless representations of Poincare group mean representations of the little group for $$p^{\mu} = (1,1,0,0)$$.(first coordinate is the energy).

This is the second question: If we are given a Lagrangian for a field(say Electromagnetic field), then we can study the representations of the relevant field. To study these representations of these, we can study the representations of the little group for various cases and using the method of induced representations, we can get the representations. Did I get the picture right ?

My question is

1.We know oen of the Casimir elements of the poicare group is $$p^{\mu}p_{\mu} = m^2$$ . How does one identify the m with mass of the field we are studying ?

2.If the Lagrangian involved mass-term,(like in Proca equation). How do the representations change? I don't see any reason why the representations should change because of the mass. (relevant explanation for the scalar field is enough for me)

The connection is through the four-momentum operator. Doing canonical quantization starting from the Lagrangian and without ever thinking about any representations on particles, you can show that the four-momentum operator for a scalar field can be expressed in terms of creation/annihilation operators as $$P^\mu = \int p^\mu a^\dagger(\vec p) a(\vec p) \frac{\mathrm{d}^3 p}{(2\pi)^3}$$ so that $$P^2 = P_\mu P^\mu = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{\mathrm{d}^3 q}{(2\pi)^3}\left(q_\mu p^\mu a^\dagger(\vec q)a(\vec q)a^\dagger(\vec p)a(\vec p)\right).$$ Using $$a(\vec q) a^\dagger(\vec p) = (2\pi)^3 \delta(\vec p - \vec q) + a^\dagger(\vec p)a(\vec q)$$, we have that $$P^2 = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\left(m^2 a^\dagger(\vec p)a(\vec p) + \text{term involving two annihilators on the right}\right)$$ and applying this operator to a one-particle state corresponding to the field (i.e. created by $$a^\dagger$$) therefore yields $$m^2$$ since $$\int a^\dagger a$$ is just a number operator.