The first operator is a Lorentz scalar formed from commuting translations. Since $[P_\mu,P_\nu]=0$,
$$
[P^2,P_\rho]=0,
$$
and the vector transformation law of $P^\mu$ gives
$$
[M_{\rho\sigma},P^\mu P_\mu]=0.
$$

The <Pauli-Lubanski pseudovector> commutes with every $P_\mu$, so $[W^2,P_\mu]=0$. Its Lorentz-vector commutator similarly gives
$$
\begin{aligned}
[M_{\rho\sigma},W^\mu W_\mu]
={}&[M_{\rho\sigma},W^\mu]W_\mu
+W^\mu[M_{\rho\sigma},W_\mu]=0,
\end{aligned}
$$
because the two tensor-rotation terms cancel after relabelling the contracted index. Thus
$$
C_1=P^2,
\qquad
C_2=W^2
$$
commute with every generator of the <Poincare algebra>. They are its two algebraically independent <Casimir operators> in four spacetime dimensions.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-305/1/a/solution