3

I have seen at various places the comment that the operator $P_\mu P^\mu$ is a Casimir operator of Lorentz algebra and thus it satisfies a on-shell condition like $P_\mu P^\mu=m^2$. Given the Poincare algebra \begin{aligned} i\left[M^{\mu \nu}, M^{\rho \sigma}\right] &=g^{\mu \sigma} M^{\nu \rho}+g^{\nu \rho} M^{\mu \sigma}-g^{\mu \rho} M^{\nu \sigma}-g^{\nu \sigma} M^{\mu \rho} \\ i\left[P^{\mu}, M^{\rho \sigma}\right] &=g^{\mu \rho} P^{\sigma}-g^{\mu \sigma} P^{\rho} \\ \left[P^{\mu}, P^{\nu}\right] &=0. \end{aligned} How does one derive its Casimir operators, especially the one $P_\mu P^\mu$? Can someone show the crucial steps? Also Does the method works for any other similar algebra? Moreover, if an operator, say, $A$ commutes with the generators $M^{\mu\nu}$, i.e., $[A,M^{\mu\nu}]=0$, can it be said that $A$ is a Casimir operator?

fogof mylife
  • 568

2 Answers2

1

Casimir operators commute with all generators. That's what you need to check. $P^\mu P_\mu$ does commute with $M$ and $P$. A fast way to say it is that

  1. $[P^2,P_\mu] \propto [P_\nu,P_\mu] =0$
  2. $P^2$ is a scalar and therefore it's annihilated by the rotation generators

But if you don't believe 2. you can just check $$ \begin{aligned} i[M_{\mu\nu},P^2] &= 2\,(g_{\rho\mu}P_\nu-g_{\rho\nu}P_\mu)P^\rho\\ &=2 P_\mu P_\nu - 2 P_\nu P_\mu \\&= 0\,. \end{aligned} $$ For other Casimirs such as $$ W^\mu W_\mu\,,\qquad W_\mu := \tfrac12 \varepsilon_{\mu\nu\rho\lambda} M^{\nu\rho}P^\lambda\,, $$ you can do the same. This is just a bit harder. The argument 2. still works because it's a scalar. Then by explicit computation $$ [P_\mu,W_\nu] = 0\,, $$ so also its square commutes with $P$.

MannyC
  • 7,150
0

As far as I know, there are no ways of deriving Casimirs. In the case of the Poincare algebra one guesses and checks that $P_{\mu}P^{\mu}$ and $W_{\mu}W^{\mu}$ are Casimirs, based on physical principles.

There are some ways of deriving the quadratic Casimir of a (simple) Lie algebra. If you calculate its Cartan metric then the quadratic Casimir is $g_{\mu \nu}x^{\mu}x^{\nu}$.

You might check the answers to this question: Explicit Quadratic Casimir for $sp(2N)$

S Thomas
  • 51