# Lorentz transformation of operator

I am reading the Zee book on group and at pg.442 when explaining the Poincare' algebra an the infinity dimensional representation, it states that since $$P_{\mu}P^{\mu}$$ is a Lorentz invariant then it commutes with all the generators of the Lorentz algebra $$so(1,3)$$.

Since $$P_{\mu}$$ are differential operator, is there any simple proof of that statement?

In the Wu-Ki Tung book pg.182 (I am using it only occasionally as a reference) an operator transform as follow under a Lorentz transformation:

$$T \to \Lambda T \Lambda^{-1}$$

Where that comes from?

I know that using element of $$SL(2,\Bbb C)$$ a Lorentz transformation can be defined as $$T \to X T X^\dagger$$ where $$X \in SL(2,\Bbb C)$$, so may be that's the way to prove that statements

• I'm not sure what you think there is to prove - the $P_\mu$ as generators of translation are part of the Poincaré algebra, and $P_\mu P^\mu$ being Lorentz invariant holds already on the level of the Poincaré algebra (it's a Casimir operator) before choosing any specific representation as differential operators, matrices or whatever else. Nov 16 '21 at 20:57
• I am fine with $P_{\mu} P^{\mu}$ beeing invariant but I don't understand how easily follow that it commutes with the generators $J^{\mu\nu}$. I have got it by doing explicit calculation using the commutators. But I have the same problem with the Pauli-Lubanski operator $W^{\mu}$ and with that doing explicitly calculation using commutators is much more challenging Nov 16 '21 at 21:03
• The $J^{\mu\nu}$ generate the Lorentz transformation - how could something that does not commute with them be invariant under them? (try writing down the infinitesimal version of $P \mapsto \Lambda P \Lambda^{-1}$) Nov 16 '21 at 21:05
• How do I check if an operator is Lorentz invariant? Sorry may be a silly question but I'm not sure how to do that with operators Nov 16 '21 at 21:09
• Zee's being silly. An operator being Lorentz invariant means, by definition, that it commutes with $J^{\mu\nu}$. So his claim is rather vacuous. What you have to do is to compute the commutator explicitly; just use the Lorentz algebra to compute $[J^{\mu\nu},P_\rho P^\rho]$ and check that, indeed, it vanishes. Nov 16 '21 at 21:51