I'm trying to prove that the generators of Poincaré group in Poisson bracket close the well-known Poincaré algebra. In particular, I don't know how to prove that $\{P_\mu,P_\nu\}=0$.

Let's take an infinitesimal translation in space-time, such that $$ x^\mu\rightarrow x'^\mu=x^\mu+\epsilon^\mu\Rightarrow\delta x^\mu=\epsilon^\mu \\\phi(x)\rightarrow\phi'(x)=\phi(x-\epsilon)=\phi(x)-\epsilon^\mu\partial_\mu\phi(x) \Rightarrow \delta_0\phi(x)=-\epsilon^\mu\partial_\mu\phi(x) $$ and suppose that our physical system has an action which is invariant under this translation $$ \delta S|_{off-shell}=0 $$ From Noether's theorem we know that we have a conserved charge, so in this case we have 4, one for each direction. And we also know that these charges are the components of the four-momentum

$$ P^\mu=\int_\sigma d\sigma_\nu T^{\nu\mu}(x) $$

where $d\sigma_\nu=n_\nu d\sigma$ ($n_\nu$ is the vector normal to the space-like surface $\sigma$) and $T^{\mu\nu}=\pi^\mu\phi^\nu-\mathcal Lg^{\mu\nu}$ is the stress-energy tensor ($\pi^\mu=\frac{\partial \mathcal L}{\partial\partial_\mu\phi}$ and $\mathcal L$ is the Lagrangian density). We also know that this quantity can be thought as the generator of the transformation, meaning that, if we call $F\equiv-\epsilon^\mu P_\mu$, we have $$ \delta_0\phi=\{\phi,F\} \Rightarrow \partial_\mu\phi=\{\phi,P_\mu\} $$

At this point I'm supposed to be able to say that $\{P_\mu,P_\nu\}=0$, but I don't know how to prove it with the means that I have. I think that, since $P_\mu$ does not depend on the particular point that I want to translate, it doesn't depend on $x$, so this could be a proof, but is it the only one? I'd like to prove it mathematically if it is possible.

Edit: In Roman, Introduction to QFT, page 77, the author says that also $J_{\mu\nu}$ (Lorentz generators) is coordinate-independent and he also uses this to prove the relation for $\{J_{\mu\nu},J_{\sigma\rho}\}$, and he claims that by saying that it follows from the definition (as he says for $P_\mu$), but

  1. Once again, how this follows from the definition? Both $P_\mu$ and $J_{\mu\nu}$ are defined as an integral in $d\sigma_\mu$ over a space-like surface of a current density. But this is true for every generator, so are all generators always coordinate-independent? Do all of them commute with $P_\mu$? It doesn't seem true to me. For example, it is not true for $\{P_\mu,J_{\sigma\rho}\}=g_{\mu\rho}P_\sigma-g_{\mu\sigma}P_\rho$.
  2. If, like Roman says, $J_{\mu\nu}$ is coordinate-independent, than $\{J_{\sigma\rho},P_\mu\}=\partial_\mu J_{\sigma\rho}=0$, but we know that $\{J_{\sigma\rho},P_\mu\}=-\{P_\mu,J_{\sigma\rho}\}\neq 0$, so how is that possible? Am I misunderstanding something?

1 Answer 1


I'm not sure if you are stymied by language or subtleties of the abstract generalization you are seeking. By way of a hint, I'll just review the free scalar field, flat-space case you no doubt learned in your first week of CFT, namely,

$$ \pi=\partial_0 \phi, \\ P^i = \int d^3x ~T^{0 i} (x)=\int d^3x ~\pi(x) \partial_i \phi(x) \\ P^0= \int d^3x ~(\pi^2 +(\partial_i \phi)^2 )/2, $$ so that, directly, given $\{ \phi(x),\pi(y)\}=\delta^3(x-y)$, $$ \{ \phi(x), P^i\}= \partial_i \phi(x), \\ \{ \phi(x), P^0\}= \pi (x) =\partial_0 \phi(x), \\ \{ P^\mu, P^\nu \} = 0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.