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I was trying to prove that: $$[P_\mu, J_{\rho \sigma}] = i(\eta_{\mu \sigma} P_\rho - \eta_{\mu \rho} P_\sigma) $$

$\textbf{Attempt}$

$$\begin{align} [P_\mu, J_{\rho \sigma}] = [P_\mu, x_\rho P_\sigma - x_\sigma P_\rho] \\ =[P_\mu, x_\rho P_\sigma] - [P_\mu, x_\sigma P_\rho] \\ =-i[x_\sigma , \partial_\mu]P_\rho+ i[x_\rho , \partial_\mu]P_\sigma \\ =-i(x_\sigma \partial_\mu - \partial_\mu x_\sigma) P_\rho + i(x_\rho \partial_\mu - \partial_\mu x_\rho)P_\sigma \end{align}$$

using $\partial_\mu x_\rho = \eta_{\mu \rho} $

$$[P_\mu, J_{\rho \sigma}] = -i(x_\sigma \partial_\mu - \eta_{\mu \sigma}) P_\rho + i(x_\rho \partial_\mu - \eta_{\mu \rho})P_\sigma \\ =i(\eta_{\mu \sigma} P_\rho - \eta_{\mu \rho}P_\sigma) $$

However in this post Commutation of position four-vector with spacetime derivatives they get:

\begin{align} [P_a,M_{bc}] =g_{ab}\partial_c - g_{ac}\partial_b \end{align}

That is the negative result of what I got, what am I doing wrong?

Qmechanic
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RKerr
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  • Why don't you apply directly that $[x_{\rho},P_{\mu}]=i\eta_{\rho\mu}$? The $P$'s commute, and the $x$'s and the $P$'s are canonically conjugate. – joigus Feb 19 '21 at 22:44
  • We didn't use that relation in class, so I was trying to prove it without it. Can you see what I'm doing wrong? – RKerr Feb 19 '21 at 22:45
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    Remember $P_{\mu}=i\partial_{\mu}$. The metric takes care of the usual sign in QM. Could that be it? – joigus Feb 19 '21 at 22:51
  • It might be! One more question isn't the relation $$[P_\mu , x_\rho] = i \eta_{\rho \mu}$$? In this paper they state that in equation (1.1) https://arxiv.org/abs/hep-th/9505152 – RKerr Feb 19 '21 at 22:57
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    Yes, that's the reason. Remember the $P_\mu$ is covariant, not contravariant, so it's minus the usual QM (non-relativistic) operator. More explicitly: $P_{\mu}=i\left(\frac{\partial}{\partial x^{0}},\boldsymbol{\nabla}\right)$ instead of $1/i$. As to the cannonical comm. rel.; it's the other way: $[x_{\rho},P_{\mu}]=i\eta_{\rho\mu}$, because $x_k$ is $-x^k$ for spatial components. Well.. with the appropriate metric, you know. – joigus Feb 19 '21 at 23:06
  • Thank you very much! – RKerr Feb 19 '21 at 23:10
  • So you may withdraw your question? It was answered here. – Cosmas Zachos Feb 21 '21 at 14:36

0 Answers0