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Considering the Lorentz algebra: $$[J_{\mu\nu},J_{\rho\sigma}] = -i(\eta_{\mu\rho}J_{\nu\sigma} - \eta_{\nu\rho}J_{\mu\sigma} + \eta_{\nu\sigma}J_{\mu\rho} - \eta_{\mu\sigma}J_{\nu\rho})$$ with $J_{0i} = K_i$, $J_{ij} = \epsilon_{ijk}J_k$ and $J_{\mu\nu} = -J_{\nu\mu}$.

For each $\Lambda \in O(1,3)$ I have that the following map is an automorphism of Lorentz algebra: $$f_\Lambda: J_{\mu\nu} \rightarrow \tilde{J}_{\mu\nu} = {\Lambda_\mu}^\rho{\Lambda_\nu}^\sigma J_{\rho\sigma}$$

Calling $\pi(J_{\mu\nu})$ an irreducible representation of the Lorentz algebra, I would like to show that if $\Lambda$ is in the connected part of Lorentz group $SO(1,3)^+$, then the representation $\pi(\tilde{J}_{\mu\nu})$ is equivalent to $\pi(J_{\mu\nu})$, that is: $$\pi(\tilde{J}_{\mu\nu}) = U^{-1}(\Lambda)\pi(J_{\mu\nu})U(\Lambda)$$

I really don't know where to start to prove it. Can anyone give me some advice?

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    $\begingroup$ 1. Since you did not give any physical context for this, might this question be more appropriate for Mathematics? 2. Your notation doesn't really make sense to me: What's the $U$ in the equation you want to show - why is that not $\pi$, too? How is $\pi(\tilde{J})$ supposed to be a different representation than $\pi(J)$? The $\pi$ is the representation map in standard notation and defines the representation, not its argument! $\endgroup$
    – ACuriousMind
    Sep 24, 2022 at 13:38
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    $\begingroup$ You do not need representations to show that JTILDE also satisfy the commutation relations. $\endgroup$
    – DanielC
    Sep 24, 2022 at 15:48

1 Answer 1

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As Daniel mentioned in the comment section, you can prove the equivalence by showing $\pi(\tilde{J}_{\mu\nu})$'s have the same Lie bracket as $\pi(J_{\mu\nu})$'s, but if you want to find the exact form of $U \in \text{GL}(V)$ where $\pi(J_{\mu\nu}) \in gl(V)$, a solution is provided in the content below.

Let $\Lambda(\theta,\hat{m})=e^{-{i \over 2}\theta^{\mu\nu} \ M_{\mu\nu}}$ where $\hat{m}$ is the axis of rotation, $\theta$ is the degree of rotation, and $M_{\mu\nu}$'s are the generators of $\text{SO}(1,3)^+$ and have the same commutation relationship as $J_{\mu\nu}$'s. We write $\Lambda(\theta,\hat{m})$ as $\Lambda(\theta)$ for notation convenience and define the unitary operator $$U(\Lambda(\theta))=e^{-{i \over 2}\theta^{\mu\nu} \ \pi(J_{\mu\nu})}=\lim_{n \rightarrow +\infty}{\big(e^{-{i \over 2}\big({1 \over n}\big)\theta^{\mu\nu} \ \pi(J_{\mu\nu})}\big)^n}=\lim_{n \rightarrow +\infty}{U\big(\Lambda\big({\theta \over n}\big)\big)^n}$$ And we are going to show $U(\Lambda(\theta))$ satisfies $\pi(\tilde{J}_{\mu\nu})=U^{-1}(\Lambda(\theta))\pi(J_{\mu\nu})U(\Lambda(\theta))$.

With the Lie bracket of $J_{\mu\nu}$'s and the definition of the representation, we know $$[\pi(J_{\mu\nu}),\pi(J_{\rho\sigma})]=-i(\eta_{\mu\rho}\pi(J_{\nu\sigma})-\eta_{\nu\rho}\pi(J_{\mu\sigma})+\eta_{\nu\sigma}\pi(J_{\mu\rho})-\eta_{\mu\sigma}\pi(J_{\nu\rho}))$$ and when $n \rightarrow +\infty$, we have $$\big(\Lambda\big({\theta \over n}\big)\big)_{\mu}^{\rho}=\delta_{\mu}^{\rho}-{i \over 2n}\theta^{\kappa\lambda}(M_{\kappa\lambda})_{\mu}^{\rho}+O\big({1 \over n^2}\big)=\delta_{\mu}^{\rho}+{1 \over 2n}\theta^{\kappa\lambda}(\eta_{\kappa\mu}\delta_{\lambda}^{\rho}-\eta_{\lambda\mu}\delta_{\kappa}^{\rho})+O\big({1 \over n^2}\big)$$ Therefore, up to the first order (the terms proportional to ${1 \over n^2}$ or smaller are discarded), \begin{align} \pi\big(\big(\Lambda\big({\theta \over n}\big)\big)_{\mu}^{\rho}\big(\Lambda\big({\theta \over n}\big)\big)_{\nu}^{\sigma}J_{\rho\sigma}\big) & \sim \pi(J_{\mu\nu})+{1 \over 2n}\big(\delta_{\mu}^{\rho}\theta^{\kappa\lambda}(\eta_{\kappa\nu}\delta_{\lambda}^{\sigma}-\eta_{\lambda\nu}\delta_{\kappa}^{\sigma})+\theta^{\kappa\lambda}(\eta_{\kappa\mu}\delta_{\lambda}^{\rho}-\eta_{\lambda\mu}\delta_{\kappa}^{\rho})\delta_{\nu}^{\sigma}\big)\pi(J_{\rho\sigma}) \\ & = \pi(J_{\mu\nu})+{1 \over 2n}\theta^{\kappa\lambda}\big(\eta_{\kappa\nu}\delta_{\lambda}^{\sigma}-\eta_{\lambda\nu}\delta_{\kappa}^{\sigma}\big)\pi(J_{\mu\sigma})+{1 \over 2n}\theta^{\kappa\lambda}\big(\eta_{\kappa\mu}\delta_{\lambda}^{\rho}-\eta_{\lambda\mu}\delta_{\kappa}^{\rho}\big)\pi(J_{\rho\nu}) \\ & = \pi(J_{\mu\nu})+{1 \over 2n}\big(\eta_{\kappa\nu}\theta^{\kappa\sigma}\pi(J_{\mu\sigma})-\eta_{\lambda\nu}\theta^{\sigma\lambda}\pi(J_{\mu\sigma})+\eta_{\kappa\mu}\theta^{\kappa\rho}\pi(J_{\rho\nu})-\eta_{\lambda\mu}\theta^{\rho\lambda}\pi(J_{\rho\nu})\big) \\ & = \pi(J_{\mu\nu})-{1 \over 2n}\big(-\eta_{\nu\rho}\theta^{\rho\sigma}\pi(J_{\mu\sigma})+\eta_{\nu\sigma}\theta^{\rho\sigma}\pi(J_{\mu\rho})-\eta_{\mu\sigma}\theta^{\rho\sigma}\pi(J_{\nu\rho})+\eta_{\mu\rho}\theta^{\rho\sigma}\pi(J_{\nu\sigma})\big) \\ & = \pi(J_{\mu\nu})-{i \over 2n}\theta^{\rho\sigma}[\pi(J_{\mu\nu}),\pi(J_{\rho\sigma})] \\ & = U^{-1}\big(\Lambda\big({\theta \over n}\big)\big)\pi(J_{\mu\nu})U\big(\Lambda\big({\theta \over n}\big)\big) \end{align} It results in \begin{align} \pi\big(\big(\Lambda(\theta)\big)_{\mu}^{\rho}\big(\Lambda(\theta)\big)_{\nu}^{\sigma}J_{\rho\sigma}\big) & = \lim_{n \rightarrow +\infty}{\pi\bigg(\Pi_{k=1}^{n}{\bigg(\big(\Lambda\big({\theta \over n}\big)\big)_{\rho_{k+1}}^{\rho_{k}}\big(\Lambda\big({\theta \over n}\big)\big)_{\sigma_{k+1}}^{\sigma_{k}}\bigg)}J_{\rho_{1}\sigma_{1}}\bigg)} \ \ \ (\rho_{n+1}=\mu,\sigma_{n+1}=\nu) \\ & = \lim_{n \rightarrow +\infty}{\Pi_{k=1}^{n}{\bigg(\big(\Lambda\big({\theta \over n}\big)\big)_{\rho_{k+1}}^{\rho_{k}}\big(\Lambda\big({\theta \over n}\big)\big)_{\sigma_{k+1}}^{\sigma_{k}}\bigg)}\pi\big(J_{\rho_{1}\sigma_{1}}\big)} \\ & = \lim_{n \rightarrow +\infty}{U^{-1}\big(\Lambda\big({\theta \over n}\big)\big)\bigg(\Pi_{k=2}^{n}{\bigg(\big(\Lambda\big({\theta \over n}\big)\big)_{\rho_{k+1}}^{\rho_{k}}\big(\Lambda\big({\theta \over n}\big)\big)_{\sigma_{k+1}}^{\sigma_{k}}\bigg)}\pi\big(J_{\rho_{2}\sigma_{2}}\big)\bigg)U\big(\Lambda\big({\theta \over n}\big)\big)} \\ & = \cdots = U^{-1}(\Lambda(\theta))\pi(J_{\mu\nu})U(\Lambda(\theta)) \end{align}

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