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when toggle format what by license comment
Sep 29, 2023 at 22:04 comment added ZeroTheHero It seems you’re working too hard. Since $X^2$ is a scalar it should be quicker to show that $[L_i,X^2]=0$ which the result on $L^2$ follows immediately.
Sep 29, 2023 at 21:49 history answered Souparna Nath CC BY-SA 4.0