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Dec 1, 2015 at 3:12 comment added Andrea Becker Thank you very much! Your idea works just fine. I would very much appreciate it if you could prove the second part, namely ${\rm R}_s {\rm R}_{s^{\prime}} = {\rm R}_{ss^{\prime}}$ if ${\lambda}_s {\lambda}_{s^{\prime}} = {\lambda}_{ss^{\prime}}$ starting from the definition of ${\rm R}_s$ transformation, namely $e^{-{\cal H}^{\prime}[\lambda]} = \ldots$.
Nov 30, 2015 at 19:21 history answered Abdelmalek Abdesselam CC BY-SA 3.0