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In the usual $SU(1,1)$ group: $$[K_+,K_-]=-2K_z,\quad [K_z,K_\pm]=\pm K_\pm.$$ Is there a decomposition exist for $e^{c(K_++K_-)^2}$?

Of course there won't exist a decomposition to $e^{K_+},e^{K_-},e^{K_z}$ but maybe to their squares:$e^{K_+^2},e^{K_-^2},e^{K_z^2}$ or a similar one exist.

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  • $\begingroup$ I think your commutation relations are wrong. What do you mean by e.g. $K_+^2$? Also, why are you interested in this question? $\endgroup$
    – Paul Levy
    Commented Feb 9, 2018 at 7:36
  • $\begingroup$ Ok there has to be $\pm$ for the second commutation relation so I have edited the question now. Thank you for that. Such operators appears in physics so I would like to now if there exist decompositions for these type of operators. $K_+^2$ and the other operators are just the square of the operator that obeys the defined group relation. $\endgroup$
    – user120129
    Commented Feb 10, 2018 at 0:10
  • $\begingroup$ That doesn't really explain why you are interested in this particular question. It looks suspiciously like a homework question. In any case, I don't think it is about research-level mathematics. $\endgroup$
    – Paul Levy
    Commented Feb 10, 2018 at 9:26
  • $\begingroup$ Ok then my next question for mathflow community is: is it possible that such a question can be a homework question? Is it that simple. If you can decompose such operator you can decompose whole range of physically important operators. Up to now we use Trotter type approximations. I have emailed to you. If you want I can give you explicit operators. I am even ready to pay for who can do it. $\endgroup$
    – user120129
    Commented Feb 10, 2018 at 9:53
  • $\begingroup$ Now I'm satisfied that you aren't a student asking for help with homework, but you haven't posed the question very well. When you say "the square of the operator", you haven't specified what operator is associated to each basis element. The standard way of thinking of ${\mathfrak su}(1,1)$ is via $2\times 2$ matrices, but I guess you have a particular unitary representation in mind. $\endgroup$
    – Paul Levy
    Commented Feb 10, 2018 at 12:23

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