In the usual $SU(1,1)$ group: $$[K_+,K_-]=-2K_z,\quad [K_z,K_\pm]=K_\pm.$$ 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.

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.

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

In the usual $SU(1,1)$ group: $$[K_+,K_-]=-2K_z,\quad [K_z,K_\pm]=K_\pm.$$ 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.