I am new to Group theory and representations and I'm having trouble with this problem in an exercise:

Given the two oscillator algebra $$[a, a^†] = 1$$ $$[b, b^†] = 1$$ $$[a, b] = 0$$ show that the forgoing generators satisfy $SU(1,1)$ algebra. $$[K_+, K_−] = 2K_z$$ $$[a, K_+] = a^†b^†$$ $$[a, K_−] = ab$$

I've read that $SU(1,1)$ algebra is defined by the the commutations: $[K_1, K_2] = −iK_0$ , $ [K_0, K_1] = iK_2$ , $[K_2, K_0] = iK_1$ .

I assumed that it was just a matter of plugging in the given generators and obtaining these relations to show that they satisfy the given algebra but I'm unable to do this.

I would really appreciate it if someone could provide me with some tips on commutator manipulation and the general strategy one would use in solving such problems.


closed as off-topic by ACuriousMind Mar 10 '18 at 16:13

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