Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 6

With $J=\mathrm{diag}(1,-1)$ these matrices obey $M^{†}JM=J$ and $\det M=1$, identifying the group as SU(1,1). Products preserve the displayed conjugate-entry pattern and determinant; the identity belongs, andalso belongs. Multiplication and inversion are smooth. The real constraint $|\alpha {|}^2-|\beta {|}^2=1$ has nonzero gradient on its level set, so the matrices form a smooth real three-dimensional manifold. Associativity is inherited from matrix multiplication. These facts establish a Lie group, which is noncompact because $|\beta |$ is unbounded.

At first order write $\alpha =1+iz$ and $\beta =x+iy$ with real $x,y,z$. The constraint forces the real part of the change in $\alpha$ to vanish. ThenThus the three supplied generators span the tangent space over the reals. Multiplying their matrices givesFor example, $m_1={\sigma }_x/2$, $m_2={\sigma }_y/2$, $m_3=i{\sigma }_z/2$ makes the noncompact signs explicit.

Set $P=a^{†}{b}^{†}$ and $Q=ab$. Independent bosonic canonical commutation relations give $[Q,P]=N_a+N_b+1=2K_3$, while $[K_3,P]=P$ and $[K_3,Q]=-Q$. Substituting $K_1=(P+Q)/2$ and $K_2=(P-Q)/(2i)$ yieldsThereforewhich matches the matrix Lie algebra. This boson-pair realization of SU(1,1) differs from the number-conserving Schwinger boson representation of $SU(2)$.

The pair products are $PQ=N_a{N}_b$ and $QP=(N_a+1)(N_b+1)$. Hencewhereas $K_3=(N_a+N_b+1)/2$. Subtraction givesThe number difference commutes with pair creation, pair annihilation and $K_3$, since each pair operation changes both occupations equally. Its boson-pair Casimir operator $C$ therefore commutes with every generator. In a fixed-difference sector $N_a-N_b=d$, the Casimir eigenvalue is $(d^2-1)/4$, including $-1/4$ for equal occupations.