# Inverse boson operator realization of $\mathfrak{so}(3)$

This is actually a homework problem.

The inverse boson operators $a^{-1}$ and $\left(a^\dagger\right)^{-1}$ are defined as $$a^{-1} |n\rangle = \frac{1}{\sqrt{n+1}} |n+1\rangle$$ $$\left(a^\dagger\right)^{-1} |n\rangle = \begin{cases} \frac{1}{\sqrt{n}} |n-1\rangle & n>0 \\ 0 & n=0 \end{cases}$$

$a^{-1}$ is in fact only right inverse of boson annihilation operator, and $\left(a^\dagger\right)^{-1}$ the left inverse of $\left(a^\dagger\right)$.

The question is how to construct a realization of Lie algebra $\mathfrak{so}(3)$ analogous to two boson realization of $\mathfrak{so}(3)$.

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What is the two boson realization of $\mathfrak{so}(3)$? What do you mean by "analogous to"? –  Qiaochu Yuan Jun 7 '12 at 2:02
@QiaochuYuan: $J_{+}=a_{1}^{\dagger}a_{2},\ J_{-}=a_{1}a_{2}^{\dagger},\ J_{z}=\frac{1}{2}\left(N_{1}-N_{2}\right)$ where $a_1$ and $a_2$ are two commuting annihilation operator. –  Siyuan Ren Jun 7 '12 at 2:04