Show that the Holstein-Primakoff transformation satisfies the spin commutation relations


 * 1) Identify the spin operators under the transformation: \(\hat{S}^z = S - \hat{b}^{\dagger}\hat{b}\) \(\hat{S}^{+} = \left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\hat{b}\) \(\hat{S}^{-} = \hat{b}^{\dagger}\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\)
 * 2) Write the commutation relation \([\hat{S}^{+},\hat{S}^{-}]\) in terms of \(\hat{b}\) and \(S\): \([\hat{S}^{+},\hat{S}^{-}] = [\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\hat{b},\hat{b}^{\dagger}\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)]\)
 * 3) Expand the commutation relation: \([\hat{S}^{+},\hat{S}^{-}] = \left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\hat{b}\hat{b}^{\dagger}\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right) - \hat{b}^{\dagger}\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\hat{b}\)
 * 4) Rearrange \(\hat{b}\hat{b}^{\dagger}\): \([\hat{S}^{+},\hat{S}^{-}] = \left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\left(\hat{b}^{\dagger}\hat{b} + 1\right)\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right) - \hat{b}^{\dagger}\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\left(\sqrt{2S- \hat{b}^{\dagger}\hat{b}}\right)\hat{b}\)
 * 5) Simplify: \([\hat{S}^{+},\hat{S}^{-}] = \left(\hat{b}^{\dagger}\hat{b} + 1\right)\left(2S- \hat{b}^{\dagger}\hat{b}\right) - \hat{b}^{\dagger}\left(2S- \hat{b}^{\dagger}\hat{b}\right)\hat{b}\)
 * 6) Expand the brackets: \([\hat{S}^{+},\hat{S}^{-}] = 2S\hat{b}^{\dagger}\hat{b} + 2S - \hat{b}^{\dagger}\hat{b} - \hat{b}^{\dagger}\hat{b}\hat{b}^{\dagger}\hat{b} - 2S\hat{b}^{\dagger}\hat{b} + \hat{b}^{\dagger}\hat{b}^{\dagger}\hat{b}\hat{b}\)
 * 7) Rearrange \(\hat{b}^{\dagger}\hat{b}^{\dagger}\hat{b}\hat{b}\): \([\hat{S}^{+},\hat{S}^{-}] = 2S\hat{b}^{\dagger}\hat{b} + 2S - \hat{b}^{\dagger}\hat{b} - \hat{b}^{\dagger}\hat{b}\hat{b}^{\dagger}\hat{b} - 2S\hat{b}^{\dagger}\hat{b} + \hat{b}^{\dagger}\hat{b}\hat{b}^{\dagger}\hat{b} - \hat{b}^{\dagger}\hat{b}\)
 * 8) Cancel like terms: \([\hat{S}^{+},\hat{S}^{-}] = 2\left(S - \hat{b}^{\dagger}\hat{b}\right) = 2\hat{S}^{z}\)