Find the eigenstate of the total spin operator in the Schwinger boson representation


 * 1) Assume that the total spin operator can be written as a Fock state: \(|S,m\rangle = \frac{\left(\hat{a}^{\dagger}\right)^{S+m}}{\sqrt{S+m!}}\frac{\left(\hat{b}^{\dagger}\right)^{S-m}}{\sqrt{S-m!}}|0\rangle = |\left(S+m\right)_a, \left(S-m\right)_b\rangle\)
 * 2) Expand the total spin operator: \(\hat{S}^2 |S,m\rangle = \left(\left(\hat{S}^x\right)^2 + \left(\hat{S}^y\right)^2 + \left(\hat{S}^z\right)^2\right)|S,m\rangle\)
 * 3) Substitute \(\left(\hat{S}^x\right)^2 + \left(\hat{S}^y\right)^2 = \frac{\hat{S}^{+}\hat{S}^{-} + \hat{S}^{-}\hat{S}^{+}}{2}\): \(\hat{S}^2 |S,m\rangle = \left(\frac{\hat{a}^{\dagger}\hat{a}\hat{b}\hat{b}^{\dagger} + \hat{a}\hat{a}^{\dagger}\hat{b}^{\dagger}\hat{b}}{2} + \left(\hat{S}^z\right)^2\right)|S,m\rangle\)
 * 4) Rearrange: \(\hat{S}^2 |S,m\rangle = \left(\frac{\hat{a}^{\dagger}\hat{a}\hat{b}^{\dagger}\hat{b} + \hat{a}^{\dagger}\hat{a} + \hat{a}^{\dagger}\hat{a}\hat{b}^{\dagger}\hat{b}+\hat{b}^{\dagger}\hat{b}}{2} + \left(\hat{S}^z\right)^2\right)|S,m\rangle\)
 * 5) Write \(\hat{S}^z\) explicitly: \(\hat{S}^2 |S,m\rangle = \left(\frac{\hat{a}^{\dagger}\hat{a}\hat{b}^{\dagger}\hat{b} + \hat{a}^{\dagger}\hat{a} + \hat{a}^{\dagger}\hat{a}\hat{b}^{\dagger}\hat{b}+\hat{b}^{\dagger}\hat{b}}{2} + \left(\frac{\hat{a}^{\dagger}\hat{a} - \hat{b}^{\dagger}\hat{b}}{2}\right)^2\right)|S,m\rangle\)
 * 6) Simplify: \(\hat{S}^2 |S,m\rangle = \left(\frac{\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}\hat{a} + \hat{b}^{\dagger}\hat{b}\hat{b}^{\dagger}\hat{b} + 2\hat{a}^{\dagger}\hat{a} + 2\hat{b}^{\dagger}\hat{b} + 2\hat{a}\hat{a}^{\dagger}\hat{b}^{\dagger}\hat{b}}{4}\right)|S,m\rangle\)
 * 7) Complete the square to factorise: \(\hat{S}^2 |S,m\rangle = \left(\frac{\left(\hat{a}^{\dagger}\hat{a} + \hat{b}^{\dagger}\hat{b}\right)^2 + 2\hat{a}^{\dagger}\hat{a} + 2\hat{b}^{\dagger}\hat{b}}{4}\right)|S,m\rangle\)
 * 8) Take a factor of \(\frac{\hat{a}^{\dagger}\hat{a}+\hat{b}^{\dagger}\hat{b}}{2}\) outside the bracket: \(\hat{S}^2 |S,m\rangle = \frac{\hat{a}^{\dagger}\hat{a}+\hat{b}^{\dagger}\hat{b}}{2}\left(\frac{\hat{a}^{\dagger}\hat{a}+\hat{b}^{\dagger}\hat{b}}{2} + 1\right)|S,m\rangle = S(S+1)|S,m\rangle\)