Use the Holstein-Primakoff transformations to find the low-lying excitation energies for a system of two interacting spins


 * 1) Identify the Hamiltonian for a system of two interacting quantum spins, \(\hat{\mathbf{S}}_1\), \(\hat{\mathbf{S}}_2\) with magnitude \(S\): \(\hat{\mathcal{H}} = -J\hat{\mathbf{S}}_1\hat{\mathbf{S}}_2 - H\left(S_1^z - S_2^z\right)\)
 * 2) Expand \(\hat{\mathbf{S}}_1\hat{\mathbf{S}}_2\): \(\hat{\mathcal{H}} = -J\left(\hat{S}^z_1 \hat{S}^z_2 + \frac{1}{2}\left(\hat{S}^+_1\hat{S}^-_2 + \hat{S}^-_1\hat{S}^+_2\right)\right) - H\left(S_1^z + S_2^z\right)\)
 * 3) Apply the harmonic approximation, \(\hat{S}^+ = \hat{b}\sqrt{2S}\), \(\hat{S}^- = \hat{b}^{\dagger}\sqrt{2S}\) and \(\hat{S}^z = S - \hat{b}^{\dagger}\hat{b}\): \(\hat{\mathcal{H}} = -J\left(\left(S_1 - \hat{b}_1^{\dagger}\hat{b}_1\right)\left(S_2 - \hat{b}_2^{\dagger}\hat{b}_2\right) + S\left(\hat{b}_1\hat{b}_2^{\dagger} + \hat{b}^{\dagger}_1\hat{b}_2\right)\right) - H\left(S_1 - \hat{b}_1^{\dagger}\hat{b}_1 + S_2 - \hat{b}_2^{\dagger}\right)\)
 * 4) Simplify: \(\hat{\mathcal{H}} = -JS^2 - 2HS + \left(JS + H\right)\left(\hat{b}_1^{\dagger}\hat{b}_1 + \hat{b}_2^{\dagger}\hat{b}_2\right) - JS\left(\hat{b}_2^{\dagger}\hat{b}_1 + \hat{b}_1^{\dagger}\hat{b}_2\right)\)
 * 5) Write in matrix form: \(\hat{\mathcal{H}} = -JS^2 - 2HS + \begin{pmatrix} \hat{b}^{\dagger}_1 & \hat{b}^{\dagger}_2 \end{pmatrix} \begin{pmatrix} JS + H & -JS \\ -JS & JS + H\end{pmatrix} \begin{pmatrix} \hat{b}_1 \\ \hat{b}_2 \end{pmatrix}\)
 * 6) Diagonalise by multiplying by \(\begin{pmatrix} \hat{\alpha} & \hat{\beta} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \hat{b}_1 \\ \hat{b}_2 \end{pmatrix}\): \(\hat{\mathcal{H}} = -JS^2 - 2HS + \begin{pmatrix} \hat{\alpha}^{\dagger} & \hat{\beta}^{\dagger} \end{pmatrix} \begin{pmatrix} H & 0 \\ 0 & 2JS + H\end{pmatrix} \begin{pmatrix} \hat{\alpha} \\ \hat{\beta} \end{pmatrix}\)
 * 7) Take out of diagonal form: \(\hat{\mathcal{H}} = -JS^2 - 2HS + \hat{\alpha}^{\dagger}\hat{\alpha}H + \left(2JS + H\right)\hat{\beta}^{\dagger}\hat{\beta}\)