Help in "commutator expansion" containing creation and annihilation operators

Question

I am trying to expand the commutator $\left[ {{H_0},S} \right]$ to get the ${H_1}$ according to the equation: $${H_1} = {H_I} + \left[ {{H_0},S} \right],$$ where, $${H_I} = {g_k}\left[ {\left( {{b_k} + b_k^\dagger } \right){\sigma ^ + } + h.c.} \right],$$ $${H_0} = \left( {{\omega _k}b_k^\dagger{b_k} + {\Omega \over 2}{\sigma _z}} \right),$$ $$S = \left( {{A_k}b_k^\dagger{\sigma ^ + } - {A_k}{b_k}{\sigma ^ - }} \right),$$ where ${b_k^\dagger }$, ${b_k}$ are the creation and annihilation operators, ${\sigma _{x,y,z}}$ are the Pauli operators I reach to: $${H_1} = {H_I} + {H_0}S - S{H_0},$$ with: $${H_0}S = {\omega _k}{A_k}b_k^\dagger{b_k}b_k^\dagger{\sigma ^ + } - {A_k}{\omega _k}b_k^\dagger{b_k}{b_k}{\sigma ^ - } + {\Omega \over 2}{A_k}{\sigma _z}b_k^\dagger{\sigma ^ + } - {\Omega \over 2}{A_k}{\sigma _z}{b_k}{\sigma ^ - },$$ $$S{H_0} = {A_k}{\omega _k}b_k^\dagger\sigma b_k^\dagger{b_k} + {A_k}{\Omega \over 2}b_k^\dagger\sigma {\sigma _z} - {A_k}{\omega _k}{b_k}{\sigma ^ - }b_k^\dagger{b_k} - {A_k}{\Omega \over 2}{b_k}{\sigma ^ - }{\sigma _z}$$ But I failed to simplify the result as in the original search, where: $${H_1} = {g_k}\left( {{b_k} + b_k^\dagger} \right){\sigma ^ + } + {A_k}\left( {{\omega _k} + \Omega } \right)b_k^\dagger{\sigma ^ + } + h.c.$$ I expect thet he used some kind of relations to get this simplified form. Please help with this simple problem?

Answer 1

We have : \begin{align} [H_0,S] &= \left [ \omega_k b^\dagger_k b_k + \frac{\Omega}{2}\sigma_z, A_kb_k^\dagger \sigma^+ - A_kb_k\sigma^-\right]\\ &=\omega_kA_k\left[b_k^\dagger b_k,b_k^\dagger\sigma^+-b_k\sigma^-\right] + \frac{A_k\Omega}{2}\left[\sigma_z,b_k^\dagger \sigma^+ - b_k\sigma^-\right] \end{align}

Now, we can use : \begin{align} \left[b_k,b_k^\dagger\right] &= 1 \\ \left[b_k,\sigma^+\right]&= \left[b_k,\sigma^-\right] = \left[b_k,\sigma_z\right] = 0 \\ \left[b_k^\dagger,\sigma^+\right]&= \left[b_k^\dagger,\sigma^-\right] = \left[b_k^\dagger,\sigma_z\right] = 0 \\ [\sigma_z,\sigma^+] &= 2\sigma^+\\ [\sigma_z,\sigma^-] &= -2\sigma^- \end{align} to get : $$[H_0,S] = \omega_k A_k b_k^\dagger \sigma^+ + \omega_k A_kb_k\sigma^- + {A_k\Omega}(b^\dagger_k \sigma^+ + b_k\sigma^-) = A_k(\omega_k + \Omega)b_k^\dagger\sigma^+ + \text{h.c}$$