Hamiltonian of Jaynes-Cummings model has in Schrodinger picture form: $$ H=\frac{1}{2}\hbar\omega\sigma_{z}+\hbar\omega a^{\dagger}a+\lambda\hbar(\sigma_{+}\hat{a}+\sigma_{-}\hat{a}^{\dagger}) $$ which we can split into two parts $$ H_{0}=\frac{1}{2}\hbar\omega\sigma_{z}+\hbar\omega a^{\dagger}a $$ and $$ V=\lambda\hbar(\sigma_{+}\hat{a}+\sigma_{-}\hat{a}^{\dagger}) $$ Any sugesstion, how to prove commutation relation? $$ [H_{0},V]=0 $$

  • $\begingroup$ You mean aside from plugging and chugging? $\endgroup$
    – Kyle Kanos
    Commented Apr 28, 2018 at 19:24
  • $\begingroup$ I meant, how to derive that comtator. When you plug in H and V in it $\endgroup$
    – marek
    Commented Apr 28, 2018 at 20:17

1 Answer 1


Use the fact that the commutator is linear $$[A,B+C]=[A,B]+[A,C]$$ and that pauli matrices $\sigma_i$ commute with $a^+$ and $a$, we only need the following commutation relations $$[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k$$ $$[N,a^+]=a^{+}$$ $$[N,a]=-a$$ where $N=a^{+}a$


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