Two Level Atom With photon Algebra

The combined operator§

There are four combination worth mentioning on, the rest is trivial / mentioned previously. Which is

$$\hat{a}\hat{\sigma}^+ \quad \hat{a}^\dagger\hat{\sigma}^+ \quad \hat{a}\hat{\sigma}^- \quad \hat{a}^\dagger\hat{\sigma}^-$$

Since photon operators and atom operators commute, the combination can be seen as a single operator as whole.

Commutation§

$$[\hat{a}\hat{\sigma}^+,\hat{a}^\dagger\hat{\sigma}^+]=1(\sigma^+)^2=0$$ $$[\hat{a}\hat{\sigma}^+, \hat{a}\hat{\sigma}^-]=(\hat{a})^2(\hat{\sigma}_e-\hat{\sigma}_g)$$ $$[\hat{a}\hat{\sigma}^+,\hat{a}^\dagger\hat{\sigma}^-]= \hat{a}\hat{a}^\dagger(\hat{\sigma}_e-\hat{\sigma}_g)+ \hat{\sigma}^-\hat{\sigma}^+=(\hat{n}+1)\hat{\sigma}_e-\hat{n}\hat{\sigma}_g$$ $$[\hat{a}^\dagger\hat{\sigma}^+,\hat{a}\hat{\sigma}^-]=\hat{n}(\hat{\sigma}_e-\hat{\sigma}_g)-\hat{\sigma}_g=\hat{n}\hat{\sigma}_e-(\hat{n}+1)\hat{\sigma}_g$$ $$[\hat{a}^\dagger\hat{\sigma}^+,\hat{a}^\dagger\hat{\sigma}^-]=(\hat{a}^\dagger)^2(\hat{\sigma}_e-\hat{\sigma}_g)$$ $$[\hat{a}\hat{\sigma}^-, \hat{a}^\dagger\hat{\sigma}^-]=0$$