# Modification of Atom-Field interaction Hamiltonian in the Knight optics textbook

I am reading Quantum optics textbook of Peter Knight.

From the Eq 4.3

$$\hat{H} = \frac{1}{2m}[\hat{P}+e\hat{A}]^2-e\Phi + V(r)$$

Using

$$\hat{H}' = \hat{R}\hat{H}\hat{R}^{\dagger} + i\hbar\frac{\partial \hat{R} }{\partial t}\hat{R}^{\dagger}$$ where $$\hat{R}$$ = exp$$(-ie\chi(\vec{r},t)/\hbar )$$

The Hamiltonian is modified by

$$\hat{H} = \frac{1}{2m}[\hat{P}+e(\hat{A}+\nabla\chi) ]^2-e(\Phi-\frac{\partial \chi}{\partial t}) + V(r).$$

This procedure is a little complicated for me.

I tried the first part of the equation

$$\hat{R}\hat{P}^2\hat{R}^{\dagger} \\ = \hat{R} *-\hbar^2\nabla\cdot\nabla(\hat{R}^{\dagger}) \\ = e\nabla\chi\cdot\hat{P}-e^2(\nabla\chi)^2$$

This catches away the $$\hat{P}^2$$ part that should be survived.

What is my fault??

Instead of developing the term $$[P+eA]^2$$ it is perhaps better to observe that $$R[P+eA]^2R^{\dagger}=R[P+eA][P+eA]R^{\dagger}=R[P+eA]I[P+eA]R^{\dagger}=R[P+eA]RR^{\dagger}[P+eA]R^{\dagger}$$ where we use $$RR^{\dagger}=I$$, since $$R$$ is unitary. We conclude that $$R[P+eA]^2R^{\dagger}=\{R[P+eA]R^{\dagger}\}^2$$ and the result follows more easilly.
• It is very helpful and I can derive the modified form by the following textbook. But I think it is impossible to get modified hamiltonian form from the calculation of hamiltonian $\hat{R}H\hat{R}^\dagger$ only Commented Jul 30, 2020 at 8:59