Dipole approximation in length gauge (quantum electric field Hamiltonian) In the beginning of the fourth chapter of Gerry's book on Quantum Optics we can find the Hamiltonian of an electron bound to an atom and interacting with an electromagnetic field of potentials $\Phi$ and $\vec{A}$:
$$ \hat{H} = \frac{1}{2m}\left[\hat{\vec{P}} + e\vec{A}\right]^2 - e\Phi + \hat{V}, $$
where $\hat{\vec{P}} = -i\hbar \vec{\nabla}$. Then, the operator $\hat{R} = \exp(-ie\chi/\hbar)$ is defined, where $\chi$ is the function that allows a gauge transformation:
$$\Phi' = \Phi + \partial_t \chi,\quad \vec{A}' = \vec{A} + \vec{\nabla}\chi.$$
Defining $\hat{H}' = \hat{R}\hat{H}\hat{R}^\dagger + i\hbar (\partial_t \hat{R})\hat{R}^\dagger$ the authors say that
$$ \hat{H}' = \frac{1}{2m}\left[\hat{\vec{P}} + e\vec{A}'\right]^2 - e\Phi' + \hat{V}. \tag{1}$$
My question is how to interpret the $\hat{R}$ definition. Where is the operator part in the exponential?
This question arises because when I compute the $\hat{H}\hat{R}^\dagger$ I get
$$-i\hbar \nabla^2\chi\hat{R}^\dagger + e^2A'^2\hat{R}^\dagger,$$
which somewhat resembles the RHS of (1), but of course does not produce the RHS.
Thanks for any clarification in these calculations.
 A: The gauge transformation operator $\hat{R} = \exp\left(-ie\chi(\hat{\mathbf r})/\hbar\right)$ is an operator because $\hat{\mathbf r}$ is. Some important facts about this operator:


*

*The derivative entailed by $\langle\mathbf r|\hat{\mathbf p}=-i\hbar\nabla_\mathbf r\langle \mathbf r|$ does not act on $\hat R$, unless you are explicitly working on the position representation (usually marked by explicit insertions of the resolution of the identity $\int |\mathbf r⟩⟨\mathbf r|\mathrm d\mathbf r=1$).

*Instead, $\hat R$ acts via the operator algebra.

*Thus, $\hat R$ commutes with $\hat{\mathbf r}$ and functions of the latter, since they share an eigenbasis (the position states, $|\mathbf r\rangle$).

*Much of this formalism, and particularly the following points, is only valid within the dipole approximation, and we require that $\nabla \chi$ be constant, i.e. that $\chi(\mathbf r) = \mathbf A_0\cdot \mathbf r$.

*The interaction between $\hat R$ and the momentum operator $\hat{\mathbf p}$, on the other hand, is somewhat more complicated. The best way to see this is as follows:


*

*For an arbitrary momentum eigenstate $|\mathbf p\rangle$, the wavefunction is $\langle\mathbf r|\mathbf p\rangle = (2\pi)^{-3/2} e^{i\mathbf r\cdot \mathbf p/\hbar}$.

*If we then apply $\hat R^\dagger$ to that momentum eigenstate, we get the wavefunction
\begin{align}
\langle \mathbf r | \hat R^\dagger | \mathbf p \rangle
& = \langle \mathbf r |\exp\left(+ie\chi(\hat{\mathbf r})/\hbar\right) | \mathbf p \rangle
\\ & = \langle \mathbf r |\exp\left(+ie\chi(\mathbf r)/\hbar\right) | \mathbf p \rangle
\\ & = e^{ie\chi(\mathbf r)/\hbar} \langle \mathbf r | \mathbf p \rangle
\\ & = e^{ie\mathbf A_0\cdot \mathbf r/\hbar} \langle \mathbf r | \mathbf p  \rangle
\\ & = e^{ie\mathbf A_0\cdot \mathbf r/\hbar} \frac{1}{(2\pi)^{3/2}} e^{ i \mathbf p \cdot \mathbf r/\hbar }
\\ & = \frac{1}{(2\pi)^{3/2}} e^{ i\left(\mathbf p+ e\mathbf A_0\right) \cdot \mathbf r/\hbar }
\\ & = \langle \mathbf r| \mathbf p + e\mathbf A_0\rangle,
\end{align}
and since their wavefunctions coincide we can 'cancel out' the $\langle \mathbf r|$ to get the operator action $\hat R^\dagger | \mathbf p \rangle = | \mathbf p + e\mathbf A_0\rangle.$ Similarly, $\hat R | \mathbf p \rangle = | \mathbf p - e\mathbf A_0\rangle.$

*We can then use this to resolve the action of $\hat R$ on any function $f(\hat{\mathbf p})$ the momentum operator via a strategically-placed resolution of the identity:
\begin{align}
\hat R f(\hat{\mathbf p})\hat R^\dagger
& = \hat R f(\hat{\mathbf p})\hat R^\dagger\int \mathrm d\mathbf p \: |\mathbf p\rangle\langle \mathbf p|
\\ & = \int \mathrm d\mathbf p \: \hat R f(\hat{\mathbf p})\hat R^\dagger|\mathbf p\rangle\langle \mathbf p|
\\ & = \int \mathrm d\mathbf p \: \hat R f(\hat{\mathbf p})|\mathbf p+ e\mathbf A_0\rangle\langle \mathbf p|
\\ & = \int \mathrm d\mathbf p \: \hat R f(\mathbf p + e\mathbf A_0)|\mathbf p+ e\mathbf A_0\rangle\langle \mathbf p|
\\ & = \int \mathrm d\mathbf p \: f(\mathbf p + e\mathbf A_0)\hat R |\mathbf p+ e\mathbf A_0r\rangle\langle \mathbf p|
\\ & = \int \mathrm d\mathbf p \: f(\mathbf p + e\mathbf A_0)|\mathbf p \rangle\langle \mathbf p|
\\ & = \int \mathrm d\mathbf p \: f(\hat{\mathbf p} + e\mathbf A_0)|\mathbf p \rangle\langle \mathbf p|
\\ & = f(\hat{\mathbf p} + e\mathbf A_0)\int \mathrm d\mathbf p \: |\mathbf p \rangle\langle \mathbf p|
\\ & =  f(\hat{\mathbf p} + e\mathbf A_0),
\end{align}
i.e. the conjugated action of $\hat R$ and its conjugate on $\hat{\mathbf p}$ acts like a displacement on the momentum operator. This is how the kinetic term of the hamiltonian gets transformed.



That's about as much as you need for the actual calculations. It is important to remark, though, on the crucial role of the linearity of $\chi$, without which we could never have re-factorized $e^{ i( \mathbf p \cdot \mathbf r +e\chi(\mathbf r))/\hbar}$ into another plane wave. This sounds trivial, but it has some huge implications, which include the following two:


*

*The length gauge is only defined within the dipole approximation. If the dipole approximation fails, you need to be very careful in specifying what exactly you mean as the length gauge. (Example in action.)

*That means that the title of your question is wrong. You can't have the dipole approximation in the length gauge - you can only have the length gauge within the dipole approximation.

A: A way to achieve the desired result by brute force calculation is to show (1) by computing $\hat{H}'\psi = \hat{R}\hat{H}\hat{R}^\dagger\psi + i\hbar\partial_t\hat{R}\hat{R}^\dagger\psi$, taking into account that $\psi$ is a function of $\vec{r}$ and $t$.
