Dipole moment in the Optical Interaction Hamiltonian I've got a simple and fundamental question regarding the atom-light interaction Hamiltonian: How do we handle the dipole moment term in this expression?
The atom-light interaction Hamiltonian, $H_{AL}$, takes the following form:
\begin{align}
    \hat{H}_{AL} &= -\hat{d} \cdot \hat{E}&&\\
 %&=d \cdot E\left(\sigma_{-} + \sigma_{+}\right) 
\end{align}
where $\hat{E}$ is the electric field and $\hat{d}$ is the dipole moment operator. The latter is the operator I'm interested in. We can evaluate the matrix elements of this operator as $\hat{d}=q\hat{r}$, where q is charge and $\hat{r}$ is the position operator. Let's take the case of a two-level system with ground and excited states $|0\rangle$ and $|1 \rangle$ respectively, if we move from position to the energy basis:
\begin{equation}
\hat{d} = |0\rangle d_{01} \langle 1| + |1 \rangle d_{10} \langle 0|
\end{equation}
where $d_{01} = \langle 0| \hat{d}|1 \rangle$. Thus, if $d_{01} = d_{10} = d$ then:
\begin{equation}
\hat{d} = d \sigma_{x}.
\end{equation}
Returning to the atom-light interaction Hamiltonian, now lets say that we have $N$ two-level systems, each treated as a dipole, coupled to a single cavity. If the electric field is treated as a quantum harmonic oscillator we find the following expression:
\begin{equation}
\hat{H}_{AL}= \sum^{N}_{i=1} d_{i} \sigma_{x} \otimes \sum_{k} g_{k} (\hat{a}_{k} + \hat{a}^{+}_{k}).
\end{equation}
Where $g$ is the coupling strength and $a_k$ are the ladder operators. Now consider if I were to physically calculate this interaction, here's my problem: In terms of the position basis I understand the dipole moment as a vector, relating to the dipoles position and charge. But it is also, as shown above, a quantum operator whose elements correspond to the coupling between two states. Say, for example, I was given a list of dipole moment vectors and wished to construct the atom-light Hamiltonian interaction Hamiltonian for that system, how would I convert those dipole moment vectors into $d_{i}$ coefficients?
If $d_{i}$ were just a number then I'd understand the maths but not the physics and if $d_{i}$ were a vector then I understand the physics but not the maths.
 A: The dipole operator is a 3 dimensional vector with regard to the cartesian coordinate system of your "lab" in which also the electro-magnetic field "lives". Then you have your electronic state basis in addition to this. Which means your operator $d$ is already a three dimensional vector before you introduce the expansion into the electronic basis. And each matrix element in the electronic basis representation is a three dimensional vector.
I guess its clearer if i write down an example.
$$
\hat d = q\left (\begin{matrix}\hat x \\ \hat y \\ \hat z \end{matrix}\right )
$$
A matrix element with regard to electronic states $|f\rangle, |i\rangle $ is then
$$\begin{aligned}
\vec d_{fi}\equiv \langle f|\hat d|i\rangle = q\left (\begin{matrix}\langle f|   \hat x| i \rangle  \\ \langle f|   \hat y| i \rangle  \\\langle f|   \hat z| i \rangle  \\ \end{matrix}\right )
\end{aligned}$$
And the scalar product with the electric field yields
$$
\vec d_{fi} \cdot \vec E = q\left( \langle f|   \hat x| i \rangle E_x + \langle f|   \hat y| i \rangle  E_y + \langle f|   \hat z| i \rangle E_z\right ) 
$$
The dipole operator matrix for two states would look like this,
$$[\hat d] =\left( \begin{matrix}
\vec d_{11} & \vec d_{12} \\
\vec d_{21} & \vec d_{22}
\end{matrix}\right )$$
The ladder operators should be associated with a polarization unit vector. Taking the scalar product of this polarization vector with a 3 dimensional matrix element gives you your single number. For example,
$$
\vec d_{fi}\cdot \vec e_x = d_{fi,x} = q\langle f|\hat x|i\rangle .
$$
When we look at light-matter interaction we know that our electric field is due to a plane wave with wavevector $\vec k$, this limits the polarization vector to a plane perpendicular to $\vec k$. We can then project the three-dimensional dipole moment operator unto this subspace to reduce its "lab dimension" from 3 to 2. This is then conform with typical expressions for the quantized field where we also limit ourselves to two polarizations. The ladder operators then come as one pair for each polarization and the scalar product of your dipole operator matrix elements with the polarization vector reduce the expressions to a number.
Another point that should be mentioned is that the diagonal matrix elements of the dipole operator are not in general zero ! They are only zero if the system under investigation has a dipole moment of zero, which is the case if the electronic states are symmetric under inversion. This is often assumed but it is a more a property of the system/states and not of the operator.
The dipole operator in a basis of two states should be written as
$$\begin{aligned}
\hat d &= |0\rangle \vec d_{00} \langle 0| + \\
&\phantom{=|} |0\rangle \vec d_{01} \langle 1|+\\
&\phantom{=|} |1\rangle \vec d_{10} \langle 0|+\\
&\phantom{=|} |1\rangle \vec d_{11} \langle 1|
\end{aligned}$$
and only when the dipole moment vanishes we have the special case with $$\vec d_{00}= \vec d_{11} = \vec 0.$$
