Rabi oscillation, electron in a box This page on Rabi oscillation says the Hamiltonian has an interaction term $d \cdot E(t)$, where $d = -$e$r$. (I'm sure what they mean is, that's what the off-diagonal terms of the Hamiltonian look like).
What is $r$? Is it the vector pointing from the (average) position of the electron while in the ground state to the (average) position of the electron while in the excited state? Or is it the vector pointing from the average position of the electron (averaged over both the ground and excited state?) and the nucleus?
I'd like to understand how Rabi oscillation* would work with an "electron in a box."** However, if the former interpretation of $r$ is correct, the average position for both the ground state and excited state are just the center of the box. If it's the latter, OTOH, then there's no positive charge to point to, just the electron. So it seems like either way, there's no interaction between the particle and the field. Is this actually the case?
*What I mean is, if you apply an electric field to an electron the same way one does in the Rabi model.
**I know that the particle-in-a-box Hamiltonian has countably many eigenstates, and the Rabi model is supposed to be for a two-state system. I guess I'm assuming if the particle begins in the ground state, the evolution will keep amplitude mostly in the ground state and first excited state, and the higher-energy states can be ignored.
 A: What is $\vec{R}$?
$\vec{R}$ is the position operator for the electron. Let's see why.
We have an interaction hamiltonian $H_I = e\vec{E} \cdot \vec{R}$, and we want to know what matrix elements of the hamiltonian are so we can right down the time-dependent schroedinger equation in matrix form (i.e. we want to be able to write equation 3.11 in the link you gave.) So what are the matrx elements? Well they are 
$$ H_{ij} = e \vec{E} \cdot \langle i | \vec{R} | j \rangle$$ where $i$ and $j$ can both take the values $e$, $g$ independently of each other (where $g$ signifies the ground state and $e$ signifies the excited state).
Now typically the ground state is symmetric and thus $\langle g | \vec{R} | g \rangle = 0$. Thus $H_{gg} = e \vec{E} \cdot \langle g | \vec{R} | g \rangle =0$. The same is usually true for the matrix element $H_{ee}$. However, we must worry about the matrix element $H_{eg} = e \vec{E} \cdot \langle e | \vec{R} | g \rangle$.
Now let's make the guess he does that $\psi = [C_g(t), C_e(t) e^{-i\omega_{eg}t}]$, where $\omega_{eg}$ is the difference in energy between the ground and excited state. The unperturbed hamiltonian is 
$\left(
  \begin{array}{ccc}
    0 & 0  \\
    0 & \hbar \omega_{eg}  \\
  \end{array}
\right)$, and the perturbation of the hamiltonian is 
$\left(
  \begin{array}{ccc}
    0 & H_{ge}  \\
    H_{eg} & 0  \\
  \end{array}
\right)$, where $H_{ge} = H_{eg}^*$.
We found that the sum of these two, the total hamiltonian is,
$\left(
  \begin{array}{ccc}
    0 & H_{ge}  \\
    H_{eg} & \hbar \omega_{eg}  \\
  \end{array}
\right)$.
Plugging this matrix for the hamiltonian as well as the guess for the wave function into the time-dependent schroedinger equation you will get his equation 3.11 (Remember I have suppresed the time dependence of $\vec{E}$ so you will have to put that in (i.e. multiply $\vec{E}$ by $\cos(\omega t)$) when trying to get his 3.11).
Now that I have said this see if you can figure out what happens with a particle in the box? The electron should leave the ground state, but how quickly does the probability of the electron being the ground state decrease? I don't know but it be a good exercise for you to do.
A: For a dipole coupling of the form $\hat H_\text{int}=-d\cdot E$, the dipole operator $d=er$ is proportional to the (vector) position of the electron, considered as an operator on the electron's wavefunction. Thus, for example, a hydrogen atom perturbed by an electric field $\mathbf E(t)$ has 
$$\hat H=-\frac{\hbar^2}{2m}\nabla^2+\frac{1}{r}-e\,\mathbf r\cdot\mathbf E(t)$$
as a hamiltonian. Note that the vector position has the same standing as an operator as the Coulomb interaction (though if the external field is small it can be treated as a perturbation).
Similarly, for a particle in a box under a dipole coupling, you will have a time-dependent Schrödinger equation of the form
$$
i\hbar\frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}-e\,x\cdot E(t)\psi
\quad\text{under}\quad\psi(0)=0=\psi(L).
$$
Note that at no point did I need to talk about eigenstates, though they can be helpful in solving both of the equations. The average value $\langle\mathbf r\rangle$, however, rarely plays a role in this story.
