# Bloch equations for a 3-level system

## Homework Statement

"Consider a system with three states, ##|1\rangle , |2\rangle ,|3\rangle ## with energies ##\hbar \omega_1 , \hbar \omega_2 , \hbar \omega_3 ##. the states are then seperated by ##\hbar \omega_3 -\hbar \omega_1 = \hbar \omega_{13}## and ## \hbar \omega_3-\hbar \omega_2= \hbar \omega_{23}##. Two seperate light fields are incident on the system with frequencies ##\omega_p## and ##\omega_c## respectively, where ##\omega_p## couples states 1 and 3 with a detuning of ##\Delta_p##, and ##\omega_c## couples states 2 and 3 with a detuning of ##\Delta_c##. The loss rates from the excited states ##|2\rangle## and ##|3\rangle## are ##\gamma_2## and ##\gamma_3## respectively.

Show that the H for the system in the rotating wave approx. is ##
\dfrac{\hbar}{2}\begin{pmatrix}
2\omega_{1}& 0 & -\Omega_{p}e^{i\omega_{p}t} \\
0& 2\omega_{2} & -\Omega_{c}e^{i\omega_{c}t}\\
-\Omega_{p}e^{-i\omega_{p}t}& -\Omega_{c}e^{-i\omega_{c}t} & 2\omega_{3}
\end{pmatrix}##

Where the Rabi frequencies ##\Omega_p## and ##\Omega_c## are defined naturally, for example
## \hbar\Omega_{p} = \langle 1 |\vec{d} \cdot \vec{E_{p}}| 3 \rangle ##."

Presumably ##\vec{E_p}## is the vector magnitude of the relevant electric field, i.e. the field coupling state 1 and 3 is described by ##\vec{E_p}\cos (\omega_p t)##.

## The Attempt at a Solution

The diagonal elements of the Hamiltonian are straightforward, but we don't understand why the off diagonals look the way they do. In the two-state case, the dipole interaction Hamiltonian is given by ##H_I = -\vec{d}\cdot \vec{E}## where our electric field is ##\vec{E} \cos (\omega t).## Then the off diagonal element is given by ##\langle 1 | H_I | 2 \rangle = \langle 2 | H_I | 1 \rangle ^* = - \langle 1 | \vec{d} \cdot \vec{E} | 2 \rangle \cos (\omega t). ## But in this case we have two electric fields and so ## H_I = -\vec{d} \cdot (\vec{E_p}\cos{w_pt} + \vec{E_c}\cos{w_ct})##. But when we try to take the matrix element of this w.r.t. the different states to find the entries of the Hamiltonian, things aren't working out. For example, why is it that ##\langle 1 | H_I | 2 \rangle = 0 ## ?