quantum mechanics - Deriving the optical Bloch equations from the von Neumann equations

Is it possible to derive the optical Bloch equations for a 2-level-system driven by an oscillating EM-Field from the von Neumann equation for the density operator?

I'm assuming a system consisting of the states $|g \rangle$ and $ e \rangle$. Those are eigenstates of the Hamiltonian $\hat{H}_0$ with energies $E_e$ and $E_g$. The whole hamiltonian should be a sum of $\hat{H}_0$ and $\hat{H}_E = \vec{r_z} E_0 \cos{\omega t}$. Because of the dipole operator, the diagonal matrix elements of $\hat{H}_E$ will disappear:

$$ \langle g | \hat{H}_E | g \rangle = \langle e | \hat{H}_E | e \rangle = 0 \\ $$ $$ \langle g | \hat{H}_E | e \rangle = \langle e | \hat{H}_E | g \rangle^* = \Omega_{Rabi} \hbar \cos{\omega t} \\ $$

Let's say i'm interested in the differential equation for the first density matrix element $\rho_gg$, and I know it's supposed to look like this (According to my professor): $$ \frac{d \rho_{gg}}{dt} = \frac i2 \Omega^*_{Rabi}e^{-i(\omega-\omega_0)t} \rho_{ge} - \frac i2 \Omega_{Rabi}e^{i(\omega-\omega_0)t} \rho_{eg} $$

However, if I try to derive this:

$$ \frac{d}{dt}\hat{\rho} = \frac{i}{\hbar} (\hat{\rho} (\hat{H}_0 + \hat{H}_E) - (\hat{H}_0 + \hat{H}_E)\hat{\rho}) \\ \frac{d}{dt} \rho_{gg} = \frac{i}{\hbar} \langle g |(\hat{\rho} (\hat{H}_0 + \hat{H}_E) - (\hat{H}_0 + \hat{H}_E)\hat{\rho})| g\rangle \\ = \frac{i}{\hbar} ( \langle g |\hat{\rho} \hat{H}_0| g\rangle + \langle g |\hat{\rho} \hat{H}_E| g\rangle - \langle g | \hat{H}_0 \hat{\rho}| g\rangle + \langle g | \hat{H}_E \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} (E_g \langle g |\hat{\rho}| g\rangle + \langle g |\hat{\rho} \hat{H}_E| g\rangle - E_g \langle g | \hat{\rho}| g\rangle + \langle g | \hat{H}_E \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} ( \langle g |\hat{\rho} \hat{H}_E| g\rangle - \langle g | \hat{H}_E \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} ( \langle g |\hat{\rho} |g \rangle \langle g | \hat{H}_E| g\rangle + \langle g |\hat{\rho} | e \rangle \langle e| \hat{H}_E| g\rangle - \langle g | \hat{H}_E |g \rangle \langle g | \hat{\rho}| g\rangle - \langle g | \hat{H}_E |e \rangle \langle e | \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} ( \rho_{ge} \Omega_{Rabi}^* \cos{\omega t} - \rho_{eg} \Omega_{Rabi} \cos{\omega t}) $$

So now here I stand and don't know wether I made a mistake, or wether it's not possible without additional assumptions. I don't know how possibly something like $e^{i\omega_0 t}$ should appear in this equation.

Answer

To arrive at the equations that my professor gave, I have to assume different states for $| e \rangle$ and $| g \rangle$. While I used the time independent, you can also use the same states multiplied by a phase-factor $| e \rangle =e^{-i\omega_g t}| e\rangle$ and $| g \rangle =g^{-i\omega_g t}| e\rangle$. Using them, the matrix elements of $\hat{H}_E$ are:

$$ \langle \tilde{g} | \hat{H}_E | \tilde{e} \rangle = \langle \tilde{e} | \hat{H}_E | \tilde{g} \rangle^* = \Omega_{Rabi} \hbar \cos{\omega t}e^{i \omega_0 t} \\ $$

With $\omega_0 = \omega_e - \omega_g$. The rotating wave approximation yields the desired result in the question:

$$ \frac{d \rho_{gg}}{dt} = \frac i2 \Omega^*_{Rabi}e^{-i(\omega-\omega_0)t} \rho_{eg} - \frac i2 \Omega_{Rabi}e^{i(\omega-\omega_0)t} \rho_{ge} $$