# Setting up differential equations for two-level Rabi problem

I try to follow the derivation of Rabi two-level problem but I went into trouble when attempting to set up the equations as many notes have suggested.

Using the book (Laser cooling and trapping by Metcalf and Straten) I am reading. We start by with writing down Schrodinger's equation for a two-level system where the Hamiltonian is given by $H=H_0 + H'$. And $H_0$ absorbs all diagonal terms from $H'$ so that the resulting Schrodinger equations are a coupled differential equations. $$i\hbar\frac{\partial\psi}{\partial t} = H \psi$$ and $$|\psi\rangle = c_g |g\rangle + c_e e^{-i\omega_{eg}t}|e\rangle$$ where $|g\rangle$ is the ground state, $|e\rangle$ is the excited state, and $\omega_{eg}=\omega_e-\omega_g$.

The coupled equations are \begin{align} i\hbar \dot{c}_g(t) &= c_e(t)H'_{ge}(t) e^{-i\omega_{eg}}t\\ i\hbar \dot{c}_e(t) &= c_g(t)H'_{eg}(t) e^{i\omega_{eg}}t \end{align}

In the book, the author uses $H'(t)=-e\vec{E}(\vec{r},t) \cdot \vec{r}$ and with a plane wave travelling in the $z$-direction, the electric field operator becomes $\vec{E}(\vec{r},t)=E_0\hat{\epsilon}\cos(kz-\omega_l t)$, where $\omega_l$ is the laser frequency. Now if we define Rabi frequency as $$\Omega\equiv \frac{-eE_0}{\hbar}\langle e|r|g\rangle,$$ the element of $H'$ becomes $H'_{eg}=\hbar \Omega \cos (kz-\omega_l t)$.

Here is where I run into problems, I plug in the expression for $H'_{eg}$, differentiate the second equation and use both first order equations to eliminate $c_g(t)$. However the resulting equation contains $z$ dependence that I do not know how to get rid of.

Referring to this note, I see they use the same expression for $H'$ but they also ignored the $z$ dependence. I am wondering what I have missed in considering setting up the equations.

You probably implicitly making the assumption that the wavelength of this wave is much larger than the size of the atom, so that $kz \ll 1$. Here is why I think that:

I see that you are using the interaction picture, so that $i\hbar \frac{\partial}{\partial t} \lvert \psi \rangle = H'\lvert \psi \rangle$, then, applying $\langle g \rvert$ and $\langle e \rvert$ we obtain the two equations:

$$i\hbar \frac{\partial}{\partial t}C_g(t) = C_g(t)\langle g \rvert H' \lvert g \rangle + C_e(t)e^{-i\omega_{eg}t}\langle g \rvert H' \lvert e \rangle$$ $$i\hbar \frac{\partial}{\partial t}C_e(t) = C_e(t)\langle e \rvert H' \lvert g \rangle + C_e(t)e^{-i\omega_{eg}t}\langle e \rvert H' \lvert e \rangle$$

But in your calculations $\langle g \rvert H' \lvert g \rangle = \langle e \rvert H' \lvert e \rangle = 0$. This is usually (as far as I know) because of a parity argument. Because, for g, and similarly for e:

$$\langle g \rvert H' \lvert g \rangle = \langle g \rvert -e E_0 \cos(kz-\omega t) \hat{\epsilon} \cdot \vec{r} \lvert g \rangle = -eE_0\hat{\epsilon}\cdot \int d^3 \vec{r} |\psi_ g(\vec{r})|^2\vec{r} \cos(kz-\omega t)$$

If $kz \ll 1$, we can consider the cosine to be constant for the integration, take it out of the integral and get 0 as a result because the resulting integrand will be an odd function ($|\psi|^2$ is even and $\vec{r}$ is odd).

Now, assuming that we are taking $kz \ll 1$,

$$H'_{ge} = \langle g \rvert -eE_0 \cos(kz-\omega t)\hat{\epsilon} \cdot \vec{r} \rvert e \rangle = -eE_0 \langle g \lvert \cos(kz - \omega t) \hat{\epsilon} \cdot \vec{r} \rvert e \rangle$$ $$H'_{ge} \approx -eE_0 \cos(\omega t) \langle g \rvert \hat{\epsilon} \cdot \vec{r} \lvert e \rangle$$