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.


1 Answer 1


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 $kz \ll 1$,

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


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