Deriving Rabi rotation matrix

Question

I want to understand where the matrix:

$$ \left|\psi(t)\right> = \binom{a(t)}{b(t)} = \begin{bmatrix} cos(\Omega t/2)&-ie^{i\phi_L t}sin(\Omega t/2) \\ -ie^{-i\phi_L t}sin(\Omega t/2) & cos(\Omega t/2) \end{bmatrix}\binom{a(0)}{b(0)}$$

comes from.

I've already derived $$\dot{a} = \frac{\Omega}{2}e^{i\phi_L t} b$$ and its counterpart $$\dot{b} = \frac{\Omega}{2}e^{-i\phi_L t} a$$

where $\Omega = \frac{1}{\hbar} \left<1\right|-d\cdot\varepsilon\left|0\right>$ and, I think, $\phi_L$ is the frequency of the incident radiation. But how can I get from the above two relations to the matrix?

Answer 1

This is a coupled linear differential equation. Just write it in a matrix form as $\dot{X} = M X$, where $X$ is a vector formed by $a$ and $b$, and $M$ is a matrix. The solution is similar to the one dimensional case, but you will get the exponential of the matrix $M$ in the solution. You can get the components of $\exp (M)$ by using the Taylor expansion of the exponential.