Question: In my [last Phys.SE question](https://physics.stackexchange.com/q/207008/2451) Emilio Pisanty mentioned this relationship $1 cm^{-1} = 10^n c^{-1} ns^{-1}$. I was wondering where this relationship came from. Does anybody know? Second of all, I am not sure how from this relationship, it can be derived that in natural units, $1 cm^{-1} = 30 ns^{-1}$. Can I have a clear derivation of this? I understand in natural units, the c vanishes, so is should be $1 cm^{-1} = 10^n ns^{-1}$, not $30 ns^{-1}$. So how is this derived?

Before downvoting, please tell me what you think is wrong with the question.

Answer: 
Finally I solved it using the Cayley–Hamilton theorem.

Since the characteristic polynomial of $M$ is
$$
\lambda^3=\Omega^2\lambda,
$$

where $\Omega=\sqrt{\delta^2+\omega\_{\perp}^2}$, we have that
$$M^3=\Omega^2M.$$

Then ($M^4=\Omega^2M^2$, $M^5=\Omega^4M$, $\dots$)

$$
e^{iMt}=\mathbb{I}+iMt+\frac{(iMt)^2}{2!}+...=\mathbb{I}-\frac{M^2}{\Omega^2}+\frac{M^2}{\Omega^2}\cos{(\Omega t)}+i\frac{M}{2}\sin{(\Omega t)}.
$$

One can finally compute the probabilities:

$$
\mathcal{P}(1\rightarrow 0,t)=|B(t)|^2=\frac{\delta^2\omega\_\perp^2}{\Omega^4}2\sin^4{\left(\frac{\Omega t}{2}\right)}+\frac{\omega^2\_\perp}{2\Omega^2}\sin^2{(\Omega t)},
$$
$$
\mathcal{P}(1\rightarrow -1,t)=|C(t)|^2=\frac{\omega\_\perp^4}{\Omega^4}\sin^4{\left(\frac{\Omega t}{2}\right)}.
$$
You can transform to the rotating frame as follows:
\begin{equation}
\psi\_{\mathrm{rot}} (t) = \hat{U}(t)\psi(t),
\end{equation}
where the time-dependent unitary transformation $U(t)$ is defined by
\begin{equation}
U(t) \equiv \exp\left(i\omega t S\_{z}/\hbar\right) = \begin{pmatrix} e^{i\omega t}&0&0\\ 0&1&0\\0&0&e^{-i\omega t}\end{pmatrix}.
\end{equation}
One can verify that $\psi\_{\mathrm{rot}} (t)$ satisfies
\begin{equation}
i\hbar\dot{\psi}\_{\!\mathrm{rot}}(t) = H\_{\mathrm{rot}}\psi\_{\mathrm{rot}}(t),
\end{equation}
where
\begin{equation}
H\_{\mathrm{rot}} = (\omega\_{0}-\omega)S\_{z} + \sqrt{2}\omega\_{\perp} S\_{x}.
\end{equation}
In the rotating frame, the Hamiltonian is time-independent.