Rabi problem for a spin-1 particle I'm trying to solve, analitically, the transition probability of a spin-1 particle in a magnetic field
$$
\vec{B}=B_0\hat{k}+b(\cos{\omega t}\hat{i}+sin{\omega t}\hat{j}).
$$
In particular I want to find
$$
\mathcal{P}(S_z=\hbar\rightarrow S_z=0)\;\;\;\;\text{and}\;\;\;\;\mathcal{P}(S_z=\hbar\rightarrow S_z=-\hbar).
$$
I know that the hamiltonian is of the form
$$
H(t)=-\hbar\omega_0\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1 \end{array} \right)
-\frac{\hbar\omega_\perp}{\sqrt2}\left( \begin{array}{ccc}
0 & e^{-i\omega t} & 0 \\
e^{i\omega t} & 0 & e^{-i\omega t} \\
0 & e^{i\omega t} & 0 \end{array} \right),
$$
since the definition of $S_x$, $S_y$ and $S_z$.
I followed the spin-1/2 case. I find the system of differential equations:
$$
i\hbar\dot a(t)=-\hbar\omega_0 a(t)-\frac{\hbar\omega_\perp}{\sqrt{2}}e^{-i\omega t}b(t)
$$
$$
i\hbar\dot b(t)=-\frac{\hbar\omega_\perp}{\sqrt 2}[e^{i\omega t}a(t)+e^{-i\omega t}c(t)]
$$
$$
i\hbar\dot c(t)=\hbar\omega_0 c(t)-\frac{\hbar\omega_\perp}{\sqrt{2}}e^{i\omega t}b(t)
$$
Since it's easier with time independent coefficients I can define
$$
A(t)=a(t)e^{i\beta t},\;\;\;\;B(t)=b(t)\;\;\;\;\text{and}\;\;\;\;C(t)=c(t)e^{-i\beta t}
$$
and (setting $\beta=\omega$) rewrite the previous system:
$$
i\dot A(t)=-\delta A(t)-\frac{\omega_\perp}{\sqrt2}B(t)
$$
$$
i\dot B(t)=-\frac{\omega_\perp}{\sqrt2}(A(t)+C(t))
$$
$$
i\dot C(t)=\delta C(t)-\frac{\omega_\perp}{\sqrt2}B(t),
$$
where $\delta=\omega-\omega_0$; or in matrix form:
$$
i\left( \begin{array}{c}
\dot A(t)\\
\dot B(t)\\
\dot C(t) \end{array} \right)
=-
\left( \begin{array}{ccc}
\delta & \frac{\omega_\perp}{\sqrt2} & 0 \\
\frac{\omega_\perp}{\sqrt2} & 0 & \frac{\omega_\perp}{\sqrt2} \\
0 & \frac{\omega_\perp}{\sqrt2} & -\delta \end{array} \right)
\left( \begin{array}{c}
A(t)\\
B(t)\\
C(t)\end{array} \right).
$$
Now, the solution of an equation like this, where the coefficient matrix is $M$, sohould be
$$
\left( \begin{array}{c}
A(t)\\
B(t)\\
C(t)\end{array} \right)=e^{iMt}
\left( \begin{array}{c}
A(0)\\
B(0)\\
C(0)\end{array} \right).
$$
I should find now an explicit solution in order to evaluate the probability (I need $|B(t)|^2$ and $|C(t)|^2$).
The problem is that in the spin-1/2 case one can use the trick of writing the exponential of the matrix as a sum of a sine and a cosine (since all odd powers $\propto$ $M$ and all the even $\propto$ $\mathbb{I}$), but now it's not possible.
How should I proceed?
I can diagonalize $M$ and use the property $e^{-iMt}=Pe^{-iDt}P^{-1}$, then with the exponential of the diagonal matrix there shouldn't be any problem, but it's long and tedious. Isn't there any other method that I can use?
Or did I make any mistake before?
 A: 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:
$$
\mathcal{P}(1\rightarrow 0,t)=|B(t)|^2=\frac{\delta^2\omega_\perp^2}{\Omega^4}2\sin^4{\left(\frac{\Omega t}{2}\right)}+\frac{\omega^2_\perp}{2\Omega^2}\sin^2{(\Omega t)},
$$
$$
\mathcal{P}(1\rightarrow -1,t)=|C(t)|^2=\frac{\omega_\perp^4}{\Omega^4}\sin^4{\left(\frac{\Omega t}{2}\right)}.
$$
A: 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.
