# 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?

• This problem should be solvable by transforming to a rotating frame. See pp. 391--394 of Shankar's book on quantum mechanics to understand how this method works for spin 1/2. Then, I guess you could post your own answer. Oct 1 '15 at 0:22
• I'm sorry but I don't understand how a rotation can solve my problem. For what I see Shankar isn't interested in finding a solution for the state in the perturbed field case. Oct 2 '15 at 10:13

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: $$\psi_{\mathrm{rot}} (t) = \hat{U}(t)\psi(t),$$ where the time-dependent unitary transformation $U(t)$ is defined by $$U(t) \equiv \exp\left(i\omega t S_{z}/\hbar\right) = \begin{pmatrix} e^{i\omega t}&0&0\\ 0&1&0\\0&0&e^{-i\omega t}\end{pmatrix}.$$ One can verify that $\psi_{\mathrm{rot}} (t)$ satisfies $$i\hbar\dot{\psi}_{\!\mathrm{rot}}(t) = H_{\mathrm{rot}}\psi_{\mathrm{rot}}(t),$$ where $$H_{\mathrm{rot}} = (\omega_{0}-\omega)S_{z} + \sqrt{2}\omega_{\perp} S_{x}.$$ In the rotating frame, the Hamiltonian is time-independent.

• Ok, but now if I try to find the explicit solution for $\psi_{rot}$ I have the same problem as before. Oct 2 '15 at 11:28
• @Charlie Right. But this is a Hamiltonian of the form $B\, \textbf{S}\cdot\hat{\textbf{n}}$, and there exists a unitary operator $V$ such that $V S_{z} V^{-1} = \textbf{S}\cdot \hat{\textbf{n}}$. For this particular problem, $V$ is a rotation about the $y$-axis. Oct 2 '15 at 20:15