One example of Rabi flopping is the spin flipping within a quantum system containing a spin-1/2 particle and an oscillating magnetic field. We split the magnetic field into a constant 'environment' field, and the oscillating part, so that our field may be written as\mathbf{B} = \mathbf{B}_{env} + \mathbf{B}_{osc} = B_0 \mathbf{k} + B_1 (\cos(\omega t)\mathbf{i} + \sin(\omega t)\mathbf{j}),where B_0 and B_1 are the strengths of the environment and the oscillating fields respectively, and \omega is the frequency at which the oscillating field oscillates. We can then write a Hamiltonian describing this field, yieldingH = -\vec{\mu} \cdot \mathbf{B} = \omega_0 S_z + \omega_1(\cos(\omega t)S_x + \sin(\omega t)S_y)where the magnetic moment is\hat{\mathbf{\mu}} = - \gamma \mathbf{S} = - \gamma (\hat{S_x}\mathbf + \hat{S_y}\mathbf + \hat{S_z}\mathbf),\gamma is the
gyromagnetic ratio of the particle, \omega_0 = \gamma B_0, \omega_1 = \gamma B_1, and \hat{S_x}, \hat{S_y}, \hat{S_z} are the
spin operators. The frequency \omega_1is known as the
Rabi frequency. We can substitute in the matrix forms of the spin operators in the z basis to find the matrix representing the Hamiltonian:\begin{align} H &= \omega_0 \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} + \omega_1 \left(\cos(\omega t)\frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} + \sin(\omega t)\frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\right) \\ &= \frac{\hbar}{2}\begin{bmatrix} \omega_0 & \omega_1 e^{-i\omega t} \\ \omega_1 e^{i\omega t} & -\omega_0\end{bmatrix} \end{align}where we have used \cos(\omega t) + i\sin(\omega t) = e^{i\omega t}. This Hamiltonian is a function of time, meaning we cannot use the standard prescription of Schrödinger time evolution in quantum mechanics. For time independent Hamiltonians, the time evolution of a state may be expressed using the time evolution operator U(t) = e^{-iHt/\hbar}, but this formula cannot directly be used for time dependent Hamiltonians. The main strategy in solving this problem is to transform the Hamiltonian so that the time dependence is gone, solve the problem in this transformed frame, and then transform the results back to normal. This can be done by shifting the reference frame that we work in to match the
rotating magnetic field. If we rotate along with the magnetic field, then from our point of view, the magnetic field is not rotating and appears constant. Therefore, in the
rotating reference frame, both the magnetic field and the Hamiltonian are constant with respect to time. We denote our spin-1/2 particle state to be |\psi(t)\rangle = c_+(t)|+\rangle + c_-(t)|-\rangle in the stationary reference frame, where |+\rangle and |-\rangle are spin up and spin down states in the z direction respectively, and the state is
normalized: |c_+(t)|^2 + |c_-(t)|^2 = 1. We can transform this state to the rotating reference frame by using a
rotation operatorR_z(\theta) = \begin{bmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{bmatrix}which rotates the state counterclockwise around the positive z-axis in state space, which may be visualized as a
Bloch sphere. At a time t and a frequency \omega, the magnetic field will have precessed around by an angle \omega t. To transform |\psi(t)\rangle into the rotating reference frame, we seek to represent the spin up and spin down states from the stationary frame in the rotating frame. Critically, note that the stationary x and y-axes rotate clockwise from the point of view of the rotating reference frame. Thus, because the R_z(\theta) operator rotates counterclockwise, we must negate the angle to create a clockwise rotation that will represent the spin up and down states in the stationary frame as viewed by the rotating frame. The state becomes|\tilde{\psi}(t)\rangle = R_z(-\omega t)|\psi(t)\rangle = c_+(t)e^{i\omega t/2}|+\rangle + c_-(t)e^{-i\omega t/2}|-\rangleWe may rewrite the amplitudes so that|\tilde{\psi}(t)\rangle = \alpha_+(t)|+\rangle + \alpha_-(t)|-\rangleThe
time dependent Schrödinger equation in the stationary reference frame isi\hbar \frac{d}{dt}|\psi(t)\rangle = H(t)|\psi(t)\rangleExpanding this using the matrix forms of the Hamiltonian and the state yieldsi\hbar\begin{bmatrix}\frac{dc_+}{dt} \\ \frac{dc_-}{dt}\end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix} \omega_0 & \omega_1 e^{-i\omega t} \\ \omega_1 e^{i\omega t} & -\omega_0 \end{bmatrix} \begin{bmatrix} c_+(t) \\ c_-(t) \end{bmatrix}Applying the matrix and separating the components of the vector allows us to write two coupled differential equations as follows\begin{align} & i\hbar \frac{dc_+}{dt} = \frac{\hbar\omega_0}{2} c_+(t) + \frac{\hbar\omega_1}{2} e^{-i\omega t} c_-(t) \\ & i\hbar \frac{dc_-}{dt} = \frac{\hbar\omega_1}{2} e^{i\omega t} c_+(t) - \frac{\hbar\omega_0}{2} c_-(t) \end{align}To transform this into the rotating reference frame, we may use the fact that c_+(t) = \alpha_+(t) e^{-i \omega t/2} and c_-(t) = \alpha_-(t) e^{i \omega t/2} to write the following:\begin{align} & i\hbar \frac{d\alpha_+}{dt} = - \frac{\hbar\Delta\omega}{2} \alpha_+(t) + \frac{\hbar\omega_1}{2} \alpha_-(t) \\ & i\hbar \frac{d\alpha_-}{dt} = \frac{\hbar\omega_1}{2} \alpha_+(t) + \frac{\hbar\Delta\omega}{2} \alpha_-(t) \end{align}where \Delta\omega = \omega - \omega_0. Now define\tilde{H} = \frac{\hbar}{2} \begin{bmatrix} -\Delta\omega & \omega_1 \\ \omega_1 & \Delta\omega \end{bmatrix}We now write these two new coupled differential equations back into the form of the Schrödinger equation:i\hbar \frac{d}{dt}|\tilde{\psi}(t)\rangle = \tilde{H}|\tilde{\psi}(t)\rangleIn some sense, this is a transformed Schrödinger equation in the rotating reference frame. Crucially, the Hamiltonian does not vary with respect to time, meaning in this reference frame, we can use the familiar solution to Schrödinger time evolution:|\tilde{\psi}(t)\rangle = \tilde{U}(t)|\tilde{\psi}(0)\rangle = e^{-i \tilde{H}t/\hbar} |\tilde{\psi}(0)\rangleThis transformed problem is equivalent to that of
Larmor precession of a spin state, which is the precession of a spin state around a constant stationary magnetic field, so we have solved the essence of Rabi flopping. The probability that a particle starting in the spin up state flips to the spin down state can be stated asP_{+\to -} = \frac{\omega_1^2}{\Omega^2} \sin^2\left(\frac{\Omega}{2}t\right)where \Omega = \sqrt{\Delta \omega^2 + \omega^2_1} is the
generalized Rabi Frequency. Something important to notice is that P_{+ \to -} will not reach 1 unless \Delta\omega = 0. In other words, the frequency of the rotating magnetic field \omega must match the environmental field's Larmor frequency \omega_0 in order for the spin to fully flip; they must achieve
resonance. When resonance (i.e. \omega = \omega_0) is achieved, \Omega = \omega_1. To transform the solved state back to the stationary reference frame, we reuse the rotation operator with the opposite angle|\psi(t)\rangle = R_z(\omega t)|\tilde{\psi}(t)\rangle, thus yielding a full solution to the problem. == Applications ==