Past exam of the quantum physics course of the University of Oxford / 2016 / 10

With $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $S_z=(\hslash /2){\sigma }_z$,ThusFor $B_0>0$, spin down is the lower energy eigenstate. The negative electron magnetic moment is already accounted for in the positive coefficient multiplying $\mathbf{S}\cdot \mathbf{B}$ in the given Hamiltonian.

In the ordered basis $|+⟩,|-⟩$, the transverse magnetic field producesThe off-diagonal complex exponentials follow directly from $\cos \omega t\mp i\sin \omega t$. This matrix is Hermitian at every time.

Insert the expansion with amplitudes $c_1,c_2$ and the two unperturbed dynamical phases into $i\hslash \dot{\Psi }=(H_0+V)\Psi$. Differentiation of the explicit phases cancels $H_0$, leavingEquivalently, with $\Omega =(\omega -{\omega }_0)/2$,These interaction-picture equations conserve $|c_1{|}^2+|c_2{|}^2$. They are exact for the circularly rotating magnetic drive; a rotating-wave approximation has not been used.

Differentiate ${\dot{c}}_2=-i\gamma e^{2i\Omega t}{c}_1$ and substitute the equation for ${\dot{c}}_1$. The result isPutting $c_2=e^{i\Omega t}u$ removes the first derivative and gives $\ddot{u}+{\kappa }^{2}u=0$, where $\kappa =\sqrt{{\Omega }^2+{\gamma }^2}$. For $\kappa >0$ the general solution isThe initial spinor fixes $c_1(0)=1,c_2(0)=0$, and the first-order coupled equation fixes the second initial datum ${\dot{c}}_2(0)=-i\gamma$. Thus $A=0$, $B=-i\gamma /\kappa$, andTheir squared magnitudes sum to one. If both $\Omega$ and $\gamma$ vanish, the initial amplitudes simply remain $(1,0)$; the formulas have that continuous limit.

Source convention correction. For positive $\gamma$, the printed expression for $c_2$ lacks the factor $-i$ required by the stated basis and initial vector. A concrete test is its derivative at $t=0$: the printed expression gives $+\gamma$, whereas the Schrödinger equation gives $-i\gamma$. Multiplying only $c_2$ by a phase changes the relative quantum phase. One may change the overall phase of the entire spinor, but must then transform $c_1$ and its initial amplitude as well. The printed modulus nevertheless gives the correct transition probability.

On Rabi resonance, $\omega ={\omega }_0$ and $\Omega =0$, soFor nonzero drive the probability starts at zero, reaches unity at $t=\pi /(2|\gamma |)$, and repeats with period $\pi /|\gamma |$. The population oscillation therefore has angular frequency $2|\gamma |$. This is complete resonant Rabi transfer; a drive of zero strength gives no transition.

For nonzero Rabi detuning,The sketch starts at zero and oscillates between zero and ${\gamma }^2/({\gamma }^2+{\Omega }^2)<1$, with period $\pi /\sqrt{{\gamma }^2+{\Omega }^2}$ when the coupling is nonzero. Relative to resonance, the oscillations have a smaller maximum and a shorter period at fixed coupling. The sign of the Rabi detuning changes the amplitudes' phases but leaves this transition probability unchanged. For $|\Omega |\gg |\gamma |$, its maximum is approximately ${\gamma }^2/{\Omega }^2$, so population transfer is strongly suppressed. The plotted detuned example has $|\Omega |=|\gamma |$, giving maximum $1/2$.