Past exam of the theoretical physics II course of the University of Cambridge / 2019 / 1

The Heisenberg equation of motion and the spin commutation relations give, componentwise,Thus $\boxed{d\mathbf{S}/dt=\gamma \mathbf{S}\times \mathbf{B}}$. For constant magnetic field this is Larmor precession: the component along $\mathbf{B}$ is conserved and the transverse components rotate. In particular, for $\mathbf{B}=B\hat{\mathbf{z}}$,The signed angular velocity is $-\gamma \mathbf{B}$; the positive precession rate is $|\gamma B|$.

Here the dimensionless spin one-half operator is $S_x={\sigma }_x/2$. Since the Pauli matrix obeys ${\sigma }_x^2=I$, the even and odd terms of the matrix exponential giveThis is the exponential of a Pauli vector for an $x$-axis rotation. With a physical spin angular momentum $S_x^{\mathrm{phys}}=\hslash {\sigma }_x/2$, the exponent is $-i\theta S_x^{\mathrm{phys}}/\hslash$.

Use the ordered orthonormal basis $(|1s⟩,|2p_x⟩)$. The projected Hamiltonian between pulses is $H_0=-\Delta {\sigma }_z$, so its free time-evolution operator through a period isAcross one instantaneous pulse, integrating the Schrödinger equation givesusing the exponential of a Pauli vector. The finite free Hamiltonian contributes vanishing evolution during the zero-duration kick. A period measured immediately before successive pulses therefore consists of the kick followed by free evolution. Composition givesThis is the Floquet operator at the specified phase of the pulse train. Actual hydrogenic energy levels are not opposite numbers: the displayed convention subtracts their midpoint, with $\Delta =(E_{2p}-E_{1s})/2$. Restoring that midpoint multiplies the result by an unobservable global phase. Likewise $E$ in the given Hamiltonian denotes the electric coupling per unit length; if it is instead a physical electric-field strength, its charge factor must be included in $E{x}_0$.

Put $\alpha =\Delta \tau /\hslash$ and $\beta =E{x}_0\tau /\hslash$. The Pauli matrix multiplication law gives the pulse-and-precession spin one-half mapBecause $C^2+|\mathbf{v}{|}^2=1$, a suitable angle and unit vector for the rotation areIf the denominator vanishes, $U=\pm I$ and the axis is arbitrary. The Bloch vector $\mathbf{r}=⟨\mathbf{\sigma }⟩$ rotates through $\theta$ about $\mathbf{n}$ per period; after $N$ periods the angle is $N\theta$. The eigenphases equivalently give quasienergies $\pm \hslash \theta /(2\tau )$ modulo $2\pi \hslash /\tau$.

In the two-orbital subspace the position operator is $X=x_0{\sigma }_x$. The Heisenberg equation of motion between pulses gives $P/m=(i/\hslash )[H_0,X]=(2\Delta x_0/\hslash ){\sigma }_y$, henceThus two components of the Bloch vector directly describe the mean position and momentum; its third gives the orbital population imbalance. The electric kick rotates about the position axis, leaving $⟨X⟩$ unchanged at the kick, while free evolution mixes the position and momentum components. Starting in $|1s⟩$, one complete period giveswhich exhibits the induced motion explicitly. These are projected two-orbital observables.