Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 1

The spin operators here are dimensionless: $S^j={\sigma }_j/2$, so $\Omega$ has units of inverse time and $\Phi$ is a dimensionless pulse area. Integrating the Schrödinger evolution across an infinitesimally short pulse gives $U_{\Phi }=e^{-i\Phi S^x}$. The finite free Hamiltonian contributes a vanishing integrated phase within that pulse. Between pulses the free evolution is $U_{\Omega }=e^{-i\Omega T{S}^z}$.

In the $S^z$ eigenbasis, $S^z=\mathrm{diag}(1,-1)/2$ and $({\sigma }_x{)}^2=I$. ConsequentlyStarting just before a pulse, the kick acts first and the free interval second. Operators act on states right to left, henceThere are exactly $N$ kicks, at times $0,T,\dots ,(N-1)T$; the pulse at $NT$ is excluded. Sampling just after kicks would change the displayed product order. This endpoint convention specifies the delta-kicked spin Floquet operator.

The Pauli matrix multiplication law givesFor a unit vector $\hat{\mathbf{n}}$, it follows that $(\hat{\mathbf{n}}\cdot \mathbf{\sigma }{)}^2=I$. In the exponential series the even powers therefore multiply $I$ and the odd powers multiply $\hat{\mathbf{n}}\cdot \mathbf{\sigma }$. Summing them separately,Equivalently, the eigenvalues of $\hat{\mathbf{n}}\cdot \mathbf{\sigma }$ are $\pm 1$, so its two spectral projectors acquire phases $e^{\mp i\theta /2}$. The half-angle belongs to a spinor rotation, while the associated physical spin vector rotates through $\theta$.

For the axis-angle composition of orthogonal spin rotations, set $A=\Omega T/2$ and $B=\Phi /2$. Multiplication, including ${\sigma }_z{\sigma }_x=i{\sigma }_y$, givesOne checks that $C^2+|\mathbf{v}{|}^2=1$. For $s=|\mathbf{v}|\ne 0$, chooseThis choice retains the full spinor sign and gives $U_{\Omega }{U}_{\Phi }=e^{-i\vartheta \hat{\mathbf{n}}\cdot \mathbf{S}}$. Repeated periods have the same axis:Thus the spin expectation vector rotates through $N\vartheta$, interpreted modulo $2\pi$ as a spatial rotation. The positive $y$ component comes from the stated ordering; reversing the two rotations changes that component's sign.

When $s=0$, the one-period operator is $+I$ or $-I$, and no unique rotation axis exists. The latter is a $2\pi$ spatial rotation with a minus sign on a spin-$1/2$ state. It is safer to retain this central group element than to replace the angle by zero in a spinor calculation.

For general spin $S$, use dimensionless generators $S^j$ obeying $[S^i,S^j]=i{ϵ}_{ijk}{S}^k$. The two exponentials represent the same group elements as for spin $1/2$, now in the $(2S+1)$-dimensional representation of $SU(2)$. A representation preserves products, sowith exactly the same axis and angle found in (c); after $N$ periods the angle is $N\vartheta$. One must not substitute higher-spin matrices into the two-term Pauli exponential in (b), which used a property special to spin $1/2$.

The dimension and action on spin states change, but the spin-independent axis-angle composition does not. In particular,Integer-spin states return with positive sign and half-integer-spin states with negative sign. spin expectation vectors transform as spatial vectors in either case, while a general higher-spin state also carries higher multipole information.