= Solution
Let
$$
K=\int dg\,|g\rangle\langle g|,
$$
where $dg$ is <Haar measure>. Left invariance gives $hKh^{-1}=K$ for every $h\in SU(2)$. Since the spin-$S$ representation is irreducible, <Schur lemma> implies
$$
\boxed{K=c\operatorname{Id}.}
$$
Inserting this coherent-state resolution between $N$ short time steps and taking the continuum limit gives, for a noninteracting spin with zero Hamiltonian,
$$
\boxed{
\langle g_f|g_i\rangle
=\int_{g_i}^{g_f}\mathcal Dg\,
\exp\left(\int_0^t\langle\dot g|g\rangle dt'\right),}
$$
up to the normalization generated by $c$. A Hamiltonian would add $-i\int\langle g|H|g\rangle dt$ in the exponent.
For
$$
|g\rangle=e^{-i\phi S_3}e^{-i\theta S_2}e^{-i\psi S_3}|\uparrow\rangle,
$$
direct differentiation and $S_3|\uparrow\rangle=S|\uparrow\rangle$ give
$$
\boxed{
\langle g|\dot g\rangle=-iS(\dot\psi+\cos\theta\,\dot\phi),
\qquad
\langle\dot g|g\rangle=iS(\dot\psi+\cos\theta\,\dot\phi).}
$$
The $\dot\psi$ term is a total derivative and records only the arbitrary phase used to represent a ray. It cancels against the endpoint-state phases, so the physical <Spin coherent-state path integral> depends only on the path on the two-sphere. For example, choosing the section $\psi=-\phi$ gives the <Spin coherent-state Berry phase> proportional to
$$
-S\int(1-\cos\theta)\dot\phi\,dt.
$$
Changing the surface used to fill a closed path changes its solid angle by $4\pi$. Single-valuedness of the path-integral phase requires
$$
e^{i4\pi S}=1,
\qquad
\boxed{2S\in\mathbb Z.}
$$
Thus the Wess-Zumino coefficient is quantized in integer or half-integer units.
Back to article page
