= Solution
For downward-phase reversal, the <P-SV displacement potentials> instead produce
$$
\mathbf v=\omega\begin{pmatrix}k&s\\-p&k\end{pmatrix}\begin{pmatrix}\phi_-\\\psi_-\end{pmatrix},\qquad \mathbf t=\begin{pmatrix}2\mu kp&H\\-H&2\mu ks\end{pmatrix}\begin{pmatrix}\phi_-\\\psi_-\end{pmatrix}.
$$
Eliminating the potentials as before gives
$$
\boxed{Z_-=\frac1{\omega D}\begin{pmatrix}-\rho\omega^2p&k(2\mu D-\rho\omega^2)\\-k(2\mu D-\rho\omega^2)&-\rho\omega^2s\end{pmatrix}=-Z_+^\dagger.}
$$
The <Hermitian part> is negative definite, and consequently
$$
\boxed{\langle S_z\rangle_-=-\frac{\rho\omega}{2D}\left(p|v_x|^2+s|v_z|^2\right)<0}
$$
for a nonzero backward <displacement field>. The P potential must use vertical <wavenumber> $k_\alpha$ and the S potential $k_\beta$, as in the original PDF; the TeX incorrectly gives the shear <wavenumber> for both.
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