= Solution
Use the displacement-potential convention $u_x=\partial_x\phi-\partial_z\psi$, $u_z=\partial_z\phi+\partial_x\psi$ and harmonic dependence $e^{-i\omega t}$. Choosing the opposite sign for the shear potential changes its column sign but not the <P-SV directional impedance matrix>. Write $p=k_\alpha>0$, $s=k_\beta>0$, $D=k^2+ps$, and $H=\rho\omega^2-2\mu k^2$. At a common point the forward potential <wave amplitudes> give
$$
\begin{pmatrix}v_x\\v_z\end{pmatrix}=\omega\begin{pmatrix}k&-s\\p&k\end{pmatrix}\begin{pmatrix}\phi_+\\\psi_+\end{pmatrix},\qquad \begin{pmatrix}\sigma_{xz}\\\sigma_{zz}\end{pmatrix}=\begin{pmatrix}-2\mu kp&H\\-H&-2\mu ks\end{pmatrix}\begin{pmatrix}\phi_+\\\psi_+\end{pmatrix}.
$$
The <traction> formulas follow from $\sigma_{xz}=\mu(u_{x,z}+u_{z,x})$ and $\sigma_{zz}=\lambda\nabla\cdot\mathbf u+2\mu u_{z,z}$, using $(\lambda+2\mu)\omega^2/\alpha^2=\rho\omega^2$ and $\mu(s^2-k^2)=H$. The <velocity> <matrix> has <determinant> $\omega^2D>0$, so elimination gives the <P-SV directional impedance matrix>
$$
\boxed{Z_+=\frac1{\omega D}\begin{pmatrix}\rho\omega^2p&k(2\mu D-\rho\omega^2)\\-k(2\mu D-\rho\omega^2)&\rho\omega^2s\end{pmatrix},\qquad\mathbf t=-Z_+\mathbf v.}
$$
Its off-diagonal entries are real and antisymmetric. Therefore
$$
Z_++Z_+^\dagger=\frac{2\rho\omega}{D}\begin{pmatrix}p&0\\0&s\end{pmatrix}.
$$
With real-part phasors, the outward $z$ flux is $-\tfrac12\operatorname{Re}(\mathbf v^\dagger\mathbf t)$, so
$$
\boxed{\langle S_z\rangle_+=\frac14\mathbf v^\dagger(Z_++Z_+^\dagger)\mathbf v=\frac{\rho\omega}{2D}\left(p|v_x|^2+s|v_z|^2\right)>0}
$$
for every nonzero forward <displacement field>. <Mass density> and <angular frequency> are essential to this flux coefficient; both appear correctly in the PDF.
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