= Solution
The QG thermodynamic relation makes vertical velocity proportional to the material derivative of $\psi_z$. At a flat rigid boundary, no normal flow therefore gives
$$
\boxed{\psi'_{zt}=0\quad(z=0),}
$$
or $\widehat\psi'(0)=0$ for a nonzero-frequency mode. Let $K^2=k^2+l^2$ and take the vertical structure $\widehat\psi=\psi_0\cos(mz)$, which satisfies this condition. Linearizing $q_t+\beta\psi_x=0$ gives the <Baroclinic Rossby wave> dispersion relation
$$
\boxed{
\omega=-\frac{\beta k}{K^2+f_0^2m^2/N^2}
=-\frac{\beta k}{K^2+R_d^{-2}},
\qquad R_d=\frac{N}{f_0|m|}.}
$$
For long horizontal waves, $K R_d\ll1$, so $\omega\simeq-\beta kR_d^2$: the vertical-mode deformation term controls the response. For short waves, $K R_d\gg1$, so $\omega\simeq-\beta k/K^2$, the <Barotropic Rossby wave> form. In both limits the zonal phase propagation is westward relative to the mean flow.
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