  Two model-independent results on the momentum-dependence of $\rho$-$\omega$
mixing are described. First, an explicit choice of interpolating fields for the
vector mesons is displayed for which both the mixing in the propagator and the
isospin-breaking at the nucleon-vector meson vertices (and hence also the
one-vector-meson-exchange contribution to NN charge symmetry breaking) vanish
identically at $q^2=0$. Second, it is shown, using the constraints of unitarity
and analyticity on the spectral function of the vector meson propagator, that
there is no possible choice of interpolating fields for the $\rho^0$,
$\omega^0$ mesons such that, with the $\rho\omega$ element of the propagator
defined by $\Delta^{\rho\omega}_{\mu\nu}(q^2)= (g_{\mu\nu}-q_\mu q_\nu
/q^2)$$\theta (q^2)/(q^2-m^2_\rho )(q^2- m^2_\omega )$, $\theta (q^2)$ is
independent of momentum. It follows that the standard treatment of charge
symmetry breaking in few-body systems cannot be interpreted as arising from any
realizable effective meson-baryon Lagrangian and must, therefore, be considered
purely phenomenological in content.
