In what sense does the Schrödinger equation obey the De Broglie relation? I've seen various accounts how the Schrödinger equation satisfies the De Broglie relation. However, they all show this only for the simply case of a free particle. A general solution to Schrödinger's equation in some potential would be an infinite linear combination of eigen-states for which we can't define a specific wavelength. While dwelling around the subject I thought maybe we can assign a De Broglie wavelength to the individual eigen-states which form the general solution, but that route is also problematic (for instance the eigen-states of the harmonic oscillator also don't have a well defined wavelength).
The root of the problem, I think, is that the De Broglie relation is a global relation (we can't talk about 'infinitesimal wavelength') while the Schrödinger equation is, by virtue of being a differential equation, a local relation. So, I'm struggling to see how a local relation can encode a global property like wavelength.
 A: The main issue here isn't superposition of eigenstates, it's the distinction between bound states and scattering states.  Particles in general don't have definite momenta, they exist as a probabilistic distribution that depends on the potential they live in.  Just like you noted, wavelengths don't make sense for a particle in a harmonic oscillator.  These are bound states, and aren't wavelike. 
 Scattering states, however, are wavelike, and it makes sense to define a de Broglie wavelength for particles in these states.  
Let's assume our potential energy $V$ dies off to $0$ at $\pm \infty$.  There are two types of solutions:
$E > 0$.  Scattering states.  Because the energy is positive, and at $\pm \infty$ the potential dies off, far away from the origin (where all the exciting stuff is happening) the Hamiltonian is that of a free particle, to which you can assign a de Broglie wavelength.
$E < 0$.  Bound states.  Because the energy is negative, and the potential is $0$ at $\pm \infty$, the particle cannot exist far away from the origin.  It is trapped (bound) by the potential and hence isn't a wave.
