Why are the allowed energies a continuum in the region $V_{\_} < E < V_{+}$? I'm studying quantum mechanics and I don't quite understand why there's an energy continuum in the region $V_{\_} < E < V_{+}$ in the following example:
 
It was explained that because of the continuity of the wave function and it's first derivative, in the case of $V_{min}<E<V_{\_}$ there are only certain discrete values for $E$ so that the wave function approaches $0$ as $x \to \pm \infty$.
Now, in this case ($V_{\_} < E < V_{+}$), how come there are a continuum of values for $E$ so that the wave function approaches $0$ as $x \to + \infty$? I understand that the entire x-axis to the left of $x_3$ is a classically allowed region so that the wave function exhibits an oscillatory behaviour, in contrast to the previous case where there was a classically forbidden region on both sides, but how does this affect the fact that I now have a continuum of choices rather than the discrete energy levels?
 A: In simple term when confined in $V_{min}<E<V_-$ the potential can be approximated as the potential of a harmonic oscillator and energy levels will be that for a Quantum harmonic Oscillator which is discrete.
However in the region $V_-<E<V_+$ the particle is essentially free as just beyond $x_1$ potential $V(x)$=constant and for free particle the energy is continuous.
A: When a particle is in a classically allowed region, the wavefunction will be oscillatory. When it is in a classically forbidden region, it will decay exponentially to 0. At the boundary of the 2 regions the wavefunction and its first derivative must be continuous. 
Now lets consider a step potential for simplicity, so the wavefunction (with an energy in the range $-V < E < 0$) in the allowed region is $\psi(x) = A \exp\left(\imath k x\right) + B \exp\left(-\imath k x\right)$. You can imagine 'sliding' the two waves past each other until you find a combination which fits the boundary conditions. In other words the conditions at the step are putting a constraint on the phases of waves in the allowed region. 
Now lets imagine a finite potential well, ie. two step potentials facing each other. For each step we have a boundary condition to satisfy, that is we have two independent constraints on the phase. Now we may be able to satisfy both of these, but only if we have an appropriate number of wavelengths between the two steps. This fixes the allowed wavelengths, which in turn fixes the allowed energies, giving a discrete spectrum. In other words the discrete bound spectrum is the result of the wavefunction having to go to 0 in both directions at the same time. 
