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Please explain it in the context of this task: we have a potential barrier that looks like $\prod$, with $E<U$. There are 3 regions:

1) no field
2) barrier
3) no field

Solution could be found as

1) $\psi_{1} = A\cdot e^{-ikx}+B\cdot e^{ikx} $
2) $\psi_{2} = C\cdot e^{-ik_{0}x}+D\cdot e^{ik_{0}x}$
3) $\psi_{3} = 0 +F\cdot e^{ikx}$

(cause we haven't got a reflected wave here)

where $k_{0}^{2}=2m_{0}(U_{0}-E)/h^{2}$ and $k^{2}=2m_{0}E/h^{2}$

So why in this case spectrum is continuous?

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Sorry, but i haven't got enough reputation to paste images. You can view equation symbols here – divideByZero Apr 4 '13 at 21:27

The potential that you are describing does not have any bound states--only scattering states. Scattering states are not normalizable and do not have discrete spectra.

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Well, you are not yet done! You should also fix $A,B,C,D$ by imposing that the wavefunction is $C^1$ crossing the discontinuities of the potential. Following that way you obtain that only an overall arbitrary constant remains. You have thus a function like this $k\psi_E$, where $k\neq 0$ is any complex number, can be fixed arbitrarily (without imposing normalization requirements). The eigenvalue equation to be solved by this prototype of solution $\psi_E$ is: $$-\frac{\hbar^2}{2m}\frac{d^2 \psi_E(x)}{dx^2} + \Pi(x) \psi_E(x) = E \psi_E(x)\:.$$ This equation holds outside the singularities of $\Pi(x)$, what happens at each singularity is already taken into account by the relations among the constants $A,B,C,D$.

You finally see that, for every $0<E<U$ ($U$ is the constant top of $\Pi(x)$ if I understand well), your function $\psi_E$ solves the eigenvalue equation above, in other words $E$ varies continuously in [0,U]. The spectrum is thus continuous.

ADDENDUM. I forgot to say that, in the considered case, it does not make sense to require that the wavefunction vanishes fast as $|x|\to \infty$ in order to have finite normalization. This is due to the oscillatory behaviour of $\psi_E$ far away from the barrier, already fixed in the hypotheses. So the (continuous) allowed values of $E$ are completely fixed by the eigenvalue equation (and the regularity conditions).

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