Schrödinger Operator (spectrum) I am considering the stationary Schrödinger equation,
$$
\Psi_{xx}+(\lambda-u)\Psi=0
$$
with the Schrödinger Operator
$$
L=-\frac{\partial^2}{\partial x^2}+u.
$$
Why for $\lambda>0$ the spectrum is continuous while for $\lambda<0$ the spectrum is discrete?
The condition on the potential $u(x)$ is that it decays sufficiently rapidly at infinity such that
$$
\int_{-\infty}^\infty\lvert u(x)\rvert\, dx<\infty\quad\textrm{ and }\quad\int_{-\infty}^\infty(1+\lvert x\rvert)\lvert u(x)\rvert\, dx<\infty.
$$
 A: physical answer:

*

*$\lambda$ is the total energy of a particle, and $u$ is a potential that vanishes at infinity, i.e. $u(x)\to0$ for $x\to\pm\infty$ ("sufficiently fast". but physicists dont always concern themselves with such details).

*The potential energy should be less then the total energy, because kinetic energy is always positive (at least for real particles).

*For $\lambda < 0$: The particle can only exist inside the potential well, which has a finite size. So there are only a finite number of states in there. So in fact only a finite number of negative values of $\lambda$ are possible.

*For $\lambda > 0$: The particle can exist anywhere on the real axis. For any $\lambda>0$ you can imagine a particle far away from the potential-well with kinetic energy exactly $\lambda$. When it comes close to the potential, its movement (or its wavefunction) will be disturbed with some combination of reflection and transmission. But any value of $\lambda>0$ is allowed. Due to quantum-tunneling it is actually impossible to completely trap a particle with positive total energy inside the potential well. (but for high enough barriers, such tunneling can be supressed a lot).

some semi-mathematical answer (really not rigorous, but might be enuogh for a course in theoretical physics):

*

*For $\lambda>0$:

*

*If $u=0$, the equation is simply $\psi''+\lambda\psi=0$. The two solutions of this are $e^{\pm ipx}$ with $p^2=\lambda$. (And of course linear combinations of these two solutions). These two solutions correspond to particle with momentum $p$, undisturbed by any potential. (there are two solutions because the particle can move either to the left or to the right).

*Now back to an arbitrary $u$, that still vanishes for large $|x|$. You can take "$\psi(x) \to e^{ipx}$ for $x\to-\infty$" as the initial condition for solving the differential equation $\psi''(x)=(u-\lambda)\psi(x)$. This will always give a valid solution. Note however that for $x\to\infty$, the solution will generally NOT be $e^{ipx}$ again, but some other linear combination $C_1e^{ipx}+C_2e^{-ipx}$. (And the coefficients are related to the transmission and reflection parts of the wave.). You might not be able to analytically solve this differential equaiton, but there is no problem in principle.



*For $\lambda < 0$ the situation changes somewhat.

*

*The solution for large $|x|$ is $e^{\pm\gamma x}$ with $\gamma^2=-\lambda$, so there is no oscillation but instead exponantial growth or exponential decay. BUT: the wavefunction has to be normalizable, so exponential growth is not allowed. Therefore the solution has to be
\begin{align}
\psi_x = \begin{cases}\text{const}\cdot e^{\gamma x} & \text{ for } x\to-\infty\\\text{const}\cdot e^{-\gamma x}&\text{ for }x\to\infty\end{cases}
\end{align}

*Now if you imagine again solving the differential equation $\psi''(x)=(u-\lambda)\psi(x)$ with initial condition $\psi(x\to-\infty)=e^{\gamma x}$, you will still always get a solution, which for large $x$ will be $\psi(x\to\infty)=C_1e^{\gamma x}+C_2e^{-\gamma x}$. But now this is only a valid solution to the schrödinger equation if $C_1=0$ exactly. This major coincidence will only happen for certain values of $\lambda$ (depending on the potential $u$, it can even happen that there is no value of $\lambda<0$ for which this works out).



EDIT: final note: The $\lambda<0$ states are called "bound states" (its a particle trapped inside the potential well), and the $\lambda>0$ states are called "scattering states" (because its a particle coming from infinity, scattering of the potential and flying to $\pm\infty$ again).
(well, this answer got longer than anticipated. hope some of this rambling helps^^)
