# Quantum mechanics: bounded states and scattering states

So in quantum physics, we can solve solutions of the schroedinger equation and have that the eigenvalues are either positive or negative, which means it is either a scattering state or a bound state.

The maths is all fine, but how is the physics different? Since both of them you can find a set of eigenfunctions that satisfy the equation

## 2 Answers

1. Bound states are characterized by the energy being less than the potential at $$-\infty$$ and $$+\infty$$. We typically say $$E>0$$ is a scattering state and $$E<0$$ is a bound state with the implicit assumption that the potential energy approaches zero at infinity. One canonical model which indicates this is the quantum harmonic oscillator: we generally write the energy eigenvalues to be positive, though they represent bound states; the potential approaches infinity as you move further from the origin in this system.

2. As in classical physics, a particle modeled by quantum mechanics will tend towards a lower energy; the negative energy of a bounded state implies that the particle is likely to remain localized within the relevant region: states at $$\pm\infty$$ are not favorable since by definition of a bounded state, the potentials there are greater than $$E$$. In contrast, when $$E>0$$, there is a the particle will 'scatter', or escape to infinity, since that is energetically favorable.

Quantization, i.e. the appearance of discrete eigenvalues, is a feature of bound states. It is this discreetness of the energy eigenvalues that was difficult to explain using classical physics. Setting up the problem of finding energy levels as an eigenvalue problem for an equation that did not exist in classical physics was the giant step.

Bound states are also usually normalizable, whereas scattering states are not. This leads to non-negligible complications where sums of normalized states are replaced by integrals of non-normalizable states.