Bound states and scattering states - Time-dependent Schrödinger equation

Question

If we have a quantum system described by the time-independent Schrödinger equation (TISE):

\begin{equation} -\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}=E \psi \end{equation}

We have two possible types of solutions:

• Bound states: this means localization. Represent discrete values of energy called energy levels. Are imaginary exponentials, that is, oscillating functions.

• Scattering states: this means movement. Represent particle beams. Are real exponentials, that is, decreasing or ascendant functions.

But, if we have a quantum system described by the time-dependent Schrödinger equation (TDSE):

\begin{equation} i \hbar \frac{\partial}{\partial t} \Psi(x, t)=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x, t)\right] \Psi(x, t) \end{equation}

For the free particle, the solutions are plane waves (or superpositions of plane waves, that is, wavepackets).

1. What can we say about its solutions in the terms of bound and scattering states?
2. Using Fourier analysis, can we ensure that the solutions of the TDSE with any potential $V(x,t)$ will be plane waves or superposition of plane waves?

Answer 1

In the time independent theory, the distinction is really the mathematical question of whether the Hamiltonian's eigenvalue spectrum is continuous (scattering) or discrete (bound), which in turn affects the kind of normalisation that can be applied to those eigenstates. In general, a hermitian operator's spectrum will have a combination of both (including the possibility of a discrete state in the continuum), a classic example of which being the hydrogen atom.

Note that if you restrict to a time independent Hamiltonian, you can just make the ansatz

$$\Psi(x,t) = \psi(x)T(t)$$

which transforms the TDSE into

$$\frac{\hat H \psi(x)}{\psi(x)} = E = i\hbar \frac{\partial_t T(t)}{T(t)}$$

Solving this gives you the famous result that energy eigenstates evolve in time by accumulating an energy-dependent phase, $T(t) = \exp(-iEt/\hbar)$, where $E$ came from the TISE.

The fully time dependent theory is more complicated. If the Hamiltonian depends on time, as yours does, then energy is not (locally) conserved and the Schrödinger equation is not separable, necessitating a different solution method. It's more difficult to categorise solutions to that equation without some concreteness that would allow the use of an approximation scheme: if V(t) is slowly varying, the adiabatic approximation can be used, if V(t) is weak relative to other terms, time dependent perturbation theory can be used.

Conceptually, it's easy to see how adiabatic (i.e. very slow) time evolution can render a bound state free- take, for example, $V(x,t) = \frac{-H(-t) t/\tau }{2}m\omega^2x^2$. We let $\tau \gg 1/\omega$ to say that V is changed very slowly.

If $t<0$, this is a harmonic oscillator with localised, bound energy eigenstates and a well defined ground state, while at $t>0$ the is a free particle.

This can be viewed as slowly turning off a trapping potential, where the wavefunction of the trapped particle expands to eventually occupy the entire real line. Similarly the energy spectrum when $t<0$ is given by $E_n = \hbar\frac{|t|}{\tau}\omega(n+\frac{1}{2}), n\in \mathbb{Z}$ As $|t| \to 0$, the spectrum 'condenses' into a continuum block corresponding to a free particle.