# Bounded and Unbounded (Scattering) States in Quantum Mechanics

I understand that bounded states in quantum mechanics imply that the total energy of the state, $E$, is less than the potential $V_0$ at + or - spatial infinity. Similarly, the scattering state implies that $E > V_0$.

But I do not understand why for bounded states, the complete set of wave-function solutions to the time-dependent Schrodinger equation is constructed from a discrete linear combination of solutions, whereas for scattering states, the complete set, or general solution, is instead written as an integral over the continuous value of $k$ (or more appropriately, as a Fourier-type integral).

Could anyone clarify why this difference is true for bound and scattering (unbound) states?

• This may help physics.stackexchange.com/q/68060 – user24999 Aug 6 '13 at 19:35
• This seems to cover it quite comprehensively. I'm not entirely convinced about the logic in the answer for the other question! Does anyone know what the original paper for this result was? – Edward Hughes Aug 6 '13 at 21:52
• Hi @Edward Hughes, from your link, could Theorem 8.20 be stronger? I mean the hydrogen atom satisfies the condition (8.10), but it has infinity number of bound state, not finite... – user26143 Aug 7 '13 at 0:37
• @user86111, just one remark. Bound state does not imply $E<V_0$ at $\pm \infty$. There is bound state embedded in continuum, ref Phys. Rev. A 11, 446–454 (1975) Figure 3 – user26143 Aug 7 '13 at 0:39
• @user26143 I don't quite understand your point. The theorem says that there are finitely many bound states with energy lower than any $a'$ where $a' < V_0$. Taking the limit as $a' \to V_0$ in the case of the hydrogen atom reproduces the infinity of energy levels we observe. – Edward Hughes Aug 7 '13 at 9:45

Mathematically, it is now well understood that eigenfunctions of a Schrödinger operator decay exponentially in space at infinity (under suitable conditions on the potential $V$). This is due to Agmon, see e.g. the following review and references thereof contained. In this sense one can associate to the discrete spectrum an orthonormal set of eigenfunctions localized in space, i.e. bound states. I do not know if this is true also for eigenvalues embedded in the continuous spectrum, but it may be possible.

This very deep mathematical result clarifies the concept of bound states, in relation with discrete spectrum, and to my knowledge is the only rigorous result: unbound states are not so clear mathematically, since it is not possible to define eigenfunctions for the continuous spectrum (what physicists call "eigenfunctions" in that case does not belong to the Hilbert space). Some information may be found in the context of "quasi-modes", but I do not have a good knowledge of the subject.

The Simon paper cited in another answer investigates the non-existence of eigenvalues in the region $E>V_0$, that is usually but not always true (so we may have bounded states for $E>V_0$, in some particular situation). That may be another indication that a complete characterization of existence of bound states with respect to the relation between $E$ and $V_0$ is not possible.

Though I've already given the link to my answer for a related question, I'd like to put it on perspective for this particular one.

You can guarantee that the energy spectrum will be discrete when

$$E < \lim_{|x|\rightarrow \infty}V(x)\hspace{0.75in}(\dagger)$$

and

$$\lim_{|x|\rightarrow \infty} \psi(x) = 0\hspace{0.75in}(\ddagger)$$

as mentioned in the question I linked to. Also, as said here in the comments, there could be bound states even if $(\dagger)$ is not satisfied or unbounded states when $(\dagger)$ is satisfied but $(\ddagger)$ is not. It is a one side implication stated the other way around as you did in the first sentence, and taking in count $(\ddagger)$. I didn't made emphasis on $(\ddagger)$ on the other question, but it is pretty important as well, which if not satisfied the stationary Schrödinger equation could not be reduced to a regular S-L problem because the hamiltonian would in general not be symmetric in all space and thus not self-adjoint.

To clarify this a little, suppose you have a potential well system for which $(\dagger)$ is satisfied and that you pick a particular solution $\psi$ that decreases when $x\to\infty$. Suppose for this matter that $E$ intersects the potential at $x_1$ and $x_2$ with $x_1<x_2$ and that $\psi$ is positive at $x_2$. Then $\psi$ should be appropriately convex at $x>x_2$ in order for $(\ddagger)$ to be satisfied. For this to happen, by the equation of the form $\psi^{\prime\prime}=q^2\psi$, we know that the curvature depends on energy via $q^2$, and thus there may be only a certain critical value of energy for this particular $\psi$ such that the curvature is the appropriate one given it must satisfy $(\dagger)$. This is rather an intuitive way to see it, in general from here, one could prove the discreteness of the energy spectrum taking the approach I mention in the other question.

So rather than to look out only for bounded or unbounded states, you can look for the conditions above to guarantee the discreteness of a system.

• Do you have any references for your results? I'd like to have a look at the maths if so! – Edward Hughes Aug 7 '13 at 9:52
• A scattering state wavefunction does not satisfy condition (‡), is the Hamiltonian still self-adjoint? – user26143 Aug 7 '13 at 10:25
• @EdwardHughes You can read on Sturm-Liouville theory on Mathematical Methods for Physicists by Arfken and Weber, for starters, or if you want to dig deeper, I'd say Sturm-Liouville Theory by Zettl. For the particular QM case my main reference (in spanish, sorry) is Introducción a la Mecánica Cuántica by Luis de la Peña. On the link you can find practically what I mention at pages 69-72. – user24999 Aug 7 '13 at 14:52
• @user26143 If $(\ddagger)$ is not satisfied, $\hat{H}$ won't be symmetric when acting on $\psi$, nor $\psi$ can be reduced to one that make it symmetric. Now $(\ddagger)$ can be somewhat loosely stated just by asking that $\psi$ remains finite, the thing is one can always reduce this case to the original $(\ddagger)$ one. In QM all wavefunctions satisfy this in order to be 'physically acceptable' (the hamiltonian is always self-adjoint or more specifically, hermitian), even scattering states, so the "scattering state" you mention is possible but not physically possible. – user24999 Aug 7 '13 at 15:01
• Thank you very much! Perhaps I will read about scattering state issues into more detail when I have more time – user26143 Aug 7 '13 at 15:07

I'll give an answer which hopefully clarifies all the naming conventions. First recall the Schroedinger equation takes the form

$$H\psi(x) = E\psi(x)$$

where $H$ is the Schroedinger operator

$$H = -\frac{1}{2m}\nabla^2 + V(x)$$

also known as the Hamiltonian.

In quantum mechanics a particle is represented by a wavefunction $\psi(x)$ taking values in the complex numbers.

We demand that all physical particles are in bound states where by definition

1. $\psi(x)\to 0$ as $|x|\to \infty$
2. $\int_{\mathbb{R}^3}\psi^*\psi \ d^3 x = I(\psi) < \infty$

To give a probabilistic interpretation to the theory we impose the normalisation $I(\psi) = 1$.

The energy levels of a system are given by the eigenvalues $E$ of the Schroedinger operator. These may be discrete or continuous.

$$* * *$$

*Lemma 1*$\ \$The energy levels are discrete iff the corresponding eigenfunctions are bound states.

Proof

The forward implication is given in the proof of Theorem 8.20 here. There may be an easier argument, but I can't think of one.

The converse is easier to see. Let $\psi_E(x)$ and $\psi_{E'}(x)$ be eigenfunctions for $E,E'$ in the continuous spectrum. Now recalling a standard argument (cf Landau and Lifschitz "Quantum Mechanics" $\S 3$) we find that the orthonormality relation is

$$\int \psi_E^* \psi_{E'} d^3 x = \delta(E - E')$$

Now setting $E = E'$ gives our result as required.

$$* * *$$

Now we suppose that $V(x) \to 0$ as $|x| \to \infty$. This is a physically reasonable assumption encompassing many potentials.

*Lemma 2*$\ \$The spectrum of energy eigenvalues $E$ with $E < 0$ is discrete.

*Proof*$\ \$This is exactly the statement of Theorem 8.20 again.

Note that therefore the negative energy eigenstates are bound states. This is consistent with the observations we make for the hydrogen atom.

$$***$$

*Lemma 3*$\ \$ There exist isolated (discrete) energy eigenvalues for $E > 0$.

Proof

This (surprising) result was mentioned in the comments above. A good review of the literature is here.

• I cannot see why $|\psi(x)| \to 0$ must hold if $|x|\to +\infty$. Bound states are simply states defined by normalized vectors of the Hilbert space which are eigenvectors of the Hamiltonian operator. – Valter Moretti Apr 5 '14 at 14:25
• @ValterMoretti Would it help if we said that for any bound state $B$ and any $\epsilon >0$ there is a function that goes to zero as $x$ goes to infinity that is within $\epsilon$ of $B$ in the norm generated by the inner product? – Timaeus Jun 16 '15 at 17:42

A bound (normalizable) state can have eigen-energy inside the continuum band.

It is the so-called bound state in the continuum (BIC).

See http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.109.116405 and references therein.