Bound States versus Scattering States

Shine ultraviolet light on a hydrogen atom and, photon by photon, you can walk its electron up a staircase of razor-sharp energy levels — $-13.6\ \mathrm{eV}$ , $-3.4\ \mathrm{eV}$ , $-1.5\ \mathrm{eV}$ , each rung a definite, repeatable number. Push past zero, though, and the staircase simply stops: the atom ionizes, and the freed electron may carry any energy at all. One atom, one equation — and two utterly different kinds of answer, a discrete ladder below a threshold and a seamless continuum above it. That split is not a peculiarity of hydrogen. It is the master dichotomy of quantum mechanics, and it is decided by a single comparison you can make before solving anything: the energy $E$ against the potential far away, $V(\pm\infty)$ . Below the asymptote, the wavefunction must die off exponentially in both directions, and threading that needle is possible only at special, isolated energies — bound states, trapped and normalizable. Above it, oscillating waves run to infinity unhindered, every energy is accepted, and the solutions describe beams that arrive, reflect, and transmit — scattering states. This article is the sorting hat for the whole chapter. The wells and oscillators ahead (Infinite Square Well, Finite Square Well, Quantum Harmonic Oscillator: Analytic Solution) live on the bound side; the steps and barriers (Potential Step and Reflection, Rectangular Barrier and Quantum Tunnelling) live on the scattering side. Learn to make the classification at a glance — energy line against landscape — and every potential you ever meet announces its spectrum before you touch the algebra.

Key takeaways

Compare $E$ with the asymptotic potentials: if $E<\min\{V(-\infty),V(+\infty)\}$ , the tails decay exponentially and the state can be a normalizable bound state; if $E>\max\{V(-\infty),V(+\infty)\}$ , the tails oscillate on both sides and the state is a scattering state.
Bound-state energies are discrete: demanding decay on both sides over-constrains the solution, so only isolated energies satisfy the boundary conditions. Scattering energies form a continuum because oscillatory tails are allowed.
Bound eigenfunctions are genuine Hilbert-space vectors, normalized to $\delta_{nm}$ ; scattering "eigenfunctions" are delta-normalized idealizations, made physical only when superposed into wave packets.
In one dimension bound levels are non-degenerate and their wavefunctions can be chosen real; each continuum energy is doubly degenerate (left-movers and right-movers).
If the two asymptotes differ, a bound state requires $E$ below the lower asymptote; in the energy window between the two asymptotes, the wave is oscillatory on one side and decays on the other, giving a continuum state with only one propagating channel.
A confining potential ( $V \to \infty$ both ways) has no continuum at all; a short-range well has finitely many levels plus a continuum; hydrogen's long-range Coulomb tail packs in infinitely many levels below its threshold.
Key idea: the spectrum of every 1D Hamiltonian is read off from one picture — energies below the asymptotic potential come discrete and trapped, energies above it come continuous and travelling.

The core insight

Compare the energy $E$ with the asymptotic potentials: $E<\min\{V(-\infty),V(+\infty)\}$ gives exponential decay and a normalizable bound state, while $E>\max\{V(-\infty),V(+\infty)\}$ gives oscillation to infinity and a continuous scattering state.

Where this lives today

Every semiconductor quantum-well device is an exercise in this article. Designers of quantum-well lasers and infrared photodetectors choose layer thicknesses so the well holds exactly the bound levels they want — the laser's colour is a spacing between two entries of the discrete spectrum — while solar cells and transistors care about how carriers in the continuum traverse the same structures. The bound-state count and the continuum transmission are computed from the same $V(x)$ , exactly the two families classified here. The 1929 von Neumann–Wigner idea has meanwhile become an active research area: photonic bound states in the continuum (BICs) can confine light at frequencies where ordinary modes would radiate away, producing resonant states with exceptionally high quality factors that are useful for low-threshold lasers, sensing, and related photonic devices. And at accelerator laboratories, the inverse problem is daily bread — reading bound-state structure out of scattering data, the programme that culminates in the S-matrix machinery of the scattering chapter.

The physical picture

Picture the potential as a landscape and the energy as a water level. Where the level sits below the surrounding terrain, water pools in the basin: that is a bound state, trapped, with only a thin film seeping into the banks — the exponential tails. Raise the level above the terrain at the horizon and the water runs off the edge of the map: that is a scattering state, a wave that arrives from infinity and leaves to infinity. The test is instantaneous — you never need to solve the equation to classify its solutions, only to look at where the energy line sits relative to the landscape far away.

Figure

The water-level test — one well, two energies, two families of solution

Why the trapped family is discrete is the deeper half of the picture. Inside the classically allowed region the wavefunction oscillates; in the forbidden region, where $E<V$ , the general solution is a combination of a decaying exponential and a growing exponential. Start on the left with the physically acceptable dying tail and integrate across: by the time you emerge on the right, the solution generically contains some of the growing branch, and the wavefunction explodes. Tuning the one knob you have — the energy — kills the explosion only at isolated values. Quantization is not a mysterious granularity of nature; it is the needle-threading of two exponential tails.

Figure

Threading the needle — detune the energy either way and the tail blows up

The continuum needs no such tuning. When $E$ exceeds the asymptotic potential, both far-field behaviours are innocent oscillations, $e^{+ikx}$ and $e^{-ikx}$ ; nothing blows up, nothing is forbidden, and every energy is accepted — with two independent solutions apiece, a wave coming from the left and a wave coming from the right. The price is normalizability: an oscillation that never dies has infinite $\int |\psi|^2 dx$ , so a single scattering "state" is an idealized beam, not a particle. The particle is recovered by superposing a band of them into a wave packet, exactly as Wave Packets and Their Construction builds a free electron out of plane waves.

The deciding comparison

Everything flows from the stationary equation of Time-Independent Schrödinger Equation in One Dimension read in the far field, where $V(x)$ has settled to its asymptotic value. There the equation forgets the details of the landscape and keeps only the sign of $V(\infty) - E$ . If the energy falls short of the asymptote, define the shortfall rate

\kappa = \frac{\sqrt{2m\,[V(\infty) - E]}}{\hbar}, \qquad \psi(x) \sim e^{-\kappa |x|} \quad (x \to \pm\infty) \qquad (1)

where: $\kappa$ — the inverse decay length of the forbidden-region tail; $m$ — the particle's mass. The deeper below the asymptote the state sits, the faster its tail dies: a level $1.0\ \mathrm{eV}$ under the rim has $1/\kappa \approx 0.2\ \mathrm{nm}$ — the wavefunction pokes a mere two ångströms into the wall. Only the decaying exponential is written in (1); its growing twin is the enemy the eigenvalue must slay. If instead the energy clears the asymptote, the same equation delivers travelling waves with the asymptotic wavenumber

k = \frac{\sqrt{2m\,[E - V(\infty)]}}{\hbar}, \qquad \psi(x) \sim A\,e^{ikx} + B\,e^{-ikx} \quad (x \to +\infty) \qquad (2)

where: $A, B$ — amplitudes of the right- and left-moving components, the raw material of the reflection and transmission story told in Transmission and Reflection Coefficients. Both behaviours stay bounded, so no energy is rejected. The two families answer to two different orthonormality conventions,

\langle \psi_n | \psi_m \rangle = \delta_{nm} \ \ (\text{bound}), \qquad \langle \psi_k | \psi_{k'} \rangle = \delta(k - k') \ \ (\text{scattering}) \qquad (3)

where: $\delta_{nm}$ — the Kronecker delta of a genuine discrete basis; $\delta(k - k')$ — the Dirac delta of a continuum of generalized states. Together the two families are complete: any physical state whatsoever expands as a sum over the ladder plus an integral over the continuum,

\Psi(x,t) = \sum_n c_n\,\psi_n(x)\,e^{-iE_n t/\hbar} + \int dk\;\phi(k)\,\psi_k(x)\,e^{-iE_k t/\hbar} \qquad (4)

where: $c_n$ — the amplitude on bound level $n$ ; $\phi(k)$ — the spectral density over the continuum. Drop either piece and the basis is incomplete — a state may require the continuum integral, the bound-state sum, or both. This is the master recipe of Time-Independent Schrödinger Equation in One Dimension upgraded to its full, two-family form.

Derivation AWhy trapped energies come out discrete

We want the chapter's central structural fact: energies below the asymptotic potential are allowed only at isolated points, never in a band. The strategy is to count freedoms against demands — the equation is second order, so its solution space at fixed energy is exactly two-dimensional, and we ask how many of those solutions can decay on both sides at once. No specific well is solved here; that is deliberate, because the argument works for every potential with $V(\pm\infty)$ finite, and the explicit transcendental bookkeeping for a square well belongs to Finite Square Well. Watch how the entire result hangs on a single mismatch: two tails to kill, one energy knob to turn.

Step 1. Fix $E < V(\pm\infty)$ . Far to the left the general solution is $a\,e^{+\kappa x} + b\,e^{-\kappa x}$ ; normalizability demands $b = 0$ there (the $e^{-\kappa x}$ piece explodes as $x \to -\infty$ ). Up to overall scale, that is one single acceptable solution launching from the left — the decaying tail. Why scale does not count: the equation is linear, so multiplying a solution by a constant is the same physical state; the launch direction in solution space is fixed, and only the energy remains adjustable.

Step 2. Integrate that unique solution rightward through the well — the equation marches it deterministically. Emerging on the right, it lands on some combination $c\,e^{-\kappa x} + d\,e^{+\kappa x}$ . For a generic energy $d \ne 0$ : the growing branch is present, and $\psi$ diverges.

Step 3. The matching demand $d(E) = 0$ is one equation in the one unknown $E$ . Its solutions are isolated points — where the function $d(E)$ happens to cross zero — not intervals. Each crossing is an eigenvalue; between crossings, no bound state exists at all. Why the crossings are guaranteed to be simple and finite in number for a short-range well: as $E$ sweeps upward the interior solution gains curvature and swings the emerging tail from growing-positive through zero to growing-negative — each swing is one new node and one new level, and the budget of swings is set by the well's depth and width. The counting formula for the square well is derived in Finite Square Well.

Step 4. Contrast the continuum: for $E > V(\pm\infty)$ both asymptotic behaviours oscillate and stay bounded, so no condition is imposed and both launch directions survive — every energy is allowed, twice over. Discreteness below the rim, double degeneracy above it, from one counting argument. Numerical shooting on a finite well (2m = ħ = 1, depth 1, half-width 1) confirms this vividly: detuning the energy by −0.10 or +0.12 from the eigenvalue E★ = −0.4538 sends the right tail off-scale (ψ ≈ +2.2 and −1.8 at the box edge), while at E★ the tail decays cleanly — the figure above is drawn from this computation.

The continuum, honestly

A plane wave is a useful lie: it has a perfectly definite momentum and a perfectly indefinite position, and its norm diverges. The continuum's mathematics keeps the lie consistent. Scattering solutions are not elements of the Hilbert space — they are generalized eigenfunctions, and their delta-function orthonormality (3) is exactly what makes the expansion (4) work like Fourier analysis. The derivation below pins this down and then shows why the physical particle never notices the scandal: any honest superposition of the idealized beams is a normalizable wave packet.

Go deeper · Derivation B · 1 minDelta normalization and why wave packets rescue the continuum

Derivation BDelta normalization and why wave packets rescue the continuum

The goal is to make sense of "normalizing" states that have no norm, and to show that the continuum, used correctly, produces genuine physical states. The strategy is direct computation: take the free asymptotic waves $\psi_k = e^{ikx}/\sqrt{2\pi}$ , compute their overlap, meet the Dirac delta, and then check that a band of them superposes into a square-integrable packet. The one subtlety worth respecting is that the delta is a statement about integrals, never a number — every line below is meant under an integral sign.

Step 1. Overlap two idealized beams: $\langle \psi_k | \psi_{k'} \rangle = \tfrac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(k'-k)x}\,dx$ . For $k' = k$ the integrand is $1$ and the integral diverges — the infinite norm, on display.

Step 2. Read the divergence correctly: the integral is the Fourier representation of the Dirac delta, $\int_{-\infty}^{\infty}e^{ikx}\,dx=2\pi\delta(k)$ , so $\langle\psi_k|\psi_{k'}\rangle=\delta(k-k')$ when the plane waves carry the $1/\sqrt{2\pi}$ normalization. This is what "delta-normalized" means. Why this is the right convention: with it, the continuum piece of the completeness relation reads ∫ dk ψ_k(x) ψ_k(x′) = δ(x − x′), and expansion coefficients come out of the same projection recipe as in the discrete case.*

Step 3. Build a physical state: $\Psi(x) = \int \phi(k)\,\psi_k(x)\,dk$ with a smooth, narrow profile $\phi(k)$ . Its norm is $\int |\Psi|^2 dx = \int |\phi(k)|^2 dk$ — finite whenever $\phi$ is square-integrable. The delta has done its job and disappeared from the answer.

Step 4. Note the bookkeeping shift: probabilities now attach to bands, $P(k_1 < k < k_2) = \int_{k_1}^{k_2} |\phi(k)|^2 dk$ , never to a single $k$ — the probability of exactly one continuum energy is zero, just as a dart never hits one mathematical point. As a check: the Fourier integral $\int e^{i(k'-k)x}\,dx$ evaluates to $2\pi\,\delta(k-k')$ exactly, and Parseval's theorem confirms $\int|\Psi|^2\,dx = \int|\phi|^2\,dk$ — the packet-normalization step — as an identity.

When the two horizons disagree

Nothing forces the landscape to level off at the same height on both sides — a conduction-band offset at a semiconductor junction is exactly a potential whose two asymptotes differ. The classification then sharpens into three windows, and it is worth deriving them once, because the middle window behaves like neither pure family.

Go deeper · Derivation C · 2 minUnequal asymptotes and the three energy windows

Derivation CUnequal asymptotes and the three energy windows

Take $V(-\infty) = V_L$ and $V(+\infty) = V_R$ with $V_L < V_R$ , and classify every energy by the same far-field logic as before. The payoff is a rule of thumb that settles any exam question about asymmetric potentials in one line, plus a degeneracy count that the symmetric case hides. The only tool needed is equations (1) and (2) applied to each side independently — the two horizons do not talk to each other.

Step 1. $E < V_L$ : the energy is under both rims. Decay is required on both sides, the needle-threading argument applies unchanged, and the states are bound with discrete energies. The window for bound states is $E$ below the lower asymptote, $E < \min(V_L, V_R)$ .

Step 2. $V_L < E < V_R$ : over the left rim, under the right one. The right side must carry the decaying tail $e^{-\kappa_R x}$ — one condition — but the left side is free oscillation $e^{\pm i k_L x}$ with no condition at all. One demand, two solution directions: exactly one acceptable solution survives at every energy in the window. The spectrum here is continuous but singly degenerate — a wave that comes in from the left and is entirely turned back, its exponential toe in the high side the only trace of the barrier's interior. Why total reflection is the physics of this window: there is no outgoing channel on the right — the state decays there — so whatever probability current arrives from the left must return. The quantitative version, |r| = 1 with a phase shift, is worked out in Potential Step and Reflection.

Step 3. $E > V_R$ : over both rims. No conditions anywhere, two independent solutions — left-incident and right-incident beams — and the continuum is doubly degenerate, the generic scattering situation of Transmission and Reflection Coefficients.

Step 4. Summary rule: bound below both, one-way between, two-way above both. The degeneracy ladder $0 \to 1 \to 2$ independent bounded solutions per energy is the cleanest fingerprint of the three windows. *As a concrete check: for an electron with $V_L=0$ and $V_R=2.0\,\mathrm{eV}$ , at $E=1.0\,\mathrm{eV}$ the state is in the one-way window, with oscillatory wavelength $\lambda_L\approx1.23\,\mathrm{nm}$ on the left and a decaying tail on the right; at $E=3.0\,\mathrm{eV}$ both sides oscillate, with $\lambda_L\approx0.71\,\mathrm{nm}$ and $\lambda_R\approx1.23\,\mathrm{nm}$ .

Limiting cases

Case	Result	Meaning
confining potential, $V \to \infty$ both sides	purely discrete spectrum, no continuum	there is no "above the rim": every state is trapped — Infinite Square Well, Quantum Harmonic Oscillator: Analytic Solution
short-range attractive well in 1D	finitely many bound levels + continuum; at least one level always	the guarantee and the counting formula are derived in Finite Square Well; the sharpest case — exactly one level — is Delta-Function Potential Well and Bound State
shallow well in 3D	no bound state until $V_0 a^2 \ge \pi^2\hbar^2/8m$ (spherical well)	binding is a 1D privilege; three dimensions demand a minimum depth — why the deuteron is barely bound
Coulomb tail, $V \sim -1/x$	infinitely many bound levels crowding the threshold	a long-range tail never quite lets go: hydrogen's $E_n \propto -1/n^2$ ladder accumulates at $E = 0$
$E$ exactly at the asymptote	marginal case, decided by the potential's tail	generic short-range wells: not normalizable, not bound; the borderline "half-bound" state matters in scattering resonances
free particle, $V = 0$	pure continuum, doubly degenerate	no basin anywhere — Free Particle and Plane Waves

Common trapA particle sitting inside the well right now is in a bound state

Location is not the criterion — asymptotics are. A wave packet built from continuum states can sit squarely on top of the well at $t = 0$ , yet it is a scattering state through and through: wait, and it drains away to infinity, because every component of it oscillates at the horizons. Conversely a bound state's exponential tail is outside the well, yet the state is bound forever. Classify by the energy against $V(\pm\infty)$ — never by where $|\psi|^2$ happens to peak at one instant. The same trap in reverse: a "particle temporarily trapped" above a well (a resonance) is not a bound state either, however long it lingers — it belongs to the continuum, as Resonant Transmission and Tunnelling Resonances makes precise.

Worked examples

1 · How far does a trapped electron leak? (confidence builder)

An electron occupies a bound level $1.0\ \mathrm{eV}$ below the asymptotic potential of a semiconductor quantum well. Find the decay constant of its forbidden-region tail, the distance over which the tail's probability falls by $e^{-2}$ , and comment on what this means for a neighbouring well placed $2\ \mathrm{nm}$ away.

Step 1. The tail is governed by equation (1) with shortfall $V(\infty) - E = 1.0\ \mathrm{eV}$ : $\kappa = \dfrac{\sqrt{2 m_e \times 1.0 \times 1.602\times10^{-19}\ \mathrm{J}}}{1.055\times10^{-34}\ \mathrm{J\,s}} = \boxed{5.12\ \mathrm{nm^{-1}}}$ .

Step 2. The decay length is $1/\kappa = \boxed{0.195\ \mathrm{nm}}$ — about two ångströms, one atomic layer. Beyond one decay length the probability density $|\psi|^2 \propto e^{-2\kappa x}$ has fallen to $e^{-2} \approx \boxed{13.5\%}$ of its value at the wall.

Step 3. A second well $2\ \mathrm{nm}$ away sits at $2/0.195 \approx 10$ decay lengths: the wavefunction amplitude there is attenuated by $e^{-\kappa d} = e^{-10} \approx 4.54\times10^{-5}$ , while the probability density is reduced by $e^{-2\kappa d} = e^{-20} \approx 2.06\times10^{-9}$ . Tiny — but not zero, and that overlap is precisely what allows levels in double wells to split and underlies tunnelling between wells.

2 · Sorting energies at a junction (exam level)

A conduction electron meets a heterojunction whose potential levels off at $V_L = 0$ far to the left and $V_R = 2.0\ \mathrm{eV}$ far to the right. Classify the stationary states at $E = -0.5\ \mathrm{eV}$ , $E = 1.0\ \mathrm{eV}$ , and $E = 3.0\ \mathrm{eV}$ , giving decay lengths or wavelengths for each side.

Step 1. Apply the three-window rule: bound needs $E < \min(V_L, V_R) = 0$ . At $E = -0.5\ \mathrm{eV}$ the energy is under both rims — bound window (whether a level actually sits there depends on the junction's interior profile; the window is where levels are allowed).

Step 2. At $E = 1.0\ \mathrm{eV}$ : above $V_L$ , below $V_R$ — the one-way window. On the left the electron oscillates with $k_L = \sqrt{2m_e E}/\hbar = 5.12\ \mathrm{nm^{-1}}$ , wavelength $\lambda_L = 2\pi/k_L = \boxed{1.23\ \mathrm{nm}}$ . On the right it can only decay, with shortfall $2.0 - 1.0 = 1.0\ \mathrm{eV}$ , tail $1/\kappa_R = \boxed{0.195\ \mathrm{nm}}$ . One independent solution: the incident wave is totally reflected, with a sub-nanometre evanescent toe past the junction.

Step 3. At $E = 3.0\ \mathrm{eV}$ : above both rims — two-way continuum. Left: $\lambda_L = 2\pi\hbar/\sqrt{2m_e \times 3.0\ \mathrm{eV}} = \boxed{0.71\ \mathrm{nm}}$ . Right: the kinetic energy is only $3.0 - 2.0 = 1.0\ \mathrm{eV}$ , so $\lambda_R = \boxed{1.23\ \mathrm{nm}}$ — the wave stretches as it climbs onto the plateau. Two independent solutions exist (left- and right-incident); how much is reflected at the step is the business of Potential Step and Reflection.

Step 4. Notice what never entered: the junction's interior shape. The classification — and the number of open scattering channels, $0, 1, 2$ — is determined by the asymptotic potentials.

Origins · 1 minJohn von Neumann and how this was discovered

John von Neumann

1903–1957 · Berlin, Princeton

Quantum mechanics used the bound/scattering split for six years before anyone proved it was mathematically sound. Von Neumann did — his 1932 Mathematische Grundlagen der Quantenmechanik built the spectral theory of unbounded self-adjoint operators and showed that a Hamiltonian's spectrum genuinely decomposes into a point part (the normalizable bound states) and a continuous part (the delta-normalized continuum), with completeness holding for the two together. The "generalized eigenfunctions" physicists had been using on faith became theorems. Characteristically, he also built the counterexample: with Wigner in 1929 he engineered a potential holding a normalizable state embedded inside the continuum — a bound state above the rim — showing the energy criterion is the rule but not an iron law. Eight decades later such "bound states in the continuum" were built for real in photonics, where they now sharpen lasers and sensors.

History

1926

— Schrödinger solves hydrogen: one equation yields both the discrete ladder below zero and the ionization continuum above it.

1928

— Gamow's alpha-decay theory shows a "bound" nuclear state can be a long-lived resonance leaking into the continuum.

1929

— von Neumann and Wigner construct a bound state embedded in the continuum — the exception that proves the classification.

1932

— von Neumann's Grundlagen makes the point-vs-continuous spectral split rigorous for self-adjoint Hamiltonians.

2013

— photonic bound states in the continuum are observed experimentally, turning the 1929 curiosity into a device principle.

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.1–2.5 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Cohen-Tannoudji, Quantum Mechanics, Vol. I, Ch. I & Complement H_I · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.5 · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6 The finite-well matching conditions are $\kappa = k\tan(ka)$ (even) and $\kappa = -k\cot(ka)$ (odd); the Dirac normalization integral is $\int e^{i(k'-k)x}dx = 2\pi\delta(k-k')$ ; no empirical constants are used beyond standard $\hbar$ and $m$ . Key numerical results in this article: the needle-threading eigenvalue $E^\star = -0.4538$ in well units (2m = ħ = 1, depth 1, half-width 1); decay constant $\kappa = 5.12\ \mathrm{nm^{-1}}$ and tail length $1/\kappa = 0.195\ \mathrm{nm}$ for an electron 1.0 eV below the rim; asymptotic wavelengths $\lambda = 1.23\ \mathrm{nm}$ at 1.0 eV kinetic energy and $\lambda = 0.71\ \mathrm{nm}$ at 3.0 eV kinetic energy. The finite-well transcendental derivation is treated in full in Finite Square Well.

Prerequisites. Time-independent Schrödinger equation in one dimension · free particle and plane waves · wave packets. Leads to. Infinite square well · finite square well · delta-function potential well · the harmonic oscillator.

If a wave packet is sitting right on top of the well, why is it not a bound state?

Because classification is about energy content, not position. Expand the packet in the two-family basis (4): if its amplitudes live in the continuum, every component oscillates at infinity, and the packet inevitably disperses off the well — its stay is temporary. A state is bound only if its expansion lives on the discrete levels. A packet can also be a mixture — partly trapped, partly escaping — which is exactly what happens when a well is suddenly switched on around a free particle.

What happens exactly at threshold, E = V(∞)?

The marginal case, and the potential's tail decides it. For a generic short-range well the threshold solution approaches a constant — bounded but not normalizable, so not a bound state. But it is not innocent either: a well tuned so that a level sits exactly at threshold ("zero-energy resonance") scatters anomalously strongly at low energy. Long-range tails rewrite the rules — the Coulomb potential's $-1/x$ tail supports infinitely many bound levels accumulating at threshold from below.

Why is the continuum doubly degenerate when bound states never are?

Count conditions. A bound state must kill growing exponentials on two sides; that pins the solution uniquely (and the Wronskian argument of Non-Degeneracy of One-Dimensional Bound States makes it a theorem). A scattering energy above both asymptotes imposes no condition, so both members of the two-dimensional solution space survive — physically, a beam may arrive from the left or from the right, and those are independent experiments. In the one-way window between unequal asymptotes exactly one condition survives, so the degeneracy there is one: the counting 0, 1, 2 tracks the number of open horizons.

If plane waves are not in the Hilbert space, why are we allowed to calculate with them?

Because they are scaffolding, not states. Delta-normalized continuum functions are a basis for integrals: every legal manipulation — expansions, matrix elements, transition rates — leaves them under an integral over $k$ , where the Dirac delta is perfectly well defined. Trouble arises only if you treat one plane wave as a literal particle (infinite norm, undefined position statistics). The wave-packet construction is the licence: smooth amplitude in, normalizable state out, and every plane-wave calculation reappears as the narrow-band limit of a packet calculation.

Does every potential well have at least one bound state?

In one dimension, yes — any purely attractive well, however feeble, binds at least one state (the counting and the guarantee are derived via the circle construction in Finite Square Well). In three dimensions, no: a spherical well must exceed the threshold depth $V_0 a^2 = \pi^2\hbar^2/8m$ before its first level appears, which is why the deuteron — bound by a whisker — needs the strong force to clear a real threshold. The dimensional difference traces to the boundary condition at the origin that 3D imposes and 1D does not.

The two asymptotes of my potential are different. Which one decides whether a state is bound?

The lower one. Bound means decay on both sides, so the energy must sit below both asymptotes — that is, below $\min[V(-\infty), V(+\infty)]$ . Energies between the two asymptotes give the one-way window: continuous spectrum, single degeneracy, total reflection off the higher side with an evanescent toe. It is a common exam trap to test the energy against the asymptote on the side where the particle "lives" — always use the global minimum of the two horizons.

Hydrogen has infinitely many bound levels, but the finite square well has only a handful. What makes the difference?

The reach of the tail. A short-range well is felt only over a finite region: it can bend the wavefunction a limited number of times, so it supports finitely many levels. The Coulomb potential falls off so slowly ( $-1/x$ ) that arbitrarily far out it is still attractive enough to trap arbitrarily low-shortfall states — the $E_n \propto -1/n^2$ ladder crowds ever closer to threshold without end. Rule of thumb: potentials dying faster than $1/x^2$ hold finitely many levels; slower tails can hold infinitely many.

Are tunnelling resonances bound states hiding in the continuum?

Almost — and the "almost" is the physics. A resonance is a continuum state at an energy where the potential's interior traps probability for a long but finite time; it decays, so it is not a true bound state, and its energy is best described as complex, $E_0 - i\Gamma/2$ , with lifetime $\hbar/\Gamma$ . True bound states in the continuum exist only in engineered exceptions (von Neumann–Wigner potentials, photonic BICs). The sharp transmission spikes that resonances cause are taken up in Resonant Transmission and Tunnelling Resonances.