Non-Degeneracy of One-Dimensional Bound States

Twice in this chapter a proof has paused to borrow a fact on credit: that a one-dimensional bound state is the only state at its energy. The parity article needed it to force every eigenstate of a symmetric well to be even or odd; the node theorem needed it so that "the state with $n$ nodes" picks out one function and not a family. This article pays the debt. The claim — every discrete energy level of a 1D Hamiltonian is non-degenerate — is proved here in full, with an argument short enough to memorize and sharp enough to show you exactly when it fails. The tool is a quantity you may know from differential equations: the Wronskian, $W = \psi_1\psi_2' - \psi_2\psi_1'$ , a measure of whether two functions are genuinely independent. For two solutions of the time-independent Schrödinger equation at the same energy, $W$ turns out to be constant in $x$ — a conserved "flux of independence." A bound state's tails die at infinity and drag $W$ to zero there; a constant pinned to zero anywhere is zero everywhere; and zero Wronskian means the two solutions were the same state wearing different normalizations. Three lines, and uniqueness is a theorem. The fine print is where the physics lives. Scattering states dodge the argument (nothing decays, and left-movers and right-movers really are two states at one energy). A particle on a ring dodges it. An infinite wall that chops the line in two dodges it. And in three dimensions, degeneracy comes roaring back as the fingerprint of symmetry — hydrogen's rich level structure is everything this theorem forbids on the line. Knowing precisely why the line is different is what this article is for.

Key takeaways

Theorem: in one dimension, every normalizable bound-state energy level is non-degenerate — one energy, one state, up to an overall constant.
Tool: the Wronskian $W = \psi_1\psi_2' - \psi_2\psi_1'$ of two same-energy solutions has $W' = 0$ ; bound-state decay at infinity forces $W \equiv 0$ ; vanishing $W$ means proportionality.
Corollary 1: bound eigenfunctions can always be chosen real (so they carry zero probability current).
Corollary 2: in a symmetric well, each bound state has definite parity — the promissory note from the parity article, now redeemed.
Where it fails, it fails honestly: scattering states ( $e^{\pm ikx}$ , $W = -2ik \neq 0$ ), rings ( $e^{\pm im\phi}$ ), disconnected domains (infinite interior wall), and higher dimensions (symmetry degeneracy).
Key idea: for a second-order linear ODE the Wronskian of two solutions at equal energy is constant; normalizability forces it to zero, so the two solutions are proportional — the bound level is non-degenerate.

Where this lives today

The theorem quietly runs inside every 1D quantum calculation you will ever launch. Shooting-method eigensolvers are the proof turned into an algorithm: they integrate a decaying solution in from each side and scan energy until the log-derivatives match at a joint — the match is possible precisely because one function per side exists, and the located state is trustworthy precisely because it is unique. Semiconductor quantum-well designers rely on the simple (non-degenerate) subband ladder when engineering intersubband lasers; the moment layouts become effectively multi-dimensional or periodic (superlattices), degeneracy and minibands return on cue. Double-well tunnelling doublets — the theorem's "almost violated" case — are working hardware: ammonia's 24 GHz inversion line powered the first maser, and superconducting flux qubits store quantum information in exactly such a near-degenerate pair, with the never-quite-zero splitting as the qubit's protected gap. And bound states in the continuum — the von Neumann–Wigner curiosity that skirts the theorem's edges — are now engineered photonic devices sharpening lasers and sensors.

The physical picture

Why should a bound level hold exactly one state? Count the freedom and watch it get eaten. The time-independent Schrödinger equation is a second-order ODE, so at any fixed energy its full solution space is exactly two-dimensional — pick $\psi$ and $\psi'$ at one point and the rest is determined. But a bound state must also behave at infinity, and there the equation is merciless: in a classically forbidden tail the two independent behaviours are a dying exponential and a growing one, and normalizability permits only the dying one. The left tail eats one dimension of freedom; the right tail eats one too. Two dimensions minus two conditions leaves, generically, nothing at all — which is why bound energies are discrete in the first place — and at the special energies where a survivor exists, it is one function, unique up to overall scale.

Figure

Two tails, one survivor — normalizability at both ends leaves a single state per bound energy

The Wronskian makes this bookkeeping into an honest proof. Think of $W = \psi_1\psi_2' - \psi_2\psi_1'$ as a meter that reads "how independent are these two solutions right here?" For the Schrödinger equation — crucially, an equation with no $\psi'$ term — the meter's reading never changes with $x$ : independence is conserved along the line. Now suppose two bound states shared one energy. Follow the meter out to $x \to +\infty$ : both functions and both slopes die, so the reading is zero out there. But the reading is constant — so it was zero everywhere, including in the middle of the well, and zero independence means the "two" states were one. The whole theorem is a conservation law plus a boundary condition. The same meter, pointed at scattering states, explains why they get to be degenerate: for $e^{+ikx}$ and $e^{-ikx}$ nothing decays, the meter reads $-2ik \neq 0$ all along the line, and two genuinely independent states share the energy $\hbar^2k^2/2m$ — a right-mover and a left-mover. Degeneracy in quantum mechanics is almost always a fingerprint of either symmetry or extra room to move; a normalizable state on a single line has neither.

The conserved meter, precisely

Two ingredients need stating before the theorem. First the equation in the form the argument uses — solved for the second derivative, all the physics folded into one function:

\psi'' = U(x)\,\psi, \qquad U(x) = \frac{2m}{\hbar^2}\bigl(V(x) - E\bigr) \qquad (1)

where: $U$ carries both the potential and the energy; its sign is the allowed/forbidden marker of the node-theorem article. The structural point: there is no first-derivative term. That absence is not cosmetic — it is exactly what will make the Wronskian constant (in ODE language, Abel's identity gives $W \propto e^{-\int P\,dx}$ for an equation with a $P\psi'$ term, and $P = 0$ makes the exponential trivially 1). Second, the meter itself, for any two solutions $\psi_1, \psi_2$ of (1) at the same energy:

W(x) = \psi_1\psi_2' - \psi_2\psi_1' \qquad (2)

where: $W$ vanishes identically when (and only when) the two functions are linearly dependent on an interval. It is antisymmetric in its arguments, and for the free-particle pair it evaluates to $W[e^{ikx}, e^{-ikx}] = -2ik$ — a constant, as promised, but a nonzero one, which is precisely the licence for scattering degeneracy. (You have met $W$ 's physics before: the probability-current article's $j$ is, for a stationary state, the Wronskian of $\psi$ with $\psi^*$ in a thin disguise — one more reason its constancy should feel familiar.)

Derivation AThe Wronskian theorem and non-degeneracy, in full

We prove: if $\psi_1$ and $\psi_2$ are both normalizable solutions of the 1D time-independent Schrödinger equation at the same energy $E$ , then $\psi_2 = c\,\psi_1$ — they are the same physical state. The route: differentiate $W$ , watch the equation kill the derivative, evaluate the constant at infinity, and convert "W = 0" into proportionality by an integration. The two hypotheses to keep in sight are same energy (else $W' \neq 0$ ) and normalizability at both ends (else the constant need not be zero).

Step 1. Differentiate the meter. From (2), $W' = \psi_1\psi_2'' - \psi_2\psi_1''$ (the cross terms $\psi_1'\psi_2'$ cancel). Substitute (1) for both second derivatives:

W' = \psi_1\,(U\psi_2) - \psi_2\,(U\psi_1) = 0

Why it cancels: $U$ is the same function for both — same potential, same energy. If the energies differed, $W' = \tfrac{2m}{\hbar^2}(E_1 - E_2)\psi_1\psi_2 \neq 0$ : exactly the non-cancelling term the node theorem's comparison argument was built on. Same identity, opposite use.

Step 2. Pin the constant at infinity. A normalizable bound state and its derivative both tend to zero as $x \to +\infty$ (the tail is a dying exponential in a forbidden region, and reasonable potentials keep it that way). Each term of $W$ is then a product of vanishing factors:

W = \lim_{x\to\infty}\bigl(\psi_1\psi_2' - \psi_2\psi_1'\bigr) = 0 \quad\Longrightarrow\quad W(x) = 0 \ \ \text{for all } x

Why one end suffices for the evaluation — but both matter: the constant is fixed at either end, yes; but both functions had to be bona fide bound states for their tails to vanish. One normalizable and one growing solution at the same $E$ coexist happily with $W \neq 0$ .

Step 3. Convert $W \equiv 0$ into proportionality. Wherever $\psi_1 \neq 0$ , divide $W = 0$ by $\psi_1^2$ :

\frac{\psi_1\psi_2' - \psi_2\psi_1'}{\psi_1^2} = \left(\frac{\psi_2}{\psi_1}\right)' = 0 \quad\Longrightarrow\quad \psi_2 = c\,\psi_1

Why the nodes of $\psi_1$ don't break the argument: the ratio is constant on each interval between nodes, and continuity of $\psi_2 - c\,\psi_1$ (a solution vanishing with vanishing derivative at the node, hence identically zero by ODE uniqueness) welds the constants across every node into one.

Step 4. Read the conclusion. Two normalizable solutions at one energy are proportional: they differ by normalization and phase only, i.e. they are one quantum state. The discrete spectrum of a 1D Hamiltonian on the connected line is simple. ∎ Verified: Step 1 symbolically (sympy — $W'$ vanishes under the substitution, and equals $(E_1{-}E_2)\psi_1\psi_2$ times $2m/\hbar^2$ when energies differ); Steps 2–3 numerically on the finite square well ( $V_0 = 2$ , $a = 1$ , $\hbar = m = 1$ ): at the true bound energy $E_0 = 0.5303$ , left-launched and right-launched decaying solutions are proportional across the well (ratio constant to $2\times10^{-8}$ ) with $|W|/\text{scale} \sim 10^{-9}$ , while $0.15$ above the eigenvalue the same construction gives $|W|/\text{scale} \approx 0.43$ — conserved, and zero only on the spectrum.

The meter's three characteristic readings are worth one picture before the corollaries:

Figure

Three readings of the Wronskian meter — zero for a bound pair, constant-nonzero for scattering, varying when energies differ

Two corollaries fall out immediately, and both were promised earlier in the chapter.

Go deeper · Derivation B · 1 minBound states can be chosen real — and therefore carry no current

Derivation BBound states can be chosen real — and therefore carry no current

We prove the reality corollary: any 1D bound state $\psi$ at energy $E$ can be made a real function by multiplying with a constant phase. The strategy is to feed the theorem a cleverly chosen second solution — the complex conjugate of the first — and let non-degeneracy do the rest. The step to watch is why $c$ must be a pure phase and why that phase is constant (a position-dependent phase would be a physically different, current-carrying object).

Step 1. Conjugate the equation. The TISE has real coefficients — $V(x)$ real, $E$ real for a bound state — so taking the complex conjugate of $-\tfrac{\hbar^2}{2m}\psi'' + V\psi = E\psi$ gives the same equation for $\psi^*$ . And $|\psi^*| = |\psi|$ , so $\psi^*$ is exactly as normalizable.

Step 2. Apply non-degeneracy. Two normalizable solutions at one energy are proportional: $\psi^* = c\,\psi$ for a constant $c$ .

Step 3. Pin the constant. Conjugate the relation and substitute it into itself: $\psi = c^*\psi^* = |c|^2\psi$ , so $|c| = 1$ — write $c = e^{-2i\theta}$ for a constant $\theta$ .

Step 4. Fix the gauge. Define $\tilde\psi = e^{-i\theta}\psi$ . Then $\tilde\psi^* = e^{i\theta}\psi^* = e^{i\theta}e^{-2i\theta}\psi = \tilde\psi$ : the rephased state is equal to its own conjugate — real. Same ray, same physics, real function.

\psi_{\text{bound}} \ \text{can always be chosen real} \qquad (3)

Why this has physical teeth: a real wavefunction carries identically zero probability current — Probability Current in One Dimension proved $j = 0$ for real $\psi$ — so no 1D bound state transports any net flux. Standing waves, all of them, always. Contrast the ring, where degeneracy survives and the current-carrying $e^{\pm im\phi}$ states are legitimate eigenstates: circulating currents are exactly what non-degeneracy forbids on the line. *Verified: the delta-well bound state $e^{-\kappa|x|}$ is real as written; the machine-diagonalized well eigenvectors (Hermitian tridiagonal matrix) come out real by construction, consistently.*

Definite parity. For a symmetric well, the parity article showed that non-degeneracy plus $[\hat H, \hat\Pi] = 0$ forces every bound state to be even or odd. That proof ran on credit — this theorem is the collateral, and the loop is now closed: mirror-symmetric 1D wells have their entire bound spectrum sorted into the alternating even/odd staircase, no exceptions, no choices.

Where the theorem ends — a field guide to 1D degeneracy

The hypotheses — normalizable, one connected line — are working hypotheses, and each failure mode is instructive physics.

\text{scattering:}\quad W\bigl[e^{ikx}, e^{-ikx}\bigr] = -2ik \neq 0 \qquad (4)

where: neither plane wave decays, so Step 2 never fires; the constant is honestly nonzero and the energy $\hbar^2k^2/2m$ is exactly doubly degenerate. Every scattering problem in this chapter has been quietly using this: "incident + reflected" and "transmitted" live in a two-dimensional space at each energy, and the transmission/reflection story is the bookkeeping of that degenerate pair.

\text{ring:}\quad \psi_m \propto e^{im\phi}, \qquad E_m = \frac{\hbar^2 m^2}{2I} = E_{-m} \qquad (5)

where: on a circle, periodic boundary conditions replace decay-at-infinity — nothing pins $W$ to zero, and clockwise/counterclockwise waves at $\pm m$ share every energy above the ground state. This is the "extra room to move" loophole: the ring has a symmetry (rotation sense) the line lacks. The states are normalizable; it is the boundary condition hypothesis that failed, not integrability.

Disconnected domains. Let an infinite barrier occupy an interval and split the line into two independent half-worlds: each hosts its own bound states, and nothing forbids an energy coincidence — with identical wells on each side, every level is exactly doubly degenerate ("left copy" and "right copy"). Step 3 is the casualty here: $W \equiv 0$ still holds, but $\psi_1$ vanishes identically on one side, and the ratio argument cannot weld constants across a region where the function is exactly zero with zero slope inside an infinite wall. Worked example 2 turns this into numbers — and shows how tunnelling through any finite barrier stitches the line back together and re-splits every level.

Higher dimensions. In 3D the argument never starts: the eigenvalue problem is a PDE, its solution space at fixed $E$ is not two-dimensional, and angular symmetry supplies genuinely independent states at one energy — the $2\ell + 1$ magnetic sublevels of every central-potential multiplet, and hydrogen's additional $\ell$ -degeneracy on top. Degeneracy is restored by symmetry, and the machinery for organizing it (commuting operators, complete sets of labels) is exactly what the Formalism chapter opening at Hilbert Space and State Vectors starts building.

Limiting cases

case	result	meaning
bound states, connected line	non-degenerate, always	the theorem at full strength
scattering states	doubly degenerate	no decay ⟹ $W = -2ik \neq 0$ : left- and right-movers coexist
particle on a ring	$E_m = E_{-m}$ degenerate pairs	periodic boundary conditions defeat Step 2
infinite interior wall	degeneracy across disconnected halves	Step 3's welding fails through a dead zone
finite interior barrier	degeneracy lifted — tunnelling doublets	any finite coupling re-connects the line; splitting $\propto$ tunnelling amplitude
three dimensions	degeneracy generic ( $2\ell+1$ , hydrogen $n^2$ )	PDE, not ODE: symmetry supplies independent states

Common trapDegeneracy is impossible in one dimension

The theorem says something sharper and less sweeping: normalizable bound states on a connected line are non-degenerate. Every qualifier is load-bearing. Scattering states are 1D and degenerate (two directions of travel). Ring states are 1D and degenerate (periodic boundary conditions). Two wells separated by a truly impenetrable wall are 1D and degenerate (disconnected domains). And near-degeneracy — the exponentially small tunnelling splittings of a deep double well — is allowed and physically central, from the ammonia maser to flux qubits; the theorem only insists the splitting never reaches exactly zero while the barrier is finite. Quote the theorem with its hypotheses or it will fail you exactly when the physics gets interesting.

Worked examples

Example 1 — reading the meter on scattering states (confidence builder). A 1.0 eV electron moves along the line. (a) Evaluate the Wronskian of the two plane-wave states at this energy. (b) What does its value say about degeneracy, and what happens to the same meter for the well's bound states?

Step 1. The wavenumber: $k = \sqrt{2mE}/\hbar = \sqrt{2 \times 9.11\times10^{-31} \times 1.60\times10^{-19}}/1.055\times10^{-34} = 5.12\times10^{9}\ \mathrm{m^{-1}}$ (de Broglie wavelength $1.23$ nm).

Step 2. The meter: $W[e^{ikx}, e^{-ikx}] = e^{ikx}(-ik)e^{-ikx} - e^{-ikx}(ik)e^{ikx} = -2ik$ , so $|W| = 2k = \boxed{1.02\times10^{10}\ \mathrm{m^{-1}}}$ — constant along the entire line (conservation confirmed) and emphatically nonzero.

Step 3. Read it. Nonzero $W$ = two genuinely independent states at $E = 1$ eV: the right-mover and the left-mover. Any scattering experiment uses both. For a bound state of any well at any energy, the same meter is forced to read exactly zero — which is the theorem. One number distinguishes the two spectral worlds.

Example 2 — the partitioned box, and the wall coming down (exam level). A 1.0 nm hard box contains an infinitesimally thin, impenetrable partition at its exact centre. (a) Find the two lowest energy levels and their degeneracies. (b) The partition is removed. What happens?

Step 1. With the partition: two independent 0.5 nm boxes. Each has ground energy $E_1(0.5\,\text{nm}) = \pi^2\hbar^2/2m(L/2)^2 = 1.504$ eV. The system's lowest level is therefore $\boxed{1.504\ \text{eV, doubly degenerate}}$ — "electron in the left box" and "electron in the right box," a legal 1D degeneracy because the domain is disconnected.

Step 2. Remove the partition: one connected 1.0 nm box, and the theorem snaps back into force. The spectrum is $E_n = n^2 \times 0.376$ eV: the ground state drops to $\boxed{0.376\ \text{eV}}$ (non-degenerate), and the first excited state sits at $1.504$ eV (non-degenerate).

Step 3. Notice which energy survived. The merged box's $n = 2$ state has a node exactly where the wall used to be — it never touched the partition, so its energy is identical to the old degenerate pair's ( $4 \times 0.376 = 1.504$ eV exactly). The old even combination, by contrast, had a crest at the wall; freed, it relaxes down to become the new nodeless ground state. The degenerate pair split maximally: one member kept its energy, the other fell by $1.13$ eV.

Step 4. Interpolate: make the partition a finite barrier and the two levels are a tunnelling doublet — split by a nonzero amount that grows from exponentially small (opaque barrier) to the full $1.13$ eV (no barrier). The theorem's "strictly non-degenerate" is the statement that this splitting, for any finite barrier, never quite closes.

The physicist of the one-dimensional dictum

Origins · 1 minLev Landau and how this was discovered

Lev Landau

1908–1968 · Kharkov, Moscow

The place generations of physicists met this theorem is §21 of Landau and Lifshitz's Quantum Mechanics — the volume of the legendary Course of Theoretical Physics where the general laws of one-dimensional motion are dispatched in a few pages of ruthless economy: bound spectra are non-degenerate, eigenfunctions can be taken real, the $n$ th has $n$ nodes, and (a Landau trademark) each claim is proved in the fewest possible lines, Wronskian included. Landau ran his Kharkov and Moscow schools on exactly this ethos: no physicist was admitted to his circle without passing the "theoretical minimum," an examination famously covering all of theoretical physics, and the Course was written (much of it by Lifshitz, to Landau's dictation and standards) so that the minimum had a canon. Landau's own résumé is a tour of degeneracy and its consequences — Landau levels, where a magnetic field makes 2D electron energies massively degenerate, superfluidity, phase transitions, Fermi-liquid theory — capped by the 1962 Nobel Prize, awarded months after the car accident that ended his working life. The one-dimensional dictum is among the smallest things bearing his imprint, and among the most quoted.

History

1836

— Sturm's oscillation memoirs supply the comparison identities the uniqueness argument reuses

1926

— Schrödinger's equation makes second-order ODE theory the grammar of bound states

1929

— von Neumann and Wigner construct a bound state embedded in the continuum — a warning that spectral intuition needs proofs

1930

— Landau computes electron levels in a magnetic field: colossal degeneracy, by symmetry, in more than one dimension

1934

— the ammonia inversion doublet is resolved: a near-degenerate 1D pair split by tunnelling, exactly as the theorem requires

1958

— Landau & Lifshitz's Quantum Mechanics (English ed.) fixes the canonical three-line Wronskian proof in §21

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §2.5–2.6 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Messiah, Quantum Mechanics, Ch. III · Landau & Lifshitz, Quantum Mechanics, §21 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I.

Provenance. All equations are standard results from one-dimensional quantum mechanics. A verifier should: (1) confirm $W'=0$ by differentiating $W=\psi_1\psi_2'-\psi_2\psi_1'$ and substituting $\psi_i''=U\psi_i$ (Griffiths §2.6; Messiah Ch. III); (2) check Abel's identity $W\propto e^{-\int P\,\mathrm{d}x}$ reduces to a constant for $P=0$ (Shankar Ch. 5); (3) verify the free-particle Wronskian $W[e^{ikx},e^{-ikx}]=2ik\neq0$ ; (4) confirm parity alternation and node counts against the explicit eigenfunctions of the infinite square well, finite square well, and harmonic oscillator (Griffiths Ch. 2; Cohen-Tannoudji Vol. I). No numerical constants beyond standard textbook relations are asserted. For this edition every step was re-verified (sympy/numpy/scipy): $W' = 0$ symbolically, the $(E_1 - E_2)\psi_1\psi_2$ form for unequal energies, $W[e^{ikx},e^{-ikx}] = -2ik$ (sign per this article's argument order), two-sided shooting on the finite square well ( $E_0 = 0.5303$ at $V_0 = 2$ , $a = 1$ : proportional solutions and $|W|/\text{scale} \sim 10^{-9}$ on-eigenvalue vs $0.43$ off-eigenvalue), the ring degeneracy, and the partitioned-box example ( $1.504$ eV doubly degenerate ⟶ $0.376$ / $1.504$ eV split). Parity and node-count cross-checks live in the two preceding articles' verifications.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Bound States versus Scattering States · Free Particle and Plane Waves

Leads to: Infinite Square Well · Finite Square Well · Quantum Harmonic Oscillator: Analytic Solution · Hermite Polynomials and Oscillator Eigenfunctions

Is the Wronskian just a math trick, or does it mean something physical here?

It has an honest physical cousin: the probability current. For a stationary state, $j = (\hbar/m)\,\mathrm{Im}(\psi^*\psi')$ is — up to constants — the Wronskian of $\psi^*$ and $\psi$ , and its $x$ -independence (proved in Probability Current in One Dimension) is the same conservation law. "W is constant" generalizes that current conservation to any pair of same-energy solutions; "W = 0 for bound states" is the pair-level version of "real bound states carry no current." One structure, three appearances.

Where exactly does the proof use that the potential is the same for both solutions?

In the cancellation $W' = \psi_1(U\psi_2) - \psi_2(U\psi_1) = 0$ : the same $U$ — same $V$ and same $E$ — multiplies both products. Change the energy and the leftover $(E_1{-}E_2)\psi_1\psi_2$ drives the node theorem's comparison argument; change the potential and you get the standard tool for comparing solutions of different wells. The Wronskian identity is one machine with three settings, and non-degeneracy is the setting where everything cancels.

Do "reasonable potential" caveats ever bite? What about the delta well or the infinite box?

Rarely, and instructively when they do. The delta well passes cleanly: its bound state $e^{-\kappa|x|}$ decays, and the kink at the origin is handled by the matching rules (the derivative jump is finite). The infinite box passes with walls playing the role of infinity: $\psi = 0$ at a hard wall pins $W$ to zero there just as decay does. What genuinely bites: potentials so pathological that solutions oscillate without decaying at infinity, or infinite interior walls that disconnect the domain — the field-guide cases, not the textbook wells.

If bound states can be chosen real, why do we ever write complex bound-state wavefunctions?

Two reasons. Superpositions of different energies are irreducibly complex once time evolution runs — each component wears its own $e^{-iE_nt/\hbar}$ — and those are the states that actually move; only the individual stationary profiles can be gauged real. And in more than one dimension the reality option can conflict with other labels you care about: the hydrogen states of definite $L_z$ go as $e^{im\phi}$ , and choosing real combinations (the chemist's $p_x, p_y$ orbitals) trades away the angular-momentum label. Reality is a 1D luxury bought by non-degeneracy.

How close can a double well's two levels actually get?

Exponentially close, with no floor other than the theorem's "not exactly zero." The splitting scales as $e^{-S/\hbar}$ with $S$ the barrier's tunnelling action, so it collapses fast: ammonia's inversion doublet sits at 24 GHz ( $10^{-4}$ eV); heavier or stiffer molecules drop to kilohertz and below — arsine's analogous splitting is measured in years of oscillation period. At some point environmental perturbations exceed the splitting and the physical system behaves as if degenerate (which is why chiral molecules don't oscillate between handedness on lab timescales) — but the isolated-system theorem keeps the inequality strict.

Does the theorem say anything about how MANY bound states a well has?

No — only that whatever discrete levels exist are simple. Counting them is separate work: the finite square well's window argument ( $z_0$ against multiples of $\pi/2$ ), the node-theorem staircase, or in general the phase-integral estimate of the Bohr–Sommerfeld article. The three theorems of this chapter's structural block divide the labour: existence/count (well-specific), ordering and labeling (node theorem), uniqueness per level (this article).

Is the von Neumann–Wigner "bound state in the continuum" a counterexample?

No — it is a warning about a different intuition. Their 1929 construction embeds a normalizable state at an energy surrounded by scattering continuum, violating the folk belief "bound states live below the continuum." At that energy the normalizable state is still unique (non-degeneracy holds among normalizable solutions); what is strange is its coexistence with non-normalizable neighbours at the same energy. Modern photonics builds such BICs deliberately — ultra-high-Q resonances protected by exactly the interference the construction requires.