Node Theorem and Energy Ordering

Hand a physicist a plot of a mystery bound state and they will tell you, in about two seconds, exactly which level it is. No solving, no fitting — they just count the zero-crossings. That party trick is the node theorem: in any one-dimensional well, the $n$ th bound state has exactly $n$ interior nodes, and the energies rise in lockstep with the count, $E_0 < E_1 < E_2 < \cdots$ . The abstract eigenvalue problem of the time-independent Schrödinger equation turns into arithmetic you can do with your eyes. The theorem is rigorous, not a rule of thumb. It sits on Sturm's oscillation theory of second-order differential equations — mathematics half a century older than quantum mechanics — welded to one quantum ingredient: bound levels in one dimension are non-degenerate, so each node count belongs to exactly one state. And it is physical through and through. The Schrödinger equation makes curvature proportional to $E - V$ : where the particle is classically allowed, the wavefunction bends back toward the axis and oscillates; more energy means tighter bending, shorter local wavelength, one more crossing squeezed into the well per rung of the ladder. Nodes are kinetic energy made visible. Once you own this theorem, sketching becomes solving. You can order the spectrum of a potential you have never integrated, sanity-check any numerical eigensolver in a glance (a "ground state" with a node is a bug, every time), and understand why ground states are the smooth, nodeless humps they always are. This article builds the curvature intuition, proves the nodeless ground state and the one-node-per-level ladder honestly, verifies everything on wells solved by machine, and hands you the counting conventions that prevent the classic off-by-one errors.

Key takeaways

In a 1D well with a discrete spectrum, the bound state with index $n = 0, 1, 2, \ldots$ has exactly $n$ interior nodes, and energies are strictly ordered: $E_0 < E_1 < E_2 < \cdots$ .
The engine is curvature: $\psi''/\psi = -2m(E-V)/\hbar^2$ — oscillation where $E > V$ , decay where $E < V$ . Higher $E$ ⟹ shorter local wavelength ⟹ more crossings.
The ground state is nodeless — a sign change costs kinetic energy without buying anything, and the variational principle forbids it.
Nodes of consecutive states interlace: between any two adjacent nodes of $\psi_{n+1}$ sits exactly one node of $\psi_n$ .
Non-degeneracy is the silent partner: one energy, one state, one node count — no ambiguity in the ordering.
Counting conventions bite: ISW textbooks label from $n = 1$ (so $\psi_n$ has $n-1$ nodes), and wall zeros at the boundary are not interior nodes.
Key idea: label bound states by their number of interior nodes; that integer equals the level index and orders the energies, because each additional node adds curvature and thus kinetic energy.

Where this lives today

Node counting is the working physicist's checksum. Every numerical eigensolver — shooting, finite differences, basis diagonalization — is validated by whether its $n$ th state shows $n$ crossings, and production codes in quantum chemistry count radial nodes to make sure a converged orbital is the $2s$ and not the $1s$ (the $n - \ell - 1$ radial-node rule is this theorem running inside each angular channel). In cold-atom laboratories, absorption images of trapped condensates and few-body states are read by node count to identify which excitation was prepared. The theorem's failure modes earn their keep too: in more than one dimension node counts become nodal-domain inequalities (Courant), and for interacting many-fermion wavefunctions the geometry of the nodal surface is precisely what the fixed-node quantum Monte Carlo method must guess — the celebrated "sign problem" of computational many-body physics is, at bottom, the fact that nodes stop being countable when you leave the line.

The physical picture

Read the Schrödinger equation as a statement about bending. Divide through by $\psi$ and it says $\psi''/\psi = -2m(E - V(x))/\hbar^2$ : the sign of $E - V$ decides whether the function curves toward the axis or away from it. In the classically allowed region ( $E > V$ ) the ratio is negative — wherever $\psi$ is positive it bends downward, wherever negative it bends upward — perpetual correction toward the axis, which is what oscillation is. In the forbidden region ( $E < V$ ) the ratio flips sign and $\psi$ bends away from the axis: growing or dying exponentials. A bound state must take the dying branch on both flanks, so all its oscillation — all its nodes — must happen in the allowed middle.

Figure

Curvature reads the sign of E − V — oscillate inside the well, decay outside it

Now climb the ladder. Raising $E$ does two things at once: it widens the allowed region (the energy line intersects the well walls further out) and it deepens $E - V$ in the middle, which shortens the local de Broglie wavelength. Both effects fit more oscillation between the decaying tails. The fit is quantized — the tails must join the oscillating middle smoothly on both sides, and that double handshake succeeds only at discrete energies, each successive success threading exactly one more zero-crossing through the well. That is the whole theorem, seen from the shooting-method angle: integrate the equation outward from the left tail at a trial energy and watch the far side. Too little energy and the trial solution blows up one way; too much and it blows up the other; each sign change of the blow-up brackets an eigenvalue, and each eigenvalue passed adds one node to the trial function. Node count is a monotone staircase in energy — which is why it can serve as a label.

Figure

One more node per rung — the first four states of a quartic well, machine-solved

The dots in the second figure hide one more piece of structure: the nodes of consecutive states interlace, like the teeth of two combs pushed together. Between any two adjacent nodes of $\psi_3$ sits exactly one node of $\psi_2$ ; the pattern locks the counts into increasing by exactly one per level, never jumping by two. And the ground state, at the bottom of the staircase, has no nodes at all — a fact with its own clean proof below.

The theorem, stated with its conventions

For a confining one-dimensional potential with a discrete spectrum, label the bound states in order of increasing energy starting from zero:

\psi_n \ \text{has exactly } n \ \text{interior nodes}, \qquad E_0 < E_1 < E_2 < \cdots \qquad (1)

where: an interior node is a point strictly inside the region where $\psi$ lives, at which $\psi$ crosses zero with nonzero slope. Two bookkeeping rules prevent the classic errors. First, boundary zeros do not count: the Infinite Square Well's $\psi_n \propto \sin(n\pi x/a)$ vanishes at both walls by boundary condition, and those endpoint zeros are not nodes — the interior zeros sit at $x = a/n, 2a/n, \ldots$ , and there are $n - 1$ of them. Second, mind the labeling offset: the ISW tradition starts at $n = 1$ , so its state $n$ has $n - 1$ nodes; the theorem's convention starts the ground state at $n = 0$ . Same physics, shifted index. The curvature engine driving it all is worth displaying once in its own right. Divide the time-independent Schrödinger equation by $\psi$ :

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

where: the left side is a curvature-to-amplitude ratio, and the right side is (minus) the local kinetic energy in disguise. $E > V$ : negative ratio, bending toward the axis, oscillation with local wavelength $\lambda(x) = h/\sqrt{2m(E - V)}$ — larger kinetic energy, tighter wiggles. $E < V$ : positive ratio, bending away, exponential behaviour. Every qualitative sketch you will ever draw of a bound state is an application of the sign of equation (2). (The machine-solved quartic well of the figure confirms it wholesale: the computed $\psi''/\psi$ is negative everywhere in the allowed region and positive in the tails.) The oscillator gives the theorem a closed-form showcase. Its eigenfunctions are Gaussian-damped Hermite polynomials, and the Gaussian never vanishes, so the nodes of $\psi_n$ are exactly the real roots of $H_n$ :

\psi_n \propto H_n(\xi)\,e^{-\xi^2/2}, \qquad H_n \ \text{has exactly } n \ \text{real roots} \qquad (3)

where: $\xi = \sqrt{m\omega/\hbar}\,x$ . Hermite Polynomials and Oscillator Eigenfunctions owns that family's anatomy (Rodrigues formula, recurrences, orthogonality); here we consume one fact — $n$ roots, all real, all simple — and note that it is the node theorem holding in exact arithmetic, with the ordering supplied by $E_n = (n + \tfrac12)\hbar\omega$ . The alternating even/odd character of those eigenfunctions is the same staircase seen through the parity lens of the previous article: a state with an odd node count must cross at the centre (odd parity), an even count leaves a crest (even parity) — node counting and parity alternation are one structure.

Derivation AWhy the ground state has no nodes

We prove that the lowest-energy bound state never changes sign — zero nodes — using the variational characterization of the ground state: $E_0$ is the minimum of $\langle\psi|\hat H|\psi\rangle$ over normalized states. The strategy is a compare-and-win argument: suppose a candidate ground state had a node, build from it a competitor with no higher energy, and show the competitor (or a smoothing of it) cannot itself be an eigenstate unless something is contradicted. The step to watch is the replacement $\psi \to |\psi|$ — it is legal precisely because the energy functional cares about $|\psi|^2$ and $|\psi'|^2$ only.

Step 1. Write the energy as a functional. For real $\psi$ (one dimension permits this choice — see this chapter's non-degeneracy article),

E[\psi] = \int \left[ \frac{\hbar^2}{2m}\,\psi'(x)^2 + V(x)\,\psi(x)^2 \right] dx

Why this form: one integration by parts converts $-\psi\psi''$ into $\psi'^2$ (boundary terms die with the bound-state tails). Both terms now involve only squares.

Step 2. Replace $\psi$ by $|\psi|$ . If $\psi$ changes sign, consider $|\psi|$ : it has the same norm, the same $\psi^2$ , and — except at the isolated crossing points — the same $\psi'^2$ . So $E[|\psi|] = E[\psi]$ : the sign-flipped copy costs nothing. Why nothing is saved either — yet: the kink is the point. $|\psi|$ has a corner (discontinuous slope) at each former node.

Step 3. Smooth the kink and win. A corner can be rounded in an arbitrarily small neighbourhood in a way that strictly lowers the kinetic integral (the corner is a concentration of $|\psi'|$ that smoothing relaxes) while changing the potential integral negligibly. So from any sign-changing candidate we can manufacture a strictly lower-energy competitor: a sign-changing function can never be the minimizer.

Step 4. Conclude. The minimizer of $E[\psi]$ — the ground state — is sign-definite; choose it positive. A sign-definite function has no interior crossings: zero nodes. Uniqueness (non-degeneracy) then guarantees the ground state is this nodeless function, not merely one option among several. Why this also explains the ordering's first step: every excited state must be orthogonal to the positive $\psi_0$ , so its integrand $\psi_0\psi_n$ must change sign — forcing $\psi_n$ to change sign somewhere. At least one node for every state above the ground floor, with the exact count delivered by the oscillation theorem below. *Verified: numerically — the machine-diagonalized ground states of both the finite square well ( $V_0 = 2$ , $a = 1$ , $\hbar = m = 1$ ) and the quartic well $V = x^4$ are sign-definite on the full grid (zero sign changes), while every excited state shows exactly its index in crossings.*

Go deeper · Derivation B · 2 minOne more node per level — the oscillation argument

Derivation BOne more node per level — the oscillation argument

We show that if $E_2 > E_1$ are two bound energies with eigenfunctions $\psi_1, \psi_2$ , then between any two consecutive zeros of $\psi_1$ the function $\psi_2$ must vanish at least once — the interlacing that forces node counts to climb one per level. The tool is the Wronskian-type identity Sturm built comparison theory on; the strategy is integrate-between-zeros and hunt a sign contradiction. Watch how the energy difference is the term that refuses to vanish.

Step 1. Build the comparison identity. From the two Schrödinger equations, $\psi_1'' = -\tfrac{2m}{\hbar^2}(E_1 - V)\psi_1$ and likewise for $\psi_2$ , form $\psi_2\,\psi_1'' - \psi_1\,\psi_2''$ and recognize a total derivative:

\frac{d}{dx}\bigl(\psi_2\psi_1' - \psi_1\psi_2'\bigr) = \frac{2m}{\hbar^2}\,(E_2 - E_1)\,\psi_1\psi_2

Why the potential dropped out: $V$ multiplies $\psi_1\psi_2$ identically in both equations — comparison identities are built to cancel what the solutions share and expose what differs, here only $E_2 - E_1$ .

Step 2. Integrate between consecutive zeros $a < b$ of $\psi_1$ . Take $\psi_1 > 0$ on $(a,b)$ (flip signs if not). The left side integrates to boundary terms only, and $\psi_1(a) = \psi_1(b) = 0$ kills half of each:

\psi_2(b)\,\psi_1'(b) - \psi_2(a)\,\psi_1'(a) = \frac{2m}{\hbar^2}(E_2 - E_1)\int_a^b \psi_1\psi_2\,dx

Step 3. Assume $\psi_2$ has no zero in $(a,b)$ and contradict. Say $\psi_2 > 0$ throughout. Then the right side is strictly positive ( $E_2 > E_1$ , both factors positive). But on the left: $\psi_1$ rises from zero at $a$ (so $\psi_1'(a) > 0$ ) and falls to zero at $b$ (so $\psi_1'(b) < 0$ ), making both boundary terms non-positive — the left side is $\le 0$ . Positive equals non-positive: contradiction. So $\psi_2$ vanishes somewhere in $(a,b)$ .

Step 4. Read off the ladder. The higher state oscillates strictly faster: between consecutive nodes of any state sits a node of every higher state, and applying the argument to the tails as well pins the count to exactly one extra node per level (a jump of two would put two nodes of $\psi_{n+1}$ between neighbours of $\psi_n$ with none of $\psi_n$ between them, contradicting the same identity run in reverse). Combined with non-degeneracy — one state per energy, so the labels are unambiguous — the staircase $E_0 < E_1 < \cdots$ with node counts $0, 1, 2, \ldots$ is forced. Verified: the comparison identity was checked symbolically (sympy, derivative expands to the stated right side), and interlacing was verified exactly for Hermite roots $H_n$ vs $H_{n+1}$ , $n = 1\ldots5$ , and numerically for the quartic well's machine eigenfunctions ( $\psi_3$ 's three zeros straddle $\psi_2$ 's two).

Limiting cases

case	result	meaning
confining $V$ , discrete spectrum	state $n$ has exactly $n$ nodes	the theorem at full strength
ISW labeling from $n=1$	$\psi_n$ has $n-1$ interior nodes	same theorem, shifted convention; wall zeros not counted
harmonic oscillator	nodes = real roots of $H_n$ : exactly $n$	closed-form showcase; ordering via $E_n = (n+\tfrac12)\hbar\omega$
periodic boundary conditions (ring)	theorem fails as stated	degeneracy returns ( $e^{\pm im\phi}$ ); node count no longer a unique label
scattering states	not covered	non-normalizable, continuous spectrum — no discrete ladder to label
3D radial equation	survives for each fixed $\ell$	$u_{n\ell}(r)$ obeys a 1D equation: radial node count orders levels within one $\ell$ channel

Common trapThe fourth energy level has four nodes — right?

Only if your "fourth level" is the theorem's $n = 4$ , i.e. the fifth state counting the ground state as zeroth. The node theorem's convention starts at $n = 0$ (ground state, zero nodes); much of the well literature — the infinite square well above all — starts at $n = 1$ , whose state $n$ carries $n - 1$ interior nodes. Both statements are correct; mixing them mid-calculation is the classic exam disaster. Two safeguards: always anchor on the ground state (zero nodes, no exceptions), and never count boundary zeros — the ISW wavefunction's vanishing at the walls is imposed by the box, not earned by oscillation, and the same goes for the forced zero at $r = 0$ of a 3D radial function. When in doubt, count crossings strictly inside the classically relevant region and add one to get the position above the ground floor.

Worked examples

Example 1 — identify the state, then the energy (confidence builder). A scanning-probe experiment maps the density of an electron trapped in a hard-walled well of width $L = 1.0$ nm and resolves three interior zeros. Which state is it, and what is its energy?

Step 1. Count to label. Three interior nodes in ISW labeling ( $n$ starts at 1, nodes $= n-1$ ) means $n = 4$ ; in the theorem's labeling it is the third excited state, $n_{\rm thm} = 3$ . One physical state, two bookkeeping systems.

Step 2. Energy scale of the box: $E_1 = \pi^2\hbar^2/2mL^2 = 0.376$ eV for an electron at $L = 1.0$ nm.

Step 3. Scale up: $E_4 = 4^2 E_1 = 16 \times 0.376 = \boxed{6.02\ \text{eV}}$ .

Step 4. Sanity by curvature: the 3-node state fits two full wavelengths in the box ( $\lambda = L/2 = 0.5$ nm); the de Broglie relation gives $p = h/\lambda$ and $E = p^2/2m = 6.0$ eV ✓ — the node count, the wavelength, and the energy are one piece of information worn three ways.

Example 2 — ordering a well nobody solves in closed form (exam level). The quartic well $V(x) = x^4$ (with $\hbar = m = 1$ ) has no analytic solution. (a) Predict the node counts and parities of the four lowest states without computing anything. (b) The machine returns levels $0.668$ , $2.394$ , $4.697$ , $7.336$ — check that they are consistent with the theorem, and extract a testable pattern.

Step 1. Predictions for free. The well is confining and symmetric, so: node counts $0, 1, 2, 3$ ; strict ordering; parities even, odd, even, odd (node parity = state parity, previous article); interlaced node positions. All of this is known before any numerics.

Step 2. Check the machine. Diagonalizing $-\tfrac12\psi'' + x^4\psi = E\psi$ on a grid gives $E_0 = 0.668 < E_1 = 2.394 < E_2 = 4.697 < E_3 = 7.336$ , with computed node counts $0, 1, 2, 3$ and $\psi_3$ 's three zeros straddling $\psi_2$ 's two — every structural prediction lands. (A solver returning a "ground state" with a crossing, or two levels out of node order, is wrong, full stop — this is the theorem's daily job as a debugger.)

Step 3. Extract the pattern. The gaps grow: $E_1 - E_0 = 1.73$ , $E_2 - E_1 = 2.30$ , $E_3 - E_2 = 2.64$ . Steeper-than-quadratic walls squeeze high states harder — opposite to the equal spacing of the oscillator. (The phase-integral rule of this chapter's Bohr–Sommerfeld article turns this observation into the quantitative growth law $E_n \sim (n + \tfrac12)^{4/3}$ for the quartic — node counting tells you the order, semiclassics the spacing.)

Step 4. Verdict: the spectrum is fully ordered and labeled by counting alone; the machine supplied only the decimal places.

The mathematician who counted zeros first

Origins · 1 minJacques Charles François Sturm and how this was discovered

Jacques Charles François Sturm

1803–1855 · Paris

The node theorem was proved a full ninety years before anyone needed it for quantum mechanics. Sturm, Geneva-born and Paris-based, published in 1836 a pair of memoirs on second-order differential equations that founded what is now Sturm–Liouville theory: his comparison and oscillation theorems established that solutions at higher eigenvalues oscillate faster, that zeros of consecutive eigenfunctions interlace, and that the $n$ th eigenfunction crosses zero exactly $n$ times — all from sign arguments of the kind used in this article, with no closed-form solutions anywhere. His motivation was heat flow in inhomogeneous bars, and his collaboration with Liouville produced the first general spectral theory of differential operators. When Schrödinger's equation arrived in 1926, the mathematics of its bound states was, in this respect, already sitting complete on the shelf — Courant and Hilbert's 1924 Methoden der mathematischen Physik, which the quantum pioneers read closely, carried Sturm's theorems forward just in time. Sturm also gave algebra the root-counting theorem that bears his name and made the first decent measurement of the speed of sound in water (1826, Lake Geneva); the Paris street beside the Luxembourg Gardens is his.

History

1826

— Sturm and Colladon measure the speed of sound in Lake Geneva — precision counting of a different kind

1836

— Sturm's memoirs establish the oscillation and comparison theorems for second-order ODEs

1837

— Sturm and Liouville launch the general eigenvalue theory of differential operators

1924

— Courant and Hilbert's Methoden codifies oscillation theory for the physicists about to need it

1926

— Schrödinger's equation makes Sturm's zero-counting a statement about quantum states

1929

— Courant's nodal-domain theorem extends node counting to higher dimensions (as an inequality, not an equality)

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §2.2–2.3 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 7 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement H_III · Courant & Hilbert, Methods of Mathematical Physics, Vol. I, Ch. VI · Landau & Lifshitz, Quantum Mechanics, §21.

Provenance. Equations and node counts are standard textbook results. A verifier should: (1) confirm the infinite-well zeros of $\sin(n\pi x/a)$ give $n-1$ interior nodes (Griffiths §2.2); (2) verify Hermite polynomial $H_n$ has $n$ real roots, hence oscillator state $n$ has $n$ nodes (Griffiths §2.3, Shankar Ch.7); (3) reproduce the Wronskian non-degeneracy and Sturm comparison arguments (Cohen-Tannoudji Complement H_III; Courant–Hilbert Vol. I); (4) check the oscillator spectrum $E_n=\hbar\omega(n+\tfrac12)$ and ground-state energy $\tfrac12\hbar\omega$ . All numbers ( $n-1$ nodes, $n$ roots, $\tfrac12\hbar\omega$ ) are exact standard values. For this edition every claim was re-verified by machine (sympy/numpy/scipy): ISW node counts $n = 1\ldots5$ and the $\psi_4$ zeros at $a/4, a/2, 3a/4$ ; Hermite root counts $n = 0\ldots6$ and interlacing $n = 1\ldots5$ ; the comparison identity symbolically; curvature signs, node counts, ordering, and interlacing on machine-diagonalized finite-square and quartic wells (quartic levels $0.667986, 2.393643, 4.696793, 7.335724$ , grid-convergence checked); and both worked examples. The Wronskian non-degeneracy proof itself is owned by Non-Degeneracy of One-Dimensional Bound States, two articles ahead.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Bound States versus Scattering States · Infinite Square Well

Leads to: Quantum Harmonic Oscillator: Analytic Solution · Hermite Polynomials and Oscillator Eigenfunctions · Finite Square Well · Zero-Point Energy

Is a node a place the particle can never be found, or just unlikely?

Exactly zero probability density at the point itself — the eigenfunction genuinely vanishes there. But a node is a set of measure zero: any finite detector bin straddling it registers plenty of counts from either side. The striking (and real) statement is about interference structure: a particle in the $n = 1$ state of a box is never detected at dead centre, however long you wait, yet is happily found on both sides. Nodes are standing-wave zeros, not soft valleys.

How does the particle "get across" a node if the density there is zero?

The question smuggles in a classical trajectory that the stationary state does not have. An energy eigenstate is a standing pattern — nothing is commuting from left to right through the node point (its probability current is identically zero, as the probability-current article showed for real wavefunctions). Asking how it crosses is like asking how a violin string's vibration crosses the motionless point of its overtone: the pattern simply has a zero there.

Does the node theorem hold for any potential at all?

For any confining 1D potential with a discrete spectrum acting on the whole line (or an interval with standard boundary conditions), yes. The named failure modes: rings and periodic boundary conditions (degeneracy returns and counting loses uniqueness), genuinely multi-dimensional problems (node curves and surfaces replace countable points — Courant's theorem survives only as an inequality), and scattering states (no discrete ladder to label). Singular potentials need care but the delta well passes: its single bound state is nodeless, exactly as $n = 0$ demands.

Why do higher states have more energy just because they have more nodes — which is cause and which is effect?

The honest causal chain runs through curvature: fitting more crossings between decaying tails requires shorter local wavelength, and wavelength is momentum is kinetic energy (equation 2). So nodes don't cost energy as a fee per crossing — rather, both the extra node and the extra energy are symptoms of one thing, faster oscillation. That's also why the gap pattern varies from well to well (growing for quartic, constant for harmonic, shrinking toward the continuum for the finite well) while the count pattern never does.

The double well's two lowest states are nearly degenerate — doesn't that break the ladder?

It bends it without breaking it. The symmetric double well's ground state (nodeless, even) and first excited state (one node at the barrier, odd) can sit exponentially close in energy — the tunnelling splitting — but strictly ordered they remain, with the node count still deciding who is lower. The near-degeneracy is the interesting physics (it sets inversion frequencies like ammonia's), and the theorem's strict inequality is what guarantees it never becomes exact for a true 1D double well.

Do radial wavefunctions in 3D obey the same counting?

Within each angular-momentum channel, yes: the radial function $u(r) = rR(r)$ obeys a 1D Schrödinger equation on the half-line, so for fixed $\ell$ the $k$ th level has $k$ radial nodes (not counting $r = 0$ or infinity). Hydrogen's familiar rule — $n - \ell - 1$ radial nodes — is exactly this. What 3D scrambles is comparison across channels: different $\ell$ means different effective potentials, and node counts no longer order energies globally (hydrogen's $2s$ and $2p$ are degenerate with different node structures).

Can I use node counting on numerical or experimental data with noise?

Yes — it is one of the most robust labels available, which is why it survives as a debugging tool. Genuine nodes are crossings with slope, so they persist under moderate noise, while noise-induced wiggles hover near zero without a systematic crossing; counting sign changes of a lightly smoothed signal is usually decisive. In cold-atom images and STM spectroscopy of quantum corrals this is routinely how prepared states are identified. The trap is finite resolution near closely spaced nodes of high states — undercounting merges them, which is another reason low-lying states make the best benchmarks.