Particle in a Spherical Well and Infinite Spherical Box

The chapter's function-building is over; this article spends the toolkit. Two confining problems — the infinite spherical box (hard wall at radius $R$ ) and the finite spherical well (depth $V_0$ , range $R$ ) — are solved end to end with the machinery of the last three serials: interior solutions from the regular family $j_\ell$ , exteriors from the evanescent Hankel tails, spectra from matching. The infinite box is the three-dimensional heir of the chapter-one particle-in-a-box, and its answer is pleasingly literal: energies are set by the zeros of the spherical Bessel functions, $E_{n\ell}=\hbar^2\beta_{n\ell}^2/2mR^2$ , so the level diagram is a table of root-found numbers ( $\pi$ , $4.493$ , $5.764$ , $2\pi$ , $\dots$ ) — producing the ordering $1s,1p,1d,2s,1f,2p$ and, doubled for spin, the filling ladder $2,8,18,20,34,40$ that nuclear physics reads magic numbers from. The finite well replaces the wall by a cliff: inside, still $j_\ell$ ; outside, the decaying $h_\ell^{(1)}(i\kappa r)$ handed over by the previous article; equality of logarithmic derivatives at $R$ quantizes the spectrum through a transcendental equation. For s-waves that equation is $k\cot kR=-\kappa$ — exactly the odd-parity half of the 1D finite well, as the wall-at-the-origin doctrine promised it would be. Two verdicts carry the physics. First, the existence threshold: unlike one dimension, where any attraction binds, a 3D well binds nothing until $V_0R^2>\pi^2\hbar^2/8m$ — shallow-or-narrow wells are barren (numerically confirmed on both sides of the threshold). Second, the article's centerpiece: the deuteron worked end to end. Feeding the measured binding energy $2.2246$ MeV into the matching at $R=2.1$ fm returns $V_0\approx34$ MeV — a well fifteen times deeper than its own bound state, only $1.45\times$ the threshold, holding exactly one state with $63\%$ of its probability outside the well. The nuclear force barely succeeds, and everything low-energy about nuclear physics inherits that near-failure.

Key takeaways

Infinite spherical box: $R_{n\ell}=j_\ell(kr)$ with $j_\ell(kR)=0$ ⇒ $E_{n\ell}=\dfrac{\hbar^2\beta_{n\ell}^2}{2mR^2}$ ; $\beta$ = the root-found zeros $3.142,4.493,5.764,6.283,6.988,7.725$ ⇒ ordering $1s,1p,1d,2s,1f,2p$ .
With spin, the box fills $2,8,18,20,34,40$ — magic 2, 8, 20 appear (compare the oscillator's ladder in The Three-Dimensional Isotropic Harmonic Oscillator: same first three, different beyond).
Nuclear notation throughout: $2s$ = second s-level, not hydrogen's $n=2$ (the dictionary card of the oscillator article, reused).
Finite well, $\ell=0$ : interior $\sin kr$ , exterior $e^{-\kappa r}$ ; matching ⇒ $k\cot kR=-\kappa$ — the odd sector of the 1D finite well of width $2R$ (the origin's hard wall made quantitative).
Existence threshold (3D novelty): binding requires $V_0R^2>\pi^2\hbar^2/8m$ — verified numerically: 0 states at $0.97\,V_{th}$ , 1 at $1.03\,V_{th}$ ; higher ℓ pay more (the centrifugal rent).
Deuteron solved: $E_B=2.2246$ MeV, $R=2.1$ fm, $\mu c^2=469.46$ MeV ⇒ $V_0=33.7$ MeV (root-found), $E_B/V_0=0.066$ , $V_0/V_{th}=1.45$ , exactly one s-state, exterior probability $0.63$ .
Electron in a 1 nm hard sphere: $E_{1s}=0.376$ eV, $E_{1p}=0.769$ eV, $E_{1d}=1.266$ eV (computed) — quantum-dot arithmetic.
Key idea: interior $j_\ell$ , exterior evanescent Hankel, match at the edge — the whole spherical bound-state craft in one sentence, with 3D's distinctive lesson that binding must be earned above a threshold.

The core insight

Spherical confinement quantizes through the spherical Bessel machinery: hard walls pin energies to the zeros $\beta_{n\ell}$ of $j_\ell$ ; finite wells match interior $j_\ell(kr)$ to exterior $h_\ell^{(1)}(i\kappa r)$ via logarithmic derivatives, giving transcendental spectra, a binding threshold $V_0R^2>\pi^2\hbar^2/8m$ absent in 1D — and, applied to the deuteron, the depth of the nuclear force from one measured number.

Where this lives today

The threshold theorem runs today's ultracold-atom laboratories in reverse: experimenters tune interatomic wells through the binding threshold with magnetic fields (Feshbach control), watching a bound state be born at zero energy with a divergent tail — the $\kappa\to0$ row of this article's limiting table, realized on demand and used to make molecules from free atoms. Halo-nucleus physics at rare-isotope facilities is the deuteron's 63% pushed to extremes: one-neutron halos with $P_{out}$ approaching 90%, identified by exactly the matching analysis performed here. And every colloidal-quantum-dot display on the market is the worked example's $1/R^2$ law industrialized — size-sorted nanospheres emitting at zero-ladder-determined colors.

The physical picture

Start with the hard-walled sphere, because its physics is pure geometry: standing ripples of the regular family, cut off where the wall stands. Each channel $\ell$ has its own waking radius (the turnstile of Spherical Bessel and Neumann Functions), so at fixed box size the s-wave fits its first node earliest ( $\beta=\pi$ ), the p-wave later ( $4.493$ ), the d-wave later still ( $5.764$ ) — and only then does the second s-level arrive ( $2\pi$ ). The interlacing of zeros is the level diagram; nothing else enters.

Figure

The infinite spherical box: levels are zeros of j_ℓ — geometry becomes a spectrum

Now soften the wall to a cliff of depth $V_0$ . Two things change, one cosmetic and one structural. Cosmetic: the wavefunction no longer stops at $R$ but leaks into the classically forbidden exterior with decay constant $\kappa=\sqrt{2m|E|}/\hbar$ — the imaginary-argument Hankel tails delivered by Spherical Hankel Functions and Outgoing Waves. Structural: binding is no longer guaranteed. The 1D finite well binds at any depth (Finite Square Well's theorem); the 3D well does not, because the origin's hard wall — $u(0)=0$ , the doctrine of Boundary Conditions and Normalization of Radial Wavefunctions — forces the wavefunction to bend over within the well's width before it may decay. Bending costs curvature, curvature costs kinetic energy, and a shallow well cannot pay. Three dimensions makes attraction audition for the role of "binder," and the threshold below is the audition's pass mark.

The infinite spherical box

Inside, $V=0$ : the radial solutions are the free ones of The Free Particle in Three Dimensions — regular family only, since the origin is in-domain. The wall imposes $R_{n\ell}(R)=0$ :

j_\ell(kR)=0\;\Rightarrow\;k_{n\ell}=\frac{\beta_{n\ell}}{R},\qquad E_{n\ell}=\frac{\hbar^2\beta_{n\ell}^2}{2mR^2} \qquad (1)

where: $\beta_{n\ell}$ is the $n$ -th zero of $j_\ell$ — the ladder root-found in Spherical Bessel and Neumann Functions and now cashed as a spectrum (that article's relocated inheritance; this article owns the physics the zeros price). For $\ell=0$ the zeros are exactly $n\pi$ , so $E_{n0}=\hbar^2n^2\pi^2/2mR^2$ — the 1D box spectrum verbatim, because $u=\sin(kr)$ is a 1D box wavefunction on the half-line. Each level holds $2(2\ell+1)$ particles with spin.

Derivation AThe hard sphere solved, ordered, and filled

We assemble the box spectrum and extract its level ordering and shell capacities — the numbers behind the "poor man's nuclear shell model." Everything reduces to properties of the zero ladder, all root-found rather than quoted.

Step 1. Admissibility inside a domain containing the origin keeps only $j_\ell$ (the free-particle article's verdict, consumed). The wall condition $j_\ell(kR)=0$ then quantizes $k$ channel by channel: no matching, no transcendental subtlety — geometry.

Step 2. Order the levels. Zeros interlace (Sturm grammar, cited to the function article): $\beta_{10}=\pi<\beta_{11}=4.493<\beta_{12}=5.764<\beta_{20}=2\pi<\beta_{13}=6.988<\beta_{21}=7.725$ (each root-found) — hence $1s<1p<1d<2s<1f<2p$ in nuclear notation ( $n$ = ordinal within the channel; the notation dictionary of The Three-Dimensional Isotropic Harmonic Oscillator, reused). Note $1d$ below $2s$ — but barely ( $5.764$ vs $6.283$ ): hard walls push high-ℓ down relative to the oscillator, which put them degenerate ( $N=2$ shell). The two caricatures bracket real nuclear wells.

Step 3. Fill with spin-½ fermions: capacities $2,6,10,2,14,6$ , cumulative $2,8,18,20,34,40$ (computed). Closures at 2, 8, 20 match the observed magic numbers; 18 and 34 are false closures the oscillator does not predict (its $1d$ and $2s$ merge into one $N=2$ shell of 12) — and nature sides with neither caricature beyond 20, where the spin-orbit force takes over the bookkeeping (one credit line; the shell-model story lives with the portraits of Effective Potential and the Centrifugal Barrier and the oscillator article). Why keep both caricatures: their disagreements localize the physics. Where box and oscillator agree (2, 8, 20), mean-field geometry alone explains nuclei; where they disagree (18/34 vs 20/40), the answer is sensitive to well shape and interactions — a bracketing technique worth more than either model alone. *Verified: all six $\beta$ root-found; ordering strict; capacities and cumulative sums; $E_{n0}\propto n^2$ exact; electron-in-1nm energies below (article verifier).*

The finite well: matching and the threshold

For $V(r)=-V_0$ inside $R$ , zero outside, and a bound state at $E=-E_B<0$ : interior wavenumber $k=\sqrt{2m(V_0-E_B)}/\hbar$ , exterior decay $\kappa=\sqrt{2mE_B}/\hbar$ . Per channel, the admissible pieces are forced:

u_\ell(r<R)\propto kr\,j_\ell(kr),\qquad u_\ell(r>R)\propto \kappa r\,h_\ell^{(1)}(i\kappa r) \qquad (2)

where: the interior keeps the regular family (origin in-domain), the exterior keeps the decaying continuation (the previous article's handover, cashed). Continuity of $u$ and $u'$ at $R$ — equivalently of the logarithmic derivative $u'/u$ , which cancels the free normalizations — is the entire quantization condition. This is precisely the matching craft of Wavefunction Matching Conditions, promoted to three dimensions with new alphabet.

Go deeper · Derivation B · 2 minThe s-wave transcendental, its 1D twin, and the existence threshold

Derivation BThe s-wave transcendental, its 1D twin, and the existence threshold

We match the $\ell=0$ channel explicitly, recognize the result as an old friend, and derive three-dimensions' signature theorem: no binding below a computable depth. The strategy is logarithmic-derivative matching plus one limit ( $E_B\to0$ ) taken honestly.

Step 1. At $\ell=0$ : $u_{in}=\sin kr$ (the wall $u(0)=0$ already enforced), $u_{out}=e^{-\kappa r}$ . Log-derivatives at $R$ : $k\cot kR$ inside, $-\kappa$ outside (both computed symbolically). The spectrum condition:

k\cot kR=-\kappa \qquad (3)

Step 2. The 1D twin. Equation (3) is identical to the quantization condition of the odd-parity states of a 1D finite well of width $2R$ — as it must be: the half-line problem with a hard wall at the origin is the odd sector of the symmetric 1D problem (the doctrine of Effective Potential and the Centrifugal Barrier, now cashed as a working identity). Every graphical-solution technique from Finite Square Well imports unchanged; only the even 1D solutions — the ones with $u(0)\ne0$ — are missing. And there hangs the physics: the 1D well's always-bound state is an even state. Discard the even sector and you discard the guarantee.

Step 3. The threshold. A bound state appears when (3) first admits a solution as $E_B\to0^+$ ( $\kappa\to0$ ): then $\cot kR\to0^-$ , i.e. $kR\to\pi/2$ with $k\to\sqrt{2mV_0}/\hbar$ . Existence therefore requires

V_0R^2>\frac{\pi^2\hbar^2}{8m} \qquad (4)

— the well must be deep and wide enough that a quarter wave fits inside with room to bend over. Numerically confirmed on both sides: at $0.97\,V_{th}$ the matching equation has no root; at $1.03\,V_{th}$ , exactly one (computed by root-counting sweeps). The $n$ -th s-state needs $(2n-1)^2$ times the threshold; ℓ-channels pay extra rent on top (the centrifugal barrier raising the effective floor — Effective Potential and the Centrifugal Barrier's landscape, one line). Why 1D and 3D disagree about weak binding: in 1D an arbitrarily weak well binds because the wavefunction can spread symmetrically with almost no curvature penalty — the even state's flat top. In 3D the origin wall forbids the flat top; the state must rise from zero and turn over inside the well. The threshold (4) is the price of the turn. (Two dimensions, for the curious, is the marginal case — any attraction binds, but exponentially weakly — an honesty line beyond this article's domain.) *Verified: matching (3) symbolic; threshold value $23.20$ MeV for the deuteron's parameters; root counts $0/1$ at $0.97/1.03\,V_{th}$ ; excited-state absence check below (article verifier).*

Figure

The deuteron matching: sine inside, exponential outside, one smooth joint — and most of the state in the tail

Limiting cases

case	result	meaning
$V_0\to\infty$	(3) forces $\sin kR\to0$ : box spectrum recovered	the hard sphere as a limit, continuously
$V_0R^2\to\pi^2\hbar^2/8m^+$	one s-state at $E_B\to0$ , tail $1/\kappa\to\infty$	threshold birth: infinite halo first, binding second
$V_0R^2<\pi^2\hbar^2/8m$	no bound state at all	3D novelty — attraction below the pass mark binds nothing
$\ell\ge1$	higher thresholds (centrifugal rent)	p-wave binding needs a visibly deeper well
$E_B\ll V_0$	exterior dominates; physics set by $\kappa$ alone	the shallow-bound universality the deuteron exemplifies
$\ell=0$ spectrum condition	$\equiv$ odd sector of 1D well, width $2R$	the origin wall as parity selection

Common trapThe well radius is a model choice — and the missing even states are the most common error

Two traps with real consequences. First: for the deuteron (and any shallow state), the extracted depth $V_0$ depends strongly on the assumed radius $R$ — this article's $33.7$ MeV holds at $R=2.1$ fm; choose $1.7$ fm and the matching returns a noticeably deeper well. Only combinations like $V_0R^2$ near threshold are robust; quoting "the" nuclear well depth without its radius is meaningless, and honest fits say so. Second: students importing 1D intuition write the even 1D solutions ( $\cos kr$ inside) for the spherical s-wave — but $u(0)=0$ kills them; they never existed in 3D. Every "the well always binds" claim and every $\tan$ -instead-of- $-\cot$ transcendental in a homework solution traces to this single illegal import. The half-line is the odd sector only.

Worked examples

Example 1 — an electron in a hard nanosphere. A colloidal quantum dot is crudely a hard sphere of radius $R=1$ nm. Ladder the spectrum.

Step 1. Scale: $E_{1s}=\hbar^2\pi^2/2m_eR^2=0.376$ eV (computed) — conveniently at the electron-volt scale that makes dots optically active.

Step 2. Ladder by zeros (all computed): $E_{1p}=(4.493/\pi)^2E_{1s}=0.769$ eV; $E_{1d}=1.266$ eV; $E_{2s}=4E_{1s}=1.504$ eV exactly (the $\ell=0$ tower's $n^2$ law). Level spacings of a few tenths of an eV — visible-light differences: the crude hard sphere already explains why dot emission color tracks size, with $E\propto1/R^2$ as the tuning law.

Step 3. Honesty: real dots have finite barriers, effective masses, and Coulomb corrections — the hard sphere is the geometry-only skeleton. Answer: $0.376,\,0.769,\,1.266,\,1.504$ eV for $1s,1p,1d,2s$ — the zero ladder in electron-volts.

Example 2 — the deuteron, end to end (the portrait's calculation). Given $E_B=2.2246$ MeV and force range $R=2.1$ fm, find the nuclear well.

Step 1. Inputs: $\mu c^2=469.46$ MeV (reduced mass, computed); $\kappa=\sqrt{2\mu E_B}/\hbar c=0.2316$ fm $^{-1}$ — the tail the previous article computed, now the matching's right-hand side.

Step 2. Solve (3) for $V_0$ (root-found): $V_0=33.7$ MeV, with interior wavenumber $k=0.872$ fm $^{-1}$ . Sanity: threshold $V_{th}=\pi^2(\hbar c)^2/8\mu R^2=23.2$ MeV (computed), so $V_0/V_{th}=1.45$ — the nuclear force clears its audition by less than half. Shallowness index $E_B/V_0=0.066$ : the state sits just under the rim of a deep well.

Step 3. Consequences, all computed: exactly one s-state (the second would need $9V_{th}=209$ MeV — root-count confirms none); exterior probability $P_{out}=0.63$ — most of the deuteron is outside the nuclear force, in the pure $e^{-\kappa r}$ tail; and no excited deuteron exists, in agreement with experiment. Answer: $V_0\approx34$ MeV at $R=2.1$ fm; one barely-bound, mostly-outside state — the nuclear force's first quantum portrait, and a threshold pass at 1.45.

Origins · 1 minHans Bethe and how this was discovered

Hans Bethe

1906–2005 · Cornell

The deuteron computation above is, in its essentials, the 1935 paper "Quantum Theory of the Diplon" by Bethe and Rudolf Peierls — written, legend has it, substantially on a train ride, months after Chadwick's neutron made the proton-neutron nucleus thinkable and Urey's deuterium gave it a name. (Bethe's electromagnetic side — the multipole-radiation systematics of his famous reviews — is told in an E&M-track portrait; this card is the nuclear-force debut.) The pair's insight was strategic modesty: nobody knew the nuclear force's shape, but a barely-bound state doesn't care. Take a square well of plausible range, feed in the measured binding energy, and everything observable at low energy — the tail, the photodisintegration cross section they computed, the scattering length — follows from $\kappa$ alone, because the state lives mostly outside the well, exactly the 63% this article computes. That "shape-independence" became the effective-range doctrine organizing all of low-energy nuclear physics. Bethe drove the program relentlessly through the 1930s — colleagues called the resulting trio of encyclopedic review articles by a nickname this card will not repeat, since their radiation chapters belong to the other track — and the habit of extracting model-independent physics from one measured number stayed his signature through stellar burning (Nobel 1967) and beyond. Peierls' own portrait duties lie elsewhere in this track; here he is the essential co-author on the calculation that taught physicists how to interrogate a force they could not see.

History

1932

— Chadwick discovers the neutron; Urey isolates deuterium — the two-body nucleus exists

1935

— Bethe and Peierls solve the deuteron as a finite spherical well: depth from binding energy

1936

— Shape-independence sharpens into effective-range theory (Bethe, Schwinger and others follow)

1949

— Shell models built on spherical-well and oscillator level ladders explain magic numbers with spin-orbit

1990

s — Halo nuclei discovered: extreme spherical-well tails at the drip lines

Going further

Griffiths & Schroeter §4.1.3 — the infinite spherical well at the source register.
Bethe & Peierls, Proc. R. Soc. A 148, 146 (1935) — the deuteron paper this article's example reruns.
Sakurai & Napolitano Ch. 6 — the matching machinery in its scattering continuation.
Krane, Introductory Nuclear Physics Ch. 4 — the deuteron and the well, at nuclear-physics register.

Provenance note: Equations are standard central-potential and spherical Bessel results; a verifier should confirm: (i) the radial-to-spherical-Bessel reduction (Griffiths §4.1.3, Sakurai §3.7); (ii) $j_0(x)=\sin x/x$ zeros at $n\pi$ and $j_1$ first zero $\approx4.4934$ solving $\tan x=x$ (Abramowitz & Stegun §10.1.1, Table 10.6); (iii) the infinite-box formula $E_{n\ell}=\hbar^2x_{n\ell}^2/(2ma^2)$ ; (iv) the $\ell=0$ finite-well condition $k\cot(ka)=-\kappa$ and its binding threshold $z_0=\pi/2$ i.e. $V_0a^2=\pi^2\hbar^2/(8m)$ (Cohen-Tannoudji Complement A_VII, Shankar Ch.12.6). All numbers quoted ( $\pi$ , $4.493$ , $5.763$ , $2\pi$ ) are tabulated spherical Bessel zeros. — Item (i) is the reduction owned by Spherical Bessel and Neumann Functions (its verifier); items (ii)–(iv) are executed in this article's verify script (zeros root-found rather than read from tables — $\pi$ , $4.4934$ , $5.7635$ , $2\pi$ , $6.9879$ , $7.7253$ ; the box formula evaluated for the 1 nm electron; the matching condition derived symbolically; the threshold bracketed numerically at $0.97/1.03\,V_{th}$ ), plus the deuteron solve ( $V_0=33.7$ MeV at $R=2.1$ fm, $\mu c^2=469.46$ MeV), the exterior-probability integral $0.63$ , and the no-excited-state check. Edition note: the infinite-well spectrum derivation formerly resident in the base of Spherical Bessel and Neumann Functions is owned HERE per the chapter map (relocation noted at both ends); scattering off the same well — phase shifts, resonances, the zero-energy story — belongs BY NAME to the scattering chapter (Phase Shifts, Low-Energy Scattering and the Scattering Length, Bound States and Zero-Energy Resonances), which inherits this article's matching setup at $E>0$ .

Why does the s-wave condition involve −cot rather than tan, like half the 1D solutions?

Because only the odd 1D sector survives. The symmetric 1D well's even states satisfy $k\tan(kR)=\kappa$ , odd states $k\cot(kR)=-\kappa$ . The spherical s-wave's $u$ must vanish at the origin — that is oddness on the doubled line — so the $\tan$ branch never appears. If your spherical-well solution contains a $\tan$ , you have accidentally counted states that violate $u(0)=0$ ; they are the ghosts of the missing even sector.

Does the threshold theorem contradict the delta-well, which binds despite being infinitely narrow?

No — the 1D delta well binds (its bound state is even, exploiting exactly the sector 3D lacks); a 3D delta-shell or contact interaction is genuinely different and does not bind below threshold strength. Take the well limit $R\to0$ , $V_0\to\infty$ with $V_0R^2$ fixed: binding survives only if $V_0R^2$ stays above $\pi^2\hbar^2/8m$ — the threshold is scale-invariant under this limit, which is why "zero-range" nuclear models must carry an auxiliary parameter (the scattering length) rather than a bare delta strength.

How sensitive is the deuteron's 63%-outside figure to the model?

Weakly — that is the shape-independence point. The exterior fraction is controlled by $\kappa R$ (here $0.49$ ): any potential of similar range reproducing the same binding energy gives a comparable split, because the exterior form $e^{-\kappa r}$ is model-independent and the interior merely has to join onto it smoothly. Realistic potentials (with repulsive cores and tensor forces) shift the number by some percent, not qualitatively. What is model-sensitive: $V_0$ itself, interior node structure, and anything probing $r<R$ physics — the warning box's caveat.

What happens to these bound states as I make the well shallower through the threshold?

The state's energy rises toward zero, its tail $1/\kappa$ stretches without bound, and at threshold the "state" becomes a zero-energy solution flat at infinity — not normalizable, no longer a bound state, but not nothing either: it leaves a fingerprint as a divergent scattering length and a huge low-energy cross section (the zero-energy resonance of the scattering chapter, named). Just below threshold you have a "quantum halo": arbitrarily large, arbitrarily weakly bound, universal in properties. Feshbach-tuned atoms walk this path in the laboratory daily.

Why do the box's magic numbers fail after 20 when the calculation seems exact?

The calculation is exact — for the model. The model errs twice: real nuclear wells are neither hard-walled nor oscillator-shaped (their flat bottom and diffuse edge reorder high levels), and, decisively, the nuclear spin-orbit interaction is strong and inverted, splitting each ℓ-level by MeV-scale amounts that regroup the ladder entirely. Box and oscillator agree on 2, 8, 20 — geometry-robust closures — and both fail beyond, which is precisely the evidence that pointed Goeppert Mayer and Jensen to the missing spin-orbit ingredient.

Can I use these solutions for a particle confined by something soft, like a trap that isn't square?

As brackets and as bases. Brackets: box (stiffest) and oscillator (softest reasonable) spectra sandwich most smooth traps' low levels — compare both before any serious computation. Bases: the box eigenfunctions $j_\ell(\beta_{n\ell}r/R)Y_\ell^m$ form a complete orthogonal set on the sphere's interior (Sturm-Liouville with weight $r^2$ — the orthogonality doctrine of Boundary Conditions and Normalization of Radial Wavefunctions), so soft-trap Hamiltonians are routinely diagonalized in exactly this basis. The zero ladder then reappears as grid points of a spectral method rather than as physical energies.