The Three-Dimensional Isotropic Harmonic Oscillator

Hydrogen has a twin. Among all central potentials, exactly two are special enough that every classical orbit closes and every quantum spectrum stacks into degenerate towers: the Coulomb funnel $-k/r$ , and this article's subject — the isotropic harmonic well $V(r)=\tfrac12 m\omega^2 r^2$ . The hydrogen serials solved the first. Here we solve the second, and the payoff is a problem you can solve twice, by two routes that disagree about everything except the answers. Route one ignores the spherical machinery entirely: $x^2+y^2+z^2$ splits the Hamiltonian into three independent one-dimensional oscillators, so the whole chapter on Hermite functions imports wholesale and the energy is $E_N=\hbar\omega(N+\tfrac32)$ with $N=n_1+n_2+n_3$ . Counting how many ways three integers add to $N$ gives the degeneracy $\tfrac12(N+1)(N+2)$ — the triangular staircase $1,3,6,10,15,\dots$ Route two runs the central-potential program of this chapter: separate off $Y_\ell^m$ , reduce to the radial equation, tame the origin and the Gaussian tail, and find $E=\hbar\omega(2n_r+\ell+\tfrac32)$ with associated Laguerre polynomials of half-integer index doing the bookkeeping. The two ladders must be the same ladder, and reconciling them — $N=2n_r+\ell$ , with $\ell$ stepping down from $N$ by twos — is the article's centerpiece derivation. The count $\sum(2\ell+1)$ over that strange ℓ-menu lands exactly on $\tfrac12(N+1)(N+2)$ : rotational symmetry explains only $2\ell+1$ of it, and the excess is the fingerprint of a hidden symmetry, named here and prosecuted in the next serial. Every degeneracy, energy, radial function, and nuclear-scale number below is machine-verified.

Key takeaways

$V=\tfrac12 m\omega^2r^2$ is the other closed-orbit central potential (Coulomb is the first): $E_N=\hbar\omega(N+\tfrac32)$ , $N=0,1,2,\dots$ , zero-point energy $\tfrac32\hbar\omega$ — three half-quanta, one per dimension.
Cartesian route: three independent 1D oscillators; degeneracy = ways to write $N=n_1+n_2+n_3$ = $\binom{N+2}{2}=\tfrac12(N+1)(N+2)$ : $1,3,6,10,15,\dots$ (verified through $N=8$ ).
Spherical route: $R_{n_r\ell}\propto r^\ell e^{-r^2/2b^2}L_{n_r}^{\ell+1/2}(r^2/b^2)$ with oscillator length $b=\sqrt{\hbar/m\omega}$ ; $E=\hbar\omega(2n_r+\ell+\tfrac32)$ (each form checked against the radial equation symbolically).
The dictionary: $N=2n_r+\ell$ , so a shell contains $\ell=N,N-2,\dots,1$ or $0$ — all of the same parity $(-1)^N$ ; $\sum(2\ell+1)$ over that menu $=\tfrac12(N+1)(N+2)$ exactly (proved both parities).
Only $2\ell+1$ of the degeneracy is symmetry's due (Degeneracy and the Role of the Magnetic Quantum Number); the rest is "accidental" — owned by the SU(3) serial next door, the oscillator's answer to hydrogen's SO(4).
Doubled by spin, the shell sizes $2,6,12,20$ fill cumulatively to $2,8,20,40$ : the first three nuclear magic numbers fall out of a bare harmonic cage (with $\hbar\omega\approx41A^{-1/3}$ MeV — $16.3$ MeV for $^{16}$ O, computed).
Key idea: one potential, two exact solutions — and the mismatch between the degeneracy symmetry demands and the degeneracy the oscillator delivers is a measurement of hidden symmetry.

The core insight

The isotropic well $\tfrac12 m\omega^2r^2$ solves in Cartesian coordinates as three 1D oscillators ( $E_N=\hbar\omega(N+\tfrac32)$ , degeneracy $\tfrac12(N+1)(N+2)$ ) and in spherical coordinates as a central-potential problem ( $N=2n_r+\ell$ , Laguerre radial functions) — the same tower described by two incompatible-looking quantum-number grammars, forced to agree state by state.

Where this lives today

Ultracold-atom traps are this article realized in the laboratory: magnetic and optical traps are parabolic near their minima, so every degenerate-gas experiment begins by writing $\hbar\omega(N+\tfrac32)$ for its confined cloud, and "trap frequency" is the first number quoted in any BEC paper. Quantum-dot spectroscopy fills oscillator shells with electrons two at a time and watches the $1,3,6$ degeneracies appear as conductance plateaus — artificial atoms with tunable $\omega$ . And in nuclear structure the oscillator basis remains the working expansion basis of ab-initio many-body calculations (the "model space" of no-core shell model computations is counted in oscillator quanta $N_{\max}$ ), so the functions verified in this article are, quite literally, the coordinates in which modern supercomputers store nuclei.

The physical picture

Picture the potential first: a bowl whose steepness is the same in every direction — equipotentials are perfect spheres, and the spring constant $m\omega^2$ has no favorite axis. That isotropy is a choice, and it is what makes this article central-potential physics; a general anisotropic trap $\tfrac12m(\omega_x^2x^2+\omega_y^2y^2+\omega_z^2z^2)$ still separates in Cartesian coordinates but has no conserved $\hat{L}^2$ and none of the ℓ-structure below.

Figure

One bowl, two coordinate grammars: Cartesian slicing vs spherical shells

Now picture the spectrum. Equally spaced rungs $\hbar\omega$ apart — the oscillator's signature, inherited intact from one dimension — but each rung is a shell of states: one state on the floor, three on the first rung, six on the second, ten on the third. The triangular numbers are a Cartesian counting fact. But re-sort each shell by angular momentum and a stranger pattern appears: the $N=2$ shell is one $\ell=0$ state plus five $\ell=2$ states; $N=3$ is three $\ell=1$ plus seven $\ell=3$ . Different ℓ — states with visibly different radial landscapes in the sense of Effective Potential and the Centrifugal Barrier — land on exactly the same energy. Hydrogen pulled the same trick with $2s/2p$ , and there the culprit was a hidden conserved vector. The oscillator's culprit is a hidden conserved tensor, and the next serial arraigns it.

Figure

The shell ladder: triangular degeneracies, re-sorted into ℓ-blocks of one parity

The Cartesian solution

The potential is a sum of one-variable pieces, so the Hamiltonian splits:

\hat{H}=\underbrace{\frac{\hat{p}_x^2}{2m}+\frac{m\omega^2\hat{x}^2}{2}}_{\hat{H}_x}+\hat{H}_y+\hat{H}_z \qquad (1)

where: each piece is exactly the one-dimensional oscillator of Quantum Harmonic Oscillator: Analytic Solution, with the same frequency $\omega$ because the well is isotropic. Commuting pieces mean product eigenstates and additive eigenvalues — the separation logic of the ONED chapter, run three times in parallel. Nothing about this route is spherical; it would work just as well (better, in fact) for an anisotropic trap.

\psi_{n_1n_2n_3}(x,y,z)=\psi_{n_1}(x)\,\psi_{n_2}(y)\,\psi_{n_3}(z), \qquad E=\hbar\omega\Big(N+\tfrac32\Big),\quad N=n_1+n_2+n_3 \qquad (2)

where: each factor carries a Hermite polynomial and a Gaussian (their anatomy belongs to Hermite Polynomials and Oscillator Eigenfunctions — never re-derived here), and each contributes its own $\tfrac12\hbar\omega$ zero-point toll: the $\tfrac32$ is three half-quanta, one per confined dimension, the confinement-cost arithmetic of Zero-Point Energy tripled. Only the sum $N$ enters the energy — and that immediately makes levels degenerate, because many triples $(n_1,n_2,n_3)$ share a sum.

Derivation AEnergy ladder and the triangular degeneracy count

We derive the spectrum and count each level's states from the Cartesian factorization alone. The strategy: additivity gives the ladder for free; the degeneracy is a counting problem — how many ordered triples of non-negative integers sum to $N$ — solved by the stars-and-bars argument, then cross-checked programmatically through $N=8$ . This count is the number the spherical route must reproduce, so we prove it carefully.

Step 1. Since $[\hat{H}_x,\hat{H}_y]=[\hat{H}_y,\hat{H}_z]=[\hat{H}_x,\hat{H}_z]=0$ , simultaneous eigenstates of the three pieces exist and span (each piece's eigenbasis is complete in its own variable; products span the tensor space — the expansion license of Completeness and Resolutions of the Identity). On the product state (2), $\hat{H}$ returns $\hbar\omega(n_1+\tfrac12)+\hbar\omega(n_2+\tfrac12)+\hbar\omega(n_3+\tfrac12)=\hbar\omega(N+\tfrac32)$ .

Step 2. Count states at fixed $N$ . Distribute $N$ identical quanta among 3 distinguishable axes: writing the quanta as $N$ stars and separating the axes by 2 bars, every arrangement of $N+2$ symbols choosing which 2 are bars is one state — $\binom{N+2}{2}$ arrangements. Why the trick is legitimate: a state is exactly an ordered triple $(n_1,n_2,n_3)$ ; the stars-and-bars string encodes the triple bijectively (stars before the first bar = $n_1$ , between bars = $n_2$ , after = $n_3$ ). Nothing quantum is being smuggled — the physics ended when the energy became a function of the sum alone.

Step 3. Evaluate: $g(N)=\binom{N+2}{2}=\tfrac12(N+1)(N+2)$ , giving $1,3,6,10,15,21,\dots$ — the triangular numbers. For $N=2$ the six states are $(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)$ , listed exhaustively because the reconciliation derivation below must find these same six in spherical clothing. Verified: $g(N)=\binom{N+2}{2}$ against brute-force enumeration for $N\le8$ ; the $N=2$ census enumerated explicitly (article verifier).

The spherical solution

Run the chapter's program instead. Isotropy means $[\hat{H},\hat{L}^2]=[\hat{H},\hat{L}_z]=0$ (Separation of Variables in Spherical Coordinates), so eigenstates factor as $R(r)Y_\ell^m$ and the radial function obeys the effective 1D problem of Effective Potential and the Centrifugal Barrier, with $u=rR$ :

-\frac{\hbar^2}{2m}u''+\Big[\frac{m\omega^2r^2}{2}+\frac{\ell(\ell+1)\hbar^2}{2mr^2}\Big]u=Eu \qquad (3)

where: the landscape now rises at both ends — the centrifugal wall at the origin (for $\ell\ge1$ ) and the quadratic wall at infinity. Every state is bound; there is no continuum at all, the cleanest possible contrast with the Coulomb well's finite ladder plus continuum. The natural length is the oscillator length $b=\sqrt{\hbar/m\omega}$ — where zero-point kinetic pressure balances the spring, the same balance-point logic that produced $a_0$ in The Bohr Radius and Characteristic Atomic Scales — and in units of $\xi=r/b$ the equation cleans up completely.

Go deeper · Derivation B · 2 minThe radial solution: Gaussian tail, centrifugal entrance, Laguerre ladder

Derivation BThe radial solution: Gaussian tail, centrifugal entrance, Laguerre ladder

We solve equation (3) by the same three-act structure as the hydrogen solve of The Hydrogen Atom: Coulomb Radial Equation — peel the asymptotics, then demand the middle terminate — and land on associated Laguerre polynomials of half-integer upper index, with $E=\hbar\omega(2n_r+\ell+\tfrac32)$ . Each printed radial form is verified by direct substitution into (3).

Step 1. Asymptotics. As $r\to\infty$ the spring dominates: $u''\approx(r/b^2)^2u$ , so $u\sim e^{-r^2/2b^2}$ — a Gaussian tail, faster than hydrogen's exponential because the confinement stiffens with distance (the decaying branch selected, the growing one discarded, by exactly the normalizability doctrine of Boundary Conditions and Normalization of Radial Wavefunctions). As $r\to0$ the centrifugal term rules and the indicial analysis of The Radial Equation (cited, not re-run) forces $u\sim r^{\ell+1}$ .

Step 2. Peel both behaviors: $u=\xi^{\ell+1}e^{-\xi^2/2}v(\xi)$ with $\xi=r/b$ . Substituting and changing variable to $s=\xi^2$ turns the equation for $v$ into Kummer's (confluent hypergeometric) equation — a two-term recursion in powers of $s$ , exactly parallel to the Coulomb recursion, with one decisive difference: the recursion steps in $\xi^2$ , not $\xi$ , because the potential is even. Steps of two in a series are parity bookkeeping made visible.

Step 3. Terminate or perish. The non-terminating series grows like $e^{+\xi^2}$ , overwhelming the Gaussian and violating normalizability — the same catastrophe-then-quantization mechanism as hydrogen's Derivation B, cited, not re-proved. Termination after $n_r$ terms demands

E=\hbar\omega\Big(2n_r+\ell+\tfrac32\Big),\qquad n_r=0,1,2,\dots \qquad (4)

and the surviving polynomial is the associated Laguerre $L_{n_r}^{\ell+1/2}(\xi^2)$ :

R_{n_r\ell}(r)\;\propto\;\xi^{\ell}\,e^{-\xi^2/2}\,L_{n_r}^{\ell+1/2}(\xi^2),\qquad \xi=r/b \qquad (5)

Why the upper index is half-integer: hydrogen's recursion ran in $\rho$ and produced $L^{2\ell+1}$ with integer index; here the $s=\xi^2$ substitution halves every exponent's bookkeeping, leaving $\ell+\tfrac12$ upstairs. The convention hazard flagged in Hydrogenic Wavefunctions and Quantum Numbers (whose Laguerre convention card this article adopts wholesale) bites twice as hard at half-integer index — older tables differ by normalization factors — so every form below is checked by substitution, not table lookup.

Step 4. Bookkeeping. $n_r$ counts radial nodes (polynomial degree, the Sturm grammar of The Node Theorem in one line); parity is $(-1)^\ell$ from $Y_\ell^m$ times an even radial factor. The lowest few verified forms: $R_{00}\propto e^{-\xi^2/2}$ (no node, pure Gaussian); $R_{01}\propto\xi e^{-\xi^2/2}$ ; $R_{10}\propto(\tfrac32-\xi^2)e^{-\xi^2/2}$ (one radial node at $\xi=\sqrt{3/2}$ ); $R_{02}\propto\xi^2e^{-\xi^2/2}$ . Verified: forms (5) substituted into (3) with $E$ from (4) — residual identically zero, symbolically, for $(n_r,\ell)=(0,0),(1,0),(0,1),(0,2),(1,1),(2,0),(0,3),(1,2)$ ; normalization integrals finite by quadrature; node counts by inspection of the polynomial (article verifier).

Read equation (4) against the Cartesian ladder $E=\hbar\omega(N+\tfrac32)$ and the dictionary is forced:

N=2n_r+\ell \qquad (6)

where: a shell of the Cartesian ladder decomposes, in spherical language, into all pairs $(n_r,\ell)$ with $2n_r+\ell=N$ . Radial quanta cost two units each; angular quanta cost one. That 2:1 exchange rate is the oscillator's signature (hydrogen's rate, $n=n_r+\ell+1$ , is 1:1) and it dictates which ℓ live together: $\ell=N,\,N-2,\,\dots$ down to 1 or 0 — one parity per shell.

Go deeper · Derivation C · 1 minReconciling the two counts: the ℓ-menu and the identity that certifies it

Derivation CReconciling the two counts: the ℓ-menu and the identity that certifies it

We prove that the spherical census of shell $N$ reproduces the Cartesian count exactly: $\sum_{\ell\in\{N,N-2,\dots\}}(2\ell+1)=\tfrac12(N+1)(N+2)$ . This one identity certifies that no state was lost or double-counted in the translation between quantum-number grammars — and it is where the "accidental" degeneracy becomes quantitative. (The base article for the next serial opens with this same equivalence; by chapter ownership the derivation lives here, once, and is cited there.)

Step 1. Fix the menu. From (6), $\ell=N-2n_r$ with $n_r=0,1,\dots,\lfloor N/2\rfloor$ : the shell contains $\ell=N,N-2,\dots$ , ending at $0$ (even $N$ ) or $1$ (odd $N$ ). Each $\ell$ appears exactly once — there is no $(n_r,\ell)$ collision, because $n_r$ is determined by $\ell$ within a shell.

Step 2. Even case $N=2M$ : $\sum_{k=0}^{M}\big(2(2k)+1\big)=4\cdot\tfrac{M(M+1)}2+(M+1)=(M+1)(2M+1)=\tfrac12(2M+1)(2M+2)$ . ✔

Step 3. Odd case $N=2M+1$ : $\sum_{k=0}^{M}\big(2(2k+1)+1\big)=4\cdot\tfrac{M(M+1)}2+3(M+1)=(M+1)(2M+3)=\tfrac12(2M+2)(2M+3)$ . ✔ Both close to $\tfrac12(N+1)(N+2)$ . Why this is a measurement and not a formality: rotational symmetry guarantees only that each $(n_r,\ell)$ level is $(2\ell+1)$ -fold degenerate — Degeneracy and the Role of the Magnetic Quantum Number's theorem, consumed. It says nothing about why $(n_r,\ell)=(1,0)$ and $(0,2)$ coincide. The ratio of actual to guaranteed degeneracy — $6$ versus $1$ or $5$ separately in the $N=2$ shell — is the size of the hidden invariance group beyond rotations, and the next serial identifies it as SU(3).

Step 4. The Cartesian↔spherical translation is a unitary change of basis within each shell. Exhibit ( $N=1$ ): $\psi_{001}\propto z\,e^{-r^2/2b^2}\propto rY_1^0$ , and $(\psi_{100}\pm i\psi_{010})/\sqrt2\propto(x\pm iy)e^{-r^2/2b^2}\propto rY_1^{\pm1}$ — the three Cartesian states reassemble into the three $m$ -states of $\ell=1$ , nothing left over. Verified: the identity programmatically for $N\le8$ and symbolically for both parities (sympy summation); the $N=1$ combinations against the Cartesian-form definitions of $Y_1^m$ (article verifier).

Limiting cases

case	result	meaning
$N=0$	one state, $E=\tfrac32\hbar\omega$ , pure Gaussian	isotropic zero-point ball; no node, no angular structure
$\ell=N$ ( $n_r=0$ )	nodeless "circular" states $\propto r^Ne^{-r^2/2b^2}$	maximal angular momentum per shell — the oscillator's Rydberg-like states
$\ell=0$ tower	$E=\hbar\omega(2n_r+\tfrac32)$ : every other rung	s-states climb by $2\hbar\omega$ — radial quanta cost double
$\omega\to0$	$b\to\infty$ , spectrum collapses toward continuum	free particle recovered; shells merge into the $E>0$ continuum of the next article
anisotropic $\omega_x\ne\omega_y\ne\omega_z$	Cartesian route survives; $\ell$ dies	degeneracy shattered — isotropy, not oscillation, made the shells
one dimension	$g(N)=1$ , zero-point $\tfrac12\hbar\omega$	degeneracy is a 3D luxury: it needs directions to trade quanta between

Common trapThe oscillator quantum numbers are not hydrogen quantum numbers

Three standing confusions, all bookkeeping. First: the oscillator's shell label $N$ counts quanta, not hydrogen's $n$ — the menu here is $\ell\le N$ in steps of two with one parity per shell, versus hydrogen's all $\ell<n$ , both parities mixed. Second: spectroscopic labels like $1p$ or $2s$ in nuclear physics count "the $k$ -th level of that ℓ" ( $1p$ = lowest $\ell=1$ ), so nuclear $2s$ means $n_r=1,\ell=0$ — hydrogen's $2s$ means something else entirely; when reading shell-model diagrams, translate before comparing. Third: the degeneracy $\tfrac12(N+1)(N+2)$ does not include spin — double it for electrons or nucleons. Sloppy translation between these three conventions is the leading source of wrong shell-filling answers on exams.

Worked examples

Example 1 — the N=2 shell, censused in both languages. Verify state-by-state that the six Cartesian states re-sort into $\ell=0\oplus\ell=2$ .

Step 1. Cartesian census (from the first derivation): $(2,0,0),(0,2,0),(0,0,2)$ — two quanta on one axis — and $(1,1,0),(1,0,1),(0,1,1)$ — one quantum on each of two axes. Six states, all at $E=\tfrac72\hbar\omega$ .

Step 2. Spherical census from $N=2n_r+\ell$ : $(n_r,\ell)=(1,0)$ contributes $2\ell+1=1$ state; $(0,2)$ contributes $5$ . Total $1+5=6$ . ✔

Step 3. Make the translation explicit for two members. The trace combination $\psi_{200}+\psi_{020}+\psi_{002}$ is proportional to $(2r^2/b^2-3)e^{-r^2/2b^2}$ — spherically symmetric, one radial node: this is $R_{10}Y_0^0$ , the $\ell=0$ state. The difference $\psi_{200}-\psi_{020}\propto(x^2-y^2)e^{-r^2/2b^2}$ carries the angular factor of $Y_2^2+Y_2^{-2}$ : an $\ell=2$ member. The remaining four independent traceless-quadratic combinations complete the $\ell=2$ quintet. Answer: $6=1+5$ , and the sorting is the decomposition of a symmetric $3\times3$ tensor into trace + traceless parts — a preview of exactly the operator structure the SU(3) serial builds.

Example 2 — a nuclear cage and an electron trap, same ladder, twelve orders of magnitude apart. Compute the shell structure's real-world scales.

Step 1. Nuclei. The empirical mean field of a nucleus is oscillator-like with $\hbar\omega\approx41\,A^{-1/3}$ MeV (fitted to nuclear radii); for $^{16}$ O, $\hbar\omega=41/16^{1/3}=16.3$ MeV (computed) — rung spacing on the MeV scale. Doubling each shell for spin: capacities $2,6,12,20$ , cumulative $2,8,20,40$ . The first three magic numbers — 2, 8, 20 — fall out of a bare harmonic cage (computed ladder). Beyond 20 the bare oscillator predicts 40 where nature says 28: the fix is the strong spin-orbit term, one sentence of credit to the shell-model story told in Effective Potential and the Centrifugal Barrier's portrait and this article's, and owned by the fine-structure/nuclear serials.

Step 2. Electrons. A semiconductor quantum dot traps electrons in a nearly parabolic well with $\hbar\omega$ of a few meV. At $\hbar\omega=3$ meV (free-electron mass for honesty; real dots renormalize $m$ ), the oscillator length is $b=\sqrt{\hbar/m\omega}=5.04$ nm and the ground-state cloud has $\sqrt{\langle r^2\rangle}=\sqrt{3/2}\,b=6.17$ nm (both computed); the ladder reads $4.5,\,7.5,\,10.5$ meV for $N=0,1,2$ — and dot spectroscopy sees exactly the $1,3,6$ shell-filling pattern, earning dots the name "artificial atoms." Answer: $\hbar\omega=16.3$ MeV and magic $2,8,20$ for the nucleus; $b=5.04$ nm and a meV ladder for the dot — one Hamiltonian, both ends of the energy axis.

Origins · 1 minJ. Hans D. Jensen and how this was discovered

J. Hans D. Jensen

1907–1973 · Hamburg and Heidelberg

Jensen ran this article's ladder in reverse: he started from the experimental magic numbers and climbed down to the oscillator. In 1948-49, reviewing the stubborn stability data — nuclei with 2, 8, 20, 28, 50, 82, 126 protons or neutrons refusing to behave like the liquid drops of prevailing theory — the Heidelberg theorist took seriously what almost everyone dismissed: that nucleons fill shells of a mean potential, the bottom of any smooth well being harmonic. The bare cage delivers 2, 8, 20 (this article's arithmetic) and then fails; Jensen's 1949 insight — arrived at independently and nearly simultaneously with Maria Goeppert Mayer in Chicago, whose own route through the level ordering is told in Effective Potential and the Centrifugal Barrier — was that a strong inverted spin-orbit splitting slices the high-ℓ oscillator levels and regroups them, turning 40 into 28 and rescuing the entire sequence. The two never raced: they wrote the definitive textbook Elementary Theory of Nuclear Shell Structure together and split the 1963 Nobel Prize (with Wigner). Jensen's half-joking summary of the discovery — that the nucleus is "a harmonic oscillator with a twist" — is literally this article plus one term. He spent the rest of his career at Heidelberg, where the physics institute's address is now Im Neuenheimer Feld — on a street grid as regular as his level scheme.

History

1926

— Schrödinger's second paper solves the 1D oscillator; the 3D isotropic case follows by separation

1940

— Jauch and Hill work out the 2D and 3D oscillator's hidden degeneracy algebra explicitly

1948

— Magic-number data sharpen: liquid-drop theory cannot explain 2, 8, 20, 28, 50, 82, 126

1949

— Goeppert Mayer (Chicago) and Haxel, Jensen, Suess (Heidelberg) publish the spin-orbit shell model

1958

— Elliott builds nuclear rotation out of oscillator SU(3) multiplets

1963

— Nobel Prize to Goeppert Mayer and Jensen for the shell model

Going further

Griffiths & Schroeter §4.1 and Problem 4.38 — the 3D oscillator at the source register.
Sakurai & Napolitano §3.7 — the oscillator among central potentials, with the degeneracy discussion this chapter's next serial expands.
Cohen-Tannoudji Vol. I, Complement $G_{VII}$ — the radial Laguerre solution and the $N=2n_r+\ell$ relation in full detail.
Shankar §12.6 — both routes side by side.

Provenance note: Spectrum $E_N=\hbar\omega(N+3/2)$ and degeneracy $\tfrac12(N+1)(N+2)$ are standard: Griffiths 3rd ed. Problem 4.38; Sakurai & Napolitano 2nd ed. §3.7; Shankar 2nd ed. §12.6. Radial Laguerre solution and $N=2n_r+\ell$ relation: Cohen-Tannoudji Vol. I Complement $G_{VII}$ . The $SU(3)$ dynamical symmetry is treated in Sakurai and in J.-M. Lévy-Leblond. A verifier should recompute: (1) the stars-and-bars degeneracy count; (2) the consistency identity $\sum(2\ell+1)=\tfrac12(N+1)(N+2)$ for several $N$ ; (3) the radial-equation reduction to associated Laguerre form; (4) $\langle r^2\rangle$ via the virial theorem. All numbers (degeneracies 1,3,6,10,15; zero-point $\tfrac32\hbar\omega$ ) are exact textbook values. — All four items are executed in this article's verify script (counts through $N=8$ plus both-parity symbolic identity; radial forms by direct ODE substitution; $\langle r^2\rangle=(N+\tfrac32)\hbar/m\omega$ confirmed by quadrature on the $N=2$ s-state, which is the virial statement $E=m\omega^2\langle r^2\rangle$ ). Edition note: the equivalence-of-counts derivation that the next serial's base sketches is owned HERE by chapter ownership (this article's third derivation); Degeneracy and SU(3) Symmetry of the 3D Oscillator consumes it by citation. The SU(3) algebra itself is deferred BY NAME to that article; the Cartesian 3D box (a different separable problem) is a gap-queue topic, teased in prose only.

Why does the same potential solve in two coordinate systems when hydrogen only solves in one?

Hydrogen separates in spherical (and, famously, parabolic) coordinates but not Cartesian — $1/r$ is not a sum of one-variable functions. The oscillator's $r^2=x^2+y^2+z^2$ is such a sum, which is a rare algebraic accident. Multi-separability is itself a symptom of extra symmetry: potentials separable in more than one coordinate system carry conserved quantities beyond $\hat{L}$ , which is precisely why both closed-orbit potentials — Coulomb (spherical + parabolic) and oscillator (spherical + Cartesian) — have degeneracies beyond $2\ell+1$ .

Is the zero-point energy 3/2 ħω physically observable?

Its consequences are. You cannot extract it, but confinement energetics are real: compressing a trapped cloud (raising $\omega$ ) costs $\tfrac32\hbar\,\Delta\omega$ per particle in zero-point alone, measurable in ultracold-gas thermodynamics. The deeper point is per-dimension additivity: each confined direction charges $\tfrac12\hbar\omega$ , so a pancake trap (one stiff axis) pays most of its zero-point toll in that one direction — the arithmetic behind engineering quasi-2D quantum gases.

Where exactly does the Cartesian-to-spherical dictionary live? Is there a formula for the overlap coefficients?

Within each shell the two bases are connected by a unitary matrix whose entries are computable (they are essentially Clebsch-Gordan-like brackets, tabulated in nuclear-structure references as "Moshinsky brackets" in the two-particle case). This article exhibits the $N=1$ and two $N=2$ rows explicitly; the general machinery is specialist bookkeeping — the physics point is only that the matrix is unitary and block-diagonal in energy, so nothing depends on which grammar you count in.

Why does ℓ skip by two within a shell instead of taking every value?

Parity. The radial series ran in $\xi^2$ (even potential ⇒ two-term recursion in even powers), so the radial factor is even about the origin and the state's parity is carried entirely by $Y_\ell^m$ : $(-1)^\ell$ . But a Cartesian shell state has parity $(-1)^{n_1+n_2+n_3}=(-1)^N$ — every state in the shell shares it. So only $\ell$ of parity $(-1)^N$ can appear: $\ell=N,N-2,\dots$ The skip-by-two menu is parity conservation written as a selection rule on the shell's contents.

The oscillator's degenerate ℓ differ by 2 while hydrogen's differ by 1 — is that meaningful?

Yes, and it diagnoses the hidden symmetry's type. Hydrogen's degeneracy connects $\ell$ to $\ell\pm1$ : its hidden generator (the Runge-Lenz vector, a rank-1 tensor) changes ℓ by one. The oscillator's degeneracy connects $\ell$ to $\ell\pm2$ : its hidden generators are rank-2 (quadrupole-like) operators built from two ladder operators. Reading the ℓ-spacing of a degenerate multiplet tells you the tensor rank of the symmetry that created it — the next article makes this precise.

Do real nuclei actually look like harmonic wells?

Only near the middle. A real nuclear mean field is flatter at the center and cuts off at the surface (Woods-Saxon shape); the oscillator is its workable caricature — exactly solvable, right level count, increasingly wrong level spacing for high shells. The professional compromise: use oscillator wavefunctions as the expansion basis (they are complete and analytically friendly) while diagonalizing more realistic Hamiltonians in that basis. That is why $b$ and $\hbar\omega=41A^{-1/3}$ MeV survive in state-of-the-art computations that long ago abandoned the potential itself.

What happens to the shells if the trap is slightly anisotropic?

The Cartesian solution survives verbatim with $E=\hbar(\omega_xn_1+\omega_yn_2+\omega_zn_3)+\tfrac{\hbar}{2}(\omega_x+\omega_y+\omega_z)$ , but the shells split: states of the former shell $N$ now spread over energies set by how their quanta are distributed. $\hat{L}^2$ ceases to commute with $\hat{H}$ , ℓ stops being a good label, and the $(2\ell+1)$ blocks dissolve too — a clean illustration that all the degeneracy here, rotational and accidental alike, is symmetry's child. Perturbative splitting patterns for weak anisotropy are a standard exercise in the approximation chapter ahead.