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In the Wu experiment, Chien-Shiung Wu and collaborators studied beta-minus decay of polarized cobalt-60,A thin cobalt-60 layer on a cerium magnesium nitrate crystal was cooled by adiabatic demagnetization to a few millikelvin. A polarizing magnetic field produced nuclear polarization. An anthracene scintillation counter and photomultiplier detected the emitted electrons; angular gamma ray observations from the daughter de-excitation monitored whether the nuclei remained polarized.^[ref]

The electrons were emitted preferentially opposite the parent nuclear spin. Reversing the nuclear polarization reversed the asymmetry, while warming the sample removed both polarization and the effect. Those controls distinguished a spin-correlated asymmetry from a fixed detector bias.

The relevant angular correlation is a polarized beta-decay angular distribution of the formUnder parity, ${\mathbf{p}}_e\to -{\mathbf{p}}_e$ but $\mathbf{J}\to \mathbf{J}$, so the spin-momentum correlation changes sign. Parity conservation would require equal rates for these two directions relative to the same spin. The observed nonzero coefficient therefore demonstrated parity violation in beta decay. An unpolarized source would average this correlation to zero and could not provide the same test.

For both modes, a leading weak quark transition in $K^{+}=u\bar{s}$ is $\bar{s}\to \bar{u}+W^{+}$, followed by $W^{+}\to u\bar{d}$, with the original $u$ acting as a spectator quark. This produces the four quarks needed for ${\pi }^{+}{\pi }^0$, using the $u\bar{u}$ component of ${\pi }^0$. Producing three pions additionally requires a $d\bar{d}$ pair from strong hadronization. A representative quark-line diagram draws it as gluon emission and splitting.

The three-pion assignment is $u\bar{d}$, $u\bar{d}$, and $d\bar{u}$, matching ${\pi }^{+}{\pi }^{+}{\pi }^{-}$. The extra soft pair creation is a schematic representation of hadronization, not a controlled perturbative prediction of the relative decay rates. Both processes are first order in the effective weak interaction.

There is an explicit error or hypothetical assumption to address: the original PDF assigns $0^{+}$ to all these mesons. Taken literally, the two-pion final state has $L=0$, because all particles have zero spin and the initial $J$ is zero. Its parity is $(+1{)}^2(-1{)}^0=+1$. For three spinless particles, use two Jacobi orbital angular momentum quantum numbers $l,L$. Coupling to total zero requires $l=L$, so the orbital parity $(-1{)}^{l+L}$ is $+1$ and the three-particle parity is also $+1$. Thus under the printed $0^{+}$ assignment, both channels conserve parity.

The named physical pions and kaons are instead pseudoscalar mesons with $J^P=0^{-}$. With those physical assignments,ThereforeThese two opposite-parity final states from the same physical kaon are a classic illustration of parity violation in a weak decay. The physical conclusion requires correcting the printed parity assignment explicitly.

A ground-state baryon contains three quarks in a spatially symmetric $L=0$ wavefunction, giving positive parity. Each quark is a spin one-half fermion, so the complete wavefunction must be antisymmetric under interchange of its complete quark labels. A physical baryon is a colour singlet, with the antisymmetric colour factorConsequently the combined spin-and-flavour factor must be symmetric. The product notation in the question organizes the degrees of freedom; it need not mean that spin and flavour are separately unentangled. In particular the octet requires a correlated sum of products.

For the three light flavours, the flavour symmetry decomposition isThe subscripts describe permutation symmetry, with $M$ denoting mixed symmetry. The addition of angular momentum of three spin one-half particles givesThe symmetric spin quartet has total $J=3/2$; the two spin doublets have $J=1/2$.

For $J=3/2$, the spin factor is symmetric, so the flavour factor must also be symmetric: this gives the spin-$3/2$ baryon decuplet. Its flavour content is four $\Delta$ states, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states, and one $\Omega$ state.

For $J=1/2$, the mixed-symmetry spin functions combine with the mixed-symmetry flavour functions to make one fully symmetric spin-flavour factor for each octet weight: this gives the spin-$1/2$ baryon octet, containing two nucleons, three Sigma baryons, one Lambda baryon, and two Xi baryons. The two mixed-symmetry octets in the raw flavour product are not two distinct ground-state octets; their permutation partners are correlated with the spin factors. An antisymmetric flavour singlet would require a fully antisymmetric spin factor to make a symmetric product, but three spin states drawn from a two-dimensional single-quark spin space have no such factor.

Equivalently, the symmetric three-particle representation of spin-flavour symmetry has dimension $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ and decomposes under flavour and spin intoThese are approximate multiplets: the larger strange quark mass breaks exact flavour symmetry and splits their energies.

Direct formation in $e^{+}{e}^{-}$ through one virtual photon selects the vector meson quantum numbers $J^{PC}=1^{--}$. The nonzero $K_S^0{K}_L^0$ mode and absence of $K_S^0{K}_S^0$ are consistent with this assignment: the two spin-zero kaons require odd orbital angular momentum for negative parity, whereas identical $K_S^0$ bosons cannot have an odd relative wave. This argument neglects the tiny CP violation in neutral kaons.

Assuming the $\rho \pi$ decays conserve isospin, their equal rates indicate $I=0$. The Clebsch-Gordan coefficients giveAn $I=1,I_3=0$ state lacks ${\rho }^0{\pi }^0$, while $I=2,I_3=0$ has squared coefficients $1:4:1$ rather than $1:1:1$. The branching fractions alone determine neither $J$ nor these relative phases; the single-photon formation and approximate strong interaction isospin conservation are essential additional inputs.

For a neutral $c\bar{c}$ bound state, meson parity and charge conjugation give $P=(-1{)}^{L+1}$ and $C=(-1{)}^{L+S}$. In spectroscopic notation $n{}^{2S+1}{L}_J$, the integer $n$ labels radial excitation, $L$ is orbital angular momentum, $S$ the total constituent spin, and $J$ their total angular momentum. The surviving level diagram contains eight states:

The spin-dependent splittings distinguish the singlet and triplet levels and the three triplet P levels. The radial excitation changes neither parity nor charge conjugation.

An observable that reverses sign under parity but has a nonzero expectation in a suitably prepared experiment tests parity violation. A polar momentum reverses, whereas an axial spin angular momentum does not. Thus $\mathbf{J}\cdot \mathbf{p}$ and helicity are parity-odd; an apparatus asymmetry must be separated from a physical asymmetry by reversing the preparation or analyser.

In the Wu experiment, polarized cobalt-60 nuclei preferentially emit beta decay electrons in one direction relative to the nuclear spin angular momentum. The distribution contains $A\mathbf{P}\cdot {\mathbf{p}}_e/E_e$; its sign changes under spatial inversion. Reversing the nuclear polarization reverses the observed asymmetry.^[ref]

In the Goldhaber experiment, the sign of the circular polarization of selected gamma rays determines the sign of the neutrino helicity. A preference for negative helicity is parity-odd because spatial inversion would produce the positive-helicity state. This provides a different experimental observable from the electron angular distribution.

The charged two-pion and three-pion channels can be assigned to the same spin-zero kaon by their parent mass and lifetime. For two spinless pions, total angular momentum zero forces their relative orbital angular momentum to zero, so the product of their intrinsic parities givesFor three spinless pions, use Jacobi orbital angular momenta $l,L$. Coupling them to total zero requires $l=L$, and henceThus both channels cannot conserve parity in decays of one parent with a definite parity. The weak decay interaction permits the observed coexistence.

Dalitz plots identify the three-body events, determine their internal kinematics, and allow amplitude and angular-momentum analyses through their populations and resonance structures. Together with the spin and parent-identity evidence, they support the parity assignment. They are not a standalone parity-odd observable: reversing all three momenta leaves every $s_{ij}$, and therefore every point on the plot, unchanged. A flat distribution by itself cannot establish parity violation without the channel and spin information.

Interference CP violation concerns the relative phase between direct decay and decay after mixing. For a common CP eigenstate final state, it can occur with both $|q/p|=1$ and $|{\bar{A}}_f/A_f|=1$, so neither of the first two mechanisms is needed. In that limit $|\lambda |=1$, butproduces different interference terms for $\lambda$ and ${\lambda }^{-1}={\lambda }^{\ast }$. With $\Delta m=m_h-m_{\ell }$,The term proportional to $\mathrm{Im}\lambda \sin (\Delta mt)$ consequently reverses sign. In the CP-conserving limit, $\lambda$ is the real CP parity of the final state in a consistent phase convention. For non-CP-eigenstate final states, compare the full coefficients in CP-conjugate rates rather than imposing this simplified criterion.

The two Weyl spinors transform in the inequivalent left- and right-chiral representations of the proper Lorentz group. They are selected by $P_{L,R}=(1\mp {\gamma }^5)/2$ and have different boost transformation laws; parity exchanges the two. A massive Dirac equation couples them, whereas in the massless limit they can propagate independently.

For the particular $u$ printed in this paper, transforming its supplied Dirac ${\gamma }^5$ givesThus the upper block is right chiral and the lower block left chiral in this convention. Many references use the opposite block order; neither convention changes physics, but they must not be mixed. Chirality is not generally the same as helicity for a massive electron. The weak charged current selects the left-chiral electron component, reducing to negative helicity at high energy.

Useful examples with no net quark flavor are the eta c meson ${\eta }_c=c\bar{c}$ and the eta b meson ${\eta }_b=b\bar{b}$. Examples with nonzero quark flavor are the kaon $K^{+}=u\bar{s}$ and the D meson $D^{+}=c\bar{d}$. These are pseudoscalar mesons: in the lowest quark model configuration the constituent spins combine to $S=0$, the relative orbital angular momentum is $L=0$, and the total angular momentum and parity are $J^P=0^{-}$. For the self-conjugate examples, meson parity and charge conjugation also gives $C=(-1{)}^{L+S}=+1$. Electrical neutrality alone would not imply zero net quark flavor; for example $K^0=d\bar{s}$ has strangeness.

Restoring reduced Planck constant, decay width and mean lifetime satisfy $\Gamma \tau =\hslash$. Thus the representative numbers correspond toThe mean lifetimes differ by at least $3.0\times 10^{10}$. Hidden-heavy-flavor quarkonium can undergo quark-antiquark annihilation into gluons, followed by hadronization. A $0^{-+}$ color-singlet state can annihilate into two gluons. A lightest open-flavor meson cannot simply annihilate through the strong interaction, which conserves net quark flavor. With no lighter accessible final state carrying that flavor, it must undergo a weak interaction changing a constituent's flavor. The small weak interaction amplitude, CKM matrix factors and available Lorentz-invariant phase space produce a much smaller decay width.

Qualification. A decay width of order tens of MeV is not a universal property of flavor-neutral pseudoscalar mesons. The neutral pion has $\Gamma \simeq 7.8\mathrm{eV}$ and decays predominantly by the electromagnetic interaction, a concrete counterexample. Its mass lies below the lightest allowed hadronic final state. The examples above illustrate the intended annihilation-versus-open-flavor comparison; the numerical scale depends on thresholds and selection rules.^[ref]

The chiral projectors $P_{L,R}=(1\mp {\gamma }^5)/2$ obey ${\gamma }^{\mu }{P}_L=P_R{\gamma }^{\mu }$, while the Dirac adjoints are ${\bar{\psi }}_L=\bar{\psi }{P}_R$ and ${\bar{\psi }}_R=\bar{\psi }{P}_L$. Consequently the cross terms vanish and the vector current isThe vector interaction preserves fermion chirality at a vertex and couples equally to left and right components. This statement about the vertex does not assert that a fermion mass term preserves chirality. Under parity, left and right components are exchanged; their equal coefficients allow a parity-invariant contraction with a true vector field.

An axial vector must couple to the axial current to obtain the analogous parity-invariant interaction. Since ${\gamma }^5{\psi }_L=-{\psi }_L$ and ${\gamma }^5{\psi }_R={\psi }_R$,The relative sign changes; the overall sign is conventional. A pure axial interaction can itself conserve parity when $A_{\mu }$ transforms as an axial vector. Simply contracting the original vector current with an axial-vector field would instead produce a parity-odd interaction.

Assume unpolarized beams, negligible electron and muon masses, a single scattering s-channel spin-one mediator with equal left and right vector couplings, and leading-order QED. Z boson exchange and radiative corrections are omitted. The angular factors are the Wigner D-matrix elements $d_{m{m}^{\prime }}^1({\theta }^{\star })$, with $m,m^{\prime }=\pm 1$. Up to common phases and normalization they are $(1+\cos {\theta }^{\star })$ for equal pair-spin projections and $(1-\cos {\theta }^{\star })$ for opposite projections. The four orthogonal helicity channels add incoherently, givingWith $s$ the squared center-of-mass energy, the normalized differential scattering cross-sections areThe second expression includes the solid angle Jacobian: a distribution per angle is different from one per $\cos {\theta }^{\star }$. Integration gives $\sigma =4\pi {\alpha }^2/(3s)$.

Parity reverses all momenta but leaves the spin axial vector unchanged, so it interchanges $L\leftrightarrow R$. The $LR\to LR$ and $RL\to RL$ amplitudes have equal magnitudes, as do $LR\to RL$ and $RL\to LR$. Thus every configuration has the same rate as its parity transform, proving parity conservation. The summed distribution is also forward-backward symmetric. Strictly, parity reverses both the beam and outgoing momenta, leaving their relative angle unchanged; forward-backward reflection by itself is not the definition of parity.

Assume the same left and right couplings $g_l,g_r$ at the electron and muon vertices. This lepton universality assumption is necessary to describe both vertices using just two numbers. Define $a=|g_l{|}^2$, $b=|g_r{|}^2$ and $z=\cos {\theta }^{\star }$. The two same-chirality channels have squared coupling weights $a^2,b^2$ and angular factor $(1+z{)}^2$; the other two each have weight $ab$ and factor $(1-z{)}^2$. Hence, for a common positive normalization $C$,If $|g_l/g_r|\ne 1$, then $a\ne b$, so a forward-backward term survives. At the amplitude level, parity interchanges the $g_l^2$ and $g_r^2$ channels; their unequal squared magnitudes show directly that the interaction violates parity. Equivalently the chiral current contains both vector current and axial current terms, with coefficients $(g_r+g_l)/2$ and $(g_r-g_l)/2$; unequal magnitudes preclude a pure parity eigen-current.

The odd-in-$z$ term is a useful signature within these assumptions, but it should not itself be called a parity-odd scalar: the unpolarized relative angle is unchanged when all momenta are reversed. If the two species have independent couplings, the coefficient is instead proportional to $(a_e-b_e)(a_{\mu }-b_{\mu })$, and a parity-violating coupling at only one vertex need not produce an unpolarized forward-backward asymmetry.

Explicit reference to laws is not necessary for every good scientific explanation. Laws remain indispensable in many explanations, but their occurrence alone is not sufficient either. The issue turns partly on whether a law means a contingent regularity of nature, any supporting generalization, or even a mathematical necessity.

The deductive-nomological model offers a strong positive case. Laws and circumstances entail the phenomenon, and at least one law is essential to the derivation. For example, Newton's second law and an inverse-square gravitational force, with appropriate initial conditions, yield predictions for an idealized two-body orbit. A universal relation unifies cases that otherwise look like unrelated observations. Carl Hempel and Paul Oppenheim explicitly develop adequacy conditions for such deductive explanations; their 1948 paper should not be read as a finished treatment of every statistical explanation.^[ref]

However, derivability does not always give the right explanatory direction. In a simplified geometrical optics model, a vertical pole of height $h$ on level ground casts a shadow of length $\ell =h/\tan \theta$ at solar elevation $\theta$. Given $h$ and $\theta$, one derives $\ell$; given $\ell$ and $\theta$, one derives $h$. The same geometrical relations enter both derivations. Yet the shadow does not normally explain why the pole was built to that height. The reverse calculation is a way to measure its height. The difference concerns which features produce or constrain which others, rather than deductive validity alone.

Mechanistic explanation is another challenge to a compulsory law-based format. A clock stopped because debris blocked a particular wheel, thereby preventing the mechanism from advancing. Inspecting the obstruction, removing it and observing restored operation can answer the causal question with considerable precision. The explanation describes organization and a relevant difference-making event without stating a universal law. It is more informative about this failure than the observation that every component obeyed Newtonian mechanics.

A defender of the covering-law model can reply that the causal account relies implicitly on regularities: debris that offers no resistance would not stop the wheel. This is a fair warning against treating a bare chronology as an explanation. But it does not establish that explanatory understanding requires identifying exceptionless fundamental laws, or that replacing the account with a complete microscopic derivation would improve the answer. Stable, tested relations in the actual operating regime may be sufficient.

An explanation can also turn on a mathematical constraint. Suppose an engineer asks why a connected network with four odd-degree junctions cannot be inspected by a route using every segment exactly once. Represent junctions as vertices and segments as edges of an undirected graph. Every intermediate visit pairs an incoming and an outgoing edge, so only the two endpoints of an open trail can have odd vertex degree. Four such vertices therefore exclude an Eulerian trail. The impossibility persists if segment lengths, construction materials or walking speeds change while the connections remain fixed. This is the obstruction in the Seven Bridges of Königsberg.^[ref]

The physical application requires that the graph represent the allowed routes correctly. Once that is granted, the explanatory necessity comes from parity, rather than a contingent physical law. Calling every mathematical constraint a law would preserve the original claim only by greatly widening its meaning. A critic may restrict “scientific” to causal explanations, but that restriction needs justification when applied mathematics answers a concrete engineering question.

Thus laws are powerful explanatory resources, and good explanations normally support relevant what-if judgements. Those judgements can be grounded in laws, mechanisms or mathematical constraints. Requiring explicit reference to laws imposes an unnecessarily narrow format; requiring some disciplined account of dependence is much more defensible.

The Sturm oscillation theorem requires respectively zero, one and two nodes of a wavefunction. Treating the vertical outer boundaries as hard walls, all three wavefunctions vanish there. The potential energy is asymmetric, so no even/odd parity should be imposed. At internal finite steps, both the wavefunction and its first derivative are continuous.

For $E_0$ and $E_1$, only the left well is classically allowed. Their wavefunctions oscillate there and continue as evanescent combinations through the barrier and right region, reaching zero at the far wall. The first is nodeless; the second has one node in the left well. For $E_2$ both wells are allowed, while the central barrier remains forbidden: the wavefunction oscillates in both wells, with an evanescent bridge and two nodes of a wavefunction overall. A node may lie inside a finite forbidden barrier when the matching coefficients have opposite signs. Precise node positions and relative amplitudes require dimensions and heights not specified by the qualitative drawing.

The original illustration solves an explicitly chosen model with the same ordering of allowed and forbidden regions. Its numerical parameters and detailed node positions are illustrative.

The parity operator acts as $P\psi (x)=\psi (-x)$, with $P^2=I$. For an even potential energy $[H,P]=0$, so energy eigenstates can be chosen even or odd, with parity eigenvalues $+1$ and $-1$. One-dimensional bound states are nondegenerate; the ground state is nodeless and even.

For the finite square well, square integrability excludes the growing exterior exponential, while even parity excludes the interior sine. Thus the interior solution is proportional to $\cos kx$, with matching decaying tails andContinuity at $a$ gives $B\cos ka=A{e}^{-qa}$ and $-Bk\sin ka=-qA{e}^{-qa}$. Their ratio yieldsThe conditions at $-a$ follow from the same parity.

For a narrow weak well, $ka\ll 1$, hence $q\simeq k^{2}a\simeq 2m{V}_{0}a/{\hslash }^2$ andThis is self-consistent because $|E_0|/V_0\simeq 2m{V}_0{a}^2/{\hslash }^2\ll 1$. It is the shallow finite-square-well energy, equivalently the attractive delta potential limit with integrated strength $2a{V}_0$. An arbitrarily weak attractive well of this form still has its even bound ground state.

The first excited state must be odd. Replace the interior cosine by $\sin kx$ and choose exterior tails with opposite signs. The matching condition becomes $q=-k\cot ka$. Its first root requires $\pi /2<ka<\pi$. At the binding threshold $q\to 0$, sois necessary and sufficient for the first odd bound state. Equality gives a zero-energy nonnormalizable threshold solution, not an additional bound state.

For a time-independent Hamiltonian operator, a symmetry operator $U$ satisfies $UH{U}^{†}=H$, or $[U,H]=0$. Thus applying $U$ to an energy eigenstate preserves its energy. Commuting Hermitian observables can be diagonalized within each degenerate energy eigenspace, allowing states to be labelled by symmetry quantum numbers.

A continuous unitary symmetry $U(\epsilon )=e^{-i\epsilon Q/\hslash }$ has Hermitian generator $Q$. Differentiating the invariance condition gives $[H,Q]=0$. The Heisenberg equation of motion then yieldswhen $Q$ has no explicit time dependence. spatial translations connect to momentum conservation, rotations to angular-momentum conservation, and time-translation invariance to energy conservation. A central potential energy conserves all components of angular momentum, but they cannot all be assigned sharp simultaneous values because they do not commute; $H,L^2,L_z$ are a compatible choice. Discrete reflection symmetry gives parity labels without a continuously variable generator. A quantum degeneracy can be protected by a symmetry, but a quantum degeneracy alone does not identify its physical origin.

The Sturm oscillation theorem gives three nodes of a wavefunction for the fourth bound state. reflection symmetry and nondegeneracy of one-dimensional bound states make it odd: one node of a wavefunction is at the centre, and the other two lie at opposite positions. In each flat allowed region the wavefunction is a sinusoid, with $k=\sqrt{2m(E-V)}/\hslash$; its wavelength is shortest in the central, deepest region. Outside the well it decays exponentially, with opposite signs on the two sides. Both $\Psi$ and ${\Psi }^{\prime }$ remain continuous at every finite step, as required by wavefunction matching at a finite potential step; only the curvature jumps. In particular, the wavefunction need not vanish at a step.

This original wavefunction matching at a finite potential step illustration uses freely chosen well dimensions and energies. It shows the required parity, node count, wavelength ordering and decaying tails; the source drawing supplies no numerical scale from which to fix exact node positions.

Let $\omega =\sqrt{k/m}$, $b=\sqrt{\hslash /(m\omega )}$ and $\xi =x/b$. With ${\psi }_0={\pi }^{-1/4}{b}^{-1/2}{e}^{-{\xi }^2/2}$, the next two normalized harmonic-oscillator wavefunctions are ${\psi }_1=\sqrt{2}\xi {\psi }_0$ and ${\psi }_2=(2{\xi }^2-1){\psi }_0/\sqrt{2}$. Their energies are $E_n=(n+1/2)\hslash \omega$, with respectively zero, one and two wavefunction nodes, alternating even and odd parity. The Schrödinger equation givesFor harmonic-oscillator wavefunction inflections, the inflections include both classical turning points and simple wavefunction wavefunction nodes. In dimensionless coordinates the complete sets areThe curvature changes sign at every listed point; none of these states has a wavefunction node coinciding with its turning points.

The anisotropic quantum harmonic oscillator is a sum of three commuting one-coordinate oscillator Hamiltonians. Substituting $\psi =X(x)Y(y)Z(z)$ and dividing by $XYZ$ gives a sum of three functions of independent coordinates, so each equals a separation constant. ConsequentlyFor a time-independent operator, conservation for arbitrary states follows from $[H,Q]=0$, since $d⟨Q⟩/dt=(i/\hslash )⟨[H,Q]⟩$. The first labelled operator is $i{L}_z/\hslash$; the second is $-L^2/{\hslash }^2$. The kinetic energy is rotationally invariant. The potential is invariant under rotations about $z$ precisely when (i) ${\omega }_x^2={\omega }_y^2$; full rotational invariance requires (ii) ${\omega }_x^2={\omega }_y^2={\omega }_z^2$. For positive oscillator frequencies these are equality of the frequencies themselves. Acting with the commutators on arbitrary test functions excludes anisotropic potentials in case (ii).

In the isotropic case let $b=m\omega /\hslash$. ThenThe polynomial $x$ is the first excited oscillator factor and the other two factors are ground-state Gaussians. The first excited spatial level has three independent choices, proportional to $x,y,z$. Odd parity gives $⟨x⟩=0$, and Gaussian integration yieldsFor oscillator uncertainty and allocation of classical energy, the $x$ mode has energy $3\hslash \omega /2$, whose classical amplitude is $A_x=\sqrt{3\hslash /(m\omega )}$, giving $\Delta x=A_x/\sqrt{2}$. If “same energy” instead means assigning the entire three-dimensional energy $5\hslash \omega /2$ to one classical coordinate, its amplitude is $\sqrt{5\hslash /(m\omega )}$ and the ratio is $\sqrt{3/10}$. A general three-dimensional classical orbit's $x$ amplitude is not fixed by total energy without specifying its allocation among modes.

The capacity of an isotropic oscillator shell, including the two electron spin states, is $(N+1)(N+2)$ electrons. Nine non-interacting electrons occupy shells as $2,6,1$ for $N=0,1,2$. ThusThe least positive excitation promotes an electron by one oscillator quantum. Rearrangements within the partly filled degenerate shell remain ground states and cost no energy.