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A 2013 RIKEN experiment used a fast calcium beam on beryllium, selected radioactive aluminium and silicon fragments, then struck carbon to produce magnesium isotopes with $N=22,24,26$. Emitted gamma rays supplied excitation energies; comparison across these isotopes supported deformation extending beyond the $N=20$ region toward $28$. This tests the limits of traditional magic numbers, not merely their arithmetic. The inferred shape depends on nuclear-structure interpretation of the energy levels, rather than direct imaging.^[ref]

Writing positive nuclear binding energy as $B(Z,N)$ and nuclear mass as $M(Z,N)$,The two-neutron nuclear separation energy is $S_{2n}(Z,N)=B(Z,N)-B(Z,N-2)$. In an independent-particle example, let the last occupied energy level have ${ϵ}_{-}=-8u$ and the next empty level ${ϵ}_{+}=-3u$, for a chosen positive energy unit $u$. With an unchanged core and no residual interactions, removal from the filled level costs $8u$, whereas removing the first particle added above it costs $3u$: the drop $5u$ equals the one-particle gap. These are invented model inputs, not measured nuclear values. Nuclear pairing, rearrangement of the nuclear mean field, and correlations complicate that identification in real masses; a finite difference of nuclear binding energies is not automatically a bare one-particle gap.

A nuclear level scheme represents nuclear energy levels and transitions among them. Gamma ray energies constrain pairwise differences; emission counts and coincidence measurements help resolve different schemes having the same difference multiset.

Nuclear magnetic resonance detects the response of atomic nuclei with a magnetic moment to a radio frequency magnetic field in a steady magnetic field. Its striking possibility is to measure a nuclear property through an ordinary electrical signal from a bulk sample: an enormous number of tiny magnetic moments can produce a measurable change in a receiving circuit.

For a nucleus with spin angular momentum $\mathbf{J}$, define its gyromagnetic ratio by $\mathbf{\mu }=\gamma \mathbf{J}$. Throughout this article $\gamma$ is angular, measured in radians per second per tesla; $\gamma /(2\pi )$ is the corresponding value in cycles per second per tesla. With ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$, the Hamiltonian is $H=-\gamma B_0{J}_z$. Adjacent energy levels differ byThus a measured resonance frequency and a calibrated magnetic field determine $|\gamma |$. Converting that into a nuclear magnetic moment also requires the spin quantum number: the maximum projection has magnitude $|\gamma |\hslash I$. This simple experiment does not by itself determine the sign of $\gamma$.

The Rabi molecular beam magnetic resonance experiment detects transitions through changes in a selected molecular beam. The bulk experiments below instead detect electrical consequences of nuclear magnetization in matter at ordinary density. A small Boltzmann distribution population imbalance is multiplied by a very large number of nuclei. For noninteracting spin one-half nuclei of number density $n$ and positive $\gamma$, their equilibrium magnetization isThis follows by weighting the two projections $\pm \gamma \hslash /2$ with their Boltzmann distribution probabilities. It explains both the smallness of the bulk polarization at ordinary temperature and the value of many nuclei contributing to one receiving circuit.

NMR spectroscopy measures resonant responses of atomic nuclei to applied electromagnetic radiation in a magnetic field. In residual-group monitoring of polyurethane formation, distinguishable signal integrals supply a relative residual-group readout; overlap or very weak signals can limit that inference.

An effective one-nucleon potential energy representing the average influence of the other nucleons. The nuclear shell model treats motion in this field first, then includes residual interactions. The field experienced by a proton also includes the repulsive Coulomb potential; proton and neutron energy levels need not have identical spacings.

Correlations that favor pairs of like nucleons coupled to zero total angular momentum in the simplest spherical picture. They help explain why an unpaired nucleon often dominates an odd nucleus's spin, and why neighboring nuclear separation energies show an odd-even staggering. Nuclear pairing belongs to the residual-interaction physics beyond merely filling independent energy levels. Mixing different occupations and deformation can also be necessary; the closed-core picture is most useful near suitable nuclear shell closures.

A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.

A nuclear shell groups single-particle energy levels for protons or neutrons in a nuclear mean field. Filling all states below a sufficiently large gap gives a nuclear shell closure. These shells describe quantum-state occupation rather than thin spatial layers. Their discrete filling effects explain departures from the smooth semi-empirical mass formula.

Why should adding one neutron be easy for one atomic nucleus and much less favorable for its neighbor? In the nuclear shell model, nucleons occupy discrete energy levels in a nuclear mean field. Filling all available states below a sufficiently large gap produces a nuclear shell closure. This connects nuclear separation energies with the counting of quantum states, and connects the spin of an odd atomic nucleus with its last unpaired nucleon.

The starting point is an approximate Hamiltonian operatorwhere $U_i$ is a confining potential energy, $a_i$ specifies nuclear spin-orbit splitting, and $V_{\mathrm{res}}$ contains interactions left out of the nuclear mean field. This is an effective organization of the nuclear many-body problem, rather than a derivation of the nuclear force. In a spherical approximation, addition of angular momentum combines the orbital angular momentum $l$ with spin $s=1/2$ to give $j=l\pm 1/2$; for $l=0$, only $j=1/2$ exists.

A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\dots ,j$. The Pauli exclusion principle permits one neutron in each such state and, separately, one proton. Thus a $j=7/2$ level holds eight neutrons, or eight protons; it does not hold sixteen identical neutrons. Here $n=1,2,\dots$ counts radial levels of the same $l$, starting with the lowest; it is not the hydrogenic principal quantum number. The letters $s,p,d,f,g,h,i$ mean $l=0,1,2,3,4,5,6$.

The splitting of one-nucleon energy levels by their spin-orbit interaction. With $\mathbf{J}=\mathbf{L}+\mathbf{S}$, the identity $2\mathbf{L}\cdot \mathbf{S}={\mathbf{J}}^2-{\mathbf{L}}^2-{\mathbf{S}}^2$ gives, in a state of definite $l,s=1/2,j$,The second branch requires $l\ge 1$. For the illustrative Hamiltonian operator $H_{\mathrm{so}}=-a\mathbf{L}\cdot \mathbf{S}/{\hslash }^2$ with $a>0$, the shifts are $-al/2$ and $+a(l+1)/2$. Thus the strong negative coefficient lowers the aligned $j=l+1/2$ branch. Its capacity is $2l+2$; the raised partner holds $2l$. Together they still hold $4l+2=2(2l+1)$ identical nucleons: the interaction rearranges states rather than creating them.

In this example the $d_{5/2}$ and $d_{3/2}$ shifts are $-a$ and $3a/2$, respectively, so the gap is $5a/2$. The capacity-weighted mean shift vanishes:This is a useful check on both the sign and state counting. A seventh identical particle cannot occupy another substate of the already full lower level. In a real atomic nucleus, other radial and orbital angular momentum levels can interleave with this doublet, and a radial coupling need not give the same effective $a$ for every orbital.

Optical pumping redistributes populations among internal quantum states by selective photon absorption followed by spontaneous emission. It can prepare atoms for measurement and participate in sub-Doppler cooling.

The practical aim is to make otherwise almost equally populated magnetic sublevels behave differently. Repeated absorption and emission can store a population imbalance in the long-lived ground state; a change in photon absorption or fluorescence then reveals a perturbation of that prepared distribution. This combines preparation and measurement: a weak radio wave can reorganize the populations, while abundant optical photons supply the readout.

An imbalance among ground state magnetic sublevels need not be a population inversion between optical energy levels. Optical pumping also supplies excitation in some lasers, where an appropriate population inversion matters for gain. The preparation problem here concerns the distribution within the ground state. A two-level system with only one lower quantum state cannot describe this redistribution.

For a bound energy eigenstate with a nondegenerate energy eigenvalue of an even one-dimensional potential, the wavefunction can be chosen with definite parity. A uniform electric field $\mathcal{E}$ gives the electric-dipole interaction $H^{\prime }=-q\mathcal{E}x$, which is odd under reflection. Hencesince $x|{\psi }_n(x){|}^2$ is odd. Second-order nondegenerate perturbation theory then gives the leading allowed Stark effect shift:Only opposite-parity intermediate states have nonzero position matrix elements. More generally, reflection maps $H(\mathcal{E})$ to $H(-\mathcal{E})$, so an analytic nondegenerate energy level is an even function of the field. This explains the absence of all odd orders and the usual quadratic Stark effect.

A level separation changes by the difference of the two quadratic coefficients. Symmetry alone does not guarantee a nonzero differential shift: in a displaced quantum harmonic oscillator every level moves by the same $-q^2{\mathcal{E}}^2/(2m{\omega }^2)$, leaving every spacing unchanged. Degenerate states would instead require degenerate perturbation theory and can split linearly; the nondegeneracy of one-dimensional bound states applies to ordinary isolated well levels.

Neglect hyperfine structure and use the electronic weak-field Zeeman splitting regime, where ${\mu }_{B}B$ is small compared with the spin-orbit interaction splitting. The sodium levels are $3s{}^2{S}_{1/2}$, $3p{}^2{P}_{1/2}$ and $3p{}^2{P}_{3/2}$. The Landé g-factor formula givesFor example, inserting $L=1,S=1/2,J=1/2$ gives $3/2+(3/4-2)/(2\cdot 3/4)=2/3$. Each sublevel with magnetic quantum number $m_J$ shifts by $g_J{m}_J{\mu }_{B}B$. Thus the lower level has shifts $\pm {\mu }_{B}B$, the upper $J=1/2$ level has shifts $\pm {\mu }_{B}B/3$, and the upper $J=3/2$ level has shifts $\pm 2{\mu }_{B}B$ and $\pm 2{\mu }_{B}B/3$.

The electric-dipole selection rule allows $\Delta J=0,\pm 1$ except $0\leftrightarrow 0$, and $m_u-m_l=0,\pm 1$. Both fine structure branches are allowed. The emitted photon energy isAll the allowed components of the sodium D-line Zeeman pattern are listed below; the last column is the photon shift in units of ${\mu }_{B}B$:These are four components of D1 and six of D2. Their strengths need not be equal; the table and arrows specify allowed transitions, not intensities.

The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so D2 has the larger zero-field photon energy. ^[ref] In the stated weak-field regime its most positive shift is $5{\mu }_{B}B/3$. Since $\lambda =hc/E_{\gamma }$, the shortest wavelength is therefore produced byThe tempting $m_u=3/2\to m_l=-1/2$ pair has a larger difference of level shifts but is forbidden because $|m_u-m_l|=2$.

LetThe normalized orbital energy eigenstates vanish outside the well and have one-particle energies $n^2\epsilon$. Define the normalized symmetric and antisymmetric combinations for $r\ne s$ byThe normalized spin functions are the singlet state ${\chi }_{00}=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ and the triplet states ${\chi }_{11}=|\uparrow \uparrow ⟩$, ${\chi }_{10}=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, ${\chi }_{1,-1}=|\downarrow \downarrow ⟩$. Their exchange energies are respectively $3J/4$ and $-J/4$.

For $0<J\ll \epsilon$, the three lowest distinct energy levels of the exchange-coupled spin-one-half particles in an infinite square well areThe spatially doubly occupied orbital can only occur with the singlet state. The next orbital sum is $1^2+2^2=5$, where either exchange symmetry is possible. The next unused sum is $2^2+2^2=8$, well separated under the small-coupling assumption. The paper's inequality is equivalent, up to a numerical factor, to $|J|\ll \epsilon$; the factor ${\hslash }^2$ in $J$ is essential because the spin operators carry units of angular momentum.

The sign of $\lambda$ is not explicitly specified. The table assumes the usual intended positive coupling in the displayed minus-sign convention. For $J<0$ with $|J|\ll \epsilon$, the same three sets of eigenfunctions occur, but the two rows at $5\epsilon$ exchange order. At $J=0$ those rows merge into a single energy level of multiplicity four; if three distinct energies are requested exactly at that point, the third is $8\epsilon$ with the single state ${\Phi }_{22}^{+}{\chi }_{00}$.

For distinguishable particles, the ordered spatial products are independent states, and neither singlet states nor triplet states are excluded from any orbital occupation. The unperturbed energy eigenvalues areFor $0<J\ll \epsilon$, the three lowest distinct energy levels arewhere $M=-1,0,1$. The first two levels do have zero first-order correction, since their spatial expectation value factorizes as $⟨1|x|1{⟩}^2=0$.

The last instruction in the original PDF is false for the third level. Although every diagonal product-state matrix element vanishes, the two orbital states have degenerate energy levels. Degenerate perturbation theory requires diagonalizing the perturbation within their span. For each $M$, its matrix in the ordered basis $(u_1{u}_2,u_2{u}_1)$ isThe eigenfunctions are ${\Phi }_{12}^{\pm }{\chi }_{1M}$, withThus the sixfold unperturbed level splits into two triplets. Its average shift is zero, and the diagonal expectation in an individual ordered product state is zero, but neither fact makes all its energy eigenvalue corrections zero. This is the degenerate orbital exchange splitting from a product perturbation; distinguishability allows both combinations rather than removing their coupling.

For negative $J$, the lowest two rows exchange order and the third unperturbed level is instead the twofold singlet orbital pair at $5\epsilon +3J/4$. It splits into ${\Phi }_{12}^{\pm }{\chi }_{00}$ with the same shifts $\pm \alpha D$, each nondegenerate. At $J=0$, the three lowest distinct unperturbed energies are $2\epsilon ,5\epsilon ,8\epsilon$, with multiplicities $4,8,4$. The first and third have zero first-order shift; the middle splits by $\pm \alpha D$, each fourfold. The printed zero-shift assertion therefore fails also in the zero-exchange limit.

By addition of angular momentum, combining the first two spin one-half particles gives $0\oplus 1$. Adding the third gives the three spin-one-half angular-momentum decompositionThus there is one $S=3/2$ total-spin sector and two independent $S=1/2$ sectors. Since the particles are distinguishable, no fermionic exchange symmetry restriction removes either copy. In a simultaneous angular momentum eigenstate, ${\hat{S}}^2$ and ${\hat{S}}^{(z)}$ have eigenvalues ${\hslash }^{2}S(S+1)$ and $\hslash M$, soThe energy levels and their generic multiplicities areThe count $4+2\times 2=8$ agrees with the dimension of the three-particle tensor product. These multiplicities must be added whenever numerical energy eigenvalues coincide. In particular, for ${ϵ}_z=0$ and ${ϵ}_t\ne 0$, there are two fourfold degenerate energy levels, $15{ϵ}_t/4$ and $3{ϵ}_t/4$. For ${ϵ}_t=0$ and ${ϵ}_z\ne 0$, the multiplicities at $M=-3/2,-1/2,1/2,3/2$ are $1,3,3,1$. If both coefficients vanish, the single energy level has multiplicity eight. Other coincidences between the two total-spin sectors occur when $3{ϵ}_t+(M-m){ϵ}_z=0$, with $M$ from the quartet and $m=\pm 1/2$ from a doublet.

Write the trap angular frequency as $\omega$ and the single-particle oscillator length as $\ell =\sqrt{\hslash /(m\omega )}$. The center of mass coordinate $X=(x_1+x_2)/2$ and relative coordinate $r=x_1-x_2$ give independent quantum harmonic oscillators, both of frequency $\omega$, with masses $2m$ and reduced mass $m/2$. Label their excitation numbers by $N$ and $n$. The unperturbed energy eigenvalue is $\hslash \omega (N+n+1)$.

Particle exchange leaves $X$ unchanged and sends $r$ to $-r$. The exchange symmetry of two identical spin-one-half fermions therefore requires even $n$ to accompany the spin-one-half singlet state, and odd $n$ to accompany a spin-one-half triplet state. For the contact shifts of two spin-one-half fermions in a harmonic trap, the delta potential depends only on $r$. For odd $n$, the relative wavefunction vanishes at $r=0$, so its contact shift is zero. Within the first-excited manifold, both perturbations are diagonal in the labels $(N,n,S,M)$: this is a perturbation diagonal within a degenerate eigenspace. For $n=0$, the relative oscillator length is $\sqrt{2}\ell$, giving the first-order nondegenerate perturbation theory shiftThe Zeeman splitting is $2{\mu }_B{B}_{z}M$ for total spin projection $\hslash M$. Thus the five lowest states, counted with multiplicity, haveunder the assumed conditions $|U|,2{\mu }_B|B_z|\ll \hslash \omega$. The three triplet states have slopes $-2{\mu }_B,0,2{\mu }_B$, while both singlet states are field independent at this order. The three triplet state branches coincide at zero magnetic field. Their crossing with the excited singlet state is allowed because the total-spin sectors do not mix. The odd relative wavefunctions are in fact unaffected by this delta potential at any coupling; the displayed singlet state shifts retain only first order.

For the magnetic flux spectrum of a quantum ring, the periodic boundary condition permits the normalized plane waves $L^{-1/2}{e}^{2\pi inx/L}$, with $n\in \mathbb{Z}$. Their energy eigenvalues areShifting $\alpha$ by an integer relabels $n$, so the ordered energy levels have period one in $\alpha$, or $2\pi \hslash /L$ in $a$. For $-1/2\le \alpha \le 1/2$, the lowest two areextended periodically. The lowest branches cross at every half-integer $\alpha$, with energy $E_L/4$. At integer $\alpha$, the first excited energy level has multiplicity two. The corresponding magnetic flux period is the electron magnetic flux quantum $h/e$.