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When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

The ground-state energy $E_0$ is the energy eigenvalue attained by a ground state. For a self-adjoint operator $H$ bounded below, distinguish this attained value from the always-defined spectral bottom $E_{\ast }=\inf \sigma (H)$. When a ground state exists, $E_0=E_{\ast }$. Some usage calls $E_{\ast }$ the ground-state energy even when it is unattained; here an unattained bottom is explicitly identified as an infimum. The quantum variational principle statesApproaching this infimum with trial wavefunctions does not ensure that a minimizing wavefunction exists. ^[ref].

For the whole-line free particle with mass $m>0$,Its spectrum is continuous and contains no energy eigenvalues represented by normalizable wavefunctions. The Fourier transform makes $H$ multiplication by ${\hslash }^2{k}^2/(2m)$; a fixed energy restricts $k$ to a set of zero length, on which a nonzero square-integrable function cannot be supported. ^[ref]. An explicit family of normalized Gaussian wave packets demonstrates how the bottom is approached:The formula follows from ${\psi }_s^{\prime }=-x{\psi }_s/(2s^2)$, integration by parts, and $⟨x^2{⟩}_s=s^2$. Every finite width gives positive kinetic energy. At zero energy, $H\psi =0$ would require an affine function on the whole line, which is never a nonzero normalizable wavefunction. Spreading a Gaussian wave packet indefinitely produces an unattained infimum, rather than a zero-energy ground state.

An additive choice of energy origin changes the numerical ground-state energy without changing the ground-state subspace. For $H^{\prime }=H+CI$, every energy eigenvalue becomes $E^{\prime }=E+C$, while all differences $E_j-E_i$ stay fixed. Time evolution gains only the global phase $e^{-iCt/\hslash }$. This makes excitation costs and spectroscopic differences more intrinsic than the isolated number $E_0$.

F. Duncan M. Haldane's 1988 tight-binding model adds complex second-neighbour hopping and opposite sublattice offsets to a honeycomb lattice. Directed phases break time-reversal symmetry while each primitive cell has zero net magnetic flux. Smaller loops can enclose magnetic flux; ordinary Bloch states remain available.^[ref]

Extend the nearest-neighbour model of graphene, ignoring physical spin and interactions. With bond length $a$, $t_1>0$, $t_2\ge 0$ and ordered sublattice basis $(A,B)$, chooseThe following Bloch Hamiltonian defines $\varphi$; reversing directed second-neighbour hopping sends $\varphi \mapsto -\varphi$:The Pauli matrices act on sublattice amplitudes, as in sublattice pseudospin in graphene. The identity matrix term $d_{0}I$ shifts both energy eigenvalues equally at each momentum; it changes neither the eigenvectors nor the direct separation $E_{+}-E_{-}$. A genuine band insulator additionally needs ${\min }_{\mathbf{k}}{E}_{+}>{\max }_{\mathbf{k}}{E}_{-}$. We use $t_2/t_1=0.08$ in the figures, within the original paper's sufficient nonoverlap restriction $|t_2/t_1|<1/3$.^[ref]

A confining Quantum Hamiltonian on $x>0$ with $a>0$, potential $a\sqrt{x}$ and a Dirichlet boundary condition at zero has natural length $\ell =({\hslash }^2/(2ma){)}^{2/5}$ and energy $E_{\ast }=({\hslash }^2{a}^4/(2m){)}^{1/5}$. Rescaling $x=\ell y$ gives $H/E_{\ast }=-d^2/d{y}^2+\sqrt{y}$, so every energy eigenvalue scales as $E_{\ast }$. The potential derivative diverges at the wall even though the potential itself is finite there.

For the infinite square well of length $a$, the energy eigenvalues are $E_n=n^2{E}_1$, with $E_1={\pi }^2{\hslash }^2/(2m{a}^2)$. At $t_{\ast }=\pi \hslash /E_1$, the dynamical phase is $e^{-i{E}_n{t}_{\ast }/\hslash }=(-1{)}^n$. Since the midpoint parity operator has eigenvalues $(-1{)}^{n+1}$ in the same complete orthonormal basis, the time-evolution operator is exactly $U(t_{\ast })=-R_a$ on every state. At $2t_{\ast }$ every phase is one, giving a full revival.

Thus a state becomes its reflected copy, up to a global phase, at the half revival. Its position probability density is reflected, while its mean momentum changes sign because $P{R}_a=-R_{a}P$. For an initially reflection-symmetric probability density, the initial and half-revival position distributions agree even when their probability currents point in opposite directions.

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

Small splittings of energy eigenvalues due to interactions involving nuclear spin.

The infinite square well is an ideal confinement model with potential energy $V(x)=0$ for $0<x<L$ and infinite walls outside. The walls impose Dirichlet boundary conditions $\varphi (0)=\varphi (L)=0$ and a vanishing exterior wavefunction. Inside, the time-independent Schrödinger equation isThe first wall removes the cosine term; the second requires $\sin (kL)=0$, hence $k=n\pi /L$. The $n=0$ solution is identically zero and cannot satisfy wavefunction normalization. Each energy eigenstate has $n-1$ interior nodes of a standing wave, at $x=jL/n$ for $j=1,\dots ,n-1$. More curvature costs more kinetic energy, explaining the $n^2$ growth. The positive ground-state energy is a confinement effect: a constant zero-energy wavefunction cannot satisfy both walls and remain normalized.

For an electron with mass $m_e=9.1093837139\times 10^{-31}\mathrm{kg}$ in a box of length $L=1\mathrm{nm}$, use $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. These constants are from the 2022 CODATA adjustment. ^[ref]. Substitution givesDoubling $L$ quarters every energy eigenvalue; doubling $m$ halves them. This scaling makes the particle in a box useful for estimating confinement energies before attempting a more realistic calculation. A finite square well permits exterior tails for bound states and has only finitely many bound energy eigenstates. For an electron, the scalar box calculation assumes fixed spin and no spin-dependent interactions.

For the ground state, the probability of finding the particle in the middle half of the box is fully computable:Thus half the available length contains about $81.8\%$ of the probability. The figure contrasts the signed wavefunctions with their nonnegative probability densities: a negative lobe is an amplitude with opposite quantum phase, not negative probability.

An equal quantum superposition of the first two energy eigenstates gives a simple example of moving density:Its density period is $T=h/(E_2-E_1)\approx 3.66608\mathrm{fs}$ for the electron box above. At $t=0$ the cross term enhances the left side; at $T/4$ it vanishes; at $T/2$ it enhances the right side. More quantitatively,which takes the values $0.924413$, $0.5$, and $0.075587$ at those times. Orthogonality keeps the total probability one throughout. An energy measurement still returns $E_1$ or $E_2$ with equal probabilities: changing position density arises from relative quantum phase, not changing populations. A mixed state describing a statistical mixture with those populations has the constant probability density $({\varphi }_1^2+{\varphi }_2^2)/2$; it lacks this matter-wave interference term.

Near a stable equilibrium bond length $r_e$, a Taylor expansion of a diatomic molecule's potential energy gives $U(r)\simeq U(r_e)+k(r-r_e{)}^2/2$, with $k=U^{\prime \prime }(r_e)>0$. Treat the internal stretch as a harmonic oscillator of reduced mass, initially omitting molecular rotation. If replacing an isotope leaves this local potential energy curvature unchanged, thenHere $c$ is the speed of light and ${\tilde{\nu }}_{\mathrm{harm}}$ a spectroscopic wavenumber; use consistent length units for $c$ and ${\tilde{\nu }}_{\mathrm{harm}}$. The quantum harmonic oscillator gives adjacent energy eigenvalues separated by $h\nu$, with $h$ the Planck constant. Consequently a vibrational photon absorption can probe the same frequency scale, subject to electric-dipole selection rules.

For an explicitly approximate isotope example, assign the atomic masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, ${}^{35}\mathrm{Cl}$, and ${}^{37}\mathrm{Cl}$ the mass-number values $1u,2u,35u,37u$, respectively, with $u$ the unified atomic mass unit. These are stipulated approximations to individual atomic masses, not measured masses of molecules or natural-abundance averages. They giveThe symbol $\mathrm{D}$ means deuterium, ${}^2\mathrm{H}$. Doubling the light partner's approximate atomic mass therefore produces a large predicted shift, whereas changing the already heavy chlorine partner by two mass units produces a much smaller one. A quantitative comparison with infrared spectroscopy requires actual isotope-specific atomic masses, fitted band parameters, and uncertainties, rather than identifying any arbitrary peak with ${\tilde{\nu }}_{\mathrm{harm}}$.

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

Consider a fixed proton and one electron, omitting spin, fine structure and the Lamb shift. Let $e>0$ be the elementary charge, $a_0$ the Bohr radius in this fixed-nucleus model, and $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$. Because the electron's electric dipole moment is $\mathbf{d}=-e\mathbf{r}$, its electric-dipole interaction isThe first excited hydrogen eigenspace has a common unperturbed energy eigenvalue $E_2$. Write $|s⟩=|2,0,0⟩$ and $|p_m⟩=|2,1,m⟩$ in the hydrogen energy eigenbasis. The electric-dipole selection rules for hydrogen leave only the $s,p_0$ coupling within this eigenspace. Each diagonal matrix element vanishes by parity, but that does not make the entire perturbation vanish. Degenerate perturbation theory requires its eigenvalues in the whole eigenspace. With the usual real wavefunction phases the off-diagonal position matrix element is $⟨s|z|p_0⟩=-3a_0$.^[ref]

The factor three can be checked directly. Set $\rho =r/a_0$ and use normalized radial wavefunctions and spherical harmonics:The angular integral is $\int Y_{00}^{\ast }\cos \theta Y_{10}d\Omega =1/\sqrt{3}$. Including $z=r\cos \theta$ and the radial volume factor $r^2$ then givesReversing the quantum phase of $|p_0⟩$ reverses this sign, with no change to the predicted energy eigenvalues.

Thus the nontrivial two-dimensional block and its determinant equation areConsequently, for the displayed sign convention,The two remaining energy eigenstates $|p_{+1}⟩,|p_{-1}⟩$ have zero first-order shifts. The slopes correspond to opposite electric dipole moments, $⟨d_z⟩=+3e{a}_0$ and $-3e{a}_0$. The electric field selects combinations of opposite parity that already have these dipole expectations at leading order. This is why applying first-order nondegenerate perturbation theory separately to $2s$ and $2p_0$, and obtaining two zero diagonal entries, misses the splitting.

The approximation requires $e{a}_0|\mathcal{E}|$ small compared with the separation to other principal quantum numbers. For real hydrogen atoms, it also requires the relevant electric-dipole interaction to dominate the small splittings omitted here, including the Lamb shift. At sufficiently small electric fields a pre-existing nonzero $2s$–$2p$ separation must be retained before taking the limit. A simple two-state illustration with unperturbed separation $\delta >0$ has shifts relative to its meanFor $3e{a}_0|\mathcal{E}|\ll \delta$, the lower branch moves by $-(3e{a}_0\mathcal{E}{)}^2/\delta +O({\mathcal{E}}^4)$ relative to its zero-field value. At $\delta =0$ the splitting is linear. This interpolation explains the importance of the hydrogen energy-level degeneracy; it is not a complete model of the fine structure multiplet.

Molecular rotation changes a molecule's orientation. Its rotational kinetic energy depends on the arrangement and masses of its chemical atoms. A rigid rotor freezes internal distances to isolate this motion; rotational spectroscopy tests the resulting energy eigenvalues through transitions between quantum states. The molecular beam electric resonance experiment provides a complementary probe of the coupling to an electric field.

Molecular vibration changes internal separations and bond angles, unlike the fixed internal molecular geometry of a rigid rotor. Its coupling to molecular rotation makes measured rotational constants depend on the vibrational quantum state.

A diatomic molecule has one internal stretching coordinate: its two atomic nuclei approach and separate about an equilibrium bond length. The diatomic bond stretch turns this motion into a particularly useful model. The local potential energy curvature supplies the restoring force, while the reduced mass accounts for both moving partners. Thus the same chemical bond can vibrate more slowly after an isotope substitution even when its electronic structure is nearly unchanged.

This separation makes infrared spectroscopy informative: transition frequencies constrain chemical bond stiffness, and isotope scaling of diatomic vibration tests the mass dependence. The University of Florida HCl and DCl FTIR experiment illustrates the measurement. Detectability also requires vibrational infrared activity; having vibrational energy eigenvalues alone does not guarantee an infrared absorption line.

An energy eigenvalue is nondegenerate when its eigenspace is one-dimensional. Its normalized eigenstate is then unique up to a complex phase.

A nuclear energy level is an energy eigenvalue of an atomic nucleus. Differences between levels give nuclear excitation energies and the approximate gamma ray energies of transitions.

For a periodic spin-one-half Heisenberg model $H=-(J/{\hslash }^2){\sum }_{n=1}^N{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, the fully aligned state has $E_0=-NJ/4$. In its single-excitation subspace, let $|n⟩$ denote a down spin at site $n$ in an otherwise up background. Direct action of each bond givesThe normalized translation eigenstates are $|q⟩=N^{-1/2}{\sum }_n{e}^{iqna}|n⟩$, with $q=2\pi \ell /(Na)$ modulo $2\pi /a$. Their energy eigenvalues are $E_0+J[1-\cos (qa)]$. This is an exact one-magnon calculation, not a many-magnon harmonic approximation. Near $q=0$ the energy above $E_0$ is $J{a}^2{q}^2/2$.

The wavenumber restriction comes from requiring a periodic plane-wave coefficient pattern, equivalently requiring this vector to be an eigenvector of a one-site translation with eigenvalue $e^{-iqa}$. The identity $T^N=I$ alone imposes no restriction on a general state: it already holds throughout the ring Hilbert space. An arbitrary finite sum with coefficients $e^{iqna}$ still returns after $N$ translations, but need not be a one-site translation eigenstate unless $e^{iqNa}=1$.

The WKB approximation, also called JWKB, describes a short local de Broglie wavelength in a slowly varying potential energy. For $p(x)=\sqrt{2m[E-V(x)]}$ in a classically allowed region, its two local waves areThe phase follows the classical momentum; its amplitude varies to conserve probability current. In a forbidden region, putting $\kappa (x)=\sqrt{2m[V(x)-E]}$ gives $\psi \simeq {\kappa }^{-1/2}[C{e}^{{\int }^x\kappa /\hslash }+D{e}^{-{\int }^x\kappa /\hslash }]$, with boundary conditions selecting the physical combination.

The WKB validity condition is $\hslash |p^{\prime }|/p^2\ll 1$, with the analogous condition on $\kappa$. It fails at a turning point where $E=V$. For a simple turning point, locally linearizing V yields an Airy function; matching fixes the relative amplitudes and a $\pi /4$ phase. Two such turning points give the two-turning-point WKB quantization conditionFor an opaque barrier, the WKB tunnelling exponent instead givesThe factor two distinguishes probability from amplitude. Matching determines the prefactor; the exponential alone is not accurate near the barrier top. These formulas connect bound state energy eigenvalues and quantum tunnelling to classical actions. ^[ref]

The Zeeman effect comes from $-\mathbf{\mu }\cdot \mathbf{B}$. For an electron, the orbital and spin magnetic moments are approximately $-{\mu }_B\mathbf{L}/\hslash$ and $-2{\mu }_B\mathbf{S}/\hslash$, respectively. This explains both the sign and the relative factor two in the term linear in $B$. The second term is the spin-orbit interaction: the electron's motion through the atomic electric field produces a magnetic interaction in its rest frame. The relativistic correction, including Thomas precession, and the radial expectation value are incorporated into $K$. This effective Hamiltonian retains the linear Zeeman splitting and spin-orbit coupling, while omitting diamagnetic and hyperfine structure corrections.

Define the angular momentum ladder operators by $L_{\pm }=L_x\pm i{L}_y$ and similarly for $S_{\pm }$. Direct expansion givesso adding $L_z{S}_z$ gives the required scalar product. The off-diagonal terms change $m_{\ell }$ and $m_s$ in opposite directions. Consequently $J_z=L_z+S_z$ commutes with $H$, and $m_j$ remains a good quantum number at every magnetic field.

In the given product eigenbasis, the diagonal matrix elements areFor $\ell =1$ and $s=1/2$, the only off-diagonal couplings are between $(1,-1/2)$ and $(0,1/2)$, and between $(0,-1/2)$ and $(-1,1/2)$. Each relevant orbital ladder factor is $\hslash \sqrt{2}$ and each spin ladder factor is $\hslash$. Multiplying by $K/{\hslash }^2$ gives $\sqrt{2}K$. ThereforeAll other off-diagonal entries vanish either by the ladder endpoints or by conservation of $m_j$. This derives every entry of the displayed matrix.

The stretched states have no mixing and yieldThe $m_j=1/2$ block is $\left(\begin{array}{cc}-K& \sqrt{2}K\\ \sqrt{2}K& a\end{array}\right)$, while the $m_j=-1/2$ block is $\left(\begin{array}{cc}-a& \sqrt{2}K\\ \sqrt{2}K& -K\end{array}\right)$. Taking the trace and determinant of each $2\times 2$ block gives the remaining four energy eigenvalues:The subscripts indicate $m_j=\pm 1/2$ and the superscripts indicate the upper or lower root. These formulas are the Zeeman spectrum for orbital angular momentum one and remain valid for either sign of $K$.

For the usual ordering $K>0$, including sodium, the zero-field states follow fromThe $j=3/2$ quartet lies at $K$ and the $j=1/2$ doublet at $-2K$, separated by $3K$. In a weak magnetic field, the Landé g-factors giveThese label the curves near $B=0$. In a strong magnetic field, the Paschen-Back effect makes $(m_{\ell },m_s)$ the appropriate approximate labels:for the mixed states, with the stretched-state formulas already exact. The branch connections for $B>0$ areWithin a fixed $m_j$, the nonzero coupling prevents a crossing; this is an avoided crossing. Curves with different $m_j$ may cross because the Hamiltonian does not couple them.

For the sodium estimate, the zero-field frequency separation fixesThe lower $3s$ doublet has shifts $\pm a$. The Sodium D-line Zeeman pattern gives photon-frequency shifts $a/h$ times $\{\pm 2/3,\pm 4/3\}$ for D${}_1$ and $\{\pm 1/3,\pm 1,\pm 5/3\}$ for D${}_2$. Thus the highest-frequency D${}_1$ component and lowest-frequency D${}_2$ component approach each other by $3a/h$. Extrapolating the weak-field Zeeman splitting givesso the crossover occurs at a field of order $10\mathrm{T}$, where $a\sim K$ and the weak-field approximation is becoming marginal.

The exact eigenvalues refine the estimate within this Hamiltonian. The relevant photon energies, with a common zero-field reference omitted, are $E_{+}^{(-)}+a$ and $E_{-}^{(+)}-a$. The corresponding electric dipole transitions have $\Delta m_j=+1$ and $-1$, respectively. Equating them givesIts positive solution is $a/K=\sqrt{10}/3$, hencein the retained model. The crossing is between optical transition frequencies; the participating upper states themselves need not be degenerate.

Use the uncoupled angular momentum eigenstates $|m_L,\uparrow ⟩$ and $|m_L,\downarrow ⟩$, with $m_L=-1,0,1$. For the Hamiltonian operator actually written in the paper, these are the energy eigenstates, withThe initial $J=3/2$ states are not all energy eigenstates of this operator. To find their energy populations, first perform the Clebsch-Gordan decomposition. The extremal states are $|3/2,3/2⟩=|1,\uparrow ⟩$ and $|3/2,-3/2⟩=|-1,\downarrow ⟩$. Applying the total angular momentum lowering operator to the first givesSimilarly,The initial preparation is an equal incoherent mixture of the four coupled states. Each extremal state contributes population $1/4$ to its one energy eigenvalue, while each nonextremal state contributes $2/3$ and $1/3$ of its own $1/4$ population to the two displayed components. The resulting Zeeman splitting is thereforefor nonzero $B_z$. The relative populations are $\boxed{3:2:2:2:3}$ in this order; the two zero-energy states each have population $1/12$. These Zeeman populations from an unpolarized angular-momentum multiplet remain constant under subsequent evolution by the stated static Hamiltonian operator, although off-diagonal quantum coherence can evolve.

There is a distinct weak-field model in which a retained spin-orbit interaction keeps $J=3/2$ isolated and one projects the magnetic perturbation into that quartet. The Landé g-factor is then $4/3$, giving four shifts ${\mu }_B{B}_z(\pm 2,\pm 2/3)$ with equal populations. That calculation requires the additional projection or dominant spin-orbit interaction; it is not diagonalization of the full magnetic Hamiltonian operator specified here, which also mixes the $J=1/2$ sector.

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.