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Use the isolated nuclear spin Zeeman effect implied by the supplied nuclear magnetic moment, with hyperfine structure neglected. The proton gyromagnetic ratio isThe two spin levels are separated by $g_p{\mu }_N{B}_0$. Resonant nuclear magnetic resonance therefore requires the Larmor frequency

The duration is a Rabi pi pulse: in a frame rotating with the resonant transverse magnetic field, the spin must rotate through $\pi$. If $B_1$ denotes the magnitude of a circularly rotating field with the correct sense of rotation, the Rabi angular frequency is ${\Omega }_R={\gamma }_p{B}_1$. The shortest pulse is then

There is an amplitude convention to specify. If the applied transverse magnetic field is instead the linearly oscillating $B_1\cos (\omega t)\hat{\mathbf{x}}$, its resonant circular component has magnitude $B_1/2$. The rotating-wave approximation gives ${\Omega }_R={\gamma }_p{B}_1/2$ and henceThe wording does not specify this polarization convention, so the two pulse durations should be distinguished. Odd multiples of either shortest Rabi pi pulse also invert the spin within the corresponding ideal model.

A uniform magnetic field quantizes the transverse cyclotron motion of an electron. With electron charge $-e$ and the Landau gauge for a uniform magnetic field $\mathbf{A}=(0,Bx,0)$, minimal coupling givesFor an eigenstate of $p_y$ with eigenvalue $\hslash k_y$, this is a quantum harmonic oscillator centered at $x_c=-\hslash k_y/(eB)$, with cyclotron frequency ${\omega }_c=eB/m$. Its Landau levels areThe oscillator width is set by the magnetic length ${\ell }_B=\sqrt{\hslash /(eB)}$. In three dimensions, the free motion parallel to the magnetic field adds ${\hslash }^2{k}_z^2/(2m)$. The electron spin adds a Zeeman splitting $\pm g_s{\mu }_{B}B/2$. In a crystal the appropriate band effective mass and effective spin factor replace the free-electron values.

Changing $k_y$ moves the orbit center without changing the energy. In a rectangle $L_x{L}_y$, adjacent allowed centers differ by $2\pi {\ell }_B^2/L_y$. Requiring the center to lie in the sample gives the degeneracy of a Landau level, per spin,This large state degeneracy, together with the increasing level spacing, rearranges the occupied states as the magnetic field changes. To resolve the Landau levels, thermal broadening $k_{B}T$ and disorder broadening should both be small compared with $\hslash {\omega }_c$.

The de Haas-van Alphen effect is an oscillation of magnetization as successive Landau levels pass the Fermi energy. The Shubnikov-de Haas effect is the corresponding oscillation in longitudinal magnetoresistance. For a simple orbit, successive crossings occur periodically in $1/B$; the general relation to the extremal orbit area is Onsager quantization. These measurements reveal the geometry and effective masses of a Fermi surface.

In a two-dimensional electron system, filling an integer number $\nu$ of resolved levels leads to the integer quantum Hall effect, withand very small longitudinal resistivity on a plateau. Disorder localization helps produce finite plateau widths. Cyclotron resonance provides another direct signature: radiation is absorbed at the adjacent-level spacing $\hslash {\omega }_c$.

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 $|m_1,m_2⟩$ for a product of the two individual spin-one spin eigenstates, with magnetic quantum numbers $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving preciselyHere these are product labels, not labels of the total angular momentum. Write their coefficients in this order as $\alpha ,\beta ,\gamma$. With physical spin ladder operators $S_{\pm }=S_{1\pm }+S_{2\pm }$, each nonzero individual step has amplitude $\sqrt{2}\hslash$, soBoth must vanish for a singlet state. The independent product states therefore require $\alpha =\gamma =-\beta$. Wavefunction normalization gives the spin-one singlet stateup to an overall quantum phase. Conversely, $S_z|\psi ⟩=S_{+}|\psi ⟩=0$ implies $S^2|\psi ⟩=0$ through $S^2=S_{-}{S}_{+}+S_z^2+\hslash S_z$, so these conditions really select total spin zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless angular momentum units; restoring its common factor $\hslash$ does not change the coefficient ratios.

Take a normalized spin one-half state in the $S_z$ eigenbasis. Since $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, direct multiplication of the Pauli matrices givesSubstitution of the half-angle parametrization givesThus $\theta$ and $\varphi$ are precisely the polar and azimuthal angles of the mean spin. Two complex coefficients contain four real parameters; normalization removes one and an unobservable common quantum phase removes another. The resulting pure states form the Bloch sphere. One may choose $\alpha$ real and nonnegative away from the patch where $\alpha =0$. At the poles the azimuth is redundant. The length $|⟨\mathbf{S}⟩|=\hslash /2$ should not be confused with $\sqrt{⟨S^2⟩}=\sqrt{3}\hslash /2$.

For a spatial plane wave, $-i\hslash \nabla$ becomes $\hslash \mathbf{k}$, so the Hamiltonian operator reduces to the Weyl Hamiltonian matrixDefine $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and ${\chi }_{+}=(\cos (\theta /2),e^{i\varphi }\sin (\theta /2){)}^{\mathsf{T}}$. The matrixsatisfies $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{+}={\chi }_{+}$ by the angle-addition identities. Consequently the displayed spinor is an energy eigenstate for either $\mathbf{k}=+k\mathbf{n}$ or $\mathbf{k}=-k\mathbf{n}$, with respective energies $+\hslash vk$ and $-\hslash vk$, for $k\ge 0$. Equivalently, at each fixed nonzero wavevector the two branches areIf $v<0$, the two energy labels exchange; the ordered eigenvalues are $\pm \hslash |v||\mathbf{k}|$. Taking $v>0$ sets the convention in the sketches. An orthogonal spinor ${\chi }_{-}=(-e^{-i\varphi }\sin (\theta /2),\cos (\theta /2){)}^{\mathsf{T}}$ obeys $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{-}=-{\chi }_{-}$ and supplies the other branch at $+k\mathbf{n}$.

The Pauli matrix multiplication law independently gives $H(\mathbf{k}{)}^2=(\hslash v{)}^2|\mathbf{k}{|}^{2}I$. The dispersion relation is therefore a pair of cones touching at zero energy and momentum; a signed straight-line cut has upper and lower branches $\pm \hslash v|k_x|$. In the $k_z=0$ plane, $\theta =\pi /2$, and the upper-branch mean spin points radially outwards, the lower-branch spin radially inwards. This is spin-momentum locking, not a tangential spin texture. At the touching point the two states are degenerate and the spin direction is not selected by the Hamiltonian operator.

Now let a spatially constant perturbation have matrix $V=\left(\begin{array}{cc}u& w\\ w^{\ast }& z\end{array}\right)$, with $u,z$ real. The Hermitian two-by-two matrix in the Pauli basis decomposition follows explicitly fromwhere $d=(u+z)/2$, $c=(u-z)/2$, $a=\mathrm{Re}w$, and $b=-\mathrm{Im}w$ are real. Equivalently, $d=\mathrm{tr}V/2$ and $(a,b,c{)}_i=\mathrm{tr}(V{\sigma }_i)/2$. The four matrices are linearly independent under the trace inner product and span the four-dimensional real vector space of Hermitian two-by-two matrices.

Spatial constancy means the perturbation does not mix different wavevectors, so each momentum sector can still be diagonalized independently. Write $\mathbf{b}=(a,b,c)$. The full matrix and its exact eigenvalues areThis constant matrix perturbation of a Weyl Hamiltonian translates the cone toIn three-dimensional momentum space the energy bands still touch; the perturbation produces no full spectral gap. The new branch spin directions are parallel and antiparallel to $\hslash v\mathbf{k}+\mathbf{b}$, so their radial texture is centered on the shifted point.

A restricted cut must be distinguished from the full dispersion relation. In the fixed plane $k_z=0$, the energies are $d\pm \sqrt{(\hslash v{k}_x+a{)}^2+(\hslash v{k}_y+b{)}^2+c^2}$, with minimum branch separation $2|c|$. This plane misses the crossing if $c\ne 0$. Along any line through ${\mathbf{k}}_{\ast }$ the dispersion relation still crosses linearly. Thus a sketch of a gapped fixed-plane cut alone would not describe the gapless three-dimensional Hamiltonian operator. The cone after a general perturbation is also no longer a function of $|\mathbf{k}|$ about the old origin alone.

For physical spin ladder operators, $S_j^{\pm }=S_j^x\pm i{S}_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and givesAdding the remaining component provesFor spin one-half, $S^{+}|\downarrow ⟩=\hslash |\uparrow ⟩$, $S^{-}|\uparrow ⟩=\hslash |\downarrow ⟩$, and raising an up spin or lowering a down spin gives zero. These formulas restore the physical $\hslash$ factors implicit in the dimensionless ladder hint.

Let $|n⟩$ abbreviate the state with the only down spin at site $n$, keeping $|{\psi }_0⟩$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up spins. The ladder terms vanish, while $S_j^z{S}_{j+1}^z$ gives ${\hslash }^2/4$. ThusFor a down spin at $j$, the $z$ product is instead $-{\hslash }^2/4$, and $S_j^{+}{S}_{j+1}^{-}$ moves it to $j+1$ with amplitude ${\hslash }^2$. For a down spin at $j+1$, the other ladder product moves it back. ThereforeEvery bond in $|{\psi }_0⟩$ is aligned, so summing its $N$ bond energies givesThis is indeed a ground-state energy: a pair of spin-one-half particles has ${\mathbf{S}}_j\cdot {\mathbf{S}}_{j+1}$ eigenvalues ${\hslash }^2/4$ and $-3{\hslash }^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic Heisenberg model has a whole maximum-total-spin ground multiplet, not a unique all-up ground state.

In the single-excitation subspace, only the two bonds adjacent to the down spin differ from the aligned background. Their diagonal energy increase is $J$, and each neighboring hop has amplitude $-J/2$. Hence, with periodic site indices,This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n⟩=|n+1⟩$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, soThere are $N$ independent such wavevectors, for example one set of $N$ consecutive integers $\ell$ in a Brillouin zone. The normalized one-magnon Bloch states on a ring are $|q⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{iqna}|n⟩$. Reindexing with the periodic coefficient condition gives $T|q⟩=e^{-iqa}|q⟩$ and hence $T^N|q⟩=|q⟩$.

There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation eigenstate, equivalently a consistently periodic plane wave. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed momentum modes.

Apply the hopping Hamiltonian operator and shift the summation index using periodic boundary condition. The coefficients from $|n-1⟩$ and $|n+1⟩$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thusand the nearest-neighbour ferromagnetic magnon dispersion isThe unnormalized state in the question has the same energy. Distinct allowed momenta are orthogonal, since $N^{-1}{\sum }_n{e}^{i(q-q^{\prime })na}={\delta }_{q{q}^{\prime }}$ over one set of representatives.

For large $N$ the sampled wavevectors become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos (qa)$ over $-2\pi <qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm \pi /a$. Near any equivalent zero,using $q$ measured from that zero. The low-momentum ferromagnetic magnon is quadratic, with group velocity ${\hslash }^{-1}dE/dq=(Ja/\hslash )\sin (qa)$. The $q=0$ state is proportional to the total spin lowering operator acting on $|{\psi }_0⟩$, so it belongs to the same ground multiplet and has exactly zero excitation energy. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.

The closed $1s^2$ and $2s^2$ subshells contribute zero total orbital angular momentum and spin, so only the two $2p$ electrons matter. The first of Hund's rules maximizes their total spin, giving $S=1$. The Pauli exclusion principle then requires an antisymmetric spatial wavefunction. Two parallel-spin electrons cannot occupy the same $m_l$ orbital: the greatest possible orbital projection is $1+0=1$, so the maximal allowed $L$ within this triplet is $L=1$.

The ground term is therefore ${}^{3}P$, in term symbol notation ${}^{2S+1}{L}_J$. Since the $p^2$ subshell contains fewer than half of its six possible electrons, the third of Hund's rules places the smallest $J$ lowest. Addition of angular momentum gives $J=0,1,2$, in that energy order within the ground term. ThusThese are the fine structure partners of the same electronic term, with magnetic multiplicities $1,3,5$, rather than different principal-shell excitations. The ordering agrees with the tabulated lowest neutral-carbon levels.^[ref]

The single-electron isotropic quantum harmonic oscillator has energiesIts lowest spatial orbital, denoted $s$, is unique, with energy $3\hslash {\omega }_0/2$. Including spin one-half gives two spin orbitals. Two identical electrons can occupy both only with opposite spin in the unique singlet state. Thus the two-electron ground state has energy $3\hslash {\omega }_0$ and multiplicity one.

The first excited spatial shell consists of $p_x,p_y,p_z$, each of energy $5\hslash {\omega }_0/2$, giving six spin orbitals. A first-excited two-electron Slater determinant occupies one of the two ground-shell spin orbitals and one of these six excited-shell spin orbitals. HenceNo further factor of two is inserted for interchanging the electrons, since the fermionic exchange symmetry is already built into a Slater determinant.

The spatial Hamiltonian operator is invariant under simultaneous rotations of both positions, so $[H_0,L_k]=0$ for each component of the total orbital angular momentum. It is also spin independent, so $[H_0,S_k]=0$ for each total spin component. Thereforeexpressing conservation of angular momentum and conservation of the total spin magnitude for this model. The Pauli exclusion principle restricts which states occur but does not break either symmetry.

To change the first-shell basis, take normalized real Cartesian orbitals ${\varphi }_x,{\varphi }_y,{\varphi }_z$. Using spherical coordinates, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$. Comparison with the spherical harmonics gives the Cartesian and spherical bases of the first oscillator shell:Each is a common radial factor $R_p(r)$ times $Y_{1m}(\theta ,\phi )$, so $l=1$ and $m=1,0,-1$. Their normalizations and mutual orthogonality follow because this is a unitary change of basis. The ground spatial orbital has $l=0$.

For two ground-shell electrons, addition of angular momentum gives $L=0$, and their symmetric spatial state requires $S=0$. For one electron in each shell, $0\otimes 1$ gives only $L=1$. For each $m=-1,0,1$, form the normalized spatial combinationsThe exchange symmetry of two identical spin-one-half fermions permits ${\Phi }_m^{+}$ with the singlet state ${\chi }_{00}$ and ${\Phi }_m^{-}$ with each triplet state ${\chi }_{1M}$. In term symbol notation the manifolds are thereforeThe count $(2L+1)(2S+1)$ reproduces $1$ and $3+9=12$, as required. Both exchange symmetries are possible for the distinct occupied orbitals; dropping either spin sector would undercount the two-electron states in the first two shells of an isotropic oscillator.

For the specified spin-orbit interaction, $\mathbf{S}$ and $\mathbf{L}$ are the total two-electron operators. Set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity givesFor the $L=1,S=0$ singlet, $J=1$ and the shift is zero. For the $L=S=1$ triplet, $J=0,1,2$. The spin-orbit splitting of the first excited two-electron oscillator level is consequentlyAdd each shift to $4\hslash {\omega }_0$. The multiplicities sum to twelve and their weighted shift sums to zero. For positive $\lambda$ the table is ordered in energy; for negative $\lambda$ the ordering reverses. Weak coupling means $|\lambda |{\hslash }^2\ll \hslash {\omega }_0$. Since this total-spin interaction commutes with $H_0$, $L^2$ and $S^2$, the shifts within this invariant manifold are in fact exact for the specified model, although only first order was requested.

Write ${\omega }_z=\gamma B_z$ and use the ordered spin basis $(|\uparrow ⟩,|\downarrow ⟩)$. The free Hamiltonian operator is $H_0=-\hslash {\omega }_z{\sigma }_z/2$, soResolving the cosine into its two complex exponentials gives the exact upper off-diagonal factorwith the complex conjugate in the lower entry. Choose the field orientation with ${\omega }_z>0$, as implicit in the printed near-resonance condition, and put $\delta =\omega -{\omega }_z$. The rotating-wave approximation discards the terms at $\omega +{\omega }_z$, retaining the slow terms at $\delta$. The resulting interaction picture equations areThe factor $1/4$ contains $1/2$ from $S_x=\hslash {\sigma }_x/2$ and $1/2$ from the cosine decomposition. This is a Driven two-level Rabi model, with signed on-resonance coupling $\Omega =\gamma B_x/2$ and generalized Rabi angular frequency ${\omega }_R=\sqrt{{\delta }^2+{\Omega }^2}$. Here the Rabi detuning uses the paper's convention, drive minus free frequency.

Small detuning alone is insufficient to derive the printed equations. One also requires $|\gamma B_x|,|\delta |\ll \omega +{\omega }_z$, and an evolution time over which neglected terms stay small. At resonance this is the weak transverse-field regime $|B_x|\ll |B_z|$. The supplied solution is thus an approximate solution for the linearly oscillating field, not an exact one at arbitrary drive strength. If $\gamma B_z<0$, the other exponential is resonant and the physical positive resonance frequency is $|\gamma B_z|$.

At resonance, retain the signed coupling $\Omega =\gamma B_x/2$. In the interaction picture, the given solution reduces to $c_1=\cos (\Omega t/2)$ and $c_2=i\sin (\Omega t/2)$. Returning to the Schrödinger picture givesFor a normalized spinor $(a,b{)}^T$, the Pauli matrices give $⟨\mathbf{\sigma }⟩=(2\mathrm{Re}{a}^{\ast }b,2\mathrm{Im}{a}^{\ast }b,|a{|}^2-|b{|}^2)$. Therefore the spin expectation value within the rotating-wave approximation isIts magnitude is $\hslash /2$: the Bloch vector stays on the unit sphere. In a frame rotating about $z$ at angular velocity $-{\omega }_z$, it makes a uniform rotation in the $yz$ plane. For $|B_x|\ll |B_z|$, this gives slow Rabi oscillations between the poles, accompanied by rapid laboratory Larmor precession about $z$.

If the same formula is formally used for $|B_z|\ll |B_x|$, it predicts fast meridional rotations whose plane turns slowly about $z$. However, the original PDF really does ask for this opposite limit, and it lies outside the assumption used in part (b). That formal description is not a controlled prediction for the stated linearly driven Hamiltonian operator.

A valid strong-drive description starts from the exact equation for the Bloch vector $\mathbf{s}=2⟨\mathbf{S}⟩/\hslash$:For times $|\gamma B_z|t\ll 1$, neglecting the weak bias gives a controlled leading approximation, since the time integral of the omitted Hamiltonian operator has small norm. The remaining field has a fixed $x$ axis, so its generators commute at different times andAt the nominal resonance $\omega =\gamma B_z$ and on the early spin-rotation scale $t\sim 1/|\gamma B_x|$, this becomes $\Theta \simeq \gamma B_{x}t$. The initial rotation rate is consequently $\gamma B_x$, twice the rate predicted by extrapolating the rotating-wave approximation. For longer times the exact differential equation must be retained. The following original calculation illustrates both regimes rather than asserting an exact strong-drive Rabi formula.

Inside the infinite square well, the orbital and spin operators act on different factors of the tensor product. Write $\mathbf{S}={\mathbf{S}}_1+{\mathbf{S}}_2$. The identityshows that the exchange interaction commutes with $S^2$ and $S_z$. The sum of the one-particle kinetic energies also commutes with them and with particle exchange. Thus one may use orbital quantum numbers together with the total-spin sector $S=0,1$ and projection $M=-S,\dots ,S$, with physical $S_z=\hslash M$.

For identical spin one-half particles the Pauli exclusion principle further restricts the states. The spin-one-half singlet state is antisymmetric in spin and must multiply a symmetric spatial wavefunction; a spin-one-half triplet state must multiply an antisymmetric spatial wavefunction. Thus $n_1,n_2$ label unordered orbital occupations in symmetry-adapted states, not arbitrary ordered product states. This gives the labels $|n_1,n_2,S,M⟩$, with the paper's last label expressed in units of $\hslash$.

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 ${\Phi }_{22}^{+}{\chi }_{00}$.

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 this spin-boson model, take $H_0=H_1+H_2$. Since $[H_0,{\sigma }_z]=0$, the interaction picture observable isThe normalized vectors $|{\varphi }_{\uparrow }⟩$ and $|{\varphi }_{\downarrow }⟩$ belong to different eigenspaces of the Hermitian operator ${\sigma }_z$, and are therefore orthogonal. This remains true even though each vector includes the oscillator degrees of freedom. Hence all cross terms in the expectation value vanish, while the diagonal terms have eigenvalues $+1$ and $-1$:Normalization also gives $|c_{\uparrow \sigma }{|}^2+|c_{\downarrow \sigma }{|}^2=1$, so these are the probabilities of the two spin outcomes.