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The microscopic electric interaction is spin independent to leading order, but the Pauli exclusion principle ties the spatial symmetry of an electronic state to its spin symmetry. Different spin combinations therefore have different Coulomb and kinetic energies. Projecting those energies onto localized low-energy moments often gives an approximately rotationally invariant bilinear exchange Hamiltonian operator $-J{\mathbf{S}}_1\cdot {\mathbf{S}}_2$. For spin one half, the spin singlet state and triplet state energies are distinguished by ${\mathbf{S}}_1\cdot {\mathbf{S}}_2=-3/4$ and $1/4$, so any isotropic two-level spin singlet state–triplet state splitting can be written in this form plus a constant. More general spins or orbital degeneracy can require extra interactions.

An antiferromagnetic oxide commonly has superexchange through a ligand such as oxygen. Virtual hopping between half-filled metal orbitals lowers the energy of antiparallel spins more strongly because the corresponding intermediate doubly occupied configurations are allowed by the Pauli exclusion principle. The simple one-orbital Hubbard model limit gives $+4t^2{\mathbf{S}}_1\cdot {\mathbf{S}}_2/U$, equivalent to negative $J$ in the question's minus-sign convention. Bond angles and orbital occupancy can change the sign in more complicated oxides, so superexchange is not universally antiferromagnetic irrespective of geometry.

For the classical XY ferromagnet choose every spin parallel within the $xy$ plane. Each nearest-neighbor and next-nearest-neighbor in-plane dot product is $S^2$. The printed sum counts one bond of each range per spin, soThere is no additional factor of two from counting both directions: the Hamiltonian operator already lists only $n\to n+1$ and $n\to n+2$. Alignment along $z$ would give zero XY exchange energy and is not the in-plane ordered state used for these expressions.

In both collinear trial states, a spin and its second neighbor are parallel. The next-nearest-neighbor energy is therefore $-J_2{S}^2$ per spin in either case. Making $J_2$ negative makes that contribution positive and raises both energies.

A negative $J_2$ prefers opposite second-neighbor spins. Neither all-parallel order nor the simple alternating nearest-neighbor pattern satisfies that preference: in the latter, alternating twice returns to the original direction. This competition is a source of frustration when the nearest-neighbor interaction favors one of those collinear patterns, and allows a noncollinear twist to lower the energy. Merely calling $J_2$ antiferromagnetic does not mean that the nearest-neighbor alternating state satisfies it.

A relative angle $qc$ between neighbors gives in-plane dot products $S^2\cos (qc)$ and $S^2\cos (2qc)$ at the two ranges. Set $x=\cos (qc)\in [-1,1]$. ThenThe classical helical minimum of the J1-J2 XY chain follows because for $J_2<0$ this is a strictly convex quadratic in $x$. Its unconstrained minimum isIt is a genuine noncollinear helix when $|J_2|>|J_1|/4$. Both chiralities $q$ and $-q$ have the same energy, and the pitch is defined modulo $2\pi /c$. At equality the helix merges into a collinear endpoint, so the strict inequality is required for an energy strictly lower than both collinear choices.

For a direct comparison write $K=|J_2|$. The minimum and the energy differences areBoth differences are strictly positive in the stated interior region. The stronger classical global argument Fourier-transforms the translationally invariant bilinear exchange: its energy is minimized by the wave vector maximizing $J_1\cos (qc)+J_2\cos (2qc)$, and the planar single-$q$ helix realizes the fixed spin length at every site. This establishes the minimum within the classical/mean-field model, not an assertion of true long-range magnetic order at positive temperature in a one-dimensional short-range quantum chain.

One orbital per primitive cell supplies two states per wave vector when spin is included. A monovalent metal therefore half-fills this band. Nearest-neighbor hopping on the bipartite rectangular lattice has $E(\mathbf{k}+\mathbf{Q})=2E_0-E(\mathbf{k})$ for $\mathbf{Q}=(\pi /a,\pi /b)$, fixing the half-filled Fermi energy atFor $t_x=2t_y>0$, the two Fermi surface obeyThey are corrugated open sheets in an extended-zone description, with $|k_{x}a|$ ranging from $\pi /3$ to $2\pi /3$. They are not a small circular pocket centered at the band bottom. The same $\mathbf{Q}$ maps one Fermi line onto its nested partner. A density-wave or Peierls-like modulation at that wave vector can open additional gaps at the Fermi energy. The unperturbed single-orbital band does not acquire those new gaps without a symmetry-breaking modulation or another coupling; ordinary lattice Bragg effects are already built into its dispersion.

For the three-dimensional tetragonal version set $t_x=t_y=t$, $t_z=t/2$, $a=b<c$. The corresponding half-filled surface isIt is a corrugated periodic surface, not a sphere. At $k_z=\pi /c$ its regular closed contour is electron-like around $\Gamma$, with $\cos (k_{x}a)+\cos (k_{y}a)=1/2$; at $k_z=0$ the closed contour is hole-like around $M$, with the sum $-1/2$. The topology switches through the square-lattice separatrix at $k_{z}c=\pm \pi /2$. Drawing one simple weakly warped cylindrical pocket across every section misses that feature of the half-filled cosine band.

Angle-resolved photoemission can directly map the dispersion and occupied Fermi surface. Angular quantum-oscillation measurements test its closed extremal cross sections and cyclotron masses; momentum-density probes such as positron annihilation or Compton scattering give a complementary bulk test. The different techniques constrain different features rather than merely providing interchangeable pictures.

A magnetic field quantizes closed semiclassical orbits into Landau levels. Their passage through the Fermi energy modulates the density of states and scattering, producing Shubnikov-de Haas oscillations of resistivity. The fundamental frequency for a regular extremal orbit perpendicular to the field isThus two distinct fundamentals ordinarily identify two inequivalent extremal areas, rather than two arbitrary points or two spin directions automatically giving two different geometric areas.

The particle-hole equality of tetragonal extremal orbit areas is a special constraint of the exact half-filled nearest-neighbor band stated here. For $B\parallel z$, its regular closed endpoint sections are the electron contour at $k_{z}c=\pi$ and the hole contour at $k_{z}c=0$ described in b. Translation $(k_{x}a,k_{y}a)\mapsto (k_{x}a+\pi ,k_{y}a+\pi )$ changes the sign of both cosines and maps one enclosed pocket onto the other. Their areas are equal. Direct integration givesConsequently this bare model predictsnot two distinct regular fundamental frequencies. The intermediate $k_{z}c=\pi /2$ contour is a separatrix with a divergent cyclotron period, not another ordinary smooth extremal pocket to insert into the Onsager formula. The larger occupied-area slice at $k_z=0$ is the complement of a hole pocket on the Brillouin torus; counting that complement as a second enclosed planar orbit would give a spurious ratio.

For comparison, an additional small-$k_x,k_y$ parabolic approximation would replace the exact surface by $(k_{x}a{)}^2+(k_{y}a{)}^2\simeq 4+\cos (k_{z}c)$ and suggest a rough belly-to-neck ratio $5/3$. That approximation is uncontrolled at half filling and erases the actual topology; it cannot be asserted as an exact prediction of the supplied cosine band. The observation of genuinely different frequencies in a real material therefore needs further information, such as longer-range hopping, a departure from half filling, extra bands, symmetry breaking or a resolved spin-dependent effect. The universal area-ratio inference is valid, but a nonunit numerical ratio is not determined consistently by the complete stated band assumptions alone.^[ref]

For a partially filled $3d$ shell, electron–electron Coulomb repulsion and Hund's rules first favour particular total spin and orbital terms. The crystal field splitting is substantial because these orbitals overlap the ligand environment; it normally exceeds the spin-orbit coupling and often quenches the orbital moment. Competition between the splitting and the electron pairing energy determines high-spin versus low-spin filling. A common hierarchy is Coulomb term energies, then crystal-field and Hund-exchange scales, then spin–orbit and weak-field Zeeman energies; the relative sizes of crystal-field splitting and pairing are material-dependent. Hyperfine coupling, if the nucleus carries spin, adds a smaller splitting. An isolated ion has no inter-ion exchange ordering field.

The tetrahedral crystal field splitting places the two $e$ orbitals below the three $t_2$ orbitals. In the weak-field, high-spin limit the occupancy is $e^2{t}_2^1$, with $S=3/2$. Its lowest cubic term is ${}^4{T}_1$: an orbital triplet, before spin degeneracy is counted. In the hypothetical sufficiently strong-field limit the occupancy is $e^3$, with $S=1/2$ and an orbital doublet ${}^{2}E$. This latter limit requires overcoming pairing/Hund energy; ordinary tetrahedral oxide complexes generally remain high-spin.

Both orbital degeneracies permit a Jahn-Teller effect, so a regular tetrahedron is not a stable static equilibrium in the elementary adiabatic treatment. The sign of a particular distortion requires an electron–lattice coupling model, not just the degeneracy count. To make that dependence explicit, let $Q>0$ stretch $z$ while contracting $x,y$. In a leading point-charge quadrupole model write the shiftsIn the high-spin configuration the two $e$ shifts cancel. For $Q>0$ the remaining electron selects $d_{xy}$, gaining $-2gQ$; for $Q<0$ it selects the $d_{xz},d_{yz}$ doublet, gaining $gQ$. Adding $K{Q}^2/2$ gives a tetragonally elongated minimum at $Q=2g/K$, with energy gain $2g^2/K$, versus only $g^2/(2K)$ along the compressed branch. The elongated branch has an orbital singlet. This is the usual elementary elongation argument; quantitative coupling includes higher crystal-field multipoles and configuration mixing.

For the strong-field $e^3$ configuration the lower $e$ orbital is doubly occupied. The shift of the three electrons is $-2g|Q|$, so the same model has both an elongated and a compressed minimum, $Q=\pm 2g/K$, with equal stabilization. Elongation favours double occupation of $d_{x^2-y^2}$; compression favours double occupation of $d_{z^2}$. This tetrahedral doublet distortion-sign ambiguity prevents a unique choice from the supplied data. For example, a small symmetry-allowed cubic elastic invariant $A{Q}_{\theta }(Q_{\theta }^2-3Q_{ϵ}^2)$ selects opposite tetragonal signs when $A$ changes sign, while leaving the regular-site electronic configuration unchanged. Higher couplings or host constraints are needed to choose. Dynamic motion between equivalent minima may also restore an average regular shape without stabilizing the undistorted adiabatic configuration.

Once a static axial distortion removes the orbital degeneracy, second-order spin-orbit coupling produces a time-reversal-even quadratic spin Hamiltonian operator. The residual tetragonal symmetry requires equal $x,y$ coefficients. Dropping a multiple of $S(S+1)$ leaves $H_{\mathrm{an}}=D{S}_z^2$ for the weak-field $S=3/2$ state. The sign and magnitude of $D$ depend on excited-state splittings. For the strong-field $S=1/2$ Kramers doublet, $S_z^2=1/4$ is constant: this term cannot cause zero-field splitting, although the Zeeman g-factor can be anisotropic.

Take dimensionless spin operators and $H=D{S}_z^2+g{\mu }_{B}B{S}_z$, with $g>0$. The four wavefunctions are $|m⟩$, $m=-3/2,-1/2,1/2,3/2$, withAt zero field they form doublets at $D/4$ and $9D/4$. A transverse microwave magnetic field drives $\Delta m=\pm 1$ transitions. When $h\nu >2D$, their positive resonance fields in ascending order areThey connect $(1/2,3/2)$, $(-1/2,1/2)$, and $(-3/2,-1/2)$, respectively. ThusHere $D/h=4.25\mathrm{GHz}$ and the zero-field doublet gap is $2D$.

At lower frequency it is essential to use absolute energy gaps, not discard a transition after its levels cross. For $0<h{\nu }^{\prime }<2D$, the distinct candidate positive fields areThe first two coincide whenThis axial spin-three-halves resonance coincidence leaves two distinct resonance positions, at $0.1275$ and $0.3825\mathrm{T}$, with two transitions superposed at the lower position. At ${\nu }^{\prime }=2D/h$ there are also only two strictly positive-field lines, but an additional zero-field resonance remains; this is a different counting convention. Observable intensities also require thermal occupation of the relevant lower levels.

For competing exchanges on a body-centred cubic antiferromagnet, use the Heisenberg model convention $H={\sum }_{⟨ij⟩}{J}_{ij}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, with each bond counted once, dimensionless spins, and positive antiferromagnetic $J$. There are eight nearest and six next-nearest neighbours. Bond counting in the two classical collinear structures givesFor type II the eight $J_1$ contributions cancel, while all six $J_2$ bonds are antiparallel. Their crossing yields $c=2/3$. One can also exclude a lower spiral within the classical bilinear model: writing $u_i=\cos (q_{i}a/2)$ givesFor a favourable sign of the product, $u_x^2+u_y^2+u_z^2\ge 3|u_x{u}_y{u}_z{|}^{2/3}\ge 3|u_x{u}_y{u}_z|$. Hence when $J_2\ge 2J_1/3$ its minimum is at $u_x=u_y=u_z=0$, the type-II ordering vectors. When $J_2\le 2J_1/3$, compare with the type-I value and use $3-\sum u_i^2\le 3(1-|u_x{u}_y{u}_z|)$ to obtain the complementary bound. ThusThis is a classical/mean-field energy comparison; an exact quantum ground state is not generally a collinear product state.

For type I a magnetic cell is the simple cube of side $a$, with opposite moments at $0$ and $a(1,1,1)/2$. Its magnetic structure factor is proportional to $1-e^{i\pi (h+k+l)}$. Allowed magnetic reflections have integer $h,k,l$ with $h+k+l$ odd. The first two families are $(100)$ and $(111)$, with symmetry-equivalent members understood.

For type II choose magnetic primitive translations $a(1,1,0)$, $a(1,0,1)$ and $a(0,1,1)$, enclosing four spins in volume $2a^3$. Equivalently, a side-$2a$ cube is a convenient nonprimitive magnetic cell. On each original simple-cubic sublattice the moment at $a\mathbf{n}$ is proportional to $(-1{)}^{n_x+n_y+n_z}$. Accordingly,The translated body-centre basis gives $1+e^{i\pi (h+k+l)}=1\pm i$, so there is no further extinction. The first two families are $(1/2,1/2,1/2)$ and $(3/2,1/2,1/2)$. The neutron polarization factor may additionally suppress a reflection for a particular moment direction; the stated rules are the positional selection rules.

Each sublattice has number density $n_A=1/a^3$ and saturation magnetization $M_s=g{\mu }_{B}S/a^3$. Replacing neighbouring spins by their averages makes their exchange energy equivalent to a molecular magnetic field. Its minus sign expresses antiferromagnetic exchange, not an externally applied field. Since $⟨S_B⟩=M_B/(n_{A}g{\mu }_B)$,For $M_B=-M_A$, $B_{\mathrm{mf}}=({\lambda }_{AB}-{\lambda }_{AA})M_A$. Linearizing the Brillouin magnetization formula givesThe g-factor cancels because exchange is specified in energy units. A convention with $2J{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ doubles these exchange energies and $T_N$.

Let $m=M_A/M_s$, $A=(S+1)/(3S)$ and let $b_B>0$ denote the cubic coefficient of the Brillouin magnetization formula. The argument is $y=(T_N/T)m/A$. Thus $m=Ay-b_B{y}^3$ impliesConsequently $\boxed{\beta =1/2}$ in mean-field theory. The positive branch falls from saturation at zero temperature to zero at $T_N$ and remains zero above it. An elastic magnetic neutron-reflection intensity scales as $I\propto M_A^2$; a log–log fit against $T_N-T$ has slope $2\beta$. Background subtraction and a suitable critical interval are needed. Fluctuations can change the measured exponent from its mean-field value; uniform bulk magnetization vanishes for a compensated antiferromagnet and is not itself this order parameter.

The conduction and valence dispersions are two cones touching at the Dirac point, with slopes $\pm \hslash v_F$. Here $\mathbf{k}$ is measured relative to one valley centre. Each cone has a density of states proportional to $|E|$, which vanishes at the touching energy. Gate-induced carriers satisfy approximately $n=C_g(V_g-V_D)/e$ per unit area. Negative and positive offsets from $V_D$ produce holes and electrons, respectively. Away from neutrality a simple charge carrier mobility model gives $\sigma \simeq e\mu |n|$, so resistance decreases on either side of a maximum near $V_D$. Disorder, electron–hole puddles, contacts and quantum transport round the peak; the cone dispersion alone fixes neither its height nor its width, nor does it require exact electron–hole symmetry in a real device.

A direct derivation of the magnetic spectrum starts from the single-valley Dirac Hamiltonian operator $H=v_F({\sigma }_x{\pi }_x+{\sigma }_y{\pi }_y)$. Introduce oscillator operators using the magnetic length ${\ell }_B=\sqrt{\hslash /(e|B|)}$. Up to exchanging sublattices or valleys,Squaring gives adjacent oscillator number operators, henceThe positive expression printed in the source describes the electron branch only; the negative-energy partners must also be included. The same square-root dependence follows from quantizing circular orbit area, with the Dirac Berry phase cancelling the usual half-integer offset.

Each spin and valley contributes $Ae|B|/h$ states in area $A$. Including two spins and two valleys gives $N_{\mathrm{LL}}=4Ae|B|/h$, or areal degeneracy $4e|B|/h$. The zero-energy level is half-filled at charge neutrality: the carrier density counts departures from that state.

For $n=10^{16}{\mathrm{m}}^{-2}$, the filling factor and fourfold graphene Hall sequence areIn a plateau, ${\rho }_{xx}\simeq 0$ and $|{\rho }_{xy}|=h/({\nu }_N{e}^2)$. The Hall sign reverses between electron and hole doping; the sketch uses its magnitude to avoid tensor-index sign conventions. Representative plateau-centre fields are $20.68$, $6.893$, $4.136$ and $2.954\mathrm{T}$ for $\nu =2,6,10,14$. Between plateaux, crossing extended states in broadened levels gives longitudinal peaks and Hall steps. Their ideal filling-centre positions have $\nu =4j$, namelyThus peaks near $10.34$, $5.17$ and $3.45\mathrm{T}$ separate the first prominent plateaux as field increases. Many weak oscillations accumulate toward zero field and are damped there. Finite-width plateaux require localized states between extended-state energies; an exactly clean Landau spectrum does not by itself prescribe plateau widths. The illustration uses an explicitly schematic broadening, not inferred experimental widths.

At $6.5\mathrm{T}$, $\nu =6.36$, close to the $\nu =6$ plateau. With $v_F=10^6\mathrm{m}{\mathrm{s}}^{-1}$, $E_1=92.5\mathrm{meV}$ and $E_2=130.8\mathrm{meV}$. The relevant adjacent gap near the chemical potential is therefore $\Delta =E_2-E_1\simeq 38.3\mathrm{meV}$, rather than the gap from the zero level. Resolving levels requires quantum-lifetime broadening $\hslash /{\tau }_q\lesssim \Delta$ and thermal broadening appreciably smaller than $\Delta$, giving the order-of-magnitude limitsThe upper temperature is an energy-scale estimate, not a guaranteed sharp-feature temperature. For oscillation amplitudes the thermal damping factor is $X/\sinh X$, $X=2{\pi }^2{k}_{B}T/(\hslash {\omega }_c^{\ast })$, with $\hslash {\omega }_c^{\ast }=\hslash eB{v}_F^2/E_F\simeq 36.7\mathrm{meV}$ and $E_F=\hslash v_F\sqrt{\pi n}$. Requiring $X\lesssim 1$ gives the more conservative scale $T\lesssim 22\mathrm{K}$ for little thermal damping. Visible weaker features may persist above that scale; their actual cutoff depends on resolution, disorder and temperature-dependent scattering. The transport lifetime and quantum lifetime also need not coincide.

In BCS theory, an electron distorts the positively charged lattice, and the retarded response can attract another electron. Within a phonon-energy shell around the Fermi surface, the effective attraction can overcome the screened Coulomb repulsion and destabilize the normal Fermi sea toward Cooper pairs. In conventional elemental superconductors the pairs predominantly join time-reversed states of opposite momentum and spin, forming a spin singlet.

In the weak-coupling approximation,Here $N(0)$ is the appropriate single-spin density of states and $V_{\mathrm{eff}}>0$ the effective attractive interaction in a consistent normalization. Both the phonon scale and the exponential coupling matter. A large Fermi-level density of states, often associated with partially filled narrow $d$ bands, can raise $T_c$; weak coupling or a small density of states can suppress it even in a good conductor. With force constants and coupling fixed, ${\omega }_D\propto M^{-1/2}$ gives the conventional isotope trend. This is not a universal monotonic ranking of elements by atomic mass: phonon stiffness, electron–phonon matrix elements, Coulomb effects and magnetism vary between elements. Magnetic pair breaking helps explain why some elements do not follow a naive density-of-states expectation.

Use the explicit convention $H=J_1{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j+J_2{\sum }_{⟨⟨ij⟩⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, counting every bond once and taking dimensionless spins of length $S$. On the simple cubic lattice there are six nearest and twelve face-diagonal next-nearest neighbours.

For type I all six nearest bonds are antiparallel and all twelve next-nearest bonds parallel, so $E_I/N=S^2(-3J_1+6J_2)$. Choose the parallel direction of type II as $z$. It has two parallel and four antiparallel nearest bonds, and four parallel $xy$ face diagonals but eight antiparallel $xz,yz$ diagonals. Thus $E_{II}/N=S^2(-J_1-2J_2)$ andConsequently, for the stated antiferromagnetic exchanges,This comparison is also consistent with the full classical spiral energy. With $u_i=\cos (q_{i}a)$,It is affine in each $u_i\in [-1,1]$, so some minimum occurs at a cube vertex, giving the competing collinear states above. At the boundary, the coefficient of $u_z$ vanishes for $u_x=u_y=-1$, producing the larger degeneracy $(\pi /a,\pi /a,q_z)$ and permutations. It is not an isolated two-state crossing. The first panel of the model figure below shows the competing energies.

For type I, a magnetic unit cell uses the magnetic Bravais translations $(0,a,a)$, $(a,0,a)$ and $(a,a,0)$. They form an FCC lattice with conventional cubic parameter $2a$. A two-spin basis is an up spin at $(0,0,0)$ and a down spin at $(a,0,0)$, or fractional conventional coordinates $(0,0,0)$ and $(1/2,0,0)$. Every translation preserves the parity of $n_x+n_y+n_z$.

At $\mathbf{Q}=(2\pi /a)(h,k,l)$, the FCC translation sum allows $h,k,l$ all integers or all half-odd-integers. The spin basis contributes $S[1-e^{2\pi ih}]$, which removes the integer set and leavesEquivalently the spin field is $S(-1{)}^{n_x+n_y+n_z}$, so magnetic Bragg peaks are nuclear reciprocal vectors shifted by $(1/2,1/2,1/2)$.

For type II with parallel spins along $z$, use $(a,a,0)$, $(-a,a,0)$ and $(0,0,a)$ as magnetic translations: a primitive tetragonal cell with side lengths $\sqrt{2}a,\sqrt{2}a,a$, containing an up spin at zero and a down spin at $(a,0,0)$. The reciprocal translations have $h,k$ both integers or both half-odd-integers, with $l$ integer. The same basis factor removes the integer-$h$ set:This also follows from $S(-1{)}^{n_x+n_y}$. Other domains permute the three coordinate directions. A magnetic neutron cross section further multiplies these lattice conditions by the magnetic form factor and by the component of ordered moment perpendicular to $\mathbf{Q}$; that polarization factor can suppress an otherwise allowed peak.

Keep the once-per-bond Hamiltonian convention of part (a). Let each sublattice magnetization be its magnetic moment per total physical volume, so the density of B-sublattice spins is $1/(2a^3)$. Writing their moments as $\mathbf{\mu }=g{\mu }_B\mathbf{S}$ fixesThe exchange energy of one A spin is $6J_1{\mathbf{S}}_A\cdot ⟨{\mathbf{S}}_B⟩=-{\mathbf{\mu }}_A\cdot {\mathbf{B}}_{\mathrm{mf}}$, givingA convention using $2J_1$ per bond would double this coefficient; defining magnetization per sublattice volume instead changes its normalization. The definitions above make the total magnetization $\mathbf{M}={\mathbf{M}}_A+{\mathbf{M}}_B$.

For equal fixed sublattice magnitudes $M_0$, put ${\mathbf{M}}_A=M_0(\sin \theta ,0,\cos \theta )$ and ${\mathbf{M}}_B=M_0(\sin \theta ,0,-\cos \theta )$, with the external field along $+x$. The effective A-site field is $(B_{\perp }-\lambda M_0\sin \theta ,0,\lambda M_0\cos \theta )$. Its alignment with ${\mathbf{M}}_A$ requires $B_{\perp }=2\lambda M_0\sin \theta$, soThe last equality uses the externally imposed vacuum field $B_{\perp }={\mu }_0{H}_{\perp }$ and neglects demagnetizing corrections. Canting costs exchange energy but need not change the magnitude of either sublattice moment. The result holds before saturation and assumes the ordered axis is maintained when defining a transverse response.

Below $T_N$, a small perpendicular field produces canting of the two ordered sublattices. In the isotropic two-sublattice mean-field model, the induced component is $B_{\perp }/\lambda$: the reduced ordered moment at higher temperature tilts more easily and the two effects cancel. Thus ${\chi }_{\perp }$ remains constant until the transition.

A field parallel to the maintained ordering axis instead increases the magnitude of one sublattice magnetization and decreases the other. At zero temperature saturated localized spins have essentially no such amplitude response in this model, so ${\chi }_{\parallel }\to 0$. Thermal population changes make it easier to redistribute the longitudinal moments as temperature rises, giving the growing longitudinal susceptibility that meets ${\chi }_{\perp }$ at $T_N$. These directions presume the domain or a weak anisotropy fixes the ordered axis; in a perfectly isotropic freely rotating sample the Néel vector can reorient.

Above $T_N$ there is no chosen magnetic axis, so the response is isotropic. Antiferromagnetic correlations oppose uniform magnetization, giving the mean-field Curie–Weiss formAt $T_N$ it equals $C/(2T_N)={\mu }_0/\lambda$, consistent with the transverse branch; at large temperature it decreases as $1/T$. A negative Weiss temperature characterizes this uniform antiferromagnetic response, rather than a ferromagnetic divergence at positive $T_N$. The illustrative mean-field curves in solution make these limits visible.

A sufficiently clean two-dimensional electron gas at low temperature in a perpendicular magnetic field has quantized orbital Landau levels. In the integer quantum Hall regime, the longitudinal resistivity is nearly zero while the Hall resistance has plateauxThe Hall sign depends on carrier charge and lead convention. Fractional plateaux can occur in strongly interacting states, but the following single-particle counting and calibration use the integer effect.

One resolved spin branch of a Landau level has one state per flux quantum $h/e$: in area $A$, its guiding-centre degeneracy is $Ae{B}_z/h$ for positive $B_z$. Defining the nominal filling factor as the number of electrons per available orbital state givesThe same count follows in Landau gauge: a $2\pi /L_y$ increment in $k_y$ moves the orbit centre by $2\pi {\ell }_B^2/L_y$, giving $A/(2\pi {\ell }_B^2)=Ae{B}_z/h$ states. Include each spin branch separately; an unresolved spin-degenerate orbital level contains twice this count, and common plateau indices are then even.

Plateau qualification: $\nu$ as defined by density is continuous when $B_z$ changes at fixed $N_S$. The integer plateau index $p$ can remain unchanged over a field interval because localized states accommodate the changing occupation. Therefore $N_S=pe{B}_z/h$ is not an exact identity across an entire plateau. It is the ideal filled-level relation, often used approximately at a suitable plateau centre; confusing the nominal filling with the plateau index would contradict fixed density.

The plateau resistance realizes the von Klitzing constant $R_K=h/e^2\simeq 25812.8\Omega$, divided by a known integer. A four-terminal Hall measurement with negligible longitudinal voltage and contacts in the quantized regime supplies a reproducible resistance reference, which can be compared to other resistors using precision bridges. Its robustness comes from the quantized transport state, not from measuring the Hall bar's width accurately.

At $B_z=8\mathrm{T}$,The cyclotron spacing is $\hslash e(8\mathrm{T})/m^{\ast }$, smaller than at 10 T, and each resolved level's degeneracy per area is also reduced by $4/5$. Thus the same carrier density occupies more levels. This ratio is determined without the missing graph, but it does not specify whether the chemical potential lies in a localized tail or an extended-state centre.

TODO — missing density/plateau assignment: a numerical index and chemical-potential marker at 8 T require the unavailable density information. For example the first illustrative density in part (i) gives ${\nu }_8\simeq 2.5$, while $N_S=3.87\times 10^{15}{\mathrm{m}}^{-2}$ instead gives ${\nu }_8\simeq 2$. These produce different schematic placements even though both satisfy every surviving sample parameter. Unresolved spin degeneracy and disorder add further qualifications; no unique 8 T label should be invented.

At $B_z=10\mathrm{T}$, the nominal filling isA labelled density-of-states picture has disorder-broadened peaks at these orbital and spin energies, localized tails and extended states near their centres. A chemical potential in the localized region corresponds to a Hall plateau and small longitudinal resistivity; passing through an extended-state centre corresponds to a transition.

TODO — missing density/plateau assignment: there is no basis for choosing one of those positions at 10 T without the missing Hall graph. For example $N_S=4.84\times 10^{15}{\mathrm{m}}^{-2}$ gives nominal ${\nu }_{10}\simeq 2$, whereas $N_S=6.04\times 10^{15}{\mathrm{m}}^{-2}$ gives ${\nu }_{10}\simeq 2.5$. In an ideal resolved-level picture the first can put the chemical potential between filled levels, while the second leaves a level partly filled. Both are compatible with the surviving current and width, demonstrating why the requested placement is undetermined.

As magnetic field grows from zero, a clean two-dimensional electron gas first has an ordinary finite longitudinal resistivity, then develops Shubnikov–de Haas oscillations as broadened Landau levels pass the chemical potential. Thermal and disorder broadening obscure the low-field oscillations; their visibility grows when the level spacing exceeds those broadening scales. In a well-developed integer quantum Hall regime, ${\rho }_{xx}$ has minima close to zero on Hall plateaux and maxima during transitions between them, when extended states cross the chemical potential. Detailed peak heights and envelopes depend on disorder and scattering and are not fixed by quantization alone.

At fixed density the nominal filling is $\nu (B)=N_{S}h/(eB)$, so level-filling features are approximately periodic in $1/B$, with period $e/(N_{S}h)$ for resolved spin levels or $2e/(N_{S}h)$ for an unresolved spin pair. Increasing $B$ lowers the filling; spin splitting can create additional odd-index features. The original schematic below shows these relations without pretending to reconstruct the missing data.

TODO — missing source graph: the actual $0$–$12\mathrm{T}$ positions and integer plateau labels require $N_S$ or surviving plateau fields from part (b). Neither is provided independently. Thus a uniquely labelled field-axis sketch for this sample cannot be completed. The following DOS diagram distinguishes extended centres and localized tails generically; its chemical-potential markers are illustrative and do not stand for the requested 8 T or 10 T assignments.