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Alexander Pines coauthored a method enhancing nuclear magnetic resonance signals from dilute nuclear spins in solids. Repeated transfer of polarization from abundant spins, usually protons, strengthens the rarer species' signal. Decoupling the abundant spins during observation then improves spectral resolution. The technique separates the problem of building a detectable spin polarization from the problem of measuring a narrow resonance, making otherwise weak solid-state spectra accessible.^[ref]

In a uniform three-dimensional Hartree-Fock electron gas with repulsive contact coupling $\gamma$, a spin-independent neutralizing background can cancel the Hartree energy. The remaining same-spin exchange term is negative and quadratic in each spin density. Integrating the filled single-spin Fermi sphere gives $A=3{\hslash }^2(6{\pi }^2{)}^{2/3}/(10m)$. At fixed total density the exchange term favors spin polarization, whereas the kinetic term opposes it.

A transverse magnetic field rotates a nonequilibrium spin polarization while diffusion and spin relaxation redistribute it. The field dependence of a detector signal can constrain spin lifetime and transport parameters. A distribution of diffusion times and rotation of the injecting/detecting electrodes must be included when relevant.

The Hartree-Fock method minimizes the expectation of an interacting-electron Hamiltonian over normalized Slater determinants of orthonormal one-particle spin orbitals. It is a variational, self-consistent mean-field approximation. The determinant enforces fermionic antisymmetry exactly within that class, while each orbital responds to the mean charge density and a nonlocal exchange operator generated by the occupied orbitals.

For an electron gas, this accounts for kinetic pressure, the average electrostatic field, same-spin exchange and the associated exchange hole, and possible spin polarization. It does not account for general dynamical electron correlation, such as the correlated avoidance of opposite-spin electrons, and cannot capture all screening or collective fluctuations merely by optimizing one determinant. Uniform plane waves are an additional restriction, not the definition of Hartree-Fock itself.

In the energy functional, the one-body term is the sum of electron kinetic energy and ionic-potential energy. The direct two-body term is the Hartree energy: the classical interaction of the mean electron density with itself. Its factor $1/2$ avoids counting every pair twice. The negative Fock term comes from exchanging the two indistinguishable electron labels in the antisymmetric state. For a spin-independent interaction, spin orthogonality makes this exchange act only between equal spin components. For an orbital interacting with itself, direct and exchange contributions cancel exactly; the exchange term removes the spurious one-electron self-interaction. Ionic background terms and their prescribed cancellation must be included consistently with the direct term.

This is the charge-and-parity-conjugate reaction of part (i). There are $t$-channel Z boson and crossed W boson contributions, with all fermion arrows reversed relative to part (i). For matched unpolarized kinematics and negligible CP violation,Combining all four results in the regime specified in part (i),In particular, conjugating both initial particles restores the part-(i) rate; conjugating the neutrino alone does not. Initial spin polarization and energies near a weak-boson mass require a separate comparison.

The information in the original question is insufficient to determine a unique spin-induced change of charge density. It specifies ideal dielectric filaments with a prescribed excess linear charge density, but not the charge carrier species, their spin state, charge carrier mobility, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.

For a carrier moving with its filament, the local magnetic field in its instantaneous rest frame is, for the transverse fields here,where $\mathbf{u}$ is that carrier's laboratory velocity. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.

In Case 1 there is a common rest frame in which both filaments and the excess charges are stationary and the interfilament magnetic field vanishes. Equivalently, in the laboratory ${\mathbf{B}}_A=\mathbf{v}\times {\mathbf{E}}_A/c^2$, so B's co-moving charges have ${\mathbf{B}}_{A,\mathrm{rest}}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic spin-orbit coupling of electrons moving within their atomic binding potentials; that material physics was never specified.

In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ on mobile carriers; spin relaxation could instead produce a spin polarization. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable Thomas precession.

For example, if $\eta >0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.

Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data. In particular, the factor $\gamma$ relating proper and laboratory linear charge densities in part (b) follows from length contraction and is not an effect of spin-orbit coupling. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.

Why degenerate perturbation theory is needed. If several unperturbed states have equal or nearly equal energy, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small energy denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the eigenvalues give the leading splitting. Other remote states can then be included perturbatively.

For the linear combination of atomic orbitals description of molecular hydrogen, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal energy $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and Hamiltonian coupling $h$, one instead solves a generalized eigenvalue problem:The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin electrons occupying the bonding molecular orbital describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total binding energy; the two-state splitting alone is not the full molecular energy.

In the nearly-free electron model, the analogous degenerate states are plane waves $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$ at a Bragg scattering. A Fourier coefficient $U_{\mathbf{G}}$ of the periodic potential couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix givesThe associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting basis states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.

Origin of broad and narrow bands. Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each wavevector. Different hopping ranges and orbital symmetry determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition metal contains several d orbitals and more complicated dispersions.

Hybridized eigenvalues and eigenstates. Let $|{\psi }_{\mathbf{k}}⟩=a|{\varphi }_{\mathbf{k}}⟩+b|{\chi }_{\mathbf{k}}⟩$, in an orthonormal basis. Hermitian operator fixes the reverse coupling to be $V^{\ast }$. Projection onto the two components givesSolving this quadratic yields the band hybridization dispersionFor nonzero $V$, one eigenvector representation is $(a,b)\propto (V,E-E_1)$, followed by normalization. Thus the wavefunctions exchange broad-band and d-band character through the avoided crossing; the levels are equal mixtures at the uncoupled crossing.

At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2m{E}_d}/\hslash$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_{-}<E_d$ everywhere and tends to $E_d$ from below as $k\to \infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.

Fixed particle number and the larger Fermi sphere. Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two spin states, the broad-band number density at $V=0$ is $k_0^3/(3{\pi }^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero temperature and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. ThereforeThis is the stated factor increase. It uses fixed total particle number, equal spin counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then Fermi-volume conservation in two hybridized electron bands fixes a sum of volumes rather than this one radius. The exactly flat band is degenerate at $V=0$, so its initial occupation must be specified, as it is here.

Fermi-velocity suppression. Set $\Delta (k)=E_1(k)-E_d$. The exact derivative of the lower branch isAt $k=K$, $E_1(K)=2^{2/3}{E}_d$ and $\Delta (K)=(2^{2/3}-1)E_d>0$. For $|V|\ll \Delta (K)$, the square bracket divided by two is $|V{|}^2/\Delta (K{)}^2+O(|V{|}^4/\Delta (K{)}^4)$. Since group velocity is ${\hslash }^{-1}dE/dk$ and the broad-band reference velocity is $v_F^{(1)}=\hslash k_0/m$,so the flat-band hybridization Fermi-velocity suppression isThe reference velocity is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.

The low-temperature qualification becomes more stringent as $V\to 0$: the new chemical potential approaches $E_d$ with $E_d-\mu \simeq |V{|}^2/\Delta (K)$, and the relevant thermal energy must be small compared with this hybridization scale. At fixed nonzero temperature the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.

Density of states and magnetism. For an isotropic three-dimensional branch with spin degeneracy two,A small Fermi velocity therefore produces a large density of states. The localized, partially filled d-derived bands in iron, cobalt and nickel place many states close to the Fermi level; broad s/p-derived conduction bands in copper, silver and gold coexist with filled d bands lying below it. This explains the substantially larger Fermi-level density of states of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.

A high density of states lowers the kinetic-energy cost of spin polarization. A repulsive exchange interaction can then satisfy the Stoner criterion and stabilize ferromagnetism. With a single-spin density of states $g_s$ and an on-site interaction normalization, the criterion is $U{g}_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble metals do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.

The energy favoring spin polarization must compete with the increase in Fermi kinetic energy. Assume the usual three-dimensional zero-temperature parabolic conduction band at fixed $N$ and $V$. Write the spin polarization as $\zeta =(N_{\uparrow }-N_{\downarrow })/N$, so $N_{\uparrow ,\downarrow }=N(1\pm \zeta )/2$. The unpolarized Fermi energy isThe two spin populations have Fermi energies $E_F(1\pm \zeta {)}^{2/3}$. Summing their individual $3E_{F,\sigma }/5$ mean kinetic energies givesThe specified interaction is an energy per electron, $U(1-{\zeta }^2)/4$. Thus the energy change per electron near the unpolarized state isThe quadratic coefficient changes sign at the Stoner threshold for an interaction normalized per electron,The positive quartic term gives a continuous onset in this model. More globally, $f^{\prime \prime }(\zeta )=(E_F/3)[(1+\zeta {)}^{-1/3}+(1-\zeta {)}^{-1/3}]-U/2$. Convexity of $x^{-1/3}$ makes this nonnegative for $U\le 4E_F/3$, so no lower polarized minimum appears before that threshold.

At fixed number density $N/V$, the critical $U$ is independent of the total sample size. A comparison restricted to a fully polarized state gives a different numerical coefficient, $U>\frac{12}{5}{E}_F(2^{2/3}-1)$, but the same requested density scaling; it misses the partially polarized onset. One must also distinguish this $U$, with units of energy, from a contact coupling multiplying an interaction-energy density $g{n}_{\uparrow }{n}_{\downarrow }$: here $g=U/(N/V)$. Importing a differently normalized Stoner criterion would introduce an erroneous extra density factor.

Work at zero temperature and fixed total carrier density. Write $n_{\uparrow }=n/2+\delta$ and $n_{\downarrow }=n/2-\delta$. Let $g_s(E_F)$ be the single-spin density of states per unit volume in the unpolarised state. Since $d{\mathcal{E}}_s/d{n}_s=E_{Fs}$ and $d{E}_{Fs}/d{n}_s=1/g_s$, the linear kinetic-energy shifts cancel and the quadratic cost isThe repulsive interaction instead decreases under spin polarization:Hence the Stoner criterion for an opposite-spin contact repulsion isThe two forms differ only in whether the stated density of states counts one spin or both. For the three-dimensional free electron model, $g_{\mathrm{tot}}(E_F)=3n/(2E_F)$, soEquivalently, for fractional polarisation $\zeta =(n_{\uparrow }-n_{\downarrow })/n$, the kinetic-energy density is $\frac{3}{10}n{E}_F[(1+\zeta {)}^{5/3}+(1-\zeta {)}^{5/3}]$; its quadratic coefficient is $n{E}_F/3$, while the interaction contributes $-U{n}^2/4$. With number densities, $U$ has units of energy times volume and the displayed interaction is an energy density; multiplying by the sample volume does not change the threshold.

At $N=M$, the two spin occupations per site are $x$ and $1-x$. With $\tau =|t|>0$, the trial energy per site isIts derivative is $e^{\prime }(x)=-4\tau \cos (\pi x)+V[4x-2-(2/\pi )\sin (2\pi x)]$. Since $x=0$ is a boundary point, its local minimum test is one-sided: $e^{\prime }(0)=-4\tau -2V$. At $V=-2\tau$, where this derivative vanishes, its Taylor series begins $e(x)-e(0)=(2{\pi }^2\tau /3)x^3+O(x^4)>0$ for small positive $x$. HenceThe same applies to $x=1$ by spin interchange.

The symmetry $e(x)=e(1-x)$ makes $x=1/2$ a stationary point for every $V$. Here $e^{\prime \prime }(1/2)=4\pi \tau +8V$, so the second-derivative test settles all cases except equality. Writing $y=x-1/2$, at $V=-\pi \tau /2$ the Taylor series is $e(1/2+y)-e(1/2)=({\pi }^3\tau /6)y^4+O(y^6)$. ThusThe minimizing branches interpolate continuously. To find all of them, put $u=\pi (1/2-x)$ on $0<x<1/2$. An interior stationary point obeysThe right side decreases strictly from $-\pi /2$ to $-2$, sinceThe positivity follows from $u-\sin u\cos u={\int }_0^{u}2{\sin }^{2}vdv$. There is therefore exactly one such stationary point on each side of $1/2$ when $-2<V/\tau <-\pi /2$. The signs of $e^{\prime }$ show that both are global minima, whereas $1/2$ is then a local maximum. The complete answer isThe pair merges through a supercritical pitchfork bifurcation at $V/\tau =-\pi /2$. Comparing only $e(0)$ and $e(1/2)$ would miss the intervening ground states with partial spin polarization.

A local moment induces conduction-electron spin polarization, which mediates a second-order coupling to another moment. The static electronic spin susceptibility fixes the spatial interaction. For a three-dimensional spherical Fermi surface its leading large-distance form oscillates as $\cos (2k_{F}R)/R^3$, up to the overall sign convention and prefactor. The sign alternates with separation. Other band structures and dimensions change this form. ^[ref]

Spin-flip scattering is a collision or scattering event that changes a particle's spin projection along a specified quantization axis. For electrons, coupling to another magnetic moment or coupling of motion to spin through spin-orbit coupling can transfer angular momentum between the electron spin and other degrees of freedom.

Repeated events can relax a nonequilibrium spin polarization, making spin-flip scattering one mechanism of spin relaxation. The axis and state basis must be stated: in a spin-orbit-coupled material, the propagating energy eigenstates need not be pure up or down states along one fixed axis. Conversely, decay of transverse coherence need not imply flips; different precession phases can cancel the ensemble's transverse signal while individual spin populations remain unchanged.

Loss of transported spin polarization at an interface, distinct from bulk spin relaxation in an adjacent layer.

For a spin-one-half ensemble, spin polarization measures unequal occupation or coherent alignment of spin states. A spin-dependent energy splitting does not alone determine the ensemble's response: occupation, temperature and spin relaxation matter. A spin polarization is distinct from a net change of charge density and from the optical polarization of waves.

Spin relaxation is the approach of nonequilibrium spin populations and coherences toward their equilibrium values through coupling to other degrees of freedom or environmental fluctuations. A spin polarized away from equilibrium can lose its net polarization even when the total angular momentum of the larger closed system is conserved.

For a simple ensemble in a fixed magnetic field, longitudinal relaxation with time $T_1$ restores the population difference along the field, while transverse relaxation with time $T_2$ damps the rotating transverse spin polarization. Static field inhomogeneity can produce an additional reversible spread of precession phases, so an observed free-induction decay time need not equal the intrinsic $T_2$. Spin-flip scattering changes populations; phase-randomizing processes can also suppress transverse coherence without individual flips. A Hanle spin-precession measurement probes the competition of precession, transport and relaxation.

A spin texture describes how a spin expectation value or spin polarization varies with position or wavevector. It specifies a vector field, including its magnitude when that matters; a direction field alone is undefined wherever that magnitude vanishes. A Rashba spin texture is an example in which spin expectation values wind around circles in the plane of wavevectors.

For the auxiliary-field convention with cost $M^2/(2U)$ and coupling $-M\bar{\psi }{\sigma }_z\psi$, the homogeneous saddle point equation equates M to U times spin polarization per site. M has energy units. It is not itself a magnetic moment, and its coefficient changes if the contact interaction or density-of-states spin convention is changed. The exact three-component decoupling must be distinguished from selecting one constant spin direction as a mean-field approximation.