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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.

An isolated spin-degenerate lowest energy band accommodates two electrons per atom when there is one atom per primitive cell. Starting from one per atom, the estimate for filling the band is thereforeThis corresponds to an additional areal carrier count of about $4.77\times 10^{19}{\mathrm{m}}^{-2}$. If a genuine band gap separates the filled first band from the next band, there are no partially occupied band states at low temperature, so ordinary metallic conduction is lost.

Simply touching an edge is insufficient. In the circular approximation first contact would occur at $k_F=\pi /a$, corresponding to $n_e{a}^2=\pi /2\simeq 1.57$ electrons per cell; that is a partly filled band and need not be insulating. Near zone boundaries, the periodic potential reshapes the free-electron contours. The two-electron filling estimate assumes a global gap without overlapping bands, rather than inferring an insulator merely from one local avoided crossing.

Formal ionic counting gives La(III), N(−III), and therefore Re(VI), with a $5d^1$ configuration. An isolated localized $d^1$ center would carry a saturation magnetic moment of order $1{\mu }_{\mathrm{B}}$, much larger than the value quoted. It is not a $d^0$ ion made diamagnetic simply by an octahedral ligand field.

The spatially extended Re $5d$ orbitals overlap strongly with N $2p$ orbitals. Re–N $\sigma$ interactions raise the $e_g$-derived antibonding states, while $\pi$ interactions mix the $t_{2g}$ states with the nitrogen network. Repetition through corner-sharing octahedra turns these combinations into broad energy bands. The nominal $d$ electron occupies a delocalized, partially filled band rather than an isolated magnetic orbital.

Only electrons close to the Fermi energy change their occupations appreciably in an applied magnetic field. Consequently a Pauli paramagnetism has a small, nonzero field-induced magnetization:when $g$ is the density of states including both spin directions, expressed per formula unit and per unit energy. This band model explains a very small moment without requiring the absence of electrons in the Re-derived band. The reported material has metallic transport and a Pauli-like response.^[ref]

The description “saturated” in the question should not be interpreted as complete alignment of one localized spin per Re: a nearly linear metallic response over a laboratory field range need not be a true fully spin-polarized saturation moment.

With spin degeneracy, each energy band contains two electrons per primitive unit cell. Divalency therefore permits a filled first band, but only if its maximum lies below the second band's minimum. With negligible ionic periodic potential, the circular Fermi surface and the rectangular zone have equal enclosed area, not equal shape. Electrons occupy some states outside the first zone while some higher-energy first-zone states remain empty. Folded bands overlap, and the material is a metal.

A periodic potential mixes plane waves differing by a reciprocal lattice vector. At a two-wave Bragg point, the nearly-free electron model gives a direct gap $2|V_{\mathbf{G}}|$. Increasing appropriate Fourier coefficients can remove the band overlap. With the first band filled, the system is then a band insulator, provided the temperature is small compared with the insulating band gap in energy units.

The rectangular free-electron band-overlap scale follows from two different points in reciprocal space. Put $C={\hslash }^2/(2m_e)$. Before gap opening, the highest first-zone energy is at the corner, while the lowest second-zone energy is at the midpoint of the nearer face:Thus the downward shift ${\delta }_v$ of the first-band maximum plus the upward shift ${\delta }_c$ of the second-band minimum must overcomeIn a rigid-splitting estimate, an opening $\Delta$ lowers the first band by $\Delta /2$ and raises the second by $\Delta /2$, giving the familiar threshold $\Delta >{\hslash }^2{\pi }^2/(2m_e{b}^2)$. Equality is only band touching, not a positive insulating gap. This expression is the overlap that must be removed, under that explicit band-shift approximation.

Positive direct gaps can coexist with band overlap: a local direct splitting and the global indirect band gap are not interchangeable. The shape and Fourier coefficients of the ionic potential are unspecified, so there is no universal exact critical direct-gap width determined by $a,b$ alone. For a concrete demonstration, a separable potential $2V_x\cos (2Kx)+2V_y\cos (2Ly)$ has axial direct gaps ${\Delta }_x=2|V_x|$ and ${\Delta }_y=2|V_y|$. To first order, the corner's lowest level is $C(K^2+L^2)-|V_x|-|V_y|$, and the two candidate second-band minima are $C{K}^2+|V_x|$ and $C{L}^2+|V_y|$. Therefore, within this model and neglecting higher-order band shifts, nonoverlap requiresEqual axial gaps would give the first-order estimate $\Delta >2C{L}^2/3$, rather than the rigid-splitting estimate. Such a large opening can exceed the quantitative range of weak-potential perturbation theory, but the example identifies the missing model assumption: corner levels feel both couplings. The exact general answer is the positive indirect-gap criterion above; a geometry-only direct-gap threshold would claim more than the supplied potential information supports.

An intrinsic semiconductor has its carrier populations set primarily by thermal excitation across the band gap; in the ideal pure, neutral case its conduction-electron and hole densities are equal, $n=p=n_i$. An extrinsic semiconductor has carrier populations significantly controlled by donor or acceptor impurities, producing predominantly n-type semiconductor or p-type semiconductor behavior. Chemical purity alone does not fix conductivity independently of temperature and gap size.

The positive charge carrier mobility is the magnitude of mean drift speed per unit weak electric field. For electrons ${\mathbf{v}}_e=-{\mu }_e\mathbf{E}$, for holes ${\mathbf{v}}_h={\mu }_h\mathbf{E}$, and its SI unit is ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. In an isotropic relaxation-time model, $\mu =|q|\tau /m^{\ast }$.

A band effective mass characterizes how an electron wave packet accelerates in an energy band, rather than replacing its vacuum rest mass. In one dimension and its tensor generalization,These follow by differentiating $v_i={\hslash }^{-1}\partial E/\partial k_i$ and using $\hslash {\dot{k}}_j=F_j$. Effective masses can be anisotropic and electron band curvature can be negative.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

Let the sample area be $A$, set the band touching energy to zero, and write $m^{\ast }=m_e^{\ast }=m_h^{\ast }$. Assume one electronic valley in each energy band with two spin states. The near-edge parabolic dispersions areUnder periodic boundary conditions, each spin occupies reciprocal-space area $(2\pi {)}^2/A$. Counting the quantum states inside a circle givesIts derivative is the constant two-dimensional free-electron density of states. Defining $D=A{m}^{\ast }/(\pi {\hslash }^2)$, the signed-energy density of states isThe total electronic density of states is therefore $D$ on either side of zero, not $2D$ at every signed energy. If instead one counts positive-energy excitations, the electron and hole species each have density of states $D$, giving $2D$ when summed. A common additional valley degeneracy multiplies all these quantities by that factor. The effective mass approximation assumes thermally sampled energies remain within the near-edge regions of both energy bands.

Use the stipulated noninteracting impurity band approximation, with one donated electron per donor and two available spin states per donor orbital. The energy band is half filled. Approximating its spin-summed density of states per volume near the Fermi energy by $D_v(E_F)\simeq 2n/W$ gives the zero-temperature Pauli paramagnetismWith $\alpha =1$,A factor of order one depends on the density of states shape; using the conduction band $g$-factor rather than silently setting $g=2$ matters here.

To locate the comparison with the magnitude of the undoped Larmor-Langevin susceptibility, put $x=n^{-1/3}/a_b>1$. Numerically,For the stated atomic-scale estimate $R=a_0$ in part (b), equality with $|{\chi }_0|=2.599\times 10^{-5}$ requiresThus the dilute-branch crossover iswell below $a_b^{-3}=9.45\times 10^{23}{\mathrm{m}}^{-3}$. Within this model the positive spin response exceeds the negative background magnitude for lower dopant densities on this dilute branch: the exponential narrowing of $W$ wins over the loss of donors. Indeed $d\ln (e^x/x^3)/dx=1-3/x>0$ for $x>3$. This is not a claim of ferromagnetism.

The degenerate Pauli paramagnetism calculation requires $k_{B}T\ll W$, with an unimportant order-one filling factor. At the estimated crossing,The weak-field condition also requires $|g|{\mu }_{B}B\ll W$. For a different assumed atomic $R$, use $|{\chi }_0|=2.599\times 10^{-5}(R/a_0{)}^2$ in the displayed equation for $x$. An order-of-magnitude background $|{\chi }_0|\sim 10^{-5}$ gives the same density order and a crossover scale of order $10^{-5}\mathrm{K}$. At higher temperature than this extremely narrow bandwidth permits, the $n/W$ enhancement cannot be used; the independent-donor Curie law is the appropriate thermal scale if interactions remain negligible. At very small dopant density, disorder, localization and Coulomb interactions can also invalidate the assumed noninteracting impurity band. The formal divergence as $n\to 0$ is therefore a limit of the stated zero-temperature band model, not a prediction for isolated donors at a fixed nonzero temperature.

For zero coupling, $\sqrt{4y^2}=2|y|$, and the two folded energy bands areThey touch at $y=0$, where the underlying free plane wave branches cross. A nonzero Fourier coefficient replaces this crossing by an avoided crossing, with $E_{\pm }(0)=E_{\mathrm{ZB}}(1\pm ϵ)$. The lower energy band moves downward and the upper energy band upward. Both branches are smooth near $y=0$ for $ϵ>0$.

The figure uses the boundary-centred interval $-1\le y\le 1$ required in part (b). With the printed definition $y=1-ka/\pi$, this is the translated Brillouin zone $0\le k\le 2\pi /a$. The printed $0\le y\le 2$ instead corresponds to $-\pi /a\le k\le \pi /a$; the local two-wave formula must be folded appropriately before treating it as a dispersion over that entire zone. It must not be integrated over $0\le y\le 2$ as though it were the filled first energy band there. The subsequent calculation follows the explicit continuation prescribed in part (b).

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

At every $y$ and every $ϵ>0$,so the lower energy band lies strictly below its zero-coupling counterpart. Its state counting in a Bloch band and complete occupation are unchanged. Subtracting the two energy densities directly givesThusEquality holds at $ϵ=0$. This is the downward displacement of the filled lower energy band in the sketch. The upward displacement of the empty upper energy band contributes nothing to the zero-temperature electronic energy density.

For a long chain of length $L$ with periodic boundary conditions, each spin has $Ldk/(2\pi )$ quantum states in a wavenumber interval $dk$. Including two spin states, a full energy band contains $2L/a$ electrons. The state counting in a Bloch band therefore gives the energy densityThe integral of $1+y^2$ is $8/3$. For $ϵ>0$, the even square-root integral isFor example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end logarithm contributes $\ln ϵ$. HenceThe continuous zero-coupling limit is $\mathcal{U}(0)=2E_{\mathrm{ZB}}/(3a)$. This agrees with directly integrating the one-dimensional free electron model up to $k_F=\pi /a$ at two electrons per unit cell.

Define the dimensionless coupling $C=\pi |V_0|/E_{\mathrm{ZB}}$, so that $ϵ=C|\delta |$. Substitution into the energy density from part (b) gives, within the prescribed continuation of the two-wave nearly-free electron model,Its value at $\delta =0$ is understood by continuity. For $C|\delta |\ll 1$,Both electrons per doubled unit cell have been included. The fixed $2G$ harmonic affects the underlying energy bands beyond the two-wave approximation; the expression here is the energy density requested from part (b), rather than an exact solution for all harmonics of the full periodic potential.

Without the Peierls distortion, adjacent chemical atoms are separated by $a/2$. The primitive unit cell contains one chemical atom and contributes one electron. A energy band with two spin states accommodates two electrons per primitive unit cell, so the first energy band is half filled. Its true first Brillouin zone extends to $\pm 2\pi /a$, while the Fermi points are at $\pm \pi /a$.

Describing the same chain with a doubled unit cell of length $a$ merely folds the half-filled energy band into two branches that touch at those Fermi points. The unbroken lattice translation by $a/2$ prevents an artificial band gap there. A Peierls distortion breaks that smaller lattice translation and supplies the missing Fourier coefficient at $2k_F=G$; the lower folded energy band can then be completely filled and separated from the upper energy band. Thus the zero-band gap result at $\delta =0$ agrees with the original one-electron band filling.

Use the same normalization for the elastic energy and electronic energy density: if $\kappa {\delta }^2$ is an energy per unit cell, divide it by $a$ first; if it is already per length, use it directly. Write its coefficient per length as $K>0$ and put $A={\pi }^2{V}_0^2/(4a{E}_{\mathrm{ZB}})>0$. The small-distortion change in total energy density isNo minimization calculation is needed. For arbitrarily small nonzero $|\delta |$, the logarithm is negative with arbitrarily large magnitude. Hence the negative electronic Peierls energy gain, proportional to ${\delta }^2\ln (1/|\delta |)$, eventually exceeds any fixed positive quadratic elastic energy. There are distorted configurations with lower energy than the undistorted configuration for every finite $K$ and nonzero coupling. The ideal zero-temperature infinite-chain model therefore hasThe source's ${\delta }_{\mathrm{eq}}>0$ is correct when $\delta$ denotes a nonnegative distortion amplitude. If $\delta$ is signed, the energy density is even and the two signs describe equally stable patterns related by a lattice translation. For example, $V(x+a/2;\delta )=V(x;-\delta )$, so a negative equilibrium distortion has the same energy as the positive one. The zero-coupling case $V_0=0$ instead has its minimum at $\delta =0$. The argument assumes a coupled chain, complete lower-energy band filling and zero temperature; it does not by itself establish finite-temperature long-range order in an isolated one-dimensional system.

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.

For a nondegenerate band $E(\mathbf{k})$, define the reciprocal effective mass tensor byIt is the curvature tensor of the energy band. Near an extremum, $E\simeq E_0+{\hslash }^2\delta {\mathbf{k}}^T{M}^{-1}\delta \mathbf{k}/2$, so positive principal curvatures give positive electron effective masses. Together with the group velocity $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$ and $\hslash \dot{\mathbf{k}}=\mathbf{F}$, it yields ${\dot{v}}_i=(M^{-1}{)}_{ij}{F}_j$. Negative electron curvature at a valence band maximum is conventionally recast as a positive band effective mass for missing electrons.

There are two electrons per primitive unit cell; a spin-degenerate energy band accommodates exactly that number. A semiconductor results when the first band is full and does not overlap the second. In the leading two-wave/four-wave approximation, compare the first-band maximum at $M$ with the second-band minimum at $X$:This is an indirect band gap; a nonzero local splitting at $X$ alone is insufficient. At the onset,With $a=0.3\mathrm{nm}$, ${\hslash }^2/(2m_e)=0.03810\mathrm{eV}\mathrm{n}{\mathrm{m}}^2$ gives $E_X\simeq 4.178\mathrm{eV}$, henceFor $V_{10}$ more negative than this value the estimated indirect band gap is positive; at equality the bands just touch. This is the onset estimate within the source's nearly-free electron model truncation, not an exact critical potential after coupling all plane waves.

For a vertical interband transition, the crystal momentum is effectively unchanged because the photon momentum is small. A stimulated downward transition needs an occupied upper state and an empty lower state, with factor $f_c(E_2)[1-f_v(E_1)]$; upward photon absorption has factor $f_v(E_1)[1-f_c(E_2)]$. The two matrix elements agree by microscopic reversibility. Their difference is thereforeA common slowly varying transition-strength factor is implicit. Broadening replaces the sharp transition by a convolution; multiple valence bands contribute separate terms.

For one isotropic pair of parabolic energy bands, writeThus $E_2-E_1=E_g+{\hslash }^2{k}^2/(2m_r)$. Including the usual twofold spin degeneracy, the number of paired states per volume inside a sphere of radius $k$ is $k^3/(3{\pi }^2)$. Differentiating with respect to transition energy $E$ gives the three-dimensional parabolic interband joint density of statesFor density per angular frequency, $g_j^{(\omega )}(\omega )=\hslash g_j^{(E)}(\hslash \omega )$. Stating the density convention avoids a missing factor of $\hslash$; other degeneracies multiply the formula.

At a common carrier temperature, the Fermi-Dirac distribution is a strictly decreasing function of $E-F$. The direct-gap semiconductor gain condition follows:The lower inequality denotes a transition above the ideal band edge, where the displayed joint density of states is nonzero. Thus quasi-Fermi-level splitting must exceed the transition photon energy. Equality is transparency, rather than net amplification; at zero temperature an allowed transition must also lie within the occupied-electron and empty-valence windows.

In a direct band gap semiconductor, the small photon momentum makes optical transitions approximately vertical in wave vector. Their energy isThe joint density of states per unit volume per transition energy, including the supplied spin counting, isFor an upper electron state of occupation $f_c$ and a lower electron state of occupation $f_v$, stimulated emission has occupation factor $f_c(1-f_v)$; absorption has factor $f_v(1-f_c)$. Their difference is exactly $f_c-f_v$. Multiplication by the joint density of states therefore givesThe proportionality suppresses the optical transition matrix element and other smoothly varying optical prefactors. It assumes parabolic isotropic energy bands, an allowed direct transition, and negligible linewidth and many-body corrections. Here $f_v$ counts electrons, not electron holes.