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Magnetic neutron diffraction uses scattering of neutrons by ordered magnetic moments to infer magnetic order. Neutrons also scatter from atomic nuclei, so a pattern can contain nuclear and magnetic contributions. A Bragg scattering peak encodes periodicity; it is not a photograph of an individual spin.

A minimal original calculation makes the extra periodicity visible. Consider identical nuclear scatterers at positions $x_n=na$, with equal neighboring separation $a$, and ordered magnetic moments alternating between $+m_A$ and $-m_B$. A two-site repeating block of length $2a$ has nuclear and magnetic scattering amplitudes, apart from common geometric factors,Here $b$ is the common nuclear scattering coefficient and $q$ the component of the scattering wave vector along the row. The magnetic expression assumes equally scattering point magnetic moments perpendicular to the scattering wave vector; real magnetic scattering measures the perpendicular component and has a momentum-dependent amplitude. This one-dimensional calculation is a periodicity illustration, not a finite-temperature solution for a chain.

For even $h$, $F_N=2b$ and $F_M=m_A-m_B$. For odd $h$, $F_N=0$ and $F_M=m_A+m_B$. Thus equivalent opposing magnetic sublattices, $m_A=m_B$, cancel in forward scattering but reinforce at the odd positions of the doubled period. The intensity is proportional to the squared magnitude of the scattering amplitude, so changing an amplitude's sign does not make a negative intensity.

This calculation also exposes a useful difference between antiferromagnetism and ferrimagnetism: with $m_A=2$ and $m_B=1$, magnetic squared amplitudes are $1$ at even $h$ and $9$ at odd $h$. Extra magnetic peaks can therefore occur in a ferrimagnet too. Identifying a real magnetic structure needs the full crystal basis structure factor, multiple reflections, and the allowed moment orientations, rather than a single extra peak.

Normalize the plane waves in a periodic crystal of area or volume $\mathcal{V}$. The Fourier coefficients of the periodic potential obeyTranslation of the integral by any lattice vector $\mathbf{R}$ leaves it invariant but multiplies it by $e^{-i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{R}}$. A nonzero result therefore requires $\mathbf{k}-{\mathbf{k}}^{\prime }$ to be a reciprocal lattice vector.

Near an isolated Bragg scattering degeneracy, ordinary nondegenerate perturbation theory fails because the free energies coincide. Retain the resonant plane waves $|\mathbf{k}⟩$, $|\mathbf{k}+\mathbf{G}⟩$ and apply degenerate perturbation theory. Absorb the constant $V_0$ into the energy origin. Since a real potential has $V_{-\mathbf{G}}=V_{\mathbf{G}}^{\ast }$, the projected Schrödinger equation has coefficient matrixA nonzero eigenstate requiresOff-resonant plane waves give higher-order energy shifts omitted in this leading nearly-free electron model.

A Bravais lattice is the set of all integer combinations of linearly independent primitive translation vectors. Its reciprocal lattice consists of $g$ satisfying $g\cdot l\in 2\pi \mathbb{Z}$ for every translation $l$; equivalently $e^{ig\cdot l}=1$.

The original PDF's third family has a bare $j$, not $n_{2}j$. Taken literally, the listed set is not the claimed Bravais lattice: the claimed primitive vector $a(i+k)/2$ does not occur. Its half-integer first and third coordinates force it into that third family, whose second coordinate is always $a$, not zero. The intended repair is $j\mapsto n_{2}j$ in the third family. The following conclusions apply to that corrected set.

The four corrected families are exactlyThe allowed parity patterns are all even, or exactly two odd coordinates. An integer combination of the proposed primitive vectors has coordinates $(a/2)(m_2+m_3,m_1+m_3,m_1+m_2)$, so lies in this set. Conversely, solveAll three are integers precisely when $h+k+l$ is even. This proves that the corrected set is the face-centered cubic lattice with the proposed primitive basis.

Direct dot products give $a_i\cdot b_j=2\pi {\delta }_{ij}$ forThese vectors generate the entire reciprocal lattice, since the coefficients of an arbitrary reciprocal vector in this basis are its integer pairings with the $a_i$. Its vectors are $(2\pi /a)(h,k,l)$ with all three coordinates of the same parity, the body-centered cubic lattice.

In Bragg scattering, translating a scattering atom by $l$ changes the amplitude by $e^{i(k-k^{\prime })\cdot l}$. Constructive interference from every lattice translate requires this factor to be one, so $k^{\prime }-k=g$ is a reciprocal lattice vector. For elastic scattering, $|k^{\prime }|=|k|$; the chord between these two vectors has length $2|k|\sin (\theta /2)$. ThereforeThe shortest nonzero reciprocal vectors are the $(111)$ family of length $(2\pi /a)\sqrt{3}$, followed by the $(200)$ family of length $(2\pi /a)2$. Identical atoms on this Bravais lattice introduce no additional basis extinction. If both families are accessible at the incident wavelength, their scattering angles obey $\boxed{\sin ({\theta }_1/2)/\sin ({\theta }_2/2)=\sqrt{3}/2}$.

The reciprocal lattice is ${\Lambda }^{\ast }=\{G:G\cdot \ell \in 2\pi \mathbb{Z}\text{ for every }\ell \in \Lambda \}$. With oriented primitive-cell volume $v_c=a_1\cdot (a_2\times a_3)$, its basis is $b_1=2\pi (a_2\times a_3)/v_c$ and the cyclic variants; thus $b_i\cdot a_j=2\pi {\delta }_{ij}$.

The total potential is $V(x)={\sum }_{\ell }U(x-\ell )$. Using the Fourier transform convention $\tilde{U}(q)=\int e^{-iq\cdot x}U(x)d^{3}x$, translation gives $\tilde{V}(q)=\tilde{U}(q)\Delta (q)$, withThe given Born approximation therefore supplies the amplitude $\boxed{-m\Delta (q)\tilde{U}(q)/(2\pi {\hslash }^2)}$. A finite geometric series gives, when the denominators are nonzero,At reciprocal-lattice vectors each phase is one, so $|\Delta |=N_1{N}_2{N}_3$, the number of atoms. For large finite crystals these narrow peaks represent coherent Bragg scattering, with peak intensity proportional to the squared number of atoms; their finite width reflects the finite crystal size.

For the specified face-centered cubic lattice, $v_c=a^3/4$ andThe reciprocal lattice have coordinates $(2\pi /a)(h,k,l)$ with $h,k,l$ all even or all odd. The two shortest nonzero length families are $G_1=2\pi \sqrt{3}/a$ for $(111)$ and $G_2=4\pi /a$ for $(200)$. Elastic scattering satisfies $|q|=2|k|\sin (\theta /2)$. The assumed incident magnitude exceeds $G_2/2$, so both families are kinematically accessible, andThis is the angular prediction for suitable crystal orientations, or for a powder containing those orientations. A fixed single-crystal orientation must additionally satisfy the vector Elastic Bragg scattering condition $2k\cdot G+|G{|}^2=0$; the magnitude bound alone cannot guarantee both peaks for every incident direction.

The reciprocal lattice is the set of wavevectors $\mathbf{G}$ satisfying $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every direct Bravais lattice translation $\mathbf{R}$. Equivalently, $\mathbf{G}\cdot \mathbf{R}$ is an integer multiple of $2\pi$. It describes the spatial periodicities that can produce coherent Bragg scattering.

For direct primitive vectors ${\mathbf{a}}_i$ and oriented cell volume $\Omega ={\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)$, the primitive reciprocal vectors areThey obey ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$, and every reciprocal vector is an integer combination of them. Their units are inverse length. The reciprocal primitive cell volume is $(2\pi {)}^3/|\Omega |$, so expanding a direct lattice contracts its reciprocal scale.

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.

Both models respect Bloch's theorem and describe electrons in a periodic lattice, but they start from different useful limits.

In the nearly-free electron model, start with extended plane waves and free-electron energies ${\epsilon }_{\mathbf{k}}^0={\hslash }^2{k}^2/(2m)$. A weak periodic potential has Fourier coefficients $V_{\mathbf{G}}$ at reciprocal lattice vectors and couples $\mathbf{k}$ to $\mathbf{k}-\mathbf{G}$. Away from degeneracies its corrections are small. Near a Bragg scattering, retain the two resonant plane waves and diagonalizeThe resulting eigenvalues areAt the crossing the gap is $\boxed{2|V_{\mathbf{G}}|}$. The two standing-wave combinations place electronic density differently relative to the periodic potential and therefore acquire different energies. Folding the many free parabolas into the first Brillouin zone produces the repeated band labels; the nonzero potential opens avoided crossings between coupled states. A Fourier component forbidden by symmetry need not open its corresponding first-order gap.

In the tight-binding model, instead start with localized orbitals on well-separated atoms or sites, with discrete on-site energies. Their overlap permits tunneling represented by hopping integrals. Bloch sums of these orbitals diagonalize the lattice translation symmetry. For one orthonormal orbital per site in a one-dimensional nearest-neighbor chain,The many initially degenerate copies of one atomic orbital broaden into a band. Other atomic orbitals or multiple sites per primitive cell produce additional bands; their hybridization creates bonding and antibonding combinations and may open gaps. For two orbitals per cell, one illustrative Hamiltonian iswith eigenvalues $(E_A+E_B)/2\pm \sqrt{(E_A-E_B{)}^2/4+|t_1+t_2{e}^{-ika}{|}^2}$. This shows explicitly how two bands arise from the two cell degrees of freedom, even when the isolated sites have equal on-site energies.

For each orbital degree of freedom per primitive cell there is one band per spin; band filling and the presence of a gap at the Fermi level determine metallic or insulating behavior. The nearly-free limit emphasizes weak Bragg scattering of extended states, while tight binding emphasizes hopping between localized states. They are complementary approximations to electronic band structure, rather than two incompatible explanations of why periodic solids have bands.

Put $C={\hslash }^2/(2m)$, $v=|V_G|$, $\mathbf{G}=(2K,0)$, and $k_x=K+q$. The upper eigenvalue of the two coupled plane waves isAt fixed Fermi energy $C{k}_F^2$, its closed Fermi contour hasThus $A(v)=4{\int }_0^{q_m}y(q)dq$ can be inverted numerically to estimate the Bragg scattering coupling. The pocket disappears at $v=C(k_F^2-K^2)$. The estimate assumes a fixed chemical potential and neglects coupling to other plane waves; a measured pocket area alone does not determine all Fourier coefficients of an arbitrary lattice potential.