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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.

Noninteracting spectral line. Treat $\mathbf{k}$ as a wavevector, so physical crystal momentum is $\hslash \mathbf{k}$. Write ${\Omega }_{\mathbf{k}}={\epsilon }_{\mathbf{k}}/\hslash$; an energy zero at the chemical potential can be chosen throughout. The temporal factor of a noninteracting state is $e^{-i{\Omega }_{\mathbf{k}}t}$. With the Fourier transform convention $\tilde{\psi }(\omega )=\int dt{e}^{i\omega t}\psi (t)$,Thus its normalized frequency spectral function isOne should not square the delta distribution as if it were an ordinary finite-time transition amplitude. More precisely, the fermionic retarded Green function is $G_0^R(\mathbf{k},t)=-i\Theta (t)e^{-i{\Omega }_{\mathbf{k}}t}$. Its Fourier transform is $(\omega -{\Omega }_{\mathbf{k}}+i{0}^{+}{)}^{-1}$, whose negative imaginary part divided by $\pi$ is the same delta peak.

Interactions and lifetime. Interactions change the dispersion through the real part of the self-energy and permit decay into other excitations through its imaginary part. In the simplest quasiparticle-pole model, exponential decay of the retarded amplitude givesPerforming the causal integral,The corresponding Lorentzian quasiparticle spectral peak isIts total frequency integral is one. The exponential amplitude lifetime is $1/{\Gamma }_{\mathbf{k}}$; the probability lifetime is $1/(2{\Gamma }_{\mathbf{k}})$. Describing ${\Gamma }_{\mathbf{k}}$ as a scattering rate therefore requires the lifetime convention to be specified. The energy half-width is $\hslash {\Gamma }_{\mathbf{k}}$.

A general interacting electron is more accurately described by $G^R=Z_{\mathbf{k}}/(\omega -{\Omega }_{\mathbf{k}}^{\ast }+i{\Gamma }_{\mathbf{k}})+G_{\mathrm{inc}}^R$ near the pole. The coherent peak then integrates to the quasiparticle residue $Z_{\mathbf{k}}$ and an incoherent continuum carries the remaining weight. In an energy-unit self-energy convention,Linearizing the real denominator gives $Z=[1-{\hslash }^{-1}{\partial }_{\omega }\mathrm{Re}{\Sigma }^R{]}^{-1}$ and $\Gamma =-Z\mathrm{Im}{\Sigma }^R/\hslash$ at the pole. The requested unit-weight Lorentzian is the normalized quasiparticle model, not the most general exact interacting spectrum.

Peak position and width. At fixed $\mathbf{k}$, put $\gamma =\gamma (\mathbf{k})>0$ and $\Omega =\hslash k^2/(2m^{\ast })$. Differentiation givesIt is positive for $\omega <\Omega$ and negative for $\omega >\Omega$, so the only maximum is at $\Omega$, with height $1/(\pi \gamma )$. Setting $A$ to half this height gives $(\omega -\Omega {)}^2={\gamma }^2$. ThereforeIn energy units the maximum is ${\hslash }^2{k}^2/(2m^{\ast })$ and the half-width is $\hslash \gamma (\mathbf{k})$. Frequency dependence of the self-energy would require reconsidering both the shape and peak position.

Fermi gas and Fermi liquid. A noninteracting Fermi gas has a delta-function spectral line at its bare dispersion for every momentum. A normal three-dimensional Fermi liquid has a shifted coherent peak plus incoherent spectral weight; close to the Fermi surface its width scales as $(\epsilon -\mu {)}^2+(\pi k_{B}T{)}^2$. At zero temperature the peak becomes arbitrarily sharp at the Fermi surface, while it broadens farther away. Spectral functions include addition and removal weight; occupied-only photoemission contains an extra Fermi factor.

Photoemission measurement. A photon of known energy $h\nu$ ejects an electron. Measure its kinetic energy $E_{\mathrm{kin}}$ and emission direction. For binding energy $E_B\ge 0$ below the Fermi level and the appropriate calibrated work function $\Phi$, conservation of energy givesConservation of momentum parallel to a periodic surface giveswhere ${\mathbf{G}}_{\parallel }$ is a surface reciprocal-lattice vector. At suitable photon energies the photon momentum can be neglected; at higher energies it must be retained. The surface breaks perpendicular translation symmetry, so the perpendicular crystal momentum generally needs a final-state model. A two-dimensional band can be reconstructed directly from the parallel components.

Within the sudden approximation and electric-dipole approximation, photoemission intensity and spectral function are related byHere $E$ is measured relative to the chemical potential. Instrumental resolution convolves the product and inelastic scattering adds background. Thus ARPES maps the occupied electron removal spectral function, weighted by a matrix element; it does not measure the full unoccupied spectrum or an unweighted spectral function automatically. Bright ridges give dispersions, and intrinsic widths give decay information after instrumental effects are removed.

Reading the supplied image. Its bright ridges form a graphene-like cone. The crossing near $k_y=0$ is the Dirac point, not in general the Fermi level. In the paper's plotted energy coordinates its energy is roughly $E_D\sim 1.4$–$1.5\mathrm{eV}$. A velocity follows from a branch slope independently of the energy zero:For example, the lower branch moves by about $0.3\mathrm{eV}$ between $|k_y|\sim 0.03$ and $0.11{\mathring{\mathrm{A}}}^{-1}$. This gives $dE/dk\sim 3.8\mathrm{eV}\mathring{\mathrm{A}}$ andReading other points changes the estimate by tens of percent. The original exam image has unequal energy increments on nearly equally spaced labelled ticks, so it does not justify a precise fitted velocity. The value is an estimate of the cone slope extrapolated to the Fermi crossing, rather than a measured derivative exactly at a visibly identified Fermi cutoff.

To determine a Fermi energy, first calibrate the Fermi edge, then subtract the chosen reference energy. Neither a Fermi-level marker nor an explicit energy-zero convention is supplied in this crop. If its upper termination near $1.9\mathrm{eV}$ is intended to represent the Fermi cutoff, the conditional graph answer is $E_F\simeq 1.9\mathrm{eV}$ on that axis, or $E_F-E_D\sim 0.4\mathrm{eV}$ relative to the Dirac crossing. Identifying the crossing itself as the Fermi energy would incorrectly assume undoped graphene.

The research article cited by the exam identifies this as linearly polarized graphene ARPES and independently states $E_F-E_D\simeq 0.4\mathrm{eV}$ for the substrate-doped sample. That resolves the relative energy when this external calibration is admitted.^[ref] The crop alone does not uniquely fix an absolute Fermi energy; the source-supported relative value and the conditional $1.9\mathrm{eV}$ axis reading should be distinguished.

For state counting in a Bloch band, take a primitive cubic lattice of spacing $a$ and impose periodic boundary conditions on lengths $L_i=N_{i}a$. The allowed wavevectors are $k_i=2\pi l_i/L_i$. The first Brillouin zone has side $2\pi /a$, so it contains $N_i$ distinct values in each direction and hence $N_x{N}_y{N}_z=N$ wavevectors. Each orbital-derived energy band has one state per wavevector per spin orientation. ThereforeThe phrase “per atom” attached to $2N$ in the PDF is a counting error: $2N$ is the total, not the per-atom count.

By the Pauli exclusion principle, a monovalent crystal supplies $N$ valence electrons and half-fills a spin-degenerate band; its Fermi level lies within allowed states, giving a metal in the independent-electron picture. Two electrons per atom can fill one band. If a band gap separates it from all empty bands, the crystal is a band insulator. An even valence count only permits insulation: overlapping bands can still produce a metal. Conversely, electron interactions can produce a Mott insulator, and symmetry breaking can permit insulation outside this simple spin-degenerate electronic band structure criterion. The relevant count in a crystal with a basis is per primitive unit cell.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

On the physical density-phase contour put $\bar{\psi }=\sqrt{\rho }{e}^{-i\varphi }$. Direct differentiation givesIntegrating the Laplacian by parts with the stated bulk boundary convention, and dropping the periodic total derivative of $\rho$, produces the density-phase action of a Bose gas,Write $\rho ={\rho }_0+\delta \rho$ and $\varphi ={\varphi }_0+\phi$, where ${\varphi }_0$ is constant. To quadratic order,The uniform static density fluctuation has positive curvature $g$. Its inverse static susceptibility at wavevector $q$ is $A_q=g+q^2/(4m{\rho }_0)$, which remains nonzero at $q=0$. This is the sense in which the amplitude or density fluctuation is massive. The phase has no term proportional to ${\phi }^2$ without derivatives; a uniform phase rotation costs no action. It is the massless direction required by global $U(1)$ symmetry.

The term $i\delta \rho {\partial }_{\tau }\phi$ shows density-phase conjugacy. It is important to distinguish static density stiffness versus an excitation gap: the density and phase do not form two independent normal-mode branches with one gapped and one gapless particle. In this nonrelativistic first-order theory they supply the conjugate variables of one positive Bogoliubov branch, derived below. The omitted cubic terms include $\delta \rho |\nabla \phi {|}^2/(2m)$ and corrections from the density-gradient denominator.

There is also a background term $i{\rho }_0\int d\tau d^{d}r{\partial }_{\tau }\phi$. It vanishes for smooth fluctuations with zero temporal winding. Periodicity of $\psi$, however, permits $\phi (\beta )-\phi (0)=2\pi w$. The compact phase winding term must be retained when such sectors are included; it is not generally the same constant in all sectors. The local Gaussian phonon calculation concerns the smooth zero-winding sector and does not describe vortices or points where the density vanishes.

Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

The periodic boundary condition requires $e^{ikM}=1$, so there are exactly $M$ independent wavevectors in a Brillouin zone:The inverse discrete Fourier transform, with the sign convention of the paper, isFourier orthogonality gives the tight binding dispersion relation ${ϵ}_k=-2t\cos k$. For the interaction, first use the canonical anticommutation relations at distinct sites to write $n_{\sigma ,j}{n}_{\sigma ,j+1}=c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}^{†}{c}_{\sigma ,j+1}{c}_{\sigma ,j}$. Summing its discrete Fourier transform over $j$ enforces crystal momentum conservation, givingEvery wavevector is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing fermions changes $q$ to $-q$ with two cancelling fermionic signs. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.

Let $\Psi =k_{\alpha }{x}^{\alpha }$. For a nonzero symmetric amplitude, substituting the plane wave into the wave equation gives $-(k_{\sigma }{k}^{\sigma }){ϵ}^{\mu \nu }\cos \Psi =0$. The gauge condition gives $-k_{\sigma }{ϵ}^{\nu \sigma }\sin \Psi =0$. Thus a propagating mode satisfiesThe wavevector is null, so the gravitational wave propagates at the vacuum speed of light. The amplitude is transverse as a four-dimensional tensor in Lorenz gauge in linearized gravity. This condition alone does not set all time components or the trace to zero: residual gauge symmetry of linearized gravity remains. Of the ten symmetric components, four Lorenz constraints and four independent residual coordinate choices leave two physical gravitational wave polarizations.

Put $R=h_t-\nu {\nabla }^{2}h$, where the diffusion coefficient $\nu$ is constant. ThenOne temporal and two spatial integrations by parts givewith compactly supported variations. HenceHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. The cross terms cancel because $\nu$ is constant and spatial and temporal derivatives commute. Equivalently, $P^0=2R$ and $Q^{ij}=-2\nu R{\delta }^{ij}$ in part (b), producing $-2{\partial }_{t}R-2\nu {\nabla }^{2}R=0$.

This squared diffusion-residual action does not have only the first-order diffusion equation as its stationary equation. The zero-residual paths $h_t=\nu {\nabla }^{2}h$ have $S=0$ and are certainly solutions, but the variational equation is the higher-order one above. For a spatial Fourier mode with nonzero wavevector $\mathbf{k}$, it givesBoundary data or an additional physical selection principle are needed to retain only the decaying diffusion branch. The spatially uniform mode instead has $h_0(t)=A_0+B_{0}t$ in the stationary equation.

For elastic scattering by a real short-range scalar potential, write $C=m/(2\pi {\hslash }^2)$, ${\mathbf{k}}_i=\mathbf{k}$ and $|{\mathbf{k}}_f|=k$. In the far field, $|\mathbf{r}-{\mathbf{r}}^{\prime }|=r-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }+O(r^{-1})$ and ${\mathbf{k}}_f=k\hat{\mathbf{r}}$. The outgoing spherical-wave coefficient isSubstitution of the incident plane wave inside this integral givesThe incoming and outgoing flux factors cancel because their speeds agree. Here $\mathbf{q}={\mathbf{k}}_f-{\mathbf{k}}_i$ is a wavevector transfer; the actual momentum transfer is $\hslash \mathbf{q}$. This distinction preserves dimensions. The approximation requires the scattered correction to the interior wave to be small, and does not in general apply to an unscreened long-range potential.

Work in temporal gauge, with zero scalar potential, and use the charge sign exactly as in the given covariant spatial derivative. A spatially uniform vector potential acts on a Fourier mode asThere are no spatial derivatives of $\mathbf{A}$. Substitution into the field equation and multiplication by $c^2$ therefore givesThe canonical wavevector $\mathbf{k}$ is conserved by spatial translation symmetry; the kinetic momentum is $\hslash \mathbf{k}-e\mathbf{A}(t)$. The absence of a term involving $\dot{\mathbf{A}}$ in this mode equation follows from temporal gauge and the spatial minimal-coupling operator. The electric field still affects evolution through the changing oscillator frequency.

Fermion occupation and ordinary metallic heat capacity. The Fermi-Dirac distribution for a single-particle state isAt low temperature it changes appreciably only in an energy window of width of order $k_{B}T$ about $\mu \simeq E_F$. If the full density of states $D(E)$ is smooth and nonzero there, only about $D(E_F)k_{B}T$ states change population. Each excited particle or hole carries energy of order $k_{B}T$ relative to the Fermi surface, so $U(T)-U(0)$ is of order $D(E_F)(k_{B}T{)}^2$. Differentiation therefore gives $C_{\mathrm{el}}\propto T$.

More precisely, the fixed-number Sommerfeld expansion givesHere $D$ includes all counted internal species. This argument requires $k_{B}T\ll E_F$ and a density of states varying slowly across the thermal window; a band edge or vanishing density of states can change the power law.

Two-dimensional linear dispersion at zero chemical potential. Let $g$ count identical positive-energy branches, keeping the state-counting convention explicit. The number of wavevector states in a ring of radius $k$ is $gA(2\pi kdk)/(2\pi {)}^2$. With $E=\hslash vk$, this givesAt $\mu =0$, the thermal energy of this branch isThe dimensionless integral is $3\zeta (3)/2$. Differentiating givesThis is the heat capacity for a two-dimensional linear fermion dispersion. A physical neutral particle–hole-symmetric Dirac cone also has electron holes in its filled negative-energy band. Their excitation energy gives an equal contribution, doubling $C_{+}$ for the same cone count. The $T^2$ scaling is unchanged. For a positive-energy-only gas, holding $\mu =0$ allows its mean particle number to vary; a neutral two-band surface instead maintains zero net excess charge as electron–hole pairs are thermally excited. These conventions should not be conflated when specifying an exact coefficient.