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For real gap and the particle-hole block $\left(\begin{array}{cc}{\xi }_k& -{\Delta }_k\\ -{\Delta }_k& -{\xi }_k\end{array}\right)$, the BCS coherence factors obey $u_k^2=(1+{\xi }_k/E_k)/2$, $v_k^2=(1-{\xi }_k/E_k)/2$ and $u_k{v}_k={\Delta }_k/(2E_k)$, where $E_k=\sqrt{{\xi }_k^2+{\Delta }_k^2}$. They diagonalize the Bogoliubov--de Gennes Hamiltonian and give the empty/pair amplitudes of the BCS ground state. Their relative sign depends on the pairing convention.

For an original check, write $x={\xi }_k/\Delta$ with constant real $\Delta >0$. At $x=0$, the BCS coherence factors give $u_k^2=v_k^2=1/2$. At $x=1$, they give $(u_k^2,v_k^2)=(0.853553,0.146447)$; at $x=-1$ these values exchange places. Thus the zero-temperature occupation $v_k^2$ rounds the normal Fermi sea's step over an energy interval of order $\Delta$. This mixing is strongest at the Fermi surface, where the BCS quasiparticle spectrum has its minimum.

For a uniform isotropic weak-gap BCS state, the zero-temperature electron occupation is $n(\xi )=(1-\xi /\sqrt{{\xi }^2+{\Delta }^2})/2$. The number above the normal-state Fermi surface is approximately $D(\mu ){\int }_0^{\infty }n(\xi )d\xi$. The integral equals $\Delta /2$, while a three-dimensional parabolic band including both spins has $D(\mu )=3N/(2\mu )$. This gives the displayed leading fraction.

The result measures redistribution relative to a normal Fermi sea, not bosonic single-mode condensate depletion or quasiparticle occupation. A different band dimension or anisotropic gap requires its actual density-of-states or momentum integral. Material estimates need specified gap and chemical-potential scales.

Physical Review 104, 1189–1190 (1956). The original two-added-electron bound-state argument in a degenerate Fermi sea. doi.org/10.1103/PhysRev.104.1189

Two added electrons move above a frozen filled Fermi sea and attract within a thin energy shell. The Pauli-blocked continuum has a logarithmic pair susceptibility, so an arbitrarily weak net attractive constant interaction produces a state below the two-particle threshold at zero temperature. This two-added-particle result is not by itself a full many-electron superconducting state.

The pairing Hamiltonian is projected onto empty states in a specified shell above the inert Fermi sea. A state restricted to that shell is an eigenstate of this projected model, not necessarily of a contact interaction that remains unrestricted at all momenta.

Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

For the degenerate electron gas, let $D_g=Am/(\pi {\hslash }^2)$ include both spin states. The Sommerfeld expansion givesFor the gapless symmetric intrinsic semiconductor, part (d) gave the same coefficient with $D_s=A{m}^{\ast }/(\pi {\hslash }^2)$. Thus for equal sample areas and equal spin and valley degeneracies,The equality often intended in this comparison requires the same density of states. The source sets $m_e^{\ast }=m_h^{\ast }$ but does not explicitly equate them to the electron gas mass $m$. Without that additional assumption the ratio is $m^{\ast }/m$, or more generally $D_s/D_g$ if sample areas or degeneracy factors differ. For example, $m^{\ast }=2m$ with equal areas satisfies the stated equal-effective mass condition in the semiconductor and gives ratio two. The filled Fermi sea does not add a factor $N_e/N_c$: only the thermal shell near its Fermi energy contributes to its heat capacity.

Use $\hslash =c=k_B=1$ and count one populated helicity per neutrino or antineutrino. For massless particles, the Fermi-Dirac distribution givesWhen ${\mu }_{\nu }/T_{\nu }\gg 1$, the occupied states form an almost sharp Fermi sea. Replacing the distribution by $\Theta ({\mu }_{\nu }-p)$ givesThe relative finite-temperature correction is of order $(T_{\nu }/{\mu }_{\nu }{)}^2$.

Chemical equilibrium gives ${\mu }_{\bar{\nu }}=-{\mu }_{\nu }$. For positive large ${\mu }_{\nu }$, the corresponding antineutrino distribution is dilute, soIt is exponentially negligible. The neutrino degeneracy parameter ${\xi }_{\nu }={\mu }_{\nu }/T_{\nu }$ is conserved in the assumed adiabatic massless evolution: redshifting preserves the distribution with both $T_{\nu }$ and ${\mu }_{\nu }$ proportional to $a^{-1}$.

There is a sign qualification in the printed absolute-value formula. If ${\mu }_{\nu }<0$, the distribution of neutrinos themselves is exponentially suppressed, and the antineutrinos, with positive chemical potential, form the degenerate sea. In either case the degenerate neutrino and antineutrino energy density isThus the formula with $|{\mu }_{\nu }|$ describes the dominant member of the pair, or their leading total, rather than the named neutrino distribution for both signs. No extra factor of two is present: only one member is densely occupied. Additional populated internal states multiply the result by their degeneracy.

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

A useful convention for the semi-empirical mass formula iswhere $B>0$ is the binding energy. The nuclear mass is $Z{m}_p+(A-Z)m_n-B/c^2$. Atomic-mass comparisons instead use hydrogen-atom masses and account for electron binding at the desired accuracy. No numerical fit coefficients are specified here.

The volume term represents short-range nuclear attraction with saturation: each nucleon has a limited number of close neighbours. The surface term subtracts the binding lost by nucleons at the boundary. Coulomb repulsion penalizes a large proton fraction. The asymmetry term penalizes unequal neutron and proton populations, reflecting the cost of filling different Fermi seas and nuclear interactions. Pairing favours even-even nuclei and penalizes odd-odd nuclei; odd-$A$ nuclei have zero pairing term in this approximation. For example, one may choose $\delta =\pm a_p{A}^{-1/2}$; an alternative fitted power requires a correspondingly different coefficient.

At fixed $A$, the smooth mass dependence on $Z$ is approximately parabolic. Neglecting the neutron-proton mass difference, pairing and small linear-in-$Z$ Coulomb corrections, minimization givesThus the preferred neutron excess grows with mass number. The surface penalty disfavors very light systems, while the growing Coulomb penalty reduces the binding per nucleon of very heavy systems; this explains broad fusion and fission energy trends. The formula is a bulk approximation: shell closures, deformation and small separation-energy differences can give important exceptions. A fitted average binding trend is not a reliable substitute for a measured decay threshold near a boundary of stability.

identical particles have the same intrinsic properties. Exchanging their bookkeeping labels cannot distinguish a physical preparation or change a measurable prediction. This is stronger than merely failing to track classical trajectories. For ordinary bosons and fermions, the many-particle wavefunction has respectively bosonic exchange symmetry or fermionic exchange symmetry under every particle exchange operator, acting on spatial and spin coordinates together. The Spin-statistics theorem connects integer spin to bosons and half-integer spin to fermions; this connection uses relativistic quantum-field assumptions, rather than following from the nonrelativistic Schrödinger equation alone.

For one-particle states $u,v$ satisfying orthonormality, the two alternatives form a quantum superposition:Its probability density contains an exchange interference term, absent from an incoherent mixture of labelled alternatives. The antisymmetric expression vanishes when both one-particle states are identical: one complete one-particle state can contain at most one fermion. This Pauli exclusion principle explains electron shell filling, the electronic Fermi sea and degeneracy pressure. Opposite spins are different one-particle states, so two electrons may share an atomic orbital. Spatially separated wave packets with negligible overlap can often be treated as distinguishable for local measurements; the underlying exchange symmetry remains.

In thermal equilibrium, occupation of a level of energy $ϵ$ obeysbosons can accumulate macroscopically in one state, giving Bose-Einstein condensation; fermions instead fill successive states up to the Fermi energy at zero temperature. When occupation is dilute, both expressions reduce to the Maxwell-Boltzmann distribution. The identical-particle classical partition function contains $1/N!$, avoiding spurious entropy from permutations and the Gibbs paradox. Exchange effects become conspicuous when the thermal de Broglie wavelength is comparable to the interparticle spacing.

An operational test is the Hong-Ou-Mandel effect. Down-converted photon pairs enter separate ports of a balanced beam splitter; indistinguishable alternatives for one photon per output cancel. Scanning the relative arrival delay restores coincidences away from overlap. Matching polarization, spectral filtering and spatial alignment controls distinguishability; measuring delayed accidental coincidences controls background. The observed dip tests two-photon interference, and its finite depth must not automatically be interpreted as a failure of bosonic symmetry. The original apparatus varied beam-splitter position and subtracted a separately measured accidental background.^[ref]

Indistinguishable particles have the same intrinsic properties and cannot be assigned physically observable permanent labels. Exchanging their mathematical coordinates leaves all physical predictions unchanged. In ordinary three-dimensional quantum mechanics the total state, including spin, obeysThe Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions. Exchange does not mean physically transporting two particles along some prescribed trajectories; it is the symmetry of the many-particle state under permutation of its arguments.

For orthonormal one-particle states $u,v$, normalized two-particle spatial combinations are $[u(1)v(2)\pm v(1)u(2)]/\sqrt{2}$. They illustrate coherent exchange amplitudes. If both fermions have the same complete one-particle state, the antisymmetric combination vanishes. This is the Pauli exclusion principle: a fermionic mode has occupation zero or one. Bosonic modes can have arbitrarily large occupation. A many-fermion product basis is antisymmetrized by a Slater determinant.

It is the full spin-spatial state that must have the appropriate symmetry. For two electrons, an antisymmetric spin singlet state can accompany a symmetric spatial orbital, so opposite-spin electrons can occupy the same spatial orbital. A symmetric spin triplet instead needs an antisymmetric spatial state. Thus exclusion concerns identical complete states, not an unconditional prohibition on sharing a position or spatial orbital.

Counting allowed occupations gives Bose-Einstein statistics and Fermi-Dirac statistics. For an ideal gas in thermal equilibrium,Bosonic accumulation permits Bose-Einstein condensation; exclusion gives a filled Fermi sea and degeneracy pressure even at zero temperature. In the dilute classical limit the occupations are small and both expressions approach the Maxwell-Boltzmann distribution. Exchange effects become important when wavepackets overlap appreciably, often characterized by number density times the cube of the thermal wavelength.

Composite particles follow their total exchange statistics while remaining intact. The composite statistics of neutral atoms depend on electrons as well as nucleons, as in the two lithium isotopes above. Symmetrization can produce bunching or antibunching in joint detection probabilities even in the absence of an interaction potential; it is not an additional classical force.

The length, area and volume of a first Brillouin zone are $(2\pi /a{)}^d$. The corresponding occupied $k$-space measures areHenceThis is state counting, independent of the interval/disk/ball geometry. One Brillouin zone contains one spatial state per unit cell in a single energy band, and that band can hold two electrons per cell because of spin. One electron per cell therefore occupies half of it. The “volume enclosed” is the occupied Fermi sea, not the lower-dimensional area of the Fermi surface itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.

In a fixed fermion-parity sector use lattice spacing one and the unitary discrete Fourier transformThe allowed momenta satisfy $e^{ikN}=-P$: they are $k=2\pi (\ell +1/2)/N$ for $P=+1$ and $k=2\pi \ell /N$ for $P=-1$, with $N$ distinct values modulo $2\pi$. The hopping and number sums become ${\sum }_{k}2\cos k{c}_k^{†}{c}_k$ and ${\sum }_k{c}_k^{†}{c}_k$. ThusHere $E$ is the energy of the all-up spin configuration, not generally the ground-state energy. The dispersion has minimum $h-J$ at $k=0$, maximum $h+J$ at $k=\pi$, and positive curvature at its minimum. Since the fermions do not interact, each mode can be occupied zero or one time. A negative-energy occupation lowers the many-body energy, so in the thermodynamic limitThis is a filled Fermi sea containing many fermions; it is not obtained by putting a single particle into the band minimum. For $h>J$ every mode has positive energy and the vacuum is the ground state. Its spin interpretation is complete longitudinal polarization. At $h=J$ the zero-momentum one-particle excitation first becomes gapless.

For an exact finite ring use parity-consistent filling of a free-fermion band. In each momentum grid fill negative-energy levels, check that their particle number has the chosen parity, and if necessary toggle a mode with the smallest $|{ϵ}_k|$. Compare the two sector energies. A zero-energy mode allows degenerate occupations. This restores the physical spin Hilbert space without changing the stated band.

The shifted Fermi circle satisfies $|\mathbf{p}-p_0\hat{\mathbf{x}}|=p_F$. Its radial boundary at polar angle $\theta$ is the positive solution of $p^2-2p{p}_0\cos \theta +p_0^2=p_F^2$. Thus, for $p_0<p_F$,and its leading Taylor expansion isForward-pointing momenta gain occupied states and backward-pointing momenta lose them. With the paper's explicitly reversed Heaviside step function convention, ${\Theta }^{\prime }(x)=-\delta (x)$; consequentlywith a positive leading sign.

For the later quadratic energy calculation it is useful to distinguish the first-order shape from exact particle number conservation. The particle-conserving displacement of a Fermi circle hasThe second-order inward correction cancels the apparent area increase one would obtain by treating $\nu =p_0\cos \theta$ as exact.

Use the total density of states $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ and $C=A/(2\pi \hslash {)}^2$ from part (b). In the paper's double-sum convention, $f(\varphi )$ has units of energy and scales as $A^{-1}$. Define its first angular moment byThis convention matters: the Landau quasiparticle energy functional written with an explicit $1/A$ instead uses the intensive kernel $\tilde{f}=Af$.

At zero temperature, the displayed functional is the change in $E-\mu N$; a true Galilean boost preserves $N$, so $\Delta \mathcal{F}=\Delta E$. For a radial displacement $\nu (\theta )$, the first term in the Landau quasiparticle energy functional becomesThe oriented radial integral also works for negative $\nu$: removing occupied levels below the chemical potential costs positive energy. Substituting $\nu =p_0\cos \theta +O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ givesThe exact second-order correction to the boundary is unnecessary here because the unperturbed excitation energy vanishes on the Fermi surface; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta \mathcal{F}$ with a fixed-particle number energy change.

For the quadratic interaction term, the first-order occupation change alone suffices. The radial Dirac delta function integrals givePut $\varphi =\theta -{\theta }^{\prime }$. The remaining angular integral usesso only the first cosine component of the interaction survives. Since $D^{\ast }=2\pi C{m}^{\ast }$,Adding both contributions and matching the bare-mass result of part (a) yields the two-dimensional Landau effective-mass relation:Equivalently, using the density of states per unit area $g^{\ast }=D^{\ast }/A$,If dimensionless Landau parameters are defined through the Fourier series $D^{\ast }f(\varphi )=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, then $D^{\ast }{\overline{f}}_1=F_1/2$ andDefining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-integral formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving Fermi sea compensate the renormalized quasiparticle velocity, so the total boost energy is governed by the bare mass. For an interaction independent of angle, ${\overline{f}}_1=0$ and this consistency condition requires $m^{\ast }=m$.

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.

For $V\gg |t|$, the leading objective is to avoid adjacent electrons of the same spin. The unpolarized Fermi sea, although optimal within the preceding restricted family, has interaction energy per site $V(1/2-2/{\pi }^2)>0$. Its cost grows linearly with $V$, while states with no forbidden neighbouring pairs have zero interaction energy.

There are two independent spinless fermion t-V chains, one for each spin. On an even ring, avoiding all neighbouring pairs allows at most $M/2$ particles in each chain. Since the total is $M$, both bounds must be saturated, and each chain must alternate occupied and empty sites. Their two independent choices give four limiting occupation patterns:The first two are Néel states with one electron per site; the last two are charge density waves alternating doubly occupied and empty sites. Opposite-spin double occupation costs nothing in this particular Hamiltonian. Each species has alternating occupation in all four patterns.

These are the exact zero-interaction ground states at $t=0$, and the limiting ordered configurations as $V/|t|\to \infty$. At large finite $V/|t|$, virtual hopping admixes configurations with forbidden pairs, producing an energy correction of order $-M|t{|}^2/V$. A finite ring can have superpositions of the ordered configurations, with possible quantum tunnelling splittings; four exactly degenerate product ground states are not implied at finite hopping. The four ordered choices describe the strong-coupling thermodynamic limit, with spontaneous symmetry breaking understood.

Interpret the displayed semi-empirical mass formula as an equation for mass energy, using $c=1$ for the mass terms. Its seven contributions are as follows.

The term $Z(m_p+m_e)$ is the rest energy of the $Z$ protons and $Z$ atomic electrons; $(A-Z)m_n$ is the rest energy of the neutrons. Including $Z{m}_e$ makes this an atomic-mass approximation. Small atomic electronic binding energies are neglected.

The volume contribution $-a_{v}A$ represents the short-range attractive strong interaction. Saturation gives each nucleon roughly a fixed number of neighbors, so the nuclear binding energy is approximately proportional to $A$; binding lowers the mass. The surface contribution $+a_s{A}^{2/3}$ restores the binding missing from nucleons near the surface. Approximately constant nuclear density gives radius proportional to $A^{1/3}$ and area proportional to $A^{2/3}$, explaining that power and the positive mass correction.

The Coulomb potential energy contribution $+a_c{Z}^2/A^{1/3}$ is positive because proton charges repel. A sphere's electrostatic self-energy scales as $Z^2/R$, and $R\propto A^{1/3}$. The more refined proton-pair count $Z(Z-1)$ differs from the $Z^2$ approximation explicitly used here.

The nuclear symmetry energy contribution $+a_a(Z-A/2{)}^2/A$ penalizes unequal neutron and proton populations. By the Pauli exclusion principle, filling one species' higher-momentum states is energetically more expensive than balancing the two Fermi seas; nuclear interactions also contribute to the empirical coefficient. The stated convention is equivalent to $23.285(N-Z{)}^2/A$ MeV, since $Z-A/2=(Z-N)/2$. It must not be combined with the usual coefficient near $23$ MeV without this factor of four.

Finally, nuclear pairing favors paired protons and paired neutrons. In the mass formula,The signs reverse when this correction is written as a contribution to positive nuclear binding energy. This smooth liquid-drop model omits detailed nuclear shell corrections.