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Use $N_0={\nu }_s$ for the normal-state single-spin density of states per unit energy and volume at the Fermi surface. Write the attraction as $-V/\mathcal{V}$, with $V>0$, and define $\lambda =N_{0}V$.^[ref]

Since $V/\mathcal{V}$ has units of energy, $V$ has units of energy times volume, making $\lambda$ dimensionless. Set ${\Omega }_D=\hslash {\omega }_D$, an energy cutoff using the Debye frequency ${\omega }_D$, rather than a frequency itself. These conventions also fix the meaning of the earlier constant-shell BCS gap equation.

Here is the reduction of the BCS gap equation in these conventions. Substituting the zero-temperature BCS anomalous average givesThe $k$ sum counts one spin species. For $\Delta >0$, cancelling $\Delta$ and replacing this sum by an integral with constant $N_0$ givesThe factor two from the symmetric integral cancels the $1/2$ in the BCS anomalous average; it is not an extra spin degeneracy. Substituting $\xi =\Delta \sinh t$ evaluates the integral as $\mathrm{arsinh}({\Omega }_D/\Delta )$. Solving givesThe first equality is exact within the constant-density of states, sharp-cutoff mean-field approximation; the last is the weak coupling asymptote. Extending the first equality to large $\lambda$ does not make this simplified model an accurate description of a strongly coupled material.

For an original numerical example, choose $\hslash {\omega }_D=20\mathrm{meV}$ and $\lambda =0.25$. The exact constant-shell BCS gap equation gives $\Delta (0)=0.732871\mathrm{meV}$; the weak coupling formula gives $0.732626\mathrm{meV}$. Its fractional underestimate, measured against the exact result, is exactly $e^{-2/\lambda }=e^{-8}\simeq 0.03355\%$. At $\lambda =1$ that error is $e^{-2}\simeq 13.53\%$, so the two formulas are no longer interchangeable. No material parameters have been fitted here.

At nonzero temperature, substitute the finite-temperature BCS anomalous average instead. The same cancellation yieldsAt the superconducting transition temperature take the continuous $\Delta \to 0$ limit. The integrand at $\xi =0$ then has the finite limiting value $1/(2k_B{T}_c)$, so no divergent endpoint needs to be inserted into a numerical quadrature. For $k_B{T}_c\ll {\Omega }_D$, the resulting integral gives $k_B{T}_c\simeq 1.134{\Omega }_D{e}^{-1/\lambda }$ and $2\Delta (0)/(k_B{T}_c)\simeq 3.528$. These ratios concern the isotropic weak coupling model.^[ref]

An attractive interaction $V_{k,k^{\prime }}=-V/\mathcal{V}$ within $|{\xi }_k|,|{\xi }_{k^{\prime }}|<{\Omega }_D$ gives a constant gap inside that energy shell and zero outside. With approximately constant single-spin density of states ${\nu }_s$ per volume,The weak coupling limit is $\Delta \simeq 2{\Omega }_D{e}^{-1/({\nu }_{s}V)}$. A density counting both spin species is twice ${\nu }_s$ and must be divided by two. The shell cutoff is a relative-energy condition, even when every Cooper pair has zero total momentum.

Differentiate the homogeneous spin-field effective action:At a stationary field, $M=(U/N){\sum }_{\mathbf{k},\sigma }\sigma n_F({ϵ}_{\mathbf{k}}-\sigma M)$. Define the spin-summed density of states per lattice site byThen the requested form isThe factor one half fixes the spin convention. With a single-spin density of states ${\nu }_s=\nu /2$, the prefactor would instead be $U$. In terms of site occupations, the same equation is $M=U(n_{\uparrow }-n_{\downarrow })=2U⟨S^z⟩$. The Fermi distribution includes the chemical potential in its argument definition; it must not be subtracted a second time.

In each factor of the BCS ground state, the BCS anomalous average is $⟨b_k⟩=u_k{v}_k={\Delta }_k/(2E_k)$. Self-consistency therefore gives the zero-temperature BCS gap equationPut ${\Omega }_D=\hslash {\omega }_D$. The specified constant attractive interaction makes ${\Delta }_k$ independent of $k$ inside the energy shell and zero outside it:The printed ${\delta }_{k,0}$ is incompatible with this interaction: $k$ labels relative pair momentum, not the total momentum of a Cooper pair. Every pair here has zero total momentum, while many relative momenta contribute.

For the nonzero solution, the constant-shell BCS gap equation becomes $1=(V/L^3){\sum }_{|{\xi }_k|<{\Omega }_D}(2\sqrt{{\xi }_k^2+{\Delta }^2}{)}^{-1}$. Let ${\nu }_s$ denote the approximately constant single-spin density of states per unit volume at the Fermi level. ThenIn weak coupling, $\Delta \simeq 2\hslash {\omega }_D{e}^{-1/({\nu }_{s}V)}$. The displayed answer in the question uses $\nu ={\nu }_s$. If “total electronic density of states” includes both spin species, ${\nu }_{\mathrm{tot}}=2{\nu }_s$, the argument instead reads $2/({\nu }_{\mathrm{tot}}V)$. If the density counts the whole box rather than unit volume, divide it by $L^3$ before using this formula.

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.

Specify the spin counting first. Let $g_s=g_s(E_F)$ denote the unpolarized single-spin density of states per atom per energy, and put $\delta n=n_{\uparrow }-n_{\downarrow }$. At fixed total occupation $n$, the two Hubbard model mean-field energies can be writtenThe common shift $Un/2$ can be absorbed into the chemical potential. Lowering the up-spin band by $\Delta$ adds $g_s\Delta$ carriers, while raising the down-spin band removes the same number. ThusIf $n_a$ is the atom number density, the magnetization is $M=n_a{\mu }_B\delta n$. The Stoner-enhanced magnetic susceptibility in SI volume units is thereforeHere ${\chi }_P$ is the noninteracting Pauli paramagnetism result. If the paper's $g(E_F)$ counts both spins, $g=2g_s$ and the same result isA susceptibility per atom omits $n_a$. Stating the spin and volume conventions avoids an otherwise spurious factor of two or a dimensional mismatch.

The Stoner criterion is $U{g}_s(E_F)>1$, equivalently $Ug(E_F)>2$ with spin-summed $g$. To see the instability, transfer $\eta$ electrons per atom from one spin band to the other. The band-filling cost is ${\eta }^2/g_s$, whereas the interaction energy changes by $-U{\eta }^2$. Their sum $(1/g_s-U){\eta }^2$ becomes negative at the threshold. The unpolarized state then ceases to be a local minimum, and nonlinear band filling determines the spontaneous ferromagnetism. The negative continuation of the displayed linear susceptibility beyond the threshold is not a stable paramagnetic response.

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

In three dimensions an ideal Bose gas can saturate its excited-state population at chemical potential zero, with the remaining particles forming a Bose-Einstein condensate. The transition follows from the Bose-Einstein distribution integrated against its density of states; below the ideal-gas transition, $n_0/n=1-(T/T_c{)}^{3/2}$. Condensation alone does not supply the finite Landau critical velocity of an interacting superfluid: the ideal quadratic spectrum has ${\inf }_{p}E(p)/p=0$.

By contrast, a weak attractive interaction near a Fermi surface produces the Cooper instability and the BCS gap equation. Its logarithmic enhancement permits pairing at arbitrarily weak attraction in the ideal model. The low-temperature excitations have $E=\sqrt{{\xi }^2+{\Delta }^2}$, and in weak coupling $k_B{T}_c=(e^{{\gamma }_E}/\pi )\Delta (0)\simeq 0.567\Delta (0)$. This transition is controlled by pair formation when phase fluctuations are small; deeply bound pairs instead motivate a Bose-condensation description with their own density and effective mass.

For the given ellipsoidal dispersion, set $q_x=k_x$, $q_y=k_y$ and $q_z=k_z/\sqrt{\gamma }$. Then $\epsilon ={\hslash }^2{q}^2/(2m)$ while the integration measure is $d^{3}k=\sqrt{\gamma }{d}^{3}q$. Thus the anisotropic ideal-gas density of states per unit volume and per internal state isThe full continuum momentum range is assumed; this is an anisotropic three-dimensional gas, not yet a discrete stack of two-dimensional bands.

At the Bose transition, substitute $x=\epsilon /(k_{B}T)$:The integral is $\Gamma (3/2)\zeta (3/2)\simeq 2.315$. Solving for the Bose-Einstein condensation temperature for an anisotropic effective mass givesThe heavier $z$-direction effective mass increases the available excited-state count and lowers the condensation temperature at fixed number density.

For a balanced two-spin Fermi gas, let $n_F$ denote total fermion number density and take $N_s$ to be the single-spin density of states used in the conventional gap equation. Counting the filled ellipsoid givesIf a spin-summed density of states is used instead, the coupling convention in the supplied integral must change accordingly; inserting a factor two into only one side would change the exponent incorrectly.

For ${\epsilon }_c\ll {\epsilon }_F$, replace the density of states across the interaction window by $N_s({\epsilon }_F,\gamma )$. Writing ${\lambda }_0=|g|N_{s0}$, the zero-temperature integral becomesand thereforeThe second expression additionally requires ${\lambda }_0{\gamma }^{1/3}\ll 1$. The explicit density dependence is ${\lambda }_0=|g|m(3{\pi }^2{n}_F{)}^{1/3}/(2{\pi }^2{\hslash }^2)$. This mass-anisotropy enhancement of a weak-coupling BCS gap initially raises the pairing scale, while the ideal Bose condensation scale decreases as ${\gamma }^{-1/3}$.

Large anisotropy weakens transport and coherence in the heavy direction. The separation of pair formation and phase ordering then matters, so a large mean-field gap need not imply a large actual $T_c$. A qualitative sketch has $T_c$ initially rising with the pairing scale and eventually turning down when coherence is the smaller scale. If preformed fermion pairs behave as bosons, use $n_B=n_F/2$ and pair mass about $2m$; in this continuum model this gives $k_B{T}^{\mathrm{BEC}}\simeq 0.218{\epsilon }_{F0}{\gamma }^{-1/3}$.

The turnover cannot be located reliably by taking the weak-coupling formulas to $\gamma \to \infty$: ${\epsilon }_F$ falls, ${\lambda }_0{\gamma }^{1/3}$ grows, the narrow-window assumption fails and the pair chemical potential and mass require a self-consistent treatment. The smooth curve in the figure is explicitly an illustrative interpolation of two extrapolated scales, not such a calculation. If actual decoupled two-dimensional layers are approached, ordinary ideal-gas finite-temperature condensation is absent and interacting phase order requires a different analysis, including the Berezinskii–Kosterlitz–Thouless transition when appropriate.

The cuprates make the distinction between pairing and phase coherence relevant: they are layered and have anisotropic motion, but their strong correlations, lattice band structure and predominantly d-wave superconductivity are outside this attractive continuum $s$-wave model. Anisotropy by itself is therefore not an explanation or quantitative fit of cuprate transition temperatures.

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.

Density-of-states formula for the susceptibility. Here $g$ is the zero-field single-spin density of states. Regard its argument $E$ as the full spin-shifted energy. The shifted functions are $g_{\pm }(E)=g(E\mp {\mu }_{B}B)$, soThe total number is even in $B$; thus $\mu (B)=\mu (0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from ${\mu }^{\prime }(0)$ cancel in the spin difference because the two densities coincide at zero field. To first order,It follows thatFor a physical energy band, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, integration by parts also givesThis form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional derivative in the first expression. The susceptibility defined here is the derivative of total magnetic moment with respect to $B$, with units $\mathrm{J}{\mathrm{T}}^{-2}$; it is not automatically a dimensionless SI volume susceptibility. A density-of-states convention per unit volume likewise gives a response per unit volume. These are the field conventions for magnetic susceptibility.

Chemical potential at fixed number. Define the integrated single-spin state countThe probability kernel $K_T$ has unit integral, mean $\mu$, and variance $a(k_{B}T{)}^2$. Using integration by parts,Expand $G$ about $\mu$. The linear term integrates to zero, while $G^{\prime \prime }=g^{\prime }$ givesThis is the number form of the Sommerfeld expansion. Since $N/2=G({\epsilon }_F)$ at zero temperature, set $\mu ={\epsilon }_F+\delta \mu$ with $\delta \mu =O(T^2)$:ThereforeThe coefficient can be obtained directly. With $y=(E-\mu )/(k_{B}T)$,For example, the sum follows by expanding $e^{-y}/(1+e^{-y}{)}^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2={\pi }^2/6$. A density of states increasing with energy gives a downward quadratic chemical-potential shift. For the constant two-dimensional density of states that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.

Low-temperature susceptibility at fixed number. Apply the same kernel expansion to $g$ itself:The shift of $\mu$ must be retained in the first term. Writing $g_F=g({\epsilon }_F)$ and similarly for its derivatives,ConsequentlyThis is the fixed-number low-temperature Pauli susceptibility. The $g_F^{\prime \prime }$ contribution is thermal broadening at fixed chemical potential; the $-g_F^{\prime 2}/g_F$ contribution enforces fixed total number. Both are needed. If the density of states were instead defined with both spin directions included, the prefactor $2$ would be absorbed into that definition. Only the spin response has been calculated; no orbital magnetic response is included.

The expansion assumes $g_F>0$, sufficient smoothness near the Fermi energy, and a temperature small compared with the energy scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular density of states. For the ideal two-dimensional parabolic band, $g_F^{\prime }=g_F^{\prime \prime }=0$, so the leading Pauli response is temperature independent, consistently with the earlier state count.

The single-spin density of states counts momentum states for one spin species. For degenerate up and down species, the spin-summed density of states is twice this value. Specify separately whether the density is per unit volume or extensive. In the BCS gap equation, the momentum sum labels opposite-spin pairs once and therefore uses the single-spin density per volume when the interaction has normalization $V/\mathcal{V}$.

For a spin-degenerate band, the total density of states includes both spin species and is twice the single-spin density of states. Per lattice site, $\nu (ϵ)=2N^{-1}{\sum }_k\delta (ϵ-{ϵ}_k)$. Using a separate explicit spin sum then requires a compensating factor one half when converting momentum sums. The normalization matters in the Stoner criterion.

For an interaction-energy density $U{n}_{\uparrow }{n}_{\downarrow }$, let $n_{\uparrow }=n/2+\delta$, $n_{\downarrow }=n/2-\delta$ at fixed total carrier density. If $g_s(E_F)$ is the unpolarized single-spin density of states per unit volume, the kinetic-energy cost of the two occupation shifts is ${\delta }^2/g_s(E_F)$, whereas the interaction changes by $-U{\delta }^2$. ThereforeFor the three-dimensional free electron model, $g_{\mathrm{tot}}(E_F)=3n/(2E_F)$ and $U_c=4E_F/(3n)$. Both the interaction normalization and the spin counting in the density of states must be specified when quoting a Stoner criterion.

Let $g_s(E_F)$ be the single-spin density of states per site per energy, and let the on-site Hubbard model repulsion have energy $U$. For a weak uniform magnetic field $H$, the half-splitting of the two spin bands is $\Delta =U\delta n/2+{\mu }_0{\mu }_{B}H$, where $\delta n=n_{\uparrow }-n_{\downarrow }$. Constant density of states gives $\delta n=2g_s\Delta$. ThereforeThis is the response per site; multiplying by the site number density gives the dimensionless SI volume magnetic susceptibility. With spin-summed density of states $g=2g_s$, the denominator is $1-Ug/2$. Its vanishing is the mean-field Stoner criterion. The unpolarized linear-response expression is stable only for $U{g}_s<1$.