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In a uniform three-dimensional Hartree-Fock electron gas with repulsive contact coupling $\gamma$, a spin-independent neutralizing background can cancel the Hartree energy. The remaining same-spin exchange term is negative and quadratic in each spin density. Integrating the filled single-spin Fermi sphere gives $A=3{\hslash }^2(6{\pi }^2{)}^{2/3}/(10m)$. At fixed total density the exchange term favors spin polarization, whereas the kinetic term opposes it.

Use normalized plane-wave spin orbitals $V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}{\chi }_{\sigma }$ and fill a separate Fermi sphere for each spin species. Each spin state counts once, soIts kinetic energy per volume isThe spin-independent contact interaction has uncancelled direct energy $E_H/V=\gamma n^2/2$, with $n=n_{\uparrow }+n_{\downarrow }$. By the imposed background assumption this direct contribution cancels. For the Fock term the contact delta function sets the two spatial positions equal; its spin factors then select equal spins. The local one-body density matrix for each spin has diagonal value $n_{\sigma }$, givingCombining these results gives the contact-interaction Hartree-Fock spin energyHere $\gamma$ has units of energy times volume. The result is the model with the stated background subtraction; an isolated contact-interacting gas without that subtraction retains its direct contribution.

At fixed total density, the unpolarized and fully polarized candidate energies arePolarization increases the kinetic energy because one spin species must occupy a larger Fermi sphere, but gains negative exchange energy. The fully polarized candidate wins this endpoint comparison ifUsing the defined $r_s$ givesThus the specified two-state comparison indeed predicts a polarized state below the unpolarized state's energy for $r_s>1.175$.

The endpoint and continuous polarization thresholds of a contact electron gas are different. To check the actual onset within the same uniform Hartree-Fock model, allow $P=(n_{\uparrow }-n_{\downarrow })/n$ to vary continuously. In units $2A(n/2{)}^{5/3}$,The unpolarized state loses local stability already at $r_s=10/9\simeq 1.111$. Full polarization becomes the minimizing endpoint only when $f^{\prime }(1)\le 0$, or $r_s\ge (5/6)2^{2/3}\simeq 1.323$. Between those values the optimum is partly polarized. Consequently $1.175$ is the requested endpoint crossing, not the exact onset of spontaneous polarization after all spin populations are allowed.

The Fermi sphere volume in reciprocal space isThe primitive real-space cell has volume $a^3/2$, so the first Brillouin zone has volumeEach band accommodates two electrons per primitive cell when spin is counted. Six electrons therefore require three such zone-volumes of occupied reciprocal-space states in the free-electron counting picture. The extended-zone sphere is not confined to the first zone; folding it produces multiple bands. It does not imply that the actual tungsten Fermi surface is spherical.

Count two electron spin states in each reciprocal-space cell of volume $(2\pi {)}^3/V$. Filling a Fermi sphere at zero temperature givesThe free electron model then givesThe energy zero is the bottom of the model's conduction band. The supplied carrier density already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.

A spin-half gas has two spin states. Filling the Fermi sphere at zero temperature gives $n=k_F^3/(3{\pi }^2)$, soWith $m_3\simeq 5.008\times 10^{-27}\mathrm{kg}$, the number density is $1.62\times 10^{28}{\mathrm{m}}^{-3}$ and $k_F\simeq 7.82\times 10^9{\mathrm{m}}^{-1}$. ThusThe associated Fermi temperature isFermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective mass and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.