Keeping the same two spin states but incorrectly imposing boson exchange symmetry would require a symmetric total wavefunction. The spin-antisymmetric singlet state would then accompany ${\Phi }_{-}$, while the spin-symmetric triplet state would accompany ${\Phi }_{+}$. Thusand $\boxed{E_T^{\mathrm{hypothetical}}-E_S^{\mathrm{hypothetical}}=2K}$. The ordering reverses: the triplet would be destabilized by $2K$ relative to the singlet. A genuinely spinless boson would have no singlet/triplet spin distinction; this comparison changes exchange symmetry while retaining the question's spin degrees of freedom.

In atomic units, define the Coulomb integral and exchange integral byExpanding the four products in the expectation of $r_{12}^{-1}$ yieldsThe one-electron terms have the same value for these two states. Within this common-orbital approximation,so $\boxed{E_S-E_T=2K}$. The exchange integral is nonnegative: it is the positive Coulomb quadratic form of the transition density $a^{\ast }b$, or equivalently has a nonnegative Fourier representation. The triplet is lower. Actual orbital relaxation and correlation make the common-orbital $2K$ an approximate physical splitting.

Let $a={\varphi }_{1s}$ and $b={\varphi }_{2s}$ be orthonormal atomic orbitals. The normalized spatial factors areFor the singlet state, use the antisymmetric spin factorFor the three components of the triplet state, useEach total wavefunction obeys the Pauli principle. At coincident spatial coordinates,Consequently the triplet pair probability density vanishes at coincidence, even for its $M=0$ component. This zero is stronger than the trivial statement that any exact point has zero integrated probability in a continuum.

The Pauli principle requires the complete wavefunction of identical fermions to be antisymmetric under exchanging the full coordinates, including spin:It is the total spatial-and-spin state that changes sign; the spatial factor alone need not be antisymmetric.

For fixed atom number, volume scales as the cube of the touching-sphere radius. Compressing silver to the stated gold lattice size therefore requiresa volume decrease of about $0.415\%$. If the bulk modulus is constant,Using the silver value derived above givesFor this small compression, $\Delta p\simeq -B\Delta V/V$ gives almost the same estimate. This is an additional pressure relative to the initial standard-pressure state, using the question's free-electron model modulus rather than an empirical silver modulus.

At fixed $N$, the Fermi energy scales as $E_{\mathrm{F}}\propto (N/V{)}^{2/3}$, so $\partial E_{\mathrm{F}}/\partial V=-(2/3)E_{\mathrm{F}}/V$. From $U=3N{E}_{\mathrm{F}}/5$,This is the nonrelativistic electron degeneracy pressure. Because $n{E}_{\mathrm{F}}\propto V^{-5/3}$,The units are energy per volume, hence pressure. Converting the Fermi energies to joules gives the ideal Fermi gas bulk modulusThese are free-electron model values. The complete mechanical bulk modulus of a real metal also involves the ionic lattice, electronic bands and interactions; the model's positive electron pressure does not by itself explain the equilibrium crystal volume.

At zero temperature, states are occupied up to $E_{\mathrm{F}}$. Write the density of states as $g(E)=C\sqrt{E}$, including any fixed volume and spin factors in $C$. ThenwhileDividing eliminates $C$:The mean is positive even at absolute zero because the Pauli exclusion principle fills a range of momenta; electrons cannot all occupy the zero-momentum state.

The reciprocal-space volume inside a spherical Fermi surface is ${\mathcal{V}}_{\mathrm{F}}=4\pi k_{\mathrm{F}}^3/3$. The first Brillouin zone has volume $(2\pi {)}^3$ divided by the primitive real-space cell volume:Since $k_{\mathrm{F}}^3=3{\pi }^2(4/a^3)$,Numerically,The half-zone volume reflects one electron per primitive cell and two spin states per Bloch wavevector: a completely filled single band would require two electrons per cell. Volume alone does not establish containment inside the polyhedral zone. Here the nearest reciprocal-lattice points are of type $(111)$, so their zone faces are a distance $\pi \sqrt{3}/a$ from the origin. The free-electron radius obeysso these ideal monovalent spheres fit inside the first zone. Real band structure can distort a Fermi surface into zone-face contacts or necks; that is additional physics beyond the free sphere.

With one valence electron per atom, the electron number density is $n=4/a^3$. For a spin-degenerate, three-dimensional free electron model at zero temperature,Substitution givesCopper has the larger Fermi wavevector and Fermi energy because its smaller cell gives a larger electron density. The energies are measured from the bottom of the free-electron dispersion, not from vacuum. Treating these real metals as monovalent free-electron systems is the model assumption, not a statement that their filled d bands are absent.

In a face-centered cubic lattice, nearest neighbours touch along a face diagonal. Four radii span that diagonal, so $\sqrt{2}a=4r$ andThusThe primitive vectors enclose volume $a^3/4$, containing one atom; the conventional cubic cell of volume $a^3$ contains four. Keeping these cell conventions distinct avoids a factor-of-four error in the electron density.

For an ordinary fixed field satisfying $\mu B\ll \Delta$, the ground manifold remains lowest. The magnetization therefore vanishes as $T\to 0$: the molecules are overwhelmingly in the stipulated nonmagnetic states. On warming, thermal excitation creates magnetic molecules and initially increases the response. At higher temperature, orientational randomization wins over the growing magnetic population, giving a maximum and then an inverse-temperature tail.

In the weak-field regime,Its maximum occurs near $T=\theta /1.2785\simeq 135$ K. At high temperature the excited fraction tends to one half, while each magnetic molecule's alignment decreases as $1/T$, so $M\to 0$. The low-temperature suppression must be evaluated from the full partition function; extending the weak-field expansion to $T=0$ at a fixed field would exceed its domain.

The sketch assumes the simplified moments stated in the question. For an arbitrarily strong field with $\mu B>\Delta$, an aligned magnetic level crosses below the nonmagnetic states and the low-temperature conclusion changes. Real NO has additional rotational and magnetic structure beyond this deliberately restricted model.

For the two field-sensitive orientations alone, choose the unsplit excited-manifold energy as reference and put $x=\mu B/(k_{\mathrm{B}}T)$. The restricted single-molecule partition function isIf the ground-manifold zero used above is retained, this restricted contribution is instead $2e^{-\beta \Delta }\cosh x$. The extra factor is independent of $B$, so it does not change magnetization conditional on occupying the excited manifold.

There are two magnetic states total; multiplying this expression by another factor of two would double count them. For the full thermal ensemble, the nonmagnetic ground states must be added:Although the omitted ground-state term has no direct field dependence, it affects normalization inside $\ln q$. Thus the restricted expression describes molecules constrained to the excited manifold; it cannot by itself give the magnetization of all thermally equilibrated NO molecules.

Differentiate the electronic energy at fixed volume:At $T\to 0$, $C_{V,\mathrm{el}}\sim N{k}_{\mathrm{B}}(\theta /T{)}^2{e}^{-\theta /T}\to 0$: thermal energy is insufficient to populate the upper manifold. At $T\to \infty$, $C_{V,\mathrm{el}}\sim N{k}_{\mathrm{B}}{\theta }^2/(4T^2)\to 0$: the two populations are already approaching equality and the electronic energy has a finite ceiling.

Between these limits the heat capacity has a Schottky anomaly, peaking near $T=\theta /2.399\simeq 72.0$ K at $C_{V,\mathrm{el}}\simeq 0.439N{k}_{\mathrm{B}}$. The energy rises monotonically toward $N\Delta /2$; its slope is largest at the heat-capacity peak. At 600 K it has not reached its infinite-temperature limit exactly.