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Choose the one-particle states of a hydrogen atom, or a central-field description of a chemical atom, with labelsThe principal quantum number $n$ identifies the electron shell. The orbital angular momentum label $\ell$ identifies the electron subshell; the magnetic quantum number $m_{\ell }$ labels $2\ell +1$ independent spatial orbitals. Each has two spin one-half along an axis choices, hence $2(2\ell +1)$ spin orbitals. Applying the Pauli exclusion principle once per spin orbital givesThese quantum numbers label wavefunctions and spin projections, not electron trajectories. They need not resolve exact energy eigenstates in the presence of spin-orbit interaction; changing to coupled angular momentum eigenstates preserves the count of independent one-particle states within this shell model.

For example, $n=3$ allows $\ell =0,1,2$, conventionally $3s,3p,3d$. Their spatial orbital counts are $1,3,5$; their spin orbital counts are $2,6,10$. The sum is eighteen. The algebra behind “one per complete state” is short: two-particle antisymmetrization gives $u\wedge u=(u\otimes u-u\otimes u)/\sqrt{2}=0$, and the zero vector cannot be a normalized quantum state. Opposite-spin electrons in the same spatial orbital have distinct complete states, so their antisymmetric product survives.

This supplies a structural link to the periodic table: finite electron subshell capacities make repeated patterns of occupied electron configurations possible. It does not predict the exact ground state electron configuration or energy of a many-electron chemical atom. Different electron subshells overlap in energy, and electron interactions change their ordering. In particular, the periodic table period lengths are not simply successive $2n^2$ values: the third periodic table period has eight chemical elements, although the full $n=3$ electron shell has eighteen available states. A capacity counts available spin orbitals; an actual filling order requires a Quantum Hamiltonian.

A density matrix has unit trace. The Pauli matrices are traceless, so $\mathrm{Tr}\hat{\rho }=C$ and hence $C=1$. Using $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$ in the expectation value givesConsequently the required Bloch vector and density matrix areThe two eigenvalues are $3/4$ and $1/4$, both positive, so this is a valid density operator. Its purity of a density operator isTherefore it is a mixed state. Equivalently, its spectral decomposition is $\rho =(3/4)|+x⟩⟨+x|+(1/4)|-x⟩⟨-x|$, an ensemble of the two distinct spin one-half along an axis states, rather than a single pure state.

A $3d^9$ shell has one hole in the otherwise filled $d$ shell. Hund's rules give $S=1/2$, $L=2$, and, for a more-than-half-filled shell, $J=L+S=5/2$: the free-ion term is ${}^2{D}_{5/2}$. Its Landé factor isDistinguish the Curie effective moment from a saturated projection:The question then explicitly asks us to use orbital quenching in the crystal. Ligand fields split and mix orbital states, suppressing their freely orientable angular momentum. The spin-only approximation has $g\simeq 2$, $S=1/2$, so ${\mu }_{\mathrm{eff}}=\sqrt{3}{\mu }_B$ but the two field-axis projections are $\pm {\mu }_B$. The latter, not $\sqrt{3}{\mu }_B$, belong in the tanh argument and saturation magnetization.

The sulfate and water environment supplies mainly closed-shell, weakly magnetic surroundings, stabilizes localized Cu2+ sites and separates their moments so that exchange can be weak on the temperature scale considered. Paramagnetism here is an approximation of weakly interacting local spins, not a claim that dilution guarantees no magnetic ordering at any temperature. Elemental copper has a filled $3d$ band and itinerant conduction electrons rather than one localized Cu2+ spin per atom. Pauli spin response and orbital/core contributions then replace a local-spin Curie law; ordinary pure copper has a weak net diamagnetic response.^[ref]

There is one Cu2+ per formula unit, hence $n_s=N_A/V_m=5.5148\times 10^{27}$ m${}^{-3}$. The spin-only weak-field SI volume susceptibility isThese neglect exchange, anisotropic $g$ factors and background diamagnetism. At high field and low temperature, saturation requires the nonlinear response in a; a field-independent Curie value then describes only the zero-field slope.

Since $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$,Its normalized eigenstates areand

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

When spatial and spin factors separate, a spin orbital combines a spatial orbital ${\varphi }_a$ with a normalized spin state ${\chi }_{\sigma }$. For an electron, choose spin one-half along an axis with $\sigma =\uparrow ,\downarrow$. The two spin orbitals $u_{a\uparrow }$ and $u_{a\downarrow }$ are orthogonal even though their spatial orbital is identical, since their inner product contains $⟨{\chi }_{\uparrow }|{\chi }_{\downarrow }⟩=0$. When spatial and spin variables do not separate, the complete one-particle state is instead a multicomponent wavefunction $u(\mathbf{r},s)$; exclusion still applies to that complete quantum state.

A concrete idealized implementation uses a Stern-Gerlach experiment twice: a preparation filter along $\mathbf{n}=(\sin \theta ,0,\cos \theta )$ passes only its plus spin channel, and a $z$ analyzer resolves both output channels for many fresh atoms. This prepares the spin one-half along an axis state $|{\psi }_{\theta }⟩$. The authored apparatus description uses slit collimation, a nonuniform magnetic field with a pointed pole, and counters at the separated outputs.^[ref] The two-channel spin one-half preparation and analyzer are described separately; the source's recombining “improved” apparatus is explicitly idealized, so the setup here is a proposed estimation model, not a report of observed counts.^[ref]

Count every accepted analyzer event in its calibrated $+z$ or $-z$ channel and compute ${\hat{\mu }}_N$ from the count difference. The magnetic field gradient separates trajectories through their magnetic moments; calibrated channel labels convert positions into spin eigenvalues. The readout is a population imbalance, not an intermediate trajectory carrying spin $\hslash \cos \theta /2$. Beam position also depends on flight time and field geometry, so averaging raw positions alone does not establish a calibrated spin expectation value.

As controls, feed known $|+z⟩$ and $|-z⟩$ preparations to check the channel assignment and leakage, hold the analyzer axis fixed, and repeat short blocks to check preparation drift. Measure background counts and relative channel efficiencies. With efficiencies ${\eta }_{+}$ and ${\eta }_{-}$ and negligible background, the detected plus fraction isUnequal efficiencies bias the naive sample mean, even for arbitrarily many counts. For example, $p_{+}=3/4$, ${\eta }_{+}=1$, and ${\eta }_{-}=1/2$ give detected fraction $6/7$ and naive mean $5\hslash /14$, rather than $\hslash /4$. Background, overlapping beams, detector saturation, angle misalignment, unwanted internal states, and preparation correlations require a better measurement in quantum mechanics model. Increasing $N$ only reduces the independent sampling contribution; it cannot remove these systematic errors. The selected preparation beam is the population being estimated, so losses in preparing it must be distinguished from losses in reading its two outcomes.