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An observable is represented by a self-adjoint operator. For an ideal projective measurement, let $P_a$ be its spectral projector for outcome $a$. The Born rule gives $p(a)=⟨\psi |P_a|\psi ⟩$, or $p(a)=\mathrm{tr}(\rho P_a)$ for a density matrix $\rho$. Conditional on obtaining $a$, the Lüders rule givesFor a degenerate outcome this projects onto the whole associated eigenspace, rather than arbitrarily selecting a vector within it. An immediate repeated ideal measurement returns the same outcome, provided there is no intervening evolution or disturbance.

Commuting finite-dimensional observables admit a common orthonormal eigenbasis; their projectors commute and ideal joint measurements are compatible. Noncommuting observables generally have no complete common eigenbasis, and measurement order can affect the subsequent statistics. Particular common eigenvectors can nevertheless exist, so a nonzero commutator does not mean that every state has a nonzero uncertainty in both quantities.

For a quantitative constraint, apply the Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$. The imaginary part of their inner product is $⟨[A,B]⟩/(2i)$, giving the Robertson uncertainty principleThe canonical commutation relation $[x,p]=i\hslash$ gives $\Delta x\Delta p\ge \hslash /2$. Retaining the real part of the same inner product gives the stronger Robertson-Schrödinger uncertainty relation, including a covariance term. These are constraints on the preparation of a quantum state, not simply statements about a poorly calibrated instrument.

A sequential spin one-half example makes the change in statistics explicit. Prepare $|+z⟩$. An immediate $S_z$ measurement is certainly positive. If an $S_x$ measurement is inserted first, its outcomes each have probability $1/2$ and leave states $|\pm x⟩$. Each then yields either $S_z$ result with probability $1/2$, so the final positive fraction is $1/2$. The change reflects projection onto incompatible spin eigenstates, even with ideal measuring apparatus. The theory of measurement in quantum mechanics allows more general state-update rules, but the operator description and the Born rule still constrain their probabilities.

A Stern-Gerlach experiment sends a collimated beam through an inhomogeneous magnetic field and then measures its transverse position after a drift region. The neutron is electrically neutral but has a nonzero magnetic moment proportional to its spin, $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A predominantly $z$-directed field gives the interaction Hamiltonian operator and transverse forceSince a neutron has spin one-half, its spin component along the local field has only two eigenvalues, $S_z=\pm \hslash /2$. Hence there are two opposite force branches, corresponding to ${\mu }_z=\pm {\gamma }_n\hslash /2$, and a beam containing both spin components gives two displaced detector spots. This is neutron Stern-Gerlach splitting, not an electric-charge Lorentz force. A uniform magnetic field can cause Larmor precession but does not provide the required separating magnetic field gradient.

The neutron's gyromagnetic ratio is negative, so its magnetic moment points opposite to its spin; this reverses which spin label goes to which spot without changing the two-way splitting. Equal populations occur for an initially unpolarized beam. A beam prepared entirely in one field-axis spin eigenstate gives only its corresponding branch, explaining the qualification “in general”.

For the isotropic three-dimensional quantum harmonic oscillator, $V(r)=m{\omega }^2{r}^2/2$, hence $r^{-1}dV/dr=m{\omega }^2$ is constant. No radial integral remains, so the spin-orbit energy shift in a central potential reduces toFor $l=0$, the only possibility is $j=1/2$, and the bracket vanishes. For $l=1$, coupling to spin one-half permits $j=1/2$ and $j=3/2$, whose brackets are $-2$ and $+1$. ThusThese spin-orbit shifts of an isotropic harmonic oscillator are a zero line, a downward parabola, and an upward parabola of half the downward magnitude. They are independent of the radial excitation number, and their magnetic multiplicities are respectively two, two and four. Their weighted mean within the $l=1$ manifold is zero: $2\Delta E_{1,1/2}+4\Delta E_{1,3/2}=0$.

Identical spin one-half particles are fermions. By fermionic exchange symmetry, the complete spatial-and-spin wavefunction must change sign when the two particle labels are exchanged. Define the normalized antisymmetric spin-one-half singlet stateand the three symmetric spin-one-half triplet statesThe Hamiltonian is independent of spin, so these spin functions carry no additional energy.

Both particles can occupy $u_0$ if their spin factor is antisymmetric. The unique ground state wavefunction isTwo equal-spin particles could not share this orbital, by the Pauli exclusion principle.

At the first excited energy, $E=ϵ={\hslash }^2/(2m{R}^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functionsFor each $s$, all allowed wavefunctions are spanned byThere are $2(1+3)$ orthonormal states, hence first-excited-state degeneracy $g=8$. In each product, exactly one factor is antisymmetric, ensuring the required total fermionic exchange symmetry.

For the Transverse-field Ising model, use dimensionless spin one-half operators, so $S^x$ has eigenvalues $\pm 1/2$. At zero transverse field every aligned bond contributes $-J$. The two ground states areThey span an exactly two-dimensional ground space on the finite periodic ferromagnetic chain. Any linear combination is also a ground state; parity eigenstates are the symmetric and antisymmetric combinations. Choosing one ordered product state in the thermodynamic limit is spontaneous symmetry breaking.

For large positive field the leading term is $-2Jg{\sum }_n{S}_n^z$. Its unique minimizing state isThe exchange has zero first-order expectation in this state and produces virtual spin-pair fluctuations at finite g. Thus the polarized product is the limiting wavefunction, not an exact finite-field ground state. Bulk perturbation theory gives $E_0/N=-Jg-J/(4g)+O(J/g^3)$. The field-polarized phase has no twofold ordered-state degeneracy.

For two spin one-half particles, $s_1^2=s_2^2=3{\hslash }^2/4$. Squaring $\mathbf{S}={\mathbf{s}}_1+{\mathbf{s}}_2$ givesTake $a,b$ real so that $f$ is Hermitian. Since $[S^2,S_z]=0$, the displayed polynomial implies $[f,S^2]=[f,S_z]=0$. There is a common orthonormal eigenbasis, so all three observables can be simultaneously certain in a common eigenstate. Commutation does not mean that every quantum superposition has definite values of all three.

For $\lambda =3/4$, the adjoint operator relation $J_{+}^{\ast }=J_{-}$ and the given product identity implyA vector of zero norm is zero, so $J_{+}\psi =0$. Part (i) givesThe commutator $[J^2,J_{-}]=0$ also gives $J^2\phi =(3/4){\hslash }^2\phi$, while $J_3\phi =-(\hslash /2)\phi$. Applying the other product identity to this unnormalized vector,ThereforeThese are the two angular momentum ladder endpoints of a spin one-half pair. In particular, ${\psi }_2=\phi /\hslash$ has unit norm, and $J_{+}{\psi }_2=\hslash {\psi }_1$ because $J_{+}{J}_{-}\psi ={\hslash }^2\psi$.

In the ordered orthonormal basis $(|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩)$, the spin one-half angular momentum operators are $J_i=(\hslash /2){\sigma }_i$, with Pauli matricesThe ladder operators areExplicit multiplication gives $J_{+}{J}_{-}={\hslash }^2\mathrm{diag}(1,0)$ and $J_{-}{J}_{+}={\hslash }^2\mathrm{diag}(0,1)$. Their commutator is therefore

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]

Specify the magnetic coupling before fixing its sign: let the magnetic moment be $\mathbf{\mu }=\gamma \mathbf{S}$, where $\gamma$ is a signed gyromagnetic ratio. The Zeeman effect gives $H_Z=-\gamma \mathbf{B}\cdot \mathbf{S}$. For spin one-half, $\mathbf{S}=\hslash \mathbf{\sigma }/2$, so with $B=|\mathbf{B}|$ and $u=\beta \gamma \hslash B/2$,This is the quantum Gibbs state. Its eigenvalues are $(1\pm \tanh u)/2$, so its purity is $(1+{\tanh }^{2}u)/2$. For nonzero level splitting it approaches a pure state as $|u|\to \infty$: positive inverse temperature selects the lower energy, while negative inverse temperature selects the upper one. It is a maximally mixed state at $u=0$, including infinite temperature or zero magnetic splitting. At zero field the Gibbs state remains $I/2$; lowering temperature alone does not select a unique pure state from that degeneracy.

Negative temperature is meaningful for an internally equilibrated spin sector with a bounded energy spectrum and population inversion. It must be sufficiently isolated from unbounded motional degrees of freedom: the oscillator's unbounded spectrum, for instance, makes its partition sum diverge at negative inverse temperature. The two-level spin partition sum itself stays finite for either sign.

For $\mathbf{B}=B{\mathbf{e}}_z$, the quantum Gibbs state is diagonal in the $S_z$ basis. Its two probabilities are proportional to $e^{\pm \beta \gamma \hslash B/2}$. ThusAt small inverse temperature, $⟨S_z⟩\simeq \beta \gamma {\hslash }^{2}B/4$. As a function of $u=\beta \gamma \hslash B/2$, the normalized spin component is the smooth odd curve $\tanh u$: it crosses zero at $u=0$ and saturates at $\pm 1$. As a function of $\beta$ its direction is reversed if $\gamma B<0$. For $\gamma B>0$, positive temperature favors $S_z=+\hslash /2$, while negative temperature gives the inverted population.

Here the dimensionless spin one-half operator is $S_x={\sigma }_x/2$. Since the Pauli matrix obeys ${\sigma }_x^2=I$, the even and odd terms of the matrix exponential giveThis is the exponential of a Pauli vector for an $x$-axis rotation. With a physical spin angular momentum $S_x^{\mathrm{phys}}=\hslash {\sigma }_x/2$, the exponent is $-i\theta S_x^{\mathrm{phys}}/\hslash$.

For two localized energy levels $0,\delta$ with degeneracies $d_0,d_1$, let $x=\delta /(k_{B}T)$. Canonical equilibrium givesHere $p_1$ is the total upper-level occupation probability and $u,s,c$ are per unit. For nondegenerate levels the entropy rises from zero to $k_B\ln 2$, while the heat capacity vanishes at both low and high temperatures and has a Schottky maximum near $k_{B}T=0.417\delta$. A spin one-half in a field, or a localized defect with two configurations, supplies an example. Higher levels must be thermally inaccessible; unequal degeneracies change the populations and peak. Distinguishable localized units have total partition function $z^N$, unlike a gas of indistinguishable particles sharing just two orbitals.

For the magnetic atom, choose level labels with $E_m=-mg{\mu }_{B}B$, $m=-J,\dots ,J$. Put $x=T_B/T=g{\mu }_{B}B/(k_{B}T)$. Its partition function and exact entropy of a finite magnetic multiplet areThe ratio form has removable limit $Z_J(0)=2J+1$. Since $s_J$ is a function of $x$ only, it depends on field and temperature solely through $B/T$. For fixed $J$ at small $x$, symmetry removes the odd powers and the supplied sum of $m^2$ givesFor large $J$, the expansion requires $J|x|\ll 1$, not merely $|x|\ll 1$.

For demagnetization with a linear-heat-capacity load, integrate its specified $C_L/T$:Isothermal magnetization at $T_i$ removes spin entropy; the bath receives the corresponding heat. During the following reversible, thermally isolated demagnetization, it is the combined entropy of spins and load that stays constant. At the minimum-field endpoint $B_f=0$, the ideal independent spins reach $s_{\max }=k_B\ln (2J+1)$. ThereforeThe bracket is the initial magnetic entropy deficit. The PDF's intermediate equality uses the initial entropy itself instead; with its preceding definition of $s$, that equality omits the maximum-entropy term. The corrected balance recovers its final temperature formula:Without the small-field expansion, the exact independent-spin prediction is $T_i-(N\theta /M)[\ln (2J+1)-s_J(T_i,B_i)/k_B]$. As a check, zero initial field gives zero entropy deficit and no cooling. The incorrectly printed intermediate balance would fail this check.

The result requires the load law throughout the temperature interval, hence $T_i\ll \theta$; a small initial Zeeman splitting, quantitatively $J{T}_{B,i}/T_i\ll 1$; positive final temperature; negligible additional heat capacities and heat leaks; thermal equilibration between subsystems; and a reversible sweep slow enough to maintain equilibrium. A negative extrapolated temperature is a breakdown of this approximation, not a cooling prediction. Spins alone would keep $B/T$ fixed, but the attached load changes that adiabat by transferring entropy to them as it cools.

Interactions invalidate the product of independent one-site partition functions. The entropy must be computed from the many-body energy spectrum or a justified interacting approximation. Correlations and possible magnetic ordering introduce an interaction scale and can remove the independent-spin entropy available for cooling near zero field. Internal fields and ordering commonly limit practical demagnetization. The detailed correction and any residual entropy depend on the interaction Hamiltonian, dimensionality and frustration; neither a universal ordering temperature nor a numerical final-temperature correction is specified here.

A two-state degree of freedom represented mathematically like spin one-half, without necessarily being physical spin. The sublattice pseudospin in graphene labels amplitudes on two sites per primitive cell.

A qubit, or quantum bit, is a quantum system whose state space is a two-dimensional complex Hilbert space. Choose an orthonormal basis $|0⟩,|1⟩$, called the computational basis. A pure state has a normalized representative $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, where $\alpha ,\beta$ are complex probability amplitudes and $|\alpha {|}^2+|\beta {|}^2=1$. General qubit states, including mixed states, require a density matrix. ^[ref].

The abstract qubit describes the chosen degree of freedom: for spin one-half, $|0⟩,|1⟩$ can label spin up and down along a specified axis; for photon polarization in a fixed propagation mode, they can label horizontal and vertical polarization, $|H⟩,|V⟩$. These implementations share the same two-dimensional linear algebra, but their preparation and measurement in quantum mechanics use different apparatus. A qubit therefore need not be an entire two-level particle. ^[ref].

The usefulness of a qubit lies in controlling quantum superpositions and their relative quantum phases, then converting them into outcome probabilities with quantum gates and an appropriate measurement. One ideal computational-basis measurement returns one label, zero or one. It does not reveal both probability amplitudes, their relative quantum phase, or all the bits of a larger register's expansion. Quantum state tomography instead estimates a qubit state using many repetitions of the preparation and suitable measurement settings. ^[ref].

An actual preparation and measurement example is Dederick and Beck's 2014 teaching-laboratory experiment. A 405-nm laser pumped two nonlinear crystals, producing signal and idler photons near 810 nm. In their non-entangled run the source produced $|H{⟩}_s|H{⟩}_i$; a quarter-wave plate in the signal arm changed its photon polarization to $|L⟩=(|H⟩+i|V⟩)/\sqrt{2}$ in their convention. The plate supplies a relative quantum phase between polarization components. An idler detection selected the signal trials; signal detectors measured horizontal/vertical, diagonal, and circular polarization bases on separate runs. Before the polarization analyzer, a half-wave plate at $22.5$ degrees selected the diagonal basis, and a quarter-wave plate at $45$ degrees selected the circular basis. Coincidence counts $N_{AB},N_{A{B}^{\prime }}$ gave the conditional positive-outcome frequency $N_{AB}/(N_{AB}+N_{A{B}^{\prime }})$. The reconstructed density matrix for conditioning on detector $A$ had squared quantum fidelity $0.96$ with $|L⟩$ and purity of a density operator $0.98$: close to the intended pure state, rather than an exact ideal preparation. ^[ref].

The model diagram explains why those settings matter. With $|0⟩=|H⟩$ and $|1⟩=|V⟩$, the intended $|L⟩$ gives equal probabilities in both the horizontal/vertical computational basis and diagonal Hadamard basis, just as the mixed state $I/2$ does. In the circular basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, however, the ideal $|L⟩=|+i⟩$ always gives the plus outcome, whereas $I/2$ still gives equal probabilities. These are deductions from the Born rule, not the paper's measured counts. Unequal detector efficiencies and finite samples must be accounted for when interpreting experimental frequencies; the ideal comparison assumes equally efficient detection.

A spin eigenstate along a specified axis is a simultaneous eigenstate of the squared spin operator $S^2$ and its component $S_{\hat{n}}$. Its eigenvalues are ${\hslash }^{2}s(s+1)$ and $\hslash m$, where $m=-s,-s+1,\dots ,s$. The axis must be specified because components along different axes generally do not commute. For spin one-half, the two field-axis eigenstates have projections $\pm \hslash /2$ and form the branches of a Stern-Gerlach experiment.

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.

For a spin one-half system, an ideal projective measurement along a unit vector $\mathbf{n}$ measures ${\sigma }_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{\sigma }$. Its eigenvalues are $\pm 1$, corresponding to physical spin angular momentum components $\pm \hslash /2$, and its orthogonal projections are ${\Pi }_{\pm }=(I\pm {\sigma }_{\mathbf{n}})/2$. For density matrix $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, the Born rule and the Pauli matrix multiplication law give the displayed probabilities. Thus a spin measurement along one direction tests only that component of the Bloch vector; measurements along three independent directions determine the entire density matrix.

For a central-potential Hamiltonian operator with the spin-orbit interaction $H_{\mathrm{so}}=[r^{-1}{V}^{\prime }(r)/(2m^2{c}^2)]\mathbf{L}\cdot \mathbf{S}$, use $\mathbf{J}=\mathbf{L}+\mathbf{S}$ andA coupled angular momentum eigenstate makes this angular factor a number. The radial operator commutes with all angular factors, giving the energy shift in the title. It is independent of the magnetic projection $m_j$, so a fixed-$j$ multiplet retains its $2j+1$ degeneracy.

For spin one-half, the angular bracket is $l$ when $j=l+1/2$ and $-(l+1)$ when $j=l-1/2$. Only the former exists at $l=0$, where it is zero. For the Coulomb potential energy, $r^{-1}{V}^{\prime }(r)=e^2/(4\pi {ϵ}_0{r}^3)$; the radial normalization convention must be respected when computing its expectation value.

For an isotropic three-dimensional quantum harmonic oscillator, $V(r)=m{\omega }^2{r}^2/2$ makes $r^{-1}{V}^{\prime }(r)=m{\omega }^2$. The spin-orbit energy shift in a central potential is therefore independent of the radial excitation. For spin one-half, the $l=0,j=1/2$ shift is zero, while the two $l=1$ shifts are $-{\hslash }^2{\omega }^2/(2m{c}^2)$ for $j=1/2$ and $+{\hslash }^2{\omega }^2/(4m{c}^2)$ for $j=3/2$.

These shifts are quadratic in trap frequency. Their degeneracy-weighted sum in the $l=1$ manifold is zero, since the multiplicities are two and four. They are perturbative corrections rather than full oscillator energies; a nonrelativistic treatment requires them to be small compared with the orbital spacings.