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For a charged spin one-half particle with the Dirac equation reference value $g=2$, the anomalous magnetic moment is conventionally expressed by the dimensionless quantity $a=(g-2)/2$. For the electron, write $a_e=(g_s-2)/2$. A stipulated $g_s=2.0024$ would therefore mean $a_e=0.0012$: a $0.12\%$ excess of the spin-component moment over one Bohr magneton. This is a teaching input, not a quoted measurement.

A Berry phase is the geometric phase acquired when a nondegenerate energy eigenstate follows a closed cycle of parameters of a Hamiltonian operator in the ideal limit of the adiabatic theorem. The energy eigenstate must stay separated from other energy eigenstates by a nonzero spectral gap, and the parameters must change sufficiently slowly. Returning the Hamiltonian operator to its starting value returns the state to the same ray, but its quantum phase can remember the journey. The total quantum phase contains both the dynamical phase and the Berry phase.

The beauty of the Berry phase is that, within the adiabatic theorem's limit, it depends on the oriented path of the parameters rather than how long the journey takes. For a spin one-half qubit, the Berry phase of a Bloch-sphere latitude turns that memory into half a solid angle. Berry-phase interference with a reference path converts this geometry into a detection probability.

Michael Berry's 1984 paper formulated the geometric phase for cycles governed by the adiabatic theorem and derived the solid angle result for spin. Its bibliographic details are Proceedings of the Royal Society of London A, volume 392, pages 45–57, DOI 10.1098/rspa.1984.0023. ^[ref]; ^[ref]. The wider concept of geometric phase does not require every cycle to obey the adiabatic theorem: Aharonov and Anandan formulated a geometric phase for general cyclic quantum evolution in 1987. ^[ref].

An isolated qubit that acquires only a global phase has unchanged measurement probabilities. To reveal the Berry phase, use two coherent paths of a Mach-Zehnder interferometer, represented by a path qubit. In one arm the internal spin one-half qubit follows the loop; in the reference arm it stays in its starting energy eigenstate. Assume equal arm amplitudes, identical final internal states, cancelled relative dynamical phases, and calibrated fixed apparatus phases. In the model above, hold the reference Hamiltonian operator at its initial value for the same duration. Both arms then have energy eigenvalue $-\Delta /2$ and acquire the same dynamical phase $\Delta T/(2\hslash )$, so their relative dynamical phase vanishes in the ideal limit of the adiabatic theorem.

The joint quantum state before recombination then factors asThe Berry phase is now a relative quantum phase between the two paths. The calibrated plus output projects onto $|+⟩=(|\mathrm{ref}⟩+|\mathrm{loop}⟩)/\sqrt{2}$. The Born rule givesFor the Berry phase of a Bloch-sphere latitude, the plus-port probabilities at $\theta =0,60^{\circ },90^{\circ },180^{\circ }$ are respectively $1,1/2,0,1$. This ideal Mach-Zehnder interferometer measurement cannot distinguish $\gamma$ from $-\gamma$ or $\gamma +2\pi$ by itself; scanning a known reference phase can resolve the sign. An uncancelled relative dynamical phase would shift the fringe as well.

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

Let $a,b$ be distinct orthonormal spatial orbitals occupied by two identical spin one-half particles, with a weak repulsive Coulomb interaction. The symmetric and antisymmetric spatial combinations have first-order energy shifts $J+K$ and $J-K$, where $J$ is the direct Coulomb integral andThe Coulomb kernel has positive Fourier transform $4\pi /|\mathbf{k}{|}^2$, so $K\ge 0$. Fermionic exchange symmetry pairs the symmetric spatial function with the spin-one-half singlet state and the antisymmetric one with the spin-one-half triplet state. For these fixed orbitals the triplet lies lower by $2K$ when $K>0$. This does not compare arbitrary singlets and triplets with different orbital occupations; for negligible orbital overlap the splitting can vanish.

Coupling orbital angular momentum $L>0$ to spin one-half produces $J=L\pm 1/2$. The dimensions add as $(2L+2)+2L=2(2L+1)$. A change between coupled and product bases changes labels, not the dimension of the Hilbert space.

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.

For $g=\mathrm{diag}(1,-1,-1,-1)$ and the block-diagonal standard representation of the gamma matrices, write a positive-energy Dirac spinor as $(\xi ,\eta {)}^T$. The Dirac equation gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. Choosing two orthonormal two-spinors ${\chi }_s$ yields the displayed positive-frequency solutions. Negative-frequency solutions are $v_s=\sqrt{E+m}((\mathbf{\sigma }\cdot \mathbf{p}){\eta }_s/(E+m),{\eta }_s{)}^T$ multiplying $e^{ip\cdot x}$. They obey $(/p+m)v_s=0$. Both have norm $2E$; their Dirac adjoints give ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$ and ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$. Spatial rotations act on each two-spinor with generators ${\sigma }_j/2$, explaining spin one-half.

The University of Iowa's Electron Spin Resonance, University of Iowa laboratory protocol (revised July 2007, Experiment and Resonance Data Analysis, printed pp. 2–6) specifies DPPH on cotton in a tube inside an RF coil, between Helmholtz coils. A bridge monitors RF absorption; an oscilloscope records this output and the voltage across a $1\Omega$ current-monitor resistor. Sweeping the coil current passes through the resonant magnetic field. The protocol describes two dips when the sweep revisits resonance; these are repeated crossings, not two distinct spectral lines.

Its calibration replaces the sample with a Hall effect probe, maps the magnetic field in $2\mathrm{mm}$ steps, and accounts for sample dimensions. At least four RF frequencies test the resonance relation. Sweep speed is adjusted to obtain comparable passages. Only a local peak-plus-background fit is prescribed, because the full shape is not Gaussian. The protocol also flags unexplained trailing spikes and possible electronics contributions. These controls constrain field and sweep errors; an empty-tube/cotton blank would be an additional background control, not a reported measurement in this protocol.^[ref]

For an isolated effective spin one-half with dimensionless positive factor $g$, the basic resonance model isHere ${\mu }_{\mathrm{B}}$ is the Bohr magneton, $h$ is the Planck constant, $\nu$ is the applied frequency, and $B_{\mathrm{res}}$ is the calibrated resonant magnetic field. A change in frequency should shift the resonance field proportionally in this model. The factor $g$ describes the electronic magnetic moment coupling, not the species' net electric charge.

A resonance supports the presence of detectable paramagnetism. It does not, by itself, identify an unknown radical: other species exhibiting paramagnetism, impurities, overlapping responses, and interactions with atomic nuclei can complicate interpretation. Conversely, absence of a detectable resonance need not mean absence of free radicals; a low concentration, a short lifetime, or an unsuitable measurement range can conceal them. The occupancy and chain diagrams are original bookkeeping illustrations; neither is a measured or simulated DPPH spectrum.

An electron has spin one-half. Its intrinsic spin angular momentum has two possible components, $\pm \hslash /2$, along a specified unit vector $\mathbf{n}$. These are the two eigenvalues of the corresponding spin measurement; the spin is distinct from the electron's orbital angular momentum.

The dimensionless coefficient in the spin part of the electron magnetic moment. It is distinct from a gyromagnetic ratio, which has units: in the angular convention, ${\gamma }_e=-g_{s}e/(2m_e)$. It is also distinct from the Landé g-factor $g_J$ of a combined atomic angular momentum and from a nuclear or hyperfine g-factor.

For spin one-half, the largest magnitude of a spin-component magnetic dipole moment is $g_s{\mu }_B/2$, not $g_s{\mu }_B$. Thus the ideal value $g_s=2$ gives one Bohr magneton. This follows directly from the spin angular momentum projections $S_z=\pm \hslash /2$.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

For a real unit vector $\mathbf{n}$, the Pauli matrix multiplication law gives $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Separating the even and odd powers in the matrix exponential proves the displayed identity. For spin one-half, the physical rotation through angle $\theta$ has $a=\theta /2$.

The Fermi energy is the highest occupied one-particle energy in the ground state of a noninteracting Fermi gas at zero temperature, measured here relative to the lowest one-particle energy. The Pauli exclusion principle allows only one fermion in each complete one-particle quantum state, including its spin, so adding more particles fills progressively higher energy eigenstates. A cold gas can therefore have substantial kinetic energy even when its temperature is zero. In the thermodynamic limit, with the same energy reference, $E_F=\mu (T=0)$ where $\mu$ is the chemical potential.^[ref]

For a uniform, three-dimensional gas of noninteracting electrons, with spin one-half, equal populations of the two spin states, no applied magnetic field, and nonrelativistic particles of mass $m=m_e$,Here $n$ is the total number density, $V$ the volume, $p_F$ the Fermi momentum, $\hslash =h/(2\pi )$ the reduced Planck constant, and $k_B$ the Boltzmann constant. The electron Fermi temperature $T_F$ expresses the same energy scale in kelvin; it is not the actual gas temperature. Strong degeneracy of a degenerate electron gas means $T\ll T_F$. The momentum formula counts states and survives changing the energy–momentum relation, whereas $E_F=p_F^2/(2m)$ requires $p_F\ll mc$.^[ref]

The reference matters: the displayed $E_F$ excludes rest energy and sets the zero at $p=0$. Including rest energy shifts both the one-particle energy and the zero-temperature chemical potential by $m{c}^2$, leaving occupations unchanged. More generally, adding a constant $C$ to every one-particle energy also adds $C$ to $\mu$; the Fermi-Dirac distribution depends only on their difference. In a finite quantum system the chemical potential at zero temperature can lie between the last occupied and first empty energy eigenstates; its equality with the filling edge is a continuum statement.

At positive temperature, $E_F$ remains the zero-temperature scale associated with the specified number density. The Fermi level $\mu (T)$ must instead satisfy the particle-number conditionwhere $g(E)$ is the extensive density of states, including both spin states. Smearing the Fermi-Dirac distribution while arbitrarily holding $\mu =E_F$ generally changes $N$. This distinction matters when interpreting electron heat capacity and any experiment that infers temperature from an occupation profile.^[ref]

The $S=1$ Heisenberg antiferromagnet chain realizes the Haldane phase: a gapped symmetry-protected topological phase with effective spin one-half degrees of freedom at open ends. Time-reversal symmetry is one possible protecting symmetry. Integer spin alone does not imply the same protected class: odd- and even-integer chains differ. The broader gap distinction is the Haldane conjecture.^[ref]

For two spin one-half systems, the dimensionless isotropic Heisenberg model coupling is $C=X\otimes X+Y\otimes Y+Z\otimes Z$, where $X,Y,Z$ are Pauli matrices. In the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$,where $S$ is the SWAP gate. This follows by acting on each basis state: the $XX+YY$ terms exchange the two opposite-spin states with coefficient two and annihilate the parallel-spin states. The symmetric three-dimensional sector has eigenvalue $1$ and the antisymmetric sector has eigenvalue $-3$. HenceThis identity explains how isotropic exchange interaction generates a SWAP gate at $a=\pi /2$ up to a global phase.

The Jordan–Wigner transformation represents a one-dimensional spin one-half chain by fermions with a parity string preceding each site. With $P_n={\prod }_{m<n}(1-2n_m)$, one has $S_n^{+}=P_n{c}_n$ and $S_n^z=1/2-n_n$. The string converts the off-site canonical anticommutation relations into commuting spin operators. Adjacent exchange terms become quadratic fermion expressions, while a periodic end bond depends on global fermion parity.

The neutron is the electrically neutral spin-one-half baryon found with protons in most atomic nuclei.

A neutron is a nucleon: it contributes to an atomic nucleus without adding electric charge. Its spin one-half makes it a fermion, and it belongs to the baryons. Electrical neutrality does not mean an absence of interactions: a neutron can transfer momentum and kinetic energy to an atomic nucleus, producing a charged nuclear recoil that a particle detector can register. This indirect route is central to neutron detection by proton recoil.

The beautiful feature of the Chadwick neutron discovery experiment is that an unseen projectile acquired a measurable mass through the motion of things it struck. Penetration alone left competing explanations open; combining nuclear recoils from different target masses supplied a much stronger test.

An electrically neutral neutron still has a magnetic moment $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A collimated beam passing through an inhomogeneous magnetic field experiences the force derived from $H=-\mathbf{\mu }\cdot \mathbf{B}$. In a region whose field axis is approximately fixed, a spin one-half neutron has two field-axis spin eigenstates, producing two allowed transverse forces and two separated beam components after a drift. A homogeneous field can cause Larmor precession but cannot separate positions through this force.

The neutron's negative gyromagnetic ratio interchanges which spin projection moves toward increasing field; it does not change the number of branches. An unpolarized beam gives equal ideal populations. A beam prepared in one field-axis eigenstate populates only one branch, so splitting into two observed spots presupposes both components and sufficient spatial resolution.

Nuclear magnetic resonance detects the response of atomic nuclei with a magnetic moment to a radio frequency magnetic field in a steady magnetic field. Its striking possibility is to measure a nuclear property through an ordinary electrical signal from a bulk sample: an enormous number of tiny magnetic moments can produce a measurable change in a receiving circuit.

For a nucleus with spin angular momentum $\mathbf{J}$, define its gyromagnetic ratio by $\mathbf{\mu }=\gamma \mathbf{J}$. Throughout this article $\gamma$ is angular, measured in radians per second per tesla; $\gamma /(2\pi )$ is the corresponding value in cycles per second per tesla. With ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$, the Hamiltonian is $H=-\gamma B_0{J}_z$. Adjacent energy levels differ byThus a measured resonance frequency and a calibrated magnetic field determine $|\gamma |$. Converting that into a nuclear magnetic moment also requires the spin quantum number: the maximum projection has magnitude $|\gamma |\hslash I$. This simple experiment does not by itself determine the sign of $\gamma$.

The Rabi molecular beam magnetic resonance experiment detects transitions through changes in a selected molecular beam. The bulk experiments below instead detect electrical consequences of nuclear magnetization in matter at ordinary density. A small Boltzmann distribution population imbalance is multiplied by a very large number of nuclei. For noninteracting spin one-half nuclei of number density $n$ and positive $\gamma$, their equilibrium magnetization isThis follows by weighting the two projections $\pm \gamma \hslash /2$ with their Boltzmann distribution probabilities. It explains both the smallness of the bulk polarization at ordinary temperature and the value of many nuclei contributing to one receiving circuit.

NMR spectroscopy measures resonant responses of atomic nuclei to applied electromagnetic radiation in a magnetic field. In residual-group monitoring of polyurethane formation, distinguishable signal integrals supply a relative residual-group readout; overlap or very weak signals can limit that inference.