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DPPH is a neutral organic radical used as a spin reference in electron paramagnetic resonance (EPR), also called electron spin resonance (ESR). Its unpaired electron remains bound within the chemical species: “free radical” does not mean a freely moving electron. The DPPH electron spin resonance experiment makes its magnetic moment observable.

A spin system with two distinguishable, mutually coupled spins A and B, each coupled to three equivalent X spins. An ethoxy methylene with nonequivalent hydrogens can display this pattern; small A/B chemical-shift separation relative to their coupling produces second-order distortion.

Two equivalent $l=1$ electrons require an antisymmetric complete state. Orbital exchange has parity $(-1{)}^L$ and the spin singlet is antisymmetric, giving allowed $L+S$ even. The terms are ${}^3{P}_{0,1,2}$, ${}^1{D}_2$ and ${}^1{S}_0$, with $9+5+1=15$ states, equal to $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund ordering puts ${}^3{P}_0$ lowest for the less-than-half-filled configuration.

The ground state of an even-even nucleus has $J^P=0^{+}$, and the emitted alpha particle also has $0^{+}$. In an alpha decay to daughter spin $J_Y$, addition of angular momentum permits total zero only when the relative orbital angular momentum is $L=J_Y$. Parity conservation gives $+1={\pi }_Y(-1{)}^L$, hence the daughter has natural parity, $0^{+},1^{-},2^{+},3^{-},\dots$. This rule does not itself restrict $J_Y$ to even integers. An even-$J$ sequence such as $0^{+},2^{+},4^{+}$ requires additional nuclear structure, for example a positive-parity ground-state nuclear rotational band.

The helium-four atomic nucleus, containing two protons and two neutrons. Its ground-state spin and isospin are zero. A quantized cubic four-Skyrmion supplies a corresponding nuclear state in the soliton description.

Labels an angular momentum eigenstate through the eigenvalue ${\hslash }^{2}j(j+1)$ of the squared angular momentum operator. The allowed projections are $m\hslash$, with $m=-j,\dots ,j$. Orbital angular momentum has integer $j$; spin can also have half-integer $j$.

Annealed disorder allows the variables representing disorder to equilibrate along with the spins. With a specified normalized prior measure over couplings, its partition function is the average ${\mathbb{E}}_J{Z}_J$. This differs from preparing samples with fixed couplings and averaging their measured thermodynamics, as in quenched disorder.^[ref]

The anomalous Zeeman effect is Zeeman splitting whose pattern is not the simple orbital triplet. Spin and spin-orbit interaction make the upper and lower Landé g-factors differ. The same weak-field formula still applies when its assumptions hold: “anomalous” does not mean nonlinear in the magnetic field.

For example, take an electronic transition with $L_u=1,S_u=1/2,J_u=1/2$ and $L_l=0,S_l=1/2,J_l=1/2$, neglecting hyperfine structure. With $g_s=2$, the formula above gives $g_u=2/3$ and $g_l=2$. The two allowed $q=0$ transitions have dimensionless shifts $g_u{m}_u-g_l{m}_l=\pm 2/3$, while the $q=\pm 1$ transitions give $\pm 4/3$. Thus there are four frequencies, even though each level contains only two magnetic quantum numbers. This is an independent worked deduction; it should not be substituted for the unresolved widths in the 1896 experiment.

The antiproton is the antiparticle of the proton: it has the same rest mass and spin $1/2$, with opposite electric charge and baryon number. A negative electric charge alone cannot identify an antiproton, because a negative pion also bends in that direction in a magnetic field. Combining magnetic rigidity with velocity determines a rest mass; agreement with the proton rest mass then makes the identification much more discriminating.

The mode expansion of a Dirac field explains how independent particle and antiparticle excitations enter a relativistic field description. It supplies a theoretical language for the antiproton, while the Chamberlain-Segrè antiproton identification supplies an experimental test. The proton is a composite baryon, so a free Dirac field is not a complete account of its internal dynamics or of antiproton production in a target.

An antisymmetric exchange interaction between spins, proportional to ${\mathbf{D}}_{ij}\cdot ({\mathbf{S}}_i\times {\mathbf{S}}_j)$ with a bond-dependent vector ${\mathbf{D}}_{ij}$. The overall sign is conventional and must be specified. It favors a handed twist rather than simple parallel or antiparallel alignment. For a uniform chain term $-D({\mathbf{S}}_j\times {\mathbf{S}}_{j+1}{)}_z$, common spin rotations about $z$ and time reversal remain symmetries, although reversing the bond orientation changes the sign of $D$.

The central system of protons and neutrons in an atom. Its mass number $A$ is its baryon number in ordinary nuclear states; $Z$ is the proton number and $N$ the neutron number.

The nucleus contains almost all of an atom's mass and carries positive electric charge $Ze$, where $e$ is the elementary charge. The atomic number $Z$ identifies the chemical element; changing $N$ at fixed $Z$ gives another isotope. The mass number $A=Z+N$ counts nucleons, rather than specifying the exact nuclear mass: nuclear binding energy lowers the mass of a bound system relative to its separated constituents. The simplest nucleus is a single proton; other atomic nuclei can contain many protons and neutrons bound by the nuclear force.

A nuclear quantum state has properties such as spin and a spatial nuclear charge distribution. These are different questions from merely counting nucleons. A small-angle elastic scattering measurement can reveal a nuclear charge radius, while a wider range of momentum transfer tests the distribution's shape. The Hofstadter nuclear electron scattering experiment made this connection between scattering patterns and nuclear structure experimentally concrete.

An LS coupling multiplet with specified total orbital angular momentum $L$ and total spin $S$, usually within a specified electron configuration. Its magnetic states have $M_L=-L,\dots ,L$ and $M_S=-S,\dots ,S$. Spin-orbit coupling splits it into atomic fine-structure levels.

The Barkhausen effect consists of abrupt magnetization changes during a smooth applied magnetic field sweep. Its inductive signal is Barkhausen noise, often arising from collective magnetic domain wall motion.^[ref]

The distinction is between a smooth drive and an intermittent material response. A smooth macroscopic hysteresis curve can hide smaller jumps. By Faraday's law of induction, a pickup coil responds to the rate of change of linked magnetic flux, so a rapid jump can be conspicuous even when the total change is small. A pulse is not one chemical atom flipping: neither its height nor the number of audible clicks counts individual electron spins. Qualitative crackling alone does not establish quantum spin; collective switching also admits a classical model with trapped interfaces. Compare the distinct mechanical readout of angular momentum transfer in the Einstein-de Haas effect.

Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

Use $N_0={\nu }_s$ for the normal-state single-spin density of states per unit energy and volume at the Fermi surface. Write the attraction as $-V/\mathcal{V}$, with $V>0$, and define $\lambda =N_{0}V$.^[ref]

Since $V/\mathcal{V}$ has units of energy, $V$ has units of energy times volume, making $\lambda$ dimensionless. Set ${\Omega }_D=\hslash {\omega }_D$, an energy cutoff using the Debye frequency ${\omega }_D$, rather than a frequency itself. These conventions also fix the meaning of the earlier constant-shell BCS gap equation.

Here is the reduction of the BCS gap equation in these conventions. Substituting the zero-temperature BCS anomalous average givesThe $k$ sum counts one spin species. For $\Delta >0$, cancelling $\Delta$ and replacing this sum by an integral with constant $N_0$ givesThe factor two from the symmetric integral cancels the $1/2$ in the BCS anomalous average; it is not an extra spin degeneracy. Substituting $\xi =\Delta \sinh t$ evaluates the integral as $\mathrm{arsinh}({\Omega }_D/\Delta )$. Solving givesThe first equality is exact within the constant-density of states, sharp-cutoff mean-field approximation; the last is the weak coupling asymptote. Extending the first equality to large $\lambda$ does not make this simplified model an accurate description of a strongly coupled material.

For an original numerical example, choose $\hslash {\omega }_D=20\mathrm{meV}$ and $\lambda =0.25$. The exact constant-shell BCS gap equation gives $\Delta (0)=0.732871\mathrm{meV}$; the weak coupling formula gives $0.732626\mathrm{meV}$. Its fractional underestimate, measured against the exact result, is exactly $e^{-2/\lambda }=e^{-8}\simeq 0.03355\%$. At $\lambda =1$ that error is $e^{-2}\simeq 13.53\%$, so the two formulas are no longer interchangeable. No material parameters have been fitted here.

At nonzero temperature, substitute the finite-temperature BCS anomalous average instead. The same cancellation yieldsAt the superconducting transition temperature take the continuous $\Delta \to 0$ limit. The integrand at $\xi =0$ then has the finite limiting value $1/(2k_B{T}_c)$, so no divergent endpoint needs to be inserted into a numerical quadrature. For $k_B{T}_c\ll {\Omega }_D$, the resulting integral gives $k_B{T}_c\simeq 1.134{\Omega }_D{e}^{-1/\lambda }$ and $2\Delta (0)/(k_B{T}_c)\simeq 3.528$. These ratios concern the isotropic weak coupling model.^[ref]

A beam's unequal population of spin projections along a specified axis. For spin-one-half particles the displayed longitudinal polarization uses populations parallel and antiparallel to the momentum. An unpolarized beam has equal populations and requires an initial-spin average in a cross section; a polarized beam weights the corresponding helicity amplitudes differently. This differs from photon linear polarization.

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].

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.

A nuclear beta decay connects initial and final quantum states through vector and axial-vector parts of the weak charged current. At leading order in the lepton wavelength expansion, Fermi transitions carry lepton-pair spin $S=0$, while Gamow-Teller transitions carry $S=1$. If the lepton pair carries orbital angular momentum $l$, its total angular momentum is $J=l$ for $S=0$, or $J=|l-1|,\dots ,l+1$ for $S=1$. The nuclear states must obey $|J_i-J_f|\le J\le J_i+J_f$, and their parities satisfy ${\pi }_f/{\pi }_i=(-1{)}^l$ in this leading orbital classification. The least permitted $l$ labels the degree of forbiddenness. Real forbidden transitions involve several weak multipole operators; the elementary spin classification does not assert that only one operator or nuclear matrix element exists.

A bipartite spin rotation rotates the spin on one sublattice by $\pi$, converting an alternating classical Néel state to an all-up reference state. Rotation about $x$ sends $S^z\mapsto -S^z$ and $S^{\pm }\mapsto S^{\mp }$ on that sublattice. For an isotropic nearest-neighbor Heisenberg antiferromagnet, a bond becomes $-S_i^z{S}_j^z+(S_i^{+}{S}_j^{+}+S_i^{-}{S}_j^{-})/2$. The transformation is unitary and preserves the spin commutation relations; it changes the reference frame, not the model's spectrum.