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An atomic nucleus with odd proton number and odd neutron number. Both unpaired species can contribute to the total nuclear spin; addition of angular momentum gives possible couplings, while the nuclear residual interaction determines their ordering.

An isotope of oxygen with 8 protons and 9 neutrons. Its nuclear spin concerns the coupled nuclear quantum state, not the electronic configuration.

Parahydrogen uses the antisymmetric spin-one-half singlet state,Here $I_{\mathrm{n}}$ labels total nuclear spin; it is not the moment of inertia $I$ in the linear rigid rotor formula. There is one nuclear-spin state for every rotational quantum state. The lowest allowed rotational level is $J=0$, and the next is $J=2$. The description “opposite spins” alone is insufficient: the relative minus sign distinguishes this singlet state from the zero-projection member of the triplet state.

For a one-electron P-to-S transition with nuclear spin zero, ten electric-dipole sublevel channels have finite strength at weak nonzero field. Six retain finite strength in the complete orbital-spin decoupling limit. Channel count is distinct from spectral frequency count: retaining first-order residual spin-orbit interaction energy gives five distinct frequencies, with two central channels degenerate.

Use the isolated nuclear spin Zeeman effect implied by the supplied nuclear magnetic moment, with hyperfine structure neglected. The proton gyromagnetic ratio isThe two spin levels are separated by $g_p{\mu }_N{B}_0$. Resonant nuclear magnetic resonance therefore requires the Larmor frequency

The duration is a Rabi pi pulse: in a frame rotating with the resonant transverse magnetic field, the spin must rotate through $\pi$. If $B_1$ denotes the magnitude of a circularly rotating field with the correct sense of rotation, the Rabi angular frequency is ${\Omega }_R={\gamma }_p{B}_1$. The shortest pulse is then

There is an amplitude convention to specify. If the applied transverse magnetic field is instead the linearly oscillating $B_1\cos (\omega t)\hat{\mathbf{x}}$, its resonant circular component has magnitude $B_1/2$. The rotating-wave approximation gives ${\Omega }_R={\gamma }_p{B}_1/2$ and henceThe wording does not specify this polarization convention, so the two pulse durations should be distinguished. Odd multiples of either shortest Rabi pi pulse also invert the spin within the corresponding ideal model.

The ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

The $4f^7$ shell carries a large spin (its ground atomic term under Hund's rules has $L=0,S=7/2$), and the $6s$ electron penetrates the inner shells. Its nonzero probability density at the atomic nucleus supplies a potentially large Fermi contact interaction with the nuclear magnetic dipole moment. The $6s$ shell here is singly occupied, so the obligatory paired-$6s^2$ cancellation is absent. Both stable europium isotopes have nonzero nuclear spin, as the spectrum below also establishes.

The term symbol ${}^8{D}_{11/2}$ specifies $S=7/2$, $L=2$, $J=11/2$. Its large $J$, combined with appreciable nuclear spin, permits a wide range of total angular momentum $F$ and hence a substantial total hyperfine structure span. For $I=5/2$, $F=3,\dots ,8$ and the span is $30|a|$. The penetrating $6s$ contribution and the available range of $F$ explain the large hyperfine structure; the term symbol alone does not determine its constant, which also depends on electronic recoupling, radial wavefunctions and configuration interaction.

Write $a_{\nu }=a/h$, so the Landé interval rule for hyperfine structure intervals in ordinary frequency is ${\nu }_F-{\nu }_{F-1}=a_{\nu }F$. For the first four selected positions, the successive positive differences areTheir ratios are approximately $4:5:6$, sinceThe decreasing frequencies therefore correspond to increasing $F$, with $a_{\nu }\simeq -0.706\mathrm{GHz}$, if increasing printed position means increasing physical frequency. Continuing the intervals predicts additional positions near $10.60-7(0.706)\simeq 5.66$ and then $5.66-8(0.706)\simeq 0.01\mathrm{GHz}$. They identify peaks $i$ and $l$ unambiguously. The full assignment for ${}^{151}\mathrm{Eu}$ isWith $J=11/2$, these endpoints requireEquivalently, a complete set of six hyperfine levels gives $2I+1=6$ when $I<J$. This inference assumes that every upper hyperfine level contributes an observed position and the negligible lower hyperfine structure lets coincident transitions be grouped by their upper energy. A line count without these completeness assumptions would not alone fix nuclear spin.

The supplied positions obey the Landé interval rule only approximately. For example, ${\nu }_e-{\nu }_i=4.97\mathrm{GHz}$, whereas $7({\nu }_a-{\nu }_b)/4=4.935\mathrm{GHz}$. Even allowing each printed position a rounding error of $0.005\mathrm{GHz}$ does not fully reconcile these two relations. A least-squares dipole-only fit has a largest position residual of about $0.017\mathrm{GHz}$. Small additional hyperfine interactions, measurement uncertainty beyond the displayed precision, or an error in the tabulated data could account for this; it is inappropriate to claim an exact identity. The grouping and nuclear spin are nevertheless robust. Reversing the convention for the printed frequency axis reverses the inferred signs of both constants, without changing the following ratio.

The remaining six positions give the other isotope's sequenceThe adjacent differences divided by $4,5,6,7,8$ are $0.3125,0.3140,0.3133,0.3143,0.3138\mathrm{GHz}$, respectively. Thus $a_{\nu ,153}\simeq -0.314\mathrm{GHz}$ and the same endpoint argument gives $\boxed{I_{153}=5/2}$.

For the same electronic energy level, the magnetic-dipole hyperfine structure constant is proportional to ${\mu }_I/I$, to leading order in the small isotope-dependent change of the electronic wavefunction. Since the two nuclear spins are equal,Using the full spans averages over the approximate Landé interval rule: ${\sum }_{F=4}^{8}F=30$, soThe constants have the same sign, so the nuclear magnetic dipole moments have the same relative sign. Their absolute sign cannot be obtained from their ratio without the electronic field's sign. Finite nuclear-size corrections to the hyperfine interaction, often called the hyperfine anomaly, and electric-quadrupole corrections are neglected at this accuracy.

A good atomic qubit needs two reproducible, orthogonal internal states whose populations and relative quantum phase survive for many gate times. Both should be long lived, with weak spontaneous emission and little decoherence from fluctuating fields or atomic motion. Their transition must nevertheless permit fast, selective, coherent control with little leakage to other states. Reliable initialization and state-selective quantum measurement are also required; optical pumping and fluorescence readout can supply these functions.

For rubidium-87, use the two ground state hyperfine levels in the $5{}^2{S}_{1/2}$ electronic configuration. The electronic angular momentum is $J=1/2$, and addition of angular momentum to the given nuclear spin $I=3/2$ gives $F=|I-J|,\dots ,I+J=1,2$. A suitable Rubidium-87 clock qubit isThese are coupled electron-nucleus angular momentum eigenstates, rather than two bare nuclear-spin projections. Their hyperfine structure supplies a resolvable splitting ${\nu }_{01}\simeq 6.835\mathrm{GHz}$. Both belong to the electronic ground state, avoiding a short optical excited-state lifetime. Since the weak-field Zeeman shift is proportional to $g_F{m}_{F}B$, both have zero linear shift at $B=0$: their rubidium-87 clock transition suppresses first-order magnetic-field dephasing near zero field. A quadratic Zeeman shift of a clock transition and differential optical shifts remain. A small stable bias field can define the quantization axis and help resolve unwanted transitions, at the cost of some residual magnetic sensitivity.^[ref]

Drive the magnetic dipole transition with a resonant microwave field of suitable polarization, here $\Delta m_F=0$, or use two phase-coherent optical fields for a far-detuned stimulated Raman transition with their frequency difference resonant with ${\nu }_{01}$. The optical excited state is then only virtually populated, reducing spontaneous emission. In the rotating frame and rotating-wave approximation, define the drive phase $\varphi$ by the effective HamiltonianHere $X,Y,Z$ are the Pauli matrices on the chosen qubit, and $\Omega$ is the Rabi angular frequency. Pulse duration controls the rotation gate angle, while phase selects the axis in the equatorial plane of the Bloch sphere. A $\pi$ pulse exchanges the populations; a $\pi /2$ pulse makes an equal quantum superposition from either basis state. Controlled Rabi detuning, a differential energy shift, or compositions of these pulses give $z$ rotations. For example, $R_z(\alpha )R_y(\beta )R_z(\gamma )$ realizes any one-qubit unitary operator up to a global phase. Thus the atomic encoding admits arbitrary single-qubit quantum gates.

TODO: Biology deferred at the user's request: clinical contrast interpretation, applications, and the head-injury MRI-versus-CT comparison.

The generic physics of magnetic resonance imaging combines a main static magnet, radiofrequency transmit and receive coils, controlled magnetic field gradients, signal electronics and reconstruction. A transverse radiofrequency pulse excites nuclear spins near their Larmor frequency, ${\omega }_0=\gamma B_0$. The evolving transverse magnetization induces a received signal; phase coherence and relaxation determine its time dependence.

Gradients encode position into phase. In a simplified homogeneous receive-sensitivity model,Here $\mathbf{k}$ is in cycles per unit length and $\gamma$ is in radians per second per tesla. Echo sign changes, pulse sequence timing and receive sensitivity belong in a more complete signal model. Sampling this MRI k-space and reconstructing its data produces an image; the measured samples are not already real-space pixel values.

The physical signal depends on spin density, longitudinal recovery and transverse dephasing. Timing changes their weighting. Static field inhomogeneity adds reversible dephasing beyond intrinsic transverse relaxation, motivating echo techniques and distinguishing $T_2^{\ast }$ from $T_2$. Those are physical signal mechanisms; their biological interpretation remains deferred.

In the Wu experiment, Chien-Shiung Wu and collaborators studied beta-minus decay of polarized cobalt-60,A thin cobalt-60 layer on a cerium magnesium nitrate crystal was cooled by adiabatic demagnetization to a few millikelvin. A polarizing magnetic field produced nuclear polarization. An anthracene scintillation counter and photomultiplier detected the emitted electrons; angular gamma ray observations from the daughter de-excitation monitored whether the nuclei remained polarized.^[ref]

The electrons were emitted preferentially opposite the parent nuclear spin. Reversing the nuclear polarization reversed the asymmetry, while warming the sample removed both polarization and the effect. Those controls distinguished a spin-correlated asymmetry from a fixed detector bias.

The relevant angular correlation is a polarized beta-decay angular distribution of the formUnder parity, ${\mathbf{p}}_e\to -{\mathbf{p}}_e$ but $\mathbf{J}\to \mathbf{J}$, so the spin-momentum correlation changes sign. Parity conservation would require equal rates for these two directions relative to the same spin. The observed nonzero coefficient therefore demonstrated parity violation in beta decay. An unpolarized source would average this correlation to zero and could not provide the same test.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

For hyperfine coupling for a spin-one nucleus, let the electron spin be $s=1/2$ and the nuclear spin $I=1$. First couple $\mathbf{j}=\mathbf{L}+\mathbf{s}$, giving $j=L\pm 1/2$, then $\mathbf{F}=\mathbf{j}+\mathbf{I}$, giving $F=|j-I|,|j-I|+1,\dots ,j+I$.

For each $F$, the projection is $m_F=-F,-F+1,\dots ,F$; electron projections similarly satisfy $m_j=-j,\dots ,j$. Summing $2F+1$ gives 18 states for $L=1$ and 30 for $L=2$, agreeing with $(2L+1)(2s+1)(2I+1)$.

Notation qualification: if the printed leading “2” denotes the principal quantum number, a hydrogenic $2d$ bound state does not exist because $L\le n-1$; a physical $d$ state needs $n\ge 3$. If a doublet term symbol ${}^{2}P$ or ${}^{2}D$ was intended, its leading 2 instead means $2s+1$, and the angular-momentum table applies without fixing $n$. The answers above use the explicitly specified $L$ values, while retaining this ambiguity.^[ref]

A qubit encoded in the two $m_F=0$ ground state hyperfine levels of rubidium-87. The electronic $J=1/2$ and nuclear spin $I=3/2$ combine to $F=1,2$. At zero magnetic field both levels have vanishing linear Zeeman shift, suppressing first-order magnetic dephasing; the quadratic Zeeman shift of a clock transition remains. Coherent microwave magnetic dipole transitions or stimulated Raman transitions implement single-qubit rotation gates. These are coupled electron-nucleus states, not bare nuclear-spin states.

Spin is intrinsic angular momentum carried by a quantum particle or another effective quantum degree of freedom. It does not require a literal rotating surface. Spin components generate rotations and obey $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$; a spin-$s$ multiplet has $2s+1$ projections along a chosen quantization axis.

For spin one-half, the operators are $S_i=\hslash {\sigma }_i/2$, with outcomes $\pm \hslash /2$ for a component measurement. A Stern-Gerlach experiment resolves discrete magnetic responses to a field gradient, while magnetic-resonance experiments control and observe Larmor precession. Spin differs from orbital angular momentum, which arises from the spatial dependence of a wavefunction. Combining both is necessary for atomic fine structure and coupling the electronic result to nuclear spin gives hyperfine structure.

An elastic recoil interaction whose leading coherent nuclear amplitude does not depend on nuclear spin. At low momentum transfer, proton and neutron amplitudes add; equal couplings give an $A$-fold amplitude and the displayed cross-section scaling, with reduced masses ${\mu }_A$ and ${\mu }_p$. A finite-momentum nuclear form factor reduces coherence. Unequal nucleon couplings or other interactions change how experiments on different targets should be compared.

Spin isomers of hydrogen are two forms of molecular hydrogen, ${}^1{\mathrm{H}}_2$, distinguished by the combined nuclear spin of their two protons. They have the same chemical formula and electronic ground state, but different allowed molecular rotation states. Parahydrogen has total nuclear spin zero and even rotational quantum number $J$; orthohydrogen has total nuclear spin one and odd $J$. The difference makes heat capacity a probe of fermionic exchange symmetry, and makes cooling history matter for a tank of molecular hydrogen.

Assume the electronic ground state $X{}^1{\Sigma }_g^{+}$ and the vibrational ground state of a free molecule. The electronic factor is symmetric under exchanging the two protons, and the vibrational factor depends only on their separation, so it is also symmetric. These assumptions concern the exchange of protons; the electronic wavefunction separately obeys fermionic exchange symmetry under exchange of electrons. Neglect electronic excitation, vibrational excitation, external fields, and small interactions involving nuclear spin when calculating rotational energy eigenvalues.

The two protons are identical particles with spin $1/2$. Their complete wavefunction must change sign under their particle exchange operator. Interchanging the two ends reverses the internuclear direction $\hat{\mathbf{r}}$, and the spherical harmonic rotational factor satisfiesThe antisymmetric spin-one-half singlet state therefore requires symmetric, even-$J$ rotation; the symmetric spin-one-half triplet state requires antisymmetric, odd-$J$ rotation. This is exchange symmetry of two identical spin-one-half fermions applied inside a diatomic molecule, not an extra rotational force. A linear rigid rotor may omit spin-dependent splittings while still requiring this restriction on its allowed quantum states.