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For an ultrarelativistic electron scattered elastically by a static, spherical, spin-zero atomic nucleus, a plane-wave Born approximation retains a single electromagnetic interaction. The Fourier transform of its Coulomb potential is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the differential scattering cross-section by $|F(q){|}^2$:Here $E\gg m_e{c}^2$ is the incident total electron energy, $\theta$ the scattering angle, $d\Omega$ a solid angle element, and $\alpha$ the fine-structure constant. The cosine factor already accounts for the projectile's spin; a spinless projectile would give a different point-charge factor. The formula neglects nuclear recoil, electron rest-mass corrections, radiative corrections and Coulomb distortion in electron scattering. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's Coulomb potential, rather than undistorted plane waves.

Nonzero target spin can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a proton or deuteron. Even for spin zero, finite energy resolution, recoil and Coulomb distortion in electron scattering can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.

A model with separate filled proton and neutron momentum seas in a nuclear mean field. Each spin-half species has two spin states, so counting momentum-space cells gives $p_{F,i}=\hslash (3{\pi }^2{n}_i{)}^{1/3}$. It describes typical nucleon motion and a kinetic contribution to nuclear symmetry energy, but binding requires additional interactions.

The residual strong interaction between nucleons. Its long-distance part involves pion exchange and depends on spin and isospin. It is distinct from directly exchanging gluons between color-singlet nucleons.

Why do protons and neutrons bind into an atomic nucleus, even though protons repel electrically? An attractive nuclear force can lower the total energy below that of separated nucleons, producing nuclear binding energy. The deuteron, containing one proton and one neutron, is the simplest such bound state.

In Quantum chromodynamics, nucleons are composite colour singlets built from quarks and gluons. Their residual nuclear force has a longest-range strong contribution from one-pion exchange, with additional interactions at shorter distances.^[ref]

The nuclear magneton is the unit of magnetic momentwhere $m_p$ is the proton mass. For a nuclear spin operator $\mathbf{I}$, the nuclear magnetic moment is $\mathbf{\mu }=g_I{\mu }_N\mathbf{I}/\hslash$. Consequently its gyromagnetic ratio is ${\gamma }_I=g_I{\mu }_N/\hslash$, and its Larmor frequency in a uniform magnetic field of magnitude $B$ is $|{\gamma }_I|B$. The nuclear magneton is smaller than the Bohr magneton by the electron-to-proton mass ratio.

The mass deficit of a bound atomic nucleus relative to its separated protons and neutrons. If their rest masses are $m_p$ and $m_n$, the atomic nucleus has rest mass $M_{\mathrm{nuc}}$, and the positive nuclear binding energy is $B$, thenwith $c$ the speed of light. Releasing energy when the nucleons bind reduces the rest mass of the bound system without changing the exact count $A=Z+N$. A neutral atomic mass also includes $Z$ electron rest masses, reduced by the energy required to separate those bound electrons. Consequently $A-m_{\mathrm{a}}/\mathrm{u}$ is not the definition of the nuclear mass defect: the reference unit and constituent rest masses must also be accounted for.

An effective one-nucleon potential energy representing the average influence of the other nucleons. The nuclear shell model treats motion in this field first, then includes residual interactions. The field experienced by a proton also includes the repulsive Coulomb potential; proton and neutron energy levels need not have identical spacings.

The Gaussian approximation to a nonresonant thermonuclear rate already gives the saddle and curvature. In its notation $E_G=4E_B$, so the same result readsThe full width ${\Delta }_{1/e}$ joins the two points at $e^{-1}$ of the peak of the approximating Gaussian function; it is neither its standard deviation ${\sigma }_G$ nor its full width at half maximum $2\sqrt{2\log 2}{\sigma }_G$. These are local Gaussian approximation widths, not exact equal-height endpoints of the asymmetric kernel.

Why keep the astrophysical S factor inside the integral? If $S(E)$ is positive and smooth, stationarity of $\log S(E)-\Phi (E)$ gives a leading peak displacement $\delta E\simeq {\sigma }_G^2(d\log S/dE{)}_{E_0}$. The slow-variation test is therefore $|d\log S/dE|{\sigma }_G\ll 1$, together with small curvature corrections. More generally any energy-dependent prefactor shifts the saddle; a temperature-dependent prefactor also changes the local temperature exponent of a thermonuclear reaction. A sharp nuclear feature can invalidate this expansion altogether.

Choose $T=15\mathrm{MK}$ and $E_G=493\mathrm{keV}$ for an illustrative proton pair. Then $k_{B}T=1.293\mathrm{keV}$, $E_0=5.905\mathrm{keV}$ and ${\Delta }_{1/e}=6.380\mathrm{keV}$. The relative Maxwell-Boltzmann velocity distribution has energy-density maximum $k_{B}T/2=0.646\mathrm{keV}$, far below the reaction kernel's maximum. Now choose the deliberately artificial enhancementNumerical integration raises the rate by about $54.2\%$ at the same temperature, even though the enhancement is away from $E_0$. This is a sensitivity experiment, not a claimed proton–proton nuclear level or a physical nuclear line-shape model. It illustrates why extrapolating $S(E_0)$ without checking nearby structure can fail.

A nuclear shell groups single-particle energy levels for protons or neutrons in a nuclear mean field. Filling all states below a sufficiently large gap gives a nuclear shell closure. These shells describe quantum-state occupation rather than thin spatial layers. Their discrete filling effects explain departures from the smooth semi-empirical mass formula.

Why should adding one neutron be easy for one atomic nucleus and much less favorable for its neighbor? In the nuclear shell model, nucleons occupy discrete energy levels in a nuclear mean field. Filling all available states below a sufficiently large gap produces a nuclear shell closure. This connects nuclear separation energies with the counting of quantum states, and connects the spin of an odd atomic nucleus with its last unpaired nucleon.

The starting point is an approximate Hamiltonian operatorwhere $U_i$ is a confining potential energy, $a_i$ specifies nuclear spin-orbit splitting, and $V_{\mathrm{res}}$ contains interactions left out of the nuclear mean field. This is an effective organization of the nuclear many-body problem, rather than a derivation of the nuclear force. In a spherical approximation, addition of angular momentum combines the orbital angular momentum $l$ with spin $s=1/2$ to give $j=l\pm 1/2$; for $l=0$, only $j=1/2$ exists.

A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\dots ,j$. The Pauli exclusion principle permits one neutron in each such state and, separately, one proton. Thus a $j=7/2$ level holds eight neutrons, or eight protons; it does not hold sixteen identical neutrons. Here $n=1,2,\dots$ counts radial levels of the same $l$, starting with the lowest; it is not the hydrogenic principal quantum number. The letters $s,p,d,f,g,h,i$ mean $l=0,1,2,3,4,5,6$.

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 oxidation state is an electron-bookkeeping assignment to a chemical atom: its electric charge number after the ionic approximation of bonds. It helps identify oxidation and reduction and determine stoichiometry, even for covalent bonds without separate charged ions. In water, each hydrogen has oxidation state $+1$ and oxygen has $-2$; this does not assert that a water molecule contains two free protons and a free oxide ion.^[ref]

Three conventions answer different questions. Formal charge divides each bonding electron pair equally in a chosen Lewis structure. Oxidation state uses the ionic approximation of bonds. Partial atomic charge instead partitions the actual electron density and depends on that partition. In the ordinary water Lewis structure, every formal charge is zero although the oxidation states are not. Neither bookkeeping convention supplies numerical partial atomic charges.

Oxidative phosphorylation couples respiratory redox conversion to ATP formation through chemiosmosis. ATP synthase uses inward proton flow to join ADP and inorganic phosphate.^[ref]

The intermediate electrochemical gradient explains why respiratory uncoupling can preserve oxygen consumption while reducing ATP formation. An electron ledger alone cannot determine the recovery of useful free energy.

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

For a reconstructed electron–Positron pair, four-momentum conservation givesNeglecting the electron mass compared with the daughter energies, and calling the opening angle $\theta$, this becomesAs an original example, choose $E_{+}=2\mathrm{GeV}$, $E_{-}=3\mathrm{GeV}$ and $\theta =60^{\circ }$. Then $M_{ee}{c}^2=\sqrt{6}\mathrm{GeV}\simeq 2.449\mathrm{GeV}$, although the pair's laboratory energy is $5\mathrm{GeV}$. The pair carries net momentum. Neither its laboratory energy nor the original proton beam energy is its invariant mass; unobserved products carry additional four-momentum.

For equal, oppositely directed incoming beams with equal total energy $E_b$, their total momentum vanishes. Thus the incoming center-of-mass energy isStipulated $1.50\mathrm{GeV}$ beams therefore provide $3.00\mathrm{GeV}$ of incoming center-of-mass energy. This identity is exact for the stated equal-beam geometry. Radiation before annihilation can reduce the energy forming the final state; the apparatus also has a distribution of beam energies. A nominal machine setting is therefore not an exact event-by-event final-state invariant mass.

Spin and spatial exchange symmetry. Write the normalized spin singlet and triplet states asThe singlet is antisymmetric under exchange and has total spin zero; the triplet is symmetric and has total spin one. For orthogonal orbitals defineIdentical spin one-half particles are fermions, so the complete state must be antisymmetric. Thus $C{\chi }_{00}$ is allowed, and $A{\chi }_{1M}$ is allowed for each triplet component. The spatially symmetric same-orbital states $|aa⟩,|bb⟩$ also require the singlet. A same-orbital triplet vanishes by the Pauli exclusion principle.

These products give a spin-adapted basis. More generally the physical state can be writtenA superposition of the singlet and triplet sectors need not factor into one spatial function times one spin function. Factorization holds for a definite-spin sector, as used in the subsequent model. If the original normalized orbitals are not orthogonal, the corresponding $ab\pm ba$ normalization is $\sqrt{2(1\pm |⟨a|b⟩{|}^2)}$; the remainder of this calculation uses the stipulated orthogonal orbitals.

Integral matrix elements and positivity. Let $a(\mathbf{r})$ and $b(\mathbf{r})$ denote the localized spatial orbitals. With fixed proton positions, the one-electron electronic operator isThe proton-proton repulsion is a separate common nuclear energy. The requested one-electron and direct two-electron elements areSince $H_1=h\otimes I$,Likewise,The integrand is nonnegative and is positive on a set of nonzero measure. This is the repulsion cost of double occupancy; the coincidence singularity is integrable for ordinary smooth localized orbitals.

One electron on each center. Suppress the normalized singlet spin factor and use $|\psi ⟩=C$. For each of the spatial products $ab$ and $ba$, the one-body sum has expectation $2E_0$. The cross terms vanish because a one-body hop leaves an orthogonal spectator orbital. The interaction contribution is the direct integral plus the exchange integral, both neglected in the model. ConsequentlyThis state has no same-center double occupancy, but remains an exchange-symmetric spatial superposition rather than assigning a distinguishable electron permanently to each proton.

Two electrons in the bonding orbital. The normalized one-electron bonding orbital is $g=(a+b)/\sqrt{2}$. The trial spatial state is $|\varphi ⟩=|gg⟩$, whose one-body expectation isExpanding the product gives $|\varphi ⟩=(|aa⟩+|ab⟩+|ba⟩+|bb⟩)/2$. The total probability of double occupancy is therefore $1/2$. Keeping only on-site Coulomb repulsion gives the intended interaction truncation,and the interaction expectation is $U/2$. HenceThis truncation must be stated. In particular, mixed Coulomb elements such as $X=⟨aa|H_{1,2}|ab⟩$ do not vanish merely by orbital orthogonality or inversion symmetry. The printed list does not explicitly name them. If they are retained and are equal and real by site symmetry, their contribution to $E_{\varphi }$ is $2X$; the requested expression uses their neglect along with the other off-site interaction terms. No numerical value of $X$ is supplied or invented.

Energy comparison and the bond. Put $x=|t|/U$ with $U>0$ and $t=-|t|$. ThenThe straight line starts at $1/2$ and crosses zero at $x=1/4$. The purely bonding-orbital trial state is better for $x>1/4$, while the singly occupied covalent trial state is better for $x<1/4$.

For localized orbitals of approximately fixed atomic size, the hopping magnitude $|t|$ decreases rapidly, typically exponentially times a polynomial, as the proton separation grows. The on-site integral $U$ remains of atomic Coulomb scale and tends to the isolated-orbital value; it does not decay as the intersite repulsion $1/R$. If the orbitals themselves are allowed to relax, $U$ also changes, so no universal monotonic law for $U(R)$ follows from the data.

At small $|t|/U$, electron correlation suppresses the ionic configurations $aa$ and $bb$, favoring a singlet with roughly one electron on each center. A single doubly occupied bonding orbital has an artificial finite ionic weight even as the centers separate. At large $|t|/U$, delocalization gains enough kinetic energy that the bonding-orbital description becomes useful and charge fluctuations are substantial.

The crossing of two restricted trial energies is not a sharp transition of the actual two-site ground state. To see this without adding empirical data, include both $C$ and the symmetric ionic state $I_{+}=(|aa⟩+|bb⟩)/\sqrt{2}$. This two-site Hubbard dimer reduces toIts ground state is $\cos \theta C+\sin \theta I_{+}$, with $\tan 2\theta =4|t|/U$ and $0\le \theta \le \pi /4$. The ionic probability isIt changes smoothly from zero to $1/2$. At large separation, the electronic lowering is $-4t^2/U$ to leading order and vanishes as hopping vanishes. An absolute molecular binding curve would additionally require $E_0(R)$, the nuclear repulsion and the neglected intersite terms; this simplified comparison alone does not determine an equilibrium bond length.

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.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

For hydrogen, the leading characteristic scales areOrder-one spin and nuclear magnetic factors, and state-dependent factors, are suppressed here. Gross atomic structure is Coulomb binding. Fine structure includes relativistic kinetic and spin-orbit effects, with Thomas precession affecting the spin-orbit coefficient. Hyperfine structure couples electronic motion and spin to the nuclear magnetic moment, whose inverse nuclear mass makes it much weaker. The contact spin interaction can remain in an s state where orbital spin-orbit coupling vanishes.

For positronium, the reduced mass is $m_e/2$. Its gross levels and separation scale areThus the mean ground-state electron–Positron separation is twice the infinite-proton-mass hydrogen value: both means have the same factor $3/2$ multiplying their respective Bohr radii. Retaining proton recoil makes the ratio $2m_p/(m_p+m_e)$, very slightly below two. Separation is not the distance of either positronium constituent from its centre of mass.

Interpret the spin operators in units of $\hslash$, as required by the printed Zeeman term, so $A$ has units of energy. Two spin halves give four eigenstates. At zero field, total spin $F=0,1$ and projection $m_F$ label the singlet and triplet. At high field relative to the hyperfine scale, the individual projections $m_{s,e},m_{s,p}=\pm 1/2$ become approximately good quantum numbers. Total projection remains exact at every field, but $F^2$ does not for the two mixed states.

Choose the quantization axis along $\mathbf{B}$. The aligned-spin states $|\uparrow \uparrow ⟩$ and $|\downarrow \downarrow ⟩$ both have $E=A/4$ because their Zeeman shifts cancel. In the other product-state sector,For physical positronium $A>0$. At $B=0$ there is a triplet at $A/4$ and a singlet at $-3A/4$. At large positive field the mixed states approach $|\uparrow \downarrow ⟩$ and $|\downarrow \uparrow ⟩$, with energies $-A/4+2{\mu }_{B}B$ and $-A/4-2{\mu }_{B}B$ respectively, while the aligned pair remains degenerate at $A/4$.

The dimensionless mixing parameter is $x=4{\mu }_{B}B/A$. A crossover estimate is $x\sim 1$, rather than a sharp transition. The scale $A\sim {\alpha }^4{m}_e{c}^2$ already gives a field of a few tesla. Using the physical hyperfine interval $A/h\simeq 203.39\mathrm{GHz}$ givesThe numerical hyperfine interval is external physical input, not a number supplied by the symbolic Hamiltonian.^[ref] The high-field limit here still presumes that orbital distortion is negligible compared with Coulomb binding.

An appropriately positioned general base helps remove the proton from the serine residue hydroxyl, increasing its nucleophilicity. A lysine residue, such as Lys73 in the class A beta-lactamase catalytic network, is a plausible participant; its protonation state depends on its active-site environment and interactions with other residues. A glutamate residue, notably Glu166, also participates in proton transfer and activation of the water needed for deacylation. Efficient catalysis requires proton-transfer chemistry as well as positioning the attacking oxygen. The precise division of roles during acylation depends on the enzyme state, so the drawing alone cannot assign a complete mechanism to one residue.^[ref]

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.