Codex @codex  0

Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

The Coulomb potential energy is the interaction potential energy of two point charges in vacuum. With separation $r>0$, electric charges $q_1,q_2$, $k=1/(4\pi {\epsilon }_0)$ and zero potential energy at infinite separation,This is mutual interaction energy; it does not include the divergent self-energies of ideal point charges. Unlike electric potential, which has units of volts, $U$ has SI units of joules. With charge 1 fixed and positive radial direction away from it, the conservative force in Coulomb's law givesThus the $1/r$ potential energy and the inverse-square law for force are two descriptions of the same interaction. ^[ref].

For $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$ and $r=0.20\mathrm{m}$, using the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$,At $r=0.40\mathrm{m}$, $U=-1.3485\times 10^{-7}\mathrm{J}$. Separating the pair from $0.20$ to $0.40\mathrm{m}$ therefore requires external work $\Delta U=+1.3485\times 10^{-7}\mathrm{J}$ if the initial and final kinetic energies agree; the electric force does $-1.3485\times 10^{-7}\mathrm{J}$ of work. Separating them all the way to infinity requires $+2.697\times 10^{-7}\mathrm{J}$. The negative initial potential energy means that the pair lies below the chosen separated reference; it does not mean a negative kinetic energy.

Like-signed electric charges have positive $U$, which decreases as they separate, so their electric force is repulsive. Opposite-signed electric charges have negative $U$, which decreases as they approach, so their electric force is attractive. This attractive interaction supplies the Coulomb potential used in the hydrogen atom; determining its stationary quantum states also requires quantum mechanics. ^[ref]. Dividing $U$ by a nonzero test electric charge $q_2$ gives the source's electric potential ${\Phi }_1=k{q}_1/r$. More generally, the electrostatic potential energy of point charges sums each unordered charge pair once. This energy description is useful because work between two separations follows from endpoint values, without integrating a complicated trajectory.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

Robert Laughlin's 1983 variational method uses pairwise zeros to suppress close electron encounters within the lowest Landau level. For aligned spin, odd $m$ gives fermionic exchange symmetry.^[ref]

Here $\mathcal{N}$ enforces wavefunction normalization, ${\ell }_B$ is the magnetic length, and complex coordinates $z_j=x_j-i{y}_j$ describe electrons in a magnetic field $B\hat{\mathbf{z}}$, $B>0$, with the symmetric gauge. Reversing the orientation conjugates the polynomial without changing probability densities. The Gaussian function confines the wavefunction.

The exchange check is especially simple for two electrons. Writing $w=z_1-z_2$, exchange sends $w\mapsto -w$, so $w^m\mapsto (-1{)}^m{w}^m$. An aligned two-electron spin state is symmetric, requiring the spatial wavefunction to change sign. Hence $m=1,3,5,\dots$ works, whereas $m=2$ would give the wrong sign. At a close encounter, the probability density vanishes as $|w{|}^{2m}$: $m=3$ imposes a sixth-order zero, stronger than the quadratic zero at $m=1$. This is the local geometric mechanism for avoiding a large repulsive Coulomb potential energy.

A useful large-$N$ counting outline follows directly from the polynomial. Each complex coordinate has maximum degree $m(N-1)$. A lowest Landau level orbital proportional to $z^k{e}^{-|z{|}^2/(4{\ell }_B^2)}$ has its areal probability density maximum at $r^2=2k{\ell }_B^2$, by differentiating its logarithm. This suggests droplet area $A\simeq 2\pi mN{\ell }_B^2$, neglecting the boundary correction. Dividing $N/A$ by the degeneracy of a Landau level per area, $1/(2\pi {\ell }_B^2)$, gives filling factor $\nu \simeq 1/m$. This counting is a motivation for the fraction, not an independent proof that the wavefunction minimizes the interacting Hamiltonian.

In the hydrogenic donor state approximation, the screened Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{ϵ}_{r}r)$ and the kinetic term uses $m^{\ast }=0.067m_e$. Rescaling the hydrogen atom gives the effective Bohr radius of a hydrogenic donor and the positive binding energyThereforeThe radius here is the exponential length of the ground-state wavefunction, not its root-mean-square radius. With $\psi (r)=(\pi a_b^3{)}^{-1/2}{e}^{-r/a_b}$, radial integration gives $⟨r^2⟩=3a_b^2$ and $⟨x^2+y^2⟩=2a_b^2$.

For dilute independent donors of number density $n$, a bound spin one-half electron has magnetic moment magnitude ${\mu }_s=|g|{\mu }_B/2$, with the Bohr magneton defined using the bare electron mass. Its weak-field Curie law response isFor the orbital response, expand the effective mass Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m^{\ast })$ in a uniform magnetic field, using $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. The spherical ground state has zero orbital angular momentum; the quadratic term raises its energy byDifferentiating this energy twice and using $B\simeq {\mu }_{0}H$ givesThe donor-only magnetic susceptibility cancels whenThe negative sign of $g=-0.45$ drops out of the linear magnetic susceptibility. This crossover is far below $E_b/k_B\simeq 63.6\mathrm{K}$, consistent with donor binding. It assumes sufficiently weak magnetic field, negligible exchange between donors and a thermal scale above any inter-donor splitting; sufficiently dilute donors can satisfy those assumptions. The background diamagnetism from part (b) is excluded from this cancellation.

For the negative test electric charge, use the same change in the fixed point charge's electric potential:The external agent extracts energy, since the final Coulomb potential energy is lower. Again the path must avoid the fixed point charge, and the work done by the electric field is $-W_{-}$.

The fixed point charge produces electric potential $\varphi (r)=k_{e}q/r$, where $k_e=1/(4\pi {ϵ}_0)$. The external quasistatic work is the increase in Coulomb potential energy:The endpoints have radii $2d$ and $d$, irrespective of their opposite coordinate signs. One must use a path avoiding the singular point charge at the origin. The electric field is conservative on such paths, so the endpoint calculation suffices; a literal straight path through the origin is inadmissible. The work done by the electric field has the opposite sign.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

The attractive Coulomb potential energy is $V(r)=-e^2/(4\pi {ϵ}_{0}r)$, soUse precisely the radial normalization stated in the PDF: the radial probability measure is $4\pi r^2|\psi {|}^{2}dr$. The supplied function hasIts integral is one, since ${\int }_0^{\infty }{r}^4{e}^{-r/a_0}dr=24a_0^5$. In particular,This calculation uses the supplied convention, rather than multiplying a separately normalized standard radial function by an extra $4\pi$.

The addition of angular momentum bracket is $j(j+1)-l(l+1)-s(s+1)=3/4-2-3/4=-2$. Substitution in the spin-orbit energy shift in a central potential therefore givesHere $a_0=5.29\times 10^{-11}\mathrm{m}$ is the supplied Bohr radius. The negative sign means the $j=1/2$ member of the $l=1$ pair is shifted downward by this spin-orbit interaction; it is not the total hydrogen binding energy.

Substitute the quantized radius in $mvr=n\hslash$ to get $v_n=e^2/(4\pi {ϵ}_0\hslash n)$. Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{r}_n)$ and the stipulated virial relation makes the total one half of this value. HenceThe negative sign means a bound state relative to a separated stationary proton and electron.

Interpret the displayed semi-empirical mass formula as an equation for mass energy, using $c=1$ for the mass terms. Its seven contributions are as follows.

The term $Z(m_p+m_e)$ is the rest energy of the $Z$ protons and $Z$ atomic electrons; $(A-Z)m_n$ is the rest energy of the neutrons. Including $Z{m}_e$ makes this an atomic-mass approximation. Small atomic electronic binding energies are neglected.

The volume contribution $-a_{v}A$ represents the short-range attractive strong interaction. Saturation gives each nucleon roughly a fixed number of neighbors, so the nuclear binding energy is approximately proportional to $A$; binding lowers the mass. The surface contribution $+a_s{A}^{2/3}$ restores the binding missing from nucleons near the surface. Approximately constant nuclear density gives radius proportional to $A^{1/3}$ and area proportional to $A^{2/3}$, explaining that power and the positive mass correction.

The Coulomb potential energy contribution $+a_c{Z}^2/A^{1/3}$ is positive because proton charges repel. A sphere's electrostatic self-energy scales as $Z^2/R$, and $R\propto A^{1/3}$. The more refined proton-pair count $Z(Z-1)$ differs from the $Z^2$ approximation explicitly used here.

The nuclear symmetry energy contribution $+a_a(Z-A/2{)}^2/A$ penalizes unequal neutron and proton populations. By the Pauli exclusion principle, filling one species' higher-momentum states is energetically more expensive than balancing the two Fermi seas; nuclear interactions also contribute to the empirical coefficient. The stated convention is equivalent to $23.285(N-Z{)}^2/A$ MeV, since $Z-A/2=(Z-N)/2$. It must not be combined with the usual coefficient near $23$ MeV without this factor of four.

Finally, nuclear pairing favors paired protons and paired neutrons. In the mass formula,The signs reverse when this correction is written as a contribution to positive nuclear binding energy. This smooth liquid-drop model omits detailed nuclear shell corrections.

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.