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Hydrogen atom spectral lines terminating at principal quantum number $n=1$; see the Bohr model for the elementary energy difference calculation.

A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

Molecular orbital theory describes the electronic arrangement of a molecule using molecular orbitals: one-electron spatial wavefunctions extending over the combined atomic nuclei. Filling bonding molecular orbitals can stabilize a chemical bond; filling antibonding molecular orbitals opposes that stabilization. A molecular orbital is a building block of a many-electron description, rather than the entire many-electron wavefunction. See Robert W. Field, MIT 5.61, lecture 24, pp. 1–4: molecular orbitals and the LCAO approximation.

First make the interference visible with an exactly normalized one-dimensional toy. Replace the three-dimensional $1s$ atomic orbitals of the hydrogen atom by two Gaussian functions centred at $x=\pm a$, with dimensionless $x$ and $a>0$:The normalized sum and difference are ${\psi }_{\pm }=({\varphi }_A\pm {\varphi }_B)/\sqrt{2(1\pm S)}$, following the same linear combination of atomic orbitals algebra. At $a=1$, the orbital overlap is $S=e^{-1}\approx 0.367879$. At the midpoint, ${\varphi }_A(0)={\varphi }_B(0)\approx 0.455581$, givingBoth probability densities have total integral one: interference redistributes probability, rather than creating an extra electron. These are toy wavefunctions, not actual hydrogen atom orbitals or predictions of molecular hydrogen energies.

Now fill the two $1s$-derived molecular orbitals in the minimal model. The lower bonding molecular orbital is labelled ${\sigma }_g$; the upper antibonding molecular orbital is ${\sigma }_u^{\ast }$. The Pauli exclusion principle permits at most two electrons in each spatial molecular orbital, with opposite spin projections along a chosen axis. For the lowest electron configuration in this model, fill ${\sigma }_g$ first. The subscripts $g$ and $u$ denote even and odd behaviour under inversion through the midpoint; the star marks an antibonding molecular orbital. See Flowers et al., Chemistry 2e, 8.4: occupancy and bond order and Field, MIT 5.61, lecture 25, pp. 1–3: bonding, antibonding, and symmetry labels.

With $N_b$ electrons in bonding molecular orbitals and $N_a$ in antibonding molecular orbitals, the model's bond order is $b=(N_b-N_a)/2$. The three elementary counts are:

The zero bond order says this minimal model gives no net covalent bond for ${\mathrm{He}}_2$. It does not rule out every bound pair of helium chemical atoms: a very weak Van der Waals force can bind a neutral helium dimer, which has been observed experimentally. See Zeller et al., Imaging the He2 quantum halo state using a free electron laser, PNAS 113, 14651–14655 (2016), p. 14651: van der Waals binding and Luo et al., The weakest bond: Experimental observation of helium dimer, Journal of Chemical Physics 98, 3564–3567 (1993). The bond order count does not include that weak binding mechanism.

The useful surprise is that sharing alone does not determine a chemical bond: the relative quantum phase and the occupancy matter together. The same two shapes explain why adding an electron to a bonding molecular orbital increases the model's bond order, while adding one to an antibonding molecular orbital decreases it. Molecular orbital theory also accounts for the unpaired electrons and paramagnetism of molecular oxygen, which its all-paired Lewis structure misses. See Chemistry 2e, 8.4: the molecular orbital description of oxygen.

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.

The printed magnetic dipole moment uses dimensionless angular momentum operators: their physical counterparts are $\hslash \mathbf{L}$ and $\hslash \mathbf{S}$. Choose the magnetic field along $z$. Then the Zeeman energy is $H_Z={\mu }_{B}B(L_z+g_s{S}_z)$, whose characteristic size is ${\mu }_{B}B$. In LS coupling, treating this as a time-independent perturbation theory requires it to be much smaller than the relevant fine structure separation, rather than merely smaller than the Coulomb interaction energy.

For a concrete hydrogen atom estimate take the $2p_{1/2},2p_{3/2}$ pair. The hydrogen fine-structure correction iswhere ${\alpha }_{\mathrm{fs}}$ is the fine-structure constant, distinguished from the later potential coefficient. ThusConsequently a suitable weak-magnetic field condition for this pair isThere is no single threshold for every hydrogen energy level: the relevant fine structure separations decrease with $n$. The $1s$ energy level has no same-$n$ spin-orbit interaction partner, so it cannot itself define this comparison. This calculation neglects hyperfine structure; resolving that smaller scale requires an additional weak-magnetic field condition.

Write $k=e^2/(4\pi {ϵ}_0)$ and retain the nuclear Coulomb potential. In spherical coordinates, the total potential energy iswhen $\beta =-\alpha /2$; $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. Treat the added term using quadratic anisotropic perturbation of hydrogen and degenerate perturbation theory in the $n=2$ hydrogen energy eigenbasis. It preserves parity and $m$, so it cannot mix $2s$ with $2p$, or distinct $2p$ magnetic sublevels.

For a hydrogen atom, the hydrogen radial second moment givesThe angular expectation values of ${\cos }^2\theta$ are $1/3$ for an $s$ orbital, $3/5$ for $p_0$, and $1/5$ for $p_{\pm 1}$, obtained by integrating the squared spherical harmonics. Therefore the five lowest spatial quantum states, counting statistical weight, have first-order energy levelsEach spatial quantum state has two electron spin states when spin-orbit coupling and the magnetic field are omitted. Thus these are five spatial states but only three distinct first-order energies; five distinct energies would be an incorrect reading. The accidental $2s,2p_{\pm 1}$ equality can be lifted at higher order. Quantitatively the added shifts must also be small compared with neighboring principal-shell gaps; $\alpha \ll k/a_0^3$ is an asymptotic condition, not a guarantee for arbitrary large $n$.

For general $\beta$, the $1s$ expectation value isIt does not cancel every order. At $\beta =-2\alpha$, $V_{\mathrm{ext}}=-2\alpha r^2{P}_2(\cos \theta )$ couples $1s$ to $d$ orbitals. For example its angular matrix element with $Y_{20}$ is $1/\sqrt{5}$, and the $3d$ radial matrix element is nonzero. The second-order nondegenerate perturbation theory sum has negative denominators, so this is a concrete nonzero higher-order contribution.

There is also a global limitation: for either negative $\beta$ above, $V(0,0,z)=-k/|z|+\beta z^2\to -\infty$. A localized wavepacket translated far along $z$ has fixed kinetic energy but arbitrarily negative potential energy, proving there is no true lowest energy level. The results describe local perturbative resonance poles, or a potential completed to a confining form far from the atomic nucleus. No globally exact ground-state cancellation exists for the potential as printed.

In atomic units, $\hslash =m_e=e=4\pi {\epsilon }_0=1$, and the hydrogen atom hasThe allowed quantum numbers areHere the fixed electron spin quantum number is $s=1/2$. The energy eigenvalues are in hartrees; restoring units gives ${\hslash }^{2}l(l+1)$ and $\hslash m$ for the orbital operators and the corresponding factors for spin. The spin-independent hydrogen Hamiltonian does not split the two spin projections.

A substitutional phosphorus atom supplies one more valence electron than the silicon atom it replaces. In the shallow hydrogenic donor state model, the remaining positive impurity attracts that extra electron through the screened Coulomb potentialRescaling the hydrogen atom binding energy by the effective mass and dielectric screening gives the ionization energyThe donor level lies at $E_c-E_D^{\mathrm{bind}}$; its binding magnitude is positive. The associated effective Bohr radius of a hydrogenic donor is $a_B^{\ast }={ϵ}_r(m_e/m_e^{\ast })a_0\simeq 2.38\mathrm{nm}$, consistent with a shallow state extended over several lattice spacings.

At $1000\mathrm{K}$, $k_{B}T=0.08617\mathrm{eV}$, approximately $3.3$ times this binding energy. Dilute donors are consequently expected to be largely ionized, rather than frozen out. The actual ionized fraction also depends on the chemical potential, available conduction states and donor charge-state degeneracy; comparing $k_{B}T$ to the binding energy alone is not a normalized ionization probability. At sufficiently high temperature, intrinsic electron-hole production can dominate the total carriers even though donors are ionized. Band anisotropy and short-range impurity corrections limit the numerical hydrogenic estimate.

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 $p+e\rightleftharpoons H+\gamma$, chemical equilibrium and the vanishing photon chemical potential imply ${\mu }_H={\mu }_p+{\mu }_e$. Taking the ratio of the three nonrelativistic Maxwell--Boltzmann number densities cancels these chemical potentials:The ground-state hydrogen atom has four combined electron and proton spin states if the small hyperfine splitting is ignored, whereas $g_p=g_e=2$. Hence $g_H/(g_p{g}_e)=1$. With Hydrogen binding energy $B=M_p+M_e-M_H$ and $M_H\simeq M_p$, the Saha ionization equation givesThe approximation in the prefactor neglects relative corrections of order $(M_e+B)/M_p$; the Hydrogen binding energy must be retained in the exponential.

Write $n_b=n_p+n_H$ for the cosmological baryon density in this hydrogen-only model, and $x_e=n_p/n_b$ for the hydrogen ionization fraction. By charge neutrality, $n_e=n_p$, so the Saha ionization equation becomesThe baryon-to-photon ratio is $\eta =n_b/n_{\gamma }$, and the equilibrium photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. At an order-one hydrogen ionization fraction, neglecting factors of order one therefore givesReplacing $T$ inside the slowly varying logarithm by $B$ givesWith $B=13.6\mathrm{eV}$, $M_e=5\times 10^5\mathrm{eV}$ and $\eta =10^{-10}$, this is $T\simeq 0.351\mathrm{eV}\simeq 4.1\times 10^3\mathrm{K}$. Solving the preceding implicit approximation instead gives $T\simeq 0.306\mathrm{eV}\simeq 3.5\times 10^3\mathrm{K}$. Both are reasonable estimates of the recombination temperature at the accuracy requested. The small baryon abundance delays hydrogen recombination: even when $T\ll B$, the high-energy tail of the abundant photons can still ionize the much rarer hydrogen atoms.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

For the near-origin probability in a maximal-angular-momentum hydrogen state, integrate the normalized radial probability density:Throughout this small interval the exponential is $1+O(r_p/a)$ for fixed $l$. HenceUsing the normalization above,The high power of the small radius reflects the near-origin behavior $R\propto r^l$ of the hydrogen atom state.

The baryon-to-photon ratio is extremely small, of order $10^{-9}$. Thus even when the typical photon energy is below $I$, the high-energy tail of the blackbody radiation contains enough photons above the ionization energy to reionize almost every newly formed hydrogen atom. Cosmological recombination becomes effective only when this exponentially small tail is also small compared with the baryon-to-photon ratio. Consequently recombination occurs at a temperature much lower than $I/k$, consistent with $kT\simeq 0.03I$. The exact freeze-out history also depends on reaction rates and expansion; the equilibrium estimate explains the large suppression of the temperature.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrödinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

Write $Q={\dot{N}}_{\mathrm{ion}}$ for the constant hydrogen-ionizing photon production rate. Assume an initially neutral gas of hydrogen atoms, a sharp spherical ionization front, complete ionization inside it, and a fixed gas density. The electron and proton number densities inside are then both $n$. Take the recombination coefficient $\alpha$ to be constant, with the usual Case B recombination interpretation: photons from recombination directly to the ground state are absorbed locally. Neglect absorption by dust, pressure-driven gas motion, and light-travel delays.

The total rate of Case B recombination within radius $R$ isAt photoionization equilibrium, $Q=\mathcal{R}(R_S)$, giving the Strömgren radiusThe square of the density appears because recombination requires encounters between an electron and a proton.

During Ionization-front growth of a Strömgren sphere, each photon either replaces an ion lost through Case B recombination or ionizes an additional atom. There are $N_i=(4\pi /3)n{R}^3$ ions inside the ionization front, so photon counting givesThe recombination time is $t_{\mathrm{rec}}=(\alpha n{)}^{-1}$. Solving this first-order differential equation with $R(0)=0$ givesFor an initial radius $R_0$, the corresponding result is $R^3(t)=R_S^3+(R_0^3-R_S^3)e^{-t/t_{\mathrm{rec}}}$. Initially, the idealized photon-counting solution has $R\simeq [3Qt/(4\pi n){]}^{1/3}$; at late times, $R/R_S\simeq 1-\frac{1}{3}{e}^{-t/t_{\mathrm{rec}}}$. The initially divergent front speed in this approximation is a consequence of neglecting light-travel delays; the formula describes the growth once that approximation is applicable.

For a power law density, specify a reference radius $r_0$ so that $n(r)=n_0(r/r_0{)}^{-\beta }$. With an inner cutoff $r_{\mathrm{in}}>0$, the recombination integral for a power-law nebula isEquivalently, a logarithmic interval in radius contributes $d\mathcal{R}/d\ln r=4\pi \alpha n_0^2{r}_0^{2\beta }{r}^{3-2\beta }$. For $\beta <3/2$, the central integral converges as $r_{\mathrm{in}}\to 0$, and the recombination rate grows as $R^{3-2\beta }$. At $\beta =3/2$, each logarithmic interval contributes equally and the centre gives a logarithmic divergence. For $\beta >3/2$, the unmodified profile gives a power divergence at the centre, while its outer recombination integral converges. The change at $\beta =3/2$ is set by the competition between the spherical volume element and the squared density.

To regularize the centre, choose the following explicit cored power-law density profile, taking $r_0=r_c$ and $n_0$ as the core density:For $R\le r_c$, $\mathcal{R}(R)=(4\pi /3)\alpha n_0^2{R}^3$. For $R>r_c$, put $u=R/r_c$ and integrate the core and envelope separately:This rate is continuous and strictly increasing with $R$. If $\beta >3/2$, it has a finite supremum, the critical ionizing photon rate of a cored power-law nebula:For $0<Q<Q_{\mathrm{crit}}$ there is a unique finite radius at photoionization equilibrium. At $Q=Q_{\mathrm{crit}}$ the equality is approached only as $R\to \infty$; for $Q>Q_{\mathrm{crit}}$ even ionizing the whole profile cannot consume all photons through Case B recombination, so no equilibrium radius exists. For $\beta \le 3/2$, the envelope integral grows without bound and there is no finite critical rate in this infinite-extent model.

For a concrete choice, take any $n_0>0$, $r_c>0$, and $\beta =2$. Then $Q_{\mathrm{crit}}=16\pi \alpha n_0^2{r}_c^3/3$. In the envelope, defining $q=Q/(4\pi \alpha n_0^2{r}_c^3)$ gives $R=r_c/(4/3-q)$ for $1/3<q<4/3$, explicitly showing the divergence as the critical rate is approached. A different core shape changes the numerical coefficient of $Q_{\mathrm{crit}}$, while the outer-slope condition $\beta >3/2$ remains the same.

The anti-muon has the same charge as a proton, so the Coulomb potential seen by the electron is the same as in the hydrogen atom. The main difference is the source mass. Separating the centre-of-mass motion replaces $m_e$ by the Coulomb two-body reduced massSince $m_{\mu }\gg m_e$, this is close to $m_e$, though less close than the electron-proton reduced mass. The smaller reduced mass gives a slightly larger Bohr radius and energies of slightly smaller magnitude. Under the assumptions in the question, the Coulomb form, angular eigenfunctions and quantum numbers are otherwise unchanged; anti-muon decay and spin effects are being neglected.

Write $C=e^2/(4\pi {ϵ}_0)=m{c}^2{r}_0$. For a slowly evolving circular Coulomb orbit, force balance and the mechanical energy giveThe supplied Larmor formula therefore yieldsSet $dE/dt=-P$. Since $dE/dt=C\dot{r}/(2r^2)$,Thus the classical radiation collapse of a Coulomb orbit obeysThe decreasing mechanical energy means a more tightly bound orbit, not a smaller speed. The formal collapse time is $a_0^3/(4r_0^{2}c)$. The nonrelativistic, small-loss-per-orbit approximation eventually fails; the result demonstrates the classical instability rather than describing an actual hydrogen atom.

Spin-orbit coupling is a relativistic interaction between a particle's spin and its orbital angular momentum or, more generally, its motion through an electric field. A moving electron experiences a transformed magnetic field, which couples to its magnetic dipole moment. For an electron bound in a central electrostatic potential energy $U(r)$, the leading interaction, including the Thomas precession correction, isHere $\mathbf{L}$ is orbital angular momentum and $\mathbf{S}$ is spin. The factor of one-half relative to the simplest rest-field argument comes from Thomas precession of the accelerated electron frame.^[ref]

For example, in the hydrogen atom, the $2p$ states with total angular momentum $j=1/2$ and $j=3/2$ have different $\mathbf{L}\cdot \mathbf{S}$ eigenvalues. Their splitting is part of the atomic fine structure. The interaction redistributes and splits spin-dependent energies; it is not the same physical effect as the spin-independent convective magnetic force between charged filaments.

For bound hydrogen atom states, the radial wave function is regular and finite at the origin, and decays at infinity so that $4\pi {\int }_0^{\infty }{r}^2|\Psi (r){|}^{2}dr<\infty$. Equivalently the reduced radial function $w(r)=r\Psi (r)$ has $w(0)=0$ and is square integrable on the half-line. A singular $1/r$ branch is not an admissible regular Coulomb eigenfunction; its derivative energy diverges and it introduces an origin singularity absent from the Hamiltonian domain.

Write $\kappa =e^2/(4\pi {ϵ}_0)$, so $a={\hslash }^2/(m\kappa )$. Multiplication by $r$ turns the radial equation intoSince $w=a{e}^{-\lambda R}f(R)$ and $r=aR$, substitution givesChoose $\boxed{\lambda =\mu a>0}$ to remove the constant coefficient and factor out the decaying exponential. For $f={\sum }_{k\ge 0}{a}_k{R}^{k+c}$, the lowest power has coefficient $c(c-1)a_0$, so $\boxed{c(c-1)=0}$. The $c=0$ behavior would give $\Psi \sim a_0/R$, violating regularity; choose $c=1$.

Matching successive powers then yieldsA nonterminating tail eventually has one sign, and its coefficient ratio is asymptotic to $2\lambda /k$. More precisely, for any $0<\delta <\lambda$, the ratio is eventually at least $(2\lambda -\delta )/(k+1)$. Its same-sign tail consequently dominates a positive constant times the exponential series of rate $2\lambda -\delta$, up to finitely many terms. The factor $e^{-\lambda R}$ cannot then make $\Psi$ decay. Thus square integrability forces termination.

If the last nonzero term is $a_{n-1}{R}^n$, termination requires $\lambda n=1$. ThereforeFor $n=1$, $\lambda =1$, $f$ is proportional to $R$, andThe radial volume element is $4\pi r^{2}dr$, so the mean radius is the ratio of the radius-weighted norm integral to the norm integral. With $r=aR$, the common angular and normalization factors cancel:In the ground state $\boxed{⟨r⟩=3a/2}$.