Hydrogen atom

For a negative Coulomb energy, remove the exponential decay and regular radial power to obtain a confluent-hypergeometric series. Its recurrence terminates only when $n=q+\ell +1$, with $q$ a nonnegative integer. Regularity at the origin and normalizability at infinity then give the hydrogenic energy levels. Without those physical boundary conditions the radial differential equation has other solutions.

For a nonnegative integer $\ell$, this Coulomb potential energy eigenstate has principal quantum number $\ell +1$ and maximal magnetic quantum number. Put $b=(\ell +1)a$. Including the spherical volume element, its radial probability density is a gamma distribution proportional to $r^{2\ell +2}{e}^{-2r/b}$; its polar density is proportional to ${\sin }^{2\ell +1}\theta$, while its azimuth is uniform. Radial and polar variables are independent. Their moments satisfyThe radial identities follow by successive integration by parts; the angular identity follows from ${\int }_0^{\pi }{\sin }^{j+2}\theta d\theta =(j+1){\int }_0^{\pi }{\sin }^j\theta d\theta /(j+2)$.

Set $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality and Markov's inequality give $\mathbb{P}(|r/\mu -1|>h)\le h$ and $\mathbb{P}(|\cos \theta |>h)\le h$. By the union bound, outside an event of probability at most $2h$, the point lies at distance $O(\mu h)$ from the equatorial circle of radius $\mu$. The quantum state therefore concentrates near a classical circular orbit in relative units as $\ell$ grows. The absolute orbit radius and absolute transverse width both grow; this statement does not assert localization in a fixed narrow tube.

Writing $\rho =r/a_0$, the normalized radial wavefunctions are $R_{20}=(2-\rho )e^{-\rho /2}/(2\sqrt{2}{a}_0^{3/2})$ and $R_{21}=\rho e^{-\rho /2}/(\sqrt{24}{a}_0^{3/2})$. Radial normalization uses $\int r^2|R{|}^{2}dr=1$. The first function has one positive radial node at $2a_0$; the second has none. Their behavior at the origin differs from that of the reduced radial function $rR$ and from the radial probability density $r^2|R{|}^2$.

An axially symmetric quadratic potential energy perturbation preserves parity and the $z$ component of orbital angular momentum. For $1s,2s,2p_0,2p_{\pm 1}$, its first-order energy shifts are respectively $(2\alpha +\beta )a_0^2$, $14(2\alpha +\beta )a_0^2$, $(12\alpha +18\beta )a_0^2$, and $(24\alpha +6\beta )a_0^2$. They follow by multiplying the hydrogen radial second moment by the angular average of $\alpha +(\beta -\alpha ){\cos }^2\theta$, whose ${\cos }^2\theta$ averages are $1/3,1/3,3/5,1/5$. Negative $\beta$ makes the globally extended potential unbounded below; the shifts then describe local perturbative resonance poles or a potential stabilized at large distance.

For a nonrelativistic point-nucleus hydrogen atom, $⟨r^2{⟩}_{nl}=a_0^2{n}^2[5n^2+1-3l(l+1)]/2$. Integrating the normalized radial wavefunction squared with measure $r^{4}dr$ gives $3a_0^2,42a_0^2,30a_0^2$ for $1s,2s,2p$. For a hydrogen-like ion, replace $a_0$ by $a_0/Z$ in the infinite-nuclear-mass model. These moments determine first-order shifts under quadratic potential energy perturbations.

The nonrelativistic hydrogen ground energy is $-e^2/(8\pi {ϵ}_0{a}_0)$. Requiring nonnegative classical kinetic energy restricts radius to $r\le 2a_0$. Integrating the quantum ground-state radial probability density beyond that radius instead gives $13e^{-4}\simeq 0.238$. A bound-state tail can therefore have substantial integrated probability outside the classically allowed region without violating energy conservation.

For the point-Coulomb ground state, radial probability density is $4r^2{e}^{-2r/a_0}/a_0^3$. Integrating to $r_p$ gives $1-e^{-2u}(1+2u+2u^2)$, with $u=r_p/a_0$, whose small-$u$ expansion is the displayed result. A finite proton charge distribution changes the interior potential, and a proton radius is not literally a sharp material boundary; the expression is an estimate of a small overlap region.

For the nonrelativistic point-Coulomb hydrogen Hamiltonian, the lowest radial state at fixed orbital angular momentum $\ell$ has principal quantum number $n=\ell +1$ and no positive-radius wavefunction nodes. Substituting the displayed radial shape cancels the centrifugal inverse-square term and fixes $a_0=4\pi {ϵ}_0{\hslash }^2/(m{e}^2)$. With $b=a_0(\ell +1)$ and unit-normalized spherical harmonic, $A=(2/b{)}^{\ell +3/2}/\sqrt{(2\ell +2)!}$. Excited radial states require additional polynomial factors.

With Bohr radius $a$, define $A_l=a[i{p}_r/\hslash -(l+1)/r+1/((l+1)a)]/\sqrt{2}$. For regular radial bound states the radial inner product and compatible domains give the displayed factorization. Expanding the first-order factors produces the centrifugal term $l(l+1){\hslash }^2/(2m{r}^2)$ and Coulomb term $-{\hslash }^2/(mar)$. factorization converts a second-order spectral problem into a positivity bound and first-order terminating-state equation.

The Coulomb radial factors satisfy $[A_l,A_l^{†}]=a^2(l+1)/r^2=m{a}^2(H_{l+1}-H_l)/{\hslash }^2$. Commuting $A_l$ through its factorized Hamiltonian gives the displayed intertwining relation. A nonzero transformed radial eigenfunction has the same energy in the next angular-momentum sector; this changes the radial sector, rather than being the usual magnetic-quantum-number raising operator within a fixed $l$ multiplet.

For a normalized bound eigenstate of a factorized Coulomb radial Hamiltonian, the displayed squared norm must be nonnegative. At fixed negative $E$, increasing $l$ indefinitely would make it negative. The intertwining chain must therefore terminate with zero norm, giving $E=-{\hslash }^2/[2m{a}^2(L+1{)}^2]$. The positivity contradiction does not apply to the positive-energy continuum; domain preservation and normalizability are necessary assumptions.

In a fixed attractive Coulomb potential, neglecting electron-electron interactions leaves hydrogenic one-particle levels with spin-inclusive degeneracy $2n^2$. The Pauli exclusion principle fills these levels in increasing energy. A partially filled degenerate shell gives many states at the same total energy; the first higher energy requires promotion to a different principal shell. This model does not describe the screening and configuration-dependent energies of a real many-electron atom.

Model a proton as a uniformly charged thin spherical shell of radius $b$. Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem gives an electron potential energy $-C/b$ inside the shell and $-C/r$ outside. Relative to a point Coulomb potential, the perturbation is consequently $C(1/r-1/b)$ for $r<b$ and zero otherwise.

For the hydrogen atom ground state, time-independent perturbation theory gives the first-order energy correctionSince $E_{1s}=-C/(2a_0)$, the signed fractional change is $\Delta E/E_{1s}=-4b^2/(3a_0^2)+O((b/a_0{)}^3)$. The correction $\Delta E$ is positive: distributing the positive charge weakens the attraction near the origin and reduces the magnitude of the binding energy. Here $a_0$ is the Bohr radius and $b\ll a_0$ is the small-radius approximation.

A family of hydrogen emission lines terminating at a fixed lower principal quantum number $n_f$. In the nonrelativistic Bohr model, their inverse wavelengths are $1/\lambda =R_H(1/n_f^2-1/n_i^2)$ for integers $n_i>n_f$. Holding $n_f$ fixed and increasing $n_i$ gives lines approaching the series limit $1/\lambda =R_H/n_f^2$. Reduced mass shifts the Rydberg constant, while additional interactions produce corrections to this gross spectral pattern.

The Brackett series consists of hydrogen spectral lines with final principal quantum number four and initial principal quantum number greater than four. The Rydberg formula gives $1/\lambda =R_H(1/16-1/n^2)$. Its infrared wavelengths make these recombination lines less susceptible to interstellar extinction than optical Balmer series lines; quantitative use requires the appropriate effective recombination coefficient.

For spherical charge density ${\rho }_e=-e{e}^{-2r/a_0}/(\pi a_0^3)$, integrate $4\pi s^2{\rho }_e(s)$ from zero to $r$ to obtain the displayed enclosed charge. Gauss's law then gives $E_e=Q_e(r)/(4\pi {ϵ}_0{r}^2)$, directed inward. Near the origin $E_e\simeq -er/(3\pi {ϵ}_0{a}_0^3)$; at large radius it approaches the field of charge $-e$.

Adding a point nucleus of charge $+e$ leaves the unperturbed total fieldIt is exponentially small rather than identically zero, because the electron cloud has no compact boundary. This frozen-density field alone does not describe the dominant long-range response to another charge: charge-induced dipole attraction requires allowing the atom to polarize.

For the elementary hydrogen atom derivation, see the Bohr model.

The inverse-length constant in the infinite-nuclear-mass hydrogenic spectrum, $R_{\infty }=m_e{e}^4/[8{ϵ}_0^2{h}^{3}c]$. A nucleus of finite mass replaces it by $R_M=R_{\infty }\mu /m_e$, where $\mu$ is the electron-nucleus reduced mass. Some spectroscopic fits call $Z^2{R}_M$ an effective Rydberg constant; the charge-inclusive and reduced-mass-only conventions must be distinguished.

Hydrogen atom spectral lines terminating at principal quantum number $n=2$; see the Bohr model for the elementary wavelength calculation.

The hydrogen Balmer emission line from the transition between principal quantum numbers three and two. Recombination-line emission from gas photoionized by short-lived massive stars makes it a star-formation tracer, subject to dust attenuation, escape of ionizing photons and contamination by other excitation sources.

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

The Lyman limit is the short-wavelength limit of transitions to the hydrogen atom ground state. Its photon energy, approximately $13.6\mathrm{eV}$, is the ground-state ionization energy.

The Lyman-alpha line is the hydrogen $2p\leftrightarrow 1s$ electric-dipole transition, with photon energy about $10.2\mathrm{eV}$ and vacuum wavelength about $1216$ angstroms. Its emission cools collisionally excited gas, while absorption traces neutral hydrogen along a background-source sightline.

A neutral hydrogen atom absorbs a photon near the Lyman-alpha resonance by excitation from $1s$ to $2p$. Absorption lines in a quasar spectrum can therefore locate hydrogen along the sightline by their redshift, subject to peculiar-velocity and line-profile effects.

A damped Lyman-alpha system has a sufficiently high neutral-hydrogen column density that the Lorentzian wings of its Voigt profile are conspicuous. These wings come from the finite lifetime of the excited atomic state, not from unusually high gas temperature. Damped systems usually trace substantial neutral reservoirs associated with galaxies.

A Lyman limit system has approximately unit or greater optical depth at the neutral-hydrogen Lyman limit near $91.2\mathrm{nm}$. The threshold follows from ${\tau }_{912}=N_{\mathrm{HI}}{\sigma }_{912}$ with ${\sigma }_{912}\simeq 6.3\times 10^{-18}{\mathrm{cm}}^2$. Its strong continuum absorption distinguishes it from ordinary weak Lyman-alpha forest lines. Classification conventions often separate damped Lyman-alpha systems as the highest-column category.

The neutral-hydrogen column-density distribution counts Lyman-alpha absorption systems per unit column density and redshift. A count in logarithmic column bins is $N\ln (10)f(N,z)$. At low forest columns an approximate power law $f\propto N^{-\beta }$ with $\beta$ near $1.5$ is useful, but the slope and normalization depend on column range and epoch. Lyman limit systems are optically thick at the hydrogen photoionization edge; damped Lyman-alpha systems have strong radiative damping wings. A distribution per absorption distance instead removes the geometrical path factor.

A neutral-hydrogen column density $N$ traces total hydrogen column $N/x_{\mathrm{HI}}$ only after specifying its neutral fraction $x_{\mathrm{HI}}$. Integrating the latter against the distribution per absorption distance gives the comoving gas mass density in the displayed expression. The factor ${\mu }_H$ includes helium if desired. In photoionization equilibrium for optically thin, almost fully ionized hydrogen, $x_{\mathrm{HI}}\simeq \alpha (T)n_H/\Gamma$, so gas density, temperature, photoionization rate and absorber size are required. An observed distribution alone does not fix the total baryon inventory.

Let $f_X={\partial }^2\mathcal{N}/(\partial N\partial X)$ count neutral hydrogen absorbers per column density and absorption distance. The summed column per unit proper light-path length is the mean neutral number density. Since $d\ell =cdz/[(1+z)H(z)]$ and $dX/dz=H_0(1+z{)}^2/H(z)$, that density is $(H_0/c)(1+z{)}^3\int N{f}_{X}dN$. Dividing the proper mass density by $(1+z{)}^3$ gives the comoving density in the displayed formula. The denominator is today's critical density; including neutral helium requires a stated mass correction. For a distribution per redshift, replace $H_0{f}_X$ by $H(z)f_z/(1+z{)}^2$.

For a population with comoving number density $n_c$ and proper interception area ${\sigma }_p$, the absorption incidence is the proper density multiplied by cross-section and proper light-path length $cdz/[H(z)(1+z)]$. The displayed relation assumes one counted absorption system per interception. Completeness, covering fraction, evolution and the distinction between observed and rest-frame wavelength separation must be stated in an inference from a spectrum.

Absorption distance is a dimensionless path variable chosen so a non-evolving comoving absorber population with fixed proper cross-section has constant $d\mathcal{N}/dX$. It absorbs the geometrical factor in absorber incidence and comoving number density. A distribution per $dNdX$ and one per $dNdz$ therefore differ by $dX/dz$.

A circular, opaque, geometrically thin disk has projected area $\pi R^2|\cos i|$. For isotropically oriented normals, $|\cos i|$ is uniform on $[0,1]$, so the mean interception area is half the face-on area. A nonunit covering fraction multiplies this cross-section. Finite thickness, orientation-dependent absorption thresholds and incomplete coverage can change it; this is an average over randomly placed disks, not the inclination distribution conditional on an observed interception.

Radiative decay from the hydrogen $2p$ level to $1s$ produces a Lyman-alpha photon. In an optically thin collisionally excited gas, escape of this photon removes the excitation energy deposited by an electron collision. The excitation threshold is responsible for the rapid temperature sensitivity of this cooling channel.

The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

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.

For emission take integers $n_i>n_f\ge 1$. Applying conservation of energy to the model's internal energy difference, while neglecting the much smaller whole-atom photon recoil, givesThus Johannes Robert Rydberg's Rydberg formula is a difference of two energy levels expressed as a spectroscopic wavenumber. The same constants determine every line; there is no new parameter for each electronic transition. With the chosen inputs, $R_H=1.096775834\times 10^7{\mathrm{m}}^{-1}$. All wavelengths below are calculated in vacuum, not air, and are rounded model predictions rather than historical measurements.

The conventional names commemorate Theodore Lyman and Johann Jakob Balmer. Fixing $n_f=1$ produces the ultraviolet Lyman series, beginning with the Lyman-alpha line; fixing $n_f=2$ produces the Balmer series. As $n_i\to \infty$, the wavelength approaches $n_f^2/R_H$: $91.1763\mathrm{nm}$ at the Lyman limit and $364.705\mathrm{nm}$ for the Balmer series. Crowding follows directly from $E_{n+1}-E_n\sim 2B_H/n^3$. Discreteness therefore coexists with a series limit.

Correct line positions do not give wave intensities, lifetimes or electric-dipole selection rules for hydrogen. Those require a theory of the wavefunctions and their coupling to electromagnetic radiation. The circular model also omits spin, relativistic corrections and the Lamb shift. Its successful gross energy levels do not make its assumed orbits modern atomic orbitals.

In On the Constitution of Atoms and Molecules, Part I (1913), pp. 8–9, Niels Bohr interprets existing hydrogen discharge and astronomical spectra. Electrical discharge in a vacuum tube removes electrons; recombination supplies spectral lines. This is a theoretical interpretation, without a reported spectrograph of his own.^[ref]

An actual optical obstacle appears in Theodore Lyman's measurements: hydrogen discharge light entered an evacuated spectrograph directly through a slit, avoiding a transmitting window. A concave reflection diffraction grating dispersed the light onto photographic film. Pumping preserved the vacuum; a second slit supplied a known spectral displacement for wavelength calibration. Gas contamination and vacuum leaks limited the measurement. This apparatus resolved ultraviolet spectral lines; it did not image orbits.^[ref]

For a separate original calibration exercise, suppose an idealized spectrograph returns synthetic vacuum line positions $x_n={\lambda }_{n\to 2}+b$ for $n=3,4,5,6$, with chosen additive offset $b=0.200\mathrm{nm}$. Assume its dispersion scale is already known and only this offset is uncertain. Compare the finite-mass prediction with the fixed-nucleus prediction ${\lambda }_n^{(\infty )}=(\mu /m_e){\lambda }_{n\to 2}$. Calibrate each model using only the $3\to 2$ line: $\hat{b}=x_3-{\lambda }_3^{\mathrm{model}}$. Then hold that offset fixed and test the other lines with residualsFor the finite-mass model every $d_n=0$. The fixed-nucleus model has $\hat{b}=0.55733\mathrm{nm}$ and residuals $d_4=-0.09264\mathrm{nm}$, $d_5=-0.12100\mathrm{nm}$, $d_6=-0.13400\mathrm{nm}$. Algebraically $d_n=(1-\mu /m_e)({\lambda }_n-{\lambda }_3)$: one additive calibration cannot absorb a multiplicative change of the wavelengths. Allowing an unknown multiplicative dispersion factor as well would make the two models degenerate in this exercise, so independent scale calibration matters.

For a hydrogenic bound state, the principal quantum number is the positive integer $n$ that determines the unperturbed energy. Its allowed orbital quantum numbers are $\ell =0,\dots ,n-1$.

The ionization energy is the minimum energy needed to remove an electron from an isolated atom or molecule in a specified state. Ground-state hydrogen has ionization energy $13.6\mathrm{eV}$.

For two particles of masses $m_1,m_2$ interacting through a central Coulomb potential, separation of the centre of mass leaves the relative-coordinate Schrödinger equation with

For reduced mass $\mu$, the Coulomb length scale is

A no-radial-node Coulomb eigenstate has principal quantum number $n=l+1$ and radial factorIts energy is $-e^2/[8\pi {ϵ}_{0}a(l+1{)}^2]$.

For $R(r)\propto r^l{e}^{-r/[a(l+1)]}$, the radial probability density is proportional to $r^{2l+2}{e}^{-2r/[a(l+1)]}$ and

Neglecting spin, bound hydrogen eigenstates are labelled $|n,\ell ,m⟩$, where $\ell =0,\dots ,n-1$ and $m=-\ell ,\dots ,\ell$. Their unperturbed energy depends only on $n$.

Neglecting spin, the hydrogen level with principal quantum number $n$ has degeneracy

For the $z$ component of position,requires $\Delta m=0$ and $\Delta \ell =\pm 1$. The first condition follows from $[L_z,z]=0$; the second follows from the vector-operator angular-momentum relation together with odd parity.

The $n=2$ eigenspace has basis

A perturbation $g{L}^2$ preserves the hydrogen eigenbasis and shifts $|n,\ell ,m⟩$ byIn the first excited eigenspace, the $2s$ state is unshifted and the three $2p$ states receive the common shift $2g{\hslash }^2$.

The operators $X_3$ and $P_3$ commute with $L_3$, so they preserve $m$. Both are odd under parity, so they connect only states of opposite orbital parity. Within the $n=2$ eigenspace, only the pair $|2,0,0⟩$ and $|2,1,0⟩$ can therefore have nonzero matrix elements.

Forset $a=U(t)b$ withThen $b$ evolves under the constant Hamiltonian

For zero orbital angular momentum and Coulomb potential $-q^2/r$, the radial equation is

For the usual three-dimensional Coulomb potential Hamiltonian operator, a separated radial wavefunction $R(r)$ uses the reduced function $u=rR$, whose radial norm has measure $dr$. The regular physical operator domain imposes $u(0)=0$. Near the origin the leading radial equation gives the indicial equation $c(c-1)-\ell (\ell +1)=0$ for $u\sim r^c$, with roots $c=\ell +1$ and $c=-\ell$.

The regular root gives $R\sim r^{\ell }$. The other root gives $R\sim r^{-\ell -1}$, and its radial probability integral ${\int }_0^{\epsilon }{r}^2|R{|}^{2}dr\sim {\int }_0^{\epsilon }{r}^{-2\ell }dr$ diverges for $\ell \ge 1$. At $\ell =0$, $R\sim 1/r$ is locally square-integrable in three dimensions, but violates $u(0)=0$ and has divergent kinetic energy integral ${\int }_0^{\epsilon }{r}^2|R^{\prime }{|}^{2}dr\sim {\int }_0^{\epsilon }{r}^{-2}dr$. Thus local wavefunction normalization alone is insufficient to select the physical branch in this exceptional case; the standard Coulomb energy domain supplies the missing condition.

For three-dimensional radial wavefunctions with measure $r^{2}dr$, this formal symmetric momentum expression obeys $[f(r),p_r]=i\hslash f^{\prime }(r)$ on appropriate test functions and has square $-{\hslash }^2({\partial }_r^2+2{\partial }_r/r)$. It is useful in separating kinetic energy as $p^2=p_r^2+L^2/r^2$. Formal symmetry and these identities do not establish a self-adjoint momentum observable on the half-line; domains and origin conditions must be specified when taking adjoints or defining products.

For a normalized maximal-angular-momentum hydrogen state, the probability inside a small radius $r_p$ is the integral of $|R{|}^2{r}^2$. Since $R$ behaves as $r^l$ near zero, it equals $2^{2l+3}(r_p/a{)}^{2l+3}/[(2l+3)!(l+1{)}^{2l+3}]$ times $1+O(r_p/a)$. Increasing angular momentum strongly suppresses penetration toward the Coulomb centre.

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 $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

After the logarithmic radial coordinate $r=e^x$, the Langer replacement $l(l+1)\mapsto (l+1/2{)}^2$ turns the radial Coulomb problem into a two-turning-point WKB approximation problem. Its action integral giveswhich in this Coulomb case agrees with the exact dependence on the principal quantum number.

For a radial solution regular at the origin, writing $R=e^{-\gamma r}{\sum }_{n\ge 0}{a}_n{r}^n$ givesNormalizability at infinity forces the series to terminate, so $2\gamma N=\beta$ for a positive integer $N$ and $E_N=-m{q}^4/(2{\hslash }^2{N}^2)$.

The lowest Coulomb bound-state energy is

In atomic units, use the radial trial wavefunction $\psi (r)=\max (1-r/\beta ,0)$ with $\beta >0$. Its radial norm integral is $\int r^2{\psi }^{2}dr={\beta }^3/30$, its gradient contribution is $\int r^2{\psi }^{\prime 2}dr=\beta /3$, and its Coulomb contribution is $-2\int r{\psi }^{2}dr=-{\beta }^2/6$. In the hydrogen radial Rayleigh quotient, this gives the displayed energy, minimized at $\beta =4$. It is above the exact hydrogen ground-state energy $-1/2$, as required by the quantum variational principle. The derivative jumps at $r=\beta$, but the function has a square-integrable weak first derivative and is an admissible trial for the energy quadratic form. One must use that form rather than assume that a classical second derivative exists at the join.

With normalized $Y_{00}$, the radial ground state $R(r)=C{e}^{-r/a_0}$ hasIf the angular factor is absorbed into the spherically symmetric wavefunction, its normalization constant is instead $(\pi a_0^3{)}^{-1/2}$.

For the one-electron Coulomb potential $-KZ/r$ with fixed nucleus, the normalized spherically symmetric ground state is ${\psi }_Z=({\alpha }_Z^3/\pi {)}^{1/2}{e}^{-{\alpha }_{Z}r}$, where ${\alpha }_Z=Z/a_{\ast }$ and $a_{\ast }={\hslash }^2/(mK)$. Its radial probability density is $p_Z(r)=4{\alpha }_Z^3{r}^2{e}^{-2{\alpha }_{Z}r}$. The factorial integral givesMore generally $⟨r^q{⟩}_Z=\Gamma (q+3)/[2(2{\alpha }_Z{)}^q]$ for real $q>-3$, by the gamma function integral. The density transformation $p_Z(r)=Z{p}_1(Zr)$ expresses the full scaling, not just its mean. It applies to a hydrogen-like ion with one electron; the exact neutral helium atom radius additionally depends on the two-electron Coulomb interaction.