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Hydrogen atom

From Wikipedia, the free encyclopedia

Atom of the element hydrogen

This article is about the physics of the hydrogen atom. For a chemical description, seehydrogen. For monatomic hydrogen, seeHydrogen § Atomic hydrogen.

Hydrogen atom
General

Symbol	¹H
Names	hydrogen atom, protium
Protons(Z)	1
Neutrons(N)	0
Nuclide data
Natural abundance	99.985%
Half-life(t_1/2)	stable
Isotope mass	1.007825Da
Spin	⁠1/2⁠ ħ
Excess energy	7288.969±0.001keV
Nuclear binding energy	0.000±0.0000 keV
Isotopes of hydrogen Complete table of nuclides

Depiction of a hydrogen atom showing the diameter as about twice theBohr model radius. (Image not to scale)

Ahydrogen atom is anatom of thechemical element hydrogen. Theelectrically neutral hydrogen atom contains a single positively chargedproton in the nucleus, and a single negatively chargedelectron bound to the nucleus by theCoulomb force.Atomic hydrogen constitutesabout 75% of thebaryonic mass of the universe.^[1]

In everyday life on Earth, isolated hydrogen atoms (called "atomic hydrogen") are extremely rare. Instead, a hydrogen atom tends to combine with other atoms in compounds, or with another hydrogen atom to form ordinary (diatomic) hydrogen gas, H₂. "Atomic hydrogen" and "hydrogen atom" in ordinary English use have overlapping, yet distinct, meanings. For example, a water molecule contains two hydrogen atoms, but does not contain atomic hydrogen (which would refer to isolated hydrogen atoms).

Atomic spectroscopy shows that there is a discrete infinite set of states in which a hydrogen (or any) atom can exist, contrary to the predictions ofclassical physics. Attempts to develop a theoretical understanding of the states of the hydrogen atom have been important to thehistory of quantum mechanics, since all other atoms can be roughly understood by knowing in detail about this simplest atomic structure.

Isotopes

Main article:Isotopes of hydrogen

The mostabundant isotope, protium (¹H), or light hydrogen, contains noneutrons and is simply aproton and anelectron. Protium isstable and makes up 99.985% of naturally occurring hydrogen atoms.^[2]

Deuterium (²H) contains one neutron and one proton in its nucleus. Deuterium is stable, makes up 0.0156% of naturally occurring hydrogen,^[2] and is used in industrial processes likenuclear reactors andnuclear magnetic resonance spectroscopy.

Tritium (³H) contains two neutrons and one proton in its nucleus and is not stable, decaying with ahalf-life of 12.32 years. Because of its short half-life, tritium does not exist in nature except in trace amounts.

Heavier isotopes of hydrogen are only created artificially inparticle accelerators and have half-lives on the order of 10⁻²² seconds. They are unboundresonances located beyond theneutron drip line; this results in promptemission of a neutron.

The formulas below are valid for all three isotopes of hydrogen, but slightly different values of theRydberg constant (correction formula given below) must be used for each hydrogen isotope.

Hydrogen ion

Main articles:hydrogen cation andhydrogen anion

Lone neutral hydrogen atoms are rare under normal conditions. However, neutral hydrogen is common when it iscovalently bound to another atom, and hydrogen atoms can also exist incationic and anionic forms.

If a neutral hydrogen atom loses its electron, it becomes a cation. The resulting ion, which consists solely of a proton for the usual isotope, is written as "H⁺" and sometimes calledhydron. Free protons are common in theinterstellar medium, andsolar wind. In the context ofaqueous solutions of classicalBrønsted–Lowry acids, such ashydrochloric acid, it is actuallyhydronium,H₃O⁺, that is meant. Instead of a literal ionized single hydrogen atom being formed, the acid transfers the hydrogen to H₂O, forming H₃O⁺.

If instead a hydrogen atom gains a second electron, it becomes an anion. Thehydrogen anion is written as "H^–" and calledhydride.

Theoretical analysis

The hydrogen atom has special significance inquantum mechanics andquantum field theory as a simpletwo-body problem physical system which has yielded many simpleanalytical solutions in closed-form.

Failed classical description

Experiments byErnest Rutherford in 1909 showed the structure of the atom to be a dense, positive nucleus with a tenuous negative charge cloud around it. This immediately raised questions about how such a system could be stable.Classical electromagnetism had shown that any accelerating charge radiates energy, as shown by theLarmor formula. If the electron is assumed to orbit in a perfect circle and radiates energy continuously, the electron would rapidly spiral into the nucleus with a fall time of:^[3] $t_{\text{fall}}\approx {\frac {{a_{0}}^{3}}{4{r_{0}}^{2}c}}\approx 1.6\times 10^{-11}{\text{ s}},$ where $a_{0}$ is theBohr radius and $r_{0}$ is theclassical electron radius. If this were true, all atoms would instantly collapse. However, atoms seem to be stable. Furthermore, the spiral inward would release a smear of electromagnetic frequencies as the orbit got smaller. Instead, atoms were observed to emit only discrete frequencies of radiation. The resolution would lie in the development ofquantum mechanics.

Bohr–Sommerfeld model

Main article:Bohr model

In 1913,Niels Bohr obtained the energy levels and spectral frequencies of the hydrogen atom after making a number of simple assumptions in order to correct the failed classical model. The assumptions included:

Electrons can only be in certain, discrete circular orbits orstationary states, thereby having a discrete set of possible radii and energies.
Electrons do not emit radiation while in one of these stationary states.
An electron can gain or lose energy by jumping from one discrete orbit to another.

Bohr supposed that the electron's angular momentum is quantized with possible values: $L=n\hbar$ where $n=1,2,3,\ldots$ and $\hbar$ isPlanck constant over $2\pi$ . He also supposed that thecentripetal force which keeps the electron in its orbit is provided by theCoulomb force, and that energy is conserved. Bohr derived the energy of each orbit of the hydrogen atom to be:^[4] $E_{n}=-{\frac {m_{\text{e}}e^{4}}{2(4\pi \varepsilon _{0})^{2}\hbar ^{2}}}{\frac {1}{n^{2}}},$ where $m_{\text{e}}$ is theelectron mass, $e {\displaystyle e}$ is theelectron charge, $\varepsilon _{0}$ is thevacuum permittivity, and $n {\displaystyle n}$ is thequantum number (now known as theprincipal quantum number). Bohr's predictions matched experiments measuring thehydrogen spectral series to the first order, giving more confidence to a theory that used quantized values.

For $n=1$ , the value^[5] ${\frac {m_{\text{e}}e^{4}}{2(4\pi \varepsilon _{0})^{2}\hbar ^{2}}}={\frac {m_{\text{e}}e^{4}}{8h^{2}\varepsilon _{0}^{2}}}=1\,{\text{Ry}}=13.605\;693\;122\;994(26)\,{\text{eV}}$ is called the Rydberg unit of energy. It is related to theRydberg constant $R_{\infty }$ ofatomic physics by $1\,{\text{Ry}}\equiv hcR_{\infty }.$

The exact value of the Rydberg constant assumes that the nucleus is infinitely massive with respect to the electron. For hydrogen-1, hydrogen-2 (deuterium), and hydrogen-3 (tritium) which have finite mass, the constant must be slightly modified to use thereduced mass of the system, rather than simply the mass of the electron. This includes the kinetic energy of the nucleus in the problem, because the total (electron plus nuclear) kinetic energy is equivalent to the kinetic energy of the reduced mass moving with a velocity equal to the electron velocity relative to the nucleus. However, since the nucleus is much heavier than the electron, the electron mass and reduced mass are nearly the same. The Rydberg constantR_M for a hydrogen atom (one electron) is given by: $R_{M}={\frac {R_{\infty }}{1+m_{\text{e}}/M}},$ where $M {\displaystyle M}$ is the mass of the atomic nucleus. For hydrogen-1, the quantity $m_{\text{e}}/M,$ is about 1/1836 (i.e. the electron-to-proton mass ratio). For deuterium and tritium, the ratios are about 1/3670 and 1/5497 respectively. These figures, when added to 1 in the denominator, represent very small corrections in the value ofR, and thus only small corrections to all energy levels in corresponding hydrogen isotopes.

There were still problems with Bohr's model:

it failed to predict other spectral details such asfine structure andhyperfine structure
it could only predict energy levels with any accuracy for single–electron atoms (hydrogen-like atoms)
the predicted values were only correct to $\alpha ^{2}\approx 10^{-5}$ , where $\alpha$ is thefine-structure constant.

Most of these shortcomings were resolved byArnold Sommerfeld's modification of the Bohr model. Sommerfeld introduced two additional degrees of freedom, allowing an electron to move on an elliptical orbit characterized by itseccentricity anddeclination with respect to a chosen axis. This introduced two additional quantum numbers, which correspond to the orbitalangular momentum and its projection on the chosen axis. Thus the correct multiplicity of states (except for the factor 2 accounting for the yet unknown electron spin) was found. Further, by applyingspecial relativity to the elliptic orbits, Sommerfeld succeeded in deriving the correct expression for the fine structure of hydrogen spectra (which happens to be exactly the same as in the most elaborate Dirac theory). However, some observed phenomena, such as the anomalousZeeman effect, remained unexplained. These issues were resolved with the full development of quantum mechanics and theDirac equation. It is often alleged that theSchrödinger equation is superior to the Bohr–Sommerfeld theory in describing hydrogen atom. This is not the case, as most of the results of both approaches coincide or are very close (a remarkable exception is the problem of hydrogen atom in crossed electric and magnetic fields, which cannot be self-consistently solved in the framework of the Bohr–Sommerfeld theory), and in both theories the main shortcomings result from the absence of the electron spin. It was the complete failure of the Bohr–Sommerfeld theory to explain many-electron systems (such as helium atom or hydrogen molecule) which demonstrated its inadequacy in describing quantum phenomena.

Schrödinger equation

The Schrödinger equation is the standard quantum-mechanics model; it allows one to calculate the stationary states and also the time evolution of quantum systems. Exact analytical answers are available for the nonrelativistic hydrogen atom. Before we go to present a formal account, here we give an elementary overview.

Given that the hydrogen atom contains a nucleus and an electron, quantum mechanics allows one to predict the probability of finding the electron at any given radial distance $r {\displaystyle r}$ . It is given by the square of a mathematical function known as the "wavefunction", which is a solution of the Schrödinger equation. The lowest energy equilibrium state of the hydrogen atom is known as the ground state. The ground state wave function is known as the $1\mathrm {s}$ wavefunction. It is written as: $\psi _{1\mathrm {s} }(r)={\frac {1}{{\sqrt {\pi }}a_{0}^{3/2}}}\mathrm {e} ^{-r/a_{0}}.$

Here, $a_{0}$ is the numerical value of the Bohr radius. The probability density of finding the electron at a distance $r {\displaystyle r}$ in any radial direction is the squared value of the wavefunction: $|\psi _{1\mathrm {s} }(r)|^{2}={\frac {1}{\pi a_{0}^{3}}}\mathrm {e} ^{-2r/a_{0}}.$

The $1\mathrm {s}$ wavefunction is spherically symmetric, and the surface area of a shell at distance $r {\displaystyle r}$ is $4\pi r^{2}$ , so the total probability $P(r)\,dr$ of the electron being in a shell at a distance $r {\displaystyle r}$ and thickness $d r {\displaystyle dr}$ is $P(r)\,\mathrm {d} r=4\pi r^{2}|\psi _{1\mathrm {s} }(r)|^{2}\,\mathrm {d} r.$

It turns out that this is a maximum at $r=a_{0}$ . That is, the Bohr picture of an electron orbiting the nucleus at radius $a_{0}$ corresponds to the most probable radius. Actually, there is a finite probability that the electron may be found at any radius $r {\displaystyle r}$ , with theprobability indicated by the square of the wavefunction. Since the probability of finding the electronsomewhere in the whole volume is unity, the integral of $P(r)\,\mathrm {d} r$ is unity. Then we say that the wavefunction is properly normalized.

As discussed below, the ground state $1\mathrm {s}$ is also indicated by thequantum numbers $(n=1,\ell =0,m=0)$ . The second lowest energy states, just above the ground state, are given by the quantum numbers $(2,0,0)$ , $(2,1,0)$ , and $(2,1,\pm 1)$ . These $n=2$ states all have the same energy and are known as the $2\mathrm {s}$ and $2\mathrm {p}$ states. There is one $2\mathrm {s}$ state: $\psi _{2,0,0}={\frac {1}{4{\sqrt {2\pi }}a_{0}^{3/2}}}\left(2-{\frac {r}{a_{0}}}\right)\mathrm {e} ^{-r/2a_{0}},$ and there are three $2\mathrm {p}$ states: $\psi _{2,1,0}={\frac {1}{4{\sqrt {2\pi }}a_{0}^{3/2}}}{\frac {r}{a_{0}}}\mathrm {e} ^{-r/2a_{0}}\cos \theta ,$ $\psi _{2,1,\pm 1}=\mp {\frac {1}{8{\sqrt {\pi }}a_{0}^{3/2}}}{\frac {r}{a_{0}}}\mathrm {e} ^{-r/2a_{0}}\sin \theta ~\mathrm {e} ^{\pm \mathrm {i} \varphi }.$

An electron in the $2\mathrm {s}$ or $2\mathrm {p}$ state is most likely to be found in the second Bohr orbit with energy given by the Bohr formula.

Wavefunction

TheHamiltonian of the hydrogen atom is the radial kinetic energy operator plus the Coulomb electrostatic potential energy between the positive proton and the negative electron. Using the time-independent Schrödinger equation, ignoring all spin-coupling interactions and using thereduced mass $\mu =m_{\text{e}}M/(m_{\text{e}}+M)$ , the equation is written as: $\left(-{\frac {\hbar ^{2}}{2\mu }}\nabla ^{2}-{\frac {e^{2}}{4\pi \varepsilon _{0}r}}\right)\psi (r,\theta ,\varphi )=E\psi (r,\theta ,\varphi )$

Expanding theLaplacian in spherical coordinates: $-{\frac {\hbar ^{2}}{2\mu }}\left[{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial \psi }{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial \psi }{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}\psi }{\partial \varphi ^{2}}}\right]-{\frac {e^{2}}{4\pi \varepsilon _{0}r}}\psi =E\psi$

This is aseparable,partial differential equation which can be solved in terms of special functions. When the wavefunction is separated as a product of functions $R(r)$ , $\Theta (\theta )$ , and $\Phi (\varphi )$ three independent differential functions appear^[6] with A and B being the separation constants:

radial: ${\frac {d}{dr}}\left(r^{2}{\frac {dR}{dr}}\right)+{\frac {2\mu r^{2}}{\hbar ^{2}}}\left(E+{\frac {e^{2}}{4\pi \varepsilon _{0}r}}\right)R-AR=0$
polar: ${\frac {\sin \theta }{\Theta }}{\frac {d}{d\theta }}\left(\sin \theta {\frac {d\Theta }{d\theta }}\right)+A\sin ^{2}\theta -B=0$
azimuth: ${\frac {1}{\Phi }}{\frac {d^{2}\Phi }{d\varphi ^{2}}}+B=0.$

The normalized positionwavefunctions, given inspherical coordinates are: $\psi _{n\ell m}(r,\theta ,\varphi )={\sqrt {{\left({\frac {2}{na_{0}^{*}}}\right)}^{3}{\frac {(n-\ell -1)!}{2n(n+\ell )!}}}}\mathrm {e} ^{-\rho /2}\rho ^{\ell }L_{n-\ell -1}^{2\ell +1}(\rho )Y_{\ell }^{m}(\theta ,\varphi )$

3D illustration of the eigenstate $\psi _{4,3,1}$ . Electrons in this state are 45% likely to be found within the solid body shown.

{\displaystyle \psi _{4,3,1}} — 3D illustration of the eigenstate $\psi _{4,3,1}$ . Electrons in this state are 45% likely to be found within the solid body shown.

where:

$\rho ={2r \over {na_{0}^{*}}}$ ,
$a_{0}^{*}$ is thereduced Bohr radius, $a_{0}^{*}={{4\pi \varepsilon _{0}\hbar ^{2}} \over {\mu e^{2}}}$ ,
$L_{n-\ell -1}^{2\ell +1}(\rho )$ is ageneralized Laguerre polynomial of degree $n-\ell -1$ , and
$Y_{\ell }^{m}(\theta ,\varphi )$ is aspherical harmonic function of degree $\ell$ and order $m {\displaystyle m}$ .

Note that the generalized Laguerre polynomials are defined differently by different authors. The usage here is consistent with the definitions used by Messiah,^[7] and Mathematica.^[8] In other places, the Laguerre polynomial includes a factor of $(n+\ell )!$ ,^[9] or the generalized Laguerre polynomial appearing in the hydrogen wave function is $L_{n+\ell }^{2\ell +1}(\rho )$ instead.^[10]

The quantum numbers can take the following values:

$n=1,2,3,\ldots$ (principal quantum number)
$\ell =0,1,2,\ldots ,n-1$ (azimuthal quantum number)
$m=-\ell ,\ldots ,\ell$ (magnetic quantum number).

Additionally, these wavefunctions arenormalized (i.e., the integral of their modulus square equals 1) andorthogonal: $\int _{0}^{\infty }r^{2}\,dr\int _{0}^{\pi }\sin \theta \,d\theta \int _{0}^{2\pi }d\varphi \,\psi _{n\ell m}^{*}(r,\theta ,\varphi )\psi _{n'\ell 'm'}(r,\theta ,\varphi )=\langle n,\ell ,m|n',\ell ',m'\rangle =\delta _{nn'}\delta _{\ell \ell '}\delta _{mm'},$ where $|n,\ell ,m\rangle$ is the state represented by the wavefunction $\psi _{n\ell m}$ inDirac notation, and $\delta$ is theKronecker delta function.^[11]

The wavefunctions in momentum space are related to the wavefunctions in position space through a Fourier transform $\varphi (p,\theta _{p},\varphi _{p})=(2\pi \hbar )^{-3/2}\int \mathrm {e} ^{-i{\vec {p}}\cdot {\vec {r}}/\hbar }\psi (r,\theta ,\varphi )\,dV,$ which, for the bound states, results in^[12] $\varphi (p,\theta _{p},\varphi _{p})={\sqrt {{\frac {2}{\pi }}{\frac {(n-\ell -1)!}{(n+\ell )!}}}}n^{2}2^{2\ell +2}\ell !{\frac {n^{\ell }p^{\ell }}{(n^{2}p^{2}+1)^{\ell +2}}}C_{n-\ell -1}^{\ell +1}\left({\frac {n^{2}p^{2}-1}{n^{2}p^{2}+1}}\right)Y_{\ell }^{m}(\theta _{p},\varphi _{p}),$ where $C_{N}^{\alpha }(x)$ denotes aGegenbauer polynomial and $p {\displaystyle p}$ is in units of $\hbar /a_{0}^{*}$ .

The solutions to the Schrödinger equation for hydrogen areanalytical, giving a simple expression for the hydrogenenergy levels and thus the frequencies of the hydrogenspectral lines and fully reproduced the Bohr model and went beyond it. It also yields two other quantum numbers and the shape of the electron's wave function ("orbital") for the various possible quantum-mechanical states, thus explaining theanisotropic character of atomic bonds.

The Schrödinger equation also applies to more complicated atoms andmolecules. When there is more than one electron or nucleus the solution is not analytical and either computer calculations are necessary or simplifying assumptions must be made.

Since the Schrödinger equation is only valid for non-relativistic quantum mechanics, the solutions it yields for the hydrogen atom are not entirely correct. TheDirac equation of relativistic quantum theory improves these solutions (see below).

Results of Schrödinger equation

The solution of the Schrödinger equation (wave equation) for the hydrogen atom uses the fact that theCoulomb potential produced by the nucleus isisotropic (it is radially symmetric in space and only depends on the distance to the nucleus). Although the resultingenergy eigenfunctions (theorbitals) are not necessarily isotropic themselves, their dependence on theangular coordinates follows completely generally from this isotropy of the underlying potential: theeigenstates of theHamiltonian (that is, the energy eigenstates) can be chosen as simultaneous eigenstates of theangular momentum operator. This corresponds to the fact that angular momentum is conserved in theorbital motion of the electron around the nucleus. Therefore, the energy eigenstates may be classified by two angular momentumquantum numbers, $\ell$ and $m {\displaystyle m}$ (both are integers). The angular momentum quantum number $\ell =0,1,2,\ldots$ determines the magnitude of the angular momentum. The magnetic quantum number $m=-\ell ,\ldots ,+\ell$ determines the projection of the angular momentum on the (arbitrarily chosen) $z {\displaystyle z}$ -axis.

In addition to mathematical expressions for total angular momentum and angular momentum projection of wavefunctions, an expression for the radial dependence of the wave functions must be found. It is only here that the details of the $1/r$ Coulomb potential enter (leading toLaguerre polynomials in $r {\displaystyle r}$ ). This leads to a third quantum number, the principal quantum number $n=1,2,3,\ldots$ . The principal quantum number in hydrogen is related to the atom's total energy.

Note that the maximum value of the angular momentum quantum number is limited by the principal quantum number: it can run only up to $n-1$ , i.e., $\ell =0,1,\ldots ,n-1$ .

Due to angular momentum conservation, states of the same $\ell$ but different $m {\displaystyle m}$ have the same energy (this holds for all problems withrotational symmetry). In addition, for the hydrogen atom, states of the same $n {\displaystyle n}$ but different $\ell$ are alsodegenerate (i.e., they have the same energy). However, this is a specific property of hydrogen and is no longer true for more complicated atoms which have an (effective) potential differing from the form $1/r$ (due to the presence of the inner electrons shielding the nucleus potential).

Taking into account thespin of the electron adds a last quantum number, the projection of the electron's spin angular momentum along the $z {\displaystyle z}$ -axis, which can take on two values. Therefore, anyeigenstate of the electron in the hydrogen atom is described fully by four quantum numbers. According to the usual rules of quantum mechanics, the actual state of the electron may be anysuperposition of these states. This explains also why the choice of $z {\displaystyle z}$ -axis for the directionalquantization of the angular momentum vector is immaterial: an orbital of given $\ell$ and $m^{'} {\displaystyle m'}$ obtained for another preferred axis $z^{'} {\displaystyle z'}$ can always be represented as a suitable superposition of the various states of different $m {\displaystyle m}$ (but same $\ell$ ) that have been obtained for $z {\displaystyle z}$ .

Mathematical summary of eigenstates of hydrogen atom

Main article:Hydrogen-like atom

In 1928,Paul Dirac foundan equation that was fully compatible withspecial relativity, and (as a consequence) made the wave function a 4-component "Dirac spinor" including "up" and "down" spin components, with both positive and "negative" energy (or matter and antimatter). The solution to this equation gave the following results, more accurate than the Schrödinger solution.

Energy levels

The energy levels of hydrogen, includingfine structure (excludingLamb shift andhyperfine structure), are given by theSommerfeld fine-structure expression:^[13] ${\begin{aligned}E_{j\,n}={}&-\mu c^{2}\left[1-\left(1+\left[{\frac {\alpha }{n-j-{\frac {1}{2}}+{\sqrt {\left(j+{\frac {1}{2}}\right)^{2}-\alpha ^{2}}}}}\right]^{2}\right)^{-1/2}\right]\\\approx {}&-{\frac {\mu c^{2}\alpha ^{2}}{2n^{2}}}\left[1+{\frac {\alpha ^{2}}{n^{2}}}\left({\frac {n}{j+{\frac {1}{2}}}}-{\frac {3}{4}}\right)\right],\end{aligned}}$ where $\alpha$ is thefine-structure constant and $j {\displaystyle j}$ is thetotal angular momentum quantum number, which is equal to $\left|\ell \pm {\tfrac {1}{2}}\right|$ , depending on the orientation of the electron spin relative to the orbital angular momentum.^[14] This formula represents a small correction to the energy obtained by Bohr and Schrödinger as given above. The factor in square brackets in the last expression is nearly one; the extra term arises from relativistic effects (for details, see#Features going beyond the Schrödinger solution). It is worth noting that this expression was first obtained byA. Sommerfeld in 1916 based on the relativistic version of theold Bohr theory. Sommerfeld has however used different notation for the quantum numbers.

Visualizing the hydrogen electron orbitals

Main article:Atomic orbital

Probability densities through thexz-plane for the electron at different quantum numbers (ℓ, across top;n, down side;m = 0)

The image to the right shows the first few hydrogen atom orbitals (energy eigenfunctions). These are cross-sections of theprobability density that are color-coded (black represents zero density and white represents the highest density). The angular momentum (orbital) quantum numberℓ is denoted in each column, using the usual spectroscopic letter code (s meansℓ = 0,p meansℓ = 1,d meansℓ = 2). The main (principal) quantum numbern (= 1, 2, 3, ...) is marked to the right of each row. For all pictures the magnetic quantum numberm has been set to 0, and the cross-sectional plane is thexz-plane (z is the vertical axis). The probability density in three-dimensional space is obtained by rotating the one shown here around thez-axis.

The "ground state", i.e. the state of lowest energy, in which the electron is usually found, is the first one, the 1s state (principal quantum leveln = 1,ℓ = 0).

Black lines occur in each but the first orbital: these are the nodes of the wavefunction, i.e. where the probability density is zero. (More precisely, the nodes arespherical harmonics that appear as a result of solving theSchrödinger equation in spherical coordinates.)

Thequantum numbers determine the layout of these nodes. There are:^{[citation needed]}

$n-1$ total nodes,
$\ell$ of which are angular nodes:
- $m {\displaystyle m}$ angular nodes go around the $\varphi$ axis (in thexy plane).(The figure above does not show these nodes since it plots cross-sections through thexz-plane.)
- $\ell -m$ (the remaining angular nodes) occur on the $\theta$ (vertical) axis.
$n-\ell -1$ (the remaining non-angular nodes) are radial nodes.

Oscillation of orbitals

The oscillation of two orbitals with the same value ofm

The frequency of a state in level n is $\omega _{n}=E_{n}/\hbar$ , so in case of a superposition of multiple orbitals, they would oscillate due to the difference in frequency. For example two states, ψ₁and ψ₂: The wavefunction is given by $\psi =\psi _{1}e^{i\omega _{1}t}+\psi _{2}e^{i\omega _{2}t}$ and the probability function is $P(t)=|\psi |^{2}=(\psi _{1}e^{i\omega _{1}t}+\psi _{2}e^{i\omega _{2}t})(\psi _{1}^{*}e^{-i\omega _{1}t}+\psi _{2}^{*}e^{-i\omega _{2}t})$ $\propto |\psi _{1}\psi _{2}|\cos {[(\omega _{1}-\omega _{2})t]}$

The oscillation of two orbitals with different angular momentum numbers

The result is a rotating wavefunction. The movement of electrons and change of quantum states radiates light at a frequency of the cosine.

Features going beyond the Schrödinger solution

There are several important effects that are neglected by the Schrödinger equation and which are responsible for certain small but measurable deviations of the real spectral lines from the predicted ones:

Although the mean speed of the electron in hydrogen is only 1/137 of thespeed of light, many modern experiments are sufficiently precise that a complete theoretical explanation requires a fully relativistic treatment of the problem. A relativistic treatment results in a momentum increase of about 1 part in37000 for the electron. Since the electron's wavelength is determined by its momentum, orbitals containing higher speed electrons show contraction due to smaller wavelengths.
Even when there is no externalmagnetic field, in theinertial frame of the moving electron, the electromagnetic field of the nucleus has a magnetic component. The spin of the electron has an associatedmagnetic moment which interacts with this magnetic field. This effect is also explained by special relativity, and it leads to the so-calledspin–orbit coupling, i.e., an interaction between theelectron'sorbital motion around the nucleus, and itsspin.

Both of these features (and more) are incorporated in the relativisticDirac equation, with predictions that come still closer to experiment. Again the Dirac equation may be solved analytically in the special case of a two-body system, such as the hydrogen atom. The resulting solution quantum states now must be classified by thetotal angular momentum numberj (arising through the coupling betweenelectron spin andorbital angular momentum). States of the samej and the samen are still degenerate. Thus, direct analytical solution ofDirac equation predicts 2S(⁠1/2⁠) and 2P(⁠1/2⁠) levels of hydrogen to have exactly the same energy, which is in a contradiction with observations (Lamb–Retherford experiment).

There are alwaysvacuum fluctuations of theelectromagnetic field, according to quantum mechanics. Due to such fluctuations degeneracy between states of the samej but differentl is lifted, giving them slightly different energies. This has been demonstrated in the famousLamb–Retherford experiment and was the starting point for the development of the theory ofquantum electrodynamics (which is able to deal with these vacuum fluctuations and employs the famousFeynman diagrams for approximations usingperturbation theory). This effect is now calledLamb shift.

For these developments, it was essential that the solution of the Dirac equation for the hydrogen atom could be worked out exactly, such that any experimentally observed deviation had to be taken seriously as a signal of failure of the theory.

Alternatives to the Schrödinger theory

In the language ofHeisenberg's matrix mechanics, the hydrogen atom was first solved byWolfgang Pauli^[15] using arotational symmetry in four dimensions [O(4)-symmetry] generated by theangular momentumand theLaplace–Runge–Lenz vector. By extending the symmetry group O(4) to thedynamical group O(4,2),the entire spectrum and all transitions were embedded in a single irreducible group representation.^[16]

In 1979 the (non-relativistic) hydrogen atom was solved for the first time withinFeynman's path integral formulationofquantum mechanics by Duru and Kleinert.^[17]^[18] This work greatly extended the range of applicability ofFeynman's method.

Further alternative models areBohm mechanics and thecomplex Hamilton–Jacobi formulation of quantum mechanics.

See also

References

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^Sommerfeld, Arnold (1919).Atombau und Spektrallinien [Atomic Structure and Spectral Lines]. Braunschweig: Friedrich Vieweg und Sohn.ISBN 3-87144-484-7.{{cite book}}:ISBN / Date incompatibility (help)German English
^Atkins, Peter; de Paula, Julio (2006).Physical Chemistry (8th ed.). W. H. Freeman. p. 349.ISBN 0-7167-8759-8.
^Pauli, W (1926). "Über das Wasserstoffspektrum vom Standpunkt der neuen Quantenmechanik".Zeitschrift für Physik.36 (5):336–363.Bibcode:1926ZPhy...36..336P.doi:10.1007/BF01450175.S2CID 128132824.
^Kleinert H. (1968)."Group Dynamics of the Hydrogen Atom"(PDF).Lectures in Theoretical Physics, Edited by W.E. Brittin and A.O. Barut, Gordon and Breach, N.Y. 1968:427–482.
^Duru I.H., Kleinert H. (1979)."Solution of the path integral for the H-atom"(PDF).Physics Letters B.84 (2):185–188.Bibcode:1979PhLB...84..185D.doi:10.1016/0370-2693(79)90280-6.
^Duru I.H., Kleinert H. (1982)."Quantum Mechanics of H-Atom from Path Integrals"(PDF).Fortschr. Phys.30 (2):401–435.Bibcode:1982ForPh..30..401D.doi:10.1002/prop.19820300802.

Books

Griffiths, David J. (1995).Introduction to Quantum Mechanics.Prentice Hall.ISBN 0-13-111892-7. Section 4.2 deals with the hydrogen atom specifically, but all of Chapter 4 is relevant.
Kleinert, H. (2009).Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 4th edition,Worldscibooks.com, World Scientific, Singapore (also available onlinephysik.fu-berlin.de)

External links

Lighter:
(none, lightest possible)

Hydrogen atom is an
isotope ofhydrogen

Heavier:
hydrogen-2

Decay product of:
free neutron
helium-2

Decay chain
of hydrogen atom

Decays to:
Stable

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