Atomic Units and Scales

Hartree atomic units are a unit system adapted to nonrelativistic electrons bound by Coulomb forces. They turn the characteristic electron mass, charge, action, length, and energy into quantities of order unity, so that an equation displays its atomic structure instead of repeatedly displaying SI conversion factors.

Required background. Units and Constants explains dimensional analysis; Hydrogen Atom supplies the Coulomb Hamiltonian. The atomic-unit reference gives additional conversion factors.

Why Atomic Units Work

Consider the fixed-nucleus, nonrelativistic hydrogen Hamiltonian in SI units,

$$H=-\frac{\hbar^2}{2m_e}\nabla_r^2 -\frac{e^2}{4\pi\epsilon_0r},$$

where $e>0$ is the elementary charge magnitude. Introduce a dimensionless coordinate $\boldsymbol\rho$ through

$$\mathbf r=a_0\boldsymbol\rho.$$

Choose $a_0$ so that the kinetic and Coulomb coefficients define the same energy scale:

$$\frac{\hbar^2}{m_ea_0^2} =\frac{e^2}{4\pi\epsilon_0a_0} \equiv E_{\mathrm h}.$$

Solving these relations gives the Bohr radius and Hartree energy,

$$\begin{aligned} a_0&=\frac{4\pi\epsilon_0\hbar^2}{m_ee^2},\\ E_{\mathrm h}&=\frac{e^2}{4\pi\epsilon_0a_0},\\ E_{\mathrm h}&=\frac{\hbar^2}{m_ea_0^2}. \end{aligned}$$

After dividing by $E_{\mathrm h}$, the Schrödinger equation becomes

$$\frac{H}{E_{\mathrm h}} =-\frac12\nabla_{\rho}^2-\frac{1}{\rho}.$$

Nothing physical has been discarded. The constants have been absorbed into the definitions of length and energy. The dimensionless eigenvalues and wavefunctions can be converted back after the calculation.

The defining convention

Hartree atomic units set

$$\hbar=m_e=e=4\pi\epsilon_0=1.$$

This statement is shorthand for measuring action in units of $\hbar$, mass in units of $m_e$, charge magnitude in units of $e$, and electromagnetic quantities in the corresponding electromagnetic units. It does not mean the electron charge is positive: in this convention the electron has charge $-1$.

The convention also does not set the speed of light to one. Instead,

$$c=\alpha^{-1}$$

in atomic units, where $\alpha$ is the fine-structure constant. This fact makes powers of $\alpha$ visible as the small parameters controlling relativistic corrections.

Core Atomic Units

The base choices generate all other units by dimensional analysis. Selected 2022 CODATA values, rounded here for practical use, are:

Quantity	Atomic unit	Approximate SI value
length	$a_0$	$5.291\,772\,105\times10^{-11}\ \mathrm m$
energy	$E_{\mathrm h}$	$4.359\,744\,722\times10^{-18}\ \mathrm J$
energy	$E_{\mathrm h}$	$27.211\,386\,246\ \mathrm{eV}$
time	$t_0=\hbar/E_{\mathrm h}$	$2.418\,884\,327\times10^{-17}\ \mathrm s$
velocity	$v_0=a_0/t_0=\alpha c$	$2.187\,691\,262\times10^6\ \mathrm{m\,s^{-1}}$
dipole moment	$d_0=ea_0$	$8.478\,353\,620\times10^{-30}\ \mathrm{C\,m}$
electric field	$F_0=E_{\mathrm h}/(ea_0)$	$5.142\,206\,751\times10^{11}\ \mathrm{V\,m^{-1}}$

Parenthesized standard uncertainties and additional derived units are available from the NIST/CODATA constants database. Do not silently combine last digits from different CODATA adjustments.

Length and momentum

A length reported as $r=3.2$ a.u. means

$$r=3.2a_0.$$

The corresponding momentum unit is

$$p_0=\frac{\hbar}{a_0}=m_e\alpha c.$$

Because position and momentum use reciprocal scales, a wavefunction localized to order $a_0$ naturally contains momenta of order $\hbar/a_0$.

Energy and force

The Hartree energy is the Coulomb energy of two elementary charge magnitudes separated by one Bohr radius. It is also twice the infinite-mass Rydberg energy. The force unit is

$$F_{\mathrm{force},0}=\frac{E_{\mathrm h}}{a_0}.$$

The symbol $F$ is also widely used for electric field strength in AMO physics. Context and units must distinguish force from field; this page uses $F_0$ only for the atomic unit of electric field.

Hartree and Rydberg Are Not Synonyms

For an infinitely heavy hydrogenic nucleus, the nonrelativistic bound energies are

$$E_n=-\frac{Z^2}{2n^2}E_{\mathrm h}.$$

The ground-state binding magnitude is therefore $E_{\mathrm h}/2$. This quantity is the infinite-mass Rydberg energy:

$$\operatorname{Ry}=\frac{E_{\mathrm h}}{2}.$$

Two unit conventions consequently coexist:

• In Hartree atomic units, the hydrogen ground-state energy is $-1/2$.
• In Rydberg units, the same energy is $-1$.

Electronic-structure codes may report either convention, particularly for kinetic-energy operators, pseudopotentials, and plane-wave cutoffs. The label “a.u.” alone is insufficient when the factor of two matters. Record whether the energy unit is $E_{\mathrm h}$ or $\operatorname{Ry}$.

The Rydberg constant $R_\infty$ is a spectroscopic inverse-length constant related by

$$E_{\mathrm h}=2hcR_\infty.$$

For a real isotope, reduced mass shifts the hydrogenic scale. Do not identify $hcR_\infty$ with an exact measured ionization energy without applying recoil and higher-order corrections.

Energy, Frequency, Angular Frequency, and Wavenumber

An energy interval may be represented in several equivalent ways:

$$\begin{aligned} \Delta E&=h\nu,\\ \Delta E&=\hbar\omega,\\ \Delta E&=hc\widetilde\nu. \end{aligned}$$

Here $\nu$ is ordinary frequency in cycles per second, $\omega=2\pi\nu$ is angular frequency, and $\widetilde\nu=1/\lambda$ is spectroscopic wavenumber. In spectroscopy, $\widetilde\nu$ is commonly reported in $\mathrm{cm}^{-1}$.

One Hartree corresponds to

$$\begin{aligned} \frac{E_{\mathrm h}}{h} &=6.579\,683\,9205\times10^{15}\ \mathrm{Hz},\\ \frac{E_{\mathrm h}}{\hbar} &=4.134\,137\,3335\times10^{16}\ \mathrm{s}^{-1},\\ \frac{E_{\mathrm h}}{hc} &=219\,474.6314\ \mathrm{cm}^{-1},\\ \frac{E_{\mathrm h}}{k_B} &=315\,775.025\ \mathrm K. \end{aligned}$$

The second line is an angular frequency even when the radian is treated as dimensionless. Writing $E/\hbar$ and labeling the result “Hz” introduces a factor-of-$2\pi$ error. Writing $E/(hc)$ as a wavenumber and then multiplying by $2\pi$ without changing convention creates the same problem in another form.

The atomic unit of time

The time unit

$$t_0=\frac{\hbar}{E_{\mathrm h}}$$

is the inverse Hartree angular-frequency scale. If a time-dependent calculation returns $t=1000$ a.u., the physical duration is about $24.19\ \mathrm{fs}$. A phase factor has the same form in either system:

$$\exp\!\left(-\frac{iEt}{\hbar}\right) =\exp(-iE_{\mathrm{au}}t_{\mathrm{au}}).$$

The right-hand expression is valid only when energy is in Hartree and time is in $\hbar/E_{\mathrm h}$.

Electric Fields, Dipoles, and Polarizabilities

For a dipole in an electric field,

$$H_E=-\mathbf d\cdot\mathbf F.$$

Choosing $d_0=ea_0$ and $F_0=E_{\mathrm h}/(ea_0)$ makes their product exactly one Hartree:

$$d_0F_0=E_{\mathrm h}.$$

Thus a dipole matrix element $d_{\mathrm{au}}$ in a field $F_{\mathrm{au}}$ produces the characteristic interaction energy

$$\Delta E_{\mathrm{au}} \sim d_{\mathrm{au}}F_{\mathrm{au}}.$$

The atomic field unit is extremely large because it is the field that changes an electron’s potential energy by one Hartree across one Bohr radius. Typical laboratory fields are often small in atomic units, even when they strongly mix nearly degenerate states.

The atomic dipole unit is

$$ea_0\simeq2.54174647\ \mathrm D,$$

where $\mathrm D$ denotes the debye. The polarizability unit follows from the quadratic Stark energy $-\alpha_EF^2/2$:

$$\alpha_{E,0} =\frac{e^2a_0^2}{E_{\mathrm h}} =4\pi\epsilon_0a_0^3.$$

The factor $4\pi\epsilon_0$ matters when converting a polarizability reported as a volume in $a_0^3$ to SI. Atomic, Gaussian, and SI polarizability conventions should not be mixed by dimensional appearance alone.

The Fine-Structure Constant as a Scale Parameter

Atomic units make the nonrelativistic Coulomb problem order unity while leaving relativity parametrically visible. Since $v_0=\alpha c$,

$$\frac{v_0}{c}=\alpha.$$

The electron rest energy in Hartree is

$$m_ec^2=\alpha^{-2}E_{\mathrm h}.$$

A nonrelativistic electronic energy is typically of order $E_{\mathrm h}$, while leading fine-structure corrections are often of relative order $\alpha^2$ for light atoms. For a hydrogenic ion, the orbital velocity and correction parameter grow with $Z$:

$$\frac{v}{c}\sim Z\alpha, \qquad \frac{\Delta E_{\mathrm{fine}}}{E_{\mathrm{gross}}} \sim(Z\alpha)^2.$$

These estimates organize an expansion; they do not replace a state-specific calculation. Coefficients, cancellations, finite nuclear size, and electron correlation can control the actual uncertainty.

Nuclear Masses and Molecular Scales

Setting $m_e=1$ does not set every mass to one. Nuclear masses are expressed in electron-mass units. For a nucleus of mass $M$, the one-electron reduced mass is

$$\mu=\frac{m_eM}{m_e+M},$$

or, in atomic units,

$$\mu_{\mathrm{au}} =\frac{M_{\mathrm{au}}}{1+M_{\mathrm{au}}}.$$

The correction $1-\mu_{\mathrm{au}}$ is of order $m_e/M$. Molecular electronic energies remain naturally Hartree-scale, while nuclear vibration and rotation inherit the larger nuclear masses and become smaller. A common hierarchy is

$$E_{\mathrm{electronic}} \gg E_{\mathrm{vibrational}} \gg E_{\mathrm{rotational}},$$

but the ratios depend on the potential surface, equilibrium geometry, and isotopic masses. This scale separation motivates the Born–Oppenheimer approximation; it does not make nonadiabatic coupling identically zero.

Converting calculated quantities

For any quantity, write $Q=Q_{\mathrm{au}}Q_0$: construct the unit $Q_0$ from $a_0,E_{\mathrm h},e,m_e,\hbar$, then multiply by the numerical result. For example, dipole divided by field has unit $(ea_0)/F_0=4\pi\epsilon_0a_0^3$, which checks the polarizability conversion above. Keep extra digits only during intermediate calculations.

Example: energy in several spectroscopic units

For $\Delta E=0.0500E_{\mathrm h}$,

$$\begin{aligned} \Delta E&=1.36057\ \mathrm{eV},\\ \frac{\Delta E}{hc} &=10\,973.7\ \mathrm{cm}^{-1},\\ \frac{\Delta E}{h} &=328.984\ \mathrm{THz}. \end{aligned}$$

These intermediate values describe the same interval. Reporting three significant figures, consistent with $0.0500$, gives $1.36\ \mathrm{eV}$, $1.10\times10^4\ \mathrm{cm}^{-1}$, and $3.29\times10^2\ \mathrm{THz}$. The last is ordinary frequency because the conversion used $h$, not $\hbar$.

Example: laboratory electric field

A field of $10\ \mathrm{kV\,cm^{-1}}$ is $10^6\ \mathrm{V\,m^{-1}}$. Its atomic-unit value is

$$F_{\mathrm{au}} =\frac{10^6}{5.1422\times10^{11}} \simeq1.945\times10^{-6}.$$

This small dimensionless number can still produce strong mixing if the relevant opposite-parity level spacing is comparably small in Hartree.

Example: dipole matrix element

A calculated dipole magnitude $d=2.10$ a.u. corresponds to

$$\begin{aligned} d&=2.10ea_0,\\ d&\simeq5.34\ \mathrm D,\\ d&\simeq1.78\times10^{-29}\ \mathrm{C\,m}. \end{aligned}$$

The sign and phase of a matrix element depend on state conventions; transition probabilities use convention-invariant combinations such as squared magnitudes and angular sums.

Mass and charge conventions

The atomic unit of mass is the electron mass $m_e$, not the unified atomic mass unit $\mathrm u\simeq1822.9m_e$. Likewise, setting $e=1$ fixes the positive charge unit; the electron still has charge $-1$. These distinctions matter in nuclear kinetic energies and in electromagnetic coupling signs.

Exercises

Exercise 1: Nondimensionalize a many-electron atom

Starting from the fixed-nucleus Coulomb Hamiltonian for $N$ electrons, use $\mathbf r_i=a_0\boldsymbol\rho_i$ and divide by $E_{\mathrm h}$. Write the resulting Hamiltonian in Hartree atomic units.

Solution

The SI Hamiltonian is

$$H={}\sum_i\left[ -\frac{\hbar^2}{2m_e}\nabla_{r_i}^2 -\frac{Ze^2}{4\pi\epsilon_0r_i} \right] +\sum_{i<j} \frac{e^2}{4\pi\epsilon_0r_{ij}}.$$

Because $\nabla_{r_i}=a_0^{-1}\nabla_{\rho_i}$, each kinetic coefficient becomes $E_{\mathrm h}/2$. Each Coulomb factor $e^2/(4\pi\epsilon_0a_0)$ becomes $E_{\mathrm h}$. Therefore

$$\frac{H}{E_{\mathrm h}} ={}\sum_i\left[ -\frac12\nabla_{\rho_i}^2 -\frac{Z}{\rho_i} \right] +\sum_{i<j}\frac{1}{\rho_{ij}}.$$

After declaring atomic units, one normally renames $\boldsymbol\rho_i$ as $\mathbf r_i$ and writes the dimensionless operator without the explicit factor $H/E_{\mathrm h}$.

Exercise 2: Convert a 589 nm photon

For a vacuum wavelength $\lambda=589\ \mathrm{nm}$, find the ordinary frequency, spectroscopic wavenumber, photon energy in eV, and photon energy in Hartree. Use enough digits to show the conversions, but do not imply that the wavelength is known more precisely than stated.

Solution

Use

$$\nu=\frac{c}{\lambda}, \qquad \widetilde\nu=\frac{1}{\lambda}, \qquad E=\frac{hc}{\lambda}.$$

For $\lambda=589\ \mathrm{nm}$,

$$\begin{aligned} \nu&\simeq5.09\times10^{14}\ \mathrm{Hz},\\ \widetilde\nu&\simeq1.70\times10^4\ \mathrm{cm}^{-1},\\ E&\simeq2.10\ \mathrm{eV},\\ E&\simeq0.0774E_{\mathrm h}. \end{aligned}$$

Using unrounded intermediate values gives approximately $508.99\ \mathrm{THz}$, $16\,977.9\ \mathrm{cm}^{-1}$, $2.1050\ \mathrm{eV}$, and $0.077357E_{\mathrm h}$. Because the stated wavelength has three significant figures, the shorter rounded set is the defensible reported result.

Exercise 3: Field scale and interaction energy

A state has a dipole matrix element of one atomic unit. Estimate the interaction energy with a field of $1.00\ \mathrm{MV\,m^{-1}}$ in Hartree and eV. Is the field “weak” for every atomic transition?

Solution

The field in atomic units is

$$F_{\mathrm{au}} =\frac{1.00\times10^6}{5.1422\times10^{11}} \simeq1.945\times10^{-6}.$$

For $d=1$ a.u., the characteristic interaction energy is therefore

$$\begin{aligned} |dF|&\simeq1.945\times10^{-6}E_{\mathrm h},\\ |dF|&\simeq5.29\times10^{-5}\ \mathrm{eV}. \end{aligned}$$

This is small compared with a Hartree-scale gross-structure interval, but it may be large compared with hyperfine splittings, narrow avoided crossings, or separations between Rydberg states. “Weak” must be defined relative to the manifold and observable.

Exercise 4: Estimate the fine-structure scale

Using $\alpha^{-1}\simeq137.036$, estimate $\alpha^2E_{\mathrm h}$ in eV, $\mathrm{cm}^{-1}$, and ordinary frequency. Explain why this is not a prediction for a particular fine-structure interval.

Solution

The dimensionless factor is

$$\alpha^2\simeq5.325\times10^{-5}.$$

Multiplying the Hartree conversions gives

$$\begin{aligned} \alpha^2E_{\mathrm h} &\simeq1.45\times10^{-3}\ \mathrm{eV},\\ \frac{\alpha^2E_{\mathrm h}}{hc} &\simeq11.7\ \mathrm{cm}^{-1},\\ \frac{\alpha^2E_{\mathrm h}}{h} &\simeq350\ \mathrm{GHz}. \end{aligned}$$

This is only an order-of-magnitude scale for a relative $\alpha^2$ correction to a Hartree-scale energy. A particular interval contains powers of $Z$, quantum-number-dependent coefficients, angular factors, cancellations, reduced-mass effects, and possibly electron correlation.

References

• B. H. Bransden and C. J. Joachain, Physics of Atoms and Molecules, 2nd ed., Pearson, 2003.
• Bureau International des Poids et Mesures, The International System of Units (SI), 9th ed. (2019), updated 2026, DOI: 10.59161/AUEZ1291.
• C. J. Foot, Atomic Physics, Oxford University Press, 2005.
• T. Helgaker, P. Jørgensen, and J. Olsen, Molecular Electronic-Structure Theory, Wiley, 2000.
• I. N. Levine, Quantum Chemistry, 7th ed., Pearson, 2014.
• P. J. Mohr, D. B. Newell, B. N. Taylor, and E. Tiesinga, “CODATA recommended values of the fundamental physical constants: 2022,” Reviews of Modern Physics 97, 025002 (2025), DOI: 10.1103/RevModPhys.97.025002.
• E. Tiesinga, P. J. Mohr, D. B. Newell, and B. N. Taylor, The 2022 CODATA Recommended Values of the Fundamental Physical Constants, Web Version 9.0, National Institute of Standards and Technology, 2024.