Atomic Units in Atomic and Molecular Calculations

Hartree atomic units express electron-scale lengths and energies using the Bohr radius $a_0$ and Hartree energy $E_{\mathrm h}$. A numerical value becomes a physical quantity through $X=X_{\rm au}X_0$, where $X_0$ is the unit for that quantity. Thus “0.03 a.u.” needs a label: energy, field, time, and dipole moment have different multipliers.

This page provides the lookup tables and unit-restoration examples. Atomic units and scales explains their origin in the Coulomb problem; Constants and conversions connects laboratory energy and frequency units.

Required background. Units and constants introduces dimensional analysis and numerical precision.

The numerical tables use the 2022 CODATA values from NIST, checked on 14 October 2026. Parentheses give standard uncertainty in the final digits. The SI Brochure defines which SI constants are exact; choosing atomic units does not make their measured SI conversion factors exact.

Hartree atomic units

Hartree atomic units set the numerical values

$$\hbar=m_e=e=4\pi\varepsilon_0=1.$$

This compact statement means:

• action is measured in units of $\hbar$;
• mass is measured in units of the electron mass $m_e$;
• charge is measured in units of the positive elementary-charge magnitude $e$; and
• the Coulomb constant $1/(4\pi\varepsilon_0)$ has numerical value one.

It does not mean that the physical constants cease to exist. They define the conversion between the dimensionless numerical representation and SI. Nor does it imply

$$\varepsilon_0=1.$$

Because the convention sets $4\pi\varepsilon_0=1$, the numerical value of $\varepsilon_0$ itself is $1/(4\pi)$ when it appears separately.

Signs and constants that remain visible

The symbol $e$ is a positive magnitude. The electron charge remains

$$q_e=-e=-1$$

in Hartree atomic units. Nuclear charges are $q_A=Z_A$, and nuclear masses are large dimensionless ratios $M_A/m_e$, not unity.

The speed of light is also not set to one:

$$c=\alpha^{-1}\approx137.036.$$

This makes the fine-structure constant $\alpha$ explicit as the parameter that separates the atomic velocity scale from $c$. A convention with $c=1$ must change at least one of the Hartree-unit choices above; it cannot set all five numerical constants to one while preserving the physical value of the dimensionless constant $\alpha$.

Atomic-unit conversion tables

The tables give multipliers for converting a numerical Hartree-atomic-unit value to SI:

$$X_{\mathrm{SI}} =X_{\mathrm{au}}\times \left(\text{SI value of }X_0\right).$$

To convert SI to atomic units, divide by the same multiplier. Parenthetical digits are standard uncertainties; displayed trailing digits should not be treated as the accuracy of a physical model.

Mechanical and kinematic quantities

Quantity	Atomic unit $X_0$	2022 CODATA SI value
charge	$e$	$1.602\,176\,634\times10^{-19}\ \mathrm C$ exact
mass	$m_e$	$9.109\,383\,7139(28)\times10^{-31}\ \mathrm{kg}$
action	$\hbar$	$1.054\,571\,817\ldots\times10^{-34}\ \mathrm{J\,s}$ derived exact
length	$a_0$	$5.291\,772\,105\,44(82)\times10^{-11}\ \mathrm m$
energy	$E_{\mathrm h}$	$4.359\,744\,722\,2060(48)\times10^{-18}\ \mathrm J$
time	$t_0=\hbar/E_{\mathrm h}$	$2.418\,884\,326\,5864(26)\times10^{-17}\ \mathrm s$
velocity	$a_0/t_0=\alpha c$	$2.187\,691\,262\,16(34)\times10^6\ \mathrm{m\,s^{-1}}$
momentum	$\hbar/a_0$	$1.992\,851\,915\,45(31)\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$
force	$E_{\mathrm h}/a_0$	$8.238\,723\,5038(13)\times10^{-8}\ \mathrm N$

Electromagnetic quantities

Quantity	Atomic unit $X_0$	2022 CODATA SI value
electric potential	$E_{\mathrm h}/e$	$27.211\,386\,245\,981(30)\ \mathrm V$
electric field	$E_{\mathrm h}/(ea_0)$	$5.142\,206\,751\,12(80)\times10^{11}\ \mathrm{V\,m^{-1}}$
electric-field gradient	$E_{\mathrm h}/(ea_0^2)$	$9.717\,362\,4424(30)\times10^{21}\ \mathrm{V\,m^{-2}}$
electric dipole moment	$ea_0$	$8.478\,353\,6198(13)\times10^{-30}\ \mathrm{C\,m}$
electric quadrupole moment	$ea_0^2$	$4.486\,551\,5185(14)\times10^{-40}\ \mathrm{C\,m^2}$
electric polarizability	$e^2a_0^2/E_{\mathrm h}$	$1.648\,777\,272\,12(51)\times10^{-41}\ \mathrm{C^2\,m^2\,J^{-1}}$
magnetic flux density	$\hbar/(ea_0^2)$	$2.350\,517\,570\,77(73)\times10^5\ \mathrm T$
magnetic dipole moment	$\hbar e/m_e=2\mu_{\mathrm B}$	$1.854\,802\,013\,15(58)\times10^{-23}\ \mathrm{J\,T^{-1}}$

The atomic unit of magnetic dipole moment is $2\mu_{\mathrm B}$, not $\mu_{\mathrm B}$. Many atomic-structure tables deliberately use the Bohr magneton instead. A value labeled only “a.u.” is therefore not enough to identify a magnetic-moment convention.

One-Hartree equivalents

Representation of $E_{\mathrm h}$	2022 CODATA value	Conversion used
joules	$4.359\,744\,722\,2060(48)\times10^{-18}\ \mathrm J$	direct
electronvolts	$27.211\,386\,245\,981(30)\ \mathrm{eV}$	$E_{\mathrm h}/e$
ordinary frequency	$6.579\,683\,920\,4999(72)\times10^{15}\ \mathrm{Hz}$	$E_{\mathrm h}/h$
angular frequency	$4.134\,137\,333\,517(5)\times10^{16}\ \mathrm{rad\,s^{-1}}$	$E_{\mathrm h}/\hbar$
vacuum wavenumber	$219\,474.631\,363\,14(24)\ \mathrm{cm^{-1}}$	$E_{\mathrm h}/(hc)$
energy-equivalent temperature	$315\,775.024\,803\,98(34)\ \mathrm K$	$E_{\mathrm h}/k_{\mathrm B}$

The last displayed digits in values derived here from rounded table entries are for lookup, not a replacement for machine-readable CODATA data. For precision metrology, retrieve the value and covariance information directly from NIST.

Recognizing the Hamiltonian convention

The coefficients in a Hamiltonian are often the fastest way to identify the unit convention.

One electron in a Coulomb field

For an infinitely heavy nucleus of charge $Z$, the Hartree-unit Hamiltonian is

$$H =-\frac12\nabla^2-\frac{Z}{r}.$$

The hydrogenic bound energies are

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

Thus hydrogen has $E_{1s}=-1/2$ in Hartree. The full derivation and wavefunction normalization live in Hydrogen Atom and Atomic Units and Scales.

Fixed-nucleus many-electron system

For nuclei at fixed positions $\mathbf R_A$, the electronic Hamiltonian in Hartree units is

$$H_{\mathrm{el}} ={}-\frac12\sum_i\nabla_i^2 -\sum_{i,A}\frac{Z_A}{r_{iA}} +\sum_{i<j}\frac{1}{r_{ij}} +\sum_{A<B}\frac{Z_AZ_B}{R_{AB}}.$$

Some electronic-structure programs omit the final nuclear-repulsion term from the electronic eigenvalue and add it when forming the total Born–Oppenheimer energy. Check whether the quoted energy includes this constant term before comparing calculations.

Coupling to an electric field

In the electric-dipole approximation,

$$H_{\mathrm{int}} =-\mathbf d\cdot\boldsymbol{\mathcal E}.$$

With $\mathbf d$ in $ea_0$ and $\boldsymbol{\mathcal E}$ in $E_{\mathrm h}/(ea_0)$, their product is automatically in Hartree:

$$(ea_0) \left(\frac{E_{\mathrm h}}{ea_0}\right) =E_{\mathrm h}.$$

The field may be a static value, a peak optical amplitude, a root-mean-square amplitude, or a complex positive-frequency amplitude. Atomic units do not remove that convention choice.

Atomic time, rates, and frequency

The time unit is $t_0=\hbar/E_{\mathrm h}$. With $H=E_{\mathrm h}H_{\rm au}$ and $t=t_0t_{\rm au}$, the evolution operator is

$$e^{-iHt/\hbar}=e^{-iH_{\rm au}t_{\rm au}}.$$

An energy gap of one Hartree therefore corresponds to angular frequency $E_{\mathrm h}/\hbar=t_0^{-1}$ and ordinary frequency $E_{\mathrm h}/h=(2\pi t_0)^{-1}$. A gap of $0.01E_{\mathrm h}$ accumulates one radian of relative phase in $100t_0$, and a full cycle in $200\pi t_0$.

Inverse time is also the unit of a decay rate. If a population obeys $P(t)=P(0)e^{-\Gamma_{\rm au}t_{\rm au}}$, its rate in inverse seconds is $\Gamma=\Gamma_{\rm au}/t_0$. There is no factor of $2\pi$ in this conversion. A factor of $2\pi$ enters when converting an angular frequency to cycles per second, or when a physical line-shape relation connects that decay rate to a linewidth in hertz.

Electric Fields, Dipoles, and Polarizabilities

Electric-field scale

The atomic field

$$\mathcal E_0 =\frac{E_{\mathrm h}}{ea_0} \approx5.1422\times10^{11}\ \mathrm{V\,m^{-1}}$$

changes the potential energy of one elementary charge by one Hartree over one Bohr radius. Laboratory fields are therefore often small as atomic-unit numbers:

$$10\ \mathrm{kV\,cm^{-1}} \approx1.9447\times10^{-6}\ \text{a.u.}$$

Small in atomic units does not automatically mean perturbatively weak. The relevant comparison is between the coupling matrix element and the energy separation of the states being mixed.

Dipole scale

The atomic dipole unit is

$$d_0=ea_0.$$

A calculated transition dipole $d_{\mathrm{au}}=2.0$ therefore means

$$\begin{aligned} d &=2.0ea_0,\\ &\approx5.0835\ \mathrm D,\\ &\approx1.6957\times10^{-29}\ \mathrm{C\,m}. \end{aligned}$$

The sign or phase of an off-diagonal dipole matrix element depends on state phase conventions. Observable line strengths use phase-invariant combinations, together with angular and degeneracy factors.

Polarizability convention

For a static scalar polarizability defined by

$$\Delta E =-\frac12\alpha_E\mathcal E^2,$$

the Hartree atomic unit is

$$\alpha_{E,0} =\frac{e^2a_0^2}{E_{\mathrm h}} =4\pi\varepsilon_0a_0^3.$$

The final identity explains a persistent notation trap. A tabulated polarizability “in $a_0^3$” usually reports the same numerical value as the Hartree atomic-unit polarizability, but the SI polarizability is not obtained by multiplying by a bare volume. It is obtained from

$$\alpha_E^{(\mathrm{SI})} =\alpha_{E,\mathrm{au}} \left(4\pi\varepsilon_0a_0^3\right).$$

If a source instead tabulates the polarizability volume $\alpha_E/(4\pi\varepsilon_0)$, the conversion multiplier is $a_0^3$. Record which quantity the table means.

Optical intensity is amplitude-dependent

For a plane wave in vacuum whose electric-field peak amplitude is $\mathcal E_{\mathrm{pk}}$,

$$I =\frac12c\varepsilon_0 \mathcal E_{\mathrm{pk}}^2.$$

If the quoted field is root-mean-square,

$$I =c\varepsilon_0 \mathcal E_{\mathrm{rms}}^2.$$

An “atomic unit of intensity” can therefore differ by a factor of two when the amplitude convention is omitted. Convert the explicitly defined field amplitude first, then apply the corresponding intensity formula.

Dimensional Restoration

Memorized tables are useful, but dimensional restoration handles quantities that are not listed.

Suppose a quantity $X$ has dimensions

$$[X] =M^aL^bT^cQ^d,$$

where $Q$ denotes electric-charge dimension. Its Hartree atomic unit is

$$X_0 =m_e^a a_0^b t_0^c e^d.$$

Then

$$X=X_{\mathrm{au}}X_0.$$

Examples follow directly:

Quantity	Dimensions	Atomic unit
area or cross section	$L^2$	$a_0^2$
volume	$L^3$	$a_0^3$
number density	$L^{-3}$	$a_0^{-3}$
rate	$T^{-1}$	$t_0^{-1}$
current	$QT^{-1}$	$e/t_0$
momentum	$MLT^{-1}$	$m_ea_0/t_0=\hbar/a_0$
force	$MLT^{-2}$	$m_ea_0/t_0^2=E_{\mathrm h}/a_0$
dipole moment	$QL$	$ea_0$

Wavefunctions and densities

A normalized three-dimensional wavefunction satisfies

$$\int \lvert\psi(\mathbf r)\rvert^2\,d^3r =1.$$

If $\mathbf r=a_0\mathbf r_{\mathrm{au}}$, then

$$\psi(\mathbf r) =a_0^{-3/2} \psi_{\mathrm{au}}(\mathbf r_{\mathrm{au}}).$$

Therefore a probability density reported in atomic units carries the scale $a_0^{-3}$. A radial function may use a different normalization and power of $a_0$; inspect its defining integral before restoring units.

Rates and cross sections

A decay rate $\Gamma_{\mathrm{au}}$ converts as

$$\Gamma_{\mathrm{SI}} =\frac{\Gamma_{\mathrm{au}}}{t_0}.$$

A cross section converts as

$$\sigma_{\mathrm{SI}} =\sigma_{\mathrm{au}}a_0^2.$$

Because

$$a_0^2 \approx2.8003\times10^{-17}\ \mathrm{cm^2},$$

a cross section of $10^{-4}$ a.u. is approximately $2.80\times10^{-21}\ \mathrm{cm^2}$. This conversion says nothing about whether the underlying scattering approximation is accurate.

Hartree and Rydberg Conventions

The abbreviation “a.u.” is sometimes used for two related but numerically different systems.

In Hartree units,

$$H_{\mathrm H} =-\frac12\nabla^2-\frac{Z}{r}, \qquad E_{1s}=-\frac12.$$

Dividing the same physical Hamiltonian by $E_{\mathrm{Ry}}=E_{\mathrm h}/2$ gives Rydberg units:

$$H_{\mathrm{Ry}} =-\nabla^2-\frac{2Z}{r}, \qquad E_{1s}=-1.$$

For a many-electron Coulomb Hamiltonian, every energy coefficient doubles under the same change of energy unit. A number expressed in Rydberg converts to Hartree through

$$E[\mathrm{Ha}] =\frac12E[\mathrm{Ry}].$$

The length coordinate in both displayed Hamiltonians is in bohr. Since only the energy unit was changed, all energy coefficients double together. If a program also rescales time or other quantities, consult its definitions instead of applying the energy factor to every number.

Nuclear Mass and Isotope Conventions

The atomic unit of mass is the electron mass:

$$m_0=m_e.$$

It is not the unified atomic mass unit $\mathrm u$. A nuclear mass supplied in unified atomic mass units must be converted to electron-mass units before entering a Hartree-unit Hamiltonian:

$$M_{\mathrm{au}} =\frac{M}{m_e}.$$

For a two-body Coulomb problem, the reduced mass is

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

or numerically in electron-mass units,

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

One may introduce problem-specific reduced-mass-scaled units, but those are not the CODATA bohr and Hartree unless explicitly declared. In standard Hartree units, isotope and recoil effects belong in the Hamiltonian rather than in a silent redefinition of $a_0$.

The same care applies to molecular coordinates. Electronic coordinates are often in bohr, nuclear masses in electron masses, vibrational frequencies in $E_{\mathrm h}/\hbar$, and final spectra in $\mathrm{cm}^{-1}$. Keep the quantity and unit attached to each number when moving between these representations.

Common pitfalls

The electron charge still has a sign. The unit $e$ is positive; the electron charge is $-e$. Nuclear masses remain ratios $M/m_e$, not one.

Magnetic-moment units need a label. The Hartree unit is $2\mu_B$; many atomic tables instead use $\mu_B$. Both are useful, but their numerical values differ by two for the same physical moment.

Converted digits do not improve a calculation. An energy given to three significant figures remains a three-significant-figure result after conversion with a longer CODATA multiplier. Retain extra digits during calculation and round at the end.

Exercises

Identify the energy convention

Two programs give the hydrogen ground-state energy as $-0.5$ and $-1.0$, respectively. Program A uses

$$H_A=-\frac12\nabla^2-\frac1r,$$

while program B uses

$$H_B=-\nabla^2-\frac2r.$$

Are the physical predictions inconsistent?

Solution

No. Program A divides physical energy by $E_{\mathrm h}$, while program B divides by $E_{\mathrm{Ry}}=E_{\mathrm h}/2$. Therefore

$$-0.5E_{\mathrm h} =-1.0E_{\mathrm{Ry}}.$$

The kinetic and Coulomb coefficients change together under the factor-of-two rescaling. A valid comparison first converts both outputs to the same unit and checks that their reference zeros and physical approximations also agree.

Convert an electronic energy gap

Convert $\Delta E=0.0200E_{\mathrm h}$ to eV, ordinary frequency in THz, and vacuum wavenumber in $\mathrm{cm}^{-1}$. State an appropriate number of significant figures.

Solution

Use the one-Hartree table:

$$\begin{aligned} \Delta E &=0.0200 \left(27.211\,386\,246\ \mathrm{eV}\right),\\ &=0.544\ \mathrm{eV},\\ \frac{\Delta E}{h} &=0.0200 \left(6579.6839\ \mathrm{THz}\right),\\ &=132\ \mathrm{THz},\\ \frac{\Delta E}{hc} &=0.0200 \left(219\,474.6314\ \mathrm{cm^{-1}}\right),\\ &=4.39\times10^3\ \mathrm{cm^{-1}}. \end{aligned}$$

The input has three significant figures, so defensible reported values are $0.544\ \mathrm{eV}$, $132\ \mathrm{THz}$, and $4.39\times10^3\ \mathrm{cm^{-1}}$. Extra CODATA digits belong in the intermediate calculation, not the final claim.

Restore a cross section

A scattering calculation reports

$$\sigma=6.0\times10^{-3}$$

in Hartree atomic units. Convert the cross section to $\mathrm{cm^2}$.

Solution

A cross section has dimensions of area, so its atomic unit is $a_0^2$:

$$\sigma =6.0\times10^{-3}a_0^2.$$

Using

$$a_0^2 \approx2.8003\times10^{-17}\ \mathrm{cm^2},$$

gives

$$\sigma \approx1.7\times10^{-19}\ \mathrm{cm^2}.$$

The two-significant-figure result reflects the precision of the input. No special cross-section conversion rule was required; dimensional restoration was enough.

Convert a polarizability

A database lists a static polarizability as $120$ a.u. Write the conversion to SI polarizability and to the polarizability volume in $a_0^3$. Why are these not the same dimensional statement?

Solution

For the Stark-shift definition

$$\Delta E=-\frac12\alpha_E\mathcal E^2,$$

the SI polarizability is

$$\begin{aligned} \alpha_E^{(\mathrm{SI})} &=120 \left( 1.648\,777\,272\,12\times10^{-41} \right) \times \mathrm{C^2\,m^2\,J^{-1}}, \\ &\approx1.98\times10^{-39} \ \mathrm{C^2\,m^2\,J^{-1}}. \end{aligned}$$

The corresponding polarizability volume is

$$\frac{\alpha_E}{4\pi\varepsilon_0} =120a_0^3.$$

The numerical value $120$ is shared because

$$\alpha_{E,0}=4\pi\varepsilon_0a_0^3,$$

but SI polarizability and volume have different dimensions. A source that prints only “$120a_0^3$” should state that it is reporting $\alpha_E/(4\pi\varepsilon_0)$.

References

• Bureau International des Poids et Mesures. The International System of Units (SI), 9th edition, updated 2026. doi:10.59161/AUEZ1291.
• Mohr, P. J., 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.
• National Institute of Standards and Technology. The 2022 CODATA Recommended Values of the Fundamental Physical Constants, web version 9.0 (2024). Database and machine-readable table, accessed 14 October 2026.