# 17.4: Atomic Units

- Page ID
- 60581

The electronic Hamiltonian is expressed, in this Section, in so-called atomic units (aus)

\[ H_e = \sum\limits \left[ -\dfrac{1}{2}\nabla_j^2 - \sum\limits_a\dfrac{Z_a}{r_{j,a}} \right] + \sum\limits_{j<k}\dfrac{1}{r_{r,k}}. \]

These units are introduced to remove all \(\hbar \text{, e, and m}_e\) factors from the equations.

To effect this unit transformation, one notes that the kinetic energy operator scales as \(r_j^{-2}\) whereas the coulombic potentials scale as \(r_j^{-1}\) and as \(r_{j,k}^{-1}\). So, if each of the distances appearing in the cartesian coordinates of the electrons and nuclei were expressed as a unit of length \(a_0\) multiplied by a dimensionless length factor, the kinetic energy operator would involve terms of the form \(\left(-\frac{\hbar^2}{2(a_0)^2m_e} \right)\nabla_j^2 \) , and the coulombic potentials would appear as \( \frac{Z_ae^2}{(a_0)r_{j,a}} \text{ and } \frac{e^2}{(a_0)r_{j,k}} \). A factor of \(\frac{e^2}{a_0}\) (which has units of energy since \(a_0\) has units of length) can then be removed from the coulombic and kinetic energies, after which the kinetic energy terms appear as \( -\frac{\hbar^2}{2(e^2a_0)m_e}\nabla_j^2 \) and the potential energies appear as \( \frac{Z_a}{r_{j,a}} \text{ and } \frac{1}{r_{k,j}}\). Then, choosing \( a_0 = \frac{\hbar^2}{e^2m_e} \) changes the kinetic energy terms into \( -\frac{1}{2}\nabla_j^2 \); as a result, the entire electronic Hamiltonian takes the form given above in which no \(e^2, \text{ me, or } \hbar^2\) factors appear. The value of the so-called Bohr radius \(a_0 = \frac{\hbar^2}{e^2m_e}\) is 0.529 Å, and the so-called Hartree energy unit \( \frac{e^2}{a_0} \), which factors out of \(H_e\), is 27.21 eV or 627.51 kcal/mol.

## Contributors and Attributions

Jack Simons (Henry Eyring Scientist and Professor of Chemistry, U. Utah) Telluride Schools on Theoretical Chemistry and Jeff A. Nichols (Oak Ridge National Laboratory)