Skip to main content

# 17.4: Atomic Units

• • Contributed by Jack Simons
• Professor Emeritus and Henry Eyring Scientist (Chemistry) at University of Utah

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.