Hartree atomic units

Hartree atomic units are a system of natural units that is especially convenient for atomic physics calculations. They are named after Douglas Hartree. The system is designed to simplify the Schrödinger equation for hydrogen-like elements by setting the fundamental atomic properties to one. 

In Hartree atomic units:

1. The reduced Planck's constant (ħ = h/2π) is 1. It means the unit of angular momentum is the reduced Planck constant. 

2. The electron mass (me) is 1. This is the unit of mass.

3. The elementary charge (e) is 1. This is the unit of charge.

4. The Coulomb's constant (1/4πε0) is 1. This is the unit of electrostatic interaction.

5. The Hartree energy (Eh) is 1. This is the unit of energy. One Hartree is approximately 27.2114 electron volts (eV).

This leaves the system with a single fundamental dimension, which is length, since time can be expressed in terms of the length divided by the speed of light. The unit of length is the Bohr radius (a0), and this is approximately equal to 5.29177210903(80) x 10^-11 m. 

By setting these quantities to 1, atomic calculations are often significantly simplified, as these constants appear frequently in quantum mechanical equations involving atoms.

Possible issues

Will results obtained from Hartree atomic unit be consistent with SI unit?

One can immediately see a problem. Units of $\hbar$,$m_e$,$e$,$\frac{1}{4 \pi \epsilon_0}$ and $E_h$ is $\frac{M L^2}{T}$, $M$,$C$,$\frac{ML^3}{T^2C^2}$ and $\frac{ML^2}{T^2}$, respectively. So, in the atomic unit, there are 5 quantities claiming to have a numerical value of 1, while on the SI units, there are only 4 quantities- $M, L, T$ and $C$.

In other words, in atomic units,

$$\frac{\hbar}{m_e} = 1 = \frac{e^2}{4 \pi \epsilon_0 m_e }\sqrt{\frac{m_e}{E_h}}$$

But the question is, does this numerical equality also hold in SI units?

Surprise. 1 hartree, $E_h$, is defined to be

$$E_\mathrm{h} = m_\mathrm{e}\left(\frac{e^2}{4\pi\varepsilon_0\hbar}\right)^2 $$

So, in the above definition of the Hartree atomic unit, only 5 new quantities are required, one for each of M,L,T and C. In fact, wikipedia also tells us that

\[ E_\mathrm{h} = \frac{\hbar^2}{{m_\mathrm{e} a^2_0}} \]

So, even $a_0 = 1$ in atomic units.

Is $\omega$ of an oscillator its frequency or energy in Hartree atomic units??

Since $\hbar=1$, what does $\omega=1$ for a harmonic oscillator mean? Does it mean that the energy difference between the energy levels of the oscillator is 1 Hartree? Yes. Does it mean that the oscillator has a natural frequency of 1 $E_h/\hbar$? Yes.

The fact that in Hartree atomic units, the energy difference between two levels of a harmonic oscillator is numerically equal to its natural frequency should not be confusing, because in SI units, the energy difference between two levels of an oscillator is numerically about 1.05457182 × 10-34 times its natural frequency. One should not be more confusing than other, if at all.