Atomic Units

Atomic Units

The natural units for describing atomic systems are obtained by setting to unity the three fundamental constants that appear in the hydrogen Hamiltonian, $\hbar =m=e=1$. One thus arrives at atomic units, such as

• length: Bohr radius = $a_{0}={\frac {\hbar ^{2}}{me^{2}}}={\frac {1}{\alpha }}{\frac {\hbar }{mc}}=0.53\ \mathrm {\AA}$
• energy: 1 hartree = ${\frac {e^{4}m}{\hbar ^{2}}}=({\frac {e^{2}}{c\hbar }})^{2}mc^{2}=\alpha ^{2}mc^{2}=27.2\ {\textrm {eV}}$
• velocity: $mv^{2}={\frac {e^{4}m}{\hbar ^{2}}}\Rightarrow v={\frac {e^{2}}{\hbar }}=\alpha \cdot c=2.2\times 10^{8}\ {\textrm {cm/s}}$
• electric field: ${\frac {e}{a_{0}^{2}}}=5.142\times 10^{9}~{\rm {V/cm}}$

Note: This is the characteristic value for the $n=1$ orbit of hydrogen.

As we see above, we can express atomic units in terms of $c$ instead of $e$ by introducing a single dimensionless constant

$\alpha ={\frac {e^{2}}{\hbar c}}\approx {\frac {1}{137}}.$

The fine structure constant $\alpha$ obtained its name from the appearance of $\alpha ^{2}$ in the ratio of fine structure splitting to the Rydberg; it is the only fundamental constant in atomic physics. As such, it should ultimately be predicted by a complete theory of physics. Whereas precision measurements of other constants are made in atomic physics for purely metrological purposes , $\alpha$, as a dimensionless constant, is not defined by metrology. Rather, $\alpha$ characterizes the strength of the electromagnetic interaction, as the following example will illustrate. If energy uncertainties become become as large as $\Delta E=mc^{2}$, the concept of a particle breaks down. This upper bound on the energy uncertainty gives us, via the Heisenberg Uncertainty Principle, a lower bound on the length scale within which an electron can be localized (before e.g. spontaneous pair production may occur) $\Delta \simeq mc^{2}\Rightarrow \Delta p=mc$ $\Delta x={\frac {\hbar }{mc}}=\lambda _{c}$ Even at this short distance of $\lambda _{c}$, the Coulumb interaction---while stronger than that in hydrogen at distance $a_{0}$ --- is only:

$E_{c}={\frac {e^{2}}{\lambda _{c}}}={\frac {e^{2}mc}{\hbar }}={\frac {e^{2}}{\hbar c}}mc^{2}=\alpha mc^{2}\,,$

i.e. in relativistic units the strength of this "stronger" Coulomb interaction is $\alpha$. The fact that $\alpha ={\frac {1}{137}}$ implies that the Coulomb interaction is weak.