The Coulomb problem and atomic units · Quantum Mechanics

Intuition

An electron bound to a proton by the Coulomb force is the problem that made quantum mechanics believable: its answer matched a spectrum measured to many decimal places. Before solving it, find its scales. Squeezing the electron into a radius $a$ costs kinetic energy of order $\hbar^{2}/ma^{2}$ , and the attraction gains $e^{2}/a$ ; the balance falls at the Bohr radius, 0.053 nm, with an energy of $-13.6$ eV. In units built from these — atomic units — every hydrogen formula loses its constants.

A cat curls up just tight enough to stay warm and no tighter. The electron settles at the size where pulling in further would cost more kinetic energy than the attraction repays.

The energy $E(a)=\frac{\hbar^{2}}{2ma^{2}}-\frac{e^{2}}{4\pi\varepsilon_{0}a}$ of an electron confined to a size , in atomic units. Squeezing it raises the kinetic term; spreading it weakens the attraction. The minimum is at with hartree, eV.

The Coulomb problem and its units

An electron of mass $m_{e}$ in the field of a proton, taken as fixed. The constants $\hbar$ , $m_{e}$ and fix a length and an energy, and measured in those every formula is a pure number.

\begin{aligned}a_{0}&=\frac{4\pi\varepsilon_{0}\hbar^{2}}{m_{e}e^{2}}\approx0.0529\ \text{nm}\\ E_{h}&=\frac{e^{2}}{4\pi\varepsilon_{0}a_{0}}\approx27.2\ \text{eV}\end{aligned}

Atomic units

Atomic units set $\hbar=m_{e}=e^{2}/4\pi\varepsilon_{0}=1$ : lengths are measured in Bohr radii and energies in hartrees .

An estimate that is exact

\begin{aligned}&E(a)=\frac{\hbar^{2}}{2ma^{2}}-\frac{k}{a}\\ &\Longrightarrow\ a_{\min}=\frac{\hbar^{2}}{mk},\quad E_{\min}=-\frac{mk^{2}}{2\hbar^{2}}\end{aligned}

A state of size $a$ has momentum spread about $\hbar/a$ , so kinetic energy about $\hbar^{2}/2ma^{2}$ , and potential energy about with . Minimising the sum over gives the Bohr radius and eV. That this rough estimate lands exactly on the true ground state is a lucky feature of the Coulomb potential, confirmed by the exact solution.

Proof steps

E(a)=\frac{\hbar^{2}}{2ma^{2}}-\frac{k}{a},\qquad k=\frac{e^{2}}{4\pi\varepsilon_{0}}

Kinetic energy from a momentum of about $\hbar/a$ , potential energy at distance about $a$ .

E'(a)=-\frac{\hbar^{2}}{ma^{3}}+\frac{k}{a^{2}}=0

The minimum.

a_{\min}=\frac{\hbar^{2}}{mk}=a_{0}

The Bohr radius.

E_{\min}=\frac{\hbar^{2}}{2ma_{0}^{2}}-\frac{k}{a_{0}}=-\frac{k}{2a_{0}}=-\frac{mk^{2}}{2\hbar^{2}}

At the minimum the kinetic term is half the potential one in size.

E_{\min}=-\tfrac12E_{h}\approx-13.6\ \text{eV}

Exactly the ground-state energy the full solution will give.

Applications

Practice

The Bohr Radius

The natural size of the hydrogen atom is the Bohr radius, about 0.053 nanometres, where kinetic and potential energy balance.

a_{0}=\frac{4\pi\varepsilon_{0}\hbar^{2}}{m_{e}e^{2}}\approx0.0529\ \text{nm}

Try it

How many Bohr radii make one nanometre? Give the nearest whole number.