Most quantum chemists report the results of their calculations using atomic units.
The hydrogen-atom Hamiltonian operator (assuming infinite nuclear mass) in SI units is $-\left(\hbar^{2} / 2 m{e}\right) \nabla^{2}-e^{2} / 4 \pi \varepsilon{0} r$. The system of atomic units is defined as follows. The units of mass, charge, and angular momentum are defined as the electron's mass $m{e}$, the proton's charge $e$, and $\hbar$, respectively (rather than the kilogram, the coulomb, and the $\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}$ ); the unit of permittivity is $4 \pi \varepsilon{0}$, rather than the $\mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$. (The atomic unit of mass used in quantum chemistry should not be confused with the quantity 1 amu , which is one-twelfth the mass of a ${ }^{12} \mathrm{C}$ atom.) When we switch to atomic units, $\hbar, m{e}, e$, and $4 \pi \varepsilon{0}$ each have a numerical value of 1 . Hence, to change a formula from SI units to atomic units, we simply set each of these quantities equal to 1 . Thus, in SI atomic units, the H -atom Hamiltonian is $-\frac{1}{2} \nabla^{2}-1 / r$, where $r$ is now measured in atomic units of length rather than in meters. The ground-state energy of the hydrogen atom is given by (6.94) as $-\frac{1}{2}\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$. Since [Eq. (6.106)] $a{0}=4 \pi \varepsilon{0} \hbar^{2} / m{e} e^{2}$, the numerical value of $a{0}$ (the Bohr radius) in atomic units is 1 , and the ground-state energy of the hydrogen atom has the numerical value (neglecting nuclear motion) $-\frac{1}{2}$ in atomic units.
The atomic unit of energy, $e^{2} / 4 \pi \varepsilon{0} a{0}$, is called the hartree (symbol $E{\mathrm{h}}$ ):
1 hartree $\equiv E{\mathrm{h}} \equiv \frac{e^{2}}{4 \pi \varepsilon{0} a{0}}=\frac{m{e} e^{4}}{\left(4 \pi \varepsilon{0}\right)^{2} \hbar^{2}}=27.211385 \mathrm{eV}=4.359744 \times 10^{-18} \mathrm{~J}$
The ground-state energy of the hydrogen atom is $-\frac{1}{2}$ hartree if nuclear motion is neglected. The atomic unit of length is called the bohr:
\(
\begin{equation}
1 \text { bohr } \equiv a{0} \equiv 4 \pi \varepsilon{0} \hbar^{2} / m_{e} e^{2}=0.52917721 \AA \tag{13.30}
\end{equation}
\)
To find the atomic unit of any other quantity (for example, time) one combines $\hbar, m{e}, e$, and $4 \pi \varepsilon{0}$ so as to produce a quantity having the desired dimensions. One finds (Prob. 13.14) the atomic unit of time to be $\hbar / E{\mathrm{h}}=2.4188843 \times 10^{-17} \mathrm{~s}$ and the atomic unit of electric dipole moment to be $e a{0}=8.478353 \times 10^{-30} \mathrm{C} \mathrm{m}$.
Atomic units will be used in Chapters 13 to 17.
A more rigorous way to define atomic units is as follows. Starting with the H -atom electronic Schrödinger equation in SI units, we define (as in Section 4.4) the dimensionless reduced variables $E{r} \equiv E / A$ and $r{r} \equiv r / B$, where $A$ and $B$ are products of powers of the Schrödinger-equation constants $\hbar, m{e}, e$, and $4 \pi \varepsilon{0}$ such that $A$ and $B$ have dimensions of energy and length, respectively. The procedure of Section 4.4 shows that (Prob. 13.13) $A=m{e} e^{4} /\left(4 \pi \varepsilon{0}\right)^{2} \hbar^{2}=e^{2} / 4 \pi \varepsilon{0} a{0} \equiv 1$ hartree and $B=\hbar^{2} 4 \pi \varepsilon{0} / m{e} e^{2}=a{0} \equiv 1$ bohr. For this three-dimensional problem, the H-atom wave function has dimensions of $\mathrm{L}^{-3 / 2}$, so the reduced dimensionless $\psi{r}$ is defined as $\psi_{r} \equiv \psi B^{3 / 2}$. Also,
\(
\frac{\partial^{2} \psi{r}}{\partial r{r}^{2}}=B^{3 / 2} \frac{\partial^{2} \psi}{\partial r^{2}}\left(\frac{\partial r}{\partial r{r}}\right)^{2}=B^{3 / 2} \frac{\partial^{2} \psi}{\partial r^{2}} B^{2}=B^{3 / 2} \frac{\partial^{2} \psi}{\partial r^{2}} a{0}^{2}
\)
Introducing the reduced quantities into the Schrödinger equation, we find (Prob. 13.13) that the reduced H -atom Schrödinger equation is $-\frac{1}{2} \nabla{r}^{2} \psi{r}-\left(1 / r{r}\right) \psi{r}=E{r} \psi{r}$, where $\nabla{r}^{2}$ is given by Eq. (6.6) with $r$ replaced by $r{r}$. In practice, people do not bother to include the $r$ subscripts and instead write $-\frac{1}{2} \nabla^{2} \psi-(1 / r) \psi=E \psi$.