《Quantum Chemistry》章末总结9
隔了有半年了才接着写下一章XD，本章是多电子原子，算是量子化学正式开场了。
Chapter 9 Many-Electron Atoms
Helium is our first multielectron system, and although the helium atom may seem to be of minimal interest to chemists, we will discuss it in detail in this chapter because the solution of the helium atom illustrates the techniques used for more complex atoms.
Then, after discussing electron spin and the Pauli exclusion principle, we will discuss the Hartree-Fock theory of many-electron atoms.
Finally, we will discuss the term symbols of atoms and ions and how they are used to label electronic states.
9.1 Atomic and Molecular Calculations Are Expressed in Atomic Units
We will apply both perturbation theory and the variational method to a helium atom, but before doing so, we will introduce a system of units, called atomic units, that is widely used in atomic and molecular calculations to simplify the equations.
| Property | Atomic unit | SI equivalent |
|---|---|---|
| Mass | Mass of an electron,$m_e$ | $9.1094\times10^{-31}\ \rm kg$ |
| Charge | Charge on a proton,$e$ | $1.6022\times10^{-19}\ \rm C$ |
| Angular momentum | Planck constant divided by 2π,$\hbar$ | $1.0546\times 10^{-34}\ \rm J\cdot s$ |
| Length | Bohr radius,$a_0=\frac {4\pi\epsilon_0\hbar^2}{m_ee^2}$ | $5.2918\times 10^{-11}\ \rm m$ |
| Energy | $\frac {m_ee^4}{14\pi^2\epsilon_0^2\hbar^2} = \frac {e^2}{4\pi\epsilon_0a_0} = E_h$ | $4.3597\times 10^{-18}\ \rm J$ |
| Permittivity | $\kappa_0 = 4\pi\epsilon_0$ | $1.1127\times 10^{-10}\ \rm C^2\cdot J^{-1}\cdot m^{-1}$ |
（ 第一节就是讲原子单位，通过约定一些常量来简化算式。 ）
9.2 Both Perturbation Theory and the Variational Method Can Yield Good Results for a Helium Atom
We applied first-order perturbation theory to a helium atom in Section 8.5 and found
that the first-order correction to the energy is $5Z/8E_h$.The first-order perturbation theory gives a result that is approximately 5% in error.
Scheer and Knight calculated the energy through many orders of perturbation theory and found that $E=-2.9037E_h$, in good agreement with the experimental value of $-2.9033E_h$.
We also used the variational method to calculate the ground-state energy of a helium
atom in Section 8. 1, and $E_{min}=-2.84766E_h$.
The agreement we have found between first-order perturbation theory or our simple
variational approximation and the experimental value of the energy may appear to be
quite good, but let’s examine this agreement more closely. The ionization energy (IE)
of a helium atom is given by:
$$
IE = E_{He^{+}} - E_{He}
$$
The energy of $He^{+}$ is $-2E_h$, so we have:
$$
\begin{aligned}
IE &= -2 + \frac {11}{4} = 0.75E_h \quad (\rm first-order\ perturbation\ theory)\[3mm]
or \[3mm]
&= -2 + \left(\frac {27}{16} \right) = 0.8477E_h \quad (\rm our\ variational\ result)
\end{aligned}
$$
whereas the experimental value of the ionization energy is $0.9033E_h$. Even our variational result, with its 6% discrepancy with the experimental total energy, is not too satisfactory.
One approach is the following. The trial function assumes that both electrons have the same effective nuclear charge. This may be so in some average sense, but it will not be true at all times because there will be instants of time where one electron is far from the nucleus and the other close to it, and so the effective nuclear charges will not be the same.
We can account for this by using two variational parameters, $\zeta_1$ and $\zeta_2$. In order to treat the two electrons on an equal footing, we write $\psi(\bold {r_1}, \bold {r_2})$ as
$$
\psi(\bold {r_1}, \bold {r_2}) = N(e^{-\zeta_1r_1} e^{-\zeta_2r_2} + e^{-\zeta_2r_1} e^{-\zeta_1r_2})
$$
When $E(\zeta_1, \zeta_2)$ is minimized with respect to $\zeta_1$ and $\zeta_2$, we find that $E=-2.87566E_h$, which is a significant improvement over our simple variational treatment. The ionization energy comes out to be $0.8757E_h$.
we can do better yet. Because a suitable trial function may be almost any convenient function (that satisfies the boundary conditions), we are not restricted to using 1s hydrogen-like orbitals as we have done up to now. For example, we could use a linear combination of a 1s orbital and a 2s orbital and write:
$$
\psi(\bold {r_1}, \bold {r_2}) = \phi(\bold {r_1})\phi(\bold {r_2})
$$
where $\phi® = N[c_1e^{\zeta r}+c_2(2-r)e^{-\zeta r/2}]$ and $N$ is a normalization constant.
However, rather than using hydrogen-like orbitals, it’s customary to use a set of functions that were introduced by the American physicist John Slater in the 1930s. These functions, which are called Slaler orbitals, are of the form
$$
S_{nlm_l}(r, \theta, \phi) = N_{nl} r^{n-1} e^{-\zeta r} Y_l^{m_l}(\theta, \phi)
$$
where $N_{nl}$ is a normalization constant and the $Y_m^{m_l}(\theta, \phi)$ are the spherical harmonics. Note that the radial parts of Slater orbitals do not have nodes like hydrogen atomic orbitals do.
As we include more and more Slater orbitals, we reach a limit that is both practical
and theoretical. In this limit, $E = -2.8617E_h$ and the ionization energy is $0.8617E_h$. This limiting value is the best value of the energy that can be obtained using a trial function of the form of a product of one-electron wave functions. This limit is called the Hartree-Fock limit.
（ 由于试验函数可以任选，因此选择合适的试验函数是取得更高精度结果的一个重要因素，Hartree-Fock法的优点在于其不仅计算相对简单，而且与物理情景有着很好的对应。 ）