Chapter 5

Many-Electron Atoms

5.1Introduction

When a second electron is added to hydrogen, the problem ceases to be solvable in closed form. This is not merely a matter of difficulty: the situation is the same as that of the three-body problem in classical mechanics, and no exact expression exists for the ground-state energy of helium.

The character of the subject therefore changes at this point. From here on we work with controlled approximations, and the essential skill is to judge how good each one is. Helium is a convenient case in which to develop that skill. Three successively better treatments, each requiring about a page of algebra, bring us to within two percent of the true answer, and each of them introduces an idea that is used repeatedly later.

Four results are established in this chapter:

Antisymmetry, from which the Pauli principle follows as a theorem rather than a postulate;
Screening, which we compute rather than assume, and which breaks hydrogen's accidental $l$ -degeneracy and produces the quantum defect;
Exchange, an energy splitting of nearly an electron volt that arises from antisymmetry alone and has no classical analogue;
Term symbols, the notation $^{2 S + 1} L_{J}$ in which tables of atomic data are written.

The chapter closes with a state that can decay without absorbing radiation, namely a bound state degenerate with a continuum. Such states are the subject of the final chapters of this book, and helium provides the simplest example of one.

5.2Identical Particles and Antisymmetry

Electrons are indistinguishable, in the sense that no measurement can determine which is which. Formally, the exchange operator ${\hat{P}}_{12}$ that swaps particles $1$ and $2$ must commute with the Hamiltonian, and since ${\hat{P}}_{12}^{2} = \hat{I}$ its eigenvalues are $\pm 1$ . Fermions, and electrons among them, take the eigenvalue $- 1$ .

Definition 5.2.1Antisymmetry postulate

The state of a system of identical fermions changes sign under exchange of any two of them:

\begin{matrix} Ψ (\dots, ξ_{i}, \dots, ξ_{j}, \dots) = - Ψ (\dots, ξ_{j}, \dots, ξ_{i}, \dots), \end{matrix}

where $ξ = (\vec{r}, m_{s})$ collects position and spin.

Theorem 5.2.1Pauli exclusion principle

No two identical fermions can occupy the same one-particle spin-orbital.

Setting $ξ_{i} = ξ_{j}$ in Eq. (5.1) makes the left- and right-hand sides the same function, so that $Ψ = - Ψ$ and therefore $Ψ = 0$ . A state with two electrons in the same spin-orbital does not exist.

The exclusion principle is thus not an additional rule imposed on quantum mechanics; it is a two-line consequence of antisymmetry.

Definition 5.2.2Slater determinant

For $N$ electrons occupying spin-orbitals $ϕ_{1}, \dots, ϕ_{N}$ , the normalized antisymmetric state is

\begin{matrix} Ψ (ξ_{1}, \dots, ξ_{N}) = \frac{1}{\sqrt{N!}} | \begin{matrix} ϕ_{1} (ξ_{1}) & ϕ_{2} (ξ_{1}) & \dots & ϕ_{N} (ξ_{1}) \\ ϕ_{1} (ξ_{2}) & ϕ_{2} (ξ_{2}) & \dots & ϕ_{N} (ξ_{2}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ ϕ_{1} (ξ_{N}) & ϕ_{2} (ξ_{N}) & \dots & ϕ_{N} (ξ_{N}) \end{matrix} | . \end{matrix}

This construction works because a determinant changes sign when two rows are interchanged, which is Eq. (5.1), and vanishes when two columns are equal, which is Theorem 5.2.1.

Example 5.2.1The two-electron determinant

Write out Eq. (5.2) for $N = 2$ and verify both properties.

Solution.

For two electrons,

Ψ (ξ_{1}, ξ_{2}) = \frac{1}{\sqrt{2}} | \begin{matrix} ϕ_{a} (ξ_{1}) & ϕ_{b} (ξ_{1}) \\ ϕ_{a} (ξ_{2}) & ϕ_{b} (ξ_{2}) \end{matrix} | = \frac{1}{\sqrt{2}} [ϕ_{a} (ξ_{1}) ϕ_{b} (ξ_{2}) - ϕ_{b} (ξ_{1}) ϕ_{a} (ξ_{2})] .

Antisymmetry. Swapping $ξ_{1} \leftrightarrow ξ_{2}$ gives

\frac{1}{\sqrt{2}} [ϕ_{a} (ξ_{2}) ϕ_{b} (ξ_{1}) - ϕ_{b} (ξ_{2}) ϕ_{a} (ξ_{1})] = - Ψ (ξ_{1}, ξ_{2}) . ✓

Exclusion. Setting $ϕ_{b} = ϕ_{a}$ ,

Ψ = \frac{1}{\sqrt{2}} [ϕ_{a} (ξ_{1}) ϕ_{a} (ξ_{2}) - ϕ_{a} (ξ_{1}) ϕ_{a} (ξ_{2})] = 0. ✓

Normalization. If $ϕ_{a}$ and $ϕ_{b}$ are orthonormal, then expanding $\int | Ψ |^{2}$ gives four terms: two of the form $\int | ϕ_{a} |^{2} \int | ϕ_{b} |^{2} = 1$ , each with prefactor $\frac{1}{2}$ , and two cross terms proportional to $⟨ ϕ_{a} | ϕ_{b} ⟩ = 0$ . Hence $\int | Ψ |^{2} = 1$ , which is the reason for the prefactor $1 / \sqrt{N!}$ .

5.3Helium: The Hamiltonian

Helium consists of two electrons and a nucleus of charge $Z e$ with $Z = 2$ . Working in atomic units (Chapter 4, Sec. 4.8), the Hamiltonian is

\begin{matrix} \hat{H} = \underset{{\hat{h}}_{1}}{\underset{⏟}{- \frac{1}{2} \nabla_{1}^{2} - \frac{Z}{r_{1}}}} + \underset{{\hat{h}}_{2}}{\underset{⏟}{- \frac{1}{2} \nabla_{2}^{2} - \frac{Z}{r_{2}}}} + \underset{{\hat{V}}_{e e}}{\underset{⏟}{\frac{1}{r_{12}}}} \end{matrix}

with $r_{12} = | {\vec{r}}_{1} - {\vec{r}}_{2} |$ . In SI units the last term is $e^{2} / 4 π ϵ_{0} r_{12}$ .

Remark 5.3.1Why this cannot be separated

The operators ${\hat{h}}_{1}$ and ${\hat{h}}_{2}$ each involve one electron only, so that if ${\hat{V}}_{e e}$ were absent the equation would separate and the solution would be a product of two hydrogenic orbitals with $Z = 2$ . The term $1 / r_{12}$ couples the coordinates. It cannot be written as $f ({\vec{r}}_{1}) + g ({\vec{r}}_{2})$ , no change of variables removes it, and consequently no separable solution exists. This single term is the source of essentially all the difficulty in many-electron atoms.

The experimental values to be reproduced are

\begin{matrix} E_{exact} = - 2.9037 a.u. = - 79.01 eV, first ionization energy = 24.59 eV . \end{matrix}

5.4First Approximation: Independent Electrons

We begin by dropping ${\hat{V}}_{e e}$ entirely. The equation then separates, and each electron occupies a hydrogenic $1 s$ orbital with $Z = 2$ ,

\begin{matrix} ϕ_{1 s} (\vec{r}) = \sqrt{\frac{Z^{3}}{π}} e^{- Z r}, \end{matrix}

which is $R_{10} Y_{0}^{0}$ from Chapter 4, Eqs. (4.51) and (4.31), collected into a single expression.

Example 5.4.1Zeroth-order energy of helium

Compute the ground-state energy of helium neglecting the electron–electron repulsion.

Solution.

Each electron has the hydrogenic energy of Eq. (4.47) with $n = 1$ ,

ε_{1 s} = - \frac{Z^{2}}{2} = - \frac{4}{2} = - 2 a.u.,

so that for two independent electrons

E^{(0)} = 2 \times (- 2) = - 4 a.u. = - 108.8 eV .

Compared with the true value of $- 79.0$ eV this is too low by $29.8$ eV, an error of $38 %$ . The sign of the error is instructive. We have omitted a repulsion, and a repulsion raises the energy, so neglecting it must give a value that is too negative. An approximation whose error has the wrong sign therefore contains a mistake.

5.5Second Approximation: First-Order Perturbation Theory

We now treat ${\hat{V}}_{e e} = 1 / r_{12}$ as a perturbation and add its expectation value in the zeroth-order state:

\begin{matrix} E^{(1)} = ⟨ ϕ_{1 s} ϕ_{1 s} | \frac{1}{r_{12}} | ϕ_{1 s} ϕ_{1 s} ⟩ = \int \int \frac{| ϕ_{1 s} ({\vec{r}}_{1}) |^{2} | ϕ_{1 s} ({\vec{r}}_{2}) |^{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |} d^{3} r_{1} d^{3} r_{2} . \end{matrix}

This is a six-dimensional integral, but it can be evaluated exactly, and the method is worth following in detail because the same device is used again later.

Theorem 5.5.1The Coulomb integral for

1 s

orbitals

For the normalized hydrogenic $1 s$ orbital of Eq. (5.5),

\begin{matrix} ⟨ \frac{1}{r_{12}} ⟩ = \frac{5 Z}{8} \end{matrix}

The charge density of electron $2$ , $ρ (r) = | ϕ_{1 s} (\vec{r}) |^{2} = (Z^{3} / π) e^{- 2 Z r}$ , is spherically symmetric. By the classical shell theorem, the potential it produces at radius $R$ separates into a contribution from the charge inside, which acts as if concentrated at the origin, and one from the shells outside, each of which contributes $1 / r$ :

V (R) = \frac{1}{R} \int_{0}^{R} ρ (r) 4 π r^{2} d r + \int_{R}^{\infty} ρ (r) 4 π r d r .

Inner integral. With $ρ 4 π r^{2} = 4 Z^{3} r^{2} e^{- 2 Z r}$ and the standard result $\int_{0}^{R} r^{2} e^{- a r} d r = \frac{2}{a^{3}} - (\frac{R^{2}}{a} + \frac{2 R}{a^{2}} + \frac{2}{a^{3}}) e^{- a R}$ with $a = 2 Z$ ,

\int_{0}^{R} ρ 4 π r^{2} d r = 1 - (2 Z^{2} R^{2} + 2 Z R + 1) e^{- 2 Z R} .

Outer integral. With $\int_{R}^{\infty} r e^{- a r} d r = (\frac{R}{a} + \frac{1}{a^{2}}) e^{- a R}$ ,

\int_{R}^{\infty} ρ 4 π r d r = 4 Z^{3} (\frac{R}{2 Z} + \frac{1}{4 Z^{2}}) e^{- 2 Z R} = (2 Z^{2} R + Z) e^{- 2 Z R} .

Combining. Dividing the first by $R$ and adding,

V (R) = \frac{1}{R} - (2 Z^{2} R + 2 Z + \frac{1}{R}) e^{- 2 Z R} + (2 Z^{2} R + Z) e^{- 2 Z R} = \frac{1}{R} - (Z + \frac{1}{R}) e^{- 2 Z R} .

The $2 Z^{2} R$ terms cancel, which provides a useful check.

Averaging over electron 1. We then have

⟨ \frac{1}{r_{12}} ⟩ = \int ρ (R) V (R) 4 π R^{2} d R = 4 Z^{3} \int_{0}^{\infty} R^{2} e^{- 2 Z R} [\frac{1}{R} - (Z + \frac{1}{R}) e^{- 2 Z R}] d R .

This splits into three elementary integrals, evaluated with $\int_{0}^{\infty} r^{n} e^{- a r} d r = n! / a^{n + 1}$ :

\int_{0}^{\infty} R e^{- 2 Z R} d R = \frac{1}{4 Z^{2}}, Z \int_{0}^{\infty} R^{2} e^{- 4 Z R} d R = \frac{2 Z}{64 Z^{3}} = \frac{1}{32 Z^{2}}, \int_{0}^{\infty} R e^{- 4 Z R} d R = \frac{1}{16 Z^{2}} .

Therefore

⟨ \frac{1}{r_{12}} ⟩ = 4 Z^{3} [\frac{1}{4 Z^{2}} - \frac{1}{32 Z^{2}} - \frac{1}{16 Z^{2}}] = 4 Z^{3} \cdot \frac{8 - 1 - 2}{32 Z^{2}} = 4 Z^{3} \cdot \frac{5}{32 Z^{2}} = \frac{5 Z}{8} .

Example 5.5.1First-order energy of helium

Use Theorem 5.5.1 to correct Example 5.4.1.

Solution.

For $Z = 2$ ,

E^{(1)} = \frac{5 Z}{8} = \frac{10}{8} = 1.25 a.u. = 34.0 eV,

so that

E ≃ E^{(0)} + E^{(1)} = - 4 + 1.25 = - 2.75 a.u. = - 74.8 eV .

The estimate has moved from $38 %$ too low to $5.3 %$ too high. This overshoot is to be expected, since the orbitals were not allowed to respond to the repulsion. Each electron is therefore held closer to the nucleus than it actually is, and experiences more repulsion than it actually does. Allowing the orbitals to relax is the subject of the next approximation.

5.6Third Approximation: The Variational Method

Theorem 5.6.1Variational principle

For any normalized trial state $| Ψ_{t} ⟩$ and any Hamiltonian $\hat{H}$ with ground-state energy $E_{0}$ ,

\begin{matrix} ⟨ Ψ_{t} | \hat{H} | Ψ_{t} ⟩ \geq E_{0} . \end{matrix}

Expand $| Ψ_{t} ⟩ = \sum_{n} c_{n} | n ⟩$ in the exact eigenstates, with $\hat{H} | n ⟩ = E_{n} | n ⟩$ and $\sum_{n} | c_{n} |^{2} = 1$ . Then

⟨ Ψ_{t} | \hat{H} | Ψ_{t} ⟩ = \sum_{n} | c_{n} |^{2} E_{n} \geq \sum_{n} | c_{n} |^{2} E_{0} = E_{0},

since every $E_{n} \geq E_{0}$ .

The principle converts approximation into optimization: we choose a trial state containing adjustable parameters, compute $⟨ \hat{H} ⟩$ , and minimize the result. The answer obtained is guaranteed to be an upper bound, which is a useful property in practice, since a calculation that returns an energy below the exact value must contain an error.

The physical shortcoming of Sec. 5.5 was that each electron still experienced the full nuclear charge $Z = 2$ , although its partner partly shields it. We therefore allow the orbital exponent to be a free parameter $Z^{'}$ :

\begin{matrix} ϕ_{Z^{'}} (\vec{r}) = \sqrt{\frac{Z^{' 3}}{π}} e^{- Z^{'} r}, Ψ_{t} ({\vec{r}}_{1}, {\vec{r}}_{2}) = ϕ_{Z^{'}} ({\vec{r}}_{1}) ϕ_{Z^{'}} ({\vec{r}}_{2}) . \end{matrix}

Example 5.6.1The variational energy of helium

Minimize $⟨ \hat{H} ⟩$ over $Z^{'}$ and evaluate the result for $Z = 2$ .

Solution.

Three expectation values are needed. For a hydrogenic orbital with exponent $Z^{'}$ , Chapter 4's results Eq. (4.58) give $⟨ 1 / r ⟩ = Z^{'}$ , and the virial relation for that orbital gives $⟨ - \frac{1}{2} \nabla^{2} ⟩ = Z^{' 2} / 2$ .

Kinetic energy, two electrons:

⟨ \hat{T} ⟩ = 2 \times \frac{Z^{' 2}}{2} = Z^{' 2} .

Nuclear attraction, two electrons, with the true charge $Z$ in the operator but the trial exponent $Z^{'}$ in the state:

⟨ {\hat{V}}_{N e} ⟩ = 2 \times (- Z ⟨ \frac{1}{r} ⟩) = - 2 Z Z^{'} .

Electron repulsion. Theorem 5.5.1 was derived for orbitals of exponent $Z$ ; here the exponent is $Z^{'}$ , so that

⟨ {\hat{V}}_{e e} ⟩ = \frac{5 Z^{'}}{8} .

Adding these,

\begin{matrix} E (Z^{'}) = Z^{' 2} - 2 Z Z^{'} + \frac{5 Z^{'}}{8} . \end{matrix}

Minimizing,

\frac{d E}{d Z^{'}} = 2 Z^{'} - 2 Z + \frac{5}{8} = 0 ⟹ Z^{'} = Z - \frac{5}{16}

Substituting back, and noting that $2 Z - \frac{5}{8} = 2 Z^{'}$ ,

E_{min} = Z^{' 2} - Z^{'} (2 Z - \frac{5}{8}) = Z^{' 2} - 2 Z^{' 2} = - Z^{' 2} .

For helium, $Z = 2$ :

Z^{'} = 2 - \frac{5}{16} = \frac{27}{16} = 1.6875, E_{min} = - {(\frac{27}{16})}^{2} = - \frac{729}{256} = - 2.8477 a.u. = - 77.49 eV .

The error is now $1.9 %$ , and, as Theorem 5.6.1 requires, the result lies above the exact value of $- 79.01$ eV.

The intermediate result $Z^{'} = 1.6875$ carries the physics of the calculation. An electron in helium does not experience the full charge $2$ ; it experiences $1.69$ , its partner having screened away about $0.31$ of an elementary charge. This idea, that the nuclear charge an electron actually experiences is smaller than $Z$ , organizes the remainder of the chapter.

Remark

The three treatments give, in order, $- 108.8$ , $- 74.8$ and $- 77.5$ eV, against the true value of $- 79.0$ eV. This pattern is characteristic. A crude model can be badly wrong; first-order perturbation theory usually overcorrects; and allowing the wave function itself to relax recovers most of the remaining discrepancy. Going further requires trial functions that depend on $r_{12}$ explicitly, which is where quantum chemistry proper begins and where this course stops.

5.7Exchange: Singlet and Triplet Helium

We now excite one electron, to the configuration $1 s 2 s$ , and denote the two spatial orbitals by $ϕ_{a} = ϕ_{1 s}$ and $ϕ_{b} = ϕ_{2 s}$ . Antisymmetry of the total state can be achieved in two ways, because the state factorizes into a spatial part and a spin part, and it is sufficient that one of the two be antisymmetric.

\begin{matrix} \begin{aligned} singlet (S = 0) : & Ψ_{+} = \underset{spatially symmetric}{\underset{⏟}{\frac{1}{\sqrt{2}} [ϕ_{a} (1) ϕ_{b} (2) + ϕ_{b} (1) ϕ_{a} (2)]}} \times \underset{spin antisymmetric}{\underset{⏟}{\frac{1}{\sqrt{2}} [| ↑↓ ⟩ - | ↓↑ ⟩]}}, \\ triplet (S = 1) : & Ψ_{-} = \underset{spatially antisymmetric}{\underset{⏟}{\frac{1}{\sqrt{2}} [ϕ_{a} (1) ϕ_{b} (2) - ϕ_{b} (1) ϕ_{a} (2)]}} \times \underset{spin symmetric}{\underset{⏟}{| ↑↑ ⟩ etc.}} . \end{aligned} \end{matrix}

Both possibilities are physically admissible, and their energies differ. The calculation below shows precisely why.

Definition 5.7.1Direct and exchange integrals

\begin{aligned} J & = \int \int \frac{| ϕ_{a} ({\vec{r}}_{1}) |^{2} | ϕ_{b} ({\vec{r}}_{2}) |^{2}}{r_{12}} d^{3} r_{1} d^{3} r_{2} & (direct, or Coulomb, integral), \\ K & = \int \int \frac{ϕ_{a}^{*} ({\vec{r}}_{1}) ϕ_{b}^{*} ({\vec{r}}_{2}) ϕ_{b} ({\vec{r}}_{1}) ϕ_{a} ({\vec{r}}_{2})}{r_{12}} d^{3} r_{1} d^{3} r_{2} & (exchange integral) . \end{aligned}

The integral $J$ admits a classical interpretation: it is the electrostatic repulsion between two charge clouds $| ϕ_{a} |^{2}$ and $| ϕ_{b} |^{2}$ . The integral $K$ admits none, since the orbitals are interchanged between the two factors and no classical charge distribution corresponds to such an expression. It arises only because the wave function must be antisymmetrized.

Theorem 5.7.1Singlet–triplet splitting

For the states of Eq. (5.11),

\begin{matrix} E_{\pm} = ε_{a} + ε_{b} + J \pm K, \end{matrix}

with $+$ for the singlet and $-$ for the triplet. Since $K > 0$ for orbitals with overlapping densities, the triplet lies lower.

The one-electron parts give $ε_{a} + ε_{b}$ regardless of the symmetrization. For the repulsion, we insert Eq. (5.11) into $⟨ Ψ_{\pm} | 1 / r_{12} | Ψ_{\pm} ⟩$ and expand the product of the two brackets. Four terms result. Two of them have the form $\int \int | ϕ_{a} (1) |^{2} | ϕ_{b} (2) |^{2} / r_{12}$ , each with prefactor $\frac{1}{2}$ , and together give $J$ . The remaining two are the exchange terms, each with prefactor $\pm \frac{1}{2}$ , and together give $\pm K$ .

Remark 5.7.1The origin of the splitting

The Hamiltonian (5.3) contains no spin operator, yet the two states differ in energy by $2 K$ and are distinguished only by their spin arrangement. The spin does not cause the splitting; antisymmetry does. The requirement that the total state change sign under exchange ties the spin arrangement to the spatial arrangement, and it is the spatial arrangement to which the Coulomb repulsion is sensitive.

The mechanism can be seen directly in Eq. (5.11). Setting ${\vec{r}}_{1} = {\vec{r}}_{2}$ makes the antisymmetric spatial function vanish identically, so that the triplet electrons avoid one another, are further apart on average, and repel less strongly. This is the exchange interaction, and it has no classical counterpart.

Example 5.7.1Helium's measured splitting

The $1 s 2 s$ levels of helium are observed at $19.820$ eV ( $^{3} S$ ) and $20.616$ eV ( $^{1} S$ ) above the ground state. Extract $K$ and comment on its magnitude.

Solution.

From Eq. (5.14) the separation is $2 K$ :

2 K = 20.616 - 19.820 = 0.796 eV ⟹ K = 0.398 eV .

The triplet lies lower, which confirms that $K > 0$ .

It is instructive to compare this with the magnetic interaction between two electron spins, which at a separation of a few $a_{0}$ is of order $10^{- 4}$ eV, four orders of magnitude smaller. The splitting is therefore not magnetic in origin. It is electrostatic, acting through antisymmetry.

The consequence is chemical as well as spectroscopic. The rule that electrons in different orbitals favor parallel spins, known as Hund's first rule, is this same $- K$ , and it is the reason that the ground state of oxygen is a triplet.

Remark

The two families were at one time thought to be different substances and were named parahelium (singlet) and orthohelium (triplet); the names have survived. We note also that $1 s 2 s^{3} S$ cannot reach the $1 s^{2}^{1} S$ ground state by emission of a single photon, since the transition would have to change $S$ , which the dipole operator of Chapter 3 cannot do. The state is therefore metastable, with a lifetime of about two hours. This may be compared with hydrogen's $2 s$ state in Example 3.6.1, which is a second instance of a forbidden route producing a long-lived state.

5.8Heavier Atoms: Screening and the Quantum Defect

For atoms with more than two electrons, the practical approach is to retain the one-electron picture and to modify the potential. Each electron is taken to move in a central average field produced by the nucleus together with all the other electrons, so that the machinery of Chapter 3 applies unchanged: the states are labeled $n l m$ , and Eq. (3.43) is solved with a modified $V (r)$ .

The essential feature of this field is its behavior at the two extremes:

\begin{matrix} V_{eff} (r) ⟶ {\begin{cases} - Z / r, & r \to 0 (inside the other electrons: full nuclear charge), \\ - 1 / r, & r \to \infty (outside them: nucleus screened to + 1) . \end{cases} \end{matrix}

An electron far from the nucleus experiences a net charge of $+ 1$ and is hydrogenic, whereas one that penetrates the core experiences a considerably larger charge and is correspondingly more tightly bound. Chapter 3 has already established which orbitals penetrate: the centrifugal barrier $l (l + 1) / 2 r^{2}$ in Eq. (3.43) excludes states of high $l$ from the core entirely, while $l = 0$ has no barrier at all. It follows that

\begin{matrix} penetration: s > p > d > f ⟹ extra binding relative to hydrogen: s > p > d > f . \end{matrix}

The accidental $l$ -degeneracy of Chapter 4 is therefore broken, and in a definite order. The effect is described by a single number for each value of $l$ .

Definition 5.8.1Quantum defect and effective quantum number

The energies of a series of states of the same $l$ converging to an ionization threshold $E_{th}$ are written

\begin{matrix} E_{n} = E_{th} - \frac{ℰ_{Ryd}}{(n - δ_{l})^{2}} = E_{th} - \frac{ℰ_{Ryd}}{(n^{*})^{2}}, \end{matrix}

with $ℰ_{Ryd} = 13.606$ eV. The number $δ_{l}$ is the quantum defect and $n^{*} \equiv n - δ_{l}$ the effective quantum number.

Two properties make this description useful. First, $δ_{l}$ is nearly independent of $n$ within a series, since it is a property of the core, which is sampled only during the small fraction of the orbit spent inside it; measuring $δ_{l}$ once therefore predicts every member of the series. Second, inverting Eq. (5.17) converts a measured binding energy directly into $n^{*}$ :

\begin{matrix} n^{*} = \sqrt{\frac{ℰ_{Ryd}}{E_{th} - E_{n}}} \end{matrix}

Example 5.8.1The quantum defects of sodium

Sodium has one electron outside a closed-shell neon-like core, so that it is hydrogen-like in structure and serves as the standard illustration. Its ionization energy is $E_{th} = 5.139$ eV above the ground state. The $3 s$ ground state is bound by $5.139$ eV, the $3 p$ state by $3.035$ eV, and the $3 d$ state by $1.522$ eV. Find $n^{*}$ and $δ_{l}$ for each.

Solution.

Applying Eq. (5.18) three times,

3 s : n^{*} = \sqrt{\frac{13.606}{5.139}} = \sqrt{2.648} = 1.627, δ_{s} = 3 - 1.627 = 1.373;

3 p : n^{*} = \sqrt{\frac{13.606}{3.035}} = \sqrt{4.483} = 2.117, δ_{p} = 3 - 2.117 = 0.883;

3 d : n^{*} = \sqrt{\frac{13.606}{1.522}} = \sqrt{8.940} = 2.990, δ_{d} = 3 - 2.990 = 0.010 .

The ordering is that of Eq. (5.16), and the magnitudes are worth examining. The $3 d$ electron is hydrogenic to within a third of a percent, since its centrifugal barrier holds it so far outside the core that it barely distinguishes sodium from hydrogen. The $3 s$ electron, with $n^{*} = 1.63$ , behaves like a hydrogen state with $n$ between $1$ and $2$ , because on every orbit it passes through the neon core and briefly experiences a charge much larger than $1$ .

A single number, $δ_{l}$ , together with a single mechanism, penetration, thus accounts for the structure of the alkali spectrum. The same description applies to every Rydberg series in this book, including the molecular series of Chapter 9.

5.9Term Symbols

A configuration lists which orbitals are occupied; carbon, for example, is $1 s^{2} 2 s^{2} 2 p^{2}$ . A configuration is not a state, since the angular momenta of the electrons can couple in several ways, giving several energies. The resulting states are labeled by a term symbol.

Definition 5.9.1Atomic term symbol

\begin{matrix} ^{2 S + 1} L_{J} \end{matrix}

where $S$ is the total spin, $L$ the total orbital angular momentum written as a capital letter ( $L = 0, 1, 2, 3 \to S, P, D, F$ ), $J$ the total angular momentum obtained by coupling $L$ and $S$ , and $2 S + 1$ the multiplicity ( $1$ singlet, $2$ doublet, $3$ triplet).

Theorem 5.9.1Closed shells and single holes

A filled subshell has $S = 0$ and $L = 0$ , and therefore contributes $^{1} S$ to a term symbol. A subshell containing one hole has the same $L$ and $S$ as one containing one electron.

In a filled subshell every $m_{l}$ from $- l$ to $+ l$ is occupied twice, once with each spin. Hence $M_{S} = \sum m_{s} = 0$ and $M_{L} = \sum m_{l} = 0$ , and since this is the only state of the configuration, $S = L = 0$ . For the second statement, removing one electron leaves $M_{L} = - m_{l}$ and $M_{S} = - m_{s}$ of the removed electron, so that the available $(M_{L}, M_{S})$ values are those of a single electron with reversed signs. This is the same set, and hence gives the same $L$ and $S$ .

These two results remove most of the work. In particular, ionizing a closed-shell atom or molecule always leaves one hole and therefore always gives a doublet, $2 S + 1 = 2$ .

Example 5.9.1Reading three term symbols

Assign term symbols to helium's ground state, sodium's ground state, and sodium's first excited state.

Solution.

Helium, $1 s^{2}$ . This is a closed shell, so that by Theorem 5.9.1 $S = 0$ and $L = 0$ , hence $J = 0$ and the term is $^{1} S_{0}$ . This is a singlet, consistent with Sec. 5.7, where the ground configuration admitted only the spin-antisymmetric combination.

Sodium, $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ . All shells are closed except for one $3 s$ electron, so that $S = \frac{1}{2}$ , $L = 0$ and $J = \frac{1}{2}$ : the term is $^{2} S_{1 / 2}$ .

Sodium, $\dots 3 p^{1}$ . Here $L = 1$ and $S = \frac{1}{2}$ , so that $J$ can be $L + S = \frac{3}{2}$ or $L - S = \frac{1}{2}$ , giving the two terms $^{2} P_{3 / 2}$ and $^{2} P_{1 / 2}$ . These are split by the spin–orbit interaction, and the two transitions down to $^{2} S_{1 / 2}$ are the sodium D lines at $589.0$ and $589.6$ nm, which produce the yellow light of a sodium street lamp. The doublet structure exists only because the electron has spin.

5.10States Inside a Continuum

One further feature of many-electron atoms must be introduced, since the final chapters of this book depend on it entirely. It has no counterpart in hydrogen.

The ionization threshold of helium is $24.59$ eV. Above that energy the system can exist as He $^{+}$ together with a free electron of any kinetic energy, which is to say as a continuum. Suppose now that both electrons are excited, to the configuration $2 s 2 p$ . That state lies at $60.15$ eV, well above the threshold. It is a perfectly good bound state of the two-electron system, in that both electrons are attached and it has a definite energy, and it is at the same time degenerate with the continuum.

Such a state cannot survive. The repulsion ${\hat{V}}_{e e}$ of Eq. (5.3) connects it to the continuum at the same energy: one electron drops to $1 s$ while the other takes up the released energy and leaves. No photon is absorbed or emitted.

Definition 5.10.1Autoionization

Autoionization is the decay of a bound state that is degenerate with a continuum, driven by an interaction internal to the system rather than by absorption of radiation.

Because the state decays, it is not sharp: it has a width $Γ$ related to its lifetime by $Γ τ = ℏ$ . For helium's $2 s 2 p$ resonance $Γ \approx 37$ meV, so that

\begin{matrix} τ = \frac{ℏ}{Γ} = \frac{0.658 eV fs}{0.037 eV} \approx 18 fs . \end{matrix}

This configuration, a discrete state embedded in and coupled to a continuum, produces the asymmetric spectral profiles that Fano explained in 1961, using this same helium resonance as his example. We do not yet have the tools to treat it, since it requires the continuum normalization mentioned at the end of Chapter 3 and the time-dependent methods of Chapter 6. The treatment is given in full in Chapter 10.

Two points should be carried forward from this section. A doubly excited atom is a bound state inside a continuum, and it lives for femtoseconds rather than nanoseconds. The cause is the same ${\hat{V}}_{e e}$ that made the ground state of helium impossible to compute exactly. The term that could not be handled in Remark 5.3.1 thus turns out to drive the physics at which this course is aimed.

5.11Summary

Fermion states are antisymmetric under exchange (Def. 5.2.1); Pauli exclusion (Thm. 5.2.1) and the Slater determinant (Def. 5.2.2) follow.
Helium's Hamiltonian, Eq. (5.3), is not separable because of the single term $1 / r_{12}$ .
Three treatments: independent electrons give $- 108.8$ eV; first-order perturbation theory, using $⟨ 1 / r_{12} ⟩ = 5 Z / 8$ (Thm. 5.5.1), gives $- 74.8$ eV; the variational method with a screened exponent gives $Z^{'} = Z - \frac{5}{16} = 1.6875$ and $E = - Z^{' 2} = - 77.5$ eV, against the exact $- 79.0$ eV.
The variational principle (Thm. 5.6.1) guarantees an upper bound, so a result below the exact energy indicates an error.
Antisymmetry ties spin to spatial symmetry, giving $E_{\pm} = ε_{a} + ε_{b} + J \pm K$ (Thm. 5.7.1). The triplet lies lower by $2 K$ ; for helium's $1 s 2 s$ , $2 K = 0.796$ eV, four orders of magnitude larger than any magnetic spin interaction. This is exchange, and it has no classical analogue.
In heavier atoms each electron moves in a screened central field running from $- Z / r$ to $- 1 / r$ . Penetration, controlled by the centrifugal barrier, breaks hydrogen's $l$ -degeneracy in the order $s > p > d > f$ .
This is described by the quantum defect, $E_{n} = E_{th} - ℰ_{Ryd} / (n - δ_{l})^{2}$ . For sodium: $δ_{s} = 1.373$ , $δ_{p} = 0.883$ , $δ_{d} = 0.010$ .
Term symbols are $^{2 S + 1} L_{J}$ ; closed shells contribute $^{1} S$ and a single hole behaves like a single electron (Thm. 5.9.1), so ionizing a closed shell always gives a doublet.
A doubly excited state can be degenerate with a continuum and decay into it without radiating, a process called autoionization. Helium's $2 s 2 p$ at $60.15$ eV has $Γ \approx 37$ meV, hence $τ \approx 18$ fs.

5.12Exercises

The determinant. Write out the Slater determinant for three electrons in orbitals $ϕ_{a}, ϕ_{b}, ϕ_{c}$ . Verify that interchanging $ξ_{1}$ and $ξ_{3}$ changes the sign, and that setting $ϕ_{c} = ϕ_{a}$ gives zero.
The Coulomb integral. Reproduce the proof of Theorem 5.5.1, filling in the two standard integrals used. Then evaluate $5 Z / 8$ in eV for $Z = 1$ , $2$ and $3$ , and explain physically why the repulsion grows with $Z$ even though the electrons are being pulled closer together.
The variational calculation.
1. Derive Eq. (5.10), stating where each of the three terms comes from.
2. Minimize it to obtain $Z^{'} = Z - \frac{5}{16}$ and $E_{min} = - Z^{' 2}$ .
3. Apply the result to H $^{-}$ ( $Z = 1$ , two electrons). What $Z^{'}$ and what energy do you obtain? Compare with the energy of a hydrogen atom plus a free electron ( $- 0.5$ a.u.) and comment on whether this trial function predicts H $^{-}$ to be bound.
Exchange.
1. Carry out the expansion in the proof of Theorem 5.7.1, showing all four terms explicitly.
2. Explain in three or four sentences why the spatially antisymmetric state has lower energy, referring to what happens at ${\vec{r}}_{1} = {\vec{r}}_{2}$ .
3. The measured splitting of helium's $1 s 2 p$ levels is $0.25$ eV, smaller than the $0.796$ eV of the $1 s 2 s$ levels. Using Eq. (5.13), explain why a $2 p$ orbital gives a smaller exchange integral with $1 s$ than a $2 s$ orbital does.
Quantum defects from data. Using $E_{th} = 5.139$ eV for sodium and Eq. (5.18), compute $n^{*}$ and $δ$ for the $4 s$ state (bound by $1.947$ eV) and the $4 d$ state (bound by $0.851$ eV). Compare with the $3 s$ and $3 d$ values of Example 5.8.1 and state whether $δ_{l}$ behaves as Def. 5.8.1 claims.
Term symbols. Give the ground-state term symbol for He ( $1 s^{2}$ ), Li ( $\dots 2 s^{1}$ ), Be ( $\dots 2 s^{2}$ ) and F ( $\dots 2 p^{5}$ ). For fluorine, use Theorem 5.9.1 rather than coupling five electrons.

5.13Project: Screening and the Quantum Defect

The problem. Take the numerical radial solver built in Chapter 4 and insert a screened potential into it. Show that screening breaks the $l$ -degeneracy, obtain quantum defects from your own eigenvalues, and compare them with sodium.

A convenient model potential, crude but with the correct limits, is

\begin{matrix} V (r) = - \frac{1}{r} - \frac{Z - 1}{r} e^{- r / d}, \end{matrix}

in atomic units, with $Z = 11$ for sodium and $d$ a parameter setting the core size.

On paper. Verify that Eq. (5.21) satisfies both limits of Eq. (5.15). Sketch it against $- 1 / r$ and $- Z / r$ , and say where the crossover lies in terms of $d$ .
On paper. Before computing, predict the sign of the effect: will $3 s$ move up or down relative to hydrogen's $n = 3$ , and will $3 d$ move more or less? Justify the prediction with the centrifugal barrier in two sentences, and record it.
On the computer. Solve Eq. (3.43) with Eq. (5.21) for $l = 0, 1, 2$ , starting from $d = 0.6 a_{0}$ . Extract the eigenvalues near $n = 3$ and convert each to a quantum defect using Eq. (5.18) with $E_{th} = 0$ , so that the binding energy is simply $| E |$ .
On the computer. Compare with sodium's measured $δ_{s} = 1.373$ , $δ_{p} = 0.883$ , $δ_{d} = 0.010$ . Tune $d$ to fit $δ_{s}$ , then report what the other two do. Does fitting one spoil the others? State which features of the real atom this one-parameter model captures and which it cannot.
On the computer. Plot $| U_{3 l} (r) |^{2}$ for $l = 0, 1, 2$ in your fitted potential, with the core region $r < d$ shaded. The figure should make the mechanism visible at a glance: the $s$ state has weight inside the shaded region and the $d$ state almost none. Quantify this by computing $\int_{0}^{d} | U_{3 l} |^{2} d r$ for each $l$ .
What has to move. One animation: increase the screening smoothly from none ( $Z = 1$ , pure hydrogen) to $Z = 11$ , showing the three $n = 3$ levels starting degenerate and fanning apart with the $s$ level falling fastest. That fanning is the breaking of the accidental degeneracy.

The check. At $Z = 1$ the three levels must be degenerate at $- 1 / 18$ a.u. and all three defects must vanish. If they do not, the error is in the solver rather than in the physics, and should be corrected before the screening is switched on.

Be ready to answer. The model has one adjustable parameter and three defects to reproduce. Suppose all three can be fitted by hand. What would that tell you, and what would it not tell you, about whether Eq. (5.21) is a good description of the sodium core?