Physical Chemistry · Part 7 of 9

Many-Electron Atoms

Quantum Chemistry, Part 7 · 9 sections · about 24,415 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

The moment a second electron appears, the Schrödinger equation stops being solvable and quantum chemistry becomes the art of good approximations. This part explains what actually goes wrong, why antisymmetry is forced on us rather than assumed, and how the orbital picture every chemist uses is recovered from a problem that strictly has no orbitals in it at all. It ends with term symbols, which the exam asks for every year. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 7

  • 1Electron spin: what forced it on us, and how it is handled Free below
  • 2Indistinguishability and the antisymmetry principle
  • 3Slater determinants, exchange energy, singlets and triplets
  • 4The helium problem and the orbital approximation
  • 5The Hartree and Hartree–Fock methods, and the SCF procedure
  • 6Screening, Slater's rules, the aufbau principle and correlation
  • 7Russell–Saunders coupling and the term symbol
  • 8Microstates, the derivation of terms, and Hund's three rules
  • 9Spin–orbit coupling, j–j coupling, the Zeeman effect and selection rules

Electron spin: what forced it on us, and how it is handled

Free extract

Section G.1 of Part 7, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

The three experiments that could not be explained, and the minimum machinery needed to fix them.

Nothing in Parts D–F needed spin. The hydrogen atom was solved, the energies came out right, and the three quantum numbers n, l and ml labelled every state. So why add a fourth? Because three separate experiments refused to fit.

Experiment 1 — the alkali doublets

The bright yellow line of sodium is not one line but two, and the same doubling runs through the alkali spectra. In the Schrödinger picture the 3p → 3s transition is a single transition between two levels and there is nowhere for a second line to come from: something is splitting the 3p level in two.

Experiment 2 — the anomalous Zeeman effect

Atoms in a magnetic field split their lines. Orbital angular momentum predicts three components spaced by μBB — the normal Zeeman effect. Most atoms instead give four, six or more components at spacings that are rational fractions of μBB. It was called ‘anomalous’ for twenty years because nobody could produce the fractions. G.9 produces them, and the ingredient is spin.

Experiment 3 — Stern and Gerlach

The decisive one. Fire a beam of neutral atoms through a strongly inhomogeneous magnetic field. An atom with a magnetic moment feels a force proportional to μz∂Bz/∂z, so the beam is deflected by an amount that reports the orientation of the moment. Classically the orientation is continuous and the beam should smear out into a band. It does not. It splits into a discrete number of sharp beams — and for silver, that number is two.

The Stern–Gerlach experiment and why it forced spin on usWhat classical physics predictsWhat the experiment showsovenNS∂Bz/∂z ≠ 0ovenNS∂Bz/∂z ≠ 0μz takes every value between−|μ| and +|μ|, so the beam issmeared into a continuous band.ms = +½ms = −½Exactly TWO spots. The magnetic momenthas only two possible z-components, andthe beam is split, not smeared.Two spots means 2s + 1 = 2, so s = ½. No integer value of any orbital quantum number can give an even number ofstates, because 2l + 1 is always odd. The two-valuedness is not orbital motion; it is something new.
The single cleanest argument that spin exists and that it is half-integral. A beam of neutral atoms carrying a magnetic moment is fired through a deliberately inhomogeneous field, so the force on the atom depends on the orientation of that moment. Classically the beam should spread into a band. It does not: it splits into a discrete number of beams, and for silver that number is two. Since an angular momentum j has 2j+1 orientations, two beams require j = ½, which no orbital angular momentum can supply. Note what the experiment does not show: it does not show the electron rotating.

Two is the killer. An angular momentum of quantum number j has 2j + 1 orientations. For orbital angular momentum j = l is an integer, so 2l + 1 is always odd: one, three, five, never two. An even number of beams cannot be produced by orbital motion of any kind. The only way to get two is 2j + 1 = 2, that is j = ½, a half-integral angular momentum. Uhlenbeck and Goudsmit proposed in 1925 that the electron simply carries one.

Spin: an intrinsic angular momentum carried by the electron itself, of fixed quantum number s = ½, independent of its motion. It is not the electron spinning; it is an internal degree of freedom with the mathematics of an angular momentum. It contributes to the magnetic moment of the atom and to the angular-momentum bookkeeping, and it is not derivable from the Schrödinger equation.
s = ½ for every electron, always     ms = +½ or −½     |S| = √[s(s+1)]ℏ = (√3/2)ℏ = 0.866ℏ

Two pieces of notation you will use constantly. The two spin states are given the names α (ms = +½, ‘spin up’, ↑) and β (ms = −½, ‘spin down’, ↓). They are functions of a formal spin coordinate that we never write out; all that matters is that they are orthonormal:

∫α*α dσ = 1,   ∫β*β dσ = 1,   ∫α*β dσ = 0
Spin-orbital: the product of a spatial orbital and a spin function — 1sα, 1sβ, 2pzβ, and so on. A spin-orbital is what an electron actually occupies. Each spatial orbital supports exactly two spin-orbitals, which is why an orbital ‘holds two electrons’. The complete label of an electron in an atom is therefore (n, l, ml, ms) — four numbers, not three.

With spin included, the number of one-electron states in shell n doubles from n² to 2n², which is where the capacities 2, 8, 18, 32 of the periodic table come from. Note carefully that this is a statement about capacity, not about filling order.

Advanced / reference layer

The spin operators, the Pauli matrices, and why spin is not rotation.

Spin as an angular momentum: the algebra

Spin is defined by its algebra. We postulate a vector operator Ŝ = (Ŝx, Ŝy, Ŝz) obeying exactly the commutation relations that L̂ obeyed in Part D:

[Ŝx, Ŝy] = iℏŜz    (and cyclic)     [Ŝ², Ŝz] = 0

That is the whole definition. Everything else follows from the general theory of angular momentum, which shows that any operator set with these commutators has eigenvalues

Ŝ²|s,ms⟩ = s(s+1)ℏ²|s,ms⟩     Ŝz|s,ms⟩ = msℏ|s,ms⟩

with s = 0, ½, 1, 3/2, … and ms = −s, …, +s. The half-integral values are permitted by the algebra but were excluded for orbital angular momentum by the requirement that the wavefunction be single-valued in the angle φ — a requirement that has no meaning for an internal degree of freedom. Spin is precisely the case the orbital argument threw away. Experiment says the electron has s = ½, and s is fixed: unlike l, it cannot be changed by exciting the electron.

Explicit matrices

Since there are only two states, a spin operator is a 2 × 2 matrix in the basis {α, β}. The standard choice is Ŝk = (ℏ/2)σk with the Pauli matrices:

σx = (0 1
1 0)     σy = (0 −i
i  0)     σz = (1  0
0 −1)

Three consequences worth having. First, σk² = 1, so Ŝk² = ℏ²/4 for each component and Ŝ² = 3ℏ²/4 = s(s+1)ℏ² with s = ½, as required. Second, α and β are eigenfunctions of Ŝz but not of Ŝx or Ŝy — an electron with a definite Sz has completely indefinite Sx, which is the vector-model statement that the spin vector precesses on a cone. Third, the raising and lowering operators Ŝ± = Ŝx ± iŜy act as

Ŝ+β = ℏα,   Ŝ+α = 0,   Ŝ−α = ℏβ,   Ŝ−β = 0

You will need those in G.3 to prove that the symmetric combination αβ + βα really is a member of the S = 1 triplet and not a stray singlet.

The magnetic moment and the g-factor anomaly

A circulating charge with orbital angular momentum L has a magnetic moment μ = −(μB/ℏ)L. If spin were literally rotation, its moment would follow the same rule. It does not:

μspin = −ge(μB/ℏ)S,    ge = 2.002319  ≈  2

The factor of two is the strongest single piece of evidence that spin is not a rotating ball of charge: a classical rotating charge distribution gives g = 1 whatever its shape. The value ge ≈ 2 drops out of the Dirac equation automatically, and the small excess over 2 comes from quantum electrodynamics and is the most accurately confirmed prediction in physics. Put the observed angular momentum on a sphere of the classical electron radius and the equatorial speed exceeds the speed of light: the picture is not merely unproven but inconsistent. Use ‘spin’ as a name, not as a description.

An electron is in the state α. What are the possible results of measuring Sz, Sx and S², and with what probabilities? Medium

S² first, because it is the easy one. α is an eigenfunction of Ŝ² with eigenvalue s(s+1)ℏ² = (½)(3/2)ℏ² = 0.75ℏ². Every electron is in an Ŝ² eigenstate always, because s = ½ is not negotiable. Probability 1.
Sz. α is by definition the eigenfunction of Ŝz with eigenvalue +ℏ/2. The measurement returns +ℏ/2 with probability 1. Nothing random happens.
Sx — the interesting one. The eigenfunctions of Ŝx are (α ± β)/√2 with eigenvalues ±ℏ/2. Invert: α = (1/√2)[(α+β)/√2] + (1/√2)[(α−β)/√2].
Read off the probabilities. Each coefficient is 1/√2, so each probability is (1/√2)² = ½. A measurement of Sx on a spin-up electron returns +ℏ/2 half the time and −ℏ/2 half the time, at random.
The expectation values. ⟨Sz⟩ = +ℏ/2; ⟨Sx⟩ = ⟨Sy⟩ = 0. So ⟨Sx²⟩ + ⟨Sy²⟩ = 0.75ℏ² − 0.25ℏ² = 0.5ℏ² ≠ 0: the transverse components are not zero, only their averages are. That is the cone of the vector model, and it is why |S| = 0.866ℏ can never be lined up along z, where the largest available projection is 0.5ℏ.

⚠ Common mistakes & exam traps

  • Writing |S| = ½ℏ. The magnitude is √[s(s+1)]ℏ = 0.866ℏ. The number ½ℏ is the largest z-component, not the length.
  • Saying that spin comes out of the Schrödinger equation. It does not, at any level of care. It is put in by hand non-relativistically and emerges only from the Dirac equation. A CSIR question asking which quantum number does not arise from solving the Schrödinger equation for hydrogen wants ms.
  • Assuming the two spin states have different energies in a free atom. Without a field and without spin–orbit coupling, α and β are exactly degenerate. The singlet–triplet story of G.3 is an electrostatic effect that spin only labels.
  • Confusing s with S. Lower-case s is one electron’s spin and is always ½. Capital S is the atom’s total spin. The multiplicity is 2S + 1, never 2s + 1.
Easy
How many spin-orbitals are there in the n = 3 shell, and how many electrons can it therefore hold?
Show solution
n² = 9 spatial orbitals (one 3s, three 3p, five 3d), each supporting two spin-orbitals, so 18 spin-orbitals and a capacity of 18 electrons = 2n². Note this is capacity, not filling: the third period holds 8 elements, because 4s fills before 3d.
Med
Calculate the angle between the spin vector and the z-axis for ms = +½.
Show solution
cosθ = Sz/|S| = (ℏ/2)/(√3ℏ/2) = 1/√3, so θ = 54.74° — the same magic angle that gives the dz² nodal cone in Part E, and for the same reason (3cos²θ − 1 = 0). The spin can never point along z: if it did, Sx and Sy would both be zero, violating their commutation relation with Ŝz.
Hard
A student argues: ‘since α and β are degenerate, spin cannot affect the energy of an atom, so it is irrelevant to chemistry.’ Demolish this.
Show solution
The premise is right and the conclusion wrong. Spin contributes no term of its own to the non-relativistic Hamiltonian, but it changes the energy in three indirect ways. (i) Through antisymmetry (G.2) — the spin function’s symmetry dictates the spatial function’s symmetry, and the spatial function is what feels 1/r12. That is the exchange energy, a full electron-volt in helium. (ii) Through the exclusion principle — without it every electron would sit in 1s and there would be no periodic table. (iii) Through spin–orbit coupling (G.9) — fine structure, the sodium D doublet, and for heavy elements effects large enough to control oxidation states. Spin has no energy of its own and governs almost everything.

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The remaining 8 sections of this part — Indistinguishability and the antisymmetry principle, Slater determinants, exchange energy, singlets and triplets, The helium problem and the orbital approximation, The Hartree and… — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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