12.09.02 · quantum / identical-particles

Exchange interaction and the helium atom

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Anchor (Master): Landau, L. D. & Lifshitz, E. M., *Quantum Mechanics: Non-Relativistic Theory*, 3rd ed. (Pergamon, 1977), Ch. IX §§62–66 (exchange interaction, the helium atom, the Heisenberg spin Hamiltonian); Bethe, H. A. & Salpeter, E. E., *Quantum Mechanics of One- and Two-Electron Atoms* (Springer, 1957), §§32–33

Intuition Beginner

Two electrons are perfectly interchangeable. You cannot paint one red and the other blue and follow them around, because nature does not let you tell them apart. Quantum mechanics takes this seriously: swapping the two electrons must leave every measurable prediction unchanged. For electrons there is a sharp rule about how the swap acts on the wavefunction — it must flip the sign. This sign rule is the whole source of the strange effects in this unit, and it has a famous consequence: two electrons can never share the exact same state, the fact behind the structure of the periodic table.

The sign rule ties the electrons' positions to their spins. If the two electrons line up their spins the same way, the rule forces them to keep their positions apart. If they pair their spins as opposites, the rule lets them sit close together. So aligning the spins changes where the electrons can be, even though the spins themselves do not push or pull. This indirect link is the exchange interaction.

Why does this matter? The energy of two electrons depends on how close they get, because closer electrons repel more. Through the sign rule, their spin arrangement decides how close they get, and therefore their energy. A purely magnetic-sounding question — are the spins aligned? — turns into a question about electric repulsion energy. That hidden lever is what makes magnets magnetic and what splits the spectrum of helium into two families.

Visual Beginner

Picture two clouds, one for each electron, sharing the space around a nucleus. In the first picture the spins are opposite. The sign rule lets the two clouds overlap, piling up in the middle, so on average the electrons sit closer and feel more repulsion. In the second picture the spins are aligned. Now the rule forces a gap: the clouds pull away from each other, leaving a hole between them where neither electron likes to be. The electrons sit farther apart and feel less repulsion.

The picture shows the heart of the matter. No force acts directly between the spins. The spin pattern only sets the shape of the position cloud, and the shape sets the repulsion. The energy gap between the two pictures is the exchange energy, and its size is fixed by how much the two clouds would have overlapped.

Worked example Beginner

Take helium in a simple excited arrangement: one electron stays in the lowest orbit (call it the $1 s$ orbit) and the other sits in a higher orbit (call it $2 s$ ). We ask how the energy depends on whether the spins are aligned or opposite.

Step 1. Split the repulsion energy into two pieces. The first piece is the plain average repulsion of the two clouds, a positive number; call it $J$ . The second piece is an extra correction tied to the overlap of the two orbits; call it $K$ , also positive here.

Step 2. Write the two energies. When the spins are opposite, the energy gets $J + K$ . When the spins are aligned, it gets $J - K$ . The only difference is the sign in front of $K$ .

Step 3. Compare them. Since $K$ is positive, $J - K$ is smaller than $J + K$ . The aligned-spin arrangement has the lower energy.

Step 4. Put numbers in. Suppose for this orbit pair $J = 9.1$ and $K = 1.2$ (in electron-volts, above some baseline). Then opposite spins give $9.1 + 1.2 = 10.3$ and aligned spins give $9.1 - 1.2 = 7.9$ . The gap between the two is $2 K = 2.4$ electron-volts.

What this tells us: aligning the spins lowers the energy by $2 K$ , purely because the sign rule keeps the aligned electrons farther apart. The spins never touch; the bookkeeping of position does all the work. This $2 K$ gap is exactly the split between the two families of helium spectral lines.

Check your understanding Beginner

Exercise (easy, multiple choice).

Why do two aligned-spin electrons end up with lower repulsion energy than two opposite-spin electrons in the same pair of orbits?

A. Aligned spins attract each other magnetically.
B. The sign rule forces aligned-spin electrons to stay farther apart, so they repel less.
C. Aligned electrons have less charge.
D. The nucleus pulls aligned electrons closer.

Hint

The spins themselves exert no force here. Think about what the sign rule does to the positions.

Answer

B. The sign rule forces aligned-spin electrons to stay farther apart, so they repel less.

When the spins are aligned, the position part of the wavefunction must change sign on swapping the electrons, which carves out a gap between them. Farther apart means less repulsion and lower energy. The spins exert no direct force; they only set the shape of the position cloud.

Formal definition Intermediate+

Two electrons have the antisymmetrised state space $Λ^{2} H_{1}$ , the antisymmetric subspace of $H_{1} \otimes H_{1}$ , with one-particle space $H_{1} = L^{2} (R^{3}) \otimes C^{2}$ . A coordinate is $x = (r, s)$ , bundling position and spin. The antisymmetrisation postulate for fermions requires that the total wavefunction change sign under exchange: $$ \Psi(\mathbf x_1, \mathbf x_2) = -,\Psi(\mathbf x_2, \mathbf x_1). $$ This is the two-particle case of the spin-statistics rule fixed in the fermionic-Fock-space treatment 12.13.02. Because the non-relativistic Hamiltonian is spin-independent, eigenstates factor into a spatial part times a spin part, and antisymmetry of the product forces opposite symmetries on the factors.

Two electron spins live in $C^{2} \otimes C^{2}$ , which decomposes under the total spin $S = S_{1} + S_{2}$ into the antisymmetric singlet ( $S = 0$ ) and the symmetric triplet ( $S = 1$ ): $$ \chi_{\text{singlet}} = \tfrac{1}{\sqrt2}\big(!\uparrow\downarrow - \downarrow\uparrow\big), \qquad \chi_{\text{triplet}} \in \Big{\uparrow\uparrow,\ \tfrac{1}{\sqrt2}(\uparrow\downarrow + \downarrow\uparrow),\ \downarrow\downarrow\Big}. $$ Antisymmetry of the full state then pairs the symmetric spatial wavefunction with the singlet and the antisymmetric spatial wavefunction with the triplet: $$ \Psi_{+} = \psi_{\text{sym}}(\mathbf r_1,\mathbf r_2),\chi_{\text{singlet}}, \qquad \Psi_{-} = \psi_{\text{antisym}}(\mathbf r_1,\mathbf r_2),\chi_{\text{triplet}}. $$

Given two orthonormal spatial orbitals $ϕ_{a}, ϕ_{b}$ , the symmetric and antisymmetric spatial states are $$ \psi_{\pm}(\mathbf r_1,\mathbf r_2) = \tfrac{1}{\sqrt2}\big[\phi_a(\mathbf r_1)\phi_b(\mathbf r_2) \pm \phi_b(\mathbf r_1)\phi_a(\mathbf r_2)\big], $$ with $+$ for the singlet and $-$ for the triplet. The triplet spatial state $ψ_{-}$ is the spatial face of a Slater determinant of the two spin-orbitals; the singlet spatial state $ψ_{+}$ is its symmetric partner. For the Coulomb perturbation $V = 1/ r_{12}$ (atomic units, $r_{12} = ∣ r_{1} - r_{2} ∣$ ), define the Coulomb integral and the exchange integral: $$ J = \iint \phi_a^(\mathbf r_1)\phi_b^(\mathbf r_2),\frac{1}{r_{12}},\phi_a(\mathbf r_1)\phi_b(\mathbf r_2),d^3r_1,d^3r_2, $$ $$ K = \iint \phi_a^(\mathbf r_1)\phi_b^(\mathbf r_2),\frac{1}{r_{12}},\phi_b(\mathbf r_1)\phi_a(\mathbf r_2),d^3r_1,d^3r_2. $$ The integral $J$ is the classical electrostatic repulsion of the two charge clouds $∣ ϕ_{a} ∣^{2}$ and $∣ ϕ_{b} ∣^{2}$ ; the integral $K$ has no classical counterpart and arises from the cross term in which the two orbitals trade places between the bra and the ket. The same $K$ reappears as the exchange operator of the self-consistent field 12.09.03.

Counterexamples to common slips

The exchange integral $K$ is not a magnetic spin-spin coupling. It is an electrostatic matrix element; spin enters only by selecting which spatial symmetry is allowed. A spin-independent Hamiltonian produces a spin-dependent spectrum through this selection alone.
$K$ is real and, for the $1/ r_{12}$ kernel with real orbitals, positive. Writing $ρ_{ab} (r) = ϕ_{a}^{*} (r) ϕ_{b} (r)$ , one has $K = \iint ρ_{ab}^{*} (r_{1}) r_{12}^{- 1} ρ_{ab} (r_{2}) d^{3} r_{1} d^{3} r_{2} > 0$ , a self-energy of the overlap density. Positivity of $K$ is what makes the triplet lie below the singlet for excited helium.
The split into $J \pm K$ is exact only to first order in perturbation theory and only when $ϕ_{a}, ϕ_{b}$ are the same fixed orbitals in both states. If the orbitals are allowed to relax differently in singlet and triplet, the clean $E_{\pm} = J \pm K$ form is corrected.

Key derivation Intermediate+

Theorem (singlet/triplet splitting and the helium energies). Let two electrons occupy orthonormal spatial orbitals $ϕ_{a}, ϕ_{b}$ , perturbed by $V = 1/ r_{12}$ . To first order in degenerate perturbation theory the symmetric (singlet) and antisymmetric (triplet) spatial states have energies $$ E_{\pm} = E_a + E_b + J \pm K, $$ with $+$ for the singlet and $-$ for the triplet, where $E_{a}, E_{b}$ are the unperturbed orbital energies and $J, K$ are the Coulomb and exchange integrals.

Proof. The unperturbed two-electron Hamiltonian $\hat{H}_{0} = \hat{h} (1) + \hat{h} (2)$ has the product states $ϕ_{a} (r_{1}) ϕ_{b} (r_{2})$ and $ϕ_{b} (r_{1}) ϕ_{a} (r_{2})$ as a degenerate pair with energy $E_{a} + E_{b}$ (taking $a \neq = b$ ). The correct zeroth-order states under the perturbation $V$ are the symmetry-adapted combinations $ψ_{\pm}$ , which diagonalise $V$ within the degenerate subspace because $V$ commutes with particle exchange. Compute the first-order shift as the diagonal matrix element $$ \langle \psi_\pm | V | \psi_\pm \rangle = \tfrac12 \big\langle \phi_a\phi_b \pm \phi_b\phi_a ,\big|, V ,\big|, \phi_a\phi_b \pm \phi_b\phi_a \big\rangle. $$ Expanding the four terms and using the symmetry $⟨ ϕ_{a} ϕ_{b} ∣ V ∣ ϕ_{a} ϕ_{b} ⟩ = ⟨ ϕ_{b} ϕ_{a} ∣ V ∣ ϕ_{b} ϕ_{a} ⟩ = J$ (relabel the two integration variables) and $⟨ ϕ_{a} ϕ_{b} ∣ V ∣ ϕ_{b} ϕ_{a} ⟩ = ⟨ ϕ_{b} ϕ_{a} ∣ V ∣ ϕ_{a} ϕ_{b} ⟩ = K$ , $$ \langle \psi_\pm | V | \psi_\pm \rangle = \tfrac12\big( J + J \pm K \pm K\big) = J \pm K. $$ Adding the unperturbed energy gives $E_{\pm} = E_{a} + E_{b} + J \pm K$ . The cross terms carry the orbital swap, so the swap-induced integral $K$ enters with the sign of the spatial symmetry: $+ K$ for the symmetric (singlet) state, $- K$ for the antisymmetric (triplet) state. $□$

Bridge. This computation builds toward the entire theory of magnetic ordering, and the bridge is the observation that a spin-independent operator $V$ has produced a spin-dependent energy purely by selecting spatial symmetry. The foundational reason is the antisymmetrisation postulate: it locks spatial symmetry to total spin, so the position integral $K$ becomes a function of the spin state. This is exactly the mechanism that reappears again in the Heisenberg spin Hamiltonian derived at Master tier, where the energies $J \pm K$ are repackaged as eigenvalues of $- 2 K S_{1} \cdot S_{2}$ , and it generalises to the $N$ -electron case where the same $K$ becomes the non-local exchange operator of the Hartree-Fock field 12.09.03. The central insight, putting these together, is that exchange is not a force but a constraint, and the energy it costs or saves is the electrostatic self-energy of the overlap density $ϕ_{a}^{*} ϕ_{b}$ .

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Show that the exchange integral can be written as the self-energy of the overlap density $ρ_{ab} (r) = ϕ_{a}^{*} (r) ϕ_{b} (r)$ , namely $K = \iint ρ_{ab}^{*} (r_{1}) r_{12}^{- 1} ρ_{ab} (r_{2}) d^{3} r_{1} d^{3} r_{2}$ , and deduce $K \geq 0$ for real orbitals.

Hint

Group the integrand of $K$ as $[ϕ_{a}^{*} (r_{1}) ϕ_{b} (r_{1})] r_{12}^{- 1} [ϕ_{b}^{*} (r_{2}) ϕ_{a} (r_{2})]$ . The Coulomb kernel is positive-definite.

Answer

Regroup the factors in $K = \iint ϕ_{a}^{*} (r_{1}) ϕ_{b}^{*} (r_{2}) r_{12}^{- 1} ϕ_{b} (r_{1}) ϕ_{a} (r_{2})$ as $\iint [ϕ_{a}^{*} (r_{1}) ϕ_{b} (r_{1})] r_{12}^{- 1} [ϕ_{a} (r_{2}) ϕ_{b}^{*} (r_{2})] = \iint ρ_{ab} (r_{1}) r_{12}^{- 1} ρ_{ab}^{*} (r_{2})$ . For real orbitals $ρ_{ab}$ is real, and since the kernel $1/ r_{12}$ has positive Fourier transform $4 π / k^{2}$ , the quadratic form $\iint ρ (r_{1}) r_{12}^{- 1} ρ (r_{2}) = \int \frac{4 π}{k ^{2}} ∣ \tilde{ρ} (k) ∣^{2} \frac{d ^{3} k}{( 2 π ) ^{3}} \geq 0$ . Hence $K \geq 0$ . Rubric: full credit for the regrouping and the positivity argument via the Fourier transform of the Coulomb kernel.

Exercise 6 (medium, short-answer).

Explain why the helium ground state has no exchange splitting, while the excited $1 s 2 s$ configuration does.

Hint

Look at whether the two electrons occupy the same or different spatial orbitals, and what antisymmetry then permits.

Answer

In the ground state both electrons share the $1 s$ orbital ( $a = b$ ). The only antisymmetric total state pairs the symmetric spatial product $ϕ_{1 s} (r_{1}) ϕ_{1 s} (r_{2})$ with the spin singlet; no triplet exists because the antisymmetric spatial combination vanishes when $a = b$ . There is a single state, hence no splitting. In the $1 s 2 s$ configuration the two orbitals differ, so both the symmetric (singlet) and antisymmetric (triplet) spatial states exist, separated by $2 K$ . Rubric: full credit for the vanishing of the antisymmetric spatial state at $a = b$ and the existence of both symmetries at $a \neq = b$ .

Advanced results Master

Theorem (Dirac spin Hamiltonian for two electrons). The first-order Coulomb splitting $E_{\pm} = E_{a} + E_{b} + J \pm K$ of a two-electron configuration is reproduced exactly by the effective spin operator $$ \hat H_{\text{spin}} = \big(E_a + E_b + J - \tfrac12 K\big),\mathbb 1 ; -; 2K,\mathbf S_1\cdot\mathbf S_2, $$ acting on the four-dimensional spin space, where $S_{i}$ are spin- $\frac{1}{2}$ operators in units of $ℏ$ . The proof uses the projector identity $S_{1} \cdot S_{2} = \frac{1}{2} (S^{2} - S_{1}^{2} - S_{2}^{2})$ , with eigenvalues $+ \frac{1}{4}$ on the triplet and $- \frac{3}{4}$ on the singlet. Substituting, the triplet receives $E_{a} + E_{b} + J - \frac{1}{2} K - 2 K (\frac{1}{4}) = E_{a} + E_{b} + J - K$ and the singlet receives $E_{a} + E_{b} + J - \frac{1}{2} K - 2 K (- \frac{3}{4}) = E_{a} + E_{b} + J + K$ , matching $E_{-}$ and $E_{+}$ . Dirac obtained this spin-operator form in 1929, recognising that a spin-independent electrostatic interaction can be encoded as an effective coupling between spins whose strength is the exchange integral. The sign of $K$ decides the ground state: $K > 0$ (the usual atomic case) favours the triplet, an effective ferromagnetic coupling; $K < 0$ favours the singlet, an effective antiferromagnetic coupling.

Ortho- and para-helium. The two-electron exchange splitting partitions the helium spectrum into two non-combining systems. Para-helium is the singlet ( $S = 0$ ) ladder, including the $1 s^{2}$ ground state; ortho-helium is the triplet ( $S = 1$ ) ladder, whose lowest member is the metastable $1 s 2 s^{3} S_{1}$ state. Since electric-dipole radiation does not change total spin to good approximation, transitions between the two systems are strongly suppressed, and early spectroscopists mistook them for two distinct elements. For every excited configuration the ortho (triplet) term lies below the corresponding para (singlet) term by $2 K$ , a direct spectroscopic readout of the exchange integral. The metastability of $1 s 2 s^{3} S_{1}$ — it cannot decay to the singlet ground state without flipping a spin — gives it a lifetime of thousands of seconds, the longest of any neutral-atom excited state.

Heitler-London and the covalent bond. The same machinery applied to two hydrogen atoms gives the first quantum explanation of the chemical bond. Heitler and London in 1927 built symmetric and antisymmetric spatial combinations of two atomic $1 s$ orbitals centred on the two protons. Here the overlap makes $K$ large and, crucially, the symmetric (singlet) spatial state is lower because the relevant exchange integral is negative once the nuclear attraction and overlap normalisation are included: the bonding orbital piles electron density between the nuclei, screening their repulsion. The result is the singlet ground state of $H_{2}$ with a binding energy of order an electron-volt, with the triplet purely repulsive. The same $J \pm K$ structure thus explains both the antibonding/bonding split of a molecule and the ortho/para split of an atom; the sign of the effective exchange is what distinguishes them.

From two spins to a lattice: the Heisenberg model. Heisenberg's 1928 insight was that exchange, not magnetic dipole forces, is strong enough to order spins in a ferromagnet. Magnetic dipole-dipole energies between neighbouring atomic moments are of order $1 0^{- 4}$ eV, far too small to survive room-temperature thermal agitation ( $k_{B} T \approx 0.025$ eV). Exchange integrals are of order $0.1$ – $1$ eV. Summing the two-electron Dirac form over neighbouring pairs on a lattice gives the Heisenberg Hamiltonian $$ \hat H = -\sum_{\langle i j\rangle} 2 K_{ij},\mathbf S_i\cdot\mathbf S_j = -\sum_{\langle ij\rangle} J^{\text{ex}}{ij},\mathbf S_i\cdot\mathbf S_j, $$ writing $J^{\text{ex}}{ij} = 2K_{ij} $f or t h ee x c han g eco u pl in g . P os i t i v e$ J^{\text{ex}} $a l i g n s n e i g hb o u r in g s p in s (f er r o ma g n e t i s m); n e g a t i v e$ J^{\text{ex}}$ anti-aligns them (antiferromagnetism). This model, built atom by atom from the two-electron exchange computed in this unit, underlies the modern theory of magnetism, spin waves, and quantum phase transitions.

Synthesis. The exchange interaction is the foundational reason that a spinless electrostatic Hamiltonian governs the magnetic order of matter, and the central insight is that antisymmetry converts a question about spin alignment into a question about charge overlap, with the exchange integral $K$ as the exact conversion factor. Putting these together, the first-order energies $J \pm K$ , the Dirac operator $- 2 K S_{1} \cdot S_{2}$ , the ortho/para split of helium, and the Heitler-London bond are four faces of one computation; the sign and magnitude of $K$ select among them. This is exactly the structure that generalises to the lattice: the two-electron exchange coupling, summed over neighbours, builds toward the Heisenberg model, and the same integral that appears here as a perturbative matrix element reappears again in the Hartree-Fock self-consistent field 12.09.03 as the non-local exchange operator that cancels the electronic self-interaction. The bridge from this unit to all of magnetism and quantum chemistry is the recognition that exchange is a constraint masquerading as a force, whose energetic price is the self-energy of the overlap density.

Full proof set Master

Proposition (correct zeroth-order states diagonalise the Coulomb perturbation). Within the degenerate subspace spanned by $ϕ_{a} (r_{1}) ϕ_{b} (r_{2})$ and $ϕ_{b} (r_{1}) ϕ_{a} (r_{2})$ ( $a \neq = b$ ), the symmetry-adapted states $ψ_{\pm}$ diagonalise $V = 1/ r_{12}$ with eigenvalues $J \pm K$ .

Proof. In the ordered basis ${u_{1} = ϕ_{a} ϕ_{b}, u_{2} = ϕ_{b} ϕ_{a}}$ , the perturbation matrix has entries $V_{11} = ⟨ u_{1} ∣ V ∣ u_{1} ⟩ = J$ and $V_{22} = ⟨ u_{2} ∣ V ∣ u_{2} ⟩ = J$ by relabelling the dummy integration variables, while $V_{12} = ⟨ u_{1} ∣ V ∣ u_{2} ⟩ = K$ and $V_{21} = K$ since $V$ and the kernel are real and symmetric under $r_{1} \leftrightarrow r_{2}$ . The matrix $$ \begin{pmatrix} J & K \ K & J \end{pmatrix} $$ has eigenvalues $J \pm K$ with eigenvectors $\frac{1}{2} (1, \pm 1)^{T}$ , which are exactly the coefficient vectors of $ψ_{\pm} = \frac{1}{2} (u_{1} \pm u_{2})$ . Because $[V, P_{12}] = 0$ for the exchange operator $P_{12}$ , these eigenvectors are simultaneously exchange-symmetric ( $+$ ) and exchange-antisymmetric ( $-$ ), so the diagonalisation respects the symmetry required by the antisymmetrisation postulate. $□$

Proposition (exchange eigenvalues of $S_{1} \cdot S_{2}$ ). On two spin- $\frac{1}{2}$ particles, $S_{1} \cdot S_{2}$ has eigenvalue $- \frac{3}{4}$ on the singlet and $+ \frac{1}{4}$ on the triplet (units of $ℏ^{2}$ ), and the projectors onto singlet and triplet are $P_{0} = \frac{1}{4} - S_{1} \cdot S_{2}$ and $P_{1} = \frac{3}{4} + S_{1} \cdot S_{2}$ .

Proof. The identity $S_{1} \cdot S_{2} = \frac{1}{2} (S^{2} - S_{1}^{2} - S_{2}^{2})$ follows from expanding $S^{2} = (S_{1} + S_{2})^{2}$ . With $S_{i}^{2} = s (s + 1) ℏ^{2} = \frac{3}{4} ℏ^{2}$ and $S^{2} = S (S + 1) ℏ^{2}$ , the singlet ( $S = 0$ ) gives $\frac{1}{2} (0 - \frac{3}{4} - \frac{3}{4}) = - \frac{3}{4}$ and the triplet ( $S = 1$ ) gives $\frac{1}{2} (2 - \frac{3}{4} - \frac{3}{4}) = + \frac{1}{4}$ . For the projectors, note that an operator with two eigenvalues $λ_{0} = - \frac{3}{4}$ and $λ_{1} = + \frac{1}{4}$ has spectral projectors $P_{S} = (S_{1} \cdot S_{2} - λ_{S^{'}}) / (λ_{S} - λ_{S^{'}})$ for $S^{'} \neq = S$ . Substituting $λ_{0} - λ_{1} = - 1$ gives $P_{0} = (S_{1} \cdot S_{2} - \frac{1}{4}) / (- 1) = \frac{1}{4} - S_{1} \cdot S_{2}$ , and $P_{1} = 1 - P_{0} = \frac{3}{4} + S_{1} \cdot S_{2}$ . $□$

Proposition (uniqueness of the Dirac spin Hamiltonian on the two-spin space). Any Hermitian operator on $C^{2} \otimes C^{2}$ that is rotationally invariant (commutes with $S$ ) and assigns energy $E_{+}$ to the singlet and $E_{-}$ to the triplet is uniquely $E_{+} P_{0} + E_{-} P_{1}$ , equal to $\frac{1}{4} (E_{+} + 3 E_{-}) 1 - (E_{+} - E_{-}) S_{1} \cdot S_{2}$ .

Proof. A rotationally invariant Hermitian operator is constant on each irreducible $S$ -multiplet by Schur's lemma, so it is determined by its singlet and triplet eigenvalues and equals $E_{+} P_{0} + E_{-} P_{1}$ . Insert the projectors from the previous proposition: $E_{+} P_{0} + E_{-} P_{1} = E_{+} (\frac{1}{4} - S_{1} \cdot S_{2}) + E_{-} (\frac{3}{4} + S_{1} \cdot S_{2}) = \frac{1}{4} (E_{+} + 3 E_{-}) 1 + (E_{-} - E_{+}) S_{1} \cdot S_{2}$ . Setting $E_{+} = E_{a} + E_{b} + J + K$ and $E_{-} = E_{a} + E_{b} + J - K$ gives coefficient $E_{-} - E_{+} = - 2 K$ on $S_{1} \cdot S_{2}$ and constant $E_{a} + E_{b} + J - \frac{1}{2} K$ , the Dirac form. Uniqueness is the Schur step: no other rotational invariant exists on this four-dimensional space beyond the two projectors. $□$

Connections Master

Fermionic Fock space and the Pauli principle 12.13.02. The antisymmetrisation postulate used throughout this unit is the two-particle instance of the canonical anticommutation relations: the triplet spatial state is the configuration-space image of a single occupied state of two fermionic creation operators, and the vanishing of the antisymmetric spatial wavefunction when the two orbitals coincide is the Pauli exclusion principle expressed as $(a_{ϕ}^{†})^{2} = 0$ . The exchange integral $K$ is the matrix element of the two-body interaction between the direct and swapped occupations of the same orbital pair.
Hartree-Fock self-consistent field 12.09.03. The exchange integral $K$ defined here as a perturbative matrix element reappears in the $N$ -electron self-consistent field as the non-local exchange operator $\hat{K}$ , whose kernel is the off-diagonal density $\sum_{j} χ_{j} (r_{1}) χ_{j}^{*} (r_{2}) / r_{12}$ . The two-electron $J \pm K$ split is the smallest case of the Slater-Condon energy expression, and the cancellation $J_{ii} - K_{ii} = 0$ that removes self-interaction in Hartree-Fock is the same algebra that makes the helium ground state ( $a = b$ ) carry no exchange splitting.
Angular momentum and the $SU (2)$ structure of spin 12.05.01. The singlet/triplet decomposition $2 \otimes 2 = 1 \oplus 3$ is the Clebsch-Gordan series for two spin- $\frac{1}{2}$ representations, and the operator $S_{1} \cdot S_{2}$ is diagonalised by the same total-spin eigenstates. The eigenvalues $- \frac{3}{4}, + \frac{1}{4}$ and the projector construction are pure representation theory of $SU (2)$ , applied here to convert an electrostatic energy into a spin coupling.
Time-independent perturbation theory 12.07.01. The helium energies $E_{\pm} = E_{a} + E_{b} + J \pm K$ are a textbook application of first-order degenerate perturbation theory: the Coulomb repulsion is the perturbation, the degenerate product states are the unperturbed manifold, and symmetry-adaptation supplies the correct zeroth-order states that diagonalise the perturbation within the degenerate subspace.

Historical & philosophical context Master

Werner Heisenberg introduced the exchange interaction in 1926 in Zeitschrift für Physik ^{[Heisenberg 1926]}, treating the two-electron problem as a resonance between the configurations in which the electrons are interchanged. He found that the energy splits according to the symmetry of the wavefunction under exchange, and that this "exchange resonance" (Austauschresonanz) accounts for the previously mysterious division of the helium spectrum into ortho- and para-systems. The effect had no classical analogue: it followed entirely from the indistinguishability of identical particles and the antisymmetry that quantum mechanics demands of electrons.

Walter Heitler and Fritz London applied the same idea to two hydrogen atoms in 1927 ^{[Heitler-London 1927]}, producing the first quantum-mechanical theory of the covalent bond and showing that the symmetric spatial (singlet) state binds while the antisymmetric (triplet) state is repulsive. Paul Dirac in 1929 ^{[Dirac 1929]} recast the exchange splitting as an effective spin coupling, writing the interaction as a function of $S_{1} \cdot S_{2}$ and thereby translating an electrostatic computation into the language of spin operators. Heisenberg then turned the two-electron result into a theory of bulk magnetism in 1928 ^{[Heisenberg 1928]}, arguing that the exchange integral, not the weak magnetic dipole interaction, supplies the energy scale needed for ferromagnetic ordering, and summing the pairwise coupling into what is now the Heisenberg model. Landau and Lifshitz give the unified treatment of symmetry, the helium atom, and the spin Hamiltonian in Chapter IX of Quantum Mechanics ^{[Landau-Lifshitz 1977]}, deriving the exchange interaction from the antisymmetry postulate and the spin Hamiltonian from the projector identity.

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}

@book{BetheSalpeter1957,
  author    = {Bethe, Hans A. and Salpeter, Edwin E.},
  title     = {Quantum Mechanics of One- and Two-Electron Atoms},
  publisher = {Springer},
  year      = {1957}
}

@book{GriffithsQM2005,
  author    = {Griffiths, David J.},
  title     = {Introduction to Quantum Mechanics},
  edition   = {2},
  publisher = {Pearson Prentice Hall},
  year      = {2005}
}

@book{BransdenJoachain2003,
  author    = {Bransden, Brian H. and Joachain, Charles J.},
  title     = {Physics of Atoms and Molecules},
  edition   = {2},
  publisher = {Pearson Education},
  year      = {2003}
}

Prerequisites

12.02.01
12.02.02
12.05.01
12.07.01
12.13.02

Tier anchors

beginner: Griffiths, *Introduction to Quantum Mechanics*, 2nd ed. (Pearson, 2005), §5.1–5.2 (two-particle systems, exchange forces) and §5.2.1 (helium)
intermediate: Landau, L. D. & Lifshitz, E. M., *Quantum Mechanics: Non-Relativistic Theory*, 3rd ed. (Pergamon, 1977), Ch. IX §§62–63 (symmetry of wavefunctions, exchange interaction); Bransden, B. H. & Joachain, C. J., *Physics of Atoms and Molecules*, 2nd ed. (Pearson, 2003), Ch. 7
master: Landau, L. D. & Lifshitz, E. M., *Quantum Mechanics: Non-Relativistic Theory*, 3rd ed. (Pergamon, 1977), Ch. IX §§62–66 (exchange interaction, the helium atom, the Heisenberg spin Hamiltonian); Bethe, H. A. & Salpeter, E. E., *Quantum Mechanics of One- and Two-Electron Atoms* (Springer, 1957), §§32–33

References

Landau, L. D. & Lifshitz, E. M. — Quantum Mechanics: Non-Relativistic Theory · 3rd ed. (Pergamon, 1977), Ch. IX §§62–66 (symmetry of wavefunctions; exchange interaction; the helium atom; the spin Hamiltonian)
Heisenberg, W. — Mehrkörperproblem und Resonanz in der Quantenmechanik · Zeitschrift für Physik 38, 411–426 (1926) (exchange resonance of identical particles; the two-electron resonance splitting)
Heisenberg, W. — Zur Theorie des Ferromagnetismus · Zeitschrift für Physik 49, 619–636 (1928) (exchange origin of ferromagnetism; the Heisenberg model)
Dirac, P. A. M. — Quantum Mechanics of Many-Electron Systems · Proceedings of the Royal Society A 123, 714–733 (1929) (the spin-exchange operator and the $-2K\,\mathbf S_1\cdot\mathbf S_2$ Hamiltonian form)
Heitler, W. & London, F. — Wechselwirkung neutraler Atome und homöopolare Bindung nach der Quantenmechanik · Zeitschrift für Physik 44, 455–472 (1927) (the covalent bond from exchange; the singlet/triplet splitting of $\mathrm H_2$)

Estimated time

beginner: 17m
intermediate: 42m
master: 78m