Exchange Symmetry and the Ortho/Para Splitting

Derivation

Exchange Symmetry and the Ortho/Para Splitting

D-304 Home PU-306 Threads matter · symmetry · energy Depends on pauli-exclusion-principle, Bound-State Spectrum of the Hydrogen Atom

Statement

For two-electron helium with Hamiltonian \(\hat H=\hat h(1)+\hat h(2)+V_{12}\), \(V_{12}=\dfrac{e^2}{4\pi\varepsilon_0 r_{12}}\), the exclusion-enforced antisymmetry of the total (space\(\,\times\,\)spin) wavefunction splits an excited configuration such as \(1s2s\) into a spatially symmetric part paired with the spin singlet (parahelium) and a spatially antisymmetric part paired with the spin triplet (orthohelium). First-order degenerate perturbation theory gives \(E_\pm=\varepsilon_{1s}+\varepsilon_{2s}+J\pm K\), where the exchange integral \(K=\big\langle 1s\,2s\,\big|\,V_{12}\,\big|\,2s\,1s\big\rangle>0\) is a purely electrostatic quantity with no classical analogue, and the para–ortho separation is \(\Delta E=E_{\text{para}}-E_{\text{ortho}}=2K\), placing the triplet below the singlet even though \(V_{12}\) contains no spin operator.

Why it matters

This is the cleanest laboratory demonstration that spin controls energy without appearing in the Hamiltonian. The interaction \(V_{12}\) is spin-blind, yet helium's ortho and para levels are split by tenths of an electron-volt because antisymmetry ties the spatial correlation to the total spin. The same exchange integral \(K\), coarse-grained over neighbouring atoms, becomes the Heisenberg coupling that decides ferromagnetic versus antiferromagnetic order, and it is the microscopic content of Hund's first rule.

The result also explains a spectroscopic fact that puzzled early atomic physics: helium behaves almost like two non-combining elements. Because electric-dipole radiation cannot flip total spin, ortho and para states barely intercombine, and the lowest triplet \(2\,^3S\) is metastable with a lifetime of thousands of seconds — a direct macroscopic fingerprint of exchange symmetry.

Assumptions

The two electrons are strictly identical, so \([\hat H,\hat P_{12}]=0\).If any label distinguished them, exchange parity would not be conserved, the singlet/triplet classification would fail, and there would be no exchange splitting at all. The total two-electron state must be antisymmetric (spin–statistics for electrons).Drop it and both symmetry sectors would be allowed for each spin state; the observed one-to-one pairing of symmetric-space\(\leftrightarrow\)singlet and antisymmetric-space\(\leftrightarrow\)triplet — and hence the sign of the splitting — would be lost. \(V_{12}\) is spin-independent (no spin–orbit or spin–spin term at this order).If present, the space and spin factors no longer separate; \(S\) ceases to be a good quantum number and the clean \(\pm K\) structure smears into fine structure labelled by \(J\). The unperturbed \(1s2s\) configuration is treated by first-order degenerate perturbation theory in \(V_{12}\).The two products \(|1s\,2s\rangle\) and \(|2s\,1s\rangle\) are degenerate at zeroth order, so ordinary non-degenerate PT is singular; one must diagonalise \(V_{12}\) in the 2D subspace. Symmetry does this for free, but forgetting the degeneracy gives divergent corrections. The single-configuration (independent-particle) picture is adequate; orbitals are orthonormal hydrogenic states.Neglecting configuration mixing (e.g. \(1s2s\) with \(2p^2\)) and using unscreened orbitals makes \(J\) and \(K\) only estimates; the qualitative \(2K\) splitting survives but its magnitude is corrected.

Derivation

\[ \hat H=\underbrace{\hat h(1)+\hat h(2)}_{\hat H_0},\qquad \hat h(i)=-\frac{\hbar^2}{2m}\nabla_i^2-\frac{Ze^2}{4\pi\varepsilon_0 r_i},\qquad V_{12}=\frac{e^2}{4\pi\varepsilon_0\,r_{12}} \]

Split the helium Hamiltonian into two independent hydrogenic pieces plus the electron–electron repulsion, treated as the perturbation. \(\hat h\) has the known bound spectrum \(\varepsilon_{n}=-Z^2\,\mathrm{Ry}/n^2\). A

\[ \hat H_0\,\varphi_a(\mathbf r_1)\varphi_b(\mathbf r_2)=(\varepsilon_a+\varepsilon_b)\,\varphi_a(\mathbf r_1)\varphi_b(\mathbf r_2),\qquad (a,b)=(1s,2s) \]

Product hydrogenic orbitals diagonalise \(\hat H_0\). For \(a\neq b\) the two assignments \(|1s\,2s\rangle\) and \(|2s\,1s\rangle\) share the energy \(\varepsilon_{1s}+\varepsilon_{2s}\): the configuration is two-fold degenerate. A

\[ [\hat H,\hat P_{12}]=0\;\Rightarrow\; \psi_\pm(\mathbf r_1,\mathbf r_2)=\frac{1}{\sqrt2}\big[\varphi_a(\mathbf r_1)\varphi_b(\mathbf r_2)\pm\varphi_a(\mathbf r_2)\varphi_b(\mathbf r_1)\big] \]

Because the perturbation is symmetric under exchange, the good zeroth-order states inside the degenerate subspace are the eigenstates of \(\hat P_{12}\). Building them now pre-diagonalises \(V_{12}\): the \(2\times2\) degenerate-PT matrix is already diagonal in the \(\psi_\pm\) basis (the antisymmetric \(\psi_-\) and symmetric \(\psi_+\) cannot be connected by the symmetric operator \(V_{12}\)). B

\[ \Psi=\psi_{\text{space}}\otimes\chi_{\text{spin}},\qquad \mathbf 2\otimes\mathbf 2=\underbrace{\mathbf 3}_{\text{triplet, symmetric}}\oplus\underbrace{\mathbf 1}_{\text{singlet, antisymmetric}} \]

With \(V_{12}\) spin-independent, the state factorises into space and spin. Total antisymmetry then forces the pairing: symmetric \(\psi_+\) must carry the antisymmetric singlet (parahelium, \(S=0\)); antisymmetric \(\psi_-\) must carry the symmetric triplet (orthohelium, \(S=1\)). This is the Clebsch–Gordan decomposition of two spin-\(\tfrac12\). B

\[ E_\pm=\langle\psi_\pm|\hat H|\psi_\pm\rangle=\varepsilon_a+\varepsilon_b+\langle\psi_\pm|V_{12}|\psi_\pm\rangle \]

The first-order energy shift of each good state is the expectation of the perturbation; the \(\hat H_0\) part just returns the degenerate zeroth-order energy \(\varepsilon_a+\varepsilon_b\). A

\[ \langle\psi_\pm|V_{12}|\psi_\pm\rangle=\tfrac12\Big[\langle ab|V|ab\rangle+\langle ba|V|ba\rangle\pm\langle ab|V|ba\rangle\pm\langle ba|V|ab\rangle\Big] \]

Expand \(\psi_\pm\) inside the bracket and multiply out the four cross terms, using the shorthand \(|ab\rangle=\varphi_a(\mathbf r_1)\varphi_b(\mathbf r_2)\). No approximation yet — pure algebra. A

\[ \langle ab|V|ab\rangle=\langle ba|V|ba\rangle\equiv J,\qquad \langle ab|V|ba\rangle=\langle ba|V|ab\rangle\equiv K \]

Because \(V_{12}\) depends only on \(r_{12}=|\mathbf r_1-\mathbf r_2|\), it is symmetric under \(1\leftrightarrow2\); relabelling the integration variables equates the two direct terms and the two exchange terms. The bracket collapses to \(J\pm K\). B

\[ J=\frac{e^2}{4\pi\varepsilon_0}\!\iint\!\frac{|\varphi_a(\mathbf r_1)|^2\,|\varphi_b(\mathbf r_2)|^2}{r_{12}}\,d^3r_1\,d^3r_2 \]

\[ K=\frac{e^2}{4\pi\varepsilon_0}\!\iint\!\frac{\varphi_a^{*}(\mathbf r_1)\varphi_b^{*}(\mathbf r_2)\,\varphi_a(\mathbf r_2)\varphi_b(\mathbf r_1)}{r_{12}}\,d^3r_1\,d^3r_2 \]

Read off the two integrals. \(J\) is the classical Coulomb energy of the two charge clouds \(|\varphi_a|^2,|\varphi_b|^2\). \(K\) has the orbitals swapped between the coordinates: it is the Coulomb interaction of the overlap density \(\varrho(\mathbf r)=\varphi_a^{*}(\mathbf r)\varphi_b(\mathbf r)\) with itself, and vanishes if the orbitals do not overlap. C

\[ K=\frac{e^2}{4\pi\varepsilon_0}\!\iint\!\frac{\varrho(\mathbf r_1)\,\varrho(\mathbf r_2)}{r_{12}}\,d^3r_1\,d^3r_2=\frac{e^2}{4\pi\varepsilon_0}\,\frac{1}{2\pi^2}\!\int\!\frac{|\tilde\varrho(\mathbf k)|^2}{k^2}\,d^3k\;\ge 0 \]

For real orbitals \(\varrho=\varphi_a\varphi_b\) is real, so \(K\) is the electrostatic self-energy of the overlap charge density. Inserting the Fourier representation \(1/r_{12}=\frac{1}{2\pi^2}\int k^{-2}e^{i\mathbf k\cdot(\mathbf r_1-\mathbf r_2)}\,d^3k\) turns it into \(\int|\tilde\varrho(\mathbf k)|^2/k^2\,d^3k\), a manifestly non-negative integral. Hence \(K>0\) whenever the orbitals overlap. C

\[ E_{\text{para}}=E_+=\varepsilon_a+\varepsilon_b+J+K,\qquad E_{\text{ortho}}=E_-=\varepsilon_a+\varepsilon_b+J-K \]

\[ \Delta E=E_{\text{para}}-E_{\text{ortho}}=2K>0 \]

The singlet (symmetric space, \(\psi_+\)) sits at \(+K\); the triplet (antisymmetric space, \(\psi_-\)) at \(-K\). Their separation is \(2K\), and since \(K>0\) orthohelium lies below parahelium: the parallel-spin state is more stable. This is Hund's first rule realised in helium. A

Result

\[ \boxed{\;E_\pm=\varepsilon_{1s}+\varepsilon_{2s}+J\pm K,\qquad K=\frac{e^2}{4\pi\varepsilon_0}\!\iint\!\frac{\varphi_{1s}^{*}(\mathbf r_1)\varphi_{2s}^{*}(\mathbf r_2)\varphi_{1s}(\mathbf r_2)\varphi_{2s}(\mathbf r_1)}{r_{12}}\,d^3r_1 d^3r_2,\qquad \Delta E_{\text{para}-\text{ortho}}=2K\;}\]

Reading. Antisymmetrizing the two-electron wavefunction produces two energies differing only in the sign of the exchange integral. The spin-blind Coulomb repulsion nonetheless separates the spin singlet (para) and spin triplet (ortho) by \(2K\), because antisymmetry forces the triplet into a spatial state \(\psi_-\) that vanishes when the electrons coincide (the Fermi/exchange hole), lowering their mutual repulsion. Orthohelium is the lower, more stable term; the two ladders scarcely intercombine because photons carry no spin flip.

Units check. \(V_{12}=e^2/4\pi\varepsilon_0 r\) has units of energy (J); each \(|\varphi|^2 d^3r\) is dimensionless (normalised probability), so \(J\) and \(K\) are energy \(\times\) dimensionless \(=\) energy. \(\varepsilon_{1s},\varepsilon_{2s}\) are single-particle energies, so \(E_\pm\) and \(\Delta E=2K\) are energies. Consistent.

Limiting cases

Same orbital, \(1s^2\) ground configuration (\(a=b\)): \(\psi_-\equiv0\) — only the symmetric space state, hence only the singlet, survives. No triplet, no exchange splitting; the ground state of helium is necessarily para (\(1\,^1S\)).
Widely separated orbitals (small overlap, e.g. \(1s\,ns\) with large \(n\)): \(\varrho=\varphi_a\varphi_b\to0\), so \(K\to0\) and para/ortho become degenerate; the splitting shrinks up a Rydberg series.
\(1s2s\) vs \(1s2p\): the \(2p\) orbital overlaps the compact \(1s\) less than \(2s\) does, so \(K_{1s2p}<K_{1s2s}\); the \(2\,^{1,3}P\) splitting (\(\sim0.25\) eV) is smaller than the \(2\,^{1,3}S\) splitting (\(\sim0.80\) eV), as observed.
Bosonic replacement: for two identical spin-0 bosons the sign flips, the antisymmetric spatial state is forbidden, and only the \(+K\) branch exists — no exchange splitting of the kind seen in helium.
Turn off \(V_{12}\): \(J=K=0\), para and ortho collapse to the same hydrogenic energy \(\varepsilon_{1s}+\varepsilon_{2s}\); the entire splitting is interaction-driven.

Breaks when

Spin–orbit coupling becomes comparable to \(K\) (heavier two-electron ions, or the fine structure of the He triplet itself). Then \([\hat H,\hat S^2]\neq0\), the space–spin factorisation of step 4 fails, singlet/triplet mix, and states must be relabelled by total \(J\). Para–ortho intercombination lines appear (weakly).
Strong configuration interaction. When \(V_{12}\) is not small against the level spacing — near-degenerate configurations, doubly excited states, or two-electron ions with low \(Z\) — a single product \(|1s\,2s\rangle\) is inadequate; first-order \(J\pm K\) is only a leading estimate and multi-configuration (CI) methods are required.
Non-orthogonal or unscreened orbitals. Using bare hydrogenic \(Z=2\) orbitals ignores mutual screening and overestimates both integrals (predicted \(2K\approx2.4\) eV vs measured \(0.80\) eV for \(1s2s\)); the normalisation and clean \(J\pm K\) split also acquire overlap corrections if the orbitals are not orthonormal.
Reduced dimensionality / anyonic statistics (2D). The \(\pm1\) exchange eigenvalue is replaced by a braid phase; the singlet/triplet dichotomy underlying \(\Delta E=2K\) no longer holds.

Failure modes

"Exchange is a magnetic spin–spin force." \(K\) comes entirely from \(V_{12}=e^2/4\pi\varepsilon_0 r_{12}\), an electrostatic term with no spin operator. Spin enters only through the antisymmetry constraint on the spatial part.
Pairing symmetric space with the triplet. The symmetric \(\psi_+\) goes with the antisymmetric singlet (para); the antisymmetric \(\psi_-\) goes with the symmetric triplet (ortho). Swapping them flips the sign of the predicted splitting.
Using non-degenerate perturbation theory. Treating \(|1s\,2s\rangle\) alone gives a singular/ill-defined correction because \(|2s\,1s\rangle\) is degenerate; one must diagonalise in the 2D subspace (symmetry does it automatically).
Writing \(E_{\text{singlet}}=\varepsilon+J-K\). Sign error: the singlet takes \(+K\), the triplet \(-K\).
Forgetting spin in the exclusion count for \(1s^2\). Two electrons do share the \(1s\) orbital, with opposite spins — the antisymmetry is carried by the singlet spin factor, so the spatial part is symmetric.
Claiming ortho and para strictly never combine. Spin–orbit coupling makes intercombination lines weakly allowed (e.g. \(2\,^3P_1\to1\,^1S_0\)); "forbidden" means suppressed, not zero.
Treating \(J\) and \(K\) as free fit parameters rather than fixed integrals over the chosen orbitals; only their sizes are configuration-dependent, not their definitions.

Discussion

The physical heart of the calculation is the exchange hole. Antisymmetry makes \(\psi_-(\mathbf r_1,\mathbf r_2)\) vanish as \(\mathbf r_1\to\mathbf r_2\), so in the triplet the two electrons actively avoid each other and pay a smaller Coulomb penalty; the singlet's symmetric \(\psi_+\) has no such node and sits higher by \(2K\). Spin never enters the energy directly — it merely selects, through the antisymmetry rule, which spatial correlation the electrons are allowed to have. This is why a spin-independent repulsion produces an apparently spin-dependent energy, and it is the template for all "exchange" phenomena.

Historically this explained why helium's spectrum looked like two nearly independent term systems, once labelled parhelium and orthohelium as if they were different substances. They are one atom; the near-absence of intercombination lines is the electric-dipole selection rule \(\Delta S=0\), which follows because the dipole operator \(-e\sum_i\mathbf r_i\) acts only on space and leaves the orthogonal singlet/triplet spin factors unmixed. The \(2\,^3S\) state, having no lower triplet to decay to and being spin-forbidden from reaching \(1\,^1S\), is metastable for \(\sim7900\) s — one of the longest-lived neutral atomic states known, and a direct macroscopic consequence of exchange symmetry.

The construction is the two-electron seed of Hartree–Fock theory: the direct term \(J\) is the Hartree (mean-field) energy that a naive product state would give, and the exchange term \(K\) is exactly the Fock correction that the antisymmetrized (Slater-determinant) state adds. Scaling the same idea to a lattice of atoms, the overlap-density integral \(K\) becomes the Heisenberg exchange constant \(J_{\text{ex}}\); its sign — set here by the positivity argument \(K>0\) for a single overlap density — decides whether neighbouring spins prefer to align (ferromagnetism) or anti-align (superexchange can reverse it). Magnetism is helium's ortho/para splitting written at the scale of a solid.

A subtle point is that \(K>0\) is guaranteed only for a single-lobed real overlap density via the Fourier positivity \(K\propto\int|\tilde\varrho(\mathbf k)|^2/k^2\,d^3k\ge0\). In many-orbital or extended systems the effective coupling can change sign because the relevant "overlap density" is not a simple self-energy: kinetic-exchange and superexchange mechanisms, in which virtual hopping into occupied orbitals is Pauli-blocked, generically favour antiferromagnetic alignment. Thus Hund's-rule ferromagnetism (intra-atomic, direct exchange) and antiferromagnetic superexchange (inter-atomic, kinetic) are two faces of the same antisymmetry, distinguished only by which orbitals overlap.

Common misconceptions. There is no "exchange force" in the Hamiltonian and no spin-dependent potential; \(K\) is bookkeeping of ordinary Coulomb energy under an antisymmetry constraint. "Identical" is a strict statement — no operator distinguishes the electrons — not "similar." And the triplet is not held apart by a repulsion between parallel spins but by the node of an antisymmetric spatial wavefunction that parallel spins are forced to occupy.

Worked examples

Example 1 — Hydrogenic estimate of the \(1s2s\) exchange integral and splitting in helium (\(Z=2\)).

\[ J_{1s2s}=\frac{17}{81}\,Z\,\Big(\frac{e^2}{4\pi\varepsilon_0 a_0}\Big),\qquad K_{1s2s}=\frac{16}{729}\,Z\,\Big(\frac{e^2}{4\pi\varepsilon_0 a_0}\Big) \]

Standard closed-form results of the integrals in step 8 for unscreened hydrogenic \(1s\) and \(2s\) orbitals, in units of the Hartree \(e^2/4\pi\varepsilon_0 a_0=27.211\) eV. B

\[ K_{1s2s}=\frac{16}{729}\times2\times27.211\ \text{eV}=\frac{32}{729}\times27.211\ \text{eV} \]

Insert \(Z=2\) and the Hartree in eV; symbols before numbers. A

\[ K_{1s2s}=0.04390\times27.211\ \text{eV}=1.19\ \text{eV},\qquad J_{1s2s}=\frac{34}{81}\times27.211=11.4\ \text{eV} \]

Evaluate. The predicted para–ortho splitting is \(2K=2.39\) eV. A

\[ K_{1s2s}\approx1.19\ \text{eV},\qquad 2K\approx2.4\ \text{eV}\quad(\text{measured }0.80\ \text{eV}) \]

Reading. The unscreened hydrogenic estimate gets the sign and order of magnitude right (ortho below para by \(\sim\) eV) but overestimates the magnitude threefold because each electron partially screens the nucleus for the other, expanding the orbitals and reducing their overlap. Screening (effective \(Z_{\text{eff}}\approx1.7\) for the outer electron) brings \(2K\) down toward the observed \(0.80\) eV. Units check. Dimensionless rational \(\times\,Z\,\times\) Hartree (eV) = eV. Consistent.

Example 2 — Extracting \(K\) from the measured helium levels, \(1s2s\) and \(1s2p\).

\[ \Delta E=E(\,^1L)-E(\,^3L)=2K\;\Rightarrow\;K=\tfrac12\Delta E \]

The exchange integral is half the singlet–triplet separation of a given configuration, independent of \(J\) and of the one-electron energies, which cancel. A

\[ \Delta E_{1s2s}=E(2\,^1S)-E(2\,^3S)=20.62\ \text{eV}-19.82\ \text{eV}=0.80\ \text{eV} \]

Excitation energies above the \(1\,^1S\) ground state (NIST). Subtract. A

\[ K_{1s2s}=\tfrac12(0.80)=0.40\ \text{eV};\qquad \Delta E_{1s2p}=21.22-20.96=0.26\ \text{eV},\quad K_{1s2p}=0.13\ \text{eV} \]

Repeat for the \(1s2p\) configuration (\(2\,^1P\) at \(21.22\) eV, \(2\,^3P\) at \(20.96\) eV). A

\[ K_{1s2s}\approx0.40\ \text{eV}>K_{1s2p}\approx0.13\ \text{eV} \]

Reading. Both configurations have the triplet lower (ortho more stable), confirming \(K>0\). The \(1s2p\) exchange is about three times smaller because the \(2p\) orbital, kept away from the nucleus by its centrifugal barrier, overlaps the compact \(1s\) core much less than the penetrating \(2s\) does — and \(K\) is the self-energy of that overlap density. Units check. Differences of energies (eV), halved, give \(K\) in eV. Consistent.

Problems

Show that the ground configuration \(1s^2\) of helium admits no exchange splitting, and hence is necessarily a spin singlet.
Solution
Set \(\varphi_a=\varphi_b=\varphi_{1s}\) in \(\psi_\pm=\frac{1}{\sqrt2}[\varphi_a(\mathbf r_1)\varphi_b(\mathbf r_2)\pm\varphi_a(\mathbf r_2)\varphi_b(\mathbf r_1)]\). The antisymmetric combination gives \(\psi_-=\frac{1}{\sqrt2}[\varphi_{1s}(\mathbf r_1)\varphi_{1s}(\mathbf r_2)-\varphi_{1s}(\mathbf r_2)\varphi_{1s}(\mathbf r_1)]=0\). Only the symmetric spatial state \(\psi_+\) survives, and total antisymmetry then requires the antisymmetric singlet spin factor. There is no triplet partner, so no \(\pm K\) pair exists: the ground state is \(1\,^1S\) with energy \(2\varepsilon_{1s}+J\) (a single \(J\), no \(K\) splitting).
The measured helium \(2\,^1S\) and \(2\,^3S\) levels lie \(20.62\) eV and \(19.82\) eV above the ground state. Find \(K_{1s2s}\), the ordering of the levels, and state why the \(2\,^3S\) state is metastable.
Solution
\(\Delta E=2K=20.62-19.82=0.80\) eV, so \(K=0.40\) eV. Since \(K>0\), \(E_{\text{ortho}}=\varepsilon+J-K<E_{\text{para}}=\varepsilon+J+K\): the triplet \(2\,^3S\) lies \(0.80\) eV below the singlet \(2\,^1S\). The \(2\,^3S\) state is the lowest triplet, so it cannot decay to a lower triplet; decay to \(1\,^1S\) requires \(\Delta S=1\), which the electric-dipole operator (space-only) cannot supply. It is therefore metastable (lifetime \(\sim7900\) s, decaying by a highly suppressed magnetic-dipole/relativistic channel).
Using the hydrogenic formulas \(J_{1s2s}=\frac{17}{81}Z\) and \(K_{1s2s}=\frac{16}{729}Z\) (in Hartree), evaluate \(J\), \(K\) and \(2K\) for \(Z=2\), and compare \(2K\) with the experimental \(0.80\) eV.
Solution
Hartree \(=27.211\) eV. \(J=\frac{17}{81}(2)(27.211)=\frac{34}{81}(27.211)=0.4198\times27.211=11.4\) eV. \(K=\frac{16}{729}(2)(27.211)=\frac{32}{729}(27.211)=0.04390\times27.211=1.19\) eV, so \(2K=2.39\) eV. The unscreened estimate exceeds experiment (\(0.80\) eV) by about \(3\times\): mutual screening lowers the effective nuclear charge seen by the outer electron (\(Z_{\text{eff}}\approx1.7\)), expands the orbitals, reduces the \(1s\)–\(2s\) overlap and hence \(K\). The sign and eV scale are nonetheless correct.
For the \(1s2p\) configuration the \(2\,^1P\) and \(2\,^3P\) terms lie \(21.22\) eV and \(20.96\) eV above the ground state. Find \(K_{1s2p}\) and explain why it is smaller than \(K_{1s2s}\).
Solution
\(\Delta E=2K=21.22-20.96=0.26\) eV, so \(K_{1s2p}=0.13\) eV, versus \(K_{1s2s}=0.40\) eV. The exchange integral is the Coulomb self-energy of the overlap density \(\varrho=\varphi_{1s}\varphi_{2\ell}\). The \(2p\) orbital has an \(\ell=1\) centrifugal barrier \(\propto\ell(\ell+1)/r^2\) that keeps it out of the region near the nucleus where the compact \(1s\) orbital has its amplitude, so the \(1s\)–\(2p\) overlap is much smaller than the \(1s\)–\(2s\) overlap. Smaller overlap density \(\Rightarrow\) smaller \(K\). The triplet is still the lower term (\(K>0\)).
Show that the electric-dipole operator cannot connect an ortho (triplet) state to a para (singlet) state, i.e. establish the selection rule \(\Delta S=0\) responsible for the near-separation of the two term systems.
Solution
The E1 transition amplitude is \(\langle\Psi_f|\hat{\mathbf d}|\Psi_i\rangle\) with \(\hat{\mathbf d}=-e(\mathbf r_1+\mathbf r_2)\), which acts only on spatial coordinates and is the identity on spin. Write \(\Psi_i=\psi_i^{\text{space}}\chi_i^{\text{spin}}\), \(\Psi_f=\psi_f^{\text{space}}\chi_f^{\text{spin}}\). Then \(\langle\Psi_f|\hat{\mathbf d}|\Psi_i\rangle=\langle\psi_f^{\text{space}}|\hat{\mathbf d}|\psi_i^{\text{space}}\rangle\,\langle\chi_f^{\text{spin}}|\chi_i^{\text{spin}}\rangle\). The singlet and triplet spin states are orthogonal, \(\langle\chi_{S=0}|\chi_{S=1}\rangle=0\), so the amplitude vanishes: no first-order E1 transition changes \(S\). Hence ortho and para helium form nearly independent spectra, coupled only weakly through spin–orbit-induced mixing.