Part 5 · Identical particles

5.2 The helium atom

Source: the Hamiltonian, the independent-electron approximation, the ground state, para and ortho states and the first-order correction from the 2024/25 board #19; this year’s course has not reached it, and it may be dropped

Objectives

Understand how the symmetry rule for electrons fixes the spin of helium’s states, and why the triplet lies below the singlet.
Recipe: write two-electron states as a spatial part times a spin part of opposite symmetry.
Picture: helium’s lowest levels, from independent electrons to the measured values.
Theorem: $E = E_{n_{1}} + E_{n_{2}}$ without repulsion; the 1s2s level splits by $2 K$ , ortho below para.
Vocabulary: independent-electron approximation, parahelium, orthohelium, direct and exchange integral.

A nucleus of charge $+ Z e$ , $Z = 2$ , and two electrons: the simplest system where identical particles matter. The nucleus sits at the origin, $\overset{r}{ˉ}_{1}$ and $\overset{r}{ˉ}_{2}$ run to the electrons, and $r_{12} = ∣ \overset{r}{ˉ}_{1} - \overset{r}{ˉ}_{2} ∣$ is their distance.

The Hamiltonian

Kinetic energy of the two electrons plus three Coulomb energies, each electron with the nucleus and the electrons with each other. With $e_{n}^{2} = e^{2} /4 π ε_{0}$ , $\hat{H} = - \frac{ℏ ^{2}}{2 μ} (\nabla_{1}^{2} + \nabla_{2}^{2}) - Z e_{n}^{2} (\frac{1}{r _{1}} + \frac{1}{r _{2}}) + \frac{e _{n}^{2}}{r _{12}} .$ $\nabla_{1}^{2}$ acts on electron 1 only. $μ = m_{e} M / (m_{e} + M)$ is the reduced mass, within $1 0^{- 4}$ of $m_{e}$ for helium (beyond the boards). The nucleus terms attract, the last one repels. $\hat{H}$ is unchanged by swapping the electrons, as it must be (Theorem 5.1.2), and has no spin in it.

Independent electrons

The repulsion couples the electrons and makes the equation unsolvable exactly. ==Drop it== (the board crosses it out): each electron then feels only the nucleus.

Theorem 5.2.1Independent-electron approximation2024/25 board #19

Without $e_{n}^{2} / r_{12}$ , $\hat{H} = \hat{H}_{1} + \hat{H}_{2}$ , two hydrogen-like Hamiltonians with nuclear charge $Z e$ . Products of their orbitals are eigenstates, with $E = E_{n_{1}} + E_{n_{2}}, E_{n} = - \frac{Z ^{2} e _{n}^{2}}{2 a _{0}^{'} n ^{2}} = - Z^{2} \frac{13.6 eV}{n ^{2}} .$

Why the energies addbeyond the boards

$\hat{H}_{1}$ acts only on $\overset{r}{ˉ}_{1}$ , $\hat{H}_{2}$ only on $\overset{r}{ˉ}_{2}$ : $(\hat{H}_{1} + \hat{H}_{2}) u_{a} (1) u_{b} (2) = (\hat{H}_{1} u_{a}) u_{b} + u_{a} (\hat{H}_{2} u_{b}) = (E_{a} + E_{b}) u_{a} (1) u_{b} (2)$ . Any symmetric or antisymmetric combination of $u_{a} (1) u_{b} (2)$ and $u_{b} (1) u_{a} (2)$ has the same energy.

$a_{0}^{'} = ℏ^{2} / μ e_{n}^{2} \approx 0.529$ Å is the Bohr radius, and $e_{n}^{2} /2 a_{0} = 13.6$ eV hydrogen’s ionisation energy (beyond the boards). For helium, $E_{1} = - 54.4$ eV and $E_{2} = - 13.6$ eV.

Ground state

Both electrons in $u_{100}$ : $E_{0} = - Z^{2} e_{n}^{2} / a_{0}^{'} = - 108.8$ eV. (The board writes $e_{n}$ without the square.) The state is a spatial part times a spin part, and electrons are fermions, so the whole must be antisymmetric. Symmetries multiply like signs:

spatial	spin	total
S	A.S.	A.S. ✓
A.S.	S	A.S. ✓
S	S	S ✗

The spatial part is the same orbital twice, symmetric, so ==the spin part must be the singlet==: $ψ_{0} = u_{100} (1) u_{100} (2) \frac{1}{2} [∣ + ⟩_{1} ∣ - ⟩_{2} - ∣ - ⟩_{1} ∣ + ⟩_{2}],$ the 1s² state of §5.1 Identical particles. The ground state has total spin 0 and is non-degenerate: a triplet would need an antisymmetric spatial part, and $u_{100} (1) u_{100} (2)$ has none (beyond the boards).

First excited state: para and ortho

One electron in $n = 1$ , the other in $n = 2$ : $E = - \frac{Z ^{2} e _{n}^{2}}{a _{0}^{'}} (\frac{1}{2} + \frac{1}{8}) = - \frac{5}{8} \frac{Z ^{2} e _{n}^{2}}{a _{0}^{'}} = - 68.0$ eV. With $u_{100}$ and $u_{200}$ the spatial part can be either $\frac{1}{2} [u_{100} (1) u_{200} (2) \pm u_{200} (1) u_{100} (2)],$ and each goes with the spin part of opposite symmetry.

Theorem 5.2.2Para and ortho states2024/25 board #19

Para, spin singlet, $s = 0$ , one state: the symmetric spatial part times $\frac{1}{2} (∣ + ⟩_{1} ∣ - ⟩_{2} - ∣ - ⟩_{1} ∣ + ⟩_{2})$ .
Ortho, spin triplet, $s = 1$ , three states: the antisymmetric spatial part times $∣ + ⟩_{1} ∣ + ⟩_{2}$ ( $m = 1$ ), $\frac{1}{2} (∣ + ⟩_{1} ∣ - ⟩_{2} + ∣ - ⟩_{1} ∣ + ⟩_{2})$ ( $m = 0$ ) or $∣ - ⟩_{1} ∣ - ⟩_{2}$ ( $m = - 1$ ).

All four have the same energy while the repulsion is dropped.

The board writes the last spatial factor of the ortho rows as $u_{1, 0} (1)$ ; it is $u_{100} (2)$ . The prefactor is $\frac{1}{2}$ when both brackets are sums of two terms, $\frac{1}{2}$ when the spin part is a single product. $n = 2$ also has $ℓ = 1$ , so 1s2p gives twelve more states at the same unperturbed energy; the board and the exam use only 1s2s (beyond the boards).

The names (beyond the boards): helium’s spectrum looked like that of two gases, “parahelium” and “orthohelium”, because emitting light almost never connects a singlet to a triplet; the photon does not act on spin. So the lowest ortho state cannot decay to the singlet ground state by emitting light, and lives very long.

The repulsion put back

The board ends with how to restore the dropped term: its average in the approximate state, $Δ E = ⟨ ψ \frac{e _{n}^{2}}{r _{12}} ψ ⟩ .$ The spin part gives $⟨ χ ∣ χ ⟩ = 1$ ; only the spatial part matters. This is first-order perturbation theory, good when the dropped term is small; everything below is beyond the boards.

Ground state. $Δ E$ is the electrostatic energy between two 1s charge clouds, $\frac{5}{8} \frac{Z e _{n}^{2}}{a _{0}^{'}} = 34.0$ eV, so $E_{0} \approx - 74.8$ eV against $- 79.0$ eV measured: from 38% off to 5% off.

Key observation 5.2.3Why ortho lies below parabeyond the boards

With the spatial part $\frac{1}{2} [u_{100} (1) u_{200} (2) \pm u_{200} (1) u_{100} (2)]$ , the average of the repulsion is $J \pm K$ :

$J$ , the direct integral, is the repulsion between a 1s and a 2s cloud, as classical charges;
$K$ , the exchange integral, is a purely quantum cross term from the two orderings, and positive.

$+$ for para, $-$ for ortho: ==the triplet is lower by $2 K$ .== The antisymmetric spatial function vanishes at $\overset{r}{ˉ}_{1} = \overset{r}{ˉ}_{2}$ , so in the triplet the electrons keep apart and repel less. Spin is not a force here: the symmetry rule ties spin to spatial shape, and the shape sets the repulsion.

Loading the demo…

Figure 5.2.1 Helium’s lowest levels in three steps: independent electrons, the first-order repulsion, and the measured values, as total binding energies with He²⁺ + 2e⁻ at zero. The dashed line is He⁺ + e⁻, one electron free. Step through: the repulsion lifts everything, and splits 1s2s into

para

above and

ortho

below,

2 K

apart. The values of

J

and

K

are textbook ones, not from the course.

Measured, the 1s2s ortho level is at $- 59.2$ eV and para at $- 58.4$ eV. Chemistry’s Hund’s rule, parallel spins in separate orbitals first, is this same effect.

Exercises

Exercise 5.2.1Question c), (i)–(ii), sample questions part II

Write the Hamiltonian of He. Find the ground-state wavefunction and energy, neglecting the repulsion between the electrons, in terms of the hydrogenic $u_{n, ℓ, m_{ℓ}}$ and spin states $∣ s, m_{s} ⟩$ . (i) Find the first 4 excited states, their energy, and discuss how the degeneracy is removed. (ii) Evaluate the contribution of the repulsion to the ground-state energy.

Solution

$\hat{H}$ as above. Without the repulsion, $\hat{H}_{1} + \hat{H}_{2}$ and $E = E_{n_{1}} + E_{n_{2}}$ . Ground state $u_{100} (1) u_{100} (2) ∣0, 0 ⟩$ , the singlet because the spatial part is symmetric; $E_{0} = - 108.8$ eV. (i) 1s2s at $- 68.0$ eV: para $\frac{1}{2} [u_{100} (1) u_{200} (2) + u_{200} (1) u_{100} (2)] ∣0, 0 ⟩$ , ortho the antisymmetric spatial part times $∣1, m ⟩$ , $m = 1, 0, - 1$ . The degeneracy is only because the repulsion was dropped. Restored, para shifts by $J + K$ and ortho by $J - K$ : a singlet above and a triplet below, $2 K$ apart, because the antisymmetric spatial function keeps the electrons apart. The triplet stays threefold degenerate, $\hat{H}$ having no spin. (Strictly, 1s2p is degenerate with these too.) (ii) $Δ E = \iint ∣ u_{100} (1) ∣^{2} ∣ u_{100} (2) ∣^{2} \frac{e _{n}^{2}}{r _{12}} d^{3} r_{1} d^{3} r_{2} = \frac{5}{8} \frac{Z e _{n}^{2}}{a _{0}^{'}} = 34$ eV. Corrected $- 74.8$ eV, against $- 79.0$ measured. Positive, as a repulsion must be, and a third of the unperturbed energy: that is why first order still misses by 4 eV.

Exercise 5.2.2beyond the boards

Ionised helium, He⁺, has one electron. What is its ground-state energy, and the energy needed to remove the first electron from He (measured $- 79.0$ eV)?

Solution

He⁺ is hydrogen-like with $Z = 2$ : $- 54.4$ eV. Removing one electron from He costs $- 54.4 - (- 79.0) = 24.6$ eV, the largest first ionisation energy of any element.

Exercise 5.2.3beyond the boards

Can two electrons be in the triplet with both in 1s?

Solution

They cannot: the triplet’s spin part is symmetric, so the spatial part must be antisymmetric, $\frac{1}{2} [u_{100} (1) u_{100} (2) - u_{100} (1) u_{100} (2)] = 0$ . Pauli again, in spatial form.

Exercise 5.2.4beyond the boards

Lithium ( $Z = 3$ ) in the independent-electron approximation. Why can all three electrons not be in 1s, and what is the ground-state energy?

Solution

1s holds two states, up and down; a third electron would repeat one, and the antisymmetric state would vanish. It goes to $n = 2$ . $E = 2 (- 9 \cdot 13.6) + (- 9 \cdot 13.6/4) = - 275.4$ eV, against $- 203.5$ eV measured: dropping the repulsion overbinds even more than in helium.