Part 5 · Identical particles

5.1 Identical particles

Source: the exchange operator, fermions and bosons, two particles in two states and two electrons in 1s from the 2024/25 boards #18 (C3–D4) and #19; this year’s course has not reached it, and it may be dropped

Objectives

Understand why a state of two identical particles must be symmetric or antisymmetric under exchange, and what that does to how they share states.
Recipe: build the allowed two-particle states from two single-particle states, for distinguishable particles, fermions and bosons.
Picture: the joint density $∣ ψ (x_{1}, x_{2}) ∣^{2}$ , empty along $x_{1} = x_{2}$ for fermions and piled up there for bosons.
Theorem: the eigenvalues of exchange are $\pm 1$ ; two identical fermions never share a state.
Vocabulary: exchange operator, symmetric and antisymmetric state, fermion, boson.
Essential vocabulary: Pauli exclusion principle.

Remark

Centini’s 2023/24 programme does not list identical particles: its blocks end with angular momentum. Older programmes close with “systems of identical particles: two electrons, the helium atom, the excited states”, and the 2024/25 boards cover it in their last two lectures. Of all the carry-over, this part and §5.2 The helium atom are the most likely to be dropped.

The exchange operator

Classically, two particles each follow their own path and can always be tracked: “the one that started at 1” stays a meaningful label. In quantum mechanics there are no paths, only a wavefunction spreading over space, and once two wavefunctions overlap no measurement says which particle is which. ==Identical particles are indistinguishable: swapping them must not change anything observable.==

A two-particle state is a function of both particles’ coordinates, position and spin; the simplest is a product $ψ (1, 2) = u (1) u (2)$ , where “1” stands for all coordinates of particle 1. $∣ ψ (1, 2) ∣^{2}$ is the density of finding particle 1 at its coordinates and particle 2 at its.

Definition 5.1.1Exchange operator2024/25 board #18

$\hat{P}_{12}$ swaps the two particles: $\hat{P}_{12} ψ (1, 2) = ψ (2, 1)$ .

Theorem 5.1.2Symmetric and antisymmetric states2024/25 board #18

The eigenvalues of $\hat{P}_{12}$ are $λ = \pm 1$ :

$λ = + 1$ : symmetric states, $ψ (1, 2) = ψ (2, 1)$ ;
$λ = - 1$ : antisymmetric states, $ψ (1, 2) = - ψ (2, 1)$ .

Proof

Swapping twice gives back the original, $\hat{P}_{12}^{2} = \hat{I}$ , so $λ^{2} ψ = ψ$ . (Board 18 writes $∣ λ ∣^{2}$ , board 19 $λ^{2}$ ; $\hat{P}_{12}$ is Hermitian, so both give $\pm 1$ .)

Why a state of identical particles must be one or the other (beyond the boards): every probability is unchanged by the swap, $∣ ψ (2, 1) ∣^{2} = ∣ ψ (1, 2) ∣^{2}$ , so $ψ (2, 1)$ is $ψ (1, 2)$ times a phase. And the Hamiltonian of two identical particles is symmetric in them, $[\hat{H}, \hat{P}_{12}] = 0$ , so energy eigenstates can be chosen symmetric or antisymmetric (The Compatible Observables Theorem) and stay so.

Key observation 5.1.3Fermions and bosons2024/25 board #18

spin	states	name
half-integer, $\frac{1}{2}, \frac{3}{2}, \dots$	antisymmetric only	fermions
integer, $0, 1, 2, \dots$	symmetric only	bosons

Electrons, protons, neutrons and ³He atoms are fermions; photons, ⁴He atoms and the deuteron are bosons. A composite is a fermion if it holds an odd number of fermions (beyond the boards). The board states the rule; its proof, the spin–statistics theorem, needs relativistic quantum field theory. (The board lists integer spin from 1; spin 0 belongs too.)

Two particles, two states

Two single-particle states $k^{'}$ and $k^{''}$ , two particles. $∣ k^{'} ⟩_{1}$ means particle 1 in $k^{'}$ .

Recipe 5.1.4Allowed states of two particles2024/25 board #18

Distinguishable: four, $∣ k^{'} ⟩_{1} ∣ k^{'} ⟩_{2}$ , $∣ k^{'} ⟩_{1} ∣ k^{''} ⟩_{2}$ , $∣ k^{''} ⟩_{1} ∣ k^{'} ⟩_{2}$ , $∣ k^{''} ⟩_{1} ∣ k^{''} ⟩_{2}$ .
The two doubly occupied ones are symmetric. The two mixed ones are not eigenstates of $\hat{P}_{12}$ : exchange turns one into the other. For identical particles only their combinations are allowed, $\frac{1}{2} (∣ k^{'} ⟩_{1} ∣ k^{''} ⟩_{2} \pm ∣ k^{''} ⟩_{1} ∣ k^{'} ⟩_{2}),$ $+$ symmetric, $-$ antisymmetric.
Fermions keep the antisymmetric combination: one state.
Bosons keep the symmetric ones: the two doubly occupied states and the $+$ combination, three states.

The $\frac{1}{2}$ normalises, since the two products are orthogonal. To check the antisymmetric one, swap the labels and reorder each product with particle 1 first: the bracket changes sign.

Theorem 5.1.5Pauli exclusion principleEssential

Two identical fermions never occupy the same single-particle state.

Proof

Put $k^{''} = k^{'}$ in the antisymmetric state: $\frac{1}{2} (∣ k^{'} ⟩_{1} ∣ k^{'} ⟩_{2} - ∣ k^{'} ⟩_{1} ∣ k^{'} ⟩_{2}) = 0$ . It does not exist. (The board shows it by what is missing; the algebra is beyond the boards.)

==The doubly occupied states the bosons have, the board’s yellow “!”, are exactly what fermions lack.== For electrons a “state” includes spin: $k^{'}$ and $k^{''}$ can be the same orbital with opposite spins. That is chemistry’s rule of at most two electrons per orbital, with opposite spins, and of no two electrons sharing all four quantum numbers. The board also draws a ladder of levels filled from the bottom, at most two each, as in an atom.

Counting (beyond the boards): put two particles at random into the two states, every allowed state equally likely. Both end up in the same one with probability $\frac{2}{4}$ classically, $\frac{2}{3}$ for bosons and $0$ for fermions. ==Bosons bunch, fermions avoid each other==: this is where Bose–Einstein and Fermi–Dirac statistics come from.

Loading the demo…

Figure 5.1.1 Two particles in the same 1 nm infinite well, one in level

n

and one in

n^{'}

. The square is the plane of

(x_{1}, x_{2})

; its

shading

∣ ψ (x_{1}, x_{2}) ∣^{2}

, and the dashed diagonal is

x_{1} = x_{2}

, the two at the same place. The antisymmetric state is zero all along it: fermions with the same spin are never found together. The symmetric state piles up there. Set

n = n^{'}

and the antisymmetric state vanishes entirely: Pauli.

Two electrons in 1s

1s is the orbital $u_{100}$ , with $n = 1$ , $ℓ = 0$ , $m = 0$ . Put two electrons in it, one up and one down. The spatial part $u_{100} (1) u_{100} (2)$ is the same function twice, symmetric, so ==the antisymmetry has to sit in the spin==: $ψ = u_{100} (1) u_{100} (2) \frac{1}{2} [∣ + ⟩_{1} ∣ - ⟩_{2} - ∣ - ⟩_{1} ∣ + ⟩_{2}] .$ The spin factor is the singlet, total spin $s = 0$ (Theorem 4.4.3). The two electrons of a filled orbital are not just opposite one by one: they form a state with zero total spin. That is why a filled shell has no magnetic moment (beyond the boards).

Exercises

Exercise 5.1.1Question b) and (ii), sample questions part II

What are fermions and bosons? Discuss the differences between the wavefunctions of systems of identical fermions and identical bosons. (ii) Write, compare and examine the possible configurations of two identical particles in two-level systems for a) classical distinguishable particles, b) fermions, c) bosons.

Solution

Indistinguishable particles are in eigenstates of exchange, $λ = \pm 1$ (Theorem 5.1.2). Fermions, half-integer spin: antisymmetric, $ψ (1, 2) = - ψ (2, 1)$ . Bosons, integer spin: symmetric. So two fermions can never share a state (Pauli); bosons can, any number of them. Electrons fill atomic levels two per orbital; bosons all pile into the lowest state at low temperature, Bose–Einstein condensation and superfluid ⁴He (beyond the boards). (ii) a) four states, the mixed ones different since the particles can be told apart; b) one, $\frac{1}{2} (∣ k^{'} ⟩_{1} ∣ k^{''} ⟩_{2} - ∣ k^{''} ⟩_{1} ∣ k^{'} ⟩_{2})$ , no double occupation and no “this one in $k^{'}$ ”; c) three, the two doubly occupied and the symmetric combination. Both in the same level: $\frac{1}{2}$ , $0$ , $\frac{2}{3}$ .

Exercise 5.1.2beyond the boards

Two electrons in a 1D infinite well, levels $E_{n} = n^{2} E_{1}$ , no interaction between them. Ground state and first excited level: wavefunctions, energies, degeneracies. And for two spinless bosons?

Solution

Ground: both in $n = 1$ , spatial $ϕ_{1} (x_{1}) ϕ_{1} (x_{2})$ symmetric, spin singlet; $2 E_{1}$ , one state. First excited: $n = 1$ and $2$ , $5 E_{1}$ : symmetric spatial with the singlet (1 state) and antisymmetric spatial with the triplet (3 states), the same structure as helium’s. Two spinless bosons: $ϕ_{1} ϕ_{1}$ at $2 E_{1}$ , and only the symmetric combination at $5 E_{1}$ , one state.

Exercise 5.1.3beyond the boards

Show that $\frac{1}{2} (∣ + ⟩_{1} ∣ - ⟩_{2} + ∣ - ⟩_{1} ∣ + ⟩_{2})$ is symmetric, normalised, and orthogonal to the singlet.

Solution

Swapping gives $∣ - ⟩_{1} ∣ + ⟩_{2} + ∣ + ⟩_{1} ∣ - ⟩_{2}$ , the same. Norm $\frac{1}{2} (1 + 1) = 1$ , the cross terms containing $⟨ + ∣ - ⟩ = 0$ . Overlap with the singlet $\frac{1}{2} (1 - 1) = 0$ .