Bosons and fermions

The last section ended in a forced choice: the state of identical particles is either fully symmetric or fully antisymmetric. The question it left behind — who picks which?

Start with two experimental facts that hint the answer has to do with spin.

One neutron apart, two different worlds

Helium has two stable isotopes. $^4$He has 2 protons and 2 neutrons; $^3$He has one neutron fewer. Chemically they are all but indistinguishable — both are noble gases too lazy to react. But cool them down:

• $^4$He undergoes a superfluid transition at 2.17 K: its viscosity drops below anything measurable, and the liquid can “creep” up the wall of a cup and out over the rim.
• $^3$He does nothing at 2.17 K. Keep cooling by three more orders of magnitude, down to about 2.6 mK, before it grudgingly turns superfluid — and by an entirely different mechanism (the atoms must first pair up two by two).

One neutron of difference, and the transition temperatures differ by nearly a factor of a thousand. The mass ratio ($4:3$) comes nowhere near explaining the gap. The same contrast recurs elsewhere: photons happily pile into a single mode by the trillions (the principle of the laser), while electrons would sooner die than share a state (the protagonist of the next section). Nature’s particles evidently come in two temperaments: one loves to bunch, the other keeps its distance.

The mathematics of the last section has conveniently prepared two drawers, symmetric and antisymmetric. Now we sort the two temperaments into them.

The two great families

Bosons: particles whose wavefunction is symmetric under exchange of any two of them. Examples: photons (spin 1), $^4$He atoms, the Higgs particle (spin 0).

Fermions: particles whose wavefunction is antisymmetric under exchange of any two of them. Examples: electrons, protons, neutrons, neutrinos (all spin 1/2), $^3$He atoms.

The names honour Bose and Fermi — the statistical distributions each family obeys are named after them. And what is the criterion? Look at the spin:

The “half-integer spin” criterion also explains the fork between $^4$He and $^3$He — we will do that count shortly. First, let us build the wavefunctions.

The construction: from two particles to N

The two-particle case was written down last section:

$$\psi_\pm(\vec{r}_1,\vec{r}_2)=\frac{1}{\sqrt2}\bigl[\phi_a(\vec{r}_1)\phi_b(\vec{r}_2)\pm\phi_b(\vec{r}_1)\phi_a(\vec{r}_2)\bigr]\tag{8.2.1}$$

With $N$ particles, every pairwise exchange must produce the same sign. The recipe: list all $N!$ ways of assigning particles to single-particle states, attach a sign according to the parity of the permutation, and sum.

The antisymmetric wavefunction for three fermions, and the Slater determinantbasic~7 minexpand

Step 1: list all the permutations. Take three orthonormal single-particle states $\phi_a,\phi_b,\phi_c$, and write the coordinates (including spin) of the three particles as $x_1,x_2,x_3$. There are $3!=6$ assignments:

$$\phi_a(x_1)\phi_b(x_2)\phi_c(x_3),\quad \phi_a(x_1)\phi_c(x_2)\phi_b(x_3),\quad\dots\tag{8.2.2}$$

Step 2: attach signs by parity. Permutations reached from the identity by one transposition get a minus sign, those reached by two get a plus (one can check this notion of parity is self-consistent):

$$\psi_A=\frac{1}{\sqrt{6}}\bigl[ \phi_a(x_1)\phi_b(x_2)\phi_c(x_3)-\phi_a(x_1)\phi_c(x_2)\phi_b(x_3) +\phi_b(x_1)\phi_c(x_2)\phi_a(x_3)\tag{8.2.3}$$$$\qquad\qquad -\phi_b(x_1)\phi_a(x_2)\phi_c(x_3) +\phi_c(x_1)\phi_a(x_2)\phi_b(x_3)-\phi_c(x_1)\phi_b(x_2)\phi_a(x_3)\bigr]\tag{8.2.4}$$

Exchange any pair of particles (say $x_1\leftrightarrow x_2$) and the six terms swap positions in pairs with opposite signs, so the whole thing changes sign — antisymmetry holds.

Step 3: recognise the determinant. This signed sum of six terms is exactly the expansion of a $3\times3$ determinant:

$$\psi_A(x_1,x_2,x_3)=\frac{1}{\sqrt{3!}} \begin{vmatrix} \phi_a(x_1) & \phi_b(x_1) & \phi_c(x_1)\\ \phi_a(x_2) & \phi_b(x_2) & \phi_c(x_2)\\ \phi_a(x_3) & \phi_b(x_3) & \phi_c(x_3) \end{vmatrix}\tag{8.2.5}$$

Each row corresponds to a particle, each column to a single-particle state. Generalised to $N$ particles this becomes the $N\times N$ Slater determinant, with normalisation $1/\sqrt{N!}$.

Step 4: determinant properties translate straight into physics.

• Swapping two rows flips the sign — exchanging two particles makes the wavefunction antisymmetric. ✔
• Two equal columns make the determinant zero — when two particles occupy the same single-particle state, $\psi_A\equiv0$: no such state exists. That is the entire mathematical content of next section’s Pauli exclusion principle, delivered free of charge.
• Linear combinations of the columns leave the state unchanged (up to a constant) — there is considerable freedom in “which set of orbitals”, and Hartree–Fock in section 8.6 will exploit it.

The symmetric, bosonic version is analogous: take every sign positive (this all-plus “determinant” is called the permanent), and allow several particles to share the same state.

The division of labour between space and spin: two electrons

The spin states of two electrons (spin 1/2) were combined in section 5.6: three states with total spin $S=1$ (the triplet) and one with $S=0$ (the singlet). Check their exchange symmetry:

What about composite particles?

A proton is made of three quarks; a $^4$He atom of 12 fermions — which statistics does a composite particle obey?

The rule is simple: exchanging two composite particles = simultaneously exchanging every corresponding pair of their internal constituents. Each internal fermion pair exchanged contributes a $-1$; each boson pair a $+1$. Therefore:

A composite containing an even number of fermions is a boson; one containing an odd number is a fermion.

• $^4$He: 2 protons + 2 neutrons + 2 electrons = 6 fermions → boson. It may pile collectively into the lowest quantum state; the 2.17 K superfluid is, at bottom, exactly this “macroscopic bunching”.
• $^3$He: 2 protons + 1 neutron + 2 electrons = 5 fermions → fermion. Bunching is forbidden; to go superfluid the atoms must first form pairs (a pair being a boson), at the price of a transition temperature squeezed down to the mK scale.
• A hydrogen atom (1 proton + 1 electron) = 2 fermions → boson. The rubidium-87 atoms that achieved Bose–Einstein condensation in 1995 (37 electrons + 37 protons + 50 neutrons = 124 fermions) are bosons too, condensing at about 170 nK.

This rule is consistent with the spin-statistics theorem: an odd number of half-integer spins combines to a half-integer total spin, an even number to an integer one (the angular-momentum addition rules of chapter 5).

∎Key formulas

• Slater determinant

  $$\psi_A=\frac{1}{\sqrt{N!}}\det\bigl[\phi_j(x_i)\bigr]$$

  Rows = particles, columns = spin-orbitals; two equal columns make it vanish identically

• Spin-statistics theorem

  $$\text{integer spin}\Rightarrow\text{symmetric (bosons)};\quad \text{half-integer}\Rightarrow\text{antisymmetric (fermions)}$$

  A result of relativistic quantum field theory; stated here without proof

• Two-electron division of labour

  $$\Psi=\psi_{\text{space}}\times\chi_{\text{spin}},\quad (+)\times(-)\ \text{or}\ (-)\times(+)$$

  Singlet pairs with symmetric space, triplet with antisymmetric space

• Composite-particle rule

  $$\text{odd number of fermions}\Rightarrow\text{fermion};\quad\text{even}\Rightarrow\text{boson}$$

  ⁴He is a boson, ³He a fermion

?Self-check4 questions

1. 1.

   ⁴He turns superfluid at 2.17 K while ³He waits until about 2.6 mK. The fundamental reason is:

   • ⁴He contains an even number of fermions and is a boson, free to occupy the lowest state collectively; ³He is a fermion and must pair up before it can behave bosonically
   • ³He is lighter, so stronger quantum fluctuations destroy the superfluidity
   • The interatomic interactions of ³He are much weaker
   • ⁴He has zero nuclear spin and is undisturbed by magnetic fields

   Submit
2. 2.

   Which statements about the Slater determinant are correct? (Select all that apply.)

   Select all that apply

   • Swapping two rows of the determinant corresponds to exchanging two particles, and the wavefunction changes sign
   • Two equal columns make the determinant zero, meaning two fermions cannot occupy the same spin-orbital
   • It is the exact general solution of the many-fermion problem
   • The normalisation factor is 1/√(N!) (for orthonormal single-particle states)

   Submit
3. 3.

   Fully expand the antisymmetric wavefunction of 5 fermions into product states. How many terms are there?

   项0% relative tolerance

   Submit
4. 4.

   Two electrons are in a spin triplet. Their spatial wavefunction must be:

   • Antisymmetric, with zero probability density for the two electrons at the same position
   • Symmetric, with the electrons tending to huddle together
   • Unrestricted
   • Also a triplet

   Submit

What comes next

The Slater determinant has handed us a great gift: two equal columns make it vanish — two fermions cannot occupy the same quantum state.

The next section unfolds that sentence to its full size: why the periodic table looks the way it does, why atoms have volume, why a white dwarf that has burned its last fuel still refuses to collapse. One minus sign, holding up half the universe.