The Pauli exclusion principle

At the end of the last section, the Slater determinant handed us a gift: two equal columns make the determinant vanish. Translated into physics —

The Pauli exclusion principle: two identical fermions cannot occupy the same single-particle quantum state.

The derivation is one line. Put both electrons in the spin-orbital $\phi_a$; the antisymmetric combination gives

$$\psi_A=\frac{1}{\sqrt2}\bigl[\phi_a(x_1)\phi_a(x_2)-\phi_a(x_1)\phi_a(x_2)\bigr]\equiv0\tag{8.3.1}$$

This state does not exist — not “very high in energy”, not “forbidden to transition into”, but simply absent from the state space altogether. Strictly speaking, the deeper principle is the stronger one from last section: fermion wavefunctions must be antisymmetric; “exclusion” is merely its most famous corollary.

The principle is short. The bill it pays is long. This section tours what it buys.

Exhibit one: why does the periodic table have periods?

Start with a question chemistry teachers usually leave unanswered: why don’t all of an atom’s electrons fall into the 1s orbital?

Chapter 6 solved the hydrogen atom: levels $E_n=-13.6\,\text{eV}/n^2$, with $1s$ lowest. The principle of minimum energy would seem to say that the ground state of a many-electron atom has every electron crammed into $1s$ — in which case lithium and uranium would both have a 1s “outer shell”, chemical properties would drift monotonically, there would be no sudden inertness of the noble gases or sudden violence of the alkali metals, and the periodic table would have no periods at all.

The exclusion principle rewrites the rules. Count how many electrons each shell can hold: the shell with principal quantum number $n$ has $n^2$ spatial orbitals ($\ell=0,\dots,n-1$, each with $2\ell+1$ values of $m$), and each spatial orbital takes spin up and spin down, giving a capacity of

$$2n^2:\qquad n=1\to2,\quad n=2\to8,\quad n=3\to18,\quad n=4\to32\tag{8.3.2}$$

Electrons can only fill shell by shell: hydrogen’s single electron takes $1s$; helium’s two fill up $n=1$; lithium’s third electron is forced to move into $2s$ — so lithium’s outermost electron sits alone and loosely bound, chemically eager, remarkably like hydrogen. Periodicity has arrived: each time a shell fills, the next electron is forced to open a new one, and the chemistry starts its cycle again.

The lengths of the first few periods — 2, 8, 8, 18 — do not match $2n^2$ term by term (the third period is 8, not 18), and the reason was prepared in chapters 6 and 7: in a many-electron atom the inner electrons screen the nuclear charge, so the levels depend on $\ell$ as well as $n$ (and $4s$ fills before $3d$). But the big question, “why are there periods at all?”, has a two-part answer: hydrogen-like level structure + Pauli exclusion.

Exhibit two: pouring a great many electrons into a box

The periodic table is filling “inside one atom”. Now the opposite extreme: a metal holds about $10^{23}$ free electrons per cubic centimetre — a colossal number of fermions poured into one big box. What do they do?

A classical gas at $T\to0$ has every particle at rest with zero energy. Fermions cannot: each state seats only two (spin up and down), so latecomers must occupy ever higher levels — like filling a stadium, where once the front rows are taken you can only climb. The highest energy reached is the Fermi energy $E_F$.

The Fermi energy of a free electron gasadvanced~9 minexpand

Step 1: count the states in the box. In a cubic box of side $L$, the generalisation of the infinite well gives standing-wave solutions with quantised wavevectors $\vec{k}=\frac{\pi}{L}(n_x,n_y,n_z)$, the $n_i$ positive integers. Each triple $(n_x,n_y,n_z)$ is one spatial orbital, seating 2 electrons.

Step 2: the ground state = a filled sphere. At $T=0$, fill from low energy up according to $E=\hbar^2k^2/2m$, stopping at some maximum wavenumber $k_F$ — an eighth of a sphere in $k$-space (all $n_i$ positive). The number of states:

$$N=2\times\frac{1}{8}\times\frac{\frac{4}{3}\pi k_F^3}{(\pi/L)^3} =\frac{k_F^3}{3\pi^2}L^3\tag{8.3.3}$$

(The denominator $(\pi/L)^3$ is the $k$-space volume per state; the factor 2 is spin.) Writing the electron number density as $n=N/L^3$ and solving,

$$k_F=(3\pi^2 n)^{1/3}\tag{8.3.4}$$

Step 3: the Fermi energy.

$$E_F=\frac{\hbar^2k_F^2}{2m}=\frac{\hbar^2}{2m}(3\pi^2n)^{2/3}\tag{8.3.5}$$

Step 4: plug in real numbers (copper). Copper contributes 1 free electron per atom, $n=8.5\times10^{28}\,\text{m}^{-3}$:

$$k_F=(3\pi^2\times8.5\times10^{28})^{1/3}=1.36\times10^{10}\,\text{m}^{-1}\tag{8.3.6}$$$$E_F=\frac{(1.055\times10^{-34})^2\times(1.36\times10^{10})^2}{2\times9.11\times10^{-31}} \approx1.13\times10^{-18}\,\text{J}\approx7.0\,\text{eV}\tag{8.3.7}$$

Reading the number. 7 eV corresponds to a “temperature” $T_F=E_F/k_B\approx8\times10^4$ K — nearly 300 times room temperature (0.025 eV). Even at absolute zero, the fastest electrons in a metal fly at about $1.6\times10^6$ m/s (0.5% of the speed of light). This speed is almost independent of temperature; it is squeezed out purely by Pauli exclusion. The classical intuition “cold = slow” fails in the world of fermions.

Exhibit three: why doesn’t a white dwarf collapse?

A star spends its whole life in a tug of war: gravity pulls inward, thermal pressure pushes outward. When the fuel runs out the thermal pressure retreats, and gravity looks certain to win — the star should collapse without limit.

Yet stars like the Sun end up as white dwarfs: a solar mass squeezed into an Earth-sized sphere, density about $10^9\,\text{kg/m}^3$ (a teaspoonful weighs some five tonnes), and no matter how far it cools it shrinks no further. What holds it up?

That $P\propto n^{5/3}/m$ from above. Collapse sends the electron density soaring, degeneracy pressure climbs as $n^{5/3}$ — faster than gravity (whose effective scaling is $n^{4/3}$) — so balance is always reached at some radius. A white dwarf is a star held up by the Pauli exclusion principle — no burning needed, no temperature needed, only electrons with “nowhere to go”.

The view on the boson side

Glance across at the other family. Bosons face no exclusion, and as $T\to0$ they can drop en masse into the same lowest state — Bose–Einstein condensation (achieved in rubidium-87 in 1995, at 170 nK). Last section’s $^4$He superfluid, and the $10^{16}$ photons sharing one mode in a laser cavity, are the same “legal bunching”. One minus sign apart: fermions build the periodic table and white dwarfs; bosons light lasers and superfluids.

∎Key formulas

• Exclusion principle

  $$\phi_a=\phi_b\ \Rightarrow\ \psi_A\equiv0$$

  At most one fermion per spin-orbital; a corollary of antisymmetry, not an independent law

• Shell capacity

  $$2n^2$$

  n = 1, 2, 3, 4 → 2, 8, 18, 32; the skeleton of the periodic table

• Fermi energy

  $$E_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$$

  About 7 eV in copper, corresponding to T_F ≈ 8×10⁴ K

• Degeneracy pressure

  $$P=\frac{2}{5}nE_F\propto \frac{n^{5/3}}{m}$$

  Non-zero even at zero temperature; supports metals and white dwarfs

?Self-check4 questions

1. 1.

   "Pauli repulsion" pushes two closed-shell electron clouds apart. Which statement about its nature is correct?

   • There is no corresponding new force term in the Hamiltonian; antisymmetry forces overlapping electrons into higher orbitals, so the energy climbs steeply as the distance shrinks
   • There is a short-range repulsion between electrons independent of the Coulomb force
   • It is a manifestation of the magnetic dipole interaction
   • It is the Coulomb repulsion enhanced at short range

   Submit
2. 2.

   If electrons were bosons, which consequences would you expect? (Select all that apply.)

   Select all that apply

   • The electrons of every atom would cluster near 1s, and the chemistry of the elements would lose its periodicity
   • The Fermi energy of metals would drop to essentially zero, and degeneracy pressure would vanish
   • White dwarfs would lose their support, changing the endpoint of stellar evolution
   • The hydrogen level formula En = −13.6 eV/n² would no longer hold

   Submit
3. 3.

   Sodium contributes 1 free electron per atom, with electron density n = 2.65×10²⁸ m⁻³. Find the Fermi energy in eV. (ħ = 1.055×10⁻³⁴ J·s, m = 9.11×10⁻³¹ kg, 1 eV = 1.602×10⁻¹⁹ J)

   eV30% relative tolerance

   Submit
4. 4.

   The physical reason white dwarfs have a mass limit (about 1.4 solar masses) is:

   • Once the electrons turn relativistic, the degeneracy-pressure scaling softens from n^{5/3} to n^{4/3}, exactly matching gravity, so stable equilibrium is lost
   • The Pauli exclusion principle breaks down at large masses
   • High temperature "melts" the degeneracy pressure
   • Neutron degeneracy pressure is stronger and takes over early

   Submit

What comes next

So far the role of antisymmetry has been to forbid: delete certain states. But it does one subtler thing as well — it reprices the states that remain.

The Hamiltonian contains no spin, yet the helium atom’s levels split into two ladders by total spin, nearly 1 eV apart; the spins in a lump of iron prefer to line up as if a powerful aligning force acted between them. The next section settles that account: the exchange interaction — a “force” that does not exist, and its very real energy.