Exercises — Pauli exclusion principle

Statement: No two identical fermions can occupy the same quantum state — i.e. they cannot have all four quantum numbers $(n,\ell ,m_{\ell },m_s)$ identical. Obey it (fermions, half-integer spin): electron, proton, neutron. Do NOT obey it (bosons, integer spin): photon (spin 1), $\alpha$-particle (spin 0). Why: the principle is a fermion-only consequence of an antisymmetric wavefunction ($\psi (2,1)=-\psi (1,2)$). Bosons have symmetric wavefunctions and can pile up.

$m_{\ell }$ runs from $-\ell$ to $+\ell$ in integer steps: $m_{\ell }=-2,-1,0,+1,+2\Rightarrow 5\text{ orbitals.}$ Each orbital holds 2 electrons (spin up + spin down): $5\times 2=10\text{ electrons.}$ (This is the $d$-subshell.)

For $n=3$, $\ell =0,1,2$.

• $\ell =0$: $2\ell +1=1$ orbital $\to 2$ e⁻ ($3s$).
• $\ell =1$: $2\ell +1=3$ orbitals $\to 6$ e⁻ ($3p$).
• $\ell =2$: $2\ell +1=5$ orbitals $\to 10$ e⁻ ($3d$).

Total $=2+6+10=18$. Check against formula: $2n^2=2(3{)}^2=18$. ✓

Fill lowest energy first (Aufbau order $1s,2s,2p,\dots$): $1s^{2}2s^{2}2p^5.$ Count: $2+2+5=9=Z$. ✓ The $2p$ subshell holds up to 6 but has only 5 — it is one electron short of full. That single vacancy is why fluorine is so reactive.

Number of orbitals for $n=2$ is unchanged: $\ell =0$ gives 1, $\ell =1$ gives 3, total $4$ orbitals. Now each orbital holds 3 electrons (three spins): $4\times 3=12\text{ electrons.}$ (Normally it would be $4\times 2=8$.)

Set $b=a$ everywhere: $\Psi (1,2)=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)-{\psi }_a(1){\psi }_a(2)]=\frac{1}{\sqrt{2}}\cdot 0=0.$ Physical meaning: the probability of finding the system in any configuration is $|\Psi {|}^2=0$. A state with zero probability everywhere does not exist. So two fermions simply cannot both sit in state $a$ — that is the Pauli exclusion principle, dropping out of the minus sign, not added by hand. See the picture below.

Total capacity: $N=2{\sum }_{\ell =0}^{n-1}(2\ell +1).$ The factor 2 out front is the spin doubling Pauli allows. The terms $2\ell +1$ for $\ell =0,1,\dots ,n-1$ are exactly $1,3,5,\dots ,(2n-1)$ — the first $n$ odd numbers. Their sum is $n^2$. Hence $N=2\cdot n^2=2n^2.✓$ The staircase picture below shows why consecutive odd numbers build a perfect square.

Helium: the two electrons have quantum sets $(1,0,0,+\frac{1}{2})\text{and}(1,0,0,-\frac{1}{2}).$ They agree on $n,\ell ,m_{\ell }$ but differ in $m_s$ — so the full four-number sets are not identical. Pauli is satisfied. Lithium ($Z=3$): a third electron in $n=1$ would need $(1,0,0,m_s)$ with $m_s=\pm \frac{1}{2}$ — but both of those addresses are already taken. Any choice repeats an existing full set → forbidden. So the third electron must start the next shell, $n=2$ (the $2s$). This is exactly how shell structure — and the whole periodic table — is born.

Nature chooses (b): two different $m_{\ell }$ orbitals, parallel spins.

• Pauli permits both (a) and (b) — neither repeats a full quantum-number set.
• Hund's rule breaks the tie: electrons maximize total spin by singly occupying separate orbitals with parallel spins before pairing up, because parallel spins keep electrons spatially apart and lower the repulsion energy.

So the actual configuration is $2p_x^{1}2p_y^1$ with both spins up, e.g. quantum sets $(2,1,-1,+\frac{1}{2})$ and $(2,1,0,+\frac{1}{2})$ — different $m_{\ell }$, same $m_s$. Legal by Pauli, favoured by Hund.

Squeeze the star smaller $\Rightarrow$ each electron is confined to a smaller region. By quantum confinement, a smaller box has higher allowed momentum/energy levels. Pauli forbids electrons from all sitting in the lowest level — every added electron must occupy the next-higher unoccupied state. So there is a huge "stack" of filled high-momentum states even at $T\to 0$. Compressing further raises every occupied level, which costs energy. Resisting that compression is felt as an outward degeneracy pressure. Key nuance: this is not an electromagnetic repulsion; it is the energy price of Pauli forcing fermions up the energy ladder. (For a BEC, bosons all crash into the lowest level — no such pressure exists.)

The connection is the content of the Spin-statistics theorem: in relativistic quantum field theory, particles with half-integer spin ($\frac{1}{2},\frac{3}{2},\dots$) must have antisymmetric multiparticle wavefunctions (fermions), while integer spin ($0,1,2,\dots$) must be symmetric (bosons). The minus sign is not a choice — it is forced by requiring a consistent, causal relativistic theory. Antisymmetry $\Rightarrow$ Pauli, so half-integer spin $\Rightarrow$ Pauli.

Aufbau order: $1s2s2p3s3p4s3d\dots$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6.$ Count: $2+2+6+2+6+2+6=26=Z$. ✓ Unpaired electrons — the $3d^6$ subshell. The $3d$ has 5 orbitals. By Hund's rule, fill each singly first (5 electrons, all parallel), then the 6th must pair up in one orbital: $\underset{5\text{ orbitals}}{\underbrace{\uparrow \uparrow \uparrow \uparrow \uparrow }}\to \uparrow \downarrow \uparrow \uparrow \uparrow \uparrow .$ That leaves $5-1=4$ orbitals still holding a single (unpaired) electron. Unpaired electrons $=4$. (This is why iron is strongly magnetic.)

Every orbital of a filled subshell contains exactly one $m_s=+\frac{1}{2}$ and one $m_s=-\frac{1}{2}$ electron (Pauli forces the two occupants of each orbital to have opposite spin). So they cancel in pairs: $\sum m_s=\underset{\text{orbital 1}}{\underbrace{(+\frac{1}{2}-\frac{1}{2})}}+\cdots +\underset{\text{orbital 5}}{\underbrace{(+\frac{1}{2}-\frac{1}{2})}}=0.$ With 5 orbitals ($1s,2s,2p_{-1},2p_0,2p_{+1}$) all doubly filled, $\sum m_s=0$. Why forced: a full subshell has no room for an unpaired spin — every $m_{\ell }$ slot uses both spins. Hence closed-shell atoms (noble gases) have zero net spin, are non-magnetic, and are chemically inert.

Part 1 — one fully specified orbital. Fixing $(n,\ell ,m_{\ell })=(4,1,+1)$ leaves only $m_s=\pm \frac{1}{2}$ free. Pauli allows exactly 2 electrons here. Part 2 — all $m_{\ell }=+1$ within $n=4$. For $m_{\ell }=+1$ to be allowed we need $\ell \ge 1$. In $n=4$, $\ell$ can be $0,1,2,3$. Of these, $\ell =1,2,3$ all include $m_{\ell }=+1$ ($\ell =0$ does not, since $m_{\ell }$ can only be $0$). That's 3 qualifying orbitals, each holding 2 electrons: $3\times 2=6\text{ electrons.}$ Edge-case lesson: $m_{\ell }=+1$ is impossible for $\ell =0$ — always check that $|m_{\ell }|\le \ell$ before counting.