Exercises — Pauli exclusion principle

Statement: Koi bhi do identical fermions ek hi quantum state mein nahi reh sakte — yaani unke charo quantum numbers $(n,\ell ,m_{\ell },m_s)$ identical nahi ho sakte. Ise follow karte hain (fermions, half-integer spin): electron, proton, neutron. Ise follow NAHI karte (bosons, integer spin): photon (spin 1), $\alpha$-particle (spin 0). Kyun: ye principle ek antisymmetric wavefunction ($\psi (2,1)=-\psi (1,2)$) ka sirf fermion-wala consequence hai. Bosons ke symmetric wavefunctions hote hain aur wo pile up kar sakte hain.

$m_{\ell }$, $-\ell$ se $+\ell$ tak integer steps mein chalta hai: $m_{\ell }=-2,-1,0,+1,+2\Rightarrow 5\text{ orbitals.}$ Har orbital 2 electrons hold karta hai (spin up + spin down): $5\times 2=10\text{ electrons.}$ (Ye $d$-subshell hai.)

$n=3$ ke liye, $\ell =0,1,2$.

• $\ell =0$: $2\ell +1=1$ 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 $=2+6+10=18$. Formula se check karo: $2n^2=2(3{)}^2=18$. ✓

Pehle lowest energy fill karo (Aufbau order $1s,2s,2p,\dots$): $1s^{2}2s^{2}2p^5.$ Count karo: $2+2+5=9=Z$. ✓ $2p$ subshell 6 tak hold kar sakta hai lekin usmein sirf 5 hain — ye ek electron short hai full hone se. Yahi ek vacancy fluorine ko itna reactive banati hai.

$n=2$ ke liye orbitals ki sankhya unchanged rehti hai: $\ell =0$ se 1, $\ell =1$ se 3, total $4$ orbitals. Ab har orbital 3 electrons hold karta hai (teen spins): $4\times 3=12\text{ electrons.}$ (Normally ye $4\times 2=8$ hota.)

Har jagah $b=a$ set karo: $\Psi (1,2)=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)-{\psi }_a(1){\psi }_a(2)]=\frac{1}{\sqrt{2}}\cdot 0=0.$ Physical meaning: system ko kisi bhi configuration mein paane ki probability $|\Psi {|}^2=0$ hai. Ek aisa state jo har jagah zero probability rakhta ho, exist hi nahi karta. Toh do fermions simply dono state $a$ mein nahi baith sakte — ye Pauli exclusion principle hai, minus sign se nikalta hai, haath se dala nahi gaya. Neeche wali picture dekho.

Total capacity: $N=2{\sum }_{\ell =0}^{n-1}(2\ell +1).$ Bahar ka factor 2 spin doubling hai jo Pauli allow karta hai. $\ell =0,1,\dots ,n-1$ ke liye terms $2\ell +1$ exactly $1,3,5,\dots ,(2n-1)$ hain — pehle $n$ odd numbers. Unka sum $n^2$ hai. Isliye $N=2\cdot n^2=2n^2.✓$ Neeche wali staircase picture dikhati hai ki consecutive odd numbers ek perfect square kyun banate hain.

Helium: dono electrons ke quantum sets hain $(1,0,0,+\frac{1}{2})\text{aur}(1,0,0,-\frac{1}{2}).$ Ye $n,\ell ,m_{\ell }$ pe agree karte hain lekin $m_s$ mein differ karte hain — toh poore chaar-number sets identical nahi hain. Pauli satisfy hai. Lithium ($Z=3$): $n=1$ mein teesre electron ko $(1,0,0,m_s)$ chahiye hoga jahan $m_s=\pm \frac{1}{2}$ ho — lekin wo dono addresses already taken hain. Koi bhi choice ek existing full set ko repeat karegi → forbidden. Toh teesra electron next shell se shuru karna padta hai, $n=2$ ($2s$). Exactly aise hi shell structure — aur poori periodic table — paida hoti hai.

Nature (b) choose karta hai: do alag $m_{\ell }$ orbitals, parallel spins.

• Pauli dono (a) aur (b) ko permit karta hai — koi bhi full quantum-number set repeat nahi karta.
• Hund's rule tie todta hai: electrons pairing karne se pehle separate orbitals mein parallel spins ke saath single occupancy maximize karte hain, kyunki parallel spins electrons ko spatially alag rakhte hain aur repulsion energy kam karte hain.

Toh actual configuration hai $2p_x^{1}2p_y^1$ dono spins up ke saath, e.g. quantum sets $(2,1,-1,+\frac{1}{2})$ aur $(2,1,0,+\frac{1}{2})$ — alag $m_{\ell }$, same $m_s$. Pauli ke hisaab se legal, Hund ke hisaab se favoured.

Star ko squeeze karo $\Rightarrow$ har electron ek chhote region mein confined ho jaata hai. Quantum confinement se, ek chhote box mein higher allowed momentum/energy levels hote hain. Pauli electrons ko lowest level mein baithne se rokta hai — har added electron ko next-higher unoccupied state occupy karni padti hai. Toh $T\to 0$ pe bhi high-momentum states ki ek badi "stack" filled rehti hai. Aur compress karo toh har occupied level raise ho jaata hai, jo energy cost karti hai. Usi compression ko resist karna ek outward degeneracy pressure ki tarah feel hota hai. Key nuance: ye electromagnetic repulsion nahi hai; ye Pauli ke fermions ko energy ladder pe upar dhakelnay ki energy price hai. (Ek BEC ke liye, bosons sab lowest level mein crash kar jaate hain — aisa koi pressure exist nahi karta.)

Ye connection Spin-statistics theorem ka content hai: relativistic quantum field theory mein, half-integer spin ($\frac{1}{2},\frac{3}{2},\dots$) wale particles ke multiparticle wavefunctions necessarily antisymmetric (fermions) hone chahiye, jabki integer spin ($0,1,2,\dots$) wale necessarily symmetric (bosons) hone chahiye. Minus sign ek choice nahi hai — ye ek consistent, causal relativistic theory require karne se forced hai. Antisymmetry $\Rightarrow$ Pauli, toh half-integer spin $\Rightarrow$ Pauli.

Aufbau order: $1s2s2p3s3p4s3d\dots$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6.$ Count karo: $2+2+6+2+6+2+6=26=Z$. ✓ Unpaired electrons — $3d^6$ subshell. $3d$ mein 5 orbitals hain. Hund's rule se, pehle har ek mein single fill karo (5 electrons, sab parallel), phir 6th ko ek orbital mein pair up karna padta hai: $\underset{5\text{ orbitals}}{\underbrace{\uparrow \uparrow \uparrow \uparrow \uparrow }}\to \uparrow \downarrow \uparrow \uparrow \uparrow \uparrow .$ Isse $5-1=4$ orbitals abhi bhi ek single (unpaired) electron hold karte hain. Unpaired electrons $=4$. (Isliye iron strongly magnetic hai.)

Ek filled subshell ke har orbital mein exactly ek $m_s=+\frac{1}{2}$ aur ek $m_s=-\frac{1}{2}$ electron hota hai (Pauli har orbital ke dono occupants ko opposite spin rakhne pe majboor karta hai). Toh ye pairs mein cancel ho jaate hain: $\sum m_s=\underset{\text{orbital 1}}{\underbrace{(+\frac{1}{2}-\frac{1}{2})}}+\cdots +\underset{\text{orbital 5}}{\underbrace{(+\frac{1}{2}-\frac{1}{2})}}=0.$ 5 orbitals ($1s,2s,2p_{-1},2p_0,2p_{+1}$) ke saath sab doubly filled, $\sum m_s=0$. Kyun forced hai: ek full subshell mein ek unpaired spin ki koi jagah nahi hoti — har $m_{\ell }$ slot dono spins use karta hai. Isliye closed-shell atoms (noble gases) ka net spin zero hota hai, ye non-magnetic hote hain, aur chemically inert hote hain.

Part 1 — ek fully specified orbital. $(n,\ell ,m_{\ell })=(4,1,+1)$ fix karne pe sirf $m_s=\pm \frac{1}{2}$ free rehta hai. Pauli yahan exactly 2 electrons allow karta hai. Part 2 — $n=4$ ke andar saare $m_{\ell }=+1$ wale. $m_{\ell }=+1$ allowed hone ke liye humein $\ell \ge 1$ chahiye. $n=4$ mein, $\ell$ ho sakta hai $0,1,2,3$. Inme se, $\ell =1,2,3$ teeno $m_{\ell }=+1$ include karte hain ($\ell =0$ nahi karta, kyunki $m_{\ell }$ sirf $0$ ho sakta hai). Ye 3 qualifying orbitals hain, har ek 2 electrons hold karta hai: $3\times 2=6\text{ electrons.}$ Edge-case lesson: $\ell =0$ ke liye $m_{\ell }=+1$ impossible hai — count karne se pehle hamesha check karo ki $|m_{\ell }|\le \ell$ ho.