Question bank — Spin — intrinsic angular momentum

False. Ek field sirf har orientation ki energy set karti hai $U=-{\mu }_{z}B$ ke zariye (jahan ${\mu }_z$ moment ka $z$-projection hai aur $B$ field magnitude); length $|\vec{S}|=\hslash \sqrt{s(s+1)}$ particle ki identity se fix hoti hai, jaise uski mass ya charge.

False. $z$ ke along point karne ka matlab hoga $S_z=|\vec{S}|$ aur phir $S_x=S_y=0$ exactly jaane — lekin $[S_x,S_y]=i\hslash S_z$ dono ko sharp hone se forbid karta hai jab $S_z$ fixed ho. Nature ka compromise ek fixed tilt hai: $\cos \theta =\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}$, isliye $\theta \approx 54.7^{\circ }$ (spin-cone figure dekho).

False. $S_z=m_s\hslash$ sirf $z$ par projection (shadow) hai; poori length $|\vec{S}|=\frac{\sqrt{3}}{2}\hslash$ hai, jo badi hai kyunki $S_x,S_y$ bhi ${\vec{S}}^2$ mein contribute karte hain.

True. Count hai $2s+1=2(\frac{1}{2})+1=2$, jo $m_s=\pm \frac{1}{2}$ se match karta hai. $m_s$ ki har value ek discrete deflection hai, isliye beam exactly do mein split hoti hai.

False. $L=0$ orbital moment ko khatam karta hai, lekin lone valence electron ka spin moment bacha rehta hai — wahi residual moment exactly tha jisne beam split ki aur humein spin invent karne par majboor kiya.

False. Spin intrinsic hai: yeh ek point particle mein bhi exist karta hai jiska koi spatial extent nahi aur koi orbit nahi. Ek classical spinning sphere jo $\frac{\hslash }{2}$ deta uski surface ko light se tez move karna padta.

False. Orbital motion $g_{\ell }=1$ deta hai, lekin electron spin ka anomalous $g_s\approx 2$ hota hai. Yeh "extra factor of two" genuinely ek relativistic/quantum result hai, jo Dirac Equation se explain hota hai.

False (trap). Ek massive spin-1 particle $2s+1=3$ deta. Lekin photon massless hai, isliye woh sirf do helicity states mein dekha ja sakta hai ($m=\pm 1$, do circular polarizations); $m=0$ state forbidden hai (neeche "why" item dekho). Massless particles naive count ko tod dete hain.

Writer ne magnitude ke liye projection formula use kar liya. Magnitude $\sqrt{s(s+1)}$ use karta hai, jo $\frac{\sqrt{3}}{2}\hslash$ deta hai; sirf $S_z$ mein $m_s$ use hota hai.

$m_s$ integer steps mein $-s$ se $+s$ tak jaata hai, yaani sirf $-\frac{1}{2},+\frac{1}{2}$. Negative values zaroori hain aur exactly $2s+1=2$ hote hain, kabhi teen nahi.

Sign galat hai (electron charge negative hai, isliye ${\vec{\mu }}_s$, $\vec{S}$ ke opposite point karta hai: ek minus sign chahiye), aur factor $g_s\approx 2$ missing hai: ${\vec{\mu }}_s=-g_s\frac{e}{2m_e}\vec{S}$.

Teeno components ek saath sharp nahi ho sakte, isliye $S_x^2+S_y^2$ contribute karta hai tab bhi jab $S_z$ maximal ho. Woh extra piece $s^2$ ko $s(s+1)$ bana deta hai.

"Up/down" sirf spin-$\frac{1}{2}$ ke liye specific hai. Massive $s=1$ ke liye, $m_s=-1,0,+1$ se $2s+1=3$ spots milte hain. Orientations ki sankhya $s$ ke saath badhti hai.

Splitting ke liye gradient chahiye: $F_z={\mu }_z\partial B_z/\partial z$. Ek uniform field ($\partial B_z/\partial z=0$) zero net force lagata hai — woh sirf moment ko precess (Larmor) karwata hai, deflect nahi (precession figure dekho).

Ek classical moment kisi bhi direction mein point kar sakta tha, deflections continuously smear ho jaate; lekin $m_s$ sirf discrete values leta hai, isliye sirf discrete forces hote hain — quantization ki discreteness visible ho gayi.

Yeh particle ki khud ki ek permanent, unchangeable property hai, space mein motion ki wajah se nahi hoti — tum isse spin up nahi kar sakte, slow down nahi kar sakte, ya remove nahi kar sakte.

Agar $\vec{S}$, $z$ ke saath align hota toh teeno components simultaneously sharp hote ($S_x=S_y=0$, $S_z=|\vec{S}|$), jo $[S_x,S_y]=i\hslash S_z$ ke khilaaf hai. Nature $S_z$ aur $|\vec{S}|$ sharp rakhta hai aur $S_x,S_y$ ko ek cone ke around blur karta hai jiska half-angle fixed $\theta \approx 54.7^{\circ }$ hai.

Ek uniform field ek torque $\vec{\tau }=\vec{\mu }\times \vec{B}$ deta hai jo $\vec{S}$ ko precess (Larmor) karwata hai bina $S_z$ badle, isliye koi net translational force nahi lagti. Lekin har orientation ki energy $U=-{\mu }_{z}B$ hoti hai, aur do ${\mu }_z$ values do energies deti hain — ek energy split, jo Zeeman Effect hai.

Classical current-loop picture $g=1$ predict karta hai; observed spin magnetism ki doubling ka koi classical explanation nahi hai aur yeh sirf relativistic Dirac Equation se emerge hota hai.

Dono ek ladder rule follow karte hain jo force karta hai ki $2s$ non-negative integer ho, jisse $s=\frac{1}{2}$ allow hota hai. Orbital $\ell$ mein ek extra constraint hai — iska wavefunction ek poore $360^{\circ }$ spatial turn ke baad khud par wapas aa jaana chahiye, jo half-integers forbid karta hai. Spin, rotations ke double-cover group $SU(2)$ se tied hai (jahan $360^{\circ }$ turn sign flip karta hai aur sirf $720^{\circ }$ restore karta hai), isme aisa koi single-valuedness demand nahi hai, isliye half-integers survive karte hain.

Ek massless particle $c$ par move karta hai, isliye tum kabhi aisa frame nahi dhundh sakte jahan woh rest mein ho aur tum uska spin ghuma sako; sirf uske apne motion ke along projection (helicity) physical hai, aur gauge invariance longitudinal ($m=0$) mode ko remove karta hai. Isliye photon sirf $m=\pm 1$ dikhata hai, do transverse (circular) polarizations.

Half-integer spin electrons ko fermions banata hai, jinke total wavefunction exchange ke under antisymmetric hone chahiye; woh antisymmetry exactly wahi hai jo do electrons ko saare quantum numbers share karne se forbid karti hai.

Magnitude sirf $s$ par depend karta hai, kabhi $m_s$ par nahi; dono $m_s=\pm \frac{1}{2}$ ek hi $s=\frac{1}{2}$ share karte hain, isliye dono ka $|\vec{S}|=\frac{\sqrt{3}}{2}\hslash$ hai aur sirf orientation mein differ karte hain.

$2s+1=1$: ek single undeflected spot. Spin se koi magnetic moment nahi matlab koi splitting nahi — yeh "expected" classical-looking result hai.

$2s+1=4$ spots, jisme $m_s=-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$. Note karo ki inme se koi $0$ nahi hai, isliye pattern centre ke bare mein symmetric hai bina central spot ke.

Nahi — $s=0$ ke saath sirf $m_s=0$ hai aur $\sqrt{s(s+1)}=0$, jo $0/0$ deta hai. Tilt karne ke liye koi vector hi nahi hai; "angle" undefined hai kyunki spin simply zero hai.

$\cos \theta =\frac{s}{\sqrt{s(s+1)}}=\sqrt{\frac{s}{s+1}}\to 1$, isliye $\theta \to 0$. Large spins zyada classically behave karte hain, axis ke saath almost fully align ho jaate hain — quantum "tilt" fade ho jaata hai.

Do deflections sides badal leti hain: spin-up aur spin-down beams exchange kar leti hain ki kaun kis taraf bend karta hai, kyunki $F_z={\mu }_z\partial B_z/\partial z$ har ek ke liye sign flip karta hai. Spots ki sankhya (do) unchanged rehti hai.

Zero net translational force, isliye koi deflection nahi; moment sirf precess (Larmor) karta hai aur uski orientations alag energies $U=-{\mu }_{z}B$ acquire karti hain (Zeeman Effect ka basis).