Spin — intrinsic angular momentum: Worked examples

2.3.17 · D3 · Physics › Modern Physics › Spin — intrinsic angular momentum

Scenario matrix

Har spin problem jo tum miloge, inhi cells mein se ek mein hogi. Neeche ke examples un cells ke saath label kiye gaye hain jo woh cover karte hain.

Cell	Kya alag banata hai	Kahan covered hai
A. Magnitude, any $s$	sirf $s (s + 1)$ par depend karta hai, $m_{s}$ par kabhi nahi	Ex 1
B. Projection, $m_{s} > 0$ (spin up)	positive $S_{z}$ , force ek taraf	Ex 2
C. Projection, $m_{s} < 0$ (spin down)	negative $S_{z}$ , force doosri taraf	Ex 2, Ex 5
D. Tilt angle from $z$	kyun $θ \neq = 0$ hamesha; saare allowed $θ$	Ex 3
E. Degenerate input $s = 0$	zero magnitude, ek spot, koi splitting nahi	Ex 4
F. Bigger ladder $s = 1$ , $s = \frac{3}{2}$	$2 s + 1$ spots, saare $m_{s}$ steps	Ex 6
G. Limiting / classical check	"spinning ball" light-speed tod deta	Ex 7
H. Real-world word problem	Stern–Gerlach actual numbers/units ke saath	Ex 8
I. Exam-style twist	energy splitting $U = - μ_{z} B$ , spin-flip photon	Ex 9

Hum poore note mein yeh constants use karte hain (symbols yaad karo, digits nahi):

Cell A — kisi bhi spin ke liye magnitude

Worked example Ex 1 — Electron ke spin ki magnitude (Cell A)

Ek electron ( $s = \frac{1}{2}$ ) ke liye $∣ S ∣$ ko $ℏ$ ke multiple mein aur SI units mein nikalo.

Forecast: padhne se pehle guess karo — kya yeh $\frac{1}{2} ℏ$ hai? Apna guess likh lo. (Yeh nahi hai; woh projection hai, magnitude nahi.)

Magnitude rule likho. $∣ S ∣ = ℏ s (s + 1)$ . Yeh step kyun? Magnitude kabhi $m_{s}$ use nahi karta — yeh sirf total-spin number $s$ par depend karta hai. Yeh vector ki "size" hai, $z$ -axis par uski shadow nahi.
$s = \frac{1}{2}$ substitute karo. $s (s + 1) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$ . Yeh step kyun? Abhi plug in karna algebra ko honest rakhta hai — humne $m_{s}$ ko touch nahi kiya, jo prove karta hai ki answer spin-up aur spin-down dono ke liye same hai.
Root lo. $∣ S ∣ = ℏ \frac{3}{4} = \frac{3}{2} ℏ \approx 0.8660 ℏ$ .
SI mein convert karo. $0.8660 \times 1.0546 \times 1 0^{- 34} \approx 9.13 \times 1 0^{- 35} J\cdots$ .

Verify: $(\frac{3}{2})^{2} = \frac{3}{4} = s (s + 1)$ ✔. Units: $ℏ$ J·s mein hai, aur angular momentum ki units J·s hoti hain ✔.

Cells B & C — projection, dono signs

Worked example Ex 2 — Dono

$S_{z}$ values aur unke forces (Cells B, C) Ek electron ke liye, dono allowed $S_{z}$ list karo aur un mein se har ek ko gradient $\partial B_{z} / \partial z > 0$ mein Stern–Gerlach force ka sign batao.

Forecast: kitni values, aur kya dono forces equal-and-opposite hain ya unrelated?

Allowed $m_{s}$ . $s = \frac{1}{2}$ ke liye: $m_{s} = + \frac{1}{2}$ (up) aur $m_{s} = - \frac{1}{2}$ (down). Yeh step kyun? $m_{s}$ runs $- s, \dots, + s$ integer steps mein, isliye sirf do rungs exist karte hain.
Projection. $S_{z} = m_{s} ℏ = + \frac{1}{2} ℏ$ ya $- \frac{1}{2} ℏ$ .
Magnetic moment. $μ_{z} = - g_{s} \frac{e}{2 m _{e}} S_{z} = - g_{s} μ_{B} m_{s}$ . $g_{s} \approx 2$ ke saath: $μ_{z} = - 2 μ_{B} (\pm \frac{1}{2}) = \mp μ_{B}$ . Minus kyun? Electron charge negative hai, isliye $μ$ opposite direction mein point karta hai $S$ se. Spin-up ( $m_{s} = + \frac{1}{2}$ ) deta hai $μ_{z} = - μ_{B}$ ; spin-down deta hai $μ_{z} = + μ_{B}$ .
Force. $F_{z} = μ_{z} \frac{\partial B _{z}}{\partial z}$ . Do moments $\pm μ_{B}$ do opposite forces dete hain → two spots.

Verify: $μ_{z}^{u p} + μ_{z}^{d o w n} = (- μ_{B}) + (+ μ_{B}) = 0$ : forces exactly equal aur opposite hain ✔. Number of spots $= 2 s + 1 = 2$ ✔.

Figure — Spin — intrinsic angular momentum

Cell D — tilt angle

Vector $S$ kabhi bhi $z$ ke flat along nahi lete sakta. Yahan kyun hai, aur saare angles jo usse allowed hain.

Worked example Ex 3 —

$S$ ka $z$ -axis se angle (Cell D) Ek electron ke liye $S$ aur $+ z$ axis ke beech angle $θ$ nikalo, dono spin states ke liye.

Forecast: kya $θ = 0^{\circ}$ ho sakta hai (perfect alignment)? Pehle yes/no guess karo.

Cosine set up karo. Kisi vector ke liye, $z$ se uske angle ka cosine hai (uska $z$ -shadow) ÷ (uski length): $cos θ = \frac{S _{z}}{∣ S ∣} = \frac{m _{s} ℏ}{ℏ s ( s + 1 )} = \frac{m _{s}}{s ( s + 1 )} .$ Cosine kyun aur sine kyun nahi? Kyunki $S_{z} = ∣ S ∣ cos θ$ vector aur uski $z$ -projection se bane right triangle ka adjacent side hai — cosine hai "adjacent over hypotenuse."
Spin up. $cos θ = \frac{1/2}{3 /2} = \frac{1}{3} \approx 0.5774$ , isliye $θ = arccos (0.5774) \approx 54. 7^{\circ}$ .
Spin down. $m_{s} = - \frac{1}{2}$ deta hai $cos θ = - \frac{1}{3}$ , isliye $θ \approx 125. 3^{\circ}$ (yani $18 0^{\circ} - 54. 7^{\circ}$ ). Yeh step kyun? Down-state $x y$ -plane ke through mirror image hai; $+ z$ se uska tilt supplement hota hai.
$0^{\circ}$ kyun kabhi nahi? Perfect alignment ke liye chahiye hoga $S_{z} = ∣ S ∣$ , yani $\frac{1}{2} ℏ = \frac{3}{2} ℏ$ — jo false hai. Agar $S$ exactly $z$ along point karta, toh hum $S_{x} = S_{y} = 0$ bhi jaante, jo spin algebra mein built-in uncertainty violate karta. Isliye $S$ hamesha tilts karta hai.

Verify: $cos (54.735 6^{\circ}) = 1/ 3 \approx 0.5774$ ✔; $54.735 6^{\circ} + 125.264 4^{\circ} = 18 0^{\circ}$ ✔.

Cell E — degenerate case $s = 0$

Worked example Ex 4 — Spin-0 particle (Cell E)

Stern–Gerlach mein ek silver atom aise behave karta hai jaise $L = 0$ ; maan lo additionally ek hypothetical particle ka total spin $s = 0$ hai. $∣ S ∣$ , allowed $m_{s}$ , aur spots ki number nikalo.

Forecast: kitne spots — zero, one, ya two?

Magnitude. $∣ S ∣ = ℏ 0 (0 + 1) = ℏ 0 = 0$ . Yeh step kyun? $s = 0$ ke saath vector ki zero length hoti hai — bilkul bhi spin angular momentum nahi.
Allowed $m_{s}$ . Range $- s \dots + s$ collapse ho jaata hai single value $m_{s} = 0$ par.
Number of spots. $2 s + 1 = 2 (0) + 1 = 1$ → ek undeflected spot. Yeh kyun matter karta hai: yeh exactly wahi tha jo physicists ne 1922 se pehle silver ke liye expect kiya tha. Do spots milna is ki jagah jo cheez thi usne spin ko exist karna majboor kiya (dekho Stern-Gerlach Experiment).

Verify: $∣ S ∣ = 0$ aur spot count $= 1$ ✔; spin-0 particle ka koi magnetic moment nahi ( $μ_{z} = - g_{s} μ_{B} \cdot 0 = 0$ ) isliye zero force ✔.

Cell C phir — ek proton (positive charge ek sign flip karta hai)

Worked example Ex 5 — Positive charge ke liye sign care (Cell C)

Proton ka bhi $s = \frac{1}{2}$ hota hai. Uske spin-up state ke liye, magnetic moment $S$ ke relative kis direction mein point karta hai? (Proton ke apne g-factor ki chinta mat karo; sirf charge sign track karo generic positive charge $+ e$ ke saath.)

Forecast: electron ki same direction, ya opposite?

General moment. Charge $q$ ke liye: $μ = g \frac{q}{2 m} S$ . Yeh step kyun? Classical loop formula charge ka actual sign carry karta hai. Electrons mein $q = - e$ tha (isliye parent note mein minus); proton mein $q = + e$ hai.
$q = + e$ substitute karo. $μ_{z} = + g \frac{e}{2 m _{p}} S_{z}$ — koi minus nahi. Yeh step kyun? Positive charge $μ$ ko same taraf point karwata hai jis taraf $S$ hai, electron ki tarah nahi.
Spin up ⇒ $S_{z} = + \frac{1}{2} ℏ > 0$ ⇒ $μ_{z} > 0$ : moment points along $S$ (up).

Verify: charge sign flip karna moment direction flip karta hai — electron up-state mein $μ_{z} < 0$ hai, proton up-state mein $μ_{z} > 0$ hai; yeh opposite hain ✔. Case coverage: humne ab dono charge signs show kar diye hain.

Cell F — bade ladders

Worked example Ex 6 — Spin-1 aur spin-

$\frac{3}{2}$ (Cell F) $s = 1$ aur $s = \frac{3}{2}$ ke liye har $m_{s}$ list karo aur spots count karo. Dono ke liye $∣ S ∣$ bhi do.

Forecast: teen spots aur chaar spots? Aur kya magnitudes $ℏ$ aur $\frac{3}{2} ℏ$ hain? (Ek mein trap hai.)

$s = 1$ ladder. $m_{s} = - 1, 0, + 1$ → teen rungs → $2 s + 1 = 3$ spots. $0$ kyun? Integer $s$ se odd number of rungs milte hain, isliye $m_{s} = 0$ par ek middle rung exist karta hai (electron ke liye nahi hota).
$s = 1$ magnitude. $∣ S ∣ = ℏ 1 (1 + 1) = ℏ 2 \approx 1.414 ℏ$ ( $1ℏ$ nahi).
$s = \frac{3}{2}$ ladder. $m_{s} = - \frac{3}{2}, - \frac{1}{2}, + \frac{1}{2}, + \frac{3}{2}$ → chaar rungs → $2 s + 1 = 4$ spots.
$s = \frac{3}{2}$ magnitude. $∣ S ∣ = ℏ \frac{3}{2} \cdot \frac{5}{2} = ℏ 15 /2 \approx 1.936 ℏ$ ( $\frac{3}{2} ℏ$ nahi).

Verify: $1 \cdot 2 = 2 \approx 1.4142$ ✔; $\frac{3}{2} \cdot \frac{5}{2} = 3.75 \approx 1.9365$ ✔; spots $3$ aur $4$ , $2 s + 1$ se match karte hain ✔.

Cell G — classical limit jo fail karta hai

Worked example Ex 7 — Kya ek spinning ball light-speed tod deta? (Cell G)

Electron ko ek uniform sphere mano jiska "classical electron radius" $r_{e} = 2.82 \times 1 0^{- 15} m$ hai aur moment of inertia $I = \frac{2}{5} m_{e} r_{e}^{2}$ hai. Agar uska spin angular momentum $∣ S ∣ = \frac{3}{2} ℏ$ ke barabar ho, toh required equatorial speed $v = ω r_{e}$ nikalo aur $c$ se compare karo.

Forecast: $c$ ka fraction, roughly $c$ , ya kai baar $c$ ?

$L = I ω$ se angular speed. $ω = \frac{∣ S ∣}{I} = \frac{\frac{3}{2} ℏ}{\frac{2}{5} m _{e} r _{e}^{2}}$ . Yeh step kyun? Rigid body ke liye angular momentum $I ω$ hota hai; solve karne par woh spin rate milti hai jo "ball" ko chahiye hogi.
Equatorial speed. $v = ω r_{e} = \frac{\frac{3}{2} ℏ}{\frac{2}{5} m _{e} r _{e}^{2}} r_{e} = \frac{5 3 ℏ}{4 m _{e} r _{e}}$ .
Plug in. $v = \frac{5 \cdot 1.732 \cdot 1.0546 \times 1 0 ^{- 34}}{4 \cdot 9.109 \times 1 0 ^{- 31} \cdot 2.82 \times 1 0 ^{- 15}} \approx 8.9 \times 1 0^{10} m/s$ .
$c = 3.0 \times 1 0^{8} m/s$ se compare karo. Ratio $v / c \approx 296$ — light ki speed se 300 baar, impossible.

Verify: $v / c \approx 2.96 \times 1 0^{2} ≫ 1$ ✔ — parent note ke claim ko confirm karta hai ki spin literal rotation nahi hai. "Tiny spinning ball" picture sirf ek mnemonic hai.

Cell H — real-world Stern–Gerlach numbers

Cell I — exam-style twist (spin-flip energy)

Worked example Ex 9 — Spin flip karne ke liye photon energy (Cell I)

$z$ along $B = 1.0 T$ ke magnetic field mein, do electron spin states ki energy $U = - μ_{z} B$ hoti hai. Unke beech energy gap $Δ U$ nikalo aur us photon ki frequency $f$ nikalo jo spin flip karta hai ( $Δ U = h f$ , jahan $h = 2 π ℏ$ ). Yeh Zeeman Effect ke underlying hai.

Forecast: microwave ( $\sim 1 0^{10}$ Hz), infrared, ya visible?

Har state ki energies. $U = - μ_{z} B$ . Up-state $μ_{z} = - μ_{B} \Rightarrow U_{u p} = + μ_{B} B$ ; down-state $μ_{z} = + μ_{B} \Rightarrow U_{d o w n} = - μ_{B} B$ . Signs kyun matter karte hain: lower energy woh state hai jiska moment $B$ ke saath align hota hai. Sign galat karne par kaunsi state ground state hai, woh flip ho jaata hai.
Energy gap. $Δ U = U_{u p} - U_{d o w n} = 2 μ_{B} B$ .
Plug in. $Δ U = 2 \times 9.274 \times 1 0^{- 24} \times 1.0 = 1.855 \times 1 0^{- 23} J$ .
Photon frequency. $f = \frac{Δ U}{h} = \frac{Δ U}{2 π ℏ} = \frac{1.855 \times 1 0 ^{- 23}}{6.626 \times 1 0 ^{- 34}} \approx 2.80 \times 1 0^{10} Hz$ — ek microwave (~28 GHz).

Verify: $2 μ_{B} B = 1.855 \times 1 0^{- 23}$ J ✔; $f \approx 2.80 \times 1 0^{10}$ Hz, microwave band mein ✔. (Yeh exactly electron spin resonance hai.)

Recall

Recall Khud ko check karo

$s = \frac{3}{2}$ ke liye magnitude ::: $∣ S ∣ = ℏ 15 /2 \approx 1.94 ℏ$ ( $\frac{3}{2} ℏ$ nahi). $s = 1$ ke liye spots ki number ::: teen ( $m_{s} = - 1, 0, + 1$ ). Spin-0 ek spot kyun deta hai ::: sirf $m_{s} = 0$ exist karta hai aur $∣ S ∣ = 0$ , isliye koi moment nahi, koi splitting nahi. 1 T mein electron spin-flip photon ::: ~28 GHz microwave ( $Δ U = 2 μ_{B} B$ ). Electron vs proton spin moment ka sign difference ::: electron ka $μ$ opposes karta hai $S$ ko; proton ka $μ$ aligns karta hai $S$ ke saath (charge sign).

Connections

Parent: Spin — intrinsic angular momentum
Stern-Gerlach Experiment — Ex 2, 4, 8 uska arithmetic hai.
Orbital Angular Momentum — same $ℓ (ℓ + 1)$ magnitude rule, integer only.
Quantum Numbers — jahan $m_{s}$ full state label mein baithta hai.
Bohr Magneton — woh unit jo Ex 8, 9 mein kaam karti hai.
Zeeman Effect & Fine Structure — Ex 9 ki energy splitting.
Pauli Exclusion Principle — Ex 2 ke do $m_{s}$ chahiye.
Dirac Equation — $g_{s} \approx 2$ ka source jo upar har jagah use hua hai.