Visual walkthrough — Quantum tunneling — concept, transmission coefficient

2.3.11 · D2 · Physics › Modern Physics › Quantum tunneling — concept, transmission coefficient

Hum ek specific scene study kar rahe hain: energy $E$ ka ek particle rightward fly kar raha hai, aur ek rectangular wall se milta hai jisme height $V_0$ aur width $L$ hai. Hum $E<V_0$ lete hain — wall particle ki energy se zyada oonchi hai, isliye ek classical ball bounce kar jaati. Chalte hain dekhte hain ek wave ki jagah kya hota hai.

---

Step 0 — Vocabulary, ek picture ke saath

Kisi bhi algebra se pehle, un chaar words se milo jinhe hum baar baar use karenge. Har ek sirf ek simple idea hai jiske saath ek picture hai.

• Wavefunction $\psi (x)$ — har position $x$ par ek number attached hota hai. Iska square $|\psi {|}^2$ batata hai ki particle wahan milne ki kitni probability hai. $\psi$ ko parent ki tennis-ball story ki "fog density" samjho.
• Energy $E$ — incoming particle ke paas kitna oomph hai. Poore safar ke liye ek fixed number.
• Barrier height $V_0$ — wall ke andar rehne ki "cost". Kyunki $E<V_0$ hai, region $0<x<L$ woh jagah hai jise particle classically "afford" nahi kar sakta.
• Barrier width $L$ — wall kitni moti hai, $x=0$ se $x=L$ tak.

---

Step 1 — Woh ek equation jo sab kuch govern karti hai

Har quantum shape time-independent Schrödinger Equation follow karti hai:

$\underset{\text{kitna "stiff" hai}}{\underbrace{-\frac{{\hslash }^2}{2m}}}\underset{\text{curvature of }\psi }{\underbrace{\frac{d^2\psi }{d{x}^2}}}+\underset{\text{yahan rehne ki cost}}{\underbrace{V(x)\psi }}=\underset{\text{total energy}}{\underbrace{E\psi }}$

HAR piece kya kar raha hai:

• $\frac{d^2\psi }{d{x}^2}$ — curvature (kitni sharply $\psi$ bend karti hai). Yahi woh quantity hai jise hum solve karte hain.
• $\hslash$ — Planck's constant divided by $2\pi$; quantum scale set karta hai. $m$ — particle ki mass.
• $V(x)$ — position $x$ par potential energy ($0$ wall ke bahar, $V_0$ andar).
• $E$ — fixed total energy.

YEH tool kyun use karein aur Newton ka $F=ma$ kyun nahi? Newton ek point track karta hai. Lekin ek quantum particle ek spread-out wave hai, aur hum uski shape chahte hain. Schrödinger equation woh rule hai jise woh shape satisfy karni chahiye. Aaiye ise rearrange karte hain curvature ko isolate karne ke liye, kyunki curvature hi decide karta hai ki wave oscillate karegi ya decay karegi:

$\frac{d^2\psi }{d{x}^2}=\frac{2m(V(x)-E)}{{\hslash }^2}\psi$

PICTURE: factor $(V(x)-E)$ ka sign poori story decide karta hai. Neeche dekho.

---

Step 2 — Region I: incoming aur reflected waves

Wall ke left mein bahar, $V=0$ hai, isliye $V-E=-E$, ek negative number. Ise $-k^2$ likho:

$\frac{d^2{\psi }_I}{d{x}^2}=-\frac{2mE}{{\hslash }^2}{\psi }_I\equiv -k^2{\psi }_I,k=\frac{\sqrt{2mE}}{\hslash }$

$k$ kya hai? Yeh wavenumber hai — pratyek metre mein kitne radians of wiggle hain. Bada $E$ → bada $k$ → chhoti wavelength (yeh de Broglie Wavelength $\lambda =2\pi /k$ disguise mein hai).

Negative curvature wave kyun deta hai? Kyunki woh function jo har jagah zero ki taraf vapas curve karti hai woh sine/cosine hai — yani $e^{\pm ikx}$. Solution:

${\psi }_I=\underset{\text{right ki taraf move kar raha hai (incoming)}}{\underbrace{A{e}^{ikx}}}+\underset{\text{left ki taraf move kar raha hai (reflected)}}{\underbrace{B{e}^{-ikx}}}$

• $A$ — incoming wave ka amplitude (jo hum fire karte hain).
• $B$ — reflected wave ka amplitude (jo bounce karke vapas aata hai).

PICTURE: ek incoming ripple, jiska kuch hissa wall se aise bounce karega jaise glass se light — dekho Potential Barrier and Reflection.

---

Step 3 — Region II: wall ke andar, wave decay karti hai

Andar, $V=V_0>E$ hai, isliye $V-E=+(V_0-E)$, ek positive number. Ise $+{\kappa }^2$ likho:

$\frac{d^2{\psi }_{II}}{d{x}^2}=+\frac{2m(V_0-E)}{{\hslash }^2}{\psi }_{II}\equiv +{\kappa }^2{\psi }_{II},\boxed{\kappa =\frac{\sqrt{2m(V_0-E)}}{\hslash }}$

$\kappa$ (kappa) kya hai? Decay constant — wall ke andar $\psi$ pratyek metre mein kitni tezi se shrink hoti hai. Yeh is page par sabse important number hai.

Real exponential kyun, wave kyun nahi? Positive curvature ka matlab hai ki graph axis se door bend karta hai. Aisa karne wale functions hain $e^{+\kappa x}$ (grows) aur $e^{-\kappa x}$ (shrinks):

${\psi }_{II}=\underset{\text{decaying}}{\underbrace{C{e}^{-\kappa x}}}+\underset{\text{growing}}{\underbrace{D{e}^{+\kappa x}}}$

Dono ek finite wall ke liye allowed hain (hum abhi $D$ nahi drop kar sakte kyunki wall $x=L$ par khatam hoti hai, growth ke blow up hone se pehle).

PICTURE: fog wall par nahi rukti — woh smoothly us par fade hoti hai.

---

Step 4 — Region III: survivor bahar aata hai

Wall ke baad, $V=0$ phir se hai — Region I wali hi equation, wahi $k$. Lekin ab sirf ek tarah ki wave exist kar sakti hai:

${\psi }_{III}=\underset{\text{transmitted, right ki taraf move kar raha hai}}{\underbrace{F{e}^{ikx}}}$

Yahan leftward wave kyun nahi? $+\infty$ ke baad kuch nahi hai jo wave ko reflect kare vapas. Isliye $e^{-ikx}$ ka coefficient zero hai. Sirf transmitted amplitude $F$ bachta hai.

$F$ kya hai? Woh amplitude jo through nikla — $A$ se chhota, lekin zero nahi. Woh "not zero" hi tunneling hai.

PICTURE: poora scene ek frame mein — badi wave andar, ek faded exponential across, ek chhoti wave bahar.

---

Step 5 — Regions ko stitching karna: boundary conditions

Hamare paas teen alag solutions hain jinme paanch unknowns $A,B,C,D,F$ hain. Schrödinger's equation demand karti hai ki har join par ($x=0$ aur $x=L$) dono $\psi$ aur uska slope ${\psi }^{\prime }$ match karein — koi jumps nahi, koi kinks nahi.

${\psi }_I(0)={\psi }_{II}(0),{\psi }_I^{\prime }(0)={\psi }_{II}^{\prime }(0),{\psi }_{II}(L)={\psi }_{III}(L),{\psi }_{II}^{\prime }(L)={\psi }_{III}^{\prime }(L)$

Yeh match kyun karne chahiye? $\psi$ mein ek jump $|\psi {|}^2$ (ek probability) ko ek point par two-valued bana dega — nonsense hai yeh. Ek kink (slope mein jump) curvature ko infinite bana dega, Schrödinger's equation blow up kar dega. Isliye dono outlawed hain. Yeh exactly Wavefunction and Boundary Conditions hai.

Hamen kya milta hai: ratios ke liye chaar equations. Algebra grind karne par ($C,D$ eliminate karo, $F/A$ solve karo) neeche diya exact result milta hai.

---

Step 6 — Thick-wall limit (woh formula jo actually use hota hai)

Ek moti ya unchi wall ke liye, $\kappa L\gg 1$. Tab $e^{-\kappa L}$ tiny hai, isliye:

$\sinh (\kappa L)=\frac{1}{2}(e^{\kappa L}-e^{-\kappa L})\approx \frac{1}{2}{e}^{\kappa L}\Rightarrow {\sinh }^2(\kappa L)\approx \frac{1}{4}{e}^{2\kappa L}$

$-e^{-\kappa L}$ aur $1$ ko kyun drop karein? Kyunki $e^{2\kappa L}$ astronomically bada hai $1$ se, "$1+$" aur chhota term uske saamne gayab ho jaate hain. Substitute karne par:

$T\approx \frac{4E(V_0-E)}{V_0^2\cdot \frac{1}{4}{e}^{2\kappa L}}=\frac{16E(V_0-E)}{V_0^2}{e}^{-2\kappa L}$

---

Step 7 — Edge aur degenerate cases (inhe kabhi skip mat karo)

Formula ko har corner mein sensibly behave karna chahiye. Sab check karte hain.

• Vanishing width, $L\to 0$: $\kappa L\to 0$, $\sinh (0)=0$, toh $T\to [1+0{]}^{-1}=1$. Koi wall nahi, sab kuch pass hota hai. ✔
• Bahut thick/tall wall, $\kappa L\to \infty$: $e^{-2\kappa L}\to 0$, toh $T\to 0$. Kuch bhi through nahi jaata. ✔
• Bhaari particle, $m\to \infty$: $\kappa \propto \sqrt{m}\to \infty$, toh $T\to 0$. Bowling balls tunnel nahi karte — isliye Alpha Decay (ek bhaari $\alpha$ particle) slow aur rare hota hai, aur macroscopic objects kabhi tunnel nahi karte. ✔
• Energy top se milti hai, $E\to V_0^{-}$: $\kappa \to 0$, aur exact formula ka ratio $\frac{{\sinh }^2(\kappa L)}{4E(V_0-E)}$ finite rehta hai (numerator aur denominator dono $\to 0$ saath mein). $T$ yahan smooth hai — koi blow-up nahi. ✔
• Barrier ke upar, $E>V_0$: ab $V_0-E<0$ hai, isliye $\kappa$ imaginary ban jaata hai, $\kappa =i{k}^{\prime }$. Tab $\sinh (\kappa L)=\sinh (i{k}^{\prime }L)=i\sin (k^{\prime }L)$, decay ko ek oscillation mein badal deta hai. $T$ ab oscillate karta hai aur resonances par exactly 1 hit kar sakta hai — Ramsauer–Townsend effect. $E>V_0$ ko decay formula mein blindly mat daalo. ✔

---

Ek-picture summary

Upar ki saari cheez, ek single frame mein compress ki gayi: wave andar (amplitude $A$), wall ke across exponential fade (rate $\kappa$), wave bahar (amplitude $F$), aur final formula apni geometry par annotated.

Recall Feynman retelling — poora walkthrough simple words mein

Ek fuzzy wave-cloud left se aata hai (woh hai $A$). Yeh ek aisi wall se takrata hai jo isse zyada unchi hai jitna yeh climb kar sake. Kuch hissa bounce karke vapas aata hai (woh hai $B$). Lekin cloud simply dead stop nahi ho sakta — rules kehte hain iske smooth fade hona zaroori hai, isliye wall ke andar yeh ek rate $\kappa$ se exponentially shrink karta hai jo particle ki heaviness aur wall particle ki energy se kitni zyada unchi hai, isse set hota hai. Agar wall patli hai, toh cloud far edge tak pahunchne par abhi bhi faintly wahan hai — aur ek chhoti wave bahar crawl karti hai (woh hai $F$). Cross karne ka chance $T=|F{|}^2/|A{|}^2$ hai, aur kyunki fade width mein exponential hai, $T\approx G{e}^{-2\kappa L}$ hai. Patli wall, halka particle, neecha wall → zyada leaks through. Woh leak tunneling hai — Scanning Tunneling Microscope aur Alpha Decay ke engine.

Recall Active recall

Curvature mein sign flip kahan se aata hai? ::: $V(x)-E$ se: bahar negative ($E>V$) waves $e^{\pm ikx}$ deta hai; andar positive ($V_0>E$) decay $e^{\pm \kappa x}$ deta hai. Wall ke andar dono $e^{-\kappa x}$ aur $e^{+\kappa x}$ kyun rakhte hain? ::: Wall finite hai ($L$ par khatam hoti hai), isliye growing term blow up nahi karti aur discard nahi ki ja sakti. Region III mein sirf $F{e}^{ikx}$ kyun? ::: $+\infty$ ke baad kuch bhi wave ko reflect nahi karta, isliye koi leftward component exist nahi karta. $\psi$ aur ${\psi }^{\prime }$ ko joins par match karne ke liye kya force karta hai? ::: Ek jump probability ko two-valued banata hai; ek kink curvature ko infinite banata hai — dono Schrödinger's equation ko break karte hain. $L\to 0$ kya deta hai aur kyun? ::: $T\to 1$: $\sinh (0)=0$ barrier term ko remove kar deta hai — koi wall nahi, full transmission. $E>V_0$ hone par kya hota hai? ::: $\kappa$ imaginary ho jaata hai, $\sinh \to i\sin$, oscillating $T$ deta hai resonances ke saath (over-barrier transmission), simple decay nahi.