Quantum tunneling — concept, transmission coefficient: Question bank

2.3.11 · D5 · Physics › Modern Physics › Quantum tunneling — concept, transmission coefficient

Definition Is page par use hone wala map aur har symbol (pehle padho)

Poori problem ek particle of energy $E$ ki hai jo ek rectangular barrier se milti hai: ek aisi region jahan potential energy $V_{0}$ height tak upar jump karti hai. Space teen hisson mein bant jaata hai — neeche di gayi figure dekho:

Region I ( $x < 0$ ): seedha maidan, potential $V = 0$ . Yahan particle ek free wave hai.
Region II ( $0 < x < L$ ): barrier, potential $V = V_{0}$ , width $L$ . Yahan $E < V_{0}$ hai, isliye ye classically forbidden hai.
Region III ( $x > L$ ): phir seedha maidan, $V = 0$ .

Neeche jo symbols dikhenge wo sab Schrödinger's equation ko un teenon hisson mein solve karne se aate hain:

$k = \frac{2 m E}{ℏ}$ — free-particle wavenumber: flat regions I aur III mein wave kitni tezi se wiggle karti hai (bada $k$ = chhoti wavelength). Yahan $m$ particle ka mass hai aur $ℏ$ reduced Planck constant hai.
$κ = \frac{2 m ( V _{0} - E )}{ℏ}$ — decay constant (Greek "kappa"): barrier ke andar wavefunction kitni tezi se sikorti hai. Kyunki $V_{0} > E$ hai, square root real hai, isliye $κ$ ek real positive number hai aur $e^{- κ x}$ sachchi decay hai.
$A, B$ — region I mein incoming aur reflected waves ke amplitudes: $ψ_{I} = A e^{ik x} + B e^{- ik x}$ .
$C, D$ — region II mein do barrier terms ke amplitudes: $ψ_{I I} = C e^{- κ x} + D e^{+ κ x}$ (ek girta aur ek chadta exponential).
$F$ — region III mein akele transmitted wave ka amplitude: $ψ_{I I I} = F e^{ik x}$ .
$T = \frac{∣ F ∣ ^{2}}{∣ A ∣ ^{2}}$ — transmission coefficient: kitna fraction incident particles cross kar paate hain. $R$ reflected fraction hai, aur $T + R = 1$ .
$T \approx G e^{- 2 κ L}$ with prefactor $G = \frac{16 E ( V _{0} - E )}{V _{0}^{2}}$ (order-1 number). Note: is page par $G$ hamesha isi prefactor ko denote karta hai. Jab hum deliberately ek galat " $G e^{- ik x}$ " kisi Spot-the-error item mein likhte hain, to woh stray $G$ hi woh mistake hai jo flag ki ja rahi hai.

Figure — Quantum tunneling — concept, transmission coefficient

Upar wali figure har neeche wale question ke peeche ki mental picture hai — ise saamne rakhho.

True or false — justify

Ek tunneling particle energy udhar leti hai cross karne ke liye aur baad mein wapas kar deti hai.

False — energy poore time conserved rehti hai; particle ki total energy

E

kabhi nahi badalti, aur ye poore time barrier height

V_{0}

se neeche rehti hai. Kuch bhi udhar nahi liya jaata; wavefunction ka bas forbidden region mein nonzero amplitude hota hai.

Particle ki energy

E

, apne safar ke dauran har jagah potential se kam hai.

False —

E < V_{0}

sirf barrier ke andar (region II) hold karta hai; flat regions I aur III mein potential

0

hai, jo

E

se neeche hai. "Forbidden" comparison

E < V_{0}

sirf barrier par laagu hota hai.

Barrier ke andar particle ek decaying wave hai, isliye cross karte waqt wo dheere dheere energy khota hai.

False — amplitude

∣ ψ ∣

time ke saath nahi balki position

x

ke saath decay karta hai, aur ye energy loss nahi balki probability encode karta hai. Jo particle bahar nikalti hai wo poori energy

E

lekar aati hai.

Agar barrier infinitely wide hoti, toh

T

exactly zero hota.

True —

T \approx G e^{- 2 κ L} \to 0

jab

L \to \infty

, kyunki

ψ

ki exponential decay

e^{- κ x}

literally koi amplitude far side tak pahunchne nahi deti.

Barrier width double karne se transmission probability half ho jaati hai.

False — width ek exponent mein hai, isliye

L

double karne se tiny factor

e^{- 2 κ L}

square ho jaata hai. Concretely, ek electron ke liye jab

E = 1 eV

V_{0} = 2 eV

L = 0.5 nm

par

2 κ L \approx 5.12

;

L = 1.0 nm

par

2 κ L \approx 10.24

T

ka ratio

e^{10.24 - 5.12} = e^{5.12} \approx 167

hai — lagbhag

170 \times

chhota,

2 \times

nahi.

Ek barrier ke liye jahan

E > V_{0}

ho, transmission hamesha 100% hoti hai.

False — barrier ke upar bhi partial reflection hoti hai, aur

T

energy ke saath oscillate karta hai (Ramsauer–Townsend resonances) kyunki

κ

imaginary ho jaata hai,

sinh

sin

mein badal deta hai.

Same

E

aur

V_{0}

par ek bhaari particle kam tunnel karta hai.

True —

κ = 2 m (V_{0} - E) /ℏ \propto m

, isliye zyada mass matlab

ψ

ki tezi se decay aur chhota

e^{- 2 κ L}

; ek proton electron se bahut kam tunnel karta hai (dekho de Broglie Wavelength kyun heavy = short wavelength = "zyada classical").

Tunneling energy conservation ko violate karta hai.

False — ye sirf classical rule ko violate karta hai ki pass karne ke liye

E \geq V_{0}

zaroori hai; total energy exactly conserved hai. Jo "violate" hota hai wo hamari classical intuition hai, koi conservation law nahi.

Exact

sinh^{2}

formula aur

e^{- 2 κ L}

formula ek thin, low barrier ke liye same answer dete hain.

False — approximation assume karta hai ki

κ L ≫ 1

; thin barriers ke liye

κ L

chhota hota hai aur tumhe exact form use karni padti hai (dekho agli item mein kyun

κ L ≫ 1

zaroori hai).

Spot the error

"Barrier ke andar

ψ = C e^{- iκ x} + D e^{+ iκ x}

, ek oscillating wave hai."

Error:

i

galat hai. Kyunki

V_{0} > E

, Schrödinger's equation padha jaata hai

ψ^{''} = + κ^{2} ψ

real solutions

e^{\pm κ x}

ke saath (

C

girta,

D

chadta) — pure decay/growth, koi oscillation nahi. Dekho Potential Barrier and Reflection.

"Region III mein hum

ψ = F e^{ik x} + G e^{- ik x}

likhte hain general rehne ke liye."

Error: doosra term ek leftward wave hai jo

+ \infty

se aa rahi hai, lekin right se kuch incident nahi hai — isliye iska amplitude zero hona chahiye. (Saath hi,

G

theek se transmission prefactor ko denote karta hai, wave amplitude ko nahi; label ka ye clash khud ek red flag hai.) Sirf transmitted wave

F e^{ik x}

bachti hai.

T = ∣ F ∣^{2} /∣ A ∣^{2}

kaam karta hai kyunki energy amplitudes ka ratio hai."

Error:

T

probability fluxes ka ratio hai (probability per second jo ek plane cross karti hai), energies ka nahi. Flux hota hai speed

\times ∣ amplitude ∣^{2}

; yahan

A

region I mein incoming amplitude hai aur

F

region III mein transmitted amplitude. Ye

∣ F ∣^{2} /∣ A ∣^{2}

tab collapse hota hai sirf isliye kyunki $k$ regions I aur III mein identical hai, isliye speeds cancel ho jaati hain; dono sides par alag potentials ke saath velocity factors rakhne padte.

"Hum barrier ke andar

D e^{+ κ x}

growing term drop karte hain kyunki wo normalizable nahi hai."

Error: ek finite barrier ke liye growing term rakhna zaroori hai — region II finite hai (

0 < x < L

) isliye kuch blow up nahi hota. Tum sirf semi-infinite barrier ke liye

e^{+ κ x}

drop karte ho. Yahan char boundary conditions dono

C

aur

D

ko fix karte hain.

"Kyunki

T \approx G e^{- 2 κ L}

aur

G

order 1 hai, hum

G

ko completely ignore kar sakte hain aur

T = e^{- 2 κ L}

Error:

G = 16 E (V_{0} - E) / V_{0}^{2}

chhota ho sakta hai jab

E ≪ V_{0}

(energy bottom ke paas), isliye ise drop karne se number badal jaata hai. Scaling exponent mein hai, lekin prefactor actual value set karta hai.

e^{- 2 κ L}

mein 2 ka factor is wajah se aata hai kyunki do barrier walls hain."

Error: ye

∣ ψ ∣^{2}

ki wajah se aata hai. Amplitude

e^{- κ L}

ki tarah decay karta hai; probability amplitude squared hoti hai, jo

e^{- 2 κ L}

deta hai. Do walls se koi lena-dena nahi.

"Tunneling ke liye barrier

E

se unchi honi chahiye, isliye unchi barrier matlab aasaan tunneling."

Error: ulta hai. Uncha

V_{0}

(V_{0} - E)

badhata hai, isliye

κ

badhta hai, isliye decay rate badhti hai — tunneling mushkil ho jaati hai, aasaan nahi.

x = 0

par hume sirf

ψ

match karni hai; slope jump kar sakta hai."

Error: dono

ψ

aur iska slope

ψ^{'}

x = 0

aur

x = L

par continuous hone chahiye. Isse char equations milti hain (

ψ_{I} = ψ_{I I}

ψ_{I}^{'} = ψ_{I I}^{'}

0

par;

ψ_{I I} = ψ_{I I I}

ψ_{I I}^{'} = ψ_{I I I}^{'}

L

par) — bilkul enough

B, C, D, F

A

ke relative fix karne ke liye. Dekho Wavefunction and Boundary Conditions.

Why questions

ψ

barrier ke andar oscillatory kyun nahi balki exponential kyun hai?

Kyunki

V_{0} > E

curvature term ka sign flip karke

ψ^{''} = + κ^{2} ψ

bana deta hai; iske sirf real solutions

e^{\pm κ x}

hain. Oscillation (

e^{\pm ik x}

) ke liye negative-curvature case

E > V

chahiye, jo regions I aur III mein hota hai.

sinh (κ L)

ko thick/high barrier ke liye

\frac{1}{2} e^{κ L}

se kyun replace kar sakte hain?

Definition se

sinh (κ L) = \frac{1}{2} (e^{κ L} - e^{- κ L})

. Jab

κ L ≫ 1

to shrinking term

e^{- κ L}

growing

e^{κ L}

ke comparison mein negligible hai (e.g.

κ L = 3

par ye already uska ~

0.25%

hai), isliye

sinh (κ L) \approx \frac{1}{2} e^{κ L}

aur hence

T \approx G e^{- 2 κ L}

. Relative error roughly

e^{- 2 κ L}

hai, thick barriers ke liye tiny.

Wavefunction wall par zero tak drop kyun nahi ho sakti decay ki jagah?

Kyunki Wavefunction and Boundary Conditions require karta hai ki

ψ

aur iska slope

ψ^{'}

har jagah continuous hon; achanak zero tak drop karna slope ki continuity tod dega, jo Schrödinger's equation forbid karta hai.

STM (dekho Scanning Tunneling Microscope) atomic resolution kyun achieve karta hai?

Kyunki

T \propto e^{- 2 κ L}

tip–surface gap

L

ke liye exponentially sensitive hai; ek atom ke diameter ke fraction jitna change tunneling current ko ek bade factor se badal deta hai, isliye height variations hugely amplified ho jaati hain.

Electrons tunnel karte hain lekin bowling balls kabhi nahi karte — kyun?

κ \propto m

, isliye macroscopic object ka enormous mass

2 κ L

ko astronomically large bana deta hai aur

e^{- 2 κ L}

effectively zero ho jaata hai. Halke electrons

2 κ L

ko modest rakhte hain, isliye appreciable amplitude survive karta hai.

Alpha decay nucleus par itni violently depend kyun karta hai (dekho Alpha Decay)?

Nikalne wala alpha particle Coulomb barrier se tunnel karke nikalta hai, aur

T \sim e^{- 2 κ L}

escape probability ko — aur hence half-life ko — energy ya barrier size mein chote changes ke liye 20+ orders of magnitude tak swing karata hai.

Transmitted particle ki energy exactly

E

hi kyun rehti hai?

Region III mein

V = 0

hai bilkul region I ki tarah, aur wahi wavenumber

k = 2 m E /ℏ

dono mein appear karta hai. Same

k

matlab same kinetic energy — koi energy add ya remove nahi hui.

E < V_{0}

ke liye

T

kabhi 1 tak kyun nahi pahunchti?

Finite barrier hamesha kuch amplitude reflect karta hai (

B \neq = 0

), isliye probability transmission aur reflection mein split hoti hai

T + R = 1

ke saath aur

T < 1

. Barrier ke neeche perfect transmission impossible hai.

Edge cases

T

ka kya hoga jab

E \to V_{0}^{-}

(energy barrier height ke paas pahunche)?

κ \to 0

, decay khatam ho jaati hai, aur

sinh (κ L) \approx κ L

use karne par exact formula

T \to [1 + \frac{m V _{0} L ^{2}}{2 ℏ ^{2}}]^{- 1}

deta hai — finite aur aksar 1 ke kaafi karib, divergent nahi. Isliye yahan

κ L ≫ 1

approximation bilkul use nahi karni chahiye.

E \to 0^{+}

(bahut slow particle) par exact formula kya karta hai?

Prefactor

G = 16 E (V_{0} - E) / V_{0}^{2} \to 0

aur

sinh^{2} (κ L)

finite rehta hai, isliye

T \to 0

. Lagbhag ruka hua particle bahut kam tunnel karta hai kyunki wo almost koi incident flux nahi le jaata.

Jab

E > V_{0}

(barrier ke upar) ho toh kya hota hai?

(V_{0} - E)

negative ho jaata hai, isliye

κ = 2 m (V_{0} - E) /ℏ

imaginary ho jaati hai; likho

κ = i k^{'}

aur

sinh (κ L) \to i sin (k^{'} L)

. Tab

T = 1

jab bhi

sin (k^{'} L) = 0

(resonant full transmission) aur beech mein dip karta hai — Ramsauer–Townsend effect.

Zero width

L \to 0

ki barrier ke liye

T

kya hai?

sinh (0) = 0

isliye exact formula

T = 1

deta hai. Cross karne ke liye koi barrier hi nahi hai toh reflect hone ki koi wajah nahi — total transmission, jaise expected hai.

Agar barrier infinitely tall ho (

V_{0} \to \infty

) fixed

E

aur

L

ke saath?

κ \to \infty

, isliye

e^{- 2 κ L} \to 0

aur

T \to 0

. Ye classical hard wall recover karta hai: infinite barrier impenetrable hai.

Kya

ψ

finite barrier ke andar exactly zero ho sakti hai?

Generally nahi ground-tunneling case ke liye —

ψ_{I I} = C e^{- κ x} + D e^{+ κ x}

monotone exponentials ka sum hai jisme koi oscillation nahi, isliye iske interior mein koi nodes nahi hote; ye barrier ke across smoothly decay karta hai.

Recall Jaane se pehle ek-line self-test

Sab cover karo: batao (1) $ψ$ oscillate kyun nahi karti balki decay kyun karti hai, (2) $e^{- 2 κ L}$ mein "2" kahan se aaya, (3) $E > V_{0}$ hone par $T$ kya karta hai, aur (4) bahar nikalne wale particle ki energy badi ya nahi. Answers ::: (1) $V_{0} > E$ se $ψ^{''} = + κ^{2} ψ$ milta hai, real exponentials; (2) probability $∣ ψ ∣^{2}$ hai, $e^{- κ L}$ ko square karne se; (3) $T = 1$ resonances ke saath partial transmission (imaginary $κ$ ); (4) unchanged, abhi bhi $E$ .