--- title: "Derivation of tunneling probability through barrier" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:tunneling-probability-derivation" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-probability-derivation site: BCT Engineering --- # Derivation of tunneling probability through barrier Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2068 Shrawan, 8 marks - Video: [Quantum Tunneling |Tunnel effect | Quantum Physics | BTech | BSc | GATE | UPSC | Physics | IGNOU](https://www.youtube.com/watch?v=rdNfjN9H-14) — Gautam Varde - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-probability-derivation - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question What is tunneling phenomenon? Derive the expression for the probability of tunneling the potential barrier of width L and height V by electron. ## Short answer Tunneling is a quantum effect where an electron with energy E penetrates a potential barrier of height V > E, classically forbidden. The transmission probability is T = [1 + V² sinh²(αL)/(4E(V−E))]⁻¹; for a wide barrier, T ≈ (16E(V−E)/V²) exp[−2L√(2m(V−E))/ħ]. ## Answer **Tunnelling** is the quantum effect in which a particle with energy $E$ passes through a potential barrier of height $V > E$, which is impossible in classical mechanics. It happens because the wave function does not drop to zero inside the barrier but decays exponentially; if the barrier is thin, some of the wave reaches the far side. Examples: tunnel diode, scanning tunnelling microscope, field emission, conduction through thin oxide layers on contacts. ```text V(x) ^ __________ | V | | | E-----|- - - - - |----- | ~~~~~ | \_ | ~~~ | I | II \__ | III |_______|__________|_____> x 0 L ``` #### Wave functions in three regions Let $k = \dfrac{\sqrt{2mE}}{\hbar}$ and $\alpha = \dfrac{\sqrt{2m(V - E)}}{\hbar}$. - Region I ($x < 0$, $V = 0$): $\psi_I = A_1 e^{ikx} + B_1 e^{-ikx}$ (incident + reflected) - Region II ($0 < x < L$, potential $V$): $\psi_{II} = A_2 e^{\alpha x} + B_2 e^{-\alpha x}$ - Region III ($x > L$, $V = 0$): $\psi_{III} = A_3 e^{ikx}$ (transmitted only) #### Boundary conditions $\psi$ and $d\psi/dx$ are continuous at $x = 0$ and $x = L$: $$ \begin{aligned} A_1 + B_1 &= A_2 + B_2 \\ ik(A_1 - B_1) &= \alpha(A_2 - B_2) \\ A_2 e^{\alpha L} + B_2 e^{-\alpha L} &= A_3 e^{ikL} \\ \alpha(A_2 e^{\alpha L} - B_2 e^{-\alpha L}) &= ik A_3 e^{ikL} \end{aligned} $$ #### Transmission coefficient The tunnelling probability is the ratio of transmitted to incident flux. Since the speed is the same in regions I and III: $$ T = \frac{|A_3|^2}{|A_1|^2} $$ Eliminating $B_1, A_2, B_2$ from the four equations gives the exact result $$ T = \left[1 + \frac{V^2 \sinh^2(\alpha L)}{4E(V - E)}\right]^{-1} $$ #### Wide or high barrier ($\alpha L \gg 1$) Then $\sinh(\alpha L) \approx \tfrac{1}{2}e^{\alpha L}$, so $\sinh^2(\alpha L) \approx \tfrac{1}{4}e^{2\alpha L} \gg 1$ and the 1 can be neglected: $$ T \approx \frac{16E(V - E)}{V^2}\, e^{-2\alpha L} = T_0 \exp\left[-\frac{2L\sqrt{2m(V - E)}}{\hbar}\right] $$ with $T_0 = \dfrac{16E(V-E)}{V^2}$. #### Conclusions 1. $T$ falls **exponentially** with barrier width $L$; doubling $L$ greatly reduces tunnelling. 2. $T$ falls as $\sqrt{V - E}$ rises, so a higher barrier or a lower electron energy reduces tunnelling. 3. Heavier particles (larger $m$) tunnel far less, which is why tunnelling is noticed for electrons and only for barriers of about a nanometre or less. 4. Reflection probability is $R = 1 - T$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is tunnelling? A. A particle with E < V passing through a barrier because psi decays but is non-zero B. A particle climbing over the barrier using extra energy C. Reflection of all particles D. Heating of the barrier Answer: A. A thin barrier lets part of the decaying wave emerge on the far side. (Key point: Tunnelling definition) 2. Why does region III contain only A3 e^(ikx)? A. Only a transmitted wave moves away; nothing reflects from beyond B. Particle stops C. Wave is evanescent D. V is infinite Answer: A. No reflecting obstacle beyond the barrier, so only a transmitted wave exists. (Key point: Wave in region III) 3. What conditions at x = 0 and x = L give the four equations? A. psi and d psi/dx continuous B. psi = 0 C. Only psi continuous D. Energy conserved but psi discontinuous Answer: A. Continuity of psi and its derivative at both boundaries. (Key point: Boundary conditions) 4. For alpha L >> 1, which expression holds? A. T = 16E(V-E)/V^2 x e^(-2 alpha L) B. T = 1 - e^(-2 alpha L) C. T = e^(2 alpha L) D. T = 1 Answer: A. sinh^2 is approx e^(2 alpha L)/4, leaving T = T0 e^(-2 alpha L). 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