--- title: "Well wave functions and finite barrier decay" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:well-sketch-finite-barrier-decay" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-sketch-finite-barrier-decay site: BCT Engineering --- # Well wave functions and finite barrier decay Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2069 Asar, 2+6 marks - Video: [Finite Quantum Well Explained - Part 1](https://www.youtube.com/watch?v=Q_rBTDwaNt0) — Jordan Edmunds Chetty - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/well-sketch-finite-barrier-decay - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question Sketch energy level and wave function diagram for n=1,2,3 for infinite potential well. Show that the wave function in the finite potential barrier decays exponentially. ## Short answer In an infinite well, wavefunctions are sinusoidal with n half‑waves and energies E_n = n²h²/(8mL²). Inside a finite barrier of height V0 > E, the Schrödinger equation gives d²ψ/dx² = α²ψ with α = √[2m(V0−E)]/ħ, leading to the decaying solution ψ ∝ e^{−αx}, so probability density decays exponentially. ## Answer #### Energy levels and wave functions for n = 1, 2, 3 For an infinite well of width $L$: $E_n = \dfrac{n^2h^2}{8mL^2}$ and $\psi_n = \sqrt{\dfrac{2}{L}}\sin\dfrac{n\pi x}{L}$. So $E_2 = 4E_1$, $E_3 = 9E_1$, and $\psi_n$ has $n$ half-wavelengths with $(n-1)$ nodes inside the well. ```text V=inf V=inf n=3 | **** **** | | ** * * **| E3=9E1 |--------------------------| | ** ** | | *** | | | n=2 | ****** | | ** ** | E2=4E1 |--------------------------| | ** ** | | ****** | | | n=1 | *********** | | ***** ***** | E1 |--------------------------| | | 0 L ``` The levels get wider apart as $n$ increases ($\Delta E \propto 2n+1$). $\psi$ is zero at both walls because the electron cannot enter a region of infinite potential. #### Exponential decay inside a finite barrier Let an electron of energy $E$ meet a barrier of height $V_0 > E$ that starts at $x = 0$. ```text V(x) ^ ________________ | V0 | | E -----|- - - - - - | wave | decaying | ~~~~~~ |\__ |_________|___\___________> x Region I 0 Region II ``` **Region II** ($x > 0$, $V = V_0$): the time-independent Schrödinger equation is $$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V_0\psi = E\psi \quad\Rightarrow\quad \frac{d^2\psi}{dx^2} = \frac{2m(V_0 - E)}{\hbar^2}\psi = \alpha^2 \psi $$ where $$ \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar} $$ is real and positive because $V_0 > E$. The general solution is $$ \psi_{II}(x) = C e^{-\alpha x} + D e^{\alpha x} $$ The term $De^{\alpha x}$ grows without limit as $x \to \infty$, so $\psi$ could not be normalised and $|\psi|^2$ would not be a valid probability. Therefore $D = 0$ and $$ \psi_{II}(x) = C e^{-\alpha x}, \qquad |\psi_{II}|^2 = |C|^2 e^{-2\alpha x} $$ The constant $C$ is fixed by matching $\psi$ and $d\psi/dx$ to the oscillating wave $Ae^{ikx} + Be^{-ikx}$ of region I at $x = 0$. **Result:** the wave function inside the barrier is not zero (the electron can penetrate it), but it **decays exponentially** with distance. The penetration depth is $\delta = 1/\alpha$; it is smaller for a heavier particle or a higher barrier ($V_0 - E$ larger). This finite penetration is the basis of tunnelling through thin barriers. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How do E2 and E3 compare with E1 in an infinite well? A. 2E1 and 3E1 B. 4E1 and 9E1 C. E1/2 and E1/3 D. Equal to E1 Answer: B. E_n = n^2 E1. (Key point: Well levels) 2. How many nodes does psi_n have inside the well? A. n B. n + 1 C. n - 1 D. 2n Answer: C. psi_n has n half-wavelengths and n - 1 internal nodes. (Key point: Nodes) 3. Why is D set to 0 in psi_II = C e^(-alpha x) + D e^(alpha x)? A. D e^(alpha x) grows without limit, so psi cannot be normalised B. D is always smaller than C C. Energy equals zero D. The region is classically allowed Answer: A. An unbounded wave function gives an invalid probability, leaving only the decaying term. (Key point: Rejecting growing term) 4. What does psi_II = C e^(-alpha x) show about the electron in the barrier? A. psi is zero inside B. psi is non-zero but decays exponentially C. psi oscillates D. psi grows Answer: B. Electron can penetrate; the penetration depth is 1/alpha. 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