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Ch 40: One-Dimensional Quantum Mechanics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 40, Problema 26

CALC Suppose that ψ1(x) and ψ2(x) are both solutions to the Schrödinger equation for the same potential energy U(x). Prove that the superposition ψ(x)=Aψ1(x) + Bψ2(x) is also a solution to the Schrödinger equation.

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Start with the time-independent Schrödinger equation, which is written as: -\(\frac{\hbar^2}{2m}\)\(\frac{d^2\psi(x)}{dx^2}\) + U(x)\(\psi\)(x) = E\(\psi\)(x). Here, \(\psi\)(x) is the wavefunction, U(x) is the potential energy, E is the energy, and \(\hbar\) is the reduced Planck's constant.
Substitute the superposition wavefunction \(\psi\)(x) = A\(\psi\)_1(x) + B\(\psi\)_2(x) into the Schrödinger equation. This gives: -\(\frac{\hbar^2}{2m}\)\(\frac{d^2}{dx^2}\)[A\(\psi\)_1(x) + B\(\psi\)_2(x)] + U(x)[A\(\psi\)_1(x) + B\(\psi\)_2(x)] = E[A\(\psi\)_1(x) + B\(\psi\)_2(x)].
Use the linearity of differentiation to separate the terms: -\(\frac{\hbar^2}{2m}\)[A\(\frac{d^2\psi_1(x)}{dx^2}\) + B\(\frac{d^2\psi_2(x)}{dx^2}\)] + U(x)[A\(\psi\)_1(x) + B\(\psi\)_2(x)] = E[A\(\psi\)_1(x) + B\(\psi\)_2(x)].
Since \(\psi\)_1(x) and \(\psi\)_2(x) are both solutions to the Schrödinger equation, they individually satisfy: -\(\frac{\hbar^2}{2m}\)\(\frac{d^2\psi_1(x)}{dx^2}\) + U(x)\(\psi\)_1(x) = E\(\psi\)_1(x) and -\(\frac{\hbar^2}{2m}\)\(\frac{d^2\psi_2(x)}{dx^2}\) + U(x)\(\psi\)_2(x) = E\(\psi\)_2(x). Substitute these into the equation for the superposition.
After substitution, the terms combine to show that the superposition \(\psi\)(x) = A\(\psi\)_1(x) + B\(\psi\)_2(x) also satisfies the Schrödinger equation. This proves that the superposition of solutions is itself a solution, as the Schrödinger equation is linear.

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