Someone argues: "Consider for

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Fourier analysis and integral transforms · Function spaces and operator theory

The Structure of Quantum Mechanics, Ch Observables, spectra and the spectral theorem

Problem

Someone argues: "Consider ψp(x)=A eipx/ℏ\psi_p(x)=A\,e^{ipx/\hbar} for −∞<x<∞-\infty<x<\infty. Since ∣ψp(x)∣2=∣A∣2|\psi_p(x)|^2=|A|^2 is a constant, use the identity ∫−∞∞ei(p−p′)x/ℏdx=2πℏ δ(p−p′)\int_{-\infty}^{\infty}e^{i(p-p')x/\hbar}dx = 2\pi\hbar\,\delta(p-p'). Setting p′=pp'=p gives ∫−∞∞dx=2πℏ δ(0)\int_{-\infty}^{\infty}dx = 2\pi\hbar\,\delta(0). Choosing A=1/2πℏ δ(0)A=1/\sqrt{2\pi\hbar\,\delta(0)} then makes ∫−∞∞∣ψp(x)∣2dx=∣A∣2×2πℏ δ(0)=1\int_{-\infty}^{\infty}|\psi_p(x)|^2dx=|A|^2\times2\pi\hbar\,\delta(0)=1, so ψp\psi_p is a normalised state with a definite momentum pp." ((a)) Find the error in this argument. ((b)) State what ∫−∞∞∣ψp(x)∣2dx\int_{-\infty}^{\infty}|\psi_p(x)|^2dx actually equals for any nonzero choice of AA, and say what this means physically. ((c)) State the correct relation that ψp\psi_p does satisfy, and what it says instead.

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