A particle's state saturates the position-momentum uncertainty relation

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · The formalism of quantum mechanics · The formalism of quantum mechanics

The Structure of Quantum Mechanics, Ch Compatible observables and uncertainty as a theorem

Problem

A particle's state ∣ψ⟩|\psi\rangle saturates the position-momentum uncertainty relation, meaning it satisfies (Δx^+iλΔp^)∣ψ⟩=0(\Delta\hat x+i\lambda\Delta\hat p)|\psi\rangle=0 for some real constant λ>0\lambda>0, where ⟨x^⟩=x0\langle\hat x\rangle=x_0, ⟨p^⟩=p0\langle\hat p\rangle=p_0 are fixed but unspecified numbers, and the position spread is Δx=0.50 nm\Delta x=0.50\ \mathrm{nm}. ((a)) Working in the position representation with p^=−iℏ d/dx\hat p=-i\hbar\,d/dx, derive the first-order differential equation obeyed by ψ(x)=⟨x∣ψ⟩\psi(x)=\langle x|\psi\rangle and solve it to find ψ(x)\psi(x) up to normalisation, showing that it is Gaussian. ((b)) Matching your solution to the standard form ψ(x)∝exp⁡ ⁣[−(x−x0)24σ2]exp⁡ ⁣(ip0xℏ)\psi(x)\propto\exp\!\left[-\dfrac{(x-x_0)^2}{4\sigma^2}\right]\exp\!\left(\dfrac{ip_0x}{\hbar}\right) with σ=Δx\sigma=\Delta x, express λ\lambda in terms of Δx\Delta x and ℏ\hbar, and evaluate it numerically. ((c)) Take the norm of both sides of Δx^∣ψ⟩=−iλ Δp^∣ψ⟩\Delta\hat x|\psi\rangle=-i\lambda\,\Delta\hat p|\psi\rangle to show that Δx=λ Δp\Delta x=\lambda\,\Delta p. Hence find Δp\Delta p from your λ\lambda, without any Fourier transform, and confirm that Δx Δp=ℏ/2\Delta x\,\Delta p=\hbar/2.

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