As a crude model of a nucleon in a light nucleus

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Wave mechanics in one dimension · Quantum dynamics and its formulations

Quantum Mechanics: From the Photon to the Hydrogen Atom, Ch Stationary states and bound states in one dimension

Problem

As a crude model of a nucleon in a light nucleus, a proton is confined to a one-dimensional infinite square well of width a=4.0a = 4.0 fm: V=0V = 0 for 0<x<a0 < x < a, with impenetrable walls at x=0x = 0 and x=ax = a. The normalised stationary states are ψn(x)=2/a sin⁡(nπx/a)\psi_n(x) = \sqrt{2/a}\,\sin(n\pi x/a) and the proton is treated as non-relativistic. ((a)) Estimate the proton's least kinetic energy from the uncertainty principle, taking the position uncertainty as Δx=a/2\Delta x = a/2, the momentum uncertainty at the smallest value allowed by Δx Δp≥ℏ/2\Delta x\,\Delta p \ge \hbar/2, and the kinetic energy as (Δp)2/2mp(\Delta p)^2/2m_p. Give the answer in MeV, and note that it is an order-of-magnitude estimate only. ((b)) Now solve the well exactly: find E1E_1 and E2E_2 in MeV, and the factor by which E1E_1 exceeds the estimate of part (a). ((c)) The proton starts in Ψ(x,0)=12[ψ1(x)+ψ2(x)]\Psi(x,0) = \tfrac{1}{\sqrt2}[\psi_1(x) + \psi_2(x)]. Write Ψ(x,t)\Psi(x,t) and find the period with which ∣Ψ(x,t)∣2|\Psi(x,t)|^2 repeats, in seconds. Take ℏc=197\hbar c = 197 MeV fm, mpc2=938m_pc^2 = 938 MeV, ℏ=1.055×10−34\hbar = 1.055\times10^{-34} J s, h=6.626×10−34h = 6.626\times10^{-34} J s and 11 MeV =1.602×10−13= 1.602\times10^{-13} J.

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