137A Homework 4 SolutionsGSI: Shannon McCurdyOctober 2, 2006Problem 2.14 (Griffiths)A particle is in the ground state of the harmonic oscillator with classical frequency ω when suddenly the spring con-stant quadruples, so ω0=2ω, without initially changing the wave fun cti on (of course, Ψ will now evolve differently,because the Hamiltonian has changed).The point is to expand our initial wave function, Ψ(x,0), which happens to be a state of definite energy under theold Hamiltonian (ˆH = ~ω(a+a−+12)), in terms of a linear combination of the new Hamiltonian’s (ˆH0=~ω0(a0+a0−+12)) energy wave functions ψ0n(x) [Griffiths, 2.16]:Ψ(x,0 ) =Xncnψ0n(x)ψ0n(x) =(a0+)npn!ψ00(x)where ψ00(x) is given by:ψ00(x) =(m(2ω)π~)14e−m(2ω)x22~and the wave functions ψ0n(x) are states with definite energy E0n:E0n=(n +12)~ω0=2(n +12)~ω =~ω,3~ω,5~ω...in terms of the old classical frequency, ω. The coefficients cnmeasure the overlap between Ψ(x,0) and ψ0n(x)∗. Wecalculate cnusing Fourier’s Trick [Griffiths, 2.34], and find:cn=Z∞−∞ψ0n(x)∗Ψ(x,0 ) dxThe probability Pnof measuring energy E0nis given by:Pn=|cn|2Part (a)What is the probability that a measurement of the energy will still return the value~ω2?Since E =~ω2is not one of the allowed energies, the probability of measuring~ω2is zero.1Part (b)What is the probability of getting ~ω?Since we are interested in the probability of measuring the new ground state energy, E00= ~ω, we only need tocalculate:c0=Z∞−∞ψ00(x)∗Ψ(x,0 ) dxc0=(2)14rmωπ~Z∞−∞e−3mωx22~dx =(2)14rmωπ~2pπ(12s2~3mω)