Introduction to Quantum Mechanics · Problem 11.34

Quantum Dynamics — David J. Griffiths

Physics — Quantum Mechanics · adiabatic-theorem-growing-barrier

Problem

A particle starts out in the ground state of the infinite square well (on the interval 0≤x≤a0 \leq x \leq a). Now a wall is slowly erected, slightly off-center: V(x)=f(t) δ ⁣(x−a2−ϵ),V(x) = f(t)\,\delta\!\left(x - \frac{a}{2} - \epsilon\right), where f(t)f(t) rises gradually from 0 to ∞\infty. According to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian. (a) Find (and sketch) the ground state at t→∞t \to \infty. *Hint:* This should be the ground state of the infinite square well with an impenetrable barrier at a/2+ϵa/2 + \epsilon. Note that the particle is confined to the (slightly) larger left "half" of the well. (b) Find the (transcendental) equation for the ground state energy at time tt. *Answer:* zsin⁡z=T[cos⁡z−cos⁡(zδ)],z\sin z = T[\cos z - \cos(z\delta)], where z≡kaz \equiv ka, T≡maf(t)/ℏ2T \equiv maf(t)/\hbar^2, δ≡2ϵ/a\delta \equiv 2\epsilon/a, and k≡2mE/ℏk \equiv \sqrt{2mE}/\hbar. (c) Setting δ=0\delta = 0, solve graphically for zz, and show that the smallest zz goes from π\pi to 2π2\pi as TT goes from 0 to ∞\infty. Explain this result. (d) Now set δ=0.01\delta = 0.01 and solve numerically for zz, using T=0,1,5,20,100T = 0, 1, 5, 20, 100, and 10001000. (e) Find the probability PrP_r that the particle is in the right "half" of the well, as a function of zz and δ\delta. *Answer:* Pr=1/[1+(I+/I−)]P_r = 1/\left[1 + (I_+/I_-)\right], where I±≡[1±δ−(1/z)sin⁡(z(1±δ))]sin⁡2 ⁣[z(1∓δ)/2]I_\pm \equiv \left[1 \pm \delta - (1/z)\sin(z(1\pm\delta))\right]\sin^2\!\left[z(1\mp\delta)/2\right]. Evaluate this expression numerically for the TT's and δ\delta in part (d). Comment on your results. (e) Plot the ground state wave function for those same values of TT and δ\delta. Note how it gets squeezed into the left half of the well, as the barrier grows.

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