Introduction to Quantum Mechanics · Problem 11.35

Quantum Dynamics — David J. Griffiths

Physics — Quantum Mechanics · expanding-infinite-square-well-exact-solution

Problem

The case of an infinite square well whose right wall expands at a constant velocity (vv) can be solved exactly. A complete set of solutions is Φn(x,t)≡2wsin⁡ ⁣(nπwx)ei(mvx2−2Eniat)/2ℏw,(11.136)\Phi_n(x,t) \equiv \sqrt{\frac{2}{w}}\sin\!\left(\frac{n\pi}{w}x\right)e^{i\left(mvx^2 - 2E^i_n at\right)/2\hbar w}, \tag{11.136} where w(t)≡a+vtw(t) \equiv a + vt is the width of the well and Eni≡n2π2ℏ2/2ma2E^i_n \equiv n^2\pi^2\hbar^2/2ma^2 is the nnth allowed energy of the original well (width aa). The general solution is a linear combination of the Φ\Phi's: Ψ(x,t)=∑n=1∞cnΦn(x,t),(11.137)\Psi(x,t) = \sum_{n=1}^\infty c_n \Phi_n(x,t), \tag{11.137} whose coefficients cnc_n are independent of tt. (a) Check that Equation 11.136 satisfies the time-dependent Schrödinger equation, with the appropriate boundary conditions. (b) Suppose a particle starts out (t=0t = 0) in the ground state of the initial well: Ψ(x,0)=2asin⁡ ⁣(πax).\Psi(x,0) = \sqrt{\frac{2}{a}}\sin\!\left(\frac{\pi}{a}x\right). Show that the expansion coefficients can be written in the form cn=2π∫0πe−iαz2sin⁡(nz)sin⁡(z) dz,(11.138)c_n = \frac{2}{\pi}\int_0^\pi e^{-i\alpha z^2}\sin(nz)\sin(z)\,dz, \tag{11.138} where α≡mva/2π2ℏ\alpha \equiv mva/2\pi^2\hbar is a dimensionless measure of the speed with which the well expands. (Unfortunately, this integral cannot be evaluated in terms of elementary functions.) (c) Suppose we allow the well to expand to twice its original width, so the "external" time is given by w(Te)=2aw(T_e) = 2a. The "internal" time is the period of the time-dependent exponential factor in the (initial) ground state. Determine TeT_e and TiT_i, and show that the adiabatic regime corresponds to α≪1\alpha \ll 1, so that exp⁡(−iαz2)≈1\exp(-i\alpha z^2) \approx 1 over the domain of integration. Use this to determine the expansion coefficients, cnc_n. Construct Ψ(x,t)\Psi(x,t), and confirm that it is consistent with the adiabatic theorem. (d) Show that the phase factor in Ψ(x,t)\Psi(x,t) can be written in the form θ(t)=−1ℏ∫0tE1(t′) dt′,(11.139)\theta(t) = -\frac{1}{\hbar}\int_0^t E_1(t')\,dt', \tag{11.139} where En(t)≡n2π2ℏ2/2mw2E_n(t) \equiv n^2\pi^2\hbar^2/2mw^2 is the nnth instantaneous eigenvalue, at time tt. Comment on this result. What is the geometric phase? If the well now contracts back to its original size, what is Berry's phase for the cycle?

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