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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 12 Topics Covered: Motion in a central potential, spherical harmonic oscillator, hydrogen atom, orbital electric and magnetic dipole moments 1. [20 pts] A particle of mass M and charge q is constrained to move in a circle of radius r in the x- y plane. (a) If no forces other than the forces of constraint act on the particle, what are the energy levels and corresponding wavefunctions? If the particle is forced to remain in the x-y plane, then it can only have angular momen- tum along the z-axis, so that ~ L = L z ~e z and L 2 = L 2 z . The kinetic energy can be found two ways: Method 1: Using our knowledge of angular momentum. We start by choosing φ as our co- ordinate H = L 2 2 I = L 2 z 2 Mr 2 (1) so that the eigenstates are eigenstates of L z → - i ~ partial φ , from which we see know that the energy levels are then E m = ~ 2 m 2 2 Mr 2 , where m = , ± 1 , ± 2 , ± 3 . . . ... , and the wavefunctions are h φ | m i = 1 √ 2 π e imφ . Method 2: Solution from first principles. We start by choosing s as our coordinate, where s is the distance measured along the circle. The classical Lagrangian is then L = M ˙ s 2 2 (2) the canonical momentum is p s = ∂ s L = M ˙ s . The Hamiltonian is then H = p ˙ s- L = p 2 s 2 M (3) promoting s and p s to operators, we must have [ S, P s ] = i ~ , so that in coordinate representation, we can take S → s , and P s → - i ~ ∂ s , which gives H =- ~ 2 2 M ∂ 2 s (4) the energy eigenvalue equation is then- ~ 2 2 M ∂ 2 s ψ ( s ) = Eψ ( s ) (5) or equivalently ∂ 2 s ψ ( s ) =- 2 ME ~ 2 ψ ( s ) (6) 1 This has solutions of the form: ψ ( s ) ∝ e ± i √ 2 ME ~ s (7) single-valuedness requires ψ ( s + 2 πr ) = ψ ( s ) (8) which means √ 2 ME ~ 2 πr = 2 πm (9) where m is any integer. This gives E = ~ 2 m 2 2 Mr 2 (10) so that ψ m ( s ) = e ims/r √ 2 πr (11) Both methods agree because s = r φ . (b) A uniform, weak magnetic field of amplitude B is applied along the z-axis. What are the new energy eigenvalues and corresponding wavefunctions? Using the angular momentum method, we now need to add the term- qB 2 M L z to the Hamil- tonian to account for the orbital magnetic dipole moment, which gives H = L 2 z 2 Mr 2- qB 2 M L z (12) so that the eigenstates are still L z eigenstates, ψ m ( φ ) = e imφ √...
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