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851HW12_09Solutions

851HW12_09Solutions - PHYS851 Quantum Mechanics I Fall 2009...

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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 0 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 vector L = L z vectore 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 0 (1) so that the eigenstates are eigenstates of L z → − i planckover2pi1 partial φ , from which we see know that the energy levels are then E m = planckover2pi1 2 m 2 2 Mr 2 0 , where m = 0 , ± 1 , ± 2 , ± 3 ...... , and the wavefunctions are ( φ | m ) = 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 planckover2pi1 , so that in coordinate representation, we can take S s , and P s → − i planckover2pi1 s , which gives H = planckover2pi1 2 2 M 2 s (4) the energy eigenvalue equation is then planckover2pi1 2 2 M 2 s ψ ( s ) = ( s ) (5) or equivalently 2 s ψ ( s ) = 2 ME planckover2pi1 2 ψ ( s ) (6) 1

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This has solutions of the form: ψ ( s ) e ± i 2 ME planckover2pi1 s (7) single-valuedness requires ψ ( s + 2 πr 0 ) = ψ ( s ) (8) which means 2 ME planckover2pi1 2 πr 0 = 2 πm (9) where m is any integer. This gives E = planckover2pi1 2 m 2 2 Mr 2 0 (10) so that ψ m ( s ) = e ims/r 0 2 πr 0 (11) Both methods agree because s = r 0 φ . (b) A uniform, weak magnetic field of amplitude B 0 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 0 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 0 qB 0 2 M L z (12) so that the eigenstates are still L z eigenstates, ψ m ( φ ) = e imφ 2 π , where m = 0 , ± 1 , ± 2 ,... , but the degeneracy is lifted so that E m = planckover2pi1 2 m 2 2 Mr 2 0 qB 0 planckover2pi1 2 M m (13)
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