851HW12_09Solutions

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

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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 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 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 planckover2pi1 partial , from which we see know that the energy levels are then E m = planckover2pi1 2 m 2 2 Mr 2 , where m = , 1 , 2 , 3 ...... , and the wavefunctions are ( | m ) = 1 2 e im . 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 ( s ) = E ( s ) (5) or equivalently 2 s ( s ) = 2 ME planckover2pi1 2 ( s ) (6) 1 This has solutions of the form: ( s ) e i 2 ME planckover2pi1 s (7) single-valuedness requires ( s + 2 r ) = ( s ) (8) which means 2 ME planckover2pi1 2 r = 2 m (9) where m is any integer. This gives E = planckover2pi1 2 m 2 2 Mr 2 (10) so that m ( s ) = e ims/r 2 r (11) Both methods agree because s = r . (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, m ( ) = e im 2 , where m = 0 , 1 , 2 ,... , but the...
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851HW12_09Solutions - PHYS851 Quantum Mechanics I, Fall...

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