{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

851HW12_09Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

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 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 0 (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 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 , 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 ) = ( s ) (5) or equivalently 2 s ψ ( s ) = - 2 ME 2 ψ ( s ) (6) 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This has solutions of the form: ψ ( s ) e ± i 2 ME s (7) single-valuedness requires ψ ( s + 2 πr 0 ) = ψ ( s ) (8) which means 2 ME 2 πr 0 = 2 πm (9) where m is any integer. This gives E = 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 = 2 m
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}