7 General motion
237
Thus, they are degenerate if [H, Q] = 0 This is certainly true here. Problem 7.26 A particle with electric charge q and mass Âľ moves under the influence of a uniform magnetic field B = zˆ B. The Hamiltonian is 1 q 2 H= p− A 2Âľ c Consider the following three distinct choices of vector potential that lead to the given uniform magnetic field: A1 = x B yˆ ,
A2 = −y B xˆ ,
A3 = 12 B (−y xˆ + x yˆ )
Find the energy eigenvalues and show that the energy eigenfunctions corresponding to the different cases are related by Ďˆ2 = e−iq Bx y/ÂŻh c Ďˆ1 ,
Ďˆ3 = e−iq Bx y/2ÂŻh c Ďˆ1
Solution The Hamiltonians in all three cases have the familiar form H=
pz2 + H (⊼) , 2¾
where H1(⊼)
p 2y 1 px2 q 2 1 q 2 (⊼) + = H2 = p y − Bx , px + By + 2Âľ 2Âľ c 2Âľ c 2Âľ 2 2 1 q 1 q H3(⊼) = px + By + p y − Bx 2Âľ 2c 2Âľ 2c
In each case the transverse Hamiltonian can be put into a harmonic-oscillator form. Introducing = px and q = x − cp y /q B in the first case, we have H1(⊼) =
2 ¾ω2 2 + q , 2¾ 2
[q, ] = iÂŻh
where ω = q B/¾c. In an analogous fashion, in the second case, introducing = p y and q = y + cpx /q B, we get H2(⊼) =
2 ¾ω2 2 + q , 2¾ 2
[q , ] = iÂŻh