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Peeter Joot [email protected] Constant magnetic solenoid field In [2] the following vector potential A= Bρ2a φ̂, 2ρ (1.1) is introduced in a discussion on the Aharonov-Bohm effect, for configurations where the interior field of a solenoid is either a constant B or zero. I wasn’t able to make sense of this since the field I was calculating was zero for all ρ 6= 0 B= ∇×A Bρ2a φ̂ φ̂ = ρ̂∂ρ + ẑ∂z + ∂φ × ρ 2ρ φ̂ Bρ2a = ρ̂∂ρ + ∂φ × φ̂ ρ 2ρ Bρ2a 1 Bρ2a φ̂ = ρ̂ × φ̂∂ρ + × ∂φ φ̂ 2 ρ 2ρ ρ Bρ2 = 2a −ẑ + φ̂ × ∂φ φ̂ . 2ρ (1.2) Note that the ρ partial requires that ρ 6= 0. To expand the cross product in the second term let j = e1 e2 , and expand using a Geometric Algebra representation of the unit vector φ̂ × ∂φ φ̂ = e2 e jφ × e2 e1 e2 e jφ D E = −e1 e2 e3 e2 e jφ (−e1 )e jφ 2 = e1 e2 e3 e2 e1 = e3 = ẑ. (1.3) So, provided ρ 6= 0, B = 0. The errata [1] provides the clarification, showing that a ρ > ρ a constraint is required for this potential to produce the desired results. Continuity at ρ = ρ a means that in the interior (or at least on the boundary) we must have one of 1 A= Bρ a φ̂, 2 (1.4) A= Bρ φ̂. 2 (1.5) or The first doesn’t work, but the second does B=∇ ×A φ̂ Bρ = ρ̂∂ρ + ẑ∂z + ∂φ × φ̂ ρ 2 B Bρ φ̂ = ρ̂ × φ̂ + × ∂φ φ̂ 2 2 ρ = Bẑ. (1.6) So the vector potential that we want for a constant Bẑ field in the interior ρ < ρ a of a cylindrical space, we need ( 2 Bρ a if ρ ≥ ρ a 2ρ φ̂ A= (1.7) Bρ φ̂ if ρ ≤ ρ . a 2 This potential is graphed in fig. 1.1. Figure 1.1: Vector potential for constant field in cylindrical region. 2 Bibliography [1] Jun John Sakurai and Jim J Napolitano. Errata: Typographical Errors, Mistakes, and Comments, Modern Quantum Mechanics, 2nd Edition, 2013. URL http://www.rpi.edu/dept/phys/Courses/ PHYS6520/Spring2015/ErrataMQM.pdf. 1 [2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014. 1 3