QuantumPhysics.dvi

solenoid, the gauge potentialAmust satisfy∇×A= 0. Nonetheless, the gauge potential

cannot vanish outside, because we have in view of Stokes’ theorem,

∮

C

dl·A=

∫

d^2 s·B= ΦB (7.36)

where the surface integral

∫

d^2 sis over the domain enclosed by the curveC, which includes

the inside of the solenoid (see figure 7).

Figure 7: Aharonov-Bohm scattering set-up

A simple solution for the gauge potential is given in cylindrical coordinatesr,θ,zby

A(r) =

nθΦB

2 πr

(7.37)

wherenθis the unit vector along the direction of varyingθ, andris the distance from the

center of the solenoid. The wave function for the system is now determined by the following

Schr ̈odinger equation,

1

2 m

(

−i ̄h∇−eA(r)

) 2

ψ(r) +V(r)ψ(r) =Eψ(r) (7.38)

To solve this equation, we notice that, outside of the solenoid,Ais actually a gradient,

A(r) =∇

(

θΦB

2 π

)

(7.39)

which in particular explains whyB= 0 there. The transformation function is not, however,

a single-valued function, which is whyAis not quite an honest gauge transformation of