QuantumPhysics.dvi

can never penetrate inside. We assume vanishing electric field throughout, so that Φ = 0. On the
outside of the cylinder, the magnetic field vanishes identically. On the inside of the cylinder, there is
a time-independent magnetic fieldB, which points in thez-direction, and produces a magnetic flux
ΦB. Such a magnetic field may be produced by an infinite solenoid. Because of Stokes’ theorem,
the gauge potential outside the cylinder cannot vanish, because it must reproduce the total flux
inside the cylinder,
∮

C

A·dx= ΦB (15.19)

whereCis any curve encircling the impenetrable cylinder once.

Next, we compare the probability amplitude for the charged particle to travel from pointA
to pointBof Fig 9, either above or below the cylinder. We denote the paths above and below
respectively byC+andC−. The amplitudes are given by

〈B|U(tb−ta)|A〉C±=

∫

Dxexp

{

i

̄h

∫tb

ta

dtLC±

}

(15.20)

The actions may be compared by assuming that the pathC−is exactly the mirror image of path
C+, so that the contribution of the kinetic terms in both are identical. The only difference is then
due to the magnetic term, so that
∫tb

ta

dtLC+−

∫tb

ta

dtLC− = e

∫tb

ta

dtA(x)·

dx

dt

∣∣

∣∣

C+−C−

= e

∫tb

ta

A(x)·dx

∣∣

∣∣

C+−C−

= e

∫

C

A(x)·dx=eΦB (15.21)

where we have used the fact thatC+−C−=C. As a result,

〈B|U(tb−ta)|A〉C+=〈B|U(tb−ta)|A〉C−exp

{

ieΦB

̄h

}

(15.22)

For an arbitrary magnetic flux, ΦB, and electric chargee, there will be a non-trivial phase differ-
ence between the probability amplitudes along pathsC+andC−, and as a result, there will be
interference when these paths merge at pointB. On the other hand, the solenoid will be quantum
mechanically unobservable provided the magnetic flux isquantized, and takes on integer multiple
values of the basic magnetic flux quantum (for the electron chargee) of

Φ(0)B =

2 π ̄h

e

∼ 4. 135 × 10 −^7 Gauss×cm^2 = 4. 135 × 10 −^15 Tm^2 (15.23)

Note that the interference is purely quantum mechanical, ithas no classical counterpart. Also, we
have never needed the explicit form of the vector potential,only its role in generating a non-zero
flux. It is possible to solve this problem using the Schr ̈odinger equation, but then a specificAmust
be chosen.