FOUNDATIONS OF QUANTUM MECHANICS

A. 3. FORMULATION OF THE PROBLEM ON THE SURFACE OF A SPHERE 193
For every point s 1 on this curve north of p we have θ s1 < θ 0 , which means that according
to (A. 24) it holds that µ(s 1 ) µ(s) for every point s ∈ B(θ 0 ). Consequently, it also holds that
µ(θ s1 ) M (θ 0 ). (A. 31)
Likewise, for all points s 2 of C south of B(θ 0 ) we see that
µ(θ s2 ) m(θ 0 ). (A. 32)
This reasoning holds no matter how close to B(θ 0 ) the points s 1 and s 2 are chosen.
Because of (A. 29) we conclude that the value of µ, when traveling from north to south along
the curve C, makes a discontinuous jump of at least
M (θ 0 ) − m(θ 0 ) = ε (A. 33)
to a lower value when passing B (θ 0 ). This conclusion applies to every continuous, strictly in -
or decreasing curve intersecting B (θ 0 ), which means we can also choose the curve C to be a
meridian,
C = {s ∈ S 2 | ϕ s = ϕ 0 }, (A. 34)
which is a great circle through the north pole having its axis t at the equator, see figure A. 8.
p 0
s ⊥ B(θ 0 )
q
θ q
s
p
C
t
Figure A. 8: Great circle C, coordinate system (p, q, t), and rotating pair (s, s ⊥ )
Let q ∈ C be orthogonal to the point of intersection p of C and B (θ 0 ), such that t, p and q
are mutually orthogonal. Choose an orthogonal pair (s, s ⊥ ) ∈ C to be a rigid coordinate system.
Rotating this system around axis t, we move s from north to south through point p, whereby,
according to (A. 33), the value of µ jumps discontinuously with at least ε while crossing over the
latitude of B(θ 0 ). The pair s and s ⊥ forming a rigid system, we know that
µ(s) + µ(s ⊥ ) + µ(t) = 1, (A. 35)
where µ(t) = 0 because the axis t is on the equator.