FOUNDATIONS OF QUANTUM MECHANICS

A. 2. CONVERSION TO A 3 - DIMENSIONAL REAL PROBLEM 185
A. 2. 1 STEP 1
THEOREM 1:
If Gleason’s theorem for pure states is true for any 3 - dimensional real Hilbert space, it
is also true for any complex Hilbert space with dim H > 2.
We will prove theorem 1 using a proof by contradiction.
Proof
Let H be a complex Hilbert space with dim H > 3 for which Gleason’s theorem is not true. Since
all higher - dimensional projectors can be decomposed to 1 - dimensional projectors, it suffices to
prove this theorem for 1 - dimensional projectors.
Assume a measure µ on H exists, which is concentrated on P 0 ∈ P (H) such that µ(P 0 ) = 1,
but differs from the measure µ 0 defined by (A. 6) in the sense that there is some 1 - dimensional
projector P 1 for which the theorem does not hold,
µ 0 (P 1 ) := Tr P 0 P 1 ≠ µ(P 1 ). (A. 8)
First we will show that, if these measures differ on a higher - dimensional Hilbert space, they also
differ on a 3 - dimensional Hilbert space.
Using the projectors P 0 and P 1 , we can construct a set of three orthogonal 1 - dimensional projectors
P 0 , ˜P 1 , P 2 in the following way. With P 0 = |e 0 ⟩ ⟨e 0 | and P 1 = |e 1 ⟩ ⟨e 1 |, construct a
unit vector |ẽ 1 ⟩ in the plane spanned by |e 0 ⟩ and |e 1 ⟩ which is perpendicular to |e 0 ⟩, i.e.
|ẽ 1 ⟩ ∝ (11 − P 0 ) |e 1 ⟩, (A. 9)
as can be seen in figure A. 1. Then the projector ˜P 1 := |ẽ 1 ⟩ ⟨ẽ 1 | is perpendicular to P 0 . 1
ẽ 1
e 1
e 2 e 0
Figure A. 1: Construction of a 3 - dimensional subspace E
Let P 2 be a 1 - dimensional projector which is perpendicular to both P 0 and ˜P 1 , it is always possible
to choose such a projector because dim H > 3. With P 2 = |e 2 ⟩ ⟨e 2 |, the three orthonormal
vectors |e 0 ⟩,|ẽ 1 ⟩ and |e 2 ⟩ together span a 3 - dimensional Hilbert space, which is a subspace of H.
We will call this space E, and, by construction, P 0 , P 1 , ˜P 1 , P 2 ∈ P (E).
1 To be exact
˜P 1 = (1 − Tr P 0 P 1 ) −1 (P 1 + P 0 P 1 P 0 − P 1 P 0 − P 0 P 1 ).