FOUNDATIONS OF QUANTUM MECHANICS

V. 3 KOCHEN AND SPECKER’S THEOREM
V. 3. KOCHEN AND SPECKER’S THEOREM 115
As we already proved in section II. 4, p. 28, in quantum mechanics the next theorem holds: if the
operators A, B, C, . . . commute, there is a maximal operator O of which they are a function,
A = f (O), B = g(O), etc. (V. 20)
A measuring procedure for A, B, C, . . . would be to measure O and apply the function relation to the
result in order to find the values for A, B, C, . . . Kochen and Specker (1967, p. 64) call the quantities
corresponding to A, B, C, . . . commeasurable.
Now it seems reasonable to require, as Von Neumann did, that the HVT also has this structure, i.e.,
for B, C : Λ → R, if B = f (C), it follows that B[λ] = f ( C [λ] ) , or
f (C)[λ] = f ( C [λ] ) . (V. 21)
This function rule, (V. 21), yields the so - called sum rule for commuting operators,
[A, B] = 0 =⇒ (A + B)[λ] = A[λ] + B[λ], (V. 22)
since, with O again the maximal operator of which A and B are a function, A = f (O), B = g(O),
implying
(A + B) = h(O) with h = f + g, (V. 23)
from (V. 21) it then follows in this HVT that
(A + B)[λ] = h(O)[λ] = h ( O[λ] ) = f ( O[λ] ) + g ( O[λ] )
= (f O)[λ] + (g O)[λ] = A[λ] + B[λ]. (V. 24)
EXERCISE 31. Prove, again using (V. 21), the product rule for commuting operators,
[A, B] = 0 =⇒ (A B)[λ] = A[λ] B[λ]. (V. 25)
Now we will see how the requirement, (V. 21), which at first sight is eminently reasonable, nevertheless
renders a HVT of quantum mechanics impossible.
THEOREM :
A HVT satisfying the requirements (i) - (iii), p. 111, and the function rule (V. 21), does
not exist if dim H > 2.