Abstract
The Kochen-Specker theorem proves the inability to assign, simultaneously, noncontextual definite values to all (of a finite set of) quantum mechanical observables in a consistent manner. If one assumes that any definite values behave noncontextually, one can nonetheless only conclude that some observables (in this set) are value indefinite. In this paper, we prove a variant of the Kochen-Specker theorem showing that, under the same assumption of noncontextuality, if a single one-dimensional projection observable is assigned the definite value 1, then no one-dimensional projection observable that is incompatible (i.e., non-commuting) with this one can be assigned consistently a definite value. Unlike standard proofs of the Kochen-Specker theorem, in order to localise and show the extent of value indefiniteness, this result requires a constructive method of reduction between Kochen-Specker sets. If a system is prepared in a pure state ψ, then it is reasonable to assume that any value assignment (i.e., hidden variable model) for this system assigns the value 1 to the observable projecting onto the one-dimensional linear subspace spanned by ψ, and the value 0 to those projecting onto linear subspaces orthogonal to it. Our result can be interpreted, under this assumption, as showing that the outcome of a measurement of any other incompatible one-dimensional projection observable cannot be determined in advance, thus formalising a notion of quantum randomness.
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