Measurements and Contextuality

Start

An observable is a set of orthogonal eigenvectors, one for for each physical value of the observable. A measurement of the observable causes a quantum state (shown in green) to project onto the eigenstate corresponding to the measurement outcome with a probability that is the square of the projection.
Different observables are defined by different sets of orthogonal eigenvectors. This is all we need to show contextuality in quantum mechanics.

Next

Pr(A1, A2 both true) = 0
Pr(A2, A3 both false) = 0
Pr(A3, A4 both true) = 0
Pr(A4, A5 both false) = 0
Pr(A1 true & A5 false) = 0

Since the eigenvectors with eigenvalue “true” of A1 and A2 are orthogonal, they cannot both be true, and the answer to the question “Are A1 and A2 both true” is therefore always false.

Pr(A1, A2 both true) = 0
Pr(A2, A3 both false) = 0
Pr(A3, A4 both true) = 0
Pr(A4, A5 both false) = 0
Pr(A1 true & A5 false) = 0

The eigenvectors with eigenvalue “true” for A2 and A3 are orthogonal, and the vector for the statement “A2 and A3 are both false” (shown in red) must be orthogonal to both these eigenvectors.
If the quantum state (shown in green) is perpendicular to this vector, the answer to the question “are A2 and A3 both false?” will be “No."

Perpendicular to

Previous

Pr(A1, A2 both true) = 0
Pr(A2, A3 both false) = 0
Pr(A3, A4 both true) = 0
Pr(A4, A5 both false) = 0
Pr(A1 true & A5 false) = 0

Since the eigenvectors with eigenvalue “true” of A3 and A4 are orthogonal, they cannot both be true, and the answer to the question “Are A3 and A4 both true” is always false.

The eigenvectors with eigenvalue “true” for A4 and A5 are orthogonal, and the vector for the statement “A4 and A5 are both false” (shown in red) must be orthogonal to both these eigenvectors.

The eigenvector for “A1 is true” lies in the plane for “A5 is false” (perpendicular to A5) and can give the answer “Yes” to the question “is A1 true and A5 false?”, depending on the quantum state. Since A1 is not orthogonal to the quantum state (shown in green), there will be a small probability that A1 is true and A5 is false, contrary to the case where each statement A1—A5 has a definite truth value.

Shift the green arrow parallel to A1 to see that there is a small overlap with it. This is the quantum effect that sometimes gives “Yes” to the question “Is A1 true and A5 false?”.

Intro

Projection of A1 onto ψ

Pr(A1, A2 both true) = 0
Pr(A2, A3 both false) = 0
Pr(A3, A4 both true) = 0
Pr(A4, A5 both false) = 0
Pr(A1 true & A5 false) = 0not true in quantum mechanics!