Mathcad - GHZ-Briefer - Users.csbsju.edu

The Kochen-Specker Theorem Illustrated UsingA Three-Qubit GHZ Spin SystemFrank RiouxEmeritus Professor of ChemistryCSB | SJUIn the '90s N. David Mermin published two articles in the general physics literature (Physics Today, June1990; American Journal of Physics, August 1990) on the Greenberger-Horne-Zeilinger (GHZ) gedankenexperiment (American Journal of Physics, December 1990; Nature, 3 February 2000) involving threespin-1/2 particles that illustrated the clash between local realism and the quantum view of reality for thequantum nonspecialist. The purpose of this tutorial is to use the GHZ example to illustrate theKochen-Specker (KS) theorem by stripping away the use of the three-spin wave function in the analysis ofthe thought experiment.The KS theorem asserts that no noncontextual (NC) hidden variable (HV) model (NCHV) can agree with themeasurement predictions of quantum theory for Hilbert space dimensions greater than 2. The problemdealt with here is, of course, three-dimensional.The three spin-1/2 particles are created in a single event and move apart in the horizontal y-z plane.It will be shown that a consideration of spin measurements (in units of h/4 in the x- and y-directionsreveals the impossibility of assigning values to the spin observables independent of measurement.0 10The x- and y-direction spin operators are the Pauli matrices: x y 1 0ii0 The eigenvalues of the Pauli matrices are +/- 1:eigenvals x 11 eigenvals y 11The following operators represent the measurement protocols for spins 1, 2 and 3. 1 2 3 1 2 3 1 2 3 1 2 3x y y y x y y y x x x xThe tensor matrix product, also known as the Kronecker product, is available in Mathcad. The fouroperators in tensor format are formed as follows. xyy kronecker x kronecker y y yxy kronecker y kronecker x y yyx kronecker y kronecker y x xxx kronecker x kronecker x xThese operators mutually commute, meaning that they can be assigned simultaneous eigenstates withsimultaneous eigenvalues. xyy yxy yxy xyy 0 xyy yyx yyx xyy 0 xyy xxx xxx xyy 0 yxy yyx yyx yxy 0 yxy xxx xxx yxy 0 yyx xxx xxx yyx 0The next step is to compare the matrix for the product of the first three operators ( xyy yxy yyx ) withthat of the fourth ( xxx ).

00 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 xyy yxy yyx 0 0 0 1 0 0 0 0xxx 0 0 0 1 0 0 0 0 0 00 0 0 0 0 0 1 000 0 0 0 0 1 0 0 001100000000000000100000010000001000000100000010000001000000100000010000000This indicates the following relationship between the four operators and leads quickly to a refutation ofthe concept of noncontextual, hidden values for quantum mechanical observables. 1 2 3 1 2 3 1 2 3 1 2 3x y yyx y yyx xxxLocal realism assumes that objects have definite properties independent of measurement. In this exampleit assumes that the x- and y-components of the spin have definite values prior to measurement. Thisposition leads to a contradiction with the above result. There is no way to assign eigenvalues (+/-1) to theoperators that is consistent with the above result.Concentrating on the operator on the left side, we notice that there is a y measurement on the firstspin in the second and third term. If the spin state is well-defined before measurement those results haveto be the same, either both +1 or both -1, so that the product of the two measurements is +1. There is a y measurement on the second spin in terms one and three. By similar arguments those results will lead toa product of +1 also. Finally there is a y measurement on the third spin in terms one and two. By similararguments those results will lead to a product of +1 also. Incorporating these observations into theexpression above leads to the following contradiction. 1 2 3 1 2 3x x x x x xA brute force method can be used to confirm this result by showing that the left and right sides of theequation are not equal for any legitimate set of values for the individual spins. This is shown for severalsuch sets below.x1 1 x2 1 x3 1 y1 1 y2 1 y3 1( x1y2y3) ( y1x2y3)( y1y2x3) 1( x1x2x3) 1x1 1x2 1 x3 1 y1 1 y2 1y3 1( x1y2y3) ( y1x2y3)( y1y2x3) 1( x1x2x3) 1x1 1x2 1x3 1y1 1y2 1y3 1( x1y2y3) ( y1x2y3)( y1y2x3) 1( x1x2x3) 1The Kochen-Specker theorem demonstrates that it is, in general, impossibleto ascribe to an individual quantum system a definite value for each of a setof observables not all of which necessarily commute. N. David Mermin, PhysicalReview Letters, 65, 3373 (1990).