8.41: Entanglement Reveals a Conflict Between Local Realism and Quantum Theory

A tensor algebra approach is used to demonstrate the challenge to the local realistic position of reality that quantum mechanical entanglement creates. The example is drawn from Chapter 3 of David Z Albert's text, Quantum Mechanics and Experience.

A quon (any entity that exhibits both wave and particle aspects in the peculiar quantum manner - Nick Herbert, Quantum Reality, page 64) has a variety of properties each of which can take on two values. For example, it has the property of hardness and can be either hard or soft. It also has the property of color and can be either black or white.

In the matrix formulation of quantum mechanics these states are represented by the following vectors.

\[ \begin{matrix} \text{Hard} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \text{Soft} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \text{Black} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \\ \frac{1}{ \sqrt{2}} \end{pmatrix} & \text{White} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \\ \frac{-1}{ \sqrt{2}} \end{pmatrix} \end{matrix} \nonumber \]

Hard and Soft represent an orthonormal basis in the two-dimensional Hardness vector space.

\[ \begin{matrix} \text{Hard}^T \text{Hard} = 1 & \text{Soft}^T \text{Soft} = 1 & \text{Hard}^T \text{Hard} = 0 \end{matrix} \nonumber \]

Likewise Black and White are an orthonormal in the two-dimensional Color vector space.

\[ \begin{matrix} \text{Black}^T \text{Black} = 1 & \text{White}^T \text{White} = 1 & \text{Black}^T \text{White} = 0 \end{matrix} \nonumber \]

The relationship between the two bases is reflected in the following projection calculations.

\[ \begin{matrix} \text{Hard}^T \text{Black} = 0.707 & \text{Hard}^T \text{White} = 0.707 & \text{Soft}^T \text{Black} = 0.707 & \text{Soft}^T \text{White} = -0.707 & \frac{1}{ \sqrt{2}} = 0.707 \end{matrix} \nonumber \]

Clearly Black and White can be written as superpositions of Hard and Soft, and vice versa.

\[ \begin{matrix} \frac{1}{ \sqrt{2}} \text{(Hard + Soft)} = \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} & \frac{1}{ \sqrt{2}} \text{(Hard - Soft)} = \begin{pmatrix} 0.707 \\ -0.707 \end{pmatrix} \\ \frac{1}{ \sqrt{2}} \text{(Black + White)} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \text{(Black - White)} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber \]

Hard, Soft, Black and White are measurable properties and the vectors representing them are eigenstates of the Hardness and Color operators with eigenvalues +/- 1.

Operators

\[ \begin{matrix} \text{Hardness} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & \text{Color} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{matrix} \nonumber \]

\[ \begin{matrix} \text{Eigenvalue +1} & \text{Eigenvalue -1} \\ \text{Hardness Hard} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \text{Hardness Soft} = \begin{pmatrix} 0 \\ -1 \end{pmatrix} \\ \text{Color Black} = \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} & \text{Color White} = \begin{pmatrix} -0.707 \\ 0.707 \end{pmatrix} \end{matrix} \nonumber \]

Hard and Soft are not eigenfunctions of the Color operator, and Black and White are not eigenfunctions of the Hardness operator.

\[ \begin{matrix} \text{Hardness Black} = \begin{pmatrix} 0.707 \\ -0.707 \end{pmatrix} & \text{Hardness White} = \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} \\ \text{Color Hard} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \text{Color Soft} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{matrix} \nonumber \]

As the Hardness-Color commutator shows, the Hardness and Color operators do not commute. They represent incompatible observables; observables that cannot simultaneously have well-defined values.

\[ \text{Hardness Color} - \text{Color Hardness} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \nonumber \]

We now proceed with an analysis of the implications of the following two-quon entangled state, expressed in tensor format. A pair of quons is prepared in the following "singlet" state; one is hard and one is soft. (The Appendix shows how to set this state up using Mathcad.)

\[ | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \text{Hard} \rangle_1 | \text{Soft} \rangle_2 - | \text{Soft} \rangle_1 | \text{Hard} \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right] \nonumber \]

\[ \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} \nonumber \]

Given |Ψ> the expectation value for measuring Hardness on the first quon is 0. The same is true for the second quon. In other words, it is equally likely for either quon to be Hard or Soft. (Kronecker is Mathcad's command for tensor multiplication of matrices. See the Appendix for more detail.)

\[ \begin{matrix} \Psi^T \text{kronecker(Hardness, I)} \Psi = 0 & \Psi^T \text{kronecker(I, Hardness)} \Psi = 0 \end{matrix} \nonumber \]

However, if one quon is found to be Hard by measurement, the second will be measured Soft, and vice versa. In other words, there is perfect anti-correlation between the joint measurement of this property on the two quons.

\[ \Psi^T \text{kronecker(Hardness, Hardness)} \Psi = -1 \nonumber \]

Given |Ψ> the expectation value for measuring Color on the first quon is 0. The same is true for the second quon. In other words, it is equally likely for either quon to be Black or White.

\[ \begin{matrix} \Psi^T \text{kronecker(Color, I)} \Psi = 0 & \Psi^T \text{kronecker(I, Color)} \Psi = 0 \end{matrix} \nonumber \]

However, if one quon is found to be Black by measurement, the second will be measured White and vice versa. In other words, there is perfect anti-correlation between the joint measurement of this property on the two quons.

\[ \Psi^T \text{kronecker(Color, Color)} \Psi = -1 \nonumber \]

Furthermore, as the following calculations show, there is no correlation between the measurement outcomes on Color and Hardness.

\[ \begin{matrix} \Psi^T \text{kronecker(Hardness, Color)} \Psi = 0 & \Psi^T \text{kronecker(Color, Hardness)} \Psi = 0 \end{matrix} \nonumber \]

As the foundation for their belief in local realism, Einstein, Podolsky and Rosen (EPR) defined the concept of element of reality in their famous 1935 Physical Review paper,

"If, without in any way disturbing a system, we can predict with certainty (i.e. with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to this physical quantity."

It would seem from the above results, namely these,

\[ \begin{matrix} \Psi^T \text{kronecker(Color, Color)} \Psi = -1 & \Psi^T \text{kronecker(Hardness, Hardness)} \Psi = -1 \end{matrix} \nonumber \]

that according to EPR both hardness and color are elements of reality. If the hardness of quon 1 is measured and found to be soft, we know without measurement (given the reliability of quantum mechanical predictions) that quon 2 is hard. Likewise, if the color of quon 2 is measured and found to be white, we know without measurement that quon 1 is black. On the basis of these calculations, the realist constructs the following table which assigns well-defined hardness and color states to both quons and is consistent with all the quantum calculations.

\[ \begin{pmatrix} \text{Quon 1} & \text{Quon 2} & \text{HardnessHardness} & \text{ColorColor} & \text{HardnessColor} \\ \text{HB} & \text{SW} & -1 & -1 & -1 \\ \text{HW} & \text{SB} & -1 & -1 & 1 \\ \text{SB} & \text{HW} & -1 & -1 & 1 \\ \text{SW} & \text{HB} & -1 & -1 & -1 \\ \text{Realist} & \text{AverageValue} & -1 & -1 &0 \\ \text{Quantum} & \text{AverageValue} & -1 & -1 & 0 \end{pmatrix} \nonumber \]

The problem with this interpretation is that it has previously been shown that the Hardness and Color operators do not commute, meaning that they represent incompatible observables. Incompatible observables cannot be known (determined) simultaneously. A contradiction between the EPR reality criterion and quantum mechanics has thus been shown to exist.

Appendix

Tensor multiplication is used to construct the initial states using Mathcad commands submatrix, kronecker, and augment.

\[ \Psi = \frac{1}{ \sqrt{2}} \begin{bmatrix} \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right],~ \text{augment} \left[ \begin{pmatrix} 0 \\ 1 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right] \right], ~1,~4,~1,~1 \right] ~... \\ +- \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \begin{pmatrix} 0 \\ 1 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right],~ \text{augment} \left[ \begin{pmatrix} 1 \\ 0 \end{pmatrix},~ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right] \right], ~1,~4,~1,~1 \right] \end{bmatrix} \nonumber \]

Kronecker is the Mathcad command that carries out the tensor multiplication of matrices. For example, consider the tensor multiplication of the Hardness and Color matrix operators.

\[ Hardness \otimes Color = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 0 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ 0 & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & -1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \nonumber \]

\[ \begin{matrix} \text{kronecker(Hardness, Color)} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} & \text{kronecker} \begin{bmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},~ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{bmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber \]