10.1 Quantum correlations

Consider two entangled qubits in the singlet159 state |\psi\rangle = \frac{1}{\sqrt{2}} \left( |01\rangle-|10\rangle \right) and note that the projector |\psi\rangle\langle\psi| can be written as |\psi\rangle\langle\psi| = \frac{1}{4} \left( \mathbf{1}\otimes\mathbf{1}- \sigma_x\otimes\sigma_x - \sigma_y\otimes\sigma_y - \sigma_z\otimes \sigma_z \right) where \sigma_x, \sigma_y, and \sigma_z are our old friends the Pauli matrices.

Also recall that any single-qubit observable160 with values \pm1 can be represented by the operator \vec{a}\cdot\vec\sigma = a_x\sigma_x + a_y\sigma_y + a_z\sigma_z, where \vec{a} is a unit vector in the three-dimensional Euclidean space.

So if Alice and Bob both choose observables, then we can characterise their choice161 by vectors \vec{a} and \vec{b}, respectively. If Alice measures the first qubit in our singlet state |\psi\rangle, and Bob the second, then the corresponding observable is described by the tensor product A\otimes B = (\vec{a}\cdot\vec\sigma)\otimes(\vec{b}\cdot\vec\sigma). The eigenvalues of A\otimes B are the products of eigenvalues of A and B. Thus A\otimes B has two eigenvalues: +1, corresponding to the instances when Alice and Bob registered identical outcomes, i.e. (+1,+1) or (-1,-1); and -1, corresponding to the instances when Alice and Bob registered different outcomes, i.e. (+1,-1) or (-1,+1).

This means that the expected value162 of A\otimes B, in any state, has a simple interpretation: \langle A\otimes B\rangle = \Pr (\text{outcomes are the same}) - \Pr (\text{outcomes are different}). This expression can take any real value in the interval [-1,1], where -1 means we have perfect anti-correlations, 0 means no correlations, and +1 means perfect correlations.

We can evaluate the expectation value in the singlet state: \begin{aligned} \langle\psi|A\otimes B|\psi\rangle & = \operatorname{tr}\Big[ (\vec{a}\cdot\vec\sigma)\otimes(\vec{b}\cdot\vec\sigma) |\psi\rangle\langle\psi| \Big] \\& = -\frac{1}{4} \operatorname{tr}\Big[ (\vec{a}\cdot\vec\sigma)\sigma_x \otimes(\vec{b}\cdot\vec\sigma)\sigma_x + (\vec{a}\cdot\vec\sigma)\sigma_y \otimes(\vec{b}\cdot\vec\sigma)\sigma_y + (\vec{a}\cdot\vec\sigma)\sigma_z \otimes(\vec{b}\cdot\vec\sigma)\sigma_z \Big] \\& = -\frac{1}{4} \operatorname{tr}\Big[ 4(a_x b_x + a_y b_y + a_z b_z) \mathbf{1}\otimes\mathbf{1} \Big] \\& = -\vec{a}\cdot\vec{b} \end{aligned} where we have used the fact that \operatorname{tr}(\vec{a}\cdot\vec\sigma)\sigma_k = 2a_k (for k=x,y,z). So if Alice and Bob choose the same observable \vec{a} = \vec{b}, then the expected value \langle A\otimes B\rangle will be equal to -1, and their outcomes will always be opposite: whenever Alice registers +1 (resp. -1) Bob is bound to register -1 (resp. +1).

  1. We say that a system is singlet if all the qubits involved are entangled. This is related to the notion of singlet states in quantum mechanics, which are those with zero net angular momentum.↩︎

  2. Here we say “observable” instead of “Hermitian operator” (and “value” instead of “eigenvalue”) simply because it is a useful skill to be able to switch between being able to speak like a physicist and like a mathematician!↩︎

  3. For example, if the two qubits are spin-half particles, they may measure the spin components along the directions \vec{a} and \vec{b}.↩︎

  4. Recall Section 4.5: the expected value of an operator E in the state |\phi\rangle is equal to \langle\phi|E|\phi\rangle.↩︎