## 6.2 Quantum correlations

Consider two entangled qubits in the singlet126 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 observable127 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 choice128 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 value129 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. For example, the Bell states (Section 5.7) are all (maximally entangled) singlet states. This is related to the notion of singlet states in quantum mechanics, which are those with zero net angular momentum.↩︎

2. We say “observable” and “value” instead of “Hermitian operator” and “eigenvalue” because it’s useful to be able to switch between speaking like a mathematician and like a physicist!↩︎

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.↩︎