9.1 Quantum correlations

Consider two entangled qubits in the singlet 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 as127 |\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). Any single qubit observable with values \pm 1 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. Suppose Alice and Bob choose measurements defined by vectors \vec{a} and \vec{b}, respectively. For example, if the two qubits are spin-half particles, they may measure the spin components along the directions \vec{a} and \vec{b}. We write the corresponding observable as 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 value 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 numerical value from -1 (perfect anti-correlations) through 0 (no correlations) to +1 (perfect correlations). We now evaluate the expectation value in the singlet state: \begin{aligned} \langle\psi|A\otimes B|\psi\rangle & = \operatorname{tr}\left[ (\vec{a}\cdot\vec\sigma)\otimes(\vec{b}\cdot\vec\sigma) |\psi\rangle\langle\psi| \right] \\& = -\frac{1}{4} \operatorname{tr}\left[ (\vec{a}\cdot\vec\sigma)\sigma_x \otimes(\vec{a}\cdot\vec\sigma)\sigma_x + (\vec{a}\cdot\vec\sigma)\sigma_y \otimes(\vec{a}\cdot\vec\sigma)\sigma_y + (\vec{a}\cdot\vec\sigma)\sigma_z \otimes(\vec{a}\cdot\vec\sigma)\sigma_z \right] \\& = -\frac{1}{4} \operatorname{tr}\left[ (a_x b_x + a_y b_y + a_z b_z) \mathbf{1}\otimes\mathbf{1} \right] \\& = -\vec{a}\cdot\vec{b} \end{aligned} where we have used the fact that \operatorname{tr}(\vec{a}\cdot\vec\sigma)\sigma_k = a_k (k=x,y,z). So if Alice and Bob choose the same observable, \vec{a} = \vec{b}, then their outcomes will be always opposite: whenever Alice registers +1 (resp. -1) Bob is bound to register -1 (resp. +1).


  1. There are other, more elementary, ways of deriving this result but here I want you to hone your skills. Now that you’ve learned about projectors, traces, and Pauli operators, why not put them to good use.↩︎