3.2.6 Entanglement and correlations

Given the similarity between entangled states and correlated probability distributions, you may wonder how these two notions are related. To compare them, let us discuss more generally the relation between quantum states and probability distributions.

To start, suppose we have a single-qubit state $\left|\psi\right\rangle=\psi_{0}\left|0\right\rangle+\psi_{1}\left|1\right\rangle$ and we measure it. Then we know from Section 2.2 that we get as outcome a bit that is either zero or one, with probabilities $\psi_{0}^{2}$ and $\psi_{1}^{2}$. We can model this as a probability distribution

Intuitively, this models the situation where we measured the qubit but we did not actually look at the outcome (if we did, we would not have a probabilistic bit but a deterministic one that is either in state zero or in state one).

The same logic works just as well for two qubits. If we measure a two-qubit state $\left|\psi\right\rangle=\psi_{00}\left|00\right\rangle+\psi_{01}\left|01\right\rangle+\psi_{10}\left|10\right\rangle+\psi_{11}\left|11\right\rangle$, we can describe the measurement outcomes by the probability distribution

For example, if we prepare and measure the maximally entangled state $\left|\Phi^{+}\right\rangle$, we obtain a perfectly correlated pair of random bits in Eq. 3.28. We can verify this using Quirky:

Clearly the same is true if we measure the Bell state $\left|\Phi^{-}\right\rangle$ instead. (How about the other two Bell states $\left|\Psi^{+}\right\rangle$ or $\left|\Psi^{-}\right\rangle$? Measuring either of them produces perfectly anti-correlated bits, described by the probability distribution $\frac{1}{2}[01]+\frac{1}{2}[10]$.)

The preceding example was not an accident. In fact, the probability distribution $p$ in Eq. 3.74 that is obtained by measuring a two-qubit quantum state can be correlated only if the corresponding quantum state $\left|\psi\right\rangle$ is entangled. To see this, assume that $\left|\psi\right\rangle$ is a product state, so that $\Delta(\left|\psi\right\rangle)=0$. Then $p$ is a product distribution since

This proves the claim that correlated measurement outcomes imply presence of entanglement in the measured state.

Note that generally quantum states are at least as useful as probabilistic bits because any probability distribution can be obtained by measuring an appropriately chosen quantum state. That is, given any probability distribution $p$ we can always find a quantum state $\left|\psi\right\rangle$ whose measurement outcomes are distributed according to $p$. For example, for a two-bit distribution $p$ we can simply choose

In particular, this means that entanglement is generally at least as useful as probabilistic correlations since any correlated distribution over two probabilistic bits can be produced by measuring some entangled two-qubit state.