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Quantum Information & Bell Inequalities

Qubits, entanglement, and the experimental refutation of local hidden variables.

A qubit is a normalized vector in $\mathbb C^2$: $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, $|\alpha|^2 + |\beta|^2 = 1$. Two qubits live in $\mathbb C^4$. A two-qubit state is entangled if it can't be written as a tensor product $|\psi_A\rangle \otimes |\psi_B\rangle$. The Bell states:

$$|\Phi^\pm\rangle = \frac{1}{\sqrt 2}(|00\rangle \pm |11\rangle), \quad |\Psi^\pm\rangle = \frac{1}{\sqrt 2}(|01\rangle \pm |10\rangle).$$

For the singlet $|\Psi^-\rangle$, the correlation $E(\mathbf a, \mathbf b)$ of measurements along $\mathbf a, \mathbf b$ is $-\mathbf a \cdot \mathbf b$. The CHSH inequality:

$$S = |E(a,b) - E(a,b') + E(a',b) + E(a',b')|$$

satisfies $S \leq 2$ for any local hidden-variable theory, but quantum mechanics achieves $S = 2\sqrt 2 \approx 2.828$ — Tsirelson's bound. Experiments (Aspect 1982, loophole-free 2015) confirm the quantum prediction, ruling out local realism. 2022 Nobel.

Interactive: CHSH correlation

Quiz

1. An entangled two-qubit state is one that:
2. The classical CHSH bound is:
3. Quantum mechanics achieves $S_{\max}$ equal to:
4. Singlet correlation $E(\mathbf a, \mathbf b)$ equals:
5. Quantum teleportation requires:
6. The no-cloning theorem says: