ISEGORIA / MATH ENCYCLOPEDIA
Quantum information: entanglement and its limits
What two entangled qubits can do that no classical pair can: correlations beyond any local model, teleporting a state with two classical bits and a shared pair, and a number, the entanglement entropy, that no local operation can change.
Before you begin: Quantum mechanics and information theory
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Bell’s inequality: the CHSH game
Alice and Bob each hold one qubit of the pair Φ⁺ = (|00⟩ + |11⟩)/√2 and each measures spin along a direction in the x-z plane: Alice at angle a or a′, Bob at b or b′. Every outcome is ±1 and the correlation is E(a, b) = cos(a − b). Drag the four settings on the dials and watch S on the right. Any theory in which each outcome is fixed in advance by a shared hidden variable gives |S| ≤ 2, the grey band; the orange curve is one such model, with the hidden variable a uniformly random angle. Move b′ towards 135° and the quantum S climbs past 2 to its ceiling 2√2.
Worked example. At \(a=0^\circ,\ a^{\prime}=90^\circ,\ b=45^\circ,\ b^{\prime}=135^\circ\) every angle difference is \(45^\circ\) except \(a-b^{\prime}=-135^\circ\), so \(S=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=2\sqrt2\approx2.828\). The hidden-variable model at the same settings gives \(E=1-2\cdot45/180=\tfrac12\) three times and \(-\tfrac12\) once, so \(S=2\) exactly: the classical bound, not beyond it.
Watch out. The orange curve is one classical model, not all of them: the bound \(|S|\le2\) holds for every local deterministic or stochastic assignment, which is what makes the violation decisive. The quantum values assume perfect detectors and a perfectly prepared \(\Phi^{+}\); real experiments must also close the locality and detection loopholes, which this page does not model. Settings on the same dial that coincide give \(S=\pm2\) for either theory.
Why can the quantum S never exceed 2√2, however the four angles are chosen?
Write \(S\) as the expectation of the operator \(\mathcal B=A\otimes(B-B^{\prime})+A^{\prime}\otimes(B+B^{\prime})\) with \(A^2=B^2=I\). Then \(\mathcal B^2=4I-[A,A^{\prime}]\otimes[B,B^{\prime}]\), and each commutator has norm at most \(2\), so \(\|\mathcal B^2\|\le8\) and \(|\langle\mathcal B\rangle|\le2\sqrt2\). This is Tsirelson’s bound; equality needs anticommuting settings, which is exactly the \(90^\circ\) spacing on each dial.
2. Teleportation, step by step
Alice holds an unknown qubit |ψ⟩ = cos(θ/2)|0⟩ + e^{iφ} sin(θ/2)|1⟩ and shares the pair Φ⁺ with Bob. Set |ψ⟩ by dragging the gold tip on the Bloch sphere, then step through the circuit: a CNOT from ψ onto her half of the pair, a Hadamard on ψ, a measurement of both her qubits, and finally Bob’s correction X^{m₂} Z^{m₁}. The bar chart shows all eight amplitudes of the three-qubit state at the current step, and the orange arrow is Bob’s qubit. Choose the outcome Alice sees and notice that before the correction Bob holds |ψ⟩ turned by a Pauli operator, and after it he holds |ψ⟩ exactly, for every one of the four outcomes.
Worked example. Take \(\theta=60^\circ,\ \varphi=0\), so \(|\psi\rangle=0.866|0\rangle+0.5|1\rangle\) with Bloch vector \((0.866,0,0.5)\). If Alice reads \(m_1m_2=01\), Bob holds \(X|\psi\rangle=0.5|0\rangle+0.866|1\rangle\), Bloch vector \((0.866,0,-0.5)\), and his fidelity with \(|\psi\rangle\) is only \(|0.866\cdot0.5+0.5\cdot0.866|^2=0.75\). One \(X\) restores \(|\psi\rangle\) and the fidelity becomes \(1\). Each outcome occurs with probability \(\tfrac14\) whatever \(|\psi\rangle\) is.
Watch out. Nothing travels faster than light: until the two classical bits arrive, Bob’s reduced state is \(\tfrac12I\), computed here by tracing out Alice’s qubits, and it carries no information about \(|\psi\rangle\). The original is destroyed by Alice’s measurement, as the no-cloning theorem demands. The simulation is exact for pure states and ideal gates; noise in the shared pair lowers the final fidelity below \(1\).
Why does Alice’s measurement outcome tell her nothing about |ψ⟩?
After the CNOT and Hadamard the state is \(\tfrac12\sum|m_1m_2\rangle\otimes X^{m_2}Z^{m_1}|\psi\rangle\), and each of the four branches has norm \(\tfrac12\) because the Paulis are unitary. So every outcome has probability \(\tfrac14\) independent of \(\theta\) and \(\varphi\). The two bits say which Pauli Bob must undo, not what \(|\psi\rangle\) is.
3. Entanglement entropy
The family |ψ(θ)⟩ = cos θ |00⟩ + sin θ |11⟩ runs from a product state at θ = 0 to the maximally entangled Φ⁺ at θ = 45°. Trace out qubit B and what remains, ρ_A, has eigenvalues cos²θ and sin²θ: a pure state on the surface of the Bloch ball when θ = 0, the centre of the ball when θ = 45°. Drag θ on the dial and watch the Bloch vector shrink, the bars equalise and the entropy S(ρ_A) rise to one bit. Then turn the slider β, a rotation of qubit B alone: every amplitude changes, and neither S, nor C, nor the Bloch vector moves.
Worked example. At \(\theta=30^\circ\): \(\lambda_1=\cos^2 30^\circ=0.75\), \(\lambda_2=0.25\), so \(S=-0.75\log_2 0.75-0.25\log_2 0.25=0.311+0.5=0.811\) bits, \(C=\sin60^\circ=0.866\), and \(|\vec r_A|=\sqrt{1-0.75}=0.5=\cos60^\circ\), the Bloch vector of \(\operatorname{diag}(0.75,0.25)\). Rotating B by any \(\beta\) leaves all four numbers unchanged.
Watch out. The entropy of a reduced state measures entanglement only for pure states of the pair. For a mixed two-qubit state, \(S(\rho_A)\) can be large with no entanglement at all, and one must use the concurrence with Wootters’ spin-flip formula on the full \(\rho\); the formula \(C=2|\psi_{00}\psi_{11}-\psi_{01}\psi_{10}|\) drawn here is its pure-state case. The Bloch ball is shown as a cross-section through the \(x\)-\(z\) plane, which contains \(\vec r_A\) because every amplitude is real.
Why do ρ_A and ρ_B always have the same eigenvalues, even though the two qubits may be treated quite differently?
By the Schmidt decomposition any pure state of the pair can be written \(\sum_i\sqrt{\lambda_i}\,|u_i\rangle|v_i\rangle\) with orthonormal \(u_i\) and \(v_i\), and then \(\rho_A=\sum\lambda_i|u_i\rangle\langle u_i|\), \(\rho_B=\sum\lambda_i|v_i\rangle\langle v_i|\). The \(\lambda_i\) are the same list; only the bases differ. Here \(\sqrt{\lambda_i}\) are \(\cos\theta\) and \(\sin\theta\), and the rotation \(\beta\) merely changes the \(v_i\).