Quantum Entanglement Fidelity Calculator
Compute concurrence, entanglement entropy, or Bell-inequality violation for a two-qubit pure state defined by amplitudes α and β. Use this when analyzing entangled qubit pairs in quantum information or communication protocols.
About this calculator
A general two-qubit pure state can be written as |ψ⟩ = α|00⟩ + β e^(iφ)|11⟩, where α and β are real amplitudes satisfying α² + β² = 1 and φ is the relative phase. Concurrence C — a measure of entanglement ranging from 0 (separable) to 1 (maximally entangled) — is given by C = 2|αβ cos φ|. Entanglement entropy S quantifies the same property using von Neumann entropy: S = −α² log₂(α²) − β² log₂(β²). The CHSH–Bell parameter measures whether correlations exceed classical limits; the maximum quantum value is B = 2√2 · |αβ cos φ|, which violates the classical bound of 2 whenever B > 2. Maximum entanglement (C = 1) is achieved at α = β = 1/√2 with φ = 0, producing a Bell state. These quantities are central to evaluating resources for quantum teleportation, superdense coding, and quantum key distribution.
How to use
Suppose α = 0.6, β = 0.8, and phase φ = 0 radians. First, verify normalization: 0.6² + 0.8² = 0.36 + 0.64 = 1.00 ✓. To compute concurrence: C = 2 · |0.6 · 0.8 · cos(0)| = 2 · 0.48 · 1 = 0.96, indicating near-maximal entanglement. For entanglement entropy: S = −(0.36 · log₂(0.36)) − (0.64 · log₂(0.64)) = −(0.36 · (−1.474)) − (0.64 · (−0.644)) ≈ 0.531 + 0.412 = 0.943 bits, close to the maximum of 1 bit. The Bell parameter would be B = 2√2 · 0.48 ≈ 1.358, indicating the state alone does not violate the CHSH inequality at this phase.
Frequently asked questions
What is the difference between concurrence and entanglement entropy as measures of entanglement?
Concurrence C is an operationally motivated measure derived from the spin-flip density matrix and ranges from 0 to 1. Entanglement entropy S is defined via the von Neumann entropy of the reduced density matrix after tracing out one qubit and is measured in bits or ebits. For pure two-qubit states both measures are monotonically related, so they agree on which state is more entangled, but they differ in how they scale and in their physical interpretation. Concurrence is often easier to compute directly from state amplitudes, while entanglement entropy generalises more naturally to mixed and multi-partite states.
How does the relative phase angle affect the entanglement of a two-qubit state?
The relative phase φ between the |00⟩ and |11⟩ components modulates the concurrence through the factor cos φ. At φ = 0 or π the concurrence is maximised (for fixed α, β), while at φ = π/2 the concurrence drops to zero and the state becomes separable regardless of the amplitude values. Physically, this means that quantum correlations can be tuned or destroyed simply by adjusting the phase relationship between the superposed components. This sensitivity to phase is why decoherence — which randomises phases — is so destructive to entanglement.
When does a two-qubit state violate the CHSH Bell inequality?
A two-qubit state violates the CHSH inequality when the Bell parameter B = 2√2 · |αβ cos φ| exceeds 2. This requires |αβ cos φ| > 1/√2 ≈ 0.707. For a maximally entangled state (α = β = 1/√2, φ = 0) the parameter reaches its maximum quantum value of 2√2 ≈ 2.828, a clear violation. Bell violation confirms that the correlations cannot be explained by any local hidden-variable theory and is used as a practical test for genuine entanglement in experiments. States with too much mixing or unfavourable phase do not violate the inequality even if they have nonzero concurrence.