Holevo Bound: Classical Information from a Quantum Alphabet

Quantum Alphabet

Cone angle θ (spread) 70°

Bob's Measurement

Axis colatitude θ_m 0°

Axis azimuth φ_m 0°

Legend

Information (bits)

Holevo bound χ

0.00

Accessible I(A:B)

0.00

Unrecovered gap χ − I

0.00

How it works

Alice encodes a classical symbol i (chosen with probability p_i = 1/N) as a pure qubit state |ψ_i⟩ placed on a cone of colatitude θ around the north pole, symbols spread evenly in azimuth. Bob receives the qubit and tries to learn which symbol was sent.

ρ̄ = Σ p_i |ψ_i⟩⟨ψ_i| = (I + r̄·σ) / 2
Holevo bound:  χ = S(ρ̄) − Σ p_i S(ρ_i) = S(ρ̄)   (pure states ⇒ S(ρ_i)=0)
Qubit entropy: S(ρ) = H₂( (1+|r|)/2 ),  H₂(p) = −p·log₂p − (1−p)·log₂(1−p)

By the cone's symmetry the azimuthal components of the states' Bloch vectors cancel exactly, so r̄ = (0, 0, cos θ) — the Holevo bound depends only on θ and is maximal (1 bit) when the cone flattens onto the equator (θ = 90°, ρ̄ maximally mixed) and zero when all states coincide at a pole (θ = 0° or 180°).

Holevo's theorem says no measurement Bob performs can extract more than χ bits on average — it is an upper bound, not a recipe. This simulator lets you pick Bob's actual projective measurement (any axis m, sliders θ_m/φ_m) and computes his real accessible information exactly from the Born rule:

P(+|i) = (1 + n_i·m) / 2
I(A:B) = H₂( P(+) ) − Σ p_i H₂( P(+|i) ),   P(+) = Σ p_i P(+|i)

For N = 2 antipodal states (θ = 180° reached from θ = 90° cone... try the equator with N=2) a matched axis reaches I(A:B) = χ exactly — the states are orthogonal and perfectly distinguishable. For N ≥ 3 no single projective measurement can close the gap even at the optimal axis: three or four non-orthogonal directions can never be told apart with zero error, which is exactly why BB84-style protocols built on more than two conjugate states stay secure — an eavesdropper is bound by the same inequality Bob is.

Measurement axis ±m Bloch sphere / cone

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