advanced27 min studyLesson 428 in Physics

Superdense coding and classical capacity

Encode four orthogonal joint states with one transmitted qubit and reconcile the protocol with the Holevo bound by counting shared entanglement.

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What you’ll understand

  • Map two classical labels onto a Bell basis using local Pauli operations.
  • Decode a joint state with a concrete inverse circuit.
  • Explain which resource assumptions permit two classical bits per transmitted qubit.

A noiseless qubit without shared entanglement can convey at most one reliable classical bit per use, even though its pure states have continuously variable amplitudes. Superdense coding conveys two bits by adding a previously shared Bell pair to the resource budget. The receiver ultimately measures two qubits jointly. Keeping that joint Hilbert space and the earlier entanglement distribution visible resolves the apparent violation of an information bound and shows why the protocol cannot signal before a physical transmission.

Local operations label orthogonal joint states

Alice holds A and Bob holds B of a shared positive-Phi Bell pair. To send bits a and b, Alice applies XbZaX^bZ^a to A. The four choices give the four Bell states up to physically irrelevant global signs. The first bit controls relative phase and the second controls computational-basis parity. These labels are encoded in the joint state, not in a locally readable change of Alice's or Bob's individual qubit.

Alice then physically sends A through an ideal qubit channel. Bob, now holding both factors, can make a complete Bell-basis measurement and identify the two bits exactly. Before transmission his B marginal is I/2 for every message. Even Alice's transmitted marginal is I/2 for every message, so inspecting that particle alone cannot identify the label. Reliable decoding uses correlations with the previously shared half, which is precisely the additional resource.

∣βab⟩AB=(XbZa⊗IB)∣Φ+⟩AB,ρB(ab)=IB2,1 qubit+1 ebit ⟹ 2 cbits|\beta_{ab}\rangle_{AB}=(X^bZ^a\otimes I_B)|\Phi^+\rangle_{AB},\qquad \rho_B^{(ab)}=\frac{I_B}{2},\qquad 1\ {\rm qubit}+1\ {\rm ebit}\ \Longrightarrow\ 2\ {\rm cbits}
Bits a,b are zero or one; global signs of Bell vectors do not affect decoding. The forward quantum channel is noiseless and one pre-shared ebit is consumed per two-bit message.↔ Scroll wide equations horizontally.

The decoder can be written as a short circuit

An explicit decoder applies controlled-NOT with A as control and B as target, followed by a Hadamard gate on A, and measures both in the computational basis. The controlled-NOT converts Bell parity into B's ordinary bit. The Hadamard converts the remaining relative sign into A's bit. For our encoding convention the outcomes are a on A and b on B, with a possible overall phase that never changes their probabilities.

The resource pair does not survive as a reusable Bell pair after complete decoding. Counting only the final forward transmission can be appropriate when entanglement is supplied in advance, but it is an entanglement-assisted rate. If distributing the initial pair also costs one forward qubit, two quantum transmissions have carried two classical bits over the complete cycle. Storage errors, a restricted Bell analyzer, or a noisy forward channel require a different performance analysis; the ideal two-bit result alone supplies no engineering error model.

Iaccessible≤χ=S ⁣(∑xpxρx)−∑xpxS(ρx)≤log⁡2dI_{\rm accessible}\le\chi=S\!\left(\sum_xp_x\rho_x\right)-\sum_xp_xS(\rho_x)\le\log_2 d
This Holevo bound describes an ensemble on the receiver's full d-dimensional system, with information in bits. In ideal dense coding d=4 for AB, so χ=2 is consistent with the bound.↔ Scroll wide equations horizontally.

Capacity depends on the allowed assistance

For an ensemble of isolated qubit messages without a shared quantum reference, the Holevo quantity is at most one bit. Increasing the number of nonorthogonal signal states does not permit arbitrary recovery of their continuous coordinates from a single carrier. In dense coding the receiver's ensemble consists of four orthogonal pure states in dimension four, whose uniform average is the identity divided by four. Its Holevo quantity is two bits and the Bell measurement actually attains it.

A Holevo upper bound is not always attained by a measurement on one signal. Channel coding can use long blocks and collective decoding, so a general classical capacity question requires an asymptotic definition and an encoding class. Shared entanglement changes that capacity problem. With a nonmaximally entangled pair, local Pauli encodings need not produce four orthogonal states; their overlap reflects the unequal Schmidt weights. The maximally entangled resource is therefore doing quantitative work, beyond merely providing any correlation between laboratories.

Decode a phase-bit message

Alice and Bob share ∣Φ+⟩|\Phi^+\rangle. Alice wants to send a=1,b=0. Write the encoded state and follow the controlled-NOT–Hadamard decoder to its two classical outcomes.

  1. The encoding is X0Z1=ZX^0Z^1=Z on A. It changes the initial Bell vector into (∣00⟩−∣11⟩)/2=∣Φ−⟩(|00\rangle-|11\rangle)/\sqrt2=|\Phi^-\rangle, encoding a relative sign without changing either local marginal.

  2. Controlled-NOT flips B only on the term with A=1, giving (∣00⟩−∣10⟩)/2=∣−⟩A∣0⟩B(|00\rangle-|10\rangle)/\sqrt2=|-\rangle_A|0\rangle_B. The parity bit has become B's zero.

  3. Hadamard on A maps minus to one. The final state is ∣1⟩A∣0⟩B|1\rangle_A|0\rangle_B, so computational measurement returns the intended pair (1,0) with certainty under the ideal assumptions.

The result

A relative phase bit becomes an ordinary measurement bit only after Bob holds both qubits and performs the joint decoder.

Explore the model

Interactive lab

Interactive lab · probability

A count histogram can merge message labels

For each fresh comparison, Reset and set both controls. Reset and set p=0.5 and n=2. Interpret two independent fair classical bits as a uniform four-message alphabet. Predict bars for the number of ones with probabilities 1/4, 1/2, 1/4: the middle bar merges the distinct ordered messages 01 and 10. Compare a fresh run with both controls p=0.5 and n=1, giving only two equal bars. This is exact binomial counting, not Bell encoding, quantum transmission, or a capacity computation.

A count histogram can merge message labelsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA two-bit message needs four distinguishable ordered labels; counting only its ones loses information.

Check your understanding

Why does ideal dense coding permit two classical bits per transmitted qubit without violating the Holevo bound?

Try it yourself

Replace the Bell resource by ∣ψ⟩=0.8∣00⟩+0.2∣11⟩|\psi\rangle=\sqrt{0.8}|00\rangle+\sqrt{0.2}|11\rangle. Compare identity and Z encodings, then find χ for four equally likely Pauli encodings on A.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Calculate the overlap of the phase-flipped pair. Averaging all four Pauli choices randomizes A while preserving B’s diagonal marginal.

Compare with the solution

The identity and Z encoded vectors have inner product 0.8−0.2=0.60.8-0.2=0.6, so they are not orthogonal and four messages cannot be decoded perfectly in one use. All encoded joint states remain pure. Their uniform average is (IA/2)⊗diag⁡(0.8,0.2)B(I_A/2)\otimes\operatorname{diag}(0.8,0.2)_B, whose entropy is 1+h2(0.2)≈1.7219281+h_2(0.2)\approx1.721928 bits. Thus χ has that value, giving a single-ensemble accessible-information upper bound below two; it does not assert that a one-copy measurement attains the bound.

Take it with you

  • Dense coding spends shared entanglement to enlarge the jointly distinguishable message space.
  • The transmitted subsystem alone carries no readable message in the ideal Bell encoding.
  • Classical capacity statements must identify assistance, coding blocks, and the complete receiver system.

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