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9.6

Superdense coding

Share a Bell pair in advance and mailing one qubit carries two classical bits: the four Pauli gates tune one entangled pair through the four orthogonal Bell states — teleportation's mirror image.

Recommended first

After this section you should be able to

  • Work through the superdense coding protocol in full: four encoding operations, four Bell states, the decoding circuit, all by hand
  • Explain why "touching only your own half" is enough to reach all four Bell states
  • Show that 2 bits is the ceiling: without pre-shared entanglement, 1 qubit carries at most 1 bit (Holevo)
  • Write the resource equations of superdense coding and teleportation side by side and see the duality

At the end of the last section we read teleportation’s trade slip backwards: pre-share 1 ebit, then mail 1 qubit — how many classical bits does that exchange for? Section 9.1’s Holevo argument says one qubit yields at most 1 bit on readout, yet this section will legally read out 2. Watch entanglement cut the postage in half.

The concrete problem: one particle per shipment, four things to say

Alice needs to send Bob a two-bit message — one of 00, 01, 10, 11 (say “buy / sell / hold / liquidate”). But the channel is expensive, and this month’s quota is down to one qubit.

Without entanglement, one qubit is worth one bit

First, the ceiling on the naive scheme. Alice encodes the message into a single qubit’s state and mails it; Bob measures on arrival. Section 9.1 told us: a measurement has only two exits, and distinct non-orthogonal states cannot be reliably distinguished. To send 2 bits reliably, Bob must separate the four states unambiguously, which requires four mutually orthogonal states — but a single qubit’s Hilbert space is only two-dimensional, with room for at most 2 orthogonal states. Dead end. (This is exactly a special case of Holevo’s theorem: with no auxiliary resource, nn qubits ≤ nn classical bits.)

The way out: Bob is allowed to hold something in advance. As long as the two share a Bell pair beforehand, the “room for manoeuvre” of Alice’s half doubles.

The full protocol, by hand

Beforehand: the two share ∣Φ+⟩=∣00⟩+∣11⟩2\ket{\Phi^+}=\frac{\ket{00}+\ket{11}}{\sqrt2}, particle A with Alice, particle B with Bob.

            encode (Alice, local)     mail A          decode (Bob, local)
 A ────────[ I / X / Z / ZX ]────────────────●──[H]──[measure]→ m₁
   ╲                                         │
    Φ⁺                                       │
   ╱                                         │
 B ──────────────────────────────────────────⊕──────[measure]→ m₂
Four Pauli gates dial up four Bell statesbasic~7 min

Step 1: encode. According to the message, Alice applies one gate to her own half:

  • Message 00: do nothing (II). The state is unchanged:
(I⊗I)∣Φ+⟩=∣00⟩+∣11⟩2=∣Φ+⟩(9.6.1)(I\otimes I)\ket{\Phi^+}=\frac{\ket{00}+\ket{11}}{\sqrt2}=\ket{\Phi^+}\tag{9.6.1}
  • Message 01: apply XX (flip A). Acting term by term on the first qubit:
(X⊗I) ∣00⟩+∣11⟩2=∣10⟩+∣01⟩2=∣Ψ+⟩(9.6.2)(X\otimes I)\,\frac{\ket{00}+\ket{11}}{\sqrt2} =\frac{\ket{10}+\ket{01}}{\sqrt2}=\ket{\Psi^+}\tag{9.6.2}
  • Message 10: apply ZZ (a minus sign on A’s ∣1⟩\ket{1}):
(Z⊗I) ∣00⟩+∣11⟩2=∣00⟩−∣11⟩2=∣Φ−⟩(9.6.3)(Z\otimes I)\,\frac{\ket{00}+\ket{11}}{\sqrt2} =\frac{\ket{00}-\ket{11}}{\sqrt2}=\ket{\Phi^-}\tag{9.6.3}
  • Message 11: first XX, then ZZ:
(ZX⊗I)∣Φ+⟩=(Z⊗I) ∣10⟩+∣01⟩2=∣01⟩−∣10⟩2=∣Ψ−⟩(9.6.4)(ZX\otimes I)\ket{\Phi^+} =(Z\otimes I)\,\frac{\ket{10}+\ket{01}}{\sqrt2} =\frac{\ket{01}-\ket{10}}{\sqrt2}=\ket{\Psi^-}\tag{9.6.4}

Four messages ↦\mapsto four Bell states, and the Bell states are mutually orthogonal — the orthogonality needed for reliable discrimination is all there; it just lives in the two-qubit space.

Step 2: mail it. Alice sends particle A to Bob. Both particles are now in Bob’s hands.

Step 3: decode = Bell measurement. Run the circuit from the end of section 9.5 in reverse reading: CNOT (A controlling B) then HH on A maps the four Bell states back onto the four computational basis states. Verify one case (take ∣Ψ−⟩\ket{\Psi^-}; the rest go the same way):

∣Ψ−⟩→ CNOT ∣01⟩−∣11⟩2=∣0⟩−∣1⟩2⊗∣1⟩→ H⊗I ∣1⟩⊗∣1⟩=∣11⟩(9.6.5)\ket{\Psi^-}\xrightarrow{\ \mathrm{CNOT}\ } \frac{\ket{01}-\ket{11}}{\sqrt2} =\frac{\ket{0}-\ket{1}}{\sqrt2}\otimes\ket{1} \xrightarrow{\ H\otimes I\ }\ket{1}\otimes\ket{1}=\ket{11}\tag{9.6.5}

The complete table of all four cases:

MessageAlice’s gateState in the channelReadout m1m2m_1m_2
00II∣Φ+⟩\ket{\Phi^+}00
01XX∣Ψ+⟩\ket{\Psi^+}01
10ZZ∣Φ−⟩\ket{\Phi^-}10
11ZXZX∣Ψ−⟩\ket{\Psi^-}11

Bob reads out two bits deterministically. No probabilities, no guessing — because the four candidate states are orthogonal.

The picture

Mirror image of teleportation. The two protocols burn the same fuel (1 ebit), in exactly opposite directions:

  • Teleportation: spend 2 classical bits to ship 1 qubit;
  • Superdense coding: spend 1 qubit to ship 2 classical bits.

One hauls quantum cargo over a classical channel; the other hauls classical cargo over a quantum channel. This mirror pair shows there is no fixed exchange rate between qubits and cbits — the rate is set by the entanglement reserves on hand. The very notion of “entanglement as a consumable communication resource” was established by these two protocols of 1992/1993.

The mathematics

The resource equations side by side (consumed on the left of the arrow, produced on the right):

1 ebit+2 cbit ⟶ 1 qubit(9.6.6)1\ \text{ebit}+2\ \text{cbit}\ \longrightarrow\ 1\ \text{qubit}\tag{9.6.6}1 ebit+1 qubit ⟶ 2 cbit(9.6.7)1\ \text{ebit}+1\ \text{qubit}\ \longrightarrow\ 2\ \text{cbit}\tag{9.6.7}

Neither can be reverse-engineered into a free lunch: substitute the second into the first and one round cancels on each side, giving 2 ebit+a round trip of channel→2\,\text{ebit} + \text{a round trip of channel} \to back where you started — entanglement only dwindles with use, never appreciates out of thin air (LOCC cannot increase entanglement, section 9.3).

Key formulas

  • Encoding

    (σ⊗I)∣Φ+⟩,σ∈{I,X,Z,ZX}(\sigma\otimes I)\ket{\Phi^+},\quad \sigma\in\{I,X,Z,ZX\}

    Four Pauli operations ↦ four mutually orthogonal Bell states

  • Decoding

    (H⊗I) CNOT: ∣Bm1m2⟩↦∣m1m2⟩(H\otimes I)\,\mathrm{CNOT}:\ \ket{B_{m_1m_2}}\mapsto\ket{m_1m_2}

    The Bell-measurement circuit; deterministic readout

  • Resource equation

    1 ebit+1 qubit→2 cbit1\ \text{ebit}+1\ \text{qubit}\to 2\ \text{cbit}

    Dual to teleportation's 1 ebit + 2 cbit → 1 qubit

  • Holevo bound

    χ≤n bit (no entanglement),2n bit (pre-shared entanglement)\chi\le n\ \text{bit}\ (\text{no entanglement}),\quad 2n\ \text{bit}\ (\text{pre-shared entanglement})

    Superdense coding exactly saturates it; no "denser" coding exists

Self-check4 questions

  1. 1.

    Alice wants to send the message 10. Which gate should she apply to her particle?

  2. 2.

    An eavesdropper Eve intercepts the particle A that Alice mailed (but cannot get B). How much of the message can she read?

  3. 3.

    The two parties share 3 Bell pairs and Alice mails 3 qubits in total. Using superdense coding, how many classical bits can be reliably transmitted at most?

    bit0% relative tolerance
  4. 4.

    Which statements about the relationship between superdense coding and teleportation are correct? (Select all that apply.)

    Select all that apply

What comes next

So far, entanglement has helped us move information: sending states, carrying messages. The next question is more ambitious: can quantum mechanics help us compute faster? The Deutsch–Jozsa algorithm of 1992 delivered the first clean “yes” — a question that classically takes exponentially many queries, answered quantumly in one. Next section: how superposition and interference become computing power.

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