Teleportation spent two classical bits to move one qubit. Superdense coding is its mirror image: spend one qubit to deliver two classical bits. The trick — as always — is a Bell pair shared ahead of time.
↩ before you start · keep these handy
·From Ch. 6: there are exactly four Bell states, and they are mutually distinguishable.
·From Ch. 5: the Pauli gates I, X, Z act on a single qubit — a bit-flip, a phase-flip, or nothing.
·From No-Cloning: Bob's lone half tells him nothing until Alice's qubit physically arrives.
🔑 symbol decoder · every new mark, in plain words
|Φ⁺⟩ … |Ψ⁻⟩the four Bell states — the four distinct "channels" a 2-bit message can become.I, X, Z, ZXAlice's four encoding gates — one for each 2-bit message.orthogonalperfectly distinguishable — no overlap, so a measurement never confuses them.Bell measurementBob's read-out that names which of the four Bell states he holds.ebitone pre-shared entangled pair — the resource paid for in advance.
feel
Two bits on one carrier
Long ago, Alice and Bob split an entangled pair — she kept one qubit, he kept the other, and they walked to opposite sides of the world. Today Alice wants to send Bob a two-bit message. Normally that takes two qubits. But because they already share that pair, Alice can nudge her single qubit in one of four ways, mail just that one qubit, and Bob recovers both bits. The pre-shared entanglement did half the work in advance.
💃 everyday picture
Two dancers rehearse a routine together, then split to separate stages. Later, with a single gesture — raise a hand, turn left — one dancer can cue the audience into one of four whole scenes, because the shared rehearsal fills in everything else. Alone, the gesture is just a gesture; paired with the rehearsed partnership, it carries far more. The pre-shared Bell pair is that rehearsal.
recapPre-shared entanglement lets Alice send a single qubit yet deliver two classical bits.
play
Pick two bits, send one qubit
Choose the message. Alice applies the matching gate to her half, which silently flips the shared pair into one of four distinguishable Bell states. She sends the qubit; Bob runs a Bell measurement and reads the two bits straight back.
▸ encode · send · decode1 qubit → 2 bits
Alice encodes
applies gate
{{ gate }}
{{ gateNote }}
the pair is now
{{ bell }}
1 qubit travels →
Bob decodes
Bell measurement reads
{{ out }}
✓ both bits recovered
messageAlice’s gateresulting Bell state
{{ r.bits }}{{ r.gate }}{{ r.bell }}
recapFour encoding gates map the shared pair to four distinguishable Bell states; Bob reads both bits at once.
math
Four gates, four orthogonal states
Start from the shared |Φ⁺⟩ = (|00⟩+|11⟩)/√2. Alice acts only on her qubit; each Pauli sends the pair to a different Bell state:
I · |Φ⁺⟩ = |Φ⁺⟩ = (|00⟩+|11⟩)/√2 → 00
X · |Φ⁺⟩ = |Ψ⁺⟩ = (|10⟩+|01⟩)/√2 → 01
Z · |Φ⁺⟩ = |Φ⁻⟩ = (|00⟩−|11⟩)/√2 → 10
ZX · |Φ⁺⟩ = |Ψ⁻⟩ = (|01⟩−|10⟩)/√2 → 11
the four Bell states are mutually orthogonal — a single Bell measurement tells them apart with certainty.
Because those four outputs are perfectly distinguishable, Bob never has to guess. One travelling qubit, two bits delivered, zero error.
✎ worked example · send the message 10
1.message 10 → the table says apply Z
2.Z flips the sign of the |11⟩ term: |Φ⁺⟩ → (|00⟩ − |11⟩)/√2 = |Φ⁻⟩
3.Alice mails just her qubit; Bob's Bell measurement reads |Φ⁻⟩
4.|Φ⁻⟩ decodes uniquely to 10 — both bits, zero error. ✓
recapThe four Bell states are orthogonal, so a single Bell measurement separates them with zero error.
⚠ common misconception
“So one qubit really holds two bits — free compression!” Count honestly. The protocol spends a pre-shared entangled pair — another qubit that had to travel from Alice to Bob earlier. Two qubits move in total; two bits arrive. The clever part is only the timing: half the cost is paid in advance, when there’s no message yet to send.
And no, it isn’t faster-than-light. Until Alice’s qubit physically arrives, Bob’s half alone is still a featureless coin — exactly the lesson from entanglement and Bell.
✓ you can now
✓map each 2-bit message to Alice's encoding gate (I, X, Z, ZX)
✓explain how Bob recovers both bits with one Bell measurement
✓count the true cost (1 sent qubit + 1 pre-shared ebit) and see it is not free compression