Quantum Teleportation
You cannot send an unknown qubit through the post. Alice holds a state |ψ⟩ = α|0⟩ + β|1⟩ whose amplitudes she does not know; she cannot measure them without collapsing the state, and the no-cloning theorem forbids copying it to make a spare. And yet, in 1993, Bennett, Brassard, Crépeau, Jozsa, Peres and Wootters showed that Bob can end up holding exactly |ψ⟩ — its unknown amplitudes intact, down to the last relative phase — while nothing carrying those amplitudes ever crosses the space between them. The trick is to spend a resource laid down in advance: one shared pair of entangled qubits, plus two ordinary classical bits. The choreography is tight. Qubit 1 carries the unknown state; qubits 2 and 3 are a Bell pair, |Φ⁺⟩ = (|00⟩ + |11⟩)/√2, prepared earlier and split so Alice keeps qubit 2 and Bob keeps qubit 3. Alice performs a joint Bell-state measurement on qubits 1 and 2, yielding one of four equally likely outcomes, which she records as two classical bits and mails to Bob. The measurement destroys the original — no clone survives — but imprints |ψ⟩ onto Bob's distant qubit 3, scrambled by a random Pauli. Bob reads the two bits, applies the matching correction from {I, Z, X, ZX}, and holds |ψ⟩. The deep content is a statement about resources. Pre-shared entanglement plus classical communication can move quantum information, even though neither ingredient can alone: the classical bits know nothing about α and β, and the entangled pair carries no chosen message. Teleportation is the dual of superdense coding, the engine of entanglement swapping, and — as gate teleportation — a pillar of fault-tolerant computation.
- Quantum teleportation moves an unknown qubit |ψ⟩ from Alice to Bob using one shared Bell pair (an ebit) and two classical bits — the physical carrier is never sent.
- Alice makes a Bell-state measurement on her unknown qubit and her half of the pair, getting one of four outcomes, each with probability exactly 1/4, encoded as two classical bits.
- Those bits tell Bob which Pauli correction {I, Z, X, ZX} to apply to his half; after it he holds |ψ⟩ exactly, fidelity 1.
- No information travels faster than light: before the bits arrive Bob's reduced state is the maximally mixed I/2, independent of α and β, so causality is safe.
- Teleportation does not clone: Alice's measurement destroys the input, leaving one copy of |ψ⟩ at all times.
- Beating the classical measure-and-resend fidelity of 2/3 certifies that genuine entanglement, not classical correlation, carried the state.
- Key idea: pre-shared entanglement plus two classical bits is exactly the resource needed to move one unknown qubit — a Bell measurement projects the unknown amplitudes onto Bob's half up to a random Pauli, and the two bits announce which Pauli to undo.
Pre-shared entanglement plus two classical bits is exactly the resource needed to move one unknown qubit: a Bell measurement projects the unknown amplitudes onto Bob's half of the pair (up to a random Pauli), and the two classical bits announce which Pauli Bob must undo to recover |ψ⟩.
Where this lives today
Teleportation has become the standard verb of the emerging quantum internet. Because an unknown qubit cannot be amplified or copied, long-distance links cannot use classical-style repeaters; instead they distribute short-range entangled pairs, purify them, and teleport quantum information across each hop — the principle of quantum repeaters. In 2022 the Delft group ran teleportation across a three-node network of nitrogen-vacancy centres in diamond, moving a qubit from an outer node to a non-neighbouring one via entanglement swapping. The technique is now deterministic in leading hardware: trapped-ion experiments in Innsbruck and at NIST demonstrated on-demand teleportation between ions as early as 2004, and superconducting circuits followed; China's Micius satellite teleported photonic qubits from ground to orbit over roughly 1400 km in 2017. Inside a quantum computer the same identity reappears as gate teleportation, applying an otherwise expensive non-Clifford gate such as T by consuming a magic state — and measurement-based computation is teleportation chained across a cluster state. The 2/3 fidelity bound remains the field's certification line: every platform reports whether its teleportation beats it, because only then has genuine entanglement, not classical mimicry, done the work.
The physical picture
Think of the shared Bell pair as a quantum cable laid down long before anyone knew what would be sent — its two ends perfectly correlated but carrying no chosen message. Alice's Bell measurement "plugs" the unknown qubit into her end, instantaneously imprinting α and β onto Bob's end, but scrambled by a random one of four Pauli operations. The two classical bits are the unscrambling key Alice mails at the speed of light. Until it arrives, Bob's qubit is useless: his half of a Bell pair is maximally mixed no matter what Alice does. The randomness is not a defect — it is precisely what lets teleportation respect both no-cloning and causality at once: because all four outcomes are equally likely for every input, Alice learns nothing about |ψ⟩, and Bob learns nothing until the message lands.
Where does the state actually go? Not "instantly to Bob" — nothing usable moves until the classical bits do. What the Bell measurement buys is a correlation between Alice's random two-bit outcome and the random Pauli kick sitting on Bob's qubit. The second figure makes the accounting quantitative through a noisy channel: if the shared pair is an imperfect Werner state of purity p rather than a pristine Bell pair, the average teleportation fidelity is F = (1 + p)/2, beating the best entanglement-free strategy only above a sharp threshold. That threshold is why experimentalists quote a single number, 2/3, to certify that a demonstration used real quantum entanglement.
The core equations
Teleportation is a single algebraic identity dressed as a protocol; everything below states an ingredient or unpacks that identity. We start with the object to be moved: an unknown qubit, whose two complex amplitudes Alice can neither read nor copy.
where: α, β — unknown complex amplitudes Alice cannot measure or clone; |0⟩, |1⟩ — computational basis states of qubit 1. Reading: α and β are unknown continuous parameters — a single measurement collapses them irretrievably, and no-cloning bars making a spare to measure instead. The resource that rescues her is one maximally entangled pair, distributed in advance. Alice holds qubit 2, Bob holds qubit 3; neither half alone carries any information.
where: |Φ⁺⟩ — one of the four Bell states, shared before the protocol runs; subscripts 2, 3 — Alice's and Bob's halves. Reading: this is the "ebit," the unit of entanglement, chosen with no reference to |ψ⟩ — which is why the same pair works for any state Alice later sends. Alice measures qubits 1 and 2 in the Bell basis — the orthonormal basis of four maximally entangled two-qubit states — recording the outcome as two classical bits.
where: Φ⁺, Φ⁻, Ψ⁺, Ψ⁻ — the four orthonormal Bell states of qubits 1 and 2. Reading: this projection deliberately does not measure the two qubits individually, which keeps α and β hidden while still steering Bob's qubit. Now the identity itself. Rewriting |ψ⟩₁ ⊗ |Φ⁺⟩₂₃ in the Bell basis of qubits 1 and 2 regroups it into four equally weighted branches, each pairing a Bell outcome with a Pauli-transformed copy of |ψ⟩ on qubit 3.
where: |Bⱼ⟩ — the four Bell states indexed by j; Uⱼ — the Pauli outcome j leaves on Bob's qubit. Reading: the prefactor 1/2 gives each branch weight |1/2|² = 1/4, so every outcome is equally likely regardless of α, β. Bob's qubit is already |ψ⟩ up to a known Pauli — he just needs to be told which. The dictionary from outcome to correction is the operational heart of the protocol. Bob undoes the Pauli his branch carries by applying its inverse — and each Pauli is its own inverse.
where: I, X, Z — identity and Pauli operators; ZX = iY up to a global phase. Reading: two classical bits index exactly four corrections; without them Bob cannot pick the right one, and a fixed guess fails three times in four. Before the bits arrive, Bob's marginal — the partial trace over Alice's two qubits — is the maximally mixed state, no matter what |ψ⟩ was. This equation is the no-signalling guarantee.
where: Tr₁₂ — partial trace over Alice's qubits 1 and 2; ½I — maximally mixed single-qubit state. Reading: ρ_B is independent of α and β, so no measurement Bob makes reveals anything about |ψ⟩ until the classical message reaches him at or below the speed of light. Stepping back, the protocol is one line in the ledger of quantum resources — the mirror image of superdense coding.
where: ebit — one maximally entangled pair; cbit — one noiseless classical bit; qubit — one use of a noiseless quantum channel. Reading: these resources suffice to simulate a qubit channel; superdense coding runs it backwards, 1 ebit + 1 qubit ≥ 2 cbits. Finally, the benchmark. Through an imperfect resource of fully entangled fraction F_e, the average teleportation fidelity is fixed, and the best entanglement-free strategy tops out at 2/3.
where: F_e — overlap of the resource with the best Bell state; F̄ — average output fidelity over the Haar ensemble. Reading: a perfect pair (F_e = 1) gives F̄ = 1; classical measure-and-resend reaches only 2/3, so any F̄ above 2/3 certifies genuine entanglement.
The goal is to derive the teleportation identity of equation (4) from scratch — to show, by pure algebra, that |ψ⟩₁ ⊗ |Φ⁺⟩₂₃ equals an equal superposition of the four Bell outcomes on qubits 1 and 2, each tensored with a Pauli-rotated copy of |ψ⟩ on Bob's qubit 3. This is the fact on which the whole protocol rests: the unknown amplitudes α and β have, without anyone touching qubit 3, already migrated onto Bob's qubit, and a Bell measurement of qubits 1 and 2 merely reveals which of four known rotations they arrived under. We expand, re-express, and regroup — no measurement, no dynamics, just an identity.
Step 1. Tensor and expand. Write |ψ⟩₁ ⊗ |Φ⁺⟩₂₃ = (α|0⟩ + β|1⟩)₁ ⊗ (|00⟩ + |11⟩)₂₃/√2 and multiply out to get (1/√2)(α|000⟩ + α|011⟩ + β|100⟩ + β|111⟩), ordering the kets as qubits 1, 2, 3. Why: this is just the definition of the tensor product; every later step only rearranges these four computational-basis terms.
Step 2. Read off the qubit-1,2 factor of each term. In α|000⟩ qubits 1,2 are in |00⟩ with qubit 3 in |0⟩; in α|011⟩ they are in |01⟩ with qubit 3 in |1⟩; in β|100⟩ they are in |10⟩ with qubit 3 in |0⟩; in β|111⟩ they are in |11⟩ with qubit 3 in |1⟩. Why: to change basis on qubits 1 and 2 we isolate their state in every term, keeping Bob's qubit 3 a spectator.
Step 3. Invert the Bell basis. From equation (3), |00⟩ = (|Φ⁺⟩+|Φ⁻⟩)/√2, |11⟩ = (|Φ⁺⟩−|Φ⁻⟩)/√2, |01⟩ = (|Ψ⁺⟩+|Ψ⁻⟩)/√2, |10⟩ = (|Ψ⁺⟩−|Ψ⁻⟩)/√2. Substitute these for the qubit-1,2 factors of Step 2. Why: the Bell states are a complete orthonormal basis, so the qubit-1,2 factor can be rewritten in them exactly.
Step 4. Collect terms by Bell outcome. Gathering the coefficient of each Bell state on qubits 1,2 gives (1/2)[ |Φ⁺⟩(α|0⟩+β|1⟩) + |Φ⁻⟩(α|0⟩−β|1⟩) + |Ψ⁺⟩(β|0⟩+α|1⟩) + |Ψ⁻⟩(β|0⟩−α|1⟩) ] on qubit 3. Recognize the four qubit-3 states as I|ψ⟩, Z|ψ⟩, X|ψ⟩, and −ZX|ψ⟩ respectively, reproducing equation (4). Why: each branch is |ψ⟩ acted on by a definite Pauli, so a Bell measurement that names the branch simultaneously names the correction — the physics is entirely contained in this regrouping. *For the complex input α = cos30°, β = i·sin30°, each of the four branches has outcome probability 0.2500 and post-correction fidelity 1.000000 with |ψ⟩.*
Go deeper · Derivation B · 2 minNo superluminal signalling and Bob's maximally mixed marginal
The aim is to prove that teleportation cannot carry information faster than light, by showing that Bob's qubit, before the two classical bits arrive, is the maximally mixed state ½I for every input |ψ⟩. If it depended in any way on α or β, Alice could modulate |ψ⟩ and Bob could read the modulation instantly at any distance, contradicting relativity. We compute Bob's reduced state two ways — as an average over the four outcomes he has not yet learned, and by tracing Alice's qubits out directly — and confirm both give ½I independent of the input.
Step 1. List Bob's conditional ensemble. From the identity, outcome j occurs with probability p_j = 1/4 and leaves Bob with Uⱼ|ψ⟩, so before learning j his state is the mixture ρ_B = ¼ Σⱼ Uⱼ|ψ⟩⟨ψ|Uⱼ† over the four Paulis {I, X, Z, ZX}. Why: not yet knowing the outcome, Bob must describe his qubit by the classical average over the equally likely branches.
Step 2. Apply the Pauli twirl. For any single-qubit σ, averaging over the full Pauli set gives ¼(σ + XσX + YσY + ZσZ) = ½I. Since {I, X, Z, ZX} equals {I, X, Z, Y} up to phases that cancel in σ → UσU†, the Step 1 sum collapses to ½I. Why: the Pauli twirl symmetrizes any state into the completely depolarized one — the algebraic engine that erases α and β.
Step 3. Cross-check by direct partial trace. Because |Φ⁺⟩ is maximally entangled, tracing qubits 1 and 2 out of |ψ⟩₁ ⊗ |Φ⁺⟩₂₃ gives qubit 3 the reduced state ½I with no outcome computation at all — equation (6). Why: the marginal of a maximally entangled pair is maximally mixed, and an unentangled qubit 1 cannot change it.
Step 4. Conclude no-signalling. Since ρ_B = ½I regardless of α, β, no measurement Bob performs can distinguish one input from another until the classical bits arrive over a light-speed-limited channel; the instantaneous collapse is real but unobservable. Why: two ensembles with identical density matrices are physically indistinguishable, so the collapse carries zero accessible information. *Both the direct partial trace and the four-outcome average yield ½I; the reduced state of qubit 3 has eigenvalues 0.5, 0.5 for every input.*
Go deeper · Derivation C · 1 minRecovering the input in every one of the four branches
This derivation checks that Bob's correction table actually works — that for each Bell outcome, applying the tabulated Pauli returns |ψ⟩ exactly, up to an irrelevant global phase. It also shows why the two classical bits are indispensable: three of the four branches leave Bob with a state visibly not |ψ⟩, and only knowing the outcome tells him which of four rotations to reverse. We track the worst-looking branch, Ψ⁻, in full, then state the pattern for the rest.
Step 1. Take the branch. From the identity, outcome Ψ⁻ leaves Bob with β|0⟩ − α|1⟩ — not |ψ⟩ = α|0⟩ + β|1⟩, since the amplitudes are swapped and one carries a minus sign. Why: the Bell measurement has kicked Bob's qubit by a random Pauli, here undone by ZX.
Step 2. Apply X. Since X swaps |0⟩ ↔ |1⟩, X(β|0⟩ − α|1⟩) = β|1⟩ − α|0⟩ = −α|0⟩ + β|1⟩. The amplitudes are now in the right slots, with a stray sign on |0⟩. Why: X repairs the basis swap the Ψ⁻ branch imposed; the sign is fixed next.
Step 3. Apply Z. Since Z flips the sign of |1⟩, Z(−α|0⟩ + β|1⟩) = −α|0⟩ − β|1⟩ = −|ψ⟩. The overall minus is a global phase, undetectable, so Bob holds |ψ⟩. Why: the operator ZX inverts the Ψ⁻ kick exactly, and global phases never affect statistics.
Step 4. Repeat the pattern. Φ⁺ leaves α|0⟩+β|1⟩ = |ψ⟩ (apply I); Φ⁻ leaves α|0⟩−β|1⟩ = Z|ψ⟩ (apply Z); Ψ⁺ leaves β|0⟩+α|1⟩ = X|ψ⟩ (apply X). Each Pauli is its own inverse, so the correction is the same Pauli reapplied. Why: the protocol is deterministic given the outcome — the randomness is fully absorbed by the two-bit message, which is why it cannot be dropped. *For the complex input α = cos30°, β = i·sin30°, all four post-correction states have fidelity 1.000000 with |ψ⟩; a Bob who always applied I would succeed only on the Φ⁺ branch, one time in four.*
Limiting cases
The ideal protocol is exact, but pushing on each resource shows what teleportation actually depends on.
| Case | Result | Meaning | ||
|---|---|---|---|---|
| Perfect Bell pair, ideal operations | F = 1 | With a pristine | Φ⁺⟩ and error-free gates and measurement, the output equals the input for every | ψ⟩ — teleportation is exact, not approximate. |
| No entanglement (classical only) | F̄ = 2/3 | The best "measure-and-resend" strategy for an unknown qubit averages only 2/3 over the Haar ensemble; exceeding it certifies genuinely quantum resources. | ||
| Werner resource of purity p | F̄ = (1+p)/2 | Isotropic noise degrades the pair; the output beats the classical 2/3 bound only when p > 1/3, the certification threshold. | ||
| Classical bits lost | ρ_B = ½I | If the two bits never arrive, Bob's qubit is maximally mixed and carries zero information about | ψ⟩ — no-signalling made manifest. | |
| High-dimensional systems (qudits) | d ebits of correction | Teleporting a d-level state needs a maximally entangled qudit pair and a generalized Bell measurement with d² outcomes, corrected by generalized Pauli operators. |
Worked examples
1 · Teleporting a specific state with a complex phase (fully worked)
Alice wants to teleport |ψ⟩ = cos30°·|0⟩ + i·sin30°·|1⟩ = (√3/2)|0⟩ + (i/2)|1⟩, a state with a genuinely complex amplitude on |1⟩. Her Bell measurement returns outcome Ψ⁻. Show that Bob recovers |ψ⟩ exactly, phase and all, and find the probability of that outcome.
Step 1. Identify Bob's conditional state. For outcome Ψ⁻ the identity gives Bob β|0⟩ − α|1⟩ with α = √3/2 ≈ 0.866 and β = i/2 = 0.5i, so Bob holds 0.5i·|0⟩ − 0.866·|1⟩ — manifestly not |ψ⟩.
Step 2. Apply the correction ZX (X first, then Z). X swaps the kets: X(0.5i·|0⟩ − 0.866·|1⟩) = 0.5i·|1⟩ − 0.866·|0⟩ = −0.866·|0⟩ + 0.5i·|1⟩. Then Z flips the |1⟩ sign: Z(−0.866·|0⟩ + 0.5i·|1⟩) = −0.866·|0⟩ − 0.5i·|1⟩.
Step 3. Recognize the result. This equals −(0.866·|0⟩ + 0.5i·|1⟩) = −|ψ⟩, which is |ψ⟩ up to a global phase of −1. The complex amplitude i/2 survived intact — teleportation transports relative phase, not just probabilities. Output fidelity = 1.
Step 4. Find the outcome probability. The identity carries a prefactor 1/2 on every branch, so P(Ψ⁻) = |1/2|² = 1/4 = 0.25, independent of α and β — as for all four outcomes, which is why Alice's result reveals nothing about the state she sends.
2 · Certifying entanglement through a noisy channel
A lab teleports qubits through a shared pair modeled as a Werner state of purity p = 0.8. Compute the average fidelity, decide whether the run certifies genuine entanglement, and find the minimum p that would.
Step 1. Apply the Werner fidelity law. From equation (8) with F̄ = (1+p)/2, a purity p = 0.8 gives F̄ = (1 + 0.8)/2 = 0.90.
Step 2. Compare to the classical ceiling. The best entanglement-free measure-and-resend strategy averages 2/3 ≈ 0.667. Since 0.90 > 0.667, the demonstration beats the classical bound and certifies that genuine quantum entanglement carried the state — no shared classical correlation could reach 0.90.
Step 3. Find the threshold purity. Setting (1+p)/2 = 2/3 gives 1 + p = 4/3, so p = 1/3 ≈ 0.333. Any Werner pair with p > 1/3 certifies entanglement through teleportation; below it, the channel is classically simulable.
Step 4. Sanity-check the endpoints. At p = 1 (a perfect Bell pair) F̄ = 1, exact teleportation; at p = 0 (maximally mixed, no correlation) F̄ = 0.5, the fidelity of blind guessing. The measured 0.90 sits comfortably in the quantum regime.
Physics of the portrait
Teleportation is the moment quantum information science turned a paradox into a protocol, its lineage running through the twentieth century's deepest argument on reality.
Origins · 1 minquantum teleportation and how this was discovered
Quantum teleportation was born from the very entanglement that Einstein distrusted. In 1935 Einstein, Podolsky and Rosen used correlated pairs to argue that quantum mechanics must be incomplete, coining the "spooky action at a distance"; in 1964 John Bell turned their thought experiment into a testable inequality, and by the 1980s experiments had sided decisively with quantum mechanics. Entanglement was real — but what was it good for? The answer came in March 1993, in a Physical Review Letters paper with an unusually crowded byline: Charles Bennett, Gilles Brassard, Claude Crépeau, Richard Jozsa, Asher Peres and William Wootters. Bennett and Brassard had already co-invented quantum cryptography (BB84) in 1984; Wootters, with Zurek, had proved the no-cloning theorem in 1982. Together they showed that the correlations EPR had wielded against quantum mechanics were in fact a usable channel — provided you also sent two ordinary classical bits. That two-bit classical side-channel is the conceptual heart of the result: it is what keeps teleportation from signalling faster than light and from cloning, and why the protocol respects relativity while doing something apparently magical. Within four years the protocol left the blackboard — Zeilinger's Innsbruck group teleported photon polarization in 1997 — and within a decade it was running deterministically in trapped ions and over a thousand kilometres to a satellite.
Going further
Texts. Nielsen & Chuang, Quantum Computation and Quantum Information (10th anniv. ed.), §1.3.7 and §12.6 · Preskill, Lecture Notes for Physics 219, Ch. 4 · Watrous, The Theory of Quantum Information, §6.4 · Wilde, Quantum Information Theory (2nd ed.), Ch. 6 · Mermin, Quantum Computer Science, Ch. 6 · Kitaev, Shen & Vyalyi, Classical and Quantum Computation, Ch. 11. The teleportation identity gives four Pauli corrections |Φ⁺⟩→I, |Φ⁻⟩→Z, |Ψ⁺⟩→X, |Ψ⁻⟩→ZX, each with probability 1/4 and post-correction fidelity 1 (confirmed numerically for a complex-amplitude input). The no-signalling result ρ_B = ½I is the Pauli-twirl identity ¼Σ_U UρU† = ½I. The classical bound 2/3 and F̄ = (1+p)/2 follow from F̄ = (2F_e+1)/3 with F_e = (1+3p)/4, crossing 2/3 at p = 1/3. Original protocol: Bennett, Brassard, Crépeau, Jozsa, Peres & Wootters, Phys. Rev. Lett. 70, 1895 (1993); first photonic demonstration: Bouwmeester et al., Nature 390, 575 (1997); satellite: Ren et al., Nature 549, 70 (2017); network: Hermans et al., Nature 605, 663 (2022).
Prerequisites. Bell-State Measurement · No-Cloning Theorem · Density Matrices and Partial Trace · Pauli Operators. Leads to. Superdense Coding · Entanglement Swapping · E91 Protocol · Gate Teleportation.
Ask the tutor: Want to walk through the Bell-basis regrouping term by term, or see how gate teleportation turns a magic state into a fault-tolerant T gate?
Does teleportation move the qubit faster than light?
No usable information moves faster than light. The instant Alice completes her Bell measurement, Bob's half of the pair does change — but into a state that is maximally mixed (½I) for every possible input, so Bob cannot tell anything has happened, let alone learn |ψ⟩. Only when Alice's two classical bits arrive, over a channel limited to the speed of light, can Bob apply the right correction and reconstruct the state. The classical message is the bottleneck, and it obeys relativity. So while the entanglement produces an instantaneous correlation, the protocol as a whole delivers the qubit no faster than a classical signal could travel, which is exactly why teleportation does not let you build a superluminal telephone.
Doesn't teleportation clone the state, violating no-cloning?
No — it is careful not to. Alice's Bell measurement destroys the original: after it, qubit 1 is entangled with qubit 2 and retains no trace of the amplitudes α and β. At every instant there is exactly one copy of |ψ⟩ — first at Alice's location, then, after the correction, at Bob's — and never two at once. This is the crucial difference between teleportation and a photocopier or a fax, both of which leave the original intact. Teleportation is more like a transporter that vaporizes you as it rebuilds you elsewhere: the reconstruction cannot begin to be usable until the original is already gone. If the input somehow survived, you would have cloned it, and the protocol would violate the no-cloning theorem — so the destruction is not a side effect, it is a requirement.
Why exactly two classical bits — could one, or zero, ever work?
Two is forced by the arithmetic. Alice's Bell measurement has four equally likely outcomes, and each leaves Bob's qubit rotated by a different one of four Paulis {I, Z, X, ZX}. To undo the rotation, Bob must know which of four cases he is in, and distinguishing four possibilities requires log₂4 = 2 bits. With one bit he could resolve only two of the four and would fail half the time; with zero bits his state is ½I and he fails three times in four. You cannot do better, either: sending fewer than two bits would let some information about the outcome leak through with less than perfect reliability, and a careful accounting shows two noiseless classical bits is the exact, irreducible classical cost of teleporting one qubit.
What is a Bell-state measurement, and why is it the hard part in practice?
A Bell-state measurement is a joint projective measurement of two qubits onto the four maximally entangled Bell states, rather than measuring each qubit separately. It entangles Alice's unknown qubit with her half of the pair and reads out which of four correlated outcomes occurred. It is demanding because it requires a coherent two-qubit interaction — you cannot do it by measuring the qubits one at a time. With linear optics alone a Bell measurement distinguishes only two of the four states, capping success at 50 percent unless you add nonlinearity or ancillary photons; in matter systems (ions, atoms, superconducting qubits) a full deterministic Bell measurement is possible but needs high-fidelity entangling gates. This is why photonic teleportation was long probabilistic while trapped-ion teleportation became deterministic.
How does teleportation power fault-tolerant quantum computing?
Through gate teleportation. Instead of teleporting a state unchanged, you run the protocol through a specially prepared "magic" resource state, and the output emerges with a desired gate already applied. This matters because in fault-tolerant codes the expensive operations are the non-Clifford gates such as T; gate teleportation lets you prepare and purify a magic state offline, then consume it through a cheap teleportation-like circuit to enact the gate on your data. Measurement-based (one-way) quantum computation is teleportation chained across a large cluster state: every step is a teleport-and-correct. The identity behind moving a qubit across a room is, reused, the identity behind performing a computation.
Can teleportation move more than a single qubit, or entangled states?
Yes, on both counts. Because the protocol is linear in the input, it teleports superpositions, mixed states, and halves of entangled states without modification. Teleporting one member of an entangled pair leaves that pair's partner entangled with Bob's qubit instead — this is entanglement swapping, by which two parties who never interacted come to share entanglement. Running many single-qubit teleportations in parallel moves multi-qubit registers, and the scheme generalizes to d-level systems (qudits) using a maximally entangled qudit pair, a generalized Bell measurement with d² outcomes, and generalized Pauli corrections. The single-qubit case is the atom of a much larger construction underlying quantum networks.