Quantum Teleportation

Quantum teleportation is a protocol by which two parties, Alice and Bob, who share a maximally entangled pair of qubits (an EPR pair or Bell pair), can transfer an arbitrary and unknown quantum state $|\psi\rangle$ from Alice's system to Bob's system. The transfer uses exactly one Bell pair and exactly two classical bits of communication, and it works for any unknown state — Alice does not need to know, and never learns, what $|\psi\rangle$ actually was. The key conceptual move is that Alice never measures her unknown qubit alone. She measures it jointly with her half of the entangled pair, in the Bell basis — the four maximally entangled two-qubit states. This joint measurement entangles the unknown state's information into correlations between her measurement outcome and Bob's remaining qubit. Because Bob's qubit was entangled with Alice's half of the pair, collapsing that pair (via Alice's measurement) instantaneously determines what unitary transformation must be applied on Bob's side to make his qubit equal to $|\psi\rangle$. Alice's two classical bits tell Bob exactly which of four fixed correction operations to apply. Crucially, teleportation does not violate relativistic causality. Before Bob receives the two classical bits, his local qubit's reduced density matrix is the maximally mixed state $I/2$, regardless of what $|\psi\rangle$ is — it carries zero information about $|\psi\rangle$ on its own. Only after the classical message arrives can he apply the correct correction and recover $|\psi\rangle$. The protocol is a beautiful illustration of the interplay between quantum correlation (instantaneous, but not information-bearing by itself) and classical communication (slower-than-light, but information-bearing) — precisely the tension studied structurally through the Bell-CHSH framework in Bell Inequalities and the CHSH Game.

Setup. Alice has qubit $Q_1$ in the unknown state

Alice and Bob pre-share a Bell pair on qubits $Q_2$ (Alice) and $Q_3$ (Bob):

The full three-qubit state is

Step 1 — Alice applies a CNOT then a Hadamard. Alice applies CNOT with $Q_1$ as control and $Q_2$ as target, then a Hadamard on $Q_1$. This is algebraically equivalent to, and implements, a change of basis into the Bell basis for the pair $(Q_1,Q_2)$. Expanding (shown fully in Complexity & Correctness), the joint state becomes

Step 2 — Alice measures $Q_1, Q_2$ in the computational basis, obtaining two classical bits $(m_1, m_2)$, each uniformly random and independent of $\alpha,\beta$. Bob's qubit $Q_3$ instantly collapses to one of four states depending on $(m_1,m_2)$:

Step 3 — Alice sends $(m_1,m_2)$ to Bob over a classical channel.

Step 4 — Bob applies the correction. Bob applies $X^{m_2}Z^{m_1}$ to $Q_3$, recovering exactly $\alpha|0\rangle+\beta|1\rangle = |\psi\rangle$. ``` PROTOCOL Teleport(Alice holds |ψ⟩ on qubit q1; Alice, Bob share Bell pair q2,q3) // Precondition: q2,q3 initialized in state |Φ+> = (|00> + |11>)/sqrt(2) // Alice's local qubits: q1 (unknown state), q2 (her half of Bell pair) // Bob's local qubit: q3 (his half of Bell pair) ALICE: CNOT(control = q1, target = q2) H(q1) m1 <- MEASURE(q1) // computational basis, classical bit m2 <- MEASURE(q2) // computational basis, classical bit SEND(m1, m2) to Bob over classical channel BOB: (m1, m2) <- RECEIVE from Alice if m2 == 1: APPLY X to q3 if m1 == 1: APPLY Z to q3 // q3 is now exactly |ψ> POSTCONDITION: q3 holds |ψ⟩ (Bob's qubit) q1, q2 are in a fixed, ψ-independent post-measurement state (destroyed) ``` Note the bit-to-gate mapping: $m_2$ (Alice's measurement of $q_2$) controls the $X$ correction, and $m_1$ (measurement of $q_1$) controls the $Z$ correction — this ordering matters and is a frequent source of implementation bugs.

Resource cost. Teleportation of one qubit requires: one pre-shared Bell pair (two qubits' worth of entanglement, consumed — it cannot be reused), two classical bits of communication, one two-qubit gate (CNOT) plus one single-qubit gate (H) on Alice's side, two single-qubit measurements, and at most one two-qubit correction (a combination of $X$ and $Z$) on Bob's side. All operations are $O(1)$ — teleportation of a single qubit takes constant time and constant resources, independent of the amplitudes $\alpha,\beta$. Teleporting an $n$-qubit register requires $n$ Bell pairs and $2n$ classical bits, applied qubit-by-qubit (assuming no entanglement between the $n$ qubits requiring joint treatment; if the state is entangled across the register, the same per-qubit protocol still works because it operates on reduced single-qubit projections — teleportation is oblivious to what the rest of the state is doing, by linearity).

Correctness — full derivation. Starting from

apply CNOT$_{1\to2}$. This flips $q_2$ whenever $q_1=1$:

Apply $H$ on $q_1$, using $H|0\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}$, $H|1\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2}$:

Regroup by the four basis states of $(q_1,q_2)$:

Each term factors as (definite 2-bit outcome) $\otimes$ (a state on $q_3$ that is $|\psi\rangle$ up to a fixed, outcome-determined Pauli). This is the entire proof: the equality above is an exact algebraic identity, valid for every $\alpha,\beta$, not an approximation or an average-case claim. Measurement in the computational basis on $q_1,q_2$ simply selects one of the four orthogonal branches with the Born-rule probability $\tfrac14$ each (independent of $\alpha,\beta$, since each branch has coefficient exactly $\tfrac12$ in modulus after factoring out the normalized $q_3$ state), and Bob's residual state is exactly the corresponding row of the correction table. Applying the inverse Pauli exactly cancels the induced error, giving $q_3=\alpha|0\rangle+\beta|1\rangle$ with certainty (probability 1, given the received classical bits). This is why teleportation has perfect fidelity in the ideal, noiseless case — there is no approximation step anywhere in the derivation.

No faster-than-light signaling. Before the classical bits arrive, Bob's reduced density matrix, obtained by tracing out $q_1,q_2$ from $|\Psi'\rangle$, is

because averaging $|\psi\rangle\langle\psi|$ over the group $\{I,X,Z,XZ\}$ (which acts transitively enough on the Bloch sphere) always yields the maximally mixed state, for any $|\psi\rangle$. Since $\rho_3$ is independent of $\psi$, no local measurement Bob could perform before receiving the bits can extract any information about $\psi$ — consistent with relativistic causality.

No cloning violated? No. After the protocol, $q_1$ and $q_2$ are left in a fixed computational basis state ($|m_1 m_2\rangle$) that carries no trace of $\alpha,\beta$. There is exactly one copy of $|\psi\rangle$ in the universe after the protocol (on $q_3$), matching the count before (on $q_1$). No-cloning, which forbids a second copy coexisting with the original, is untouched.

Tradeoffs. The protocol is "free" in qubit-gate count but expensive in entanglement: each teleported qubit burns one Bell pair that must be replenished (typically via Entanglement as a Physical Resource generation and distillation) before the next teleportation. In noisy real hardware, the Bell pair is never perfect, measurements have finite fidelity, and classical communication has latency — so real teleportation fidelity is $<1$ and the speed of delivery is bounded below by the classical round-trip time, not by any exotic faster-than-light effect.

Take a concrete non-trivial input, avoiding symmetric special cases:

Suppose Alice's Bell-basis measurement (CNOT + H + measure) yields outcome $(m_1,m_2) = (1,0)$ — i.e., $Z$-branch. Trace through the identity derived above: the term with $|10\rangle_{12}$ carries $\alpha|0\rangle_3-\beta|1\rangle_3$, so immediately after Alice's measurement Bob's (unnormalized-looking but actually already normalized) qubit is

This is not yet $|\psi\rangle$ — the relative sign is flipped. Alice sends the classical bits $m_1=1,\ m_2=0$ to Bob. Per the protocol, Bob checks: $m_2=0$ so skip $X$; $m_1=1$ so apply $Z$. Since $Z|0\rangle=|0\rangle$ and $Z|1\rangle=-|1\rangle$:

Exact recovery. The table below shows what would have happened for all four possible measurement outcomes, verifying the correction table is consistent for this specific $\alpha,\beta$:

Note each of Alice's four outcomes occurs with probability exactly $\tfrac14$, regardless of $\alpha=\tfrac{1}{\sqrt5}, \beta=\tfrac{2}{\sqrt5}$ — a direct check that Alice's measurement statistics leak no information about $\psi$ to an eavesdropper who only sees the classical bits, since all four outcomes are equiprobable no matter what $\psi$ is.

Quantum teleportation was invented as a theoretical protocol, not observed in nature — it required the specific realization that a Bell-basis measurement, rather than a computational-basis one, is the right joint operation to perform. The six authors of the founding paper combined ideas from quantum cryptography, entanglement theory, and information theory into a single clean circuit.

Quantum teleportation is the elementary transport primitive for any architecture that must move quantum states between physically separated quantum processors. Quantum repeaters, proposed for long-distance quantum networks, chain together teleportation and entanglement swapping to overcome the exponential photon-loss problem of direct fiber transmission. Distributed quantum computing designs use teleportation to move logical qubits between separate chips or ion traps, avoiding the need for direct physical qubit transport. Quantum error correction protocols use teleportation-based gates (notably in the surface code and other measurement-based approaches) so that certain gates can be implemented by consuming pre-prepared entangled resource states — the "magic state" pipeline in fault-tolerant quantum computing has teleportation at its core. Experimentally, satellite-based teleportation experiments (e.g., China's Micius mission) are prototypes for a future global quantum-secure communication backbone. The protocol's mirror image, Superdense Coding, uses the same shared Bell pair to send two classical bits using one qubit of quantum communication — together the two protocols reveal a precise duality between quantum and classical channel capacity, foreshadowing deeper results in Entanglement as a Physical Resource theory about how entanglement, once quantified, can be converted into other forms of communication advantage.

Ask the tutor: Why must Alice's classical bits physically reach Bob before he can recover |ψ⟩, given that the entanglement collapse itself is instantaneous?