Why Teleportation Obeys No-Signalling

Quantum teleportation is one of quantum mechanics' most startling results: the exact quantum state of a qubit can be moved from Alice to Bob, even when neither party ever directly exchanges that qubit. Yet despite the "spooky" word, teleportation cannot transmit information faster than light. This lesson explains why — and what role the mandatory classical channel plays.

The teleportation protocol in brief

Alice holds the qubit $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ she wants to send. She and Bob each hold one qubit of a shared Bell pair

$|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr).$

Alice performs two operations on her two qubits (the data qubit and her half of the Bell pair): a CNOT followed by a Hadamard. She then measures both of her qubits and obtains one of four classical outcomes: $00$ , $01$ , $10$ , or $11$ — each occurring with probability $\tfrac{1}{4}$ , regardless of $\alpha$ and $\beta$ . She transmits these 2 classical bits to Bob.

Depending on which bits Bob receives, he applies one of four Pauli corrections to his qubit:

| Alice's result | Bob applies | |:--------------:|:-----------:| | $00$ | $I$ (nothing) | | $01$ | $X$ | | $10$ | $Z$ | | $11$ | $ZX$ |

After the correction Bob's qubit is in exactly the state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ , and Alice's original copy is destroyed by her measurement — consistent with the no-cloning theorem.

Why the classical bits are mandatory

Before Bob receives Alice's 2-bit message his qubit is described by the reduced density matrix obtained by tracing out Alice's pair. Because each measurement outcome is equally likely, that reduced state is

$\rho_B = \tfrac{1}{4}\bigl(|\psi\rangle\!\langle\psi| + X|\psi\rangle\!\langle\psi|X + Z|\psi\rangle\!\langle\psi|Z + ZX|\psi\rangle\!\langle\psi|XZ\bigr) = \tfrac{1}{2}I.$

This is the maximally mixed state — a perfect coin-flip between $|0\rangle$ and $|1\rangle$ , carrying no information about $\alpha$ or $\beta$ whatsoever. No measurement Bob can perform on his qubit tells him anything about $|\psi\rangle$ until the classical bits arrive. This is the no-signalling principle in action: the entangled link alone cannot transmit information, because the correlations only become visible once the classical results are compared.

No faster-than-light communication

Any attempt to skip the classical channel fails. From Bob's perspective, his qubit looks like random noise until the correction index arrives. He cannot distinguish "Alice has measured and the result is $01$ " from "Alice has not yet measured." Superluminal signalling would require Bob to extract $\alpha$ or $\beta$ from the maximally mixed state, which is impossible by the argument above.

Teleportation therefore obeys relativistic causality: useful information travels no faster than the classical message Alice sends, which is bounded by the speed of light. Quantum entanglement provides a resource that makes the transfer possible at all — without the pre-shared Bell pair, no classical channel of any size could transmit an arbitrary qubit state exactly. Together, one shared entangled pair (one ebit) and the 2 classical bits are both necessary and jointly sufficient: teleportation embodies the resource identity $1\ \text{qubit} \equiv 1\ \text{ebit} + 2\ \text{cbits}$ [@nielsen-chuang].