Quantum teleportation

Quantum teleportation is the most famous result in quantum information, and it comes with more cultural baggage than any other topic in this course. Let’s clear up two things upfront:

1. It’s not faster than light. Teleportation requires Alice to send Bob two classical bits. Those bits travel at most at the speed of light. Before they arrive, Bob’s qubit is in a useless, random state.
2. No matter is transported. The qubit that ends up in Bob’s hands is the one he had all along. What’s “teleported” is the state — the complete description of the qubit. Alice’s original qubit is destroyed in the process.

With those out of the way: what quantum teleportation actually does is astonishing. It’s the fact that a quantum state — which in general requires infinite classical bits to describe exactly — can be transferred from Alice to Bob using only two classical bits, plus a shared Bell pair.

The setup

• Alice has a qubit in an unknown state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$. She does not know $\alpha$ or $\beta$. She cannot measure to find out (measuring would destroy the superposition).
• Alice also has half of a Bell pair $|\Phi^+\rangle$, prepared in advance. The other half is with Bob.
• Alice can send classical information to Bob.
• Alice wants Bob to end up with a qubit in state $|\psi\rangle$.

The protocol

Let’s call Alice’s message qubit $A$, Alice’s Bell half $B$, and Bob’s Bell half $C$. The joint state is:

$|\psi\rangle_A \otimes |\Phi^+\rangle_{BC} = (\alpha|0\rangle_A + \beta|1\rangle_A) \otimes \frac{|00\rangle + |11\rangle}{\sqrt{2}}$

Step 1: Alice applies $CNOT_{A \to B}$.

This entangles her message qubit with her Bell half. The state becomes:

$\frac{1}{\sqrt{2}}\big(\alpha(|000\rangle + |011\rangle) + \beta(|110\rangle + |101\rangle)\big)$

where I’m writing kets in $|ABC\rangle$ order.

Step 2: Alice applies $H$ to $A$.

$\frac{1}{2}\big( |00\rangle_{AB}(\alpha|0\rangle_C + \beta|1\rangle_C) + |01\rangle_{AB}(\alpha|1\rangle_C + \beta|0\rangle_C) + |10\rangle_{AB}(\alpha|0\rangle_C - \beta|1\rangle_C) + |11\rangle_{AB}(\alpha|1\rangle_C - \beta|0\rangle_C)\big)$

Notice that the state has very suggestively separated into four pieces, each with a different computational-basis state on qubits $A$ and $B$ and a specific state on qubit $C$. Each version of Bob’s qubit is related to $|\psi\rangle$ by a simple Pauli transformation.

Step 3: Alice measures both of her qubits.

She gets two classical bits $(a, b)$, each $0$ or $1$. The measurement collapses the state to one of the four branches above, picking Bob’s qubit into the corresponding form:

| $(a, b)$ | Bob’s qubit | Correction to get $|\psi\rangle$ | |---|---|---| | $(0, 0)$ | $\alpha\|0\rangle + \beta\|1\rangle = \|\psi\rangle$ | $I$ (nothing) | | $(0, 1)$ | $\alpha\|1\rangle + \beta\|0\rangle$ | $X$ | | $(1, 0)$ | $\alpha\|0\rangle - \beta\|1\rangle$ | $Z$ | | $(1, 1)$ | $\alpha\|1\rangle - \beta\|0\rangle$ | $X$ then $Z$ |

Step 4: Alice sends $(a, b)$ to Bob classically. Phone call, email, carrier pigeon — whatever works. Only two bits.

Step 5: Bob applies the correction. Based on the table: $X^b$ followed by $Z^a$. After the correction, Bob’s qubit is in state $|\psi\rangle$. Teleportation complete.

Try it

Pick an initial state on the left by dragging $\theta$ and $\varphi$. Step through the protocol. At step 4, Bob’s qubit should be showing the same state you picked at the start. That state has been teleported — using only the classical bits $a$ and $b$ as communication.

What just happened

Reflect on this for a moment. Alice had a qubit containing an arbitrary amount of classical information — the amplitudes $\alpha$ and $\beta$ can be any complex numbers satisfying $|\alpha|^2 + |\beta|^2 = 1$. That’s infinitely many possible states (technically: a continuous 2-parameter family).

Alice sent Bob 2 classical bits. Just $(a, b)$ — four possible values. That’s it.

Somehow, that was enough for Bob to reconstruct the full qubit state exactly. Where did the “infinite” classical information come from?

The answer: it was already there, hidden in the shared Bell pair. The entanglement between $B$ and $C$ was a pre-existing resource holding correlations that, combined with Alice’s two classical bits, locate the specific state Alice wanted to send. Without the Bell pair, 2 classical bits are just 2 bits — you cannot teleport a qubit. Without the classical bits, the Bell pair is just noise — you cannot teleport a qubit. Together, they let you do something neither can alone.

This is the resource theory view of quantum information: entanglement is a resource, classical bits are a resource, and specific protocols consume specific combinations of these resources to do non-classical things.

The no-cloning theorem is safe

You might worry: “if Bob can reconstruct Alice’s state exactly, isn’t that cloning?” No — because Alice’s original qubit is destroyed in the process. Her measurement collapses it to a definite value, losing the original superposition. After the protocol, there is exactly one qubit in the state $|\psi\rangle$: Bob’s. Alice no longer has the state. This is why teleportation is consistent with the no-cloning theorem.

In fact, teleportation is essentially the only way to move an unknown quantum state between two distant parties. You can’t copy it (no-cloning) and you can’t measure it and send the result (measurement destroys information). Teleportation works because the shared Bell pair lets Alice “teleport” the state without learning what it was.

What’s next

You have now seen the two most famous entanglement-powered protocols: superdense coding (classical bits via quantum channel) and teleportation (quantum states via classical channel). Module 7 turns to the big question: what computations can you do with all of this that classical computers can’t match? We start with a toy example called the Deutsch-Jozsa problem, which was the first demonstration that quantum computers could beat classical ones — then build up through Grover’s search, the QFT, and Shor’s algorithm.