Chapter 09

Teleportation and dense coding, step by step

Alice holds a qubit in a state she does not know, and shares a Bell pair with Bob. What must she send him so that he ends up with that state, and why can he not have it before her message arrives?

Before you start: Circuits and the controlled-NOT (Chapter 4); averaging over unread outcomes and reduced states (Chapter 2 and Chapter 3); no-cloning (Chapter 6). Three-register basis labels extend the two-register ones: $|q\,a\,b\rangle$, leftmost first. Notation guide.

Fix the Bell-state convention

Use the four orthonormal states

$$|\beta_{ab}\rangle=(I\otimes X^bZ^a)|\Phi^+\rangle,\qquad a,b\in\{0,1\}.$$

Thus $\beta_{00}=(|00\rangle+|11\rangle)/\sqrt2$, $\beta_{01}=(|01\rangle+|10\rangle)/\sqrt2$, $\beta_{10}=(|00\rangle-|11\rangle)/\sqrt2$, and $\beta_{11}=(|01\rangle-|10\rangle)/\sqrt2$. The order in $X^bZ^a$ means $Z^a$ acts first. In particular $|\beta_{00}\rangle=|\Phi^+\rangle$, the Bell state of earlier chapters. The letter $\beta$ in $\beta_{ab}$ is part of a label; it is unrelated to the amplitude $\beta$ of the input state below.

The preparation circuit $\operatorname{CNOT}(H\otimes I)$ maps $|ab\rangle$ to $|\beta_{ab}\rangle$. Its inverse, CNOT followed by $H$ on the first qubit, maps each Bell state to its two-bit label. A Bell measurement can therefore be implemented by that inverse circuit followed by ordinary computational-basis measurements.

Teleportation: three registers and one unknown input

Alice holds an unknown input $Q$ in $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ and one half $A$ of a shared Bell pair. Bob holds the other half $B$. The figure shows the whole protocol before the algebra.

Registers $Q$, $A$, $B$, in the order used in every basis label below. Alice applies a controlled-NOT from $Q$ to $A$ and then $H$ to $Q$, and measures both, obtaining bits $a$ (from $Q$) and $b$ (from $A$). The double lines are classical: the two bits travel to Bob by an ordinary channel. Bob applies $X$ if $b=1$ and then $Z$ if $a=1$. Bob's unconditioned state, which is all he can use before the bits arrive, is $I/2$ whatever $|\psi\rangle$ is, and it stays $I/2$ after Alice has measured. Conditioned on Alice's outcome $(a,b)$, his qubit is $X^bZ^a|\psi\rangle$. After the correction it is $|\psi\rangle$.

In register order $Q,A,B$, the initial state is

$$|\psi\rangle_Q|\Phi^+\rangle_{AB}=\frac1{\sqrt2}\left(\alpha|000\rangle+\alpha|011\rangle+\beta|100\rangle+\beta|111\rangle\right).$$

$$\begin{aligned}\frac12\big(&\alpha|000\rangle+\alpha|100\rangle +\alpha|011\rangle+\alpha|111\rangle\\ &+\beta|010\rangle-\beta|110\rangle +\beta|001\rangle-\beta|101\rangle\big). \end{aligned}$$

$$\begin{aligned} 10:&\quad \alpha|0\rangle-\beta|1\rangle=Z|\psi\rangle,\\ 11:&\quad -\beta|0\rangle+\alpha|1\rangle=XZ|\psi\rangle. \end{aligned}$$

This is a regrouping of the same eight terms, with no measurement yet performed. It puts the joint state in a form where each possible result on Alice's side has an explicit conditional vector on Bob's side:

Each bracketed Bob vector has norm one, so every two-bit outcome occurs with probability $1/4$, independently of $\alpha,\beta$. If Alice obtains $(a,b)$, Bob holds $X^bZ^a|\psi\rangle$. Alice sends the two classical bits; Bob applies the inverse $Z^aX^b$, meaning $X^b$ first and then $Z^a$.

Teleportation outcomes and recovery
Alice’s bits	Bob before correction	Correction
00	$\|\psi\rangle$	$I$
01	$X\|\psi\rangle$	$X$
10	$Z\|\psi\rangle$	$Z$
11	$XZ\|\psi\rangle$	$ZX$

What Bob has before the message arrives

Without Alice's bits, Bob must average over the four outcomes. For any input density matrix,

$$\rho_B=\frac14\sum_{a,b}X^bZ^a\rho Z^aX^b=\frac I2.$$

To check it, expand $\rho=(I+r_xX+r_yY+r_zZ)/2$: conjugation by the four Paulis cancels each Bloch component. Bob's local state contains no dependence on the unknown input until the classical record permits correction. Teleportation therefore cannot signal faster than the classical channel.

After Alice's measurement, her input register has a definite computational-basis value; it no longer carries $|\psi\rangle$. The shared Bell pair is consumed. The amplitude pair $\alpha,\beta$ survives in Bob’s corrected register, while Alice’s measured registers contain only the two random outcome bits. There is no second remaining register in state $|\psi\rangle$, which is why this transfer is compatible with the no-cloning theorem. Moreover, the protocol preserves entanglement with an external reference: expand a joint input as $\sum_jc_j|j\rangle_R|\psi_j\rangle_Q$ and apply the same identity term by term. Every corrected branch transfers that joint state from $RQ$ to $RB$.

Dense coding: encode a classical label in a Bell state

Now Alice and Bob share $|\Phi^+\rangle$, and Alice wants to transmit two classical bits $(a,b)$. She applies $X^bZ^a$ to her qubit and sends that qubit to Bob. Moving a matrix across this Bell state transposes it: $(M\otimes I)|\Phi^+\rangle=(I\otimes M^T)|\Phi^+\rangle$.

Why a matrix moves across the Bell state as its transpose

To see this, write $|\Phi^+\rangle=\frac1{\sqrt2}\sum_j|j\rangle|j\rangle$; then $\sum_j M|j\rangle|j\rangle=\sum_{i,j}M_{ij}|i\rangle|j\rangle=\sum_i|i\rangle\sum_j (M^T)_{ji}|j\rangle=\sum_i|i\rangle\,M^T|i\rangle$.

For these Paulis the transpose changes at most a global sign, so the joint state is $|\beta_{ab}\rangle$ up to phase.

Bob now possesses both qubits and uses the Bell-decoding circuit to read $a,b$ with certainty. This transmits two classical bits using one transmitted qubit and one previously shared Bell pair. Both resources belong in the statement.

Each transmitted-qubit state on its own is $I/2$, for all four messages. An observer holding only that qubit learns zero bits about the chosen label, not one. The information is in joint correlations with Bob's retained qubit. Chapter 15 will calculate this using the Holevo quantity.

Where this leaves us

Alice must send two classical bits, the results of her two measurements. With them Bob applies one of four corrections and holds $|\psi\rangle$; without them his qubit is described by $I/2$ for every input, so nothing about $|\psi\rangle$ reaches him faster than the bits do. The Bell pair is used up, and Alice is left with two random bits and no copy. Dense coding runs the trade the other way: one transmitted qubit plus a shared Bell pair carries two classical bits.

Both protocols hinge on the same move. Information held in phases and correlations is invisible to a computational-basis measurement until a change of basis, here the Bell-decoding circuit, turns it into definite outcomes. The next chapter uses that move to compute: it writes a function's values into phases and reads a property of the function out with a final change of basis.

Problems

1. Teleport $|\psi\rangle=(|0\rangle+i|1\rangle)/\sqrt2$ in the branch $(a,b)=(1,1)$. Verify the correction.

Worked answer

First apply $Z$ and then $X$, in the order specified by $XZ$:

$$\frac{|0\rangle+i|1\rangle}{\sqrt2}\xrightarrow{Z}\frac{|0\rangle-i|1\rangle}{\sqrt2}\xrightarrow{X}\frac{|1\rangle-i|0\rangle}{\sqrt2}.$$

Bob’s inverse $ZX$ means applying $X$ first. That gives $(|0\rangle-i|1\rangle)/\sqrt2$; applying $Z$ then gives $(|0\rangle+i|1\rangle)/\sqrt2$, the input. The algebraic check is $ZX\,XZ=Z(X^2)Z=Z^2=I$. The two noncommuting factors must be reversed when taking the inverse.

2. Alice sends only the bit $b$, so Bob corrects $X^b$ but does not know $a$. What channel remains?

Worked answer

After Bob corrects $X^b$, his state is $Z^a\rho Z^a$. The unreported bit $a$ is still equally likely to be zero or one, so he must average these two possibilities. Write the input in matrix form and multiply:

$$Z\rho Z=\begin{pmatrix}\rho_{00}&-\rho_{01}\\-\rho_{10}&\rho_{11}\end{pmatrix},\qquad\frac{\rho+Z\rho Z}{2}=\begin{pmatrix}\rho_{00}&0\\0&\rho_{11}\end{pmatrix}.$$

The diagonal entries survive, while the off-diagonal terms cancel. Input $|+\rangle$ therefore becomes $I/2$, whereas $|0\rangle$ remains $|0\rangle$. The missing bit matters for restoring arbitrary coherence, even though it does not matter for a computational-basis input.

References and further reading

Nielsen & Chuang, §1.3.6–1.3.7 (Bell states and teleportation) and §2.3 (dense coding).