8.11: Teleportation as a Quantum Computation

This tutorial works through the following teleportation circuit provided by Gilles Brassard in "Teleportation as a Quantum Computation" (arXiv:quant-ph/9605035v1). The computational methodology employed here is similar to that used in the other teleportation examples given in this series of tutorials.

\[ \begin{matrix} \text{Initial} & ~ & 1 & 2 & 3 & 4 & ~ & 5 & ~ & 6 & 7 & ~ & \text{Final} \\ | \Psi \rangle & \triangleright & \cdots & \cdots & \cdot & \fbox{R} & \cdots & \fbox{S} & \oplus & \fbox{S} & \oplus & \triangleright & \frac{1}{ \sqrt{2}} ( | 0 \rangle + |1 \rangle ) \\ ~ & ~ & ~ & ~ & | ~ & ~ & ~ & ~ & | & ~ & | \\ | 0 \rangle & \triangleright & \fbox{L} & \cdot & \oplus & \cdots & \cdots & \cdots & | & \cdots & | & \triangleright & \frac{1}{ \sqrt{2}} (|0 \rangle + |1 \rangle ) \\ ~ & ~ & ~ & | & ~ & ~ & ~ & | & | & ~ & | \\ | 0 \rangle & \triangleright & \cdots & \oplus & \cdots & \cdots & \cdots & \oplus & \cdot & \fbox{T} & \cdot & \triangleright & | \Psi \rangle \end{matrix} \nonumber \]

The necessary quantum bits or qubit states are:

Base states:

\[ \begin{matrix} 0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & & 1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber \]

Superposition of base states:

\[ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \\ 1 \end{pmatrix} \nonumber \]

where

\[ (| \alpha |)^2 + (| \beta |)^2 = 1 \nonumber \]

Using \( \begin{pmatrix} \sqrt{ \frac{2}{3}} \\ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} 0.816 \\ 0.577 \end{pmatrix} \) as the teleportee (|Ψ>), Brassard's circuit yields the final states shown in the circuit above. In other words |Ψ > is teleported from the first wire on the left to the third wire on the right.

Initial state:

\[ \begin{pmatrix} \sqrt{ \frac{2}{3}} \\ \sqrt{ \frac{1}{3}} \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{2}{3}} \\ 0 \\ 0 \\ 0 \\ \sqrt{ \frac{1}{3}} \\ 0 \\ 0 \\ 0 \end{pmatrix} \nonumber \]

Final state:

\[ \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \\ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{1}{6}} \\ \sqrt{ \frac{1}{12}} \\ \sqrt{ \frac{1}{6}} \\ \sqrt{ \frac{1}{12}} \\ \sqrt{ \frac{1}{6}} \\ \sqrt{ \frac{1}{12}} \\ \sqrt{ \frac{1}{6}} \\ \sqrt{ \frac{1}{12}} \end{pmatrix} \nonumber \]

The identity operator and the following quantum gates are required to calculate the result of the teleportation circuit. L and R are single qubit rotations, S and T are single qubit phase shifts. The other gates (CNOT, CnNOT, and ICnNOT) are well-known in quantum circuitry.

\[ \begin{matrix} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & L = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} & S = \begin{pmatrix}i & 0 \\ 0 & 1 \end{pmatrix} & T = \begin{pmatrix} -1 & 0 \\ 0 & -i \end{pmatrix} \end{matrix} \nonumber \]

\[ \begin{matrix} \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} & \text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} & \text{ICnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber \]

\[ \begin{matrix} \text{Step1 = kronecker(I, kronecker(L, I))} & \text{Step2 = kronecker(I, CNOT)} & \text{Step3 = kronecker(CNOT, I)} \\ \text{Step4 = kronecker(R, kronecker(I, I))} & \text{Step5 = kronecker(S, CNOT)} & \text{Step6 = ICnNOT} \\ \text{Step7 = kronecker(S, kronecker(I, T))} &\text{Step8 = ICnNOT} \end{matrix} \nonumber \]

According to this result |Ψ > has indeed been teleported to the bottom wire on the right, so the goal has been achieved. However, Brassard suggests that measurements on the top wires in the |0 >-|1 > basis are also instructive. There are four possible measurement outcomes: |00>, |01>, |10 > and |11 >. The projection operators for |0 > and |1 > are as follows.

Projection operator for |0>:

\[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \nonumber \]

Projection operator for |1>:

\[ \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \nonumber \]

These calculations also show successful teleportation.

\[ \begin{matrix} \text{Measurement result for qubits x and y} & \text{Final 3-qubit state} \\ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0.816 \\ 0.577 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} & \begin{bmatrix} x & y & z \\ \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \begin{pmatrix} 0.816 \\ 0.577 \end{pmatrix} \end{bmatrix} \\ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0 \\ 0 \\ 0.816 \\ 0.577 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} & \begin{bmatrix} x & y & z \\ \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \begin{pmatrix} 0.816 \\ 0.577 \end{pmatrix} \end{bmatrix} \\ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0.816 \\ 0.577 \\ 0 \\ 0 \end{pmatrix} & \begin{bmatrix} x & y & z \\ \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \begin{pmatrix} 1 \\ 0 \end{pmatrix} & \begin{pmatrix} 0.816 \\ 0.577 \end{pmatrix} \end{bmatrix} \\ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0.816 \\ 0.577 \end{pmatrix} & \begin{bmatrix} x & y & z \\ \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \begin{pmatrix} 0 \\ 1 \end{pmatrix} & \begin{pmatrix} 0.816 \\ 0.577 \end{pmatrix} \end{bmatrix} \end{matrix} \nonumber \]

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