# Teleportation Protocol

The objective of this technique is to transmit one qubit between Alice and Bob by sending two classical bits. However, Alice and Bob must initially share one entangled state.

Alice and Bob perform the following steps:

• Alice and Bob initially share a Bell State $\beta_{00} = \frac{1}{\sqrt{2}}({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})$.

\beta_{00} .

• Alice applies Hadamardto the first of her two qubits, then performs a Bell measurement on both and sends the result of the measurement (2 classical bits) to Bob.

-->

• Alice performs a joint measurement upon the qubit she wants to transmit and her half of the Bell State. This measurement may be performed by applying a CNOT between the qubit to be transmitted and the first one on the pair β00 and then applying Hadamard to the first of her two qubits. Measuring both the qubits in the computational basis completes the Bell measurement.
• Alice then sends the results of the measurements on her two qubits (2 classical bits b1, b2) to Bob.
• Bob applies a transformation upon his qubit, according to the two received bits, based on the following table

Received bits Gate to be applied

00 I

01 X

10 Z

11 ZX

The complete circuit is the following center

where ${\left\vert{\psi}\right\rangle}$ is the qubit to be teleported.

Let's see, let ${\left\vert{\psi}\right\rangle}=\alpha{\left\vert{0}\right\rangle}+\beta{\left\vert{1}\right\rangle}$ then ${\left\vert{\psi}\right\rangle} \otimes \beta_{00}$

$=(\alpha{\left\vert{0}\right\rangle}+\beta{\left\vert{1}\right\rangle})\left(\frac{1}{\sqrt{2}}({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})\right)$

$=\frac{1}{\sqrt{2}}\left(\alpha{\left\vert{0}\right\rangle}({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})+\beta{\left\vert{1}\right\rangle}({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})\right)$

${\;{{CNOT(1,2)} \atop \longrightarrow}\;} \frac{1}{\sqrt{2}}\left(\alpha{\left\vert{0}\right\rangle}({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})+\beta{\left\vert{1}\right\rangle}({\left\vert{10}\right\rangle}+{\left\vert{01}\right\rangle})\right)$

${\;{{H(1)} \atop \longrightarrow}\;} \frac{1}{\sqrt{2}}\left(\alpha\frac{1}{\sqrt{2}}({\left\vert{0}\right\rangle}+{\left\vert{1}\right\rangle})({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})+\beta\frac{1}{\sqrt{2}}({\left\vert{0}\right\rangle}-{\left\vert{1}\right\rangle})({\left\vert{10}\right\rangle}+{\left\vert{01}\right\rangle})\right)$

$=\frac{1}{2}\left({\left\vert{00}\right\rangle}(\alpha{\left\vert{0}\right\rangle}+\beta{\left\vert{1}\right\rangle}) +{\left\vert{01}\right\rangle}(\alpha{\left\vert{1}\right\rangle}+\beta{\left\vert{0}\right\rangle}) +{\left\vert{10}\right\rangle}(\alpha{\left\vert{0}\right\rangle}-\beta{\left\vert{1}\right\rangle}) +{\left\vert{11}\right\rangle}(\alpha{\left\vert{1}\right\rangle}-\beta{\left\vert{0}\right\rangle})\right)$

$=\frac{1}{2}\sum_{b_1b_2=0}^{1}{\left\vert{b_1 b_2}\right\rangle}(X^{b_2}Z^{b_1}){\left\vert{\psi}\right\rangle}$

Therefore, appliyng Zb1Xb2 Bob will obtain the original state ${\left\vert{\psi}\right\rangle}$.