Quantum Teleportation

From Qwiki

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

The steps to get through by Alice and Bob are the following:

Alice and Bob prepare a Bell State $\beta_{00} = \frac{1}{\sqrt{2}}({\left\vert{00}\right\rangle}+{\left\vert{11}\right\rangle})$ .
Alice keeps the first qubit of the pair and Bob takes the second away.
Alice applies CNOT between the qubit to be transmitted and the first one on the pair $β 00$ .
Alice applies Hadamard to the first of her two qubits, then performs a measurement on both and sends the result of the measurement (2 classical bits) 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
$00$	$I$
$01$	$X$
$10$	$Z$
$11$	$Z X$

The complete circuit is the following

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 $Z^{b_1}X^{b_2}$ Bob will obtain the original state ${\left\vert{\psi}\right\rangle}$ .