## Tuesday, 10 September 2013

### Quantum Teleportation

Suppose Alice has a qubit in an arbitrary state $\left|\psi\right>=\alpha\left|0\right>+\beta\left|1\right>$, and Bob wants to have an identical qubit $\left|\psi\right>$ too.  If Alice knew enough information about the state $\left|\psi\right>$ to either completely specify the numbers $\alpha$ and $\beta$, then these can be communicated to Bob. This would be worthwhile if Bob knew a unitary operator $U$ such that, say,  $U\left|0\right>=\left|\psi\right>$ that allows Bob to prepare the state. If the exact state $\left|\psi\right>$ is unknown to Alice, then the situation is complicated because there is no amount of measurements Alice could make to the state $\left|\psi\right>$ that provide sufficient information to uniquely determine the state $\left|\psi\right>$ since any measurement made to $\left|\psi\right>$ would only return some computational basis state. If Alice is willing to give up her qubit there is the option of physically sending the qubit to Bob if Bob is spatially displaced. This method entails a safe means of sending $\left|\psi\right>$ through some quantum channel without any error, which may be rather difficult in practice. If the means of communicating quantum mechanically does not exist than the only alternative Alice and Bob have is to rely on classical means of communication. Fortunately, communicating classical bits is fairly easy to do in practice.

The quantum teleportation protocol allows two parties to send a qubit in an unknown state $\left|\psi\right>$ without actually physically relocating the original qubit, and only having to communicating two classical bits of information. The protocol relies on the essential ability of the two parties involved to share an entangled Bell state between them.

Suppose Alice possesses an unknown state $\left|\psi\right>=\alpha\left|0\right>+\beta\left|1\right>$ and shares an entangled pair of qubits in the Bell state $\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>)$ with Bob. In the quantum teleportation protocol, Alice first performs a Bell measurement on her two qubits of the register. This will yield information of some basis state $\left|\beta_{ab}\right>$, or equivalently some state $\left|ab\right>$ in the computational basis. Using the information from what state Alice observes, Alice communicates  two classical bits $ab$ to Bob who then performs the operation $Z^aX^b$ on his qubit based off the values of the bits $ab$  to turn it into the desired state $\left|\psi\right>$.

(A circuit for quantum teleportation.  The top two wires represent Alice's register, which begin with  some qubit in an unknown state $\left|\psi\right>$, and one qubit of a Bell state $\left|\beta_{00}\right>$ shared with Bob's register represented by the bottom wire. Alice first performs a Bell measurement obtaining a state  $\left|ab\right>$, and then communicates two classical bits $ab$ to Bob who performs the operation $X^bZ^a$ that transforms his qubit to the state $\left|\psi\right>$.)

In the circuit for the quantum teleportation protocol the top two wires represent Alice's register and the bottom wire represents Bob's register. Note that the operations and measurement that Alice performs constitutes a Bell measurement. At the start of the protocol the state representing the whole system is
$\left|\psi_0\right>=\left|\psi\right>\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\alpha\left|0\right>+\beta\left|1\right>)(\left|00\right>+\left|11\right>),$or when expanded
$\left|\psi_0\right>=\frac{1}{\sqrt2}(\alpha\left|000\right>+\alpha\left|011\right>+\beta\left|100\right>+\beta\left|111\right>),$
Since the first two qubits in the register represent Alice's system and the third qubit represents Bob's register the states are grouped with kets accordingly in order to make this more explicit in the notation as follows
$\left|\psi_0\right>=\frac{1}{\sqrt2}(\alpha\left|00\right>\left|0\right>+\alpha\left|01\right>\left|1\right>+\beta\left|10\right>\left|0\right>+\beta\left|11\right>\left|1\right>).$
Then the state after the controlled-NOT operation where the first qubit serves as the control and the second qubit as the target is
$\left|\psi_1\right>=\frac{1}{\sqrt2}(\alpha\left|00\right>\left|0\right>+\alpha\left|01\right>\left|1\right>+\beta\left|11\right>\left|0\right>+\beta\left|10\right>\left|1\right>),$
and after the Hadamard gate is applied to the first qubit the state becomes
$\left|\psi_2\right>=\frac{1}{2}(\alpha(\left|00\right>+\left|10\right>)\left|0\right>+\alpha(\left|01\right>+\left|11\right>)\left|1\right>+\beta(\left|01\right>-\left|11\right>)\left|0\right>+\beta(\left|00\right>-\left|10\right>)\left|1\right>),$
which can be regrouped as
$\left|\psi_2\right>=\frac{1}{2}(\left|00\right>(\alpha\left|0\right>+\beta\left|1\right>)+\left|01\right>(\alpha\left|1\right>+\beta\left|0\right>)+\left|10\right>(\alpha\left|0\right>-\beta\left|1\right>)+\left|11\right>(\alpha\left|1\right>-\beta\left|0\right>)).$
Examining Bob's register and noticing that
$\begin{array}{l l} \alpha\left|0\right>+\beta\left|1\right> = \left|\psi\right>, \\ \alpha\left|1\right>+\beta\left|0\right> = X\left|\psi\right>, \\ \alpha\left|0\right>-\beta\left|1\right> = Z\left|\psi\right>, \\ \alpha\left|1\right>-\beta\left|0\right> = XZ\left|\psi\right>, \\ \end{array}$
allows the state before measurement to be expressed as
$\left|\psi_2\right>=\frac{1}{2}(\left|00\right>\left|\psi\right>+\left|01\right>X\left|\psi\right>+\left|10\right>Z\left|\psi\right>+\left|11\right>XZ\left|\psi\right>),$
or more concisely as
$\left|\psi_2\right>=\frac{1}{2}\displaystyle\sum\limits_{i,j\in\{0,1\}}\left|ij\right>X^jZ^i\left|\psi\right>.$
This is an entangled state! Therefore, when Alice measures the first two qubits where some basis state $\left|ab\right>$ will be observed with equal probability,  Bob's qubit will be left in the state $X^bZ^a\left|\psi\right>$. When Alice communicates the two bits $ab$ that result as the outcome of the measurement, Bob can perform the  inverse operation $Z^aX^b$ on the state $X^bZ^a\left|\psi\right>$ turning the qubit in his register to the desired state $Z^aX^bX^bZ^a\left|\psi\right>=\left|\psi\right>$ thereby completing the protocol.

The state of the joint system at the end of the protocol is $\left|\psi_3\right>=\left|ab\right>\left|\psi\right>$, which leaves Bob's register in the state $\left|\psi\right>$. Its important to note that as a consequence of this protocol Alice no longer possess the state $\left|\psi\right>$. The original register that did contain $\left|\psi\right>$ has turned into the state $\left|a\right>$ after the measurement was performed. If this were not the case, and the joint system did happen to contain two copies of $\left|\psi\right>$, then the proposed protocol would in fact contradict the no-cloning theorem. Moreover, it may seem that this teleportation protocol is in violation of the theory of special relativity, which prohibits information to be exchanged faster than the speed of light. In this regard, the protocol is actually void of contradiction because the state $\left|\psi\right>$ is not effectively transmitted to Bob until after he receives the information of the two bits $ab$ from Alice. Since these bits are communicated classically, they are indeed constrained from traveling faster than the speed of light avoiding any violation of the principles of special relativity.

The quantum teleportation protocol is another prime example of how entanglement can be used as a resource. In this sense, it can be said that a shared entangled pair and  the exchange of two bits of classical information is equatable to the exchange of one quantum bit of information.