Suppose we have two parties, Alice and Bob, and Alice would like to communicate some information to Bob. More specifically, Alice would like to communicate the values of two classical bits \(a\) and \(b\) to Bob. Quite trivially, Alice can do so by simply sending these two bits of information to Bob, but this requires Alice to send

In the superdense coding scenario, Alice and Bob share the entangled Bell state \[\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>),\] where Alice posses the first qubit in the register and Bob the second. Depending on which of the four possible combinations of two bits \(ab\in\{00,01,10,11\}\) Alice would like to communicate, Alice performs the operation \(Z^aX^b\) to her qubit of the Bell state and sends her qubit to Bob who then performs a Bell measurement on both of the qubits to get the state \(\left|ab\right>\) encoding the two classical bits Alice originally wanted to communicate. The circuit diagram describing this procedure is presented in the figure below, where the top wire of the circuit represents Alice's qubit and the bottom wire represents Bob's. Before the Bell measurement is performed, it is assumed that Alice has somehow provided Bob with her qubit so that Bob can jointly measure both qubits.

(A circuit for superdense coding. To communicate two classical bits \(ab\) to Bob, Alice applies the operation \(X^bZ^a\) to her qubit (represented by the top wire of the circuit) of the shared bell state \(\left|\beta_{00}\right>\), and then allows Bob to perform a joint Bell measurement on both qubits to yield the state \(\left|ab\right>\) encoding the two classical bits \(ab\).)

To observe why the superdense coding circuit accomplishes what its intended to note that applying \(Z^aX^b\) to \(\left|\beta_{00}\right>\), for the different choices of \(ab\in\{00,01,10,11\}\), transforms \(\left|\beta_{00}\right>\) to the other Bell states.

To see this, let \(Z^aX^b\left|\beta_{00}\right>=\left|\psi_1\right>\), then

\[\left|\psi_1\right>=\frac{1}{\sqrt2}(\overline b\left|00\right>+b\left|01\right>+(-1)^ab\left|10\right>+(-1)^a\overline b\left|11\right>)=\left|\beta_{ab}\right>,\]

where \(\overline b=(b+1)mod2\). Thus, by applying a Bell measurement to the state \(\left|\psi_1\right>=\left|\beta_{ab}\right>\), the state \(\left|ab\right>\) will be observed with certainty.

Superdense coding shows that it is possible to send

*two*bits to completely specify both. That is, communicating only a single bit is not sufficient information to determine the state of the second bit---at least classically. Superdense coding is a quantum algorithm or protocol that allows Alice to send two bits of classical information to Bob by only sending a*single quantum bit*. This is accomplished by exploiting the entanglement that exists in the Bell state \(\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>)\).In the superdense coding scenario, Alice and Bob share the entangled Bell state \[\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>),\] where Alice posses the first qubit in the register and Bob the second. Depending on which of the four possible combinations of two bits \(ab\in\{00,01,10,11\}\) Alice would like to communicate, Alice performs the operation \(Z^aX^b\) to her qubit of the Bell state and sends her qubit to Bob who then performs a Bell measurement on both of the qubits to get the state \(\left|ab\right>\) encoding the two classical bits Alice originally wanted to communicate. The circuit diagram describing this procedure is presented in the figure below, where the top wire of the circuit represents Alice's qubit and the bottom wire represents Bob's. Before the Bell measurement is performed, it is assumed that Alice has somehow provided Bob with her qubit so that Bob can jointly measure both qubits.

(A circuit for superdense coding. To communicate two classical bits \(ab\) to Bob, Alice applies the operation \(X^bZ^a\) to her qubit (represented by the top wire of the circuit) of the shared bell state \(\left|\beta_{00}\right>\), and then allows Bob to perform a joint Bell measurement on both qubits to yield the state \(\left|ab\right>\) encoding the two classical bits \(ab\).)

To observe why the superdense coding circuit accomplishes what its intended to note that applying \(Z^aX^b\) to \(\left|\beta_{00}\right>\), for the different choices of \(ab\in\{00,01,10,11\}\), transforms \(\left|\beta_{00}\right>\) to the other Bell states.

To see this, let \(Z^aX^b\left|\beta_{00}\right>=\left|\psi_1\right>\), then

\[\left|\psi_1\right>=\frac{1}{\sqrt2}(\overline b\left|00\right>+b\left|01\right>+(-1)^ab\left|10\right>+(-1)^a\overline b\left|11\right>)=\left|\beta_{ab}\right>,\]

where \(\overline b=(b+1)mod2\). Thus, by applying a Bell measurement to the state \(\left|\psi_1\right>=\left|\beta_{ab}\right>\), the state \(\left|ab\right>\) will be observed with certainty.

Superdense coding shows that it is possible to send

*two bits of classical information*by sending only a*single qubit*. However, this can only be done provided that the qubit being sent is in an entangled state with another qubit. It is also implicit here that one party has the ability to successfully*send*a qubit to another receiving party. This is an example that shows how entanglement can be used a resource to accomplish something that would otherwise be impossible in the classical realm.
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