###### tags: `Quantum Calculation`

# Superdence Coding

In qunatum computational theory, superdence coding is a technique to send information within smaller number of qubits.

This page is about superdence doding procedures, and errors on it.

## Superdence Coding Procedures
Consider Alice who has qubit $\left|0\right>_A$, is sending 2-qubit-infromation to Bob who has qubit $\left|0\right>_B$. Superdence coding is done by four processes: sharing, encoding, sending, and decoding.

### Sharing
$\left|0\right>_A$ and $\left|0\right>_B$ are prepared to be entangled as Bell state $\left|\psi\right>$.
\begin{eqnarray}
\left|\psi\right> = \frac{1}{2}\left( \left|0\right>_ A \left|0\right>_ B + \left|0\right>_ A \left|1\right>_ B \right)
\end{eqnarray}

### Encoding
If Alice is sending $\left|00\right>,\left|01\right>,\left|10\right>,\left|11\right>$, then alice operates $I,X,Z,Y$ to her qubit.

\begin{eqnarray}
00: ~ \left|\psi\right> = \frac{1}{\sqrt{2}}\left( \left|0\right>_ A \left|0\right>_ B + \left|1\right>_ A \left|1\right>_ B \right) \\
01: ~ \left|\psi\right> = \frac{1}{\sqrt{2}}\left( \left|1\right>_ A \left|0\right>_ B + \left|0\right>_ A \left|1\right>_ B \right) \\
10: ~ \left|\psi\right> = \frac{1}{\sqrt{2}}\left( \left|0\right>_ A \left|0\right>_ B - \left|1\right>_ A \left|1\right>_ B \right) \\
11: ~ \left|\psi\right> = \frac{1}{\sqrt{2}}\left( \left|1\right>_ A \left|0\right>_ B - \left|0\right>_ A \left|1\right>_ B \right)
\end{eqnarray}

### Sending
Alice will send 2 qubits through conventional physical media.

### Decoding
Bob is decoding two qubits by measuring his qubit $\left|0\right>_B$ in computational basis, to see which 2 qubits were sent by ALice.

Decoding is done by first applying CNOT on Bob qubit as a target and Alice qubit as the control, then applying Hadamard gate on Alice qubit.

The decoding scheme of all states will be as the following.

\begin{eqnarray}
00: ~ \left|\psi\right> =&& \frac{1}{\sqrt{2}}\left( \left|0\right>_ A \left|0\right>_ B + \left|1\right>_ A \left|1\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|0\right>_ A \left|0\right>_ B + \left|1\right>_ A \left|0\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|+\right>_ A \left|0\right>_ B + \left|-\right>_ A \left|0\right>_ B \right) = \left|0\right>_ A \left|0\right>_ B \\
01: ~ \left|\psi\right> =&& \frac{1}{\sqrt{2}}\left( \left|1\right>_ A \left|0\right>_ B + \left|0\right>_ A \left|1\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|1\right>_ A \left|1\right>_ B + \left|0\right>_ A \left|1\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|-\right>_ A \left|1\right>_ B + \left|+\right>_ A \left|1\right>_ B \right) = \left|0\right>_ A \left|1\right>_ B \\
10: ~ \left|\psi\right> =&& \frac{1}{\sqrt{2}}\left( \left|0\right>_ A \left|0\right>_ B - \left|1\right>_ A \left|1\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|0\right>_ A \left|0\right>_ B - \left|1\right>_ A \left|0\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|+\right>_ A \left|0\right>_ B - \left|-\right>_ A \left|0\right>_ B \right) = \left|1\right>_ A \left|0\right>_ B \\
11: ~ \left|\psi\right> =&& \frac{1}{\sqrt{2}}\left( \left|1\right>_ A \left|0\right>_ B - \left|0\right>_ A \left|1\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|1\right>_ A \left|1\right>_ B - \left|0\right>_ A \left|1\right>_ B \right) \\
\to && \frac{1}{\sqrt{2}}\left( \left|-\right>_ A \left|1\right>_ B - \left|+\right>_ A \left|1\right>_ B \right) = -\left|1\right>_ A \left|1\right>_ B
\end{eqnarray}

Thus we can decode two qubits Alice sent to Bob.

## Errors
When the entangle state shared is rotated from Bell states, namely,
\begin{eqnarray}
\cos\theta \left|0\right>_ A \left|0\right>_ B + \sin\theta \left|1\right>_ A \left|1\right>_ B,
\end{eqnarray}
Bob can only decode Alice qubit with some probability less than one. The probability in succeding can be calculated by the final state which is derived after decoding. The final state is:
\begin{eqnarray}
00: \frac{1}{\sqrt{2}}
(\cos\theta + \sin\theta) \left|0\right>_ A \left|0\right>_ B + \frac{1}{\sqrt{2}}
(\cos\theta - \sin\theta) \left|1\right>_ A \left|1\right>_ B \\
01: \frac{1}{\sqrt{2}}
(\cos\theta + \sin\theta) \left|0\right>_ A \left|1\right>_ B - \frac{1}{\sqrt{2}}
(\cos\theta - \sin\theta) \left|1\right>_ A \left|1\right>_ B \\
10: \frac{1}{\sqrt{2}}
(\cos\theta - \sin\theta) \left|0\right>_ A \left|0\right>_ B + \frac{1}{\sqrt{2}}
(\cos\theta + \sin\theta) \left|1\right>_ A \left|0\right>_ B \\
11: \frac{1}{\sqrt{2}}
(\cos\theta - \sin\theta) \left|0\right>_ A \left|1\right>_ B - \frac{1}{\sqrt{2}}
(\cos\theta + \sin\theta) \left|1\right>_ A \left|1\right>_ B.
\end{eqnarray}

Thus, the probability $p$ for sucessful decoding will be
\begin{eqnarray}
p = \frac{1}{2}
(\cos\theta + \sin\theta)^2.
\end{eqnarray}

Preparing perfect Bell state is necessary for superdence coding.