Basic Quantum Computing - Super dense coding

Note: This is a rework of an earlier post. You can check out the original here

We are finally ready to look at some applications of quantum computation. We’re going to start by looking at super-dense coding which allows us to send 2 classical bits of information by sending one qubit.

The Setup

Alice and Bob want to be able to communicate. So a third party (Charlie) takes two qubits and puts them in the state

\[\frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}}\]

Afterwards, he sends qubit A to Alice and qubit B to Bob.

Sending information

Alice wants to send 2 classical bits of information to Bob. That means she wants to send Bob: 00, 10, 01 or 11.

Depending on which of these 4 she wants to send she’ll perform a different transformation on her qubit before sending it to Bob. (Since Bob does nothing this is represented by the identity transformation I)

Sending 00

Now if Alice wants to send 00 she does nothing so the system stays in the state:

\[\frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}}\]

Sending 01

If Alice wants to send 01 then she applies the NOT gate (aka the X gate).

This just swaps the 0 and 1 in her qubit so she gets:

\[\frac{\ket{1}_A\ket{0}_B + \ket{0}_A\ket{1}_B}{\sqrt{2}}\]

Sending 10

If Alice wants to send 10 then she applies the Z gate.

\[Z = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}\]

We can work out the matrix for this using the formula from a couple of lessons ago:

\[(Z \otimes I) = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\\end{bmatrix}\] \[(Z \otimes I)(\frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}})\] \[= \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ -1 \\\end{bmatrix} = \frac{\ket{00} - \ket{11}}{\sqrt{2}}\]

Sending 11

If Alice wants to send 11 then she applies the iY gate (this is a new gate but it’s very similar to the Y gate we’ve seen before).

\[iY = i\begin{bmatrix}0 & -i \\ i & 0 \end{bmatrix}=\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}\] \[(iY \otimes I) = \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\\end{bmatrix}\] \[(iY\otimes I)(\frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}})\] \[=\begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \\\end{bmatrix} = \frac{\ket{01}-\ket{10}}{\sqrt{2}}\]

Bob receives information

Alice now sends her qubit to Bob. Now Bob has both qubits, he applies two gates on the qubits, a controlled not (CNOT) followed by a Hadamard gate.

\[(H \otimes I)(CNOT)\] \[=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\\end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{bmatrix}\] \[=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\\end{bmatrix}\]

Bob now measures both qubits so he gets one of: 00, 01, 10 or 11. So he’s gotten two bits of classical information but Alice only had to send him one qubit.

Worked example

Let’s work through an example.

So we start by making the system:

\[\frac{\ket{0}_A\ket{0}_B + \ket{1}_A\ket{1}_B}{\sqrt{2}}\]

Now we send one to Alice and one to Bob.

Alice now wants to send 01 to Bob. So she applies the NOT gate to her qubit to get:

\[\frac{\ket{1}_A\ket{0}_B + \ket{0}_A\ket{1}_B}{\sqrt{2}}\]

She now sends the qubit to Bob.

Bob now applies the CNOT gate followed by the Hadamard gate to both qubits:

\[=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\\end{bmatrix}\frac{1}{\sqrt{2}}\begin{bmatrix}0 \\1\\1\\0\end{bmatrix}\] \[=\frac{1}{2}\begin{bmatrix}0 \\1+1\\0 \\1-1\end{bmatrix} = \begin{bmatrix}0 \\1\\0 \\0\end{bmatrix}\] \[=\ket{01}\]

So when he measures both qubits he will always get 01 which is what Alice wanted to send.