Aspiring Physicist. Studying a Maths and Philosophy degree at Durham and trying to fix payroll at Onfolk. Previously building a better bank at Monzo.
by Charles Thomas
One interesting application of Quantum Computing is the ability to send information using fewer bits. This is known as superdense coding.
Specifically, super dense coding allows us to send 2 classical bits of information by sending one qubit.
To see how this works let’s go through an example.
Alice and Bob want to be able to communicate. So a third party (Charlie) takes two qubits \(\ket{a}\) and \(\ket{b}\). And then creates a combined system of them:
\[\ket{a} \otimes \ket{b}\]He then prepares the system in a special state:
\[\ket{a} \otimes \ket{b} = \frac{\ket{00} + \ket{11}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix}\]Afterwards, he sends one qubit to Alice and the other to Bob.
We can transform a qubit by applying a unitary transformation to it. Mathematically, we multiply the vector representing the state of the system by a unitary matrix.
We can build transformations for multi-qubit systems out of transformations for single qubits.
Let A and B be transformations on individual qubits. Then the transformation that applies A to the first qubit and B to the second qubit is denoted \(A \otimes B\)
The value of \(A \otimes B\) is calculated by multiplying each element of A by the matrix B e.g.
\[\begin{bmatrix}a & b \\ c & d\end{bmatrix} \otimes \begin{bmatrix}e & f \\ g & h\end{bmatrix} = \begin{bmatrix}a \begin{bmatrix}e & f \\ g & h\end{bmatrix} & b \begin{bmatrix}e & f \\ g & h\end{bmatrix} \\ c \begin{bmatrix}e & f \\ g & h\end{bmatrix} & d \begin{bmatrix}e & f \\ g & h\end{bmatrix}\end{bmatrix}=\begin{bmatrix}ae & af & be & bf \\ag & ah & bg & bh \\ ce & cf & de & df \\ cg & ch & dg & dh \end{bmatrix}\]Alice wants to send 2 classical bits of information. 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 transform on her qubit before sending it to Bob. (Since Bob does nothing this is represented by the identity transform I)
Now if Alice wants to send 00 she does nothing so the system stays in the state:
\[\frac{\ket{00} + \ket{11}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix}\]If Alice wants to send 01 then she applies the transform X (this is also known as the Quantum NOT gate).
\[X = \begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}\] \[(X \otimes I) = \begin{bmatrix}0 & I \\ I & 0 \end{bmatrix}\] \[(X \otimes I)(\ket{a} \otimes \ket{b}) = \begin{bmatrix}0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}0 \\ 1 \\ 1 \\ 0 \\\end{bmatrix} = \frac{\ket{10} + \ket{01}}{\sqrt{2}}\]If Alice wants to send 10 then she applies the transform Z.
\[Z = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}\] \[(Z \otimes I) = \begin{bmatrix}I & 0 \\ 0 & -I \end{bmatrix}\] \[(Z \otimes I)(\ket{a} \otimes \ket{b}) = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ -1 \\\end{bmatrix} = \frac{\ket{00} - \ket{11}}{\sqrt{2}}\]If Alice wants to send 11 then she applies the transform iY.
\[iY = i\begin{bmatrix}0 & -i \\ i & 0 \end{bmatrix}=\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}\] \[(iY \otimes I) = \begin{bmatrix}0 & I \\ -I & 0 \end{bmatrix}\] \[(iY\otimes I)(\ket{a} \otimes \ket{b}) = \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\\end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \\\end{bmatrix} = \frac{\ket{01}-\ket{10}}{\sqrt{2}}\]In quantum mechanics, if states are orthogonal, it means you can do a measurement to distinguish between them with certainty.
For example, \(\begin{bmatrix}0 \\ 1\end{bmatrix}\) and \(\begin{bmatrix}1 \\ 0\end{bmatrix}\) so if I am given a qubit in one of these two states I can always work out which one I have.
However, if I am given a qubit and told it is either in the state: \(\begin{bmatrix}0.3 \\ 0.7\end{bmatrix}\) or \(\begin{bmatrix}0.6 \\ 0.4\end{bmatrix}\) there is no way for me to work out which one I have been given.
Since the four possible states of the system above form an orthogonal basis it is possible to distinguish between these four states. Therefore, Bob can measure which of the four states he has with certainty and work out what he was sent.
Bob can do the measurement above because the four states form an orthogonal basis. They form a special basis called the Bell Basis. But it is possible to put these back in the basis we’re used to (which is called the computational basis).
To do this once Bob has both qubits, does two transforms on the qubits, a controlled not (CNOT) followed by a Hadamard gate.
The controlled not (CNOT) gate is a Quantum gate that works on two qubits. Its name comes from the fact that if the first qubit (known as the control bit) is 0 it does nothing to the second qubit. But if the first qubit is 1 then it negates the second qubit.
It is represented by the matrix: \(\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{bmatrix}\)
The Hadamard gate is a common gate used in Quantum computing. It is represented as the following matrix.
\[H = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}\] \[H \otimes I = \frac{1}{\sqrt{2}}\begin{bmatrix}I & I \\ I & -I \end{bmatrix}=\begin{bmatrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\\end{bmatrix}\]This matrix will convert a vector in the Bell basis to the computational basis.
tags: