Quantum computing gates

See wikipedia page on quantum gates for more details.

One qubit

Spin \({ 1/2 }\) representation and conventions

\[\ket{ s = 1/2, {m}_{s} = \pm 1/2 } = \ket{\pm}\]

Sometimes the kets \({ \ket{\pm} }\) are used for the Hadamard transformed basis link, otherwise, the convention is to take \({ \ket{+} = \ket{0} }\) and \({ \ket{-} = \ket{1} }\), so in the matrix representation one has:

\begin{equation} \begin{aligned} & \ket{+} = \ket{0} = \begin{pmatrix} 1 \\\
0 \end{pmatrix} \\\
& \ket{-} = \ket{1} = \begin{pmatrix} 0 \\\
1 \end{pmatrix} \end{aligned} \end{equation}

One-qubit gates

The identity matrix is trivially:

\begin{equation} \begin{aligned} I =
\begin{pmatrix} 1 & 0 \\\
0 & 1 \end{pmatrix} = \ket{0}\bra{0} + \ket{1}\bra{1} \end{aligned} \end{equation}

Pauli matrices are the building blocks for all operations acting on spins \({ 1/2 }\) and so on individual qubits. In matrix representation they are defined as:

\begin{equation} \begin{aligned} {\sigma}_{1} =
\begin{pmatrix} 0 & 1 \\\
1 & 0 \end{pmatrix} = X = \ket{1}\bra{0} + \ket{0}\bra{1} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} {\sigma}_{2} =
\begin{pmatrix} 0 & -i \\\
i & 0 \end{pmatrix} = Y = i \ket{1}\bra{0} -i \ket{0}\bra{1} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} {\sigma}_{3} =
\begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} = Z = \ket{0}\bra{0} - \ket{1}\bra{1} \end{aligned} \end{equation}

The \({ X }\) matrix is also called the NOT gate.

The rotation gates along the \({ X }\), \({ Y }\), and \({ Z }\) axes are defined as:

\begin{equation} \begin{aligned} {R}_{X}(\theta) = {e}^{ -i \frac{\theta}{2} X } = cos\left( \frac{\theta}{2} \right) I - i \sin\left( \frac{\theta}{2} \right) X = \begin{pmatrix} \cos\left( \frac{\theta}{2} \right) & -i \sin\left( \frac{\theta}{2} \right) \\\
-i \sin\left( \frac{\theta}{2} \right) & \cos\left( \frac{\theta}{2} \right) \end{pmatrix} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} {R}_{Y}(\theta) = {e}^{ -i \frac{\theta}{2} Y } = cos\left( \frac{\theta}{2} \right) I - i \sin\left( \frac{\theta}{2} \right) Y = \begin{pmatrix} \cos\left( \frac{\theta}{2} \right) & - \sin\left( \frac{\theta}{2} \right) \\\
\sin\left( \frac{\theta}{2} \right) & \cos\left( \frac{\theta}{2} \right) \end{pmatrix} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} {R}_{Z}(\theta) = {e}^{ -i \frac{\theta}{2} Z } = cos\left( \frac{\theta}{2} \right) I - i \sin\left( \frac{\theta}{2} \right) Z = \begin{pmatrix} {e}^{ -i \frac{\theta}{2} } & 0 \\\
0 & {e}^{ i \frac{\theta}{2} } \end{pmatrix} \end{aligned} \end{equation}

The Hadamard gate \({ H }\) satisfies \({ {H}^{2} = I }\) and \({ {H}^{\dagger} = H }\), and writes:

\begin{equation} \begin{aligned} H = \frac{1}{ \sqrt{2} } \begin{pmatrix} 1 & 1 \\\
1 & -1 \end{pmatrix} \end{aligned} \end{equation}

This gate puts a qubit in a pure state into a mixed state (superposition):

\begin{equation} \begin{aligned} & H \ket{0} = \frac{1}{ \sqrt{2} } \begin{pmatrix} 1 & 1 \\\
1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\\
0 \end{pmatrix} = \frac{1}{ \sqrt{2} } ( \ket{0} + \ket{1} ) \\\
& H \ket{1} = \frac{1}{ \sqrt{2} } \begin{pmatrix} 1 & 1 \\\
1 & -1 \end{pmatrix} \begin{pmatrix} 0 \\\
1 \end{pmatrix} = \frac{1}{ \sqrt{2} } ( \ket{0} - \ket{1} ) \end{aligned} \end{equation}

Two qubits

two-qubit states

The two-qubit states can also be represented as matrices: \(\begin{equation} \ket{00} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad \ket{01} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad \ket{10} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad \ket{11} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{equation}\)

two-qubit gates

The CNOT gate from 1 to 2 is denoted \({ \text{CNOT}_{10} }\) while the CNOT gate from 2 to 1 is denoted \({ \text{CNOT}_{01} }\):

\begin{equation} \begin{aligned} \text{CNOT}_{10} =
\begin{pmatrix} 1 & 0 & 0 & 0 \\\
0 & 1 & 0 & 0 \\\
0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \end{pmatrix} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} \text{CNOT}_{01} =
\begin{pmatrix} 1 & 0 & 0 & 0 \\\
0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \end{pmatrix} \end{aligned} \end{equation}

The gates \({ {X}_{0} {X}_{1} }\), \({ {Y}_{0} {Y}_{1} }\), and \({ {Z}_{0} {Z}_{1} }\) can be obtained with the tensorial product \({ \otimes }\) of the one-qubit matrices:

\begin{equation} \begin{aligned} X_0 \otimes X_1 =
\begin{pmatrix} 0\begin{pmatrix} 0 & 1 \\\
1 & 0 \end{pmatrix} & 1\begin{pmatrix} 0 & 1 \\\
1 & 0 \end{pmatrix} \\\
1\begin{pmatrix} 0 & 1 \\\
1 & 0 \end{pmatrix} & 0\begin{pmatrix} 0 & 1 \\\
1 & 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
1 & 0 & 0 & 0 \end{pmatrix} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} Y_0 \otimes Y_1 =
\begin{pmatrix} 0\begin{pmatrix} 0 & -i \\\
i & 0 \end{pmatrix} & -i\begin{pmatrix} 0 & -i \\\
i & 0 \end{pmatrix} \\\
i\begin{pmatrix} 0 & -i \\\
i & 0 \end{pmatrix} & 0\begin{pmatrix} 0 & -i \\\
i & 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
-1 & 0 & 0 & 0 \end{pmatrix} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} Z_0 \otimes Z_1 =
\begin{pmatrix} 1\begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} & 0\begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} \\\
0\begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} & -1\begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\\
0 & -1 & 0 & 0 \\\
0 & 0 & -1 & 0 \\\
0 & 0 & 0 & 1 \end{pmatrix} \end{aligned} \end{equation}