Single-qubit gates
Assume we want to build a circuit using only one qubit. In this case, gates are unitary $2\times2$ matrices with complex coefficients. First, we present the simple and fundamental $\textbf{Pauli gates}$.
Pauli gates
These gates are called $X, Y, Z$ as they perform rotations along the $x, y, z$ axes on a special sphere called Bloch sphere, which gives a visual representation of states for single qubits.
$$ X = \begin{bmatrix} 0 & 1\\ 1 & 0 \\ \end{bmatrix} $$$$ Y = \begin{bmatrix} 0 & -i\\ i & 0 \\ \end{bmatrix} $$$$ Z = \begin{bmatrix} 1 & 0\\ 0 & -1 \\ \end{bmatrix} $$Remark
Let’s see how X acts on the computational basis:
$$X\ket{0}= \begin{bmatrix} 0 & 1\\ 1 & 0 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \ket{1} $$and
$$X\ket{1} = \ket{0}.$$So, this gate is the quantum version of the NOT gate.
Hadamard gate and Fourier basis
This gate will be very important in the following topics. It is defined as
$$H=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1 \\ \end{bmatrix}.$$From an algebraic perspective, it is a rotation of $45^\circ$ on the complex plane. Let’s check its action on the computational basis.
$$H\ket{0}=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}= \frac{\ket{0} + \ket{1}}{\sqrt{2}}$$$$H\ket{1}=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1 \\ \end{bmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}= \frac{\ket{0} - \ket{1}}{\sqrt{2}}$$We thus define these two particular states as:
$$\ket{+} = \frac{\ket{0} + \ket{1}}{\sqrt{2}}$$$$\ket{-} = \frac{\ket{0} - \ket{1}}{\sqrt{2}}$$and we call {$\ket{+}, \ket{-}$} the $\textbf{Fourier basis}.$
More on the Fourier basis
- It is an orthonormal basis;
- the change of basis matrix from the computational basis to the Fourier basis is the Hadamard gate (do you recall what we said about unitary transformations and o.n. bases?);
- it has a particular representation on the Bloch sphere.
Hadamard gate as a source of randomness
Remark
The Hadamard gate transforms the states $\ket{0}$ and $\ket{1}$ in the superposition states $\ket{+}$ and $\ket{-}$, which are balanced (i.e. the probabilities are both $\frac{1}{2}$). Keep in mind this property as it will be useful to understand many of the topics, Shor’s algorithm included.
Observe that $H$ is not only invertible, but it is its own inverse. A function such as this is called an involution and every transformation that is self-adjoint and unitary is an involution (prove as an exercise).
Exercise
Show that
$$Z\ket{+} = \ket{-}$$and
$$Z\ket{-} = \ket{+}.$$Circuit notation
An example of visual representation of a circuit is the following:
- The initial state is $\ket{0}$;
- the line is called “qubit register”;
- the qubit passes through an Hadamard gate and its state becomes $\ket{+}$;
- the generic-meter-shaped symbol denotes the measurement: the system collapses to either $\ket{0}$ or $\ket{1}$.
We used $\square$ as a placeholder (it is not standard, it’s just useful) of “$\ket{0}$ or $\ket{1}$, both with probability $\frac{1}{2}$.”
$\textbf{NOTE}:$ every circuit ends with a measurement. In general, the output is not predictable, but one can design the circuit in order to obtain some desired result with high probability.
A question may arise: how can we assume to start with the state $\ket{0}$? It is a result of the so called “preparation”, a physical process aimed to set the qubit in the determined state $\ket{0}$ (and so, also $\ket{1}$ by simply applying the $X$ gate). Of course, we can generalize this and say that the initial state may be assumed as every state which can be obtained starting from $\ket{0}$ and applying a given circuit.