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.