The Bloch sphere and rotations

Recall that the a complex number $z$ can be written in polar coordinates as:

$$z = re^{i\alpha}$$

where $r = |z|$ is a non-negative real number and $\alpha$ is an angle in $[0, 2\pi]$. Consider, then a qubit in a state $\lvert\psi\rangle = a\lvert0\rangle + b\lvert1\rangle$ and write $a$ and $b$ in polar coordinates as:

$$\begin{array}{ccc} a = r_{1}e^{i\alpha_{1}} &, & b = r_{2}e^{i\alpha_{2}} \end{array}$$

we also know that $r_{1}^{2}+r_{2}^{2}=|a|^{1}+|b|^{2}=1$ and, since $0\leq r_{1}, r_{2}\leq 1$, there must exist a angle $\theta$ in $[0, \pi]$ such that $\text{cos}(\theta/2) = r_{1}$ and $\text{cos}(\theta/2) = r_{2}$. By now, we have,

$$\lvert \psi\rangle = \text{cos}\frac{\theta}{2}e^{i\alpha_{1}}\lvert0\rangle + \text{sin}\frac{\theta}{2}e^{i\alpha_{2}}\lvert1\rangle$$

We can multiply $\lvert\psi\rangle$ by $e^{-i\alpha_{1}}$ to obtain the following:

$$\lvert\psi\rangle = \text{cos}\frac{\theta}{2}\lvert0\rangle + \text{sin}\frac{\theta}{2}e^{i\phi}\lvert1\rangle$$

where we have defined $\phi = \alpha_{2} - \alpha_{1}$. In this way, we can describe the state of any qubit with just two number $\theta \in[0, \pi]$ and $\phi \in [0, 2\pi]$ that we can interpret as a polar angle and an azimuthal angle, respectively.

$$(\text{sin}\theta\text{cos}\phi,\text{sin}\theta\text{sin}\phi,\text{cos}\theta)$$

This gives us a three-dimensional point to locates the state of the qubit on the surface of a sphere, called the Bloch sphere

In the Bloch sphere, $\lvert0\rangle$ is mapped to the north pole and $\lvert1\rangle$ to the South pole. In general, states that are orthogonal with repest to the inner prodcut are antipodal on the sphere. As we already know, the $X$ gate takes $\lvert0\rangle$ to $\lvert1\rangle$ and $\lvert1\rangle$ to $\lvert0\rangle$, but leaves $\lvert+\rangle$ and $\lvert-\rangle$ unchanged. This makes $X$ gate acts like a rotation of $\pi$ radians around the $X$ axis of the Bloch sphere. Imagining you are looking in to a qubit with state $\lvert\psi\rangle$ from the $X$ axis, you wil see the qubits only rotate around the $X$ axis. Therefore, for the $X$, $Y$, and $Z$ axes we may define

$$R_{X}(\theta) = e^{-i\frac{\theta}{2}X} = \text{cos}\frac{\theta}{2}I - i\text{sin}\frac{\theta}{2}Y = \begin{bmatrix} \text{cos}\frac{\theta}{2} & -i\text{sin}\frac{\theta}{2}\\ -i\text{sin}\frac{\theta}{2} & \text{cos}\frac{\theta}{2} \end{bmatrix}$$

$$R_{Y}(\theta) = e^{-i\frac{\theta}{2}Y} = \text{cos}\frac{\theta}{2}I - i\text{sin}\frac{\theta}{2}Y = \begin{bmatrix} \text{cos}\frac{\theta}{2} & -\text{sin}\frac{\theta}{2}\\ \text{sin}\frac{\theta}{2} & \text{cos}\frac{\theta}{2} \end{bmatrix}$$

$$R_{Z}(\theta) = e^{-i\frac{\theta}{2}Z} = \text{cos}\frac{\theta}{2}I - i\text{sin}\frac{\theta}{2}Z = \begin{bmatrix} e^{-i\frac{\theta}{2}} & 0\\ 0 & e^{i\frac{\theta}{2}} \end{bmatrix} \equiv \begin{bmatrix} 1 & 0\\ 0 & e^{i\frac{\theta}{2}} \end{bmatrix}$$

In general form, for any one-qubit gate $U$ there exists a unit vector $r = (r_{x},r_{y},r_{z})$ and the angle $\theta$ such that

$$U \equiv \text{cos}\frac{\theta}{2}I - i\text{sin}\frac{\theta}{2}(r_{x}X,r_{y}Y,r_{z}Z)$$

For instance, if we choose $\theta = \pi$ and $r = (1/\sqrt{2},0,1/\sqrt{2})$ we can obtain the Hadamard gate,

$$H \equiv -i\frac{1}{sqrt{2}}(X+Z)$$

More, the universal one-qubit gate used by IBM, is called the $U$-gate, which depends on three angles and is able to generate any other one-qubit gate,

$$U(\theta, \phi, \lambda) = \begin{bmatrix} \text{cos}\frac{\theta}{2} & -e^{i\lambda}\text{sin}\frac{\theta}{2}\\ e^{i\phi}\text{sin}\frac{\theta}{2} & e^{i(\phi+\lambda)}\text{cos}\frac{\theta}{2} \end{bmatrix}$$

These can be used in Chapter 9, Quantum Support Vector Machines, and Chapter 10, Quantum Neural Networks.