Quantum Computing #03: Represent single qubit on the Bloch Sphere

Introduction to Quantum Computing series

Hoa Nguyen

Published on Jun 20, 2021

The general state of a qubit $$|q\rangle$$

$$|q\rangle = \alpha|0\rangle + \beta|1\rangle$$

where $$\alpha, \beta \in \mathbb{C}$$ (complex number)

• since we cannot measure global phase, we can only measure the difference in phase between the states $$|0\rangle$$ and $$|1\rangle$$
• Instead of complex number, confine them to the real numbers and add a term to tell us the relative phase between them:

$$|q\rangle = \alpha|0\rangle + e^{i\phi}\beta|1\rangle$$

$$\alpha, \beta, \phi \in \mathbb{R}$$

• since the qubit state must be normalised: $$\sqrt{\alpha^2 + \beta^2} = 1$$

⇒ we can use the trigonometric identity: $$\sqrt{\sin^2{x} + \cos^2{x}} = 1$$ to describe the real $$\alpha$$ and $$\beta$$ in terms of one variable $$\theta$$:

$$\alpha = \cos{\tfrac{\theta}{2}}, \quad \beta=\sin{\tfrac{\theta}{2}}$$

⇒ we can describe the state of any qubit using the two variables $$\phi$$ and $$\theta$$

$$|q\rangle = \cos{\tfrac{\theta}{2}}|0\rangle + e^{i\phi}\sin{\tfrac{\theta}{2}}|1\rangle$$

$$\theta, \phi \in \mathbb{R}$$

Bloch sphere

We can write any normalized (pure) state as $$|\psi\rangle = \cos{\tfrac{\theta}{2}}|0\rangle + e^{i\phi}\sin{\tfrac{\theta}{2}}|1\rangle$$ where $$\phi \in [0, 2\pi)$$ describe the relative phase and $$\theta \in [0, \pi]$$ determines the probability to measure $$|0\rangle , |1\rangle$$: $$p(0) = \cos^2\tfrac{\theta}{2}, p(1) = \sin^2\tfrac{\theta}{2}$$

$$\Rightarrow$$ all normalized pure state can be illustrated on the surface of a sphere with radius $$|\vec{r}|=1$$, which we call the Bloch sphere

$$\Rightarrow$$ the coordinates of such a state are given by the Bloch vector $$\vec{r} = \begin{bmatrix}\sin\theta \cos\theta \\ \sin\theta\sin\phi \\ \cos\theta\end{bmatrix} = \begin{bmatrix}x \\ y \\ z \end{bmatrix}$$

For example:

• $$|0\rangle: \theta=0, \phi$$ is arbitrary $$\Rightarrow \vec{r} =\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}$$

• $$|1\rangle: \theta=\pi, \phi$$ is arbitrary $$\Rightarrow \vec{r} =\begin{bmatrix}0 \\ 0 \\ -1 \end{bmatrix}$$

• $$|+\rangle: \theta=\tfrac{\pi}{2}, \phi =0$$ $$\Rightarrow \vec{r} =\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$$

• $$|-\rangle: \theta=\tfrac{\pi}{2}, \phi =\pi \Rightarrow \vec{r} =\begin{bmatrix}-1 \\ 0 \\ 0 \end{bmatrix}$$

• $$|+i\rangle: \theta=\tfrac{\pi}{2}, \phi = \tfrac{\pi}{2} \Rightarrow \vec{r} =\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$$

• $$|-i\rangle: \theta=\tfrac{\pi}{2}, \phi = \tfrac{3\pi}{2} \Rightarrow \vec{r} =\begin{bmatrix}0 \\ -1 \\ 0 \end{bmatrix}$$

Visually Representing a Qubit State

If we interpret $$\theta$$ and $$\phi$$ as spherical coordinates (r=1, since the magnitude of the qubit state, is 1), we can plot any single-qubit state on the surface of a sphere, known as the Bloch sphere.

• we have plotted a qubit in the state $$|{+}\rangle, \theta = \pi/2, \phi=0$$

An important thing to remember about qubits is that we don't care about overall phase. Hence if we have a qubit of the form: $$|\psi\rangle = e^{i\alpha}a|0\rangle + e^{i\beta}b|1\rangle = e^{i\alpha} \left( a|0\rangle + e^{i(\beta-\alpha)}b|1\rangle \right)$$ ⇒ we find the Bloch present of $$a|0\rangle + e^{i(\beta-\alpha)}b|1\rangle$$

Example: Represent the following qubit on Bloch Sphere $$\tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

Using $$|q\rangle = \cos{\tfrac{\theta}{2}}|0\rangle + e^{i\phi}\sin{\tfrac{\theta}{2}}|1\rangle$$ ⇒ $$\cos{\tfrac{\theta}{2}}= 1/\sqrt 2 , \ e^{i\phi}\sin{\tfrac{\theta}{2}}=1/\sqrt2$$

• $$\cos{\tfrac{\theta}{2}} = \cos{\tfrac{\pi}{4}} = \tfrac{1}{\sqrt2}$$

⇒ $$\theta = \tfrac{\pi}{2}$$

• $$\ e^{i\phi}\sin{\tfrac{\theta}{2}}=1/\sqrt2$$

$$\sin{\tfrac{\theta}{2}} = \sin{\tfrac{\pi}{4}} = \tfrac{1}{\sqrt2}$$

$$e^{i\phi} = 1$$ ⇒ $$\phi =0$$ So that $$x = r \sin(\tfrac{\pi}{2}) \cos(0) \\ y = r \sin(\tfrac{\pi}{2}) \sin(0) \\ z = r * \cos(\tfrac{\pi}{2})$$ $$\Rightarrow$$ Cartesian Bloch Vector = [1.0, 0, 0]

Some common statevectors

Be careful: On the Bloch sphere, angles are twice as big as in Hilbert space e.g. $$|0\rangle , |1\rangle$$ are orthogonal, but on the Bloch sphere, their angle is 180^0. For a general state $$|\psi\rangle = \cos{\tfrac{\theta}{2}}|0\rangle + e^{i\phi}\sin{\tfrac{\theta}{2}}|1\rangle \Rightarrow \theta$$ is the angle on the Bloch sphere, while $$\tfrac{\theta}{2}$$ is the actual angle in Hibert space $$\Rightarrow Z$$-measurement corresponds to a projection onto the z-axis and analogously for X and Y