Appendix 1: Mapping of 1-Qubit: Bloch Sphere

The mathematical abstraction of quantum mechanics unfolds on the stage of Hilbert space. The Hilbert space of a 1-qubit system is $C^2$ (a 2-dimensional complex vector space).

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

Here, $\alpha$ and $\beta$ are $\alpha, \beta \in C$ (complex numbers), and it seems four real numbers (the real/imaginary parts of $\alpha$ and $\beta$) are needed. However, two constraints on quantum states—normalization and global phase—reduce this degree of freedom to just two.

Normalization: The total probability must be 1. $|\alpha|^2 + |\beta|^2 = 1$
Global Phase: $|\psi\rangle$ and $e^{i\gamma}|\psi\rangle$ are considered physically identical states.

These two remaining degrees of freedom correspond perfectly one-to-one with two angles (e.g., latitude, longitude) that describe a point on the unit sphere surface in 3-dimensional real space ($R^3$).

The Bloch Sphere visually maps the states of this 2-dimensional complex Hilbert space onto the surface of a sphere in 3-dimensional real space, providing the most powerful and standard tool to geometrically understand all states and operations of a 1-qubit.

1. Mapping from State Vector to Sphere

Applying these two constraints, all pure state vectors $|\psi\rangle$ of a 1-qubit can be uniquely expressed (ignoring the global phase) by two real angles $\theta$ (theta) and $\phi$ (phi) as follows.

\[ |\psi\rangle = \cos\left(\frac{\theta}{2}\right)|0\rangle + e^{i\phi}\sin\left(\frac{\theta}{2}\right)|1\rangle \] * $\theta$ (Polar Angle): $0 \le \theta \le \pi$. The angle descending from the positive direction of the Z-axis (the North Pole). (Determines the ‘latitude’ of the sphere) * $\phi$ (Azimuthal Angle): $0 \le \phi < 2\pi$. The angle rotated counterclockwise in the X-Y plane from the positive direction of the X-axis. (Determines the ‘longitude’ of the sphere)

These $\theta$ and $\phi$ are used as spherical coordinates in 3-dimensional space to map the state vector $|\psi\rangle$ to a point $\vec{r}$ on the surface of the sphere. > 💡 Detailed Explanation: Why $\theta/2$? (Orthogonality and 2:1 Mapping) > > In Hilbert space, the fact that two states $|\psi\rangle$ and $|\psi_\perp\rangle$ are orthogonal ($\langle\psi|\psi_\perp\rangle = 0$) is equivalent to the two states being exactly antipodal on the Bloch sphere. > > * The $|0\rangle$ state is $\theta=0$. > * The $|1\rangle$ state orthogonal to $|0\rangle$ is $\theta=\pi$ (180°). > > The angle of the state vector is $\theta/2$, but the actual angle on the sphere’s surface is $\theta$. To go from $|0\rangle$ to $|1\rangle$, the state vector must rotate $ (0^)|0(90)|0$ (i.e., a 90° rotation), but on the Bloch sphere, it must move 180° from the North Pole ($\theta=0$) to the South Pole ($\theta=\pi$). > > In this way, the $SU(2)$ operations in Hilbert space and the $SO(3)$ rotations on the Bloch sphere have a 2:1 correspondence, and this $\theta/2$ factor precisely guarantees that relationship.

2. Key Points on the Bloch Sphere (Key Geography)

According to this mapping, the six fundamental basis states of quantum computing are exactly located at the six directions (North Pole, South Pole, $\pm X$, $\pm Y$) on the Bloch sphere.

Z-axis (Computational Basis):
$|0\rangle$: $\theta=0$. $\implies$ North Pole (+Z-axis)
$|1\rangle$: $\theta=\pi$. $\implies$ South Pole (-Z-axis)
X-axis (Hadamard Basis):
$|+\rangle = \frac{|0\rangle+|1\rangle}{\sqrt{2}}$: $\theta=\pi/2, \phi=0$. $\implies$ +X-axis
$|-\rangle = \frac{|0\rangle-|1\rangle}{\sqrt{2}}$: $\theta=\pi/2, \phi=\pi$. $\implies$ -X-axis
Y-axis (Circular Basis):
$|+i\rangle = \frac{|0\rangle+i|1\rangle}{\sqrt{2}}$: $\theta=\pi/2, \phi=\pi/2$. $\implies$ +Y-axis
$|-i\rangle = \frac{|0\rangle-i|1\rangle}{\sqrt{2}}$: $\theta=\pi/2, \phi=3\pi/2$. $\implies$ -Y-axis

3. Geometry of Operations: Gates Are Rotations

The most important reason the Bloch sphere is essential is that all unitary operations (gates) applied to a single qubit correspond one-to-one with rotating the state vector about an axis on the Bloch sphere.

Complex 2x2 complex matrix multiplications are intuitively translated into “rotations” on a 3D sphere. (Refer to Chapter 1 of the text)

Pauli-X Gate (NOT):
$\hat{X} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
$|0\rangle \to |1\rangle$, $|1\rangle \to |0\rangle$.
Bloch Sphere Interpretation: Rotates 180° ($\pi$ radians) about the X-axis. (North Pole $|0\rangle$ rotates around the X-axis to the South Pole $|1\rangle$.)
Pauli-Z Gate (Phase-Flip):
$\hat{Z} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
$|0\rangle \to |0\rangle$, $|1\rangle \to -|1\rangle$ (non-observable global phase).
But $|+\rangle \to |-\rangle$ changes the state.
Bloch sphere interpretation: Rotates around the Z-axis by 180° ($\pi$ radians). (+X-axis $|+\rangle$ moves to -X-axis $|-\rangle$ after rotating around the Z-axis.)
Pauli-Y gate:
$\hat{Y} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$
Bloch sphere interpretation: Rotates around the Y-axis by 180° ($\pi$ radians).
Hadamard (H) gate:
$\hat{H} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
$|0\rangle \to |+\rangle$, $|1\rangle \to |-\rangle$.
Bloch sphere interpretation: Changes the Z-basis to the X-basis. This is equivalent to rotating around the (X+Z)-axis by 180°. (The north pole $|0\rangle$ moves to +X-axis $|+\rangle$.)
S gate ($Z^{1/2}$) / T gate ($Z^{1/4}$):
S gate: Rotates around the Z-axis by 90° ($\pi/2$ radians).
T gate: Rotates around the Z-axis by 45° ($\pi/4$ radians).

4. Inside the Sphere: Pure States and Mixed States

The Bloch sphere is a powerful visualization tool that perfectly distinguishes the pure state (Pure State) and mixed state (Mixed State) of a 1-qubit system. (See Chapter 6 and 9 of the main text)

Pure States:
$\rho = |\psi\rangle\langle\psi|$ (e.g., $\rho_{\text{pure}}$)
A state with complete information about the system.
Located on the surface of the Bloch sphere. (Vector length $r=1$)
Mixed States:
$\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|$ (e.g., $\rho_{\text{mix}}$)
A state with incomplete information due to classical probability (ignorance).
Located in the interior of the Bloch sphere. (Vector length $r < 1$)
Maximally Mixed State:
$\rho = \frac{1}{2}I = \frac{1}{2}(|0\rangle\langle0| + |1\rangle\langle1|)$
The most ignorant state, where $|0\rangle$ or $|1\rangle$ is randomly mixed with 50% probability each.
Located at the origin of the Bloch sphere. (Vector length $r=0$) > 💡 Application: Visualization of Decoherence > > The process of Decoherence (Decoherence), where a qubit loses quantum information by interacting with the external environment, can be intuitively understood through the Bloch sphere. > > * $T_1$ process (Energy decay): A qubit initially in the $|1\rangle$ state (South Pole) loses energy and moves to the $|0\rangle$ state (North Pole), causing the vector to contract along the Z-axis. > * $T_2$ process (Phase decay): When a qubit in the $|+\rangle$ state (+X-axis) loses phase information, the vector contracts in the X-Y plane and approaches the Z-axis. > > Ultimately, all decoherence can be visualized as a process that pulls the pure state vector from the surface ($r=1$) inward toward the center of the sphere ($r=0$, completely mixed state).