# The Bloch Sphere > Two real numbers describe a qubit's state — and two real numbers are the latitude and longitude of a > point on a sphere. The **Bloch sphere** turns the algebra of $\mathbb{C}^2$ into geometry you can > see: states are points, gates are rotations, orthogonal states are antipodes. It is the single most > useful picture in all of single-qubit quantum computing, and we'll lean on it constantly. ## Learning Objectives After this lesson you will be able to: 1. Derive the **Bloch-sphere parametrization** $\lvert\psi\rangle = \cos\tfrac\theta2\lvert0\rangle + e^{i\varphi}\sin\tfrac\theta2\lvert1\rangle$. 2. Compute the **Bloch vector** $\vec r = (\langle X\rangle, \langle Y\rangle, \langle Z\rangle)$ of a pure state. 3. Locate the standard states ($\lvert0\rangle,\lvert1\rangle,\lvert\pm\rangle,\lvert\pm i\rangle$) on the sphere. 4. Explain why **antipodal points are orthogonal** and why gates act as **rotations**. 5. Map states to Bloch vectors in NumPy. --- ## Intuition In [1.1.1 E5](../c01-postulates/01-state-postulate.md) we counted a qubit's degrees of freedom: four real numbers in $\mathbb{C}^2$, minus one for normalization, minus one for the unobservable global phase, leaving **two**. Two angles parametrize a sphere. So the set of pure qubit states is (geometrically) the surface of a unit ball — the Bloch sphere. The north and south poles are $\lvert0\rangle$ and $\lvert1\rangle$; the equator holds the "equal-superposition" states whose only difference is a relative phase. Once you internalize this, single-qubit gates stop being matrices and become *rotations of a sphere*, which is far easier to reason about. --- ## Theory ### The parametrization Start from a general $\lvert\psi\rangle = \alpha\lvert0\rangle + \beta\lvert1\rangle$. Write $\alpha,\beta$ in polar form and factor out the global phase of $\alpha$ (which is unobservable, [1.1.1](../c01-postulates/01-state-postulate.md)). Normalization $|\alpha|^2+|\beta|^2 = 1$ lets us set $|\alpha| = \cos\tfrac\theta2$, $|\beta| = \sin\tfrac\theta2$ for a unique $\theta\in[0,\pi]$. The only surviving phase is the **relative** phase $\varphi\in[0,2\pi)$ between $\beta$ and $\alpha$. Hence $$ \boxed{\ \lvert\psi(\theta,\varphi)\rangle = \cos\tfrac{\theta}{2}\,\lvert0\rangle + e^{i\varphi}\sin\tfrac{\theta}{2}\,\lvert1\rangle\ } $$ The half-angle $\theta/2$ is not a typo — it's what makes the map to the sphere exact (and why a $2\pi$ rotation in $\theta$ takes you only halfway, the famous spin-$\tfrac12$ sign; see Exercises). ### The Bloch vector Map $\lvert\psi(\theta,\varphi)\rangle$ to the point $$ \vec r = (\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta) \in S^2 . $$ The components are exactly the **expectation values of the Pauli operators**: $$ \vec r = \big(\langle X\rangle,\ \langle Y\rangle,\ \langle Z\rangle\big), \qquad \langle\sigma_k\rangle = \langle\psi|\sigma_k|\psi\rangle . $$ **Derivation of the $z$-component.** $\langle Z\rangle = \langle\psi|Z|\psi\rangle = |\alpha|^2 - |\beta|^2 = \cos^2\tfrac\theta2 - \sin^2\tfrac\theta2 = \cos\theta = r_z$. ✓ Similarly (Exercise E1) $\langle X\rangle = \sin\theta\cos\varphi$ and $\langle Y\rangle = \sin\theta\sin\varphi$. For a **pure** state $|\vec r| = 1$ (on the surface); **mixed** states ([1.5.3](../c05-density-matrices/03-bloch-ball-purity.md)) live *inside* the ball with $|\vec r| < 1$, and $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ ([Appendix E](../../appendix/E-math-identities.md)). ### Landmarks on the sphere ``` +z = |0> (north pole, θ=0) | |-i> ----- + ----- |+i> (±y on the equator) / | \ |-> ---/ | \--- |+> (±x on the equator, φ=0 and φ=π) | -z = |1> (south pole, θ=π) ``` | State | $(\theta,\varphi)$ | Bloch vector $\vec r$ | Axis | |---|---|---|---| | $\lvert0\rangle$ | $(0,\,-)$ | $(0,0,+1)$ | $+z$ | | $\lvert1\rangle$ | $(\pi,\,-)$ | $(0,0,-1)$ | $-z$ | | $\lvert+\rangle$ | $(\tfrac\pi2, 0)$ | $(+1,0,0)$ | $+x$ | | $\lvert-\rangle$ | $(\tfrac\pi2, \pi)$ | $(-1,0,0)$ | $-x$ | | $\lvert{+}i\rangle$ | $(\tfrac\pi2, \tfrac\pi2)$ | $(0,+1,0)$ | $+y$ | | $\lvert{-}i\rangle$ | $(\tfrac\pi2, \tfrac{3\pi}2)$ | $(0,-1,0)$ | $-y$ | ### Antipodes are orthogonal; gates are rotations Two key geometric facts: - **Orthogonal ⟺ antipodal.** A state and its orthogonal partner sit at *opposite* points of the sphere ($\theta\to\pi-\theta$, $\varphi\to\varphi+\pi$). E.g. $\lvert0\rangle\perp\lvert1\rangle$ are the two poles; $\lvert+\rangle\perp\lvert-\rangle$ are $\pm x$. (Note: a $90°$ *physical* angle between measurement axes corresponds to a $180°$ angle on the Bloch sphere — the half-angle again.) - **Unitaries are rotations.** Every single-qubit unitary acts on $\vec r$ as a rotation in $SO(3)$; in particular $R_n(\theta) = e^{-i\theta\,\hat n\cdot\vec\sigma/2}$ rotates the Bloch vector by angle $\theta$ about axis $\hat n$ ([Appendix E](../../appendix/E-math-identities.md)). So $X$ is a $180°$ flip about $\hat x$, $Z$ about $\hat z$, the Hadamard about $(\hat x+\hat z)/\sqrt2$. This is the geometric heart of Term 2's gate algebra, and the picture behind Rabi precession ([1.6.2](../c06-dynamics/02-two-level-rabi.md)). All of this section, live. Jump between the landmarks and check that each orthogonal pair sits at opposite ends of a diameter; then apply gates and watch each one turn the whole sphere about a fixed axis, drawn dashed while it turns: ```qm-demo bloch-sphere ``` --- ## Worked Examples ### Example 1 — Bloch vector of $\lvert+\rangle$ $\lvert+\rangle$ has $\alpha=\beta=\tfrac1{\sqrt2}$, so $\cos\tfrac\theta2 = \tfrac1{\sqrt2}\Rightarrow\theta = \tfrac\pi2$ and relative phase $\varphi = 0$. Then $\vec r = (\sin\tfrac\pi2\cos0,\ \sin\tfrac\pi2\sin0,\ \cos\tfrac\pi2) = (1,0,0)$ — the $+x$ axis. Directly: $\langle X\rangle = \langle+|X|+\rangle = 1$, $\langle Y\rangle = \langle Z\rangle = 0$. ✓ ### Example 2 — From Bloch angles to a state vector The point $(\theta,\varphi) = (\tfrac\pi3, \tfrac\pi2)$ corresponds to $$ \lvert\psi\rangle = \cos\tfrac\pi6\lvert0\rangle + e^{i\pi/2}\sin\tfrac\pi6\lvert1\rangle = \tfrac{\sqrt3}{2}\lvert0\rangle + \tfrac{i}{2}\lvert1\rangle . $$ Check $|\alpha|^2+|\beta|^2 = \tfrac34 + \tfrac14 = 1$ ✓, and $\langle Z\rangle = \tfrac34-\tfrac14 = \tfrac12 = \cos\tfrac\pi3$ ✓. --- ## Hands-on (Python) ```python import numpy as np X = np.array([[0, 1], [1, 0]], dtype=complex) Y = np.array([[0, -1j], [1j, 0]], dtype=complex) Z = np.array([[1, 0], [0, -1]], dtype=complex) def state(theta, phi): return np.array([np.cos(theta/2), np.exp(1j*phi)*np.sin(theta/2)], dtype=complex) def bloch_vector(psi): """(,,) for a single-qubit pure state.""" return np.array([np.real(psi.conj() @ M @ psi) for M in (X, Y, Z)]) # Landmarks: print(np.round(bloch_vector(state(0, 0)), 3)) # |0> -> [0 0 1] (north pole) print(np.round(bloch_vector(state(np.pi, 0)), 3)) # |1> -> [0 0 -1] (south pole) print(np.round(bloch_vector(state(np.pi/2, 0)), 3)) # |+> -> [1 0 0] print(np.round(bloch_vector(state(np.pi/2, np.pi/2)),3))# |+i>-> [0 1 0] # Pure states have |r| = 1: r = bloch_vector(state(1.1, 2.0)) print(round(np.linalg.norm(r), 6)) # 1.0 ``` ```python # Antipodal = orthogonal. |+> at +x, |-> at -x: plus = state(np.pi/2, 0) minus = state(np.pi/2, np.pi) print(np.round(bloch_vector(plus), 3), np.round(bloch_vector(minus), 3)) # [1 0 0] [-1 0 0] print(abs(np.vdot(plus, minus))) # 0.0 -> orthogonal # A gate is a rotation: X flips the Bloch vector 180° about x (z -> -z). psi0 = state(0, 0) # |0>, r=(0,0,1) print(np.round(bloch_vector(X @ psi0), 3)) # |1>, r=(0,0,-1) ``` > For a rendered 3-D Bloch sphere, `qutip`'s `Bloch` class is excellent (`pip install qutip`); we keep to > the lightweight `bloch_vector` here to avoid an extra dependency, as in > [Appendix B §7.2](../../appendix/B-python-numpy-refresher.md). --- ## Exercises **E1 (easy).** Derive $\langle X\rangle = \sin\theta\cos\varphi$ for $\lvert\psi(\theta,\varphi)\rangle$.
Solution $X\lvert\psi\rangle = \cos\tfrac\theta2\lvert1\rangle + e^{i\varphi}\sin\tfrac\theta2\lvert0\rangle$. Then $\langle\psi|X|\psi\rangle = \cos\tfrac\theta2\,e^{i\varphi}\sin\tfrac\theta2 + e^{-i\varphi}\sin\tfrac\theta2\cos\tfrac\theta2 = 2\sin\tfrac\theta2\cos\tfrac\theta2\cos\varphi = \sin\theta\cos\varphi$, using $\tfrac{e^{i\varphi}+e^{-i\varphi}}2=\cos\varphi$ and the double-angle identity. ∎
**E2 (easy).** Where on the sphere is $\lvert{-}i\rangle = \tfrac1{\sqrt2}(\lvert0\rangle - i\lvert1\rangle)$?
Solution $\theta=\tfrac\pi2$ (equal magnitudes), relative phase $-i = e^{-i\pi/2}$ so $\varphi = \tfrac{3\pi}2$ (or $-\tfrac\pi2$). $\vec r = (\cos\tfrac{3\pi}2,\ \sin\tfrac{3\pi}2,\ 0) = (0,-1,0)$ — the $-y$ axis.
**E3 (medium).** Show that orthogonal states map to antipodal Bloch vectors ($\vec r_\perp = -\vec r$).
Solution The orthogonal partner of $\lvert\psi(\theta,\varphi)\rangle$ is $\lvert\psi(\pi-\theta,\varphi+\pi)\rangle$ (check $\langle\psi_\perp|\psi\rangle=0$). Its Bloch vector: $\sin(\pi-\theta)=\sin\theta$ but $\cos(\varphi+\pi)=-\cos\varphi$, $\sin(\varphi+\pi)=-\sin\varphi$, and $\cos(\pi-\theta)=-\cos\theta$. So $\vec r_\perp = (-\sin\theta\cos\varphi, -\sin\theta\sin\varphi, -\cos\theta) = -\vec r$. ∎
**E4 (medium).** The gate $S = \operatorname{diag}(1,i)$ acts on $\lvert+\rangle$. Compute the new Bloch vector and describe the geometric action.
Solution $S\lvert+\rangle = \tfrac1{\sqrt2}(\lvert0\rangle + i\lvert1\rangle) = \lvert{+}i\rangle$, with $\vec r = (0,1,0)$. So $S$ rotated the Bloch vector from $+x$ to $+y$ — a $90°$ rotation about the $z$-axis (indeed $S = e^{i\pi/4}R_z(\tfrac\pi2)$, [Appendix E](../../appendix/E-math-identities.md)). Start at $\lvert+\rangle$ and press $S$ once to land on $\lvert{+}i\rangle$. Press it three more times and the state walks the equator a quarter-turn at a time, back to where it began: ```qm-demo bloch-sphere { "initial": "+", "gates": ["S"] } ```
**E5 (hard).** Show that $R_z(2\pi) = -I$ (not $+I$), and interpret: why does a "full rotation" return the Bloch vector to itself but multiply the state by $-1$?
Solution $R_z(\theta) = e^{-i\theta Z/2} = \operatorname{diag}(e^{-i\theta/2}, e^{i\theta/2})$; at $\theta=2\pi$ this is $\operatorname{diag}(e^{-i\pi}, e^{i\pi}) = \operatorname{diag}(-1,-1) = -I$. The Bloch vector is built from expectation values, which are invariant under the global phase $-1$, so it returns to itself after $2\pi$. But the *state vector* picks up the global $-1$ and only returns to itself after $4\pi$ — the defining signature of spin-$\tfrac12$ and the reason the parametrization uses the half-angle $\theta/2$. (This sign is unobservable for an isolated qubit but becomes a real *relative* phase in a controlled operation — Term 2.4.) ∎
--- ## Checkpoint 1. Write the Bloch parametrization and explain the half-angle $\theta/2$. 2. What are the three components of the Bloch vector in terms of observables? 3. Give the Bloch vectors of $\lvert0\rangle,\lvert1\rangle,\lvert\pm\rangle,\lvert\pm i\rangle$. 4. How are orthogonality and antipodality related on the sphere? 5. How do single-qubit unitaries act geometrically?
Answers 1. $\lvert\psi\rangle = \cos\tfrac\theta2\lvert0\rangle + e^{i\varphi}\sin\tfrac\theta2\lvert1\rangle$; the half-angle makes the map to $S^2$ exact and reflects the $4\pi$ periodicity of spin-$\tfrac12$. 2. $\vec r = (\langle X\rangle, \langle Y\rangle, \langle Z\rangle)$. 3. $\lvert0\rangle\!:(0,0,1)$, $\lvert1\rangle\!:(0,0,-1)$, $\lvert\pm\rangle\!:(\pm1,0,0)$, $\lvert\pm i\rangle\!:(0,\pm1,0)$. 4. Orthogonal states are exactly antipodal ($\vec r_\perp = -\vec r$); a $90°$ axis angle = $180°$ on the sphere. 5. As rotations in $SO(3)$: $R_{\hat n}(\theta)$ rotates $\vec r$ by $\theta$ about axis $\hat n$.
--- ## Further Reading - **[NC]** Nielsen & Chuang, §1.2, §4.2 — the Bloch sphere and single-qubit rotations. - **[Pre]** Preskill, *Ph219*, Ch. 2 — qubit geometry. - **[QuTiP]** QuTiP `Bloch` class docs — for 3-D visualization. --- ← Prev: [The Qubit](01-the-qubit.md) · Up: [Term 1](../README.md) · Next: [Bases & Expectation Values](03-bases-and-expectation.md) →