Bloch Sphere Geometry

Bloch sphere geometry is the coordinate geometry of a two-dimensional complex Hilbert space after normalization and global phase have been removed. It is special to qubits: the pure-state projective space $\mathbb{CP}^1$ is a two-sphere, and every one-qubit density matrix is a point in the corresponding three-dimensional ball.

This page gives the linear-algebra formulas. The spinor interpretation is Bloch Sphere, and the density-operator introduction is Bloch Sphere in Core Formalism.

Helpful background. Pauli matrices for the two-by-two operator basis; Hermitian operators for adjoints and real spectra.

Pauli Coordinates

The matrices

$$I,\sigma_x,\sigma_y,\sigma_z$$

form a real basis for Hermitian $2\times2$ matrices. Therefore every Hermitian trace-one matrix can be written uniquely as

$$\rho(\mathbf r) = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

where

$$\boldsymbol\sigma = (\sigma_x,\sigma_y,\sigma_z), \qquad \mathbf r=(r_x,r_y,r_z)\in\mathbb R^3.$$

The coordinates are Pauli expectation values:

$$r_i = \operatorname{Tr}(\rho\sigma_i), \qquad i=x,y,z.$$

Thus the Bloch vector is not an additional physical object. It is the density matrix written in the Pauli basis.

Positivity and the Bloch Ball

Using

$$(\mathbf r\cdot\boldsymbol\sigma)^2 = \lVert\mathbf r\rVert^2 I,$$

the eigenvalues of $\rho(\mathbf r)$ are

$$\lambda_\pm = \frac12 \left( 1\pm\lVert\mathbf r\rVert \right).$$

Therefore

$$\rho(\mathbf r)\ge0 \quad \Longleftrightarrow \quad \lVert\mathbf r\rVert\le1.$$

The allowed one-qubit density matrices fill the unit ball. Surface points are pure states; interior points are mixed states; the center $\mathbf r=\mathbf0$ is the maximally mixed state $I/2$.

This $r_y=0$ cross-section shows the distinction between the pure-state sphere and the mixed-state interior. The point $(0,0,1/2)$ has $\sigma_z$ probabilities $3/4$ and $1/4$; the center gives equal probabilities along every axis.

Pure States on the Sphere

A normalized qubit ray can be represented as

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

where

$$0\le\theta\le\pi, \qquad 0\le\phi<2\pi.$$

The associated unit Bloch vector is

$$\mathbf n = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).$$

The rank-one projector onto the ray is

$$\lvert\psi(\theta,\phi)\rangle \langle\psi(\theta,\phi)\rvert = \frac12 \left( I+\mathbf n\cdot\boldsymbol\sigma \right).$$

The global phase of $\lvert\psi\rangle$ has disappeared. Away from the poles, the relative phase $\phi$ remains as the azimuthal angle. At either pole one coefficient vanishes, so $\phi$ no longer labels distinct rays.

Antipodal Points and Orthogonality

Opposite points on the Bloch sphere are orthogonal pure states, not the same state. If $\mathbf n$ is a unit Bloch vector, then the projectors for the two antipodal directions are

$$P_\pm(\mathbf n) = \frac12 \left( I\pm\mathbf n\cdot\boldsymbol\sigma \right).$$

They satisfy

$$P_+P_-=0, \qquad P_++P_-=I.$$

Do not confuse $|\psi\rangle\mapsto-|\psi\rangle$ with $\mathbf n\mapsto-\mathbf n$. The first leaves the projector unchanged; the second changes it to the orthogonal projector $I-P$.

Overlap Geometry

Let $\mathbf n$ and $\mathbf m$ be two unit Bloch vectors, with angle $\gamma$ between them:

$$\mathbf n\cdot\mathbf m=\cos\gamma.$$

The transition probability between the corresponding pure states is

$$\lvert \langle\psi(\mathbf n)\vert\psi(\mathbf m)\rangle \rvert^2 = \frac12(1+\mathbf n\cdot\mathbf m) = \cos^2\frac{\gamma}{2}.$$

The angle between rays, defined by $\arccos|\langle\psi|\phi\rangle|$, is therefore $\gamma/2$. Orthogonal states have $\gamma=\pi$, while identical rays have $\gamma=0$.

Rotations

A unitary generated by a Pauli direction has the form

$$U(\hat{\mathbf a},\alpha) = \exp \left( -\frac{i\alpha}{2}\, \hat{\mathbf a}\cdot\boldsymbol\sigma \right).$$

Acting on a density matrix,

$$\rho(\mathbf r) \mapsto U\rho(\mathbf r)U^\dagger,$$

rotates the Bloch vector:

$$\mathbf r\mapsto R_{\hat{\mathbf a}}(\alpha)\mathbf r,$$

where $R_{\hat{\mathbf a}}(\alpha)$ is the ordinary $SO(3)$ rotation by angle $\alpha$ around $\hat{\mathbf a}$.

The map from $SU(2)$ spinor unitaries to $SO(3)$ rotations is two-to-one: $U$ and $-U$ rotate the Bloch vector in the same way. The group-theory background is SU(2), and the physical spin version is Spin Rotations.

Measurements as Projections

The projective measurement of a Pauli component along a unit direction $\hat{\mathbf a}$ has effects

$$P_\pm(\hat{\mathbf a}) = \frac12 \left( I\pm\hat{\mathbf a}\cdot\boldsymbol\sigma \right).$$

For a state $\rho(\mathbf r)$,

$$p_\pm = \operatorname{Tr}\bigl(\rho(\mathbf r)P_\pm(\hat{\mathbf a})\bigr) = \frac12 \left( 1\pm\mathbf r\cdot\hat{\mathbf a} \right).$$

Geometrically, measuring along $\hat{\mathbf a}$ reads the projection of the Bloch vector onto the measurement axis. A pure state at $\hat{\mathbf a}$ gives the plus outcome with probability one; a state at the center gives equal probabilities in every direction.

Small changes of a pure-state Bloch direction are tangent directions on $S^2$, not arbitrary independent changes in $\mathbb R^3$. The local tangent-space language is summarized in Tangent and Cotangent Spaces.

Why the Geometry Is Special to Qubits

The Bloch-ball formula depends on the Pauli basis and on the special identity

$$(\mathbf r\cdot\boldsymbol\sigma)^2 = \lVert\mathbf r\rVert^2 I.$$

Higher-dimensional density matrices can be expanded in generalized generator bases, but the allowed set is not simply a Euclidean ball, and pure states do not form an ordinary two-sphere. The qubit case is unusually visual.

Common Mistakes

• Treating the Bloch vector as a two-component spinor.
• Confusing the surface sphere of pure states with the full Bloch ball of density matrices.
• Calling antipodal points the same state; they represent orthogonal rays.
• Forgetting that global phase is removed but relative phase is visible as azimuth.
• Assuming the Bloch-ball picture generalizes directly to qutrits or larger systems.
• Missing the half-angle relation between spinors and Bloch-sphere directions.
• Treating $U$ and $-U$ as different rotations of the Bloch vector.

References

• F. Bloch, “Nuclear induction,” Physical Review 70, 460-474, 1946.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Find the Bloch vector for

$$\lvert+\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle+\lvert1\rangle \bigr).$$

Solution

The projector is

$$\lvert+\rangle\langle+\rvert = \frac12 \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} = \frac12(I+\sigma_x).$$

Therefore the Bloch vector is

$$\mathbf r=(1,0,0).$$

2. Show that $\rho(\mathbf r)$ is pure exactly when $\lVert\mathbf r\rVert=1$.

Solution

The eigenvalues of $\rho(\mathbf r)$ are

$$\lambda_\pm = \frac12 \left( 1\pm\lVert\mathbf r\rVert \right).$$

A density matrix is pure exactly when one eigenvalue is $1$ and the other is $0$. This happens exactly when $\lVert\mathbf r\rVert=1$.

3. Two pure states have Bloch vectors separated by angle $\gamma=\pi/3$. What is their transition probability?

Solution

Use

$$\lvert \langle\psi(\mathbf n)\vert\psi(\mathbf m)\rangle \rvert^2 = \cos^2\frac{\gamma}{2}.$$

For $\gamma=\pi/3$,

$$\cos^2\frac{\pi}{6} = \left(\frac{\sqrt3}{2}\right)^2 = \frac34.$$

4. A state has Bloch vector $\mathbf r=(0,0,1/2)$. What are the probabilities for measuring $\sigma_z$?

Solution

For measurement direction $\hat{\mathbf z}$,

$$p_\pm = \frac12 \left( 1\pm\mathbf r\cdot\hat{\mathbf z} \right) = \frac12 \left( 1\pm\frac12 \right).$$

Therefore

$$p_+=\frac34, \qquad p_-=\frac14.$$

5. Why do $U$ and $-U$ give the same Bloch-vector rotation?

Solution

The density matrix transforms by conjugation:

$$\rho\mapsto U\rho U^\dagger.$$

Replacing $U$ by $-U$ gives

$$\rho \mapsto (-U)\rho(-U)^\dagger = U\rho U^\dagger.$$

Thus the Bloch vector changes in the same way. This is the two-to-one relation between $SU(2)$ and $SO(3)$ rotations.