The Bloch Ball & Purity

In Course 1.2 you learned that every pure qubit state lives on the surface of the Bloch sphere. Density matrices complete the picture: every qubit state — pure or mixed — is a point in the solid Bloch ball. Pure states are the skin ($|\vec r| = 1$), mixed states fill the interior, and the maximally mixed state $I/2$ sits dead center. This is not a cute analogy: the entire single-qubit state space is a unit ball in $\mathbb R^3$, the algebra is an exact dictionary $\rho\leftrightarrow\vec r$, and purity is a one-line function of the radius. It is the geometry you will reason with for every qubit noise channel in Term 4.

Learning Objectives

After this lesson you will be able to:

1. Derive the single-qubit form $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ from Hermiticity and unit trace using the Pauli basis.
2. Identify the Bloch vector $\vec r = (\langle X\rangle, \langle Y\rangle, \langle Z\rangle)$ and recover it from $\rho$ via $r_k = \operatorname{Tr}(\rho\,\sigma_k)$.
3. Prove the constraint $|\vec r|\le 1$ from positivity, and that $|\vec r| = 1$ iff $\rho$ is pure.
4. Derive the purity identity $\operatorname{Tr}(\rho^2) = \tfrac12(1 + |\vec r|^2)$ and use it to classify states.
5. Connect the Bloch ball to the pure-state Bloch sphere of Course 1.2, mapping mixed states inward toward the center.

---

Intuition

A single qubit's density matrix is a $2\times2$ Hermitian matrix with trace $1$. Count its real degrees of freedom: a $2\times2$ Hermitian matrix has $4$ real parameters (two real diagonal entries, one complex off-diagonal), and fixing the trace to $1$ removes one — leaving exactly $3$. Three real numbers, one constraint to keep eigenvalues nonnegative: that is precisely the data of a point inside a ball in $\mathbb R^3$. The Pauli matrices $\{I, X, Y, Z\}$ are the natural axes for that space, and the three coordinates are the expectation values $\langle X\rangle, \langle Y\rangle, \langle Z\rangle$ — directly measurable.

Geometrically: pure states are the extreme points of the convex state set (Lesson 1), and the extreme points of a ball are its surface — the sphere of Course 1.2. Mixing pure states means taking convex combinations, which move you inside the ball. The deepest interior point, reachable from every antipodal pair $\tfrac12|\hat n\rangle\langle\hat n| + \tfrac12|{-}\hat n\rangle\langle{-}\hat n|$, is the center $I/2$ — maximal mixing, total ignorance. Purity, which Lesson 1 defined abstractly, becomes literally "how far from the center are you," squared.

---

Theory

Deriving $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$

We use the key fact from Appendix E §1: the Pauli set $\{I, X, Y, Z\}$ is an orthogonal basis for the real vector space of $2\times2$ Hermitian matrices, under the Hilbert–Schmidt inner product $\langle A, B\rangle = \operatorname{Tr}(A^\dagger B)$, with

$$\operatorname{Tr}(I) = 2,\quad \operatorname{Tr}(\sigma_j) = 0,\quad \operatorname{Tr}(\sigma_j\sigma_k) = 2\delta_{jk}\quad (j,k\in\{x,y,z\}).$$

Step 1 — Hermiticity gives a real Pauli expansion. Since $\rho = \rho^\dagger$ (Lesson 1) and any $2\times2$ Hermitian matrix is a real linear combination of the Pauli basis, write

$$\rho = a_0 I + a_x X + a_y Y + a_z Z, \qquad a_0, a_x, a_y, a_z \in \mathbb R .$$

(Reality of the coefficients is exactly what Hermiticity buys: $X, Y, Z$ are Hermitian, and a complex coefficient on a Hermitian matrix would break $\rho^\dagger = \rho$.)

Step 2 — Unit trace fixes $a_0$. Take the trace of both sides and use $\operatorname{Tr}\sigma_j = 0$:

$$1 = \operatorname{Tr}\rho = a_0\operatorname{Tr}(I) + 0 = 2a_0 \;\Longrightarrow\; a_0 = \tfrac12 .$$

Step 3 — extract the Pauli coefficients by orthogonality. Multiply $\rho$ by $\sigma_k$ and trace, using $\operatorname{Tr}(\sigma_j\sigma_k) = 2\delta_{jk}$ and $\operatorname{Tr}(I\sigma_k) = 0$:

$$\operatorname{Tr}(\rho\,\sigma_k) = a_0\underbrace{\operatorname{Tr}(\sigma_k)}_{0} + \sum_j a_j\underbrace{\operatorname{Tr}(\sigma_j\sigma_k)}_{2\delta_{jk}} = 2a_k \;\Longrightarrow\; a_k = \tfrac12\operatorname{Tr}(\rho\,\sigma_k) .$$

Step 4 — name the coordinates. By the expectation formula $\langle A\rangle = \operatorname{Tr}(\rho A)$ of Lesson 1, $\operatorname{Tr}(\rho\,\sigma_k) = \langle\sigma_k\rangle$. Define the Bloch vector $r_k := \langle\sigma_k\rangle = \operatorname{Tr}(\rho\,\sigma_k)$, so $a_k = \tfrac12 r_k$. Substituting $a_0 = \tfrac12$ and $a_k = \tfrac12 r_k$:

$$\boxed{\;\rho = \tfrac12\big(I + \vec r\cdot\vec\sigma\big) = \tfrac12\big(I + r_x X + r_y Y + r_z Z\big),\qquad \vec r = \big(\langle X\rangle, \langle Y\rangle, \langle Z\rangle\big)\;}$$

This is an exact, invertible dictionary: every qubit density matrix $\leftrightarrow$ a vector $\vec r\in\mathbb R^3$. Writing it out as a matrix,

$$\rho = \frac12\begin{pmatrix} 1 + r_z & r_x - i r_y \\ r_x + i r_y & 1 - r_z \end{pmatrix} .$$

The diagonal carries the $Z$-statistics ($P(0) = \tfrac{1+r_z}{2}$, $P(1) = \tfrac{1-r_z}{2}$) and the off-diagonal carries the $X, Y$ coherences.

The positivity constraint: $|\vec r|\le 1$

Hermiticity and unit trace are automatic in the parametrization; the remaining requirement, $\rho\succeq0$, is what carves out the ball.

Proposition. $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ is a valid density operator (i.e. $\rho\succeq0$) iff $|\vec r|\le 1$.

Proof. A $2\times2$ Hermitian matrix is positive semidefinite iff $\operatorname{Tr}\rho\ge0$ and $\det\rho\ge0$ (then both eigenvalues are $\ge0$). The trace is $1>0$, so positivity reduces to $\det\rho\ge0$. Compute the determinant from the matrix form:

$$\det\rho = \frac14\big[(1 + r_z)(1 - r_z) - (r_x - i r_y)(r_x + i r_y)\big] = \frac14\big[1 - r_z^2 - (r_x^2 + r_y^2)\big] = \frac14\big(1 - |\vec r|^2\big).$$

Hence $\det\rho\ge0 \iff |\vec r|^2\le 1 \iff |\vec r|\le 1$. ∎

So the single-qubit state space is exactly the closed unit ball ${\vec r\in\mathbb R^3 : |\vec r|\le1}$— the **Bloch ball**. The eigenvalues fall out of the same computation: from$\operatorname{Tr}\rho = \lambda_+ + \lambda_- = 1$and$\det\rho = \lambda_+\lambda_- = \tfrac14(1 - |\vec r|^2)$,

$$\lambda_\pm = \frac{1 \pm |\vec r|}{2}\;\in[0,1],$$

which are manifestly valid probabilities precisely when $|\vec r|\le1$.

Pure on the sphere, mixed inside, maximally mixed at the center

The eigenvalues $\lambda_\pm = \tfrac{1\pm|\vec r|}{2}$ instantly classify every qubit state by radius:

| $|\vec r|$ | Eigenvalues $\lambda_\pm$ | State | |---|---|---| | $|\vec r| = 1$ (surface) | $1, 0$ | pure | | $0 < |\vec r| < 1$ (interior) | $\tfrac{1\pm|\vec r|}{2}$, both in $(0,1)$ | mixed | | $|\vec r| = 0$ (center) | $\tfrac12, \tfrac12$ | maximally mixed, $\rho = I/2$ |

Pure ⟺ surface. $\rho$ is pure iff $|\vec r| = 1$.

Proof. $\rho$ pure $\iff$ one eigenvalue is $1$ and the other $0$ (Lesson 1) $\iff \lambda_+ = 1$, i.e. $\tfrac{1+|\vec r|}{2} = 1 \iff |\vec r| = 1$. ∎ At the center $\vec r = \vec 0$, the formula gives $\rho = \tfrac12 I$ directly — the maximally mixed state of Lessons 1–2 (and the Bell reduced state).

This is the Bloch sphere of Course 1.2, now seen as a boundary. There, a pure qubit $|\psi\rangle = \cos\tfrac\theta2|0\rangle + e^{i\phi}\sin\tfrac\theta2|1\rangle$ sits at the unit-sphere point $\hat r = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$; you can check $\langle X\rangle = \sin\theta\cos\phi$, etc., so that $\hat r$ is exactly the Bloch vector with $|\hat r| = 1$. Density matrices add the radial direction: the interior. A mixed state with Bloch vector $\vec r = |\vec r|\,\hat n$ ($0\le|\vec r|<1$) points in the same direction $\hat n$ as the pure state on the surface but is "shrunk" toward the center by the factor $|\vec r|$ — the geometric meaning of decoherence.

Purity as a quadratic in the radius

Finally, the abstract purity of Lesson 1 becomes a clean function of $|\vec r|$.

Purity–radius identity.

$$\operatorname{Tr}(\rho^2) = \tfrac12\big(1 + |\vec r|^2\big).$$

Proof (algebraic). Square the parametrization and expand using $(\vec r\cdot\vec\sigma)^2 = |\vec r|^2 I$ — which follows from $\sigma_j\sigma_k = \delta_{jk}I + i\varepsilon_{jkl}\sigma_l$ (Appendix E §1): the symmetric sum kills the antisymmetric $\varepsilon$ term, leaving $\sum_{jk}r_j r_k\sigma_j\sigma_k = \sum_{jk}r_j r_k\delta_{jk}I = |\vec r|^2 I$. Then

$$\rho^2 = \tfrac14\big(I + \vec r\cdot\vec\sigma\big)^2 = \tfrac14\big(I + 2\,\vec r\cdot\vec\sigma + (\vec r\cdot\vec\sigma)^2\big) = \tfrac14\big((1 + |\vec r|^2)I + 2\,\vec r\cdot\vec\sigma\big).$$

Take the trace, using $\operatorname{Tr}(I) = 2$ and $\operatorname{Tr}(\vec r\cdot\vec\sigma) = 0$:

$$\operatorname{Tr}(\rho^2) = \tfrac14\big((1 + |\vec r|^2)\cdot 2 + 0\big) = \tfrac12\big(1 + |\vec r|^2\big). \qquad\blacksquare$$

Consistency check (eigenvalues). $\operatorname{Tr}(\rho^2) = \lambda_+^2 + \lambda_-^2 = \big(\tfrac{1+|\vec r|}{2}\big)^2 + \big(\tfrac{1-|\vec r|}{2}\big)^2 = \tfrac{2 + 2|\vec r|^2}{4} = \tfrac12(1 + |\vec r|^2)$. ✓ The identity reproduces Lesson 1's bounds for$d = 2$: at$|\vec r| = 1$, purity $= 1$ (pure); at $|\vec r| = 0$, purity $= \tfrac12 = 1/d$ (maximally mixed). Purity is a monotonic dial on the radius — knowing $\operatorname{Tr}\rho^2$ is equivalent to knowing $|\vec r| = \sqrt{2\operatorname{Tr}(\rho^2) - 1}$, the distance from center.

---

Worked Examples

Example 1 — From $\rho$ to $\vec r$ to classification

Let \rho = \begin{psmallmatrix} 3/4 & 1/4 \\ 1/4 & 1/4 \end{psmallmatrix} (Hermitian, trace $1$). Read off the Bloch vector via $r_k = \operatorname{Tr}(\rho\,\sigma_k)$, or directly by matching to the matrix form \rho = \tfrac12\begin{psmallmatrix}1+r_z & r_x - ir_y\\ r_x + ir_y & 1 - r_z\end{psmallmatrix}:

$$\tfrac{1+r_z}{2} = \tfrac34 \Rightarrow r_z = \tfrac12,\qquad \tfrac{r_x - i r_y}{2} = \tfrac14 \Rightarrow r_x = \tfrac12,\ r_y = 0 .$$

So $\vec r = (\tfrac12, 0, \tfrac12)$ and $|\vec r| = \sqrt{\tfrac14 + \tfrac14} = \tfrac1{\sqrt2}\approx 0.707 < 1$ — a mixed interior state. Purity:

$$\operatorname{Tr}(\rho^2) = \tfrac12(1 + |\vec r|^2) = \tfrac12\big(1 + \tfrac12\big) = \tfrac34 .$$

Cross-check directly: $\rho^2 = \begin{psmallmatrix}3/4&1/4\1/4&1/4\end{psmallmatrix}^2 = \begin{psmallmatrix}10/16 & 4/16\ 4/16 & 2/16\end{psmallmatrix}$, trace$= \tfrac{12}{16} = \tfrac34$. ✓ Eigenvalues $\lambda_\pm = \tfrac{1\pm1/\sqrt2}{2}\approx 0.854,\,0.146$, both in $(0,1)$ — consistent with "mixed."

Example 2 — Decoherence as radial shrinkage

Start from the pure equator state $|+\rangle$, with $\vec r = (1, 0, 0)$ on the surface ($\langle X\rangle = 1$, the rest$0$). Suppose a dephasing process (Term 4 preview) damps the off-diagonal coherence by a factor $p\in[0,1]$:

$$\rho_p = \begin{pmatrix} \tfrac12 & \tfrac{p}{2}\\[2pt] \tfrac{p}{2} & \tfrac12 \end{pmatrix} = \tfrac12\big(I + p\,X\big), \qquad \vec r = (p, 0, 0).$$

The Bloch vector keeps its direction $\hat x$ but its length collapses from $1$ to $p$: the state slides straight in from the surface point $|+\rangle$ toward the center along the $x$-axis. Purity tracks the radius exactly:

$$\operatorname{Tr}(\rho_p^2) = \tfrac12(1 + p^2),$$

falling from $1$ (pure $|+\rangle$, $p = 1$) to $\tfrac12$ (maximally mixed, $p = 0$). This single picture — coherence loss = radial shrinkage = purity decrease — is the geometric heart of how noise acts on a qubit, and you will meet it formally as the dephasing channel in Term 4.

---

Hands-on (Python)

We build the exact $\rho\leftrightarrow\vec r$ dictionary both directions, compute purity from the radius, and classify states.

import numpy as np

I = np.eye(2, dtype=complex)
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)
PAULI = [X, Y, Z]

def bloch_vector(rho):
    """r_k = Tr(ρ σ_k) = ⟨σ_k⟩, for k = x, y, z.  (Real for a valid state.)"""
    return np.array([np.trace(rho @ s).real for s in PAULI])

def density_from_bloch(r):
    """ρ = ½ (I + r·σ).  Valid density matrix iff |r| ≤ 1."""
    rx, ry, rz = r
    return 0.5 * (I + rx * X + ry * Y + rz * Z)

def purity(rho):
    return np.trace(rho @ rho).real

# Round-trip the dictionary: ρ -> r -> ρ.
rho = np.array([[3/4, 1/4], [1/4, 1/4]], dtype=complex)   # Worked Example 1
r = bloch_vector(rho)
print("Bloch vector r =", np.round(r, 4))                 # [0.5  0.   0.5]
print("round-trip ok?", np.allclose(density_from_bloch(r), rho))  # True

print("|r| =", round(np.linalg.norm(r), 4))               # 0.7071
print("purity (from radius) =", 0.5 * (1 + np.linalg.norm(r)**2))  # 0.75
print("purity (direct)      =", round(purity(rho), 4))    # 0.75  -> identity confirmed

def classify(rho, tol=1e-9):
    """Classify a single-qubit state by its Bloch radius."""
    r_norm = np.linalg.norm(bloch_vector(rho))
    if r_norm > 1 + tol:
        return "INVALID (|r| > 1, not positive semidefinite)"
    if abs(r_norm - 1) <= tol:
        return "pure (on the Bloch sphere)"
    if r_norm <= tol:
        return "maximally mixed (center, I/2)"
    return "mixed (inside the Bloch ball)"

states = {
    "|0⟩":        density_from_bloch([0, 0, 1]),     # north pole, pure
    "|+⟩":        density_from_bloch([1, 0, 0]),     # equator, pure
    "I/2":        density_from_bloch([0, 0, 0]),     # center, maximally mixed
    "WE1 mixed":  np.array([[3/4, 1/4], [1/4, 1/4]], dtype=complex),
}
for name, rho in states.items():
    print(f"{name:10s} |r|={np.linalg.norm(bloch_vector(rho)):.3f}  "
          f"purity={purity(rho):.3f}  -> {classify(rho)}")

# Decoherence as radial shrinkage (Worked Example 2): |+⟩ dephasing toward center.
for p in [1.0, 0.7, 0.3, 0.0]:
    rho_p = density_from_bloch([p, 0, 0])             # ρ = ½(I + p X)
    print(f"p={p:.1f}  |r|={p:.1f}  purity={purity(rho_p):.3f}  "
          f"(= ½(1+p²) = {0.5*(1+p*p):.3f})")
# Coherence damps -> Bloch vector shrinks along x -> purity falls 1.0 → 0.5.

# A point with |r| > 1 is NOT a valid state (eigenvalue would be negative):
bad = density_from_bloch([1.0, 1.0, 0.0])             # |r| = √2 > 1
print("eigvals of an invalid 'state':", np.round(np.linalg.eigvalsh(bad), 3))  # one < 0

Geometry, not metaphor. bloch_vector and density_from_bloch are exact inverses on valid states — the Bloch ball is a faithful coordinate system for the qubit, not a visualization aid. Every single-qubit noise channel in Term 4 is, geometrically, an affine map of the ball (a shrink + shift of $\vec r$); having the dictionary in code means you can read off purity and validity for any of them with the functions above.

---

Exercises

E1 (easy). Find the Bloch vectors of $|0\rangle$, $|1\rangle$, $|+\rangle$, $|-\rangle$, $|{+}i\rangle$, $|{-}i\rangle$. Confirm each has $|\vec r| = 1$.

Solution

Using $r_k = \langle\sigma_k\rangle$ (or matching matrices): $|0\rangle\to(0,0,1)$, $|1\rangle\to(0,0,-1)$, $|+\rangle\to(1,0,0)$, $|-\rangle\to(-1,0,0)$, $|{+}i\rangle\to(0,1,0)$, $|{-}i\rangle\to(0,-1,0)$. These are the six poles of the three Pauli axes (Appendix C), each with $|\vec r| = 1$ — all pure, all on the sphere, as expected.

E2 (easy). A qubit has Bloch vector $\vec r = (0, 0, 0.6)$. Write $\rho$ as a matrix, compute its purity two ways, and classify it.

Solution

\rho = \tfrac12(I + 0.6\,Z) = \begin{psmallmatrix}0.8 & 0\\ 0 & 0.2\end{psmallmatrix}. Purity from the radius: $\tfrac12(1 + 0.36) = 0.68$. Directly: $0.8^2 + 0.2^2 = 0.64 + 0.04 = 0.68$. ✓ Since $0 < |\vec r| = 0.6 < 1$, it is a mixed interior state (eigenvalues $0.8, 0.2$).

E3 (medium). Prove that the eigenvalues of $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ are $\tfrac{1\pm|\vec r|}{2}$, and use this to re-derive $\operatorname{Tr}(\rho^2) = \tfrac12(1 + |\vec r|^2)$.

Solution

$\vec r\cdot\vec\sigma = |\vec r|(\hat n\cdot\vec\sigma)$ with $\hat n = \vec r/|\vec r|$, and $(\hat n\cdot\vec\sigma)^2 = I$ (from $(\vec r\cdot\vec\sigma)^2 = |\vec r|^2 I$), so $\hat n\cdot\vec\sigma$ has eigenvalues $\pm1$. Hence $\vec r\cdot\vec\sigma$ has eigenvalues $\pm|\vec r|$, and $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$has eigenvalues$\tfrac{1\pm|\vec r|}{2}$. Then$\operatorname{Tr}(\rho^2) = \big(\tfrac{1+|\vec r|}{2}\big)^2 + \big(\tfrac{1-|\vec r|}{2}\big)^2 = \tfrac{2 + 2|\vec r|^2}{4} = \tfrac12(1 + |\vec r|^2)$. ∎

E4 (medium). Show that the Bloch vector of a 50/50 mixture of two pure states with Bloch vectors $\vec a$, $\vec b$ (both on the sphere) is $\tfrac12(\vec a + \vec b)$, and that the result is pure iff $\vec a = \vec b$. Interpret geometrically.

Solution

If $\rho_a = \tfrac12(I + \vec a\cdot\vec\sigma)$ and $\rho_b = \tfrac12(I + \vec b\cdot\vec\sigma)$, then $\rho = \tfrac12\rho_a + \tfrac12\rho_b = \tfrac12\big(I + \tfrac{\vec a + \vec b}{2}\cdot\vec\sigma\big)$, so the mixture's Bloch vector is the midpoint $\vec r = \tfrac12(\vec a + \vec b)$ — convex mixing of states is convex combination of Bloch vectors (the map is affine). With $|\vec a| = |\vec b| = 1$, $|\vec r| = \tfrac12|\vec a + \vec b| = 1$ iff $\vec a, \vec b$ point the same way (the triangle inequality is tight only when collinear and equal), i.e. iff $\vec a = \vec b$. Geometrically: the chord between two distinct surface points lies strictly inside the ball, so any nontrivial mixture is mixed; antipodal points ($\vec b = -\vec a$) mix to the center $\vec r = 0$.

E5 (hard). Prove that $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ is a valid density operator iff $|\vec r|\le1$ without computing the determinant — instead use the eigenvalue characterization of positivity directly.

Solution

$\rho\succeq0$ iff both eigenvalues are $\ge0$. As in E3, the eigenvalues are $\lambda_\pm = \tfrac{1\pm|\vec r|}{2}$(using$(\hat n\cdot\vec\sigma)^2 = I \Rightarrow$eigenvalues$\pm1$ of $\hat n\cdot\vec\sigma$). The smaller one is $\lambda_- = \tfrac{1 - |\vec r|}{2}$, which is $\ge0$ iff $|\vec r|\le1$; $\lambda_+ = \tfrac{1+|\vec r|}{2}\ge0$ always. Hermiticity and $\operatorname{Tr}\rho = 1$hold by construction. So$\rho$is a valid state iff$|\vec r|\le1$. ∎ (This matches the determinant route, since $\det\rho = \lambda_+\lambda_- = \tfrac14(1 - |\vec r|^2)$.)

E6 (hard). The trace distance between two qubit states is $D(\rho,\sigma) = \tfrac12\operatorname{Tr}|\rho - \sigma|$(half the sum of the singular values of$\rho - \sigma$). Show that for single qubits $D(\rho,\sigma) = \tfrac12|\vec r - \vec s|$ — i.e. trace distance is just half the Euclidean distance between Bloch vectors.

Solution

$\rho - \sigma = \tfrac12((\vec r - \vec s)\cdot\vec\sigma) =: \tfrac12(\vec d\cdot\vec\sigma)$ with $\vec d = \vec r - \vec s$. This is traceless Hermitian; as in E3 its eigenvalues are $\pm\tfrac12|\vec d|$ (eigenvalues of $\vec d\cdot\vec\sigma$ are $\pm|\vec d|$). The singular values are the absolute eigenvalues $\tfrac12|\vec d|, \tfrac12|\vec d|$, so $\operatorname{Tr}|\rho - \sigma| = |\vec d|$ and $D(\rho,\sigma) = \tfrac12|\vec d| = \tfrac12|\vec r - \vec s|$. ∎ So the Bloch ball with half the Euclidean metric is the trace-distance geometry — distinguishability of qubit states is literally how far apart their Bloch vectors are. (Trace distance is developed in Term 4 / [Wil].)

---

Checkpoint

1. Derive $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ from Hermiticity and unit trace. How do you extract $r_k$ from $\rho$?
2. Why is the single-qubit state space a ball of radius $1$? Where does $|\vec r|\le1$ come from?
3. Which states are on the surface, in the interior, and at the center?
4. State and derive the purity–radius identity, and give purity at $|\vec r| = 0, \tfrac12, 1$.
5. How does the Bloch ball relate to the pure-state Bloch sphere of Course 1.2?

Answers

1. Hermiticity $\Rightarrow \rho = a_0 I + \sum_k a_k\sigma_k$ with real coefficients; unit trace fixes $a_0 = \tfrac12$; orthogonality $\operatorname{Tr}(\sigma_j\sigma_k) = 2\delta_{jk}$ gives $a_k = \tfrac12\operatorname{Tr}(\rho\sigma_k) = \tfrac12 r_k$. Recover$r_k = \operatorname{Tr}(\rho\sigma_k) = \langle\sigma_k\rangle$.
2. Positivity is the only remaining constraint; $\det\rho = \tfrac14(1 - |\vec r|^2)\ge0$ (equivalently $\lambda_\pm = \tfrac{1\pm|\vec r|}{2}\ge0$) forces $|\vec r|\le1$ — the closed unit ball.
3. Surface $|\vec r| = 1$: pure; interior $0<|\vec r|<1$: mixed; center $|\vec r| = 0$: maximally mixed $\rho = I/2$.
4. $\operatorname{Tr}(\rho^2) = \tfrac12(1 + |\vec r|^2)$ (square $\rho$, use $(\vec r\cdot\vec\sigma)^2 = |\vec r|^2 I$, trace). Values:$\tfrac12$at$|\vec r| = 0$;$\tfrac58$at$|\vec r| = \tfrac12$; $1$ at $|\vec r| = 1$.
5. The sphere ($|\vec r| = 1$) is the pure-state boundary of Course 1.2; the ball adds the radial direction for mixed states, which point the same way but are shrunk toward the center as they lose purity/coherence.

---

Further Reading

• [NC] Nielsen & Chuang, §2.4.2 and Exercise 2.72 — the Bloch-vector form $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$and the$|\vec r|\le1$ condition.
• [Pre] Preskill, Ph219, Ch. 2–3 — the Bloch ball, purity, and (forward to) qubit channels as affine maps of the ball.
• [Wil] Wilde, Quantum Information Theory, Ch. 4–9 — trace distance and the Bloch picture.
• Appendix E §1, §3 — Pauli completeness, $(\vec r\cdot\vec\sigma)^2 = |\vec r|^2 I$, and rotation exponentials.

---

← Prev: Partial Trace & Reduced States · Up: Term 1 · Next: Time Evolution →