Exercise: The Bloch Ball and the Geometry of Mixed States

Introduction

This article is a worked exercise on the geometry of the qubit state space. It is a set of six problems on one theme, and it applies the state-space results established in the companion article The Bloch Ball as the Trace-One Slice of the Future Light Cone. The reader is assumed to have read that article; the exercise exists precisely to test it, so nothing is carried over except what that article establishes.

The conventions are those of the companion articles. The biquaternion algebra is $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$, with the Hermitian subspace $\mathbb{M}_+$ and the anti-Hermitian subspace $\mathbb{M}_-$, and $\mathbb{B} = \mathbb{M}_+ \oplus \mathbb{M}_-$. The quaternion units are $e_0 = 1, e_1, e_2, e_3$ with $e_k^2 = -e_0$, and $i$ is the scalar imaginary, $i^2 = -1$. The real-quaternion subspace is $\mathbb{H}_{\mathbb{B}}$ and the center is $\mathbb{C}_{\mathbb{B}}$. The trace of an element of $\mathbb{M}_+$ is twice its scalar part, $\mathrm{Tr}(\tilde{H}) = 2\,\mathrm{Sc}(\tilde{H})$.

Assumed results. From the parent article we take, without rederivation, the following. A state is an element

$$ \tilde{\rho} = \tfrac{1}{2}\bigl(e_0 + i\mathbf{r}\bigr), \qquad \mathbf{r} = r_1 e_1 + r_2 e_2 + r_3 e_3 \in \mathbb{R}^3, $$

with $\mathbf{r}$ the Bloch vector; it is a state exactly when $|\mathbf{r}| \leq 1$, i.e. when it lies in the closed unit ball $B^3$. The biquaternion norm on the trace-one slice is $\tilde{\rho}\tilde{\rho}^{\natural} = \tfrac{1}{4}(1 - |\mathbf{r}|^2)e_0$. The eigenvalues are $\lambda_\pm = \tfrac{1}{2}(1 \pm |\mathbf{r}|)$. The purity is $\mathrm{Tr}(\tilde{\rho}^2) = \tfrac{1}{2}(1 + |\mathbf{r}|^2)$ and the linear entropy is $S_{\mathrm{lin}}(\tilde{\rho}) = 1 - \mathrm{Tr}(\tilde{\rho}^2) = \tfrac{1}{2}(1 - |\mathbf{r}|^2)$. The von Neumann entropy is $S(\tilde{\rho}) = -\lambda_+\log\lambda_+ - \lambda_-\log\lambda_-$. The trace pairing on the slice is

$$ \mathrm{Tr}(\tilde{\rho}\tilde{\sigma}) = \tfrac{1}{2}\bigl(1 + \mathbf{r}\cdot\mathbf{s}\bigr), \qquad \tilde{\sigma} = \tfrac{1}{2}\bigl(e_0 + i\mathbf{s}\bigr), $$

the squared Hilbert–Schmidt distance is $\mathrm{Tr}((\tilde{\rho}-\tilde{\sigma})^2) = \tfrac{1}{2}|\mathbf{r}-\mathbf{s}|^2$, the Uhlmann transition probability is $\tfrac{1}{2}(1 + \mathbf{r}\cdot\mathbf{s} + \sqrt{(1-|\mathbf{r}|^2)(1-|\mathbf{s}|^2)})$, and convex combinations act on Bloch vectors by the same weights. The pure states are the idempotents $\tilde\Pi_\pm(\hat{\boldsymbol{\mu}}) = \tfrac{1}{2}(e_0 \pm i\hat{\boldsymbol{\mu}})$ with $|\hat{\boldsymbol{\mu}}| = 1$, forming the boundary sphere, and the maximally mixed state is the center $\mathbf{r} = 0$.

What is to be shown. Six problems: (1) the parametrisation of a mixed state by its Bloch vector; (2) purity, linear entropy, and von Neumann entropy for explicit states; (3) the geometry of convex combinations; (4) the metric and distinguishability structure; (5) the center and the boundary; (6) when two Bloch vectors give identical or orthogonal states. Each is solved in full, and numerical values are given where they aid the check.

Problem 1: Parametrising a Mixed State by Its Bloch Vector

Problem. (a) A qubit is prepared by mixing the spin-up and spin-down states along the $z$-axis with probabilities $\tfrac34$ and $\tfrac14$. Write the state in the form $\tilde{\rho} = \tfrac{1}{2}(e_0 + i\mathbf{r})$ and read off $\mathbf{r}$. (b) Conversely, for $\mathbf{r} = \tfrac12(e_1 + e_2)$, write $\tilde{\rho}$ and its spectral decomposition. (c) Verify that under the isomorphism $e_j \mapsto -i\sigma_j$ the state maps to $\tfrac12(I + \mathbf{r}\cdot\boldsymbol{\sigma})$, and identify the components $r_k$.

Solution. (a) The two pure states along $\pm z$ are the idempotents

$$ \tilde\Pi_+(e_3) = \tfrac{1}{2}\bigl(e_0 + i e_3\bigr), \qquad \tilde\Pi_-(e_3) = \tilde\Pi_+(-e_3) = \tfrac{1}{2}\bigl(e_0 - i e_3\bigr). $$

The mixture is a convex combination, so its Bloch vector is the same weighted combination of $e_3$ and $-e_3$:

$$ \tilde{\rho} = \tfrac34 \tilde\Pi_+(e_3) + \tfrac14 \tilde\Pi_-(e_3) = \tfrac{3}{8}\bigl(e_0 + ie_3\bigr) + \tfrac{1}{8}\bigl(e_0 - ie_3\bigr) = \tfrac12 e_0 + \tfrac14 i e_3 = \tfrac12\Bigl(e_0 + i\,\tfrac12 e_3\Bigr). $$

Hence $\mathbf{r} = \tfrac12 e_3$, with $|\mathbf{r}| = \tfrac12 \leq 1$. The state is mixed, as expected from a nontrivial mixture, and its eigenvalues are $\lambda_\pm = \tfrac12(1 \pm \tfrac12) = \tfrac34, \tfrac14$: the two preparation probabilities, recovered from the Bloch vector.

(b) For $\mathbf{r} = \tfrac12(e_1 + e_2)$ we have $|\mathbf{r}|^2 = \tfrac14 + \tfrac14 = \tfrac12$, so $|\mathbf{r}| = 1/\sqrt2$. The state is

$$ \tilde{\rho} = \tfrac12\Bigl(e_0 + i\,\tfrac12(e_1 + e_2)\Bigr) = \tfrac12 e_0 + \tfrac14 i(e_1 + e_2). $$

With $\hat{\mathbf{r}} = \mathbf{r}/|\mathbf{r}| = (e_1 + e_2)/\sqrt2$ and $\lambda_\pm = \tfrac12(1 \pm 1/\sqrt2)$, the spectral decomposition is

$$ \tilde{\rho} = \lambda_+ \tilde\Pi_+(\hat{\mathbf{r}}) + \lambda_- \tilde\Pi_-(\hat{\mathbf{r}}), \qquad \tilde\Pi_\pm(\hat{\mathbf{r}}) = \tfrac12\bigl(e_0 \pm i\hat{\mathbf{r}}\bigr). $$

This is correct because the two idempotents reconstruct the state,

$$ \lambda_+ \tilde\Pi_+(\hat{\mathbf{r}}) + \lambda_- \tilde\Pi_-(\hat{\mathbf{r}}) = \tfrac12(\lambda_+ + \lambda_-)e_0 + \tfrac12(\lambda_+ - \lambda_-) i\hat{\mathbf{r}} = \tfrac12 e_0 + \tfrac{1}{2\sqrt2}\, i\,\frac{e_1 + e_2}{\sqrt2} = \tfrac12 e_0 + \tfrac14 i(e_1 + e_2), $$

where we used $\lambda_+ + \lambda_- = 1$ and $\lambda_+ - \lambda_- = 1/\sqrt2$. Every mixed state is thus a mixture of two orthogonal pure states along its own Bloch direction, with weights fixed by the radius.

(c) Since $e_0 \mapsto I$ and $e_j \mapsto -i\sigma_j$, we have $i e_j \mapsto i(-i\sigma_j) = \sigma_j$, so

$$ \tilde{\rho} = \tfrac12\bigl(e_0 + i\textstyle\sum_j r_j e_j\bigr) \;\longmapsto\; \tfrac12\Bigl(I + \sum_j r_j \sigma_j\Bigr) = \tfrac12\bigl(I + \mathbf{r}\cdot\boldsymbol{\sigma}\bigr), $$

the standard Bloch parametrisation of a qubit density matrix. The components are the Pauli expectations: using the trace pairing with the observable $i e_k \mapsto \sigma_k$,

$$ r_k = \mathrm{Tr}\bigl(\tilde{\rho}\,(i e_k)\bigr) = \langle \sigma_k\rangle_{\tilde{\rho}}, $$

so the Bloch vector is the vector of single-qubit expectation values, $\mathbf{r} = (\langle\sigma_1\rangle, \langle\sigma_2\rangle, \langle\sigma_3\rangle)$. In particular, measuring the spin along a unit direction $\hat{\mathbf{n}}$ gives outcomes with probabilities $p_\pm = \tfrac12(1 \pm \hat{\mathbf{n}}\cdot\mathbf{r})$, so the projection of $\mathbf{r}$ along $\hat{\mathbf{n}}$ is the bias of the measurement.

Problem 2: Purity, Linear Entropy, and von Neumann Entropy

Problem. For each of the following states compute the purity $\mathrm{Tr}(\tilde{\rho}^2)$, the linear entropy $S_{\mathrm{lin}}$, and the von Neumann entropy $S(\tilde{\rho})$: (a) the pure state with $\hat{\mathbf{n}} = (1,1,1)/\sqrt3$; (b) the maximally mixed state; (c) the state with $\mathbf{r} = \tfrac12 e_3$; (d) the state with $\mathbf{r} = \tfrac12(e_1 + e_2)$. Confirm that all three quantities depend on the state only through $|\mathbf{r}|$.

Solution. All three quantities are functions of the radius alone:

$$ \mathrm{Tr}(\tilde{\rho}^2) = \tfrac12\bigl(1 + |\mathbf{r}|^2\bigr), \qquad S_{\mathrm{lin}} = 1 - \mathrm{Tr}(\tilde{\rho}^2) = \tfrac12\bigl(1 - |\mathbf{r}|^2\bigr), $$

$$ S(\tilde{\rho}) = -\lambda_+\log\lambda_+ - \lambda_-\log\lambda_-, \qquad \lambda_\pm = \tfrac12\bigl(1 \pm |\mathbf{r}|\bigr), $$

so that states of equal radius — the concentric spheres of the ball — are equally pure and equally entropic.

(a) Pure state, $|\mathbf{r}| = 1$. The purity is $\tfrac12(1+1) = 1$, the linear entropy is $0$, the eigenvalues are $\lambda_\pm = 1, 0$, and $S = -1\log 1 - 0 = 0$. Purity is the boundary condition.

(b) Maximally mixed state, $|\mathbf{r}| = 0$. The purity is $\tfrac12(1+0) = \tfrac12$, the linear entropy is $\tfrac12$, the eigenvalues are $\lambda_\pm = \tfrac12, \tfrac12$, and

$$ S = -\tfrac12\log\tfrac12 - \tfrac12\log\tfrac12 = \log 2 \approx 0.693147 \ \text{nats} = 1 \ \text{bit}. $$

This is the maximum on the ball.

(c) $\mathbf{r} = \tfrac12 e_3$, $|\mathbf{r}| = \tfrac12$. The purity is $\tfrac12(1 + \tfrac14) = \tfrac58$, the linear entropy is $1 - \tfrac58 = \tfrac38$, and the eigenvalues are $\lambda_\pm = \tfrac34, \tfrac14$, so

$$ S = -\tfrac34\log\tfrac34 - \tfrac14\log\tfrac14 = \tfrac34\log\tfrac43 + \tfrac14\log 4 \approx 0.562335 \ \text{nats} \approx 0.811278 \ \text{bits}. $$

(d) $\mathbf{r} = \tfrac12(e_1+e_2)$, $|\mathbf{r}| = 1/\sqrt2$. The purity is $\tfrac12(1 + \tfrac12) = \tfrac34$, the linear entropy is $\tfrac14$, and with $\lambda_\pm = \tfrac12(1 \pm 1/\sqrt2) \approx 0.853553, 0.146447$,

$$ S \approx 0.416496 \ \text{nats} \approx 0.600876 \ \text{bits}. $$

Comparing (c) and (d) illustrates the monotonicity: as the radius grows from $\tfrac12$ to $1/\sqrt2$, the purity grows from $\tfrac58$ to $\tfrac34$ and the entropy falls from about $0.5623$ to about $0.4165$ nats. As in the parent article, the base of the logarithm is a convention; natural logarithms (nats) are used here, and the bit values are the same numbers divided by $\log 2$.

Problem 3: Convex Combinations and Where Mixtures Land

Problem. (a) Show that a convex combination of states is a state whose Bloch vector is the same convex combination of the Bloch vectors. (b) For a mixture of two pure states with Bloch directions $\hat{\boldsymbol{\mu}}, \hat{\boldsymbol{\nu}}$ at angle $\theta$, with weights $p$ and $1-p$, compute the radius and the purity of the mixture. (c) Determine exactly when such a mixture is again pure. (d) Evaluate the mixture of the pure states along $z$ and along $x$ with equal weights, and the mixture with weights $\tfrac34, \tfrac14$.

Solution. (a) Write $\tilde{\rho}_i = \tfrac12(e_0 + i\mathbf{r}_i)$ and let $p_i \geq 0$ with $\sum_i p_i = 1$. Then

$$ \sum_i p_i \tilde{\rho}_i = \tfrac12\sum_i p_i e_0 + \tfrac12 i \sum_i p_i \mathbf{r}_i = \tfrac12\Bigl(e_0 + i \sum_i p_i \mathbf{r}_i\Bigr). $$

Since each $\tilde{\rho}_i$ is Hermitian of trace one, so is the combination, and its Bloch vector is $\mathbf{r} = \sum_i p_i \mathbf{r}_i$. By convexity of the ball, $|\mathbf{r}| \leq \sum_i p_i |\mathbf{r}_i| \leq 1$, so the combination is again a state. The map $\mathbf{r} \leftrightarrow \tilde{\rho}$ is affine, so the ball is a faithful affine model of the state space.

(b) Let $\hat{\boldsymbol{\mu}}, \hat{\boldsymbol{\nu}}$ be unit vectors with $\hat{\boldsymbol{\mu}}\cdot\hat{\boldsymbol{\nu}} = \cos\theta$, and set $\mathbf{r} = p\hat{\boldsymbol{\mu}} + (1-p)\hat{\boldsymbol{\nu}}$. Then

$$ |\mathbf{r}|^2 = p^2 + (1-p)^2 + 2p(1-p)\cos\theta = 1 - 2p(1-p)\bigl(1 - \cos\theta\bigr). $$

The mixture therefore lies on the chord joining $\hat{\boldsymbol{\mu}}$ and $\hat{\boldsymbol{\nu}}$, at the point that divides it in the ratio $(1-p) : p$. Its purity and linear entropy are

$$ \mathrm{Tr}(\tilde{\rho}^2) = \tfrac12\bigl(1 + |\mathbf{r}|^2\bigr) = 1 - p(1-p)\bigl(1-\cos\theta\bigr), \qquad S_{\mathrm{lin}} = p(1-p)\bigl(1-\cos\theta\bigr). $$

(c) From the second form, $|\mathbf{r}| = 1$ requires $p(1-p)(1-\cos\theta) = 0$, i.e. $p = 0$, or $p = 1$, or $\cos\theta = 1$ (the two pure states coincide). Apart from these degenerate cases the mixture is strictly interior: $|\mathbf{r}| < 1$, and the ball is strictly convex. In particular, a nontrivial mixture of two distinct pure states is never pure.

(d) Take $\hat{\boldsymbol{\mu}} = e_3$ and $\hat{\boldsymbol{\nu}} = e_1$, so $\theta = \pi/2$ and $\cos\theta = 0$. With $p = \tfrac12$,

$$ \mathbf{r} = \tfrac12(e_1 + e_3), \qquad |\mathbf{r}|^2 = 1 - 2\cdot\tfrac14\cdot 1 = \tfrac12, \qquad |\mathbf{r}| = 1/\sqrt2, $$

so the purity is $\tfrac34$, the linear entropy is $\tfrac14$, and the entropy is about $0.416496$ nats — the value found in Problem 2(d), as it must be. With $p = \tfrac34$,

$$ \mathbf{r} = \tfrac34 e_3 + \tfrac14 e_1, \qquad |\mathbf{r}|^2 = 1 - 2\cdot\tfrac34\cdot\tfrac14 = \tfrac58, $$

so the purity is $\tfrac12(1 + \tfrac58) = \tfrac{13}{16}$ and the linear entropy is $\tfrac{3}{16}$. A useful special case is $\theta = \pi$ (complementary pure states $\hat{\boldsymbol{\nu}} = -\hat{\boldsymbol{\mu}}$): then $\mathbf{r} = (2p-1)\hat{\boldsymbol{\mu}}$, $|\mathbf{r}| = |2p-1|$, and the equal mixture $p = \tfrac12$ gives $\mathbf{r} = 0$, the maximally mixed state, independently of the axis $\hat{\boldsymbol{\mu}}$.

Problem 4: The Metric and Distinguishability Structure

Problem. (a) Compute the squared Hilbert–Schmidt distance $\mathrm{Tr}((\tilde{\rho}-\tilde{\sigma})^2)$ in terms of the Bloch vectors. (b) Compute the trace distance $D(\tilde{\rho},\tilde{\sigma}) = \tfrac12\mathrm{Tr}|\tilde{\rho}-\tilde{\sigma}|$, verify the parent's formula for the distance to the center, and relate the two distances. (c) Evaluate both for $\mathbf{r} = \tfrac12 e_3$ and $\mathbf{s} = \tfrac12 e_1$, and compute the trace distance between two pure states as a function of the angle. (d) State the range of $D$ on the ball.

Solution. (a) The difference of two states is

$$ \tilde{\rho} - \tilde{\sigma} = \tfrac12 i(\mathbf{r} - \mathbf{s}), \qquad \mathbf{r} - \mathbf{s} \in \mathbb{H}_{\mathbb{B}} \text{ a pure real quaternion.} $$

For a pure real quaternion $\mathbf{u} = \mathbf{r} - \mathbf{s}$ one has $\mathbf{u}^2 = -|\mathbf{u}|^2 e_0$, hence

$$ (\tilde{\rho} - \tilde{\sigma})^2 = \tfrac14 (i\mathbf{u})^2 = -\tfrac14 \mathbf{u}^2 = \tfrac14 |\mathbf{r}-\mathbf{s}|^2\, e_0 . $$

Taking the trace, $\mathrm{Tr}((\tilde{\rho}-\tilde{\sigma})^2) = 2\cdot\tfrac14|\mathbf{r}-\mathbf{s}|^2 = \tfrac12|\mathbf{r}-\mathbf{s}|^2$, which is the parent's Hilbert–Schmidt formula. Since $(\tilde{\rho}-\tilde{\sigma})^2$ is a scalar multiple of $e_0$, the traceless difference $\tilde{\rho}-\tilde{\sigma}$ has eigenvalues $\pm\tfrac12|\mathbf{r}-\mathbf{s}|$.

(b) The absolute value is

$$ |\tilde{\rho}-\tilde{\sigma}| = \sqrt{(\tilde{\rho}-\tilde{\sigma})^2} = \tfrac12|\mathbf{r}-\mathbf{s}|\,e_0, $$

the positive square root in $\mathbb{M}_+$. Hence

$$ D(\tilde{\rho},\tilde{\sigma}) = \tfrac12\mathrm{Tr}|\tilde{\rho}-\tilde{\sigma}| = \tfrac12\cdot\tfrac12|\mathbf{r}-\mathbf{s}|\cdot\mathrm{Tr}(e_0) = \tfrac12|\mathbf{r}-\mathbf{s}| . $$

Setting $\mathbf{s} = 0$ recovers the parent's statement $D(\tilde{\rho},\tfrac12 e_0) = \tfrac12|\mathbf{r}|$. Comparing with (a), the trace distance and the Hilbert–Schmidt distance carry the same information on a qubit,

$$ \mathrm{Tr}\bigl((\tilde{\rho}-\tilde{\sigma})^2\bigr) = 2\,D(\tilde{\rho},\tilde{\sigma})^2, \qquad D = \sqrt{\tfrac12\,\mathrm{Tr}\bigl((\tilde{\rho}-\tilde{\sigma})^2\bigr)} . $$

Both are monotone functions of the Euclidean separation $|\mathbf{r}-\mathbf{s}|$; the trace distance is the one with the operational meaning, being half the trace-norm difference of the two states.

(c) For $\mathbf{r} = \tfrac12 e_3$ and $\mathbf{s} = \tfrac12 e_1$, $|\mathbf{r}-\mathbf{s}| = \tfrac12\sqrt2 = 1/\sqrt2$, so the Hilbert–Schmidt distance squared is $\tfrac12\cdot\tfrac12 = \tfrac14$, and

$$ D = \tfrac12\cdot\tfrac1{\sqrt2} = \frac{1}{2\sqrt2} \approx 0.353553 . $$

For two pure states $\mathbf{r} = \hat{\boldsymbol{\mu}}$, $\mathbf{s} = \hat{\boldsymbol{\nu}}$ at angle $\theta$, $|\mathbf{r}-\mathbf{s}| = \sqrt{2 - 2\cos\theta} = 2\sin(\theta/2)$, so

$$ D\bigl(\tilde\Pi(\hat{\boldsymbol{\mu}}), \tilde\Pi(\hat{\boldsymbol{\nu}})\bigr) = \sin\frac{\theta}{2}. $$

Together with the transition probability $\mathrm{Tr}(\tilde\Pi(\hat{\boldsymbol{\mu}})\tilde\Pi(\hat{\boldsymbol{\nu}})) = \cos^2(\theta/2)$ this gives the clean pair

$$ D^2 + \mathrm{Tr}(\tilde{P}\tilde{Q}) = 1 \qquad \text{for pure states,} $$

so the two notions of separation are complementary on the boundary: coincident directions give $D = 0$ and transition probability $1$, antipodal directions give $D = 1$ and transition probability $0$.

(d) Since $|\mathbf{r}-\mathbf{s}| \leq |\mathbf{r}| + |\mathbf{s}| \leq 2$, we have $0 \leq D \leq 1$. The upper bound is attained exactly when $|\mathbf{r}| = |\mathbf{s}| = 1$ and $\mathbf{s} = -\mathbf{r}$, the case of orthogonal states treated in Problem 6. The trace distance is thus bounded by the information-theoretic maximum $1$, the probability of perfectly distinguishing the two states, and it reaches it only at the two ends of a diameter.

Problem 5: The Centre and the Boundary

Problem. (a) Characterize the center of the ball and compute its purity, entropies, and biquaternion norm. (b) Characterize the boundary and show that a boundary state is an extreme point that cannot be written as a nontrivial convex combination. (c) Show that the elements of the trace-one hyperplane with $|\mathbf{r}| > 1$ are not states.

Solution. (a) The center is $\mathbf{r} = 0$, i.e.

$$ \tilde{\rho} = \tfrac12 e_0, $$

the maximally mixed state. Its purity is $\mathrm{Tr}(\tilde{\rho}^2) = \tfrac12(1+0) = \tfrac12$, the minimum on the ball; its linear entropy is $\tfrac12$, the maximum; its biquaternion norm is $\tilde{\rho}\tilde{\rho}^{\natural} = \tfrac14 e_0$, whose scalar coefficient $\tfrac14$ is the largest attainable on the slice; and its von Neumann entropy is $\log 2$, the maximum. In the spectral form it is the equal mixture $\tfrac12\tilde\Pi_+(\hat{\boldsymbol{\mu}}) + \tfrac12\tilde\Pi_-(\hat{\boldsymbol{\mu}})$ of the complementary idempotents along any axis, and it is the unique state invariant under the full unitary group $U(2)$, which acts on the ball by rotations. It is also the barycenter: the uniform average of the pure states over the boundary sphere is

$$ \frac{1}{4\pi}\int_{S^2}\tilde\Pi(\hat{\boldsymbol{\mu}})\,d\Omega = \tfrac12\Bigl(e_0 + i\,\frac{1}{4\pi}\int_{S^2}\hat{\boldsymbol{\mu}}\,d\Omega\Bigr) = \tfrac12 e_0, $$

because the mean of the unit vector over the sphere vanishes.

(b) The boundary is $|\mathbf{r}| = 1$. For these states the biquaternion norm vanishes,

$$ \tilde{\rho}\tilde{\rho}^{\natural} = \tfrac14(1 - |\mathbf{r}|^2)e_0 = 0 \quad\Longleftrightarrow\quad |\mathbf{r}| = 1, $$

so a boundary state is a zero divisor of $\mathbb{B}$; the deviation from idempotency,

$$ \tilde{\rho}^2 - \tilde{\rho} = \tfrac14\bigl(|\mathbf{r}|^2 - 1\bigr)e_0, $$

vanishes exactly there, so a boundary state is an idempotent, hence a rank-one projection; and its eigenvalues are $\lambda_\pm = 1, 0$, so its purity is $1$ and its entropy $0$. The boundary is parametrized by the unit sphere $S^2$: $\mathbf{r} = \hat{\boldsymbol{\mu}}$ gives the idempotent $\tilde\Pi_+(\hat{\boldsymbol{\mu}})$, and $-\hat{\boldsymbol{\mu}}$ gives its orthogonal complement $\tilde\Pi_-(\hat{\boldsymbol{\mu}})$, with $\tilde\Pi_+ + \tilde\Pi_- = e_0$ and $\tilde\Pi_+\tilde\Pi_- = 0$.

A boundary state is an extreme point of the ball. Suppose $\tilde{P} = \lambda\tilde{\rho}_1 + (1-\lambda)\tilde{\rho}_2$ with $0 < \lambda < 1$. Then $\mathbf{r} = \lambda\mathbf{r}_1 + (1-\lambda)\mathbf{r}_2$ with $|\mathbf{r}_1|, |\mathbf{r}_2| \leq 1$ and $|\mathbf{r}| = 1$. By the strict convexity of the Euclidean norm, equality $|\mathbf{r}| = 1$ forces $\mathbf{r}_1 = \mathbf{r}_2 = \mathbf{r}$, so $\tilde{\rho}_1 = \tilde{\rho}_2 = \tilde{P}$: no boundary state is a nontrivial mixture. Conversely, every interior state is a nontrivial mixture (Problem 3), so the extreme points of the ball are exactly the pure states.

(c) On the trace-one hyperplane an element $\tilde{\rho} = \tfrac12(e_0 + i\mathbf{r})$ has eigenvalues $\lambda_\pm = \tfrac12(1 \pm |\mathbf{r}|)$. If $|\mathbf{r}| > 1$ then $\lambda_- = \tfrac12(1 - |\mathbf{r}|) < 0$, so $\tilde{\rho}$ is Hermitian of trace one but not positive semidefinite, and is not a state. Its biquaternion norm is $\tfrac14(1 - |\mathbf{r}|^2)e_0$ with negative scalar coefficient, so it lies in the spacelike region outside the future cone. For instance $\mathbf{r} = 2e_3$ gives $\lambda_\pm = \tfrac32, -\tfrac12$. The ball $|\mathbf{r}| \leq 1$ is thus exactly the positivity domain on the slice.

Problem 6: Identical and Orthogonal States

Problem. (a) Determine when two states are identical. (b) Determine when two states have orthogonal support, i.e. when $\mathrm{Tr}(\tilde{\rho}\tilde{\sigma}) = 0$. (c) Specialize to pure states and express the result in terms of the angle. (d) Show that the Uhlmann transition probability vanishes under the same condition.

Solution. (a) Two states are identical exactly when their Bloch vectors are equal: from $\tilde{\rho} - \tilde{\sigma} = \tfrac12 i(\mathbf{r}-\mathbf{s})$ we have $\tilde{\rho} = \tilde{\sigma}$ iff $\mathbf{r} = \mathbf{s}$. In that case the Hilbert–Schmidt and trace distances vanish, the transition probability is $1$, and all radius-dependent quantities (purity, entropies) coincide. Conversely, equal purity does not imply identical states: by Problem 2 all states of a given radius share the same purity and entropy, and they are rotated into one another by $U(2)$.

(b) The trace pairing on the slice is $\mathrm{Tr}(\tilde{\rho}\tilde{\sigma}) = \tfrac12(1 + \mathbf{r}\cdot\mathbf{s})$. Because $\tilde{\rho},\tilde{\sigma}$ are positive and of trace one, orthogonality of their supports is equivalent to the vanishing of this pairing. Thus

$$ \mathrm{Tr}(\tilde{\rho}\tilde{\sigma}) = 0 \quad\Longleftrightarrow\quad \mathbf{r}\cdot\mathbf{s} = -1 . $$

Since $|\mathbf{r}\cdot\mathbf{s}| \leq |\mathbf{r}|\,|\mathbf{s}| \leq 1$, the equality $\mathbf{r}\cdot\mathbf{s} = -1$ forces both Cauchy–Schwarz and the ball bounds to be saturated:

$$ |\mathbf{r}| = |\mathbf{s}| = 1, \qquad \mathbf{s} = -\mathbf{r}. $$

So two qubit states have orthogonal supports if and only if they are the two complementary pure states along a common axis.

(c) For pure states $\mathbf{r} = \hat{\boldsymbol{\mu}}$, $\mathbf{s} = \hat{\boldsymbol{\nu}}$, the transition probability is the trace pairing,

$$ \mathrm{Tr}\bigl(\tilde\Pi(\hat{\boldsymbol{\mu}})\tilde\Pi(\hat{\boldsymbol{\nu}})\bigr) = \tfrac12\bigl(1 + \hat{\boldsymbol{\mu}}\cdot\hat{\boldsymbol{\nu}}\bigr) = \cos^2\frac{\theta}{2}, $$

where $\theta$ is the angle between the Bloch directions. It equals $1$ for $\theta = 0$ (identical states) and $0$ for $\theta = \pi$ (orthogonal states, $\hat{\boldsymbol{\nu}} = -\hat{\boldsymbol{\mu}}$). Equivalently, $\hat{\boldsymbol{\mu}}\cdot\hat{\boldsymbol{\nu}} = -1$, the same condition as in (b). As a concrete check, take $\hat{\boldsymbol{\mu}} = e_3$, $\hat{\boldsymbol{\nu}} = -e_3$: then $\tilde\Pi_+(e_3)\tilde\Pi_-(e_3) = 0$ and $\mathrm{Tr}(\tilde\Pi_+(e_3)\tilde\Pi_-(e_3)) = \tfrac12(1 - 1) = 0$, while $\tilde\Pi_+ + \tilde\Pi_- = e_0$.

(d) For general states the relevant overlap is the Uhlmann transition probability,

$$ T(\tilde{\rho},\tilde{\sigma}) = \tfrac12\Bigl(1 + \mathbf{r}\cdot\mathbf{s} + \sqrt{\bigl(1-|\mathbf{r}|^2\bigr)\bigl(1-|\mathbf{s}|^2\bigr)}\Bigr), $$

which for mixed states differs from the Hilbert–Schmidt pairing $\mathrm{Tr}(\tilde{\rho}\tilde{\sigma})$. Setting $T = 0$ gives

$$ \mathbf{r}\cdot\mathbf{s} = -1 - \sqrt{\bigl(1-|\mathbf{r}|^2\bigr)\bigl(1-|\mathbf{s}|^2\bigr)} \leq -1 . $$

Combined with $\mathbf{r}\cdot\mathbf{s} \geq -|\mathbf{r}|\,|\mathbf{s}| \geq -1$, every inequality is an equality, so $|\mathbf{r}| = |\mathbf{s}| = 1$, $\mathbf{r}\cdot\mathbf{s} = -1$, and $\mathbf{s} = -\mathbf{r}$ — the same pure antipodal pair. Hence the transition probability vanishes under exactly the same condition as orthogonal support, and for mixed states it is strictly positive: a mixed state is never perfectly distinguishable from another. For example, with $\mathbf{r} = \tfrac12 e_3$ and $\mathbf{s} = \tfrac12 e_1$ one has $\mathbf{r}\cdot\mathbf{s} = 0$ and $\sqrt{(1-\tfrac14)(1-\tfrac14)} = \tfrac34$, so $T = \tfrac12(1 + \tfrac34) = \tfrac78$, consistent with the nonzero trace distance $D = 1/(2\sqrt2)$ found in Problem 4.

Summary

We have worked six problems on the geometry of the qubit state space, using only the state-space results of the parent article.

Parametrisation. A state is $\tilde{\rho} = \tfrac12(e_0 + i\mathbf{r})$ with $|\mathbf{r}| \leq 1$; the components of $\mathbf{r}$ are the single-qubit expectation values $r_k = \langle\sigma_k\rangle$, and a measurement along $\hat{\mathbf{n}}$ has outcomes $\tfrac12(1 \pm \hat{\mathbf{n}}\cdot\mathbf{r})$.

Purity and entropy. Purity, linear entropy, and von Neumann entropy are functions of the radius alone: $\mathrm{Tr}(\tilde{\rho}^2) = \tfrac12(1+|\mathbf{r}|^2)$, $S_{\mathrm{lin}} = \tfrac12(1-|\mathbf{r}|^2)$, and $S = -\sum_\pm\lambda_\pm\log\lambda_\pm$ with $\lambda_\pm = \tfrac12(1\pm|\mathbf{r}|)$.

Convex combinations. A mixture of two pure states lies on the chord joining them, with $|\mathbf{r}|^2 = 1 - 2p(1-p)(1-\cos\theta)$ and linear entropy $p(1-p)(1-\cos\theta)$; it is pure only in the degenerate cases $p \in \{0,1\}$ or coincident directions.

Metric. The trace distance is $D = \tfrac12|\mathbf{r}-\mathbf{s}|$ and the squared Hilbert–Schmidt distance is $2D^2$. For pure states $D = \sin(\theta/2)$ and the transition probability is $\cos^2(\theta/2)$, so $D^2 + \mathrm{Tr}(\tilde{P}\tilde{Q}) = 1$.

Centre and boundary. The center is the maximally mixed state $\tfrac12 e_0$, with minimal purity $\tfrac12$ and maximal entropy $\log 2$; the boundary $|\mathbf{r}| = 1$ consists of the pure states, equivalently the idempotents, the zero divisors, and the extreme points. Elements with $|\mathbf{r}| > 1$ are not states.

Identical and orthogonal. Two states are identical iff $\mathbf{r} = \mathbf{s}$; they have orthogonal supports — and the Uhlmann transition probability vanishes — iff $\mathbf{r}\cdot\mathbf{s} = -1$, which forces $|\mathbf{r}| = |\mathbf{s}| = 1$ and $\mathbf{s} = -\mathbf{r}$. Only the two ends of a diameter are perfectly distinguishable.

Summary of Notation

Further Reading

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information (Cambridge, 2000), for the Bloch ball, the trace distance, and the fidelity.
• Ingemar Bengtsson and Karol Życzkowski, Geometry of Quantum States (Cambridge, 2006), for the convex and metric geometry of the state space.
• Asher Peres, Quantum Theory: Concepts and Methods (Kluwer, 1995), for the operational meaning of distinguishability and the trace distance.
• Richard Jozsa, "Fidelity for mixed quantum states," Journal of Modern Optics 41 (1994) 2315–2323, for the closed-form Uhlmann transition probability of a qubit.
• J. J. Sakurai and Jim Napolitano, Modern Quantum Mechanics (Pearson, 2017), for the spin-1/2 formalism and the Bloch vector.
• The companion article of this series: The Bloch Ball as the Trace-One Slice of the Future Light Cone, whose state-space results are applied throughout.