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Use the optimal three-to-one qubit random access code. For Alice's bits $(b_1,b_2,b_3)$, put $s_j=(-1{)}^{b_j}$ and send the pure state with Bloch vectorThese are the eight vertices of a cube inscribed in the Bloch sphere. If an explicit ket is wanted, choose $\cos {\theta }_b=s_3/\sqrt{3}$ and $e^{i{\phi }_b}=(s_1+i{s}_2)/\sqrt{2}$, and prepare $\cos ({\theta }_b/2)|0⟩+e^{i{\phi }_b}\sin ({\theta }_b/2)|1⟩$.

Bob wanting bit $j$ performs a Pauli measurement of ${\sigma }_x,{\sigma }_y,{\sigma }_z$ respectively, reporting zero for eigenvalue $+1$ and one for $-1$. Its success for every input and every requested bit isOnly one chosen bit is extracted; the protocol does not recover all three simultaneously and uses no shared entanglement.

To prove that no one-qubit protocol gives a greater guaranteed probability, allow Alice arbitrary mixed encoding vectors $|r_b|\le 1$ and Bob any binary positive operator-valued measure. Write his decision difference as $D_j=E_{j,0}-E_{j,1}=t_{j}I+v_j\cdot \sigma$. Its eigenvalues lie in $[-1,1]$, implying $|v_j|\le 1-|t_j|\le 1$. On input $b$, success for bit $j$ is $[1+s_j(t_j+v_j\cdot r_b)]/2$. Average over all eight strings and three choices. The terms in $t_j$ cancel, leavingThe Cauchy-Schwarz inequality and cancellation of cross terms over all sign choices giveHence $p_{\mathrm{avg}}\le p_{\ast }$. Any worst-case guarantee is no greater than this average; the cube protocol attains it uniformly, proving optimality both for guaranteed success and for the uniform-input average. Biased prior information about the bits would define a different optimization problem.

The PDF's first Pauli term contains an undefined ${\rho }_x$. For the stated depolarizing channel, use ${\sigma }_x\rho {\sigma }_x$, consistently with the other two terms. This is the minimal typographical correction needed to define the intended Pauli channel.

Write $\rho =\frac{1}{2}(I+\mathbf{s}\cdot \mathbf{\sigma })$ using its Bloch vector. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k{\sigma }_j={\sigma }_k$ for $j=k$ and $-{\sigma }_k$ for $j\ne k$. Thus conjugating by each nonidentity Pauli matrix leaves one Bloch component unchanged and reverses the other two. Summing the three conjugates givesSubstitution yieldsso the Pauli-mixture parametrization of qubit depolarization, with $p$ here the total error probability, givesThere is also a genuine range error in the printed claim. For its full $0<p<1$ assumption, $0<q<4/3$, and $0<q<1$ holds exactly when $0<p<3/4$. For example $p=9/10$ gives $q=6/5$; on $|0⟩⟨0|$ the output is $\mathrm{diag}(2/5,3/5)$, with negative $z$ Bloch component. No $q\in (0,1)$ could produce this output from that input, since its component would be $1-q>0$.

The corrected channel is still completely positive in the larger range: its original Pauli probabilities are $1-p,p/3,p/3,p/3$. This is depolarizing noise weight above one; the formula with $q>1$ is not a convex mixture with weights $1-q,q$, but it has a valid random unitary channel representation. The subsequent parts can therefore be solved both in the printed $q\in (0,1)$ regime and in the actual parameter range.

From the derived formula,so the action on the Bloch vector isFor $0<q<1$, every direction is contracted by the same positive factor. The pure-state Bloch sphere becomes a concentric sphere of radius $1-q$, and the Bloch ball contracts towards the maximally mixed density operator $I/2$. The lack of a preferred direction explains the name depolarizing channel.

At $q=1$ all inputs become $I/2$. For the allowed Pauli-error range $1<q<4/3$, the factor is negative, so there is also an antipodal reversal; the radius is $q-1\le 1/3$. Thus it would be incorrect to describe the whole printed $p\in (0,1)$ range as positive radial contraction.

The orthogonal projection onto the $+1$ eigenspace of the Pauli matrix ${\sigma }_z$ is $P_{+}=(I+{\sigma }_z)/2$. By the Born rule and the Bloch vector representation,Since $s_z=1/5$,The given vector has length $19/30<1$, so it lies inside the Bloch ball and indeed defines a valid mixed qubit state. For the physical spin operator $S_z=(\hslash /2){\sigma }_z$, the corresponding outcome is $+\hslash /2$ with the same probability.

Normalize the vector defining the first effect and write $E_1=\alpha |\psi ⟩⟨\psi |$. Since the two effects sum to $I$, the eigenvalues of $E_2=I-E_1$ are $1-\alpha$ along $\psi$ and $1$ on its orthogonal complement. For $E_2$ to have rank one, $\alpha =1$. Hence this binary rank-one qubit POVM is a projective measurement:The two Bloch vectors are antipodal on the Bloch sphere; the corresponding state vectors are orthogonal. The plot below chooses one representative unit vector. Rotating both directions gives every other case. The sphere represents the effect projectors' Bloch directions, rather than plotting their matrix entries.

A centre $c$ turns position into a displacement $x-c$. If $v_1,\dots ,v_N$ are a basis of the $N$-dimensional vector space, every position has unique coordinatesThus $N$ linearly independent vectors are necessary and sufficient. Fewer vectors span a proper subspace and cannot represent arbitrary displacements. The coordinates use the field defining the space. For qubit tomography, the relevant parameter space is the real three-dimensional space of Bloch vectors, rather than the complex two-dimensional space of state amplitudes.

Take any rank-one orthogonal projection $P$ and its generalized Bloch vector $s$. The antipodal vector $-s$ is on the same sphere, but its corresponding trace-one Hermitian matrix isOn the range of $P$, this matrix has eigenvalue $2/N-1$, which is negative for $N>2$. It is therefore not a positive semidefinite matrix and cannot be a density operator. Hence the pure states do not cover the sphere when $N>2$. This also shows directly why the purity bound for generalized Bloch vectors does not by itself guarantee a physical quantum state.

Write the unitary conjugation as $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. The identity component is unchanged, and the generalized Bloch vector transforms byThese coefficients are real, by the same trace argument as in part (a). Unitary conjugation preserves the Hilbert-Schmidt inner product, so the conjugated traceless generators form another orthonormal basis. It follows that $O(U{)}^{T}O(U)=I$. The unitary group is connected, $O(I)=I$, and the determinant of an orthogonal matrix is $\pm 1$; hence $\det O(U)=1$. Thus Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$.

Equivalently, in units $\hslash =1$, the Von Neumann equation givesCyclicity of the trace shows $A_{k\ell }=-A_{\ell k}$, so $A$ is a real skew-symmetric matrix. This is the generator form of Hamiltonian rotations of Bloch vectors, valid also for a time-dependent Quantum Hamiltonian. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator spectrum, not merely its purity of a density operator.

The individual rates describe population relaxation: ${\gamma }_{12}$ transfers population from level one to level two, and ${\gamma }_{21}$ transfers it back. With $p_1+p_2=1$ and the conventional qubit coordinate $r_z=p_1-p_2$,Thus $T_1^{-1}={\gamma }_{12}+{\gamma }_{21}$ is the longitudinal relaxation rate. The parameter $\Gamma =T_2^{-1}$ is the transverse relaxation rate: it damps the off-diagonal coherence, or the $x,y$ components of the Bloch vector. In the usual completely positive two-level model,so $\Gamma$ includes both the coherence loss caused by population relaxation and additional pure dephasing.

There is a normalization switch in the displayed qubit Affine Bloch equation. Its constant term is appropriate to the conventional Pauli matrices and unit-radius coordinate $r_k=\mathrm{Tr}(\rho {\sigma }_k^{\mathrm{Pauli}})$. With the orthonormal generators used earlier, $s=r/\sqrt{2}$, the same physical rates would instead give a constant term $({\gamma }_{21}-{\gamma }_{12})/\sqrt{2}$. The drift matrix is unchanged. We interpret the supplied qubit equation in its conventional $r$ coordinates.

With zero controls and $g={\gamma }_{12}+{\gamma }_{21}>0$, $\Gamma >0$, the steady state is unique:The earlier normalized vector is $s_{\ast }=r_{\ast }/\sqrt{2}$. Indeed $r_x,r_y$ decay as $e^{-\Gamma t}$ and $r_z-r_{\ast ,z}$ as $e^{-gt}$.

For vanishing rates, the full equilibrium conditions of the Affine Bloch equation are $\Gamma r_x=\Gamma r_y=0$ and $g{r}_z={\gamma }_{21}-{\gamma }_{12}$, together with $\Vert r\Vert \le 1$. In particular, for nonnegative rates, $g=0$ means both population rates vanish. If then $\Gamma >0$, all diagonal density operators are stationary; if also $\Gamma =0$, every density operator is stationary. The formal case $g>0$, $\Gamma =0$ permits extra transverse equilibrium coordinates, but is excluded by the completely positive rate bound above.

With access to the uncontrolled physical gate, use single-qubit process tomography. Prepare the positive eigenstate of each of $X,Y,Z$, apply $V$, and on independent copies measure each of $X,Y,Z$. Arbitrary single-qubit rotation gates let one prepare these inputs and rotate each measurement into the computational basis. The nine measured means areThey determine the real rotation matrix $R$ of the Bloch vector, and hence the action of $V$ on every qubit state. To see precisely what this identifies, suppose another unitary $U$ has the same action. Then $V^{†}U$ commutes with every Pauli matrix. Commuting with $Z$ makes it diagonal; commuting with $X$ makes its diagonal entries equal. Therefore $U=e^{i\gamma }V$.

Consequently tomography determines $V$, and its eigenvalues, up to a common phase. More explicitly, write a representative as $V=e^{i\gamma }(\cos aI-i\sin a\mathbf{n}\cdot \mathbf{\sigma })$. Its eigenvalues are $e^{i(\gamma -a)}$ and $e^{i(\gamma +a)}$, and its Bloch vector action is rotation through $2a$ about $\mathbf{n}$. The measured rotation identifies the unordered relative eigenphase information; for example $\mathrm{Tr}R=1+2\cos (2a)$. Once a representative with a specified overall phase is supplied, diagonalizing that representative gives the corresponding two eigenvalues.

The common phase itself cannot be recovered with the stated uncontrolled access. Indeed, $(e^{i\gamma }V)\rho (e^{i\gamma }V{)}^{†}=V\rho V^{†}$ for every input, including inputs entangled with ancillas. Replacing $V$ by $e^{i\gamma }V$ therefore leaves every such preparation-and-measurement experiment unchanged, whereas it multiplies both eigenvalues by $e^{i\gamma }$. For example, $I$ and $e^{i\gamma }I$ are indistinguishable physical gates under these operations. This proves that the request for absolute eigenvalue phases needs an additional phase reference or controlled-gate assumption: the global phase of an uncontrolled quantum gate is unobservable. The controlled powers in the earlier parts supply that extra resource, since $\mathrm{diag}(I,e^{i\gamma }V)$ changes a relative phase between the two control branches.

The practical advantages are that no unknown eigenstate preparation, controlled powers, entangling gates or inverse quantum Fourier transform are needed: a fixed set of single-qubit preparations and measurements suffices. The disadvantages are statistical estimation and the phase ambiguity just proved. Estimating the bounded Pauli measurement means to additive accuracy $ϵ$ at fixed confidence requires order ${ϵ}^{-2}$ independent repetitions, giving the usual sampling cost for this direct procedure. The earlier quantum phase estimation circuit uses $N$ phase qubits and high controlled powers to resolve scale $2^{-N}$ in one coherent experiment with a constant success bound; constructing those powers from controlled uses of $V$ costs $2^N-1$ gate calls, so its exponential-looking precision is not obtained with only $N$ uses of $V$. Repetitions can raise confidence. Finally, full process tomography grows exponentially with the number of system qubits, whereas the eigenstate-based phase procedure can estimate a selected eigenphase without reconstructing the entire operator.

A density operator is Hermitian, positive semidefinite and has trace one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:The remaining basis vectors are traceless by orthogonality to $I/\sqrt{N}$. The Generalized Bloch representation is consequentlyUsing orthonormality to expand the squared Hilbert-Schmidt norm gives the purity bound for generalized Bloch vectors:If $p_1,\dots ,p_N$ are its nonnegative eigenvalues, then ${\sum }_j{p}_j=1$ andIt follows thatEquality requires every pairwise product $p_j{p}_k$ to vanish, so exactly one eigenvalue equals one and all others are zero. Conversely, a rank-one projector has precisely that spectrum and attains equality. Thus equality holds exactly for a pure state. This normalization is different from the commonly used unit-radius qubit Bloch vector.

For $N=2$, take the normalized Pauli operators as the traceless basis. The matrix associated with a vector $s$ is $\rho =I/2+s\cdot \sigma$, where these basis matrices are the usual Pauli matrices divided by $\sqrt{2}$. Their multiplication relations giveAt radius $1/\sqrt{2}$, these eigenvalues are one and zero. Every sphere point therefore gives a positive rank-one projector. Any other orthonormal Hermitian basis merely changes the real coordinates by an orthogonal transformation, so the result is basis independent.

For every $N>2$, choose a rank-one projector $P$ and consider the Hermitian trace-one matrixIts eigenvalues are $2/N-1$ once and $2/N$ with multiplicity $N-1$. They sum to one and satisfyBy orthonormality its traceless coordinate vector has squared length $1-1/N$, so it is on the specified sphere. But $2/N-1<0$, so $A$ is not a density operator. Thus for $N>2$, the pure states occupy a proper subset of the sphere; positivity excludes some of its points. In fact, this example is the antipodal Bloch vector to that of $P$.

The usual assertion presumes $N\ge 2$. If the trivial dimension $N=1$ is admitted, the radius-zero sphere consists of the single coordinate point, also corresponding to the single pure state, so that dimension is an additional trivial exception.

For a Hermitian Quantum Hamiltonian, $[{\sigma }_m,{\sigma }_n{]}^{†}=-[{\sigma }_m,{\sigma }_n]$. ConsequentlySwapping $m,n$ changes the sign of the commutator, so $L_{nm}=-L_{mn}$. The generator is therefore real and antisymmetric. Its last column vanishes because the identity commutes with $H$, so $c=0$ and the traceless block satisfies $A^T=-A$.

For constant $H$, the Bloch vector evolves by $s(t)=R(t)s(0)$ with $R(t)=e^{At}$. Direct differentiation givesHence $R^{T}R=I$. Its determinant is continuously connected to one and cannot change between the two possible values $\pm 1$, so $\det R=1$. This proves that the evolution is a rotation in the special orthogonal group. A time-dependent Hermitian Quantum Hamiltonian has the same conclusion from $\dot{R}=A(t)R$.

These Hamiltonian rotations of Bloch vectors preserve the Euclidean norm and purity of a density operator. For $N>2$, not every rotation of the ambient sphere is a physical unitary conjugation: the higher-dimensional positivity constraints and full spectrum of the density operator impose further restrictions.

A steady state is a time-independent physical state: its derivative vanishes. With $n=N^2-1$, the Affine Bloch equation therefore requiresAs a real linear equation, it has a solution exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. If $s_p$ is one solution, all algebraic solutions are $s_p+\ker A$, an affine set of dimension $n-\mathrm{rank}A$. Physical steady states are its intersection with the set of Bloch vectors corresponding to positive trace-one density operators. An arbitrary affine equation need not have a steady state: $A=0$, $c\ne 0$ is a counterexample.

The equation here comes from a finite-dimensional time-independent Lindblad equation, however, so it always has at least one physical steady state. To prove this, start from any density operator and average its trajectory:These averages remain positive with trace one. That set is a compact set, so some sequence has a limit ${\rho }_{\ast }$. If $\mathcal{L}$ is the time-independent generator, thenbecause density operators are bounded. Continuity gives $\mathcal{L}({\rho }_{\ast })=0$. This proves directly that Finite-dimensional Lindbladians have stationary states.

If $A$ has full rank, its algebraic solution $s_{\ast }=-A^{-1}c$ is unique and existence makes that solution physical. For these Lindblad dynamics the converse also holds, even for a boundary steady state. If $A$ is singular, choose a nonzero traceless Hermitian stationary direction $X$. The two parts of its positive-negative decomposition have the same trace $\tau >0$, so $X=\tau ({\rho }_{+}-{\rho }_{-})$ for two density operators. Time-average their trajectories along a common convergent subsequence. Their limits are stationary by the preceding argument, and their difference remains $X/\tau$, since $X$ itself is stationary. The limits are therefore distinct physical stationary states.

Thus the finite-dimensional physical Bloch equation has a unique steady state exactly when $\mathrm{rank}A=n$. These steady states of an affine Bloch equation must be distinguished from convergence to that steady state, which requires the additional spectral condition in (b). The possibly nonorthonormal coordinate basis does not affect existence or uniqueness, since it still gives an invertible parametrization of the traceless Hermitian space.

Write a qubit density operator as $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with Bloch vector $|\mathbf{r}|\le 1$. The depolarizing channel sends $\mathbf{r}$ to $p\mathbf{r}$. Its output eigenvalues are $(1\pm |p||\mathbf{r}|)/2$, so its Von Neumann entropy is minimized at $|\mathbf{r}|=1$. DefineThe minimum output entropy is $h(q)$, since binary entropy is symmetric under $q\mapsto 1-q$. The usual probabilistic-mixture convention has $0\le p\le 1$. The argument also holds throughout the physical qubit CPTP range $-1/3\le p\le 1$, where this $q$ still lies in $[0,1]$.

To handle multiple channel uses, let a classical message $x$ select a product input ${\rho }_x={⨂}_{j=1}^n{\rho }_{xj}$ with probability ${\pi }_x$. Memorylessness gives product outputs ${\sigma }_x={⨂}_j\Phi ({\rho }_{xj})$. Additivity of Von Neumann entropy on tensor products gives $S({\sigma }_x)\ge nh(q)$. The average output $\bar{\sigma }={\sum }_x{\pi }_x{\sigma }_x$ acts on $n$ qubits, so $S(\bar{\sigma })\le {\log }_2(2^n)=n$. Thus the Holevo quantity obeys the product-input block bound for a qubit depolarizing channelThe Holevo bound says the classical mutual information obtained by any measurement, including a joint measurement on all outputs, is at most $\chi$.

For a reliable code with $M$ equiprobable messages and decoding error $P_e$, Fano's inequality gives $H(X\mid \hat{X})\le h(P_e)+P_e{\log }_2(M-1)$. Consequently $(1-P_e){\log }_{2}M\le n[1-h(q)]+h(P_e)$. Letting the error tend to zero proves $\boxed{\mathcal{I}(\Phi )\le 1-h((1-p)/2)}$.

The bound is tight: equal use of $|0⟩,|1⟩$ and computational-basis output measurements produce a binary symmetric channel with crossover probability $q$. Ordinary classical coding achieves $1-h(q)$ bits per use.

A steady state is a time-independent density operator satisfying $\mathcal{L}({\rho }_{\ast })=0$, equivalently a constant physical Bloch vector $s_{\ast }$ satisfying $\boxed{A{s}_{\ast }=-c}$. Let $K=N^2-1$. By the rank criterion for a linear system, this equation has a solution exactly whenWhen it is consistent, every algebraic solution has the form $s_0+v$ with $v\in \ker A$, so the solution affine space has dimension $K-\mathrm{rank}A$ by the rank-nullity theorem. Physical steady states are its intersection with the positive density operator body. The algebraic equilibrium is unique exactly when $\mathrm{rank}A=K$, in which case $\boxed{s_{\ast }=-A^{-1}c}$.

An arbitrary affine differential equation need not have an equilibrium: $A=0$, $c\ne 0$ is inconsistent. For the finite-dimensional, time-independent Lindblad equation under discussion, however, a physical steady state always exists. Indeed, the averages ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$ remain positive with trace one. Compactness supplies a convergent subsequence as $T\to \infty$, andThe limit is a physical steady state, proving that Finite-dimensional Lindbladians have stationary states and that the rank consistency condition is automatically satisfied for this physical generator.

In this setting, full rank also characterizes uniqueness among physical steady states. To justify the converse when $A$ is singular, a nonzero $v\in \ker A$ gives a fixed traceless Hermitian operator $X={\sum }_k{v}_k{\sigma }_k$. Write its positive-negative decomposition $X=X_{+}-X_{-}$. Time-average the evolutions of the two positive operators along a common convergent subsequence; the limits $Y_{\pm }$ are positive stationary operators and satisfy $X=Y_{+}-Y_{-}$. Their traces are equal because $\mathrm{Tr}X=0$. If there were only one stationary density operator, both $Y_{\pm }$ would be this matrix times the same trace, forcing $X=0$, a contradiction. Hence $\boxed{\text{unique physical steady state}⟺\mathrm{rank}A=N^2-1}$.

A steady state is attractive here if every physical initial density operator approaches it. Subtracting its Affine Bloch equation from the trajectory equation gives $\delta \dot{s}=A\delta s$, so $\delta s(t)=e^{At}\delta s(0)$. The necessary and sufficient condition isSuch an $A$ is a Hurwitz stable matrix. To prove sufficiency without assuming diagonalizability, put it in Jordan normal form. Each Jordan block contributes $e^{\lambda t}{\sum }_{j=0}^{q-1}{t}^j{N}^j/j!$, and the negative real part makes the exponential dominate all polynomial factors. For necessity, an eigenmode with positive real part grows; an eigenmode with zero real part remains constant or oscillates, so it does not tend to zero. A nontrivial Jordan block on the imaginary axis also prevents decay.

The physical Bloch vector body has nonempty interior, so differences of physical initial vectors from $s_{\ast }$ span the full real coordinate space. Thus convergence for every physical initial state also forces $e^{At}\to 0$ on that space. This rules out hiding a nondecaying eigenmode outside the physical set. The strict spectral condition entails uniqueness and global asymptotic stability of the steady state.

For real $\alpha ,\beta$, the equilibrium equations are $-{\beta }^{2}x-\alpha z=0$, $y=0$ and $\alpha x-({\beta }^2+1)z=\beta$. The determinant of the coefficient matrix is $-D$, where $D={\alpha }^2+{\beta }^2({\beta }^2+1)>0$ whenever $(\alpha ,\beta )\ne (0,0)$. Solving the two-by-two system, including cases with one parameter zero, givesSubstitution verifies the first equation by cancellation of $-\alpha {\beta }^3$ and $+\alpha {\beta }^3$, and the third because ${\alpha }^2\beta +{\beta }^3({\beta }^2+1)=\beta D$. The nonzero determinant proves uniqueness. Its squared length is ${\beta }^2({\alpha }^2+{\beta }^4)/D^2$, andThus $\Vert s_{\ast }\Vert \le 1/2$, which is inside the qubit physical Bloch vector body; in the Hilbert-Schmidt normalization of part (b), that body has radius $1/\sqrt{2}$.

The $y$ mode has eigenvalue $-1$. The $xz$ block has characteristic polynomial $(\lambda +{\beta }^2)(\lambda +{\beta }^2+1)+{\alpha }^2$, so the other eigenvalues areIf $|\alpha |>1/2$, they have real part $-{\beta }^2-1/2<0$. If $|\alpha |\le 1/2$, the square root is at most $1/2$, with equality only when $\alpha =0$; the larger eigenvalue is still negative unless $\alpha =\beta =0$. At $|\alpha |=1/2$ an eventual Jordan factor does not change decay because its eigenvalue is negative. Therefore for every allowed parameter pair all eigenvalues have negative real part, and $\boxed{s_{\ast }\text{ is globally attractive}}$.

Iterating the Bloch-vector map of amplitude damping givesThereforeThe repeated amplitude damping limit describes irreversible relaxation of the excited state $|1⟩$ into the ground state $|0⟩$, with energy transferred to the environment and no compensating thermal excitation. At $p=1$, the ground state is reached in one action. At the endpoint $p=0$, the quantum channel is the identity and the initial Bloch vector remains unchanged; the ground-state conclusion requires nonzero damping.

In the computational basis, writeThe Kraus representation of the amplitude damping channel givesReading off its Bloch vector gives the Bloch-vector map of amplitude damping:Here $0\le p\le 1$. The shift in $s_z$ shows that this quantum channel is not unital unless $p=0$; its fixed ground state lies at the north pole, rather than the center of the Bloch sphere.