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Write a qubit density operator in Bloch vector form, $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. The quantum depolarizing channel maps $\mathbf{r}$ to $p\mathbf{r}$, so its output eigenvalues are $(1\pm p|\mathbf{r}|)/2$. The output Von Neumann entropy is minimized on pure inputs, where it equals $h_2((1+p)/2)$.

For any input ensemble, its output Holevo quantity is at most the maximum qubit entropy minus this minimum output entropy:Equality is attained by two equiprobable orthogonal pure inputs: their average output is $I/2$, while each output has the minimum entropy. ThusThis uses the question's retention parameter $p$; in the usual convex-mixture range, $0\le p\le 1$. The expression also holds throughout the full completely positive qubit range $-1/3\le p\le 1$, since $h_2((1+p)/2)=h_2((1+|p|)/2)$.

The supplied additivity of Holevo capacity gives ${\chi }^{\ast }({\Lambda }_{\mathrm{dep}}^{\otimes n})=n{\chi }^{\ast }({\Lambda }_{\mathrm{dep}})$. The Holevo-Schumacher-Westmoreland theorem expresses the unassisted classical capacity of a quantum channel as the regularized Holevo capacity, soTherefore entangled inputs across channel uses cannot increase this classical capacity.

Every density operator can be written asTherefore $⟨\varphi |\rho |\varphi ⟩={\sum }_j{p}_j|⟨{\psi }_j|\varphi ⟩{|}^2\ge 0$, so $\rho$ is a positive semidefinite operator, and $\mathrm{Tr}\rho ={\sum }_j{p}_j=1$. A pure quantum state has a rank-one density operator $|\psi ⟩⟨\psi |$, equivalently ${\rho }^2=\rho$ and $\mathrm{Tr}({\rho }^2)=1$. A mixed quantum state is not pure and has $\mathrm{Tr}({\rho }^2)<1$ in finite dimension.

The Von Neumann equation and the Pauli matrix commutator identity giveso the Bloch vector obeysWriting $\omega =|\mathbf{\omega }|$ and $\hat{\mathbf{\omega }}=\mathbf{\omega }/\omega$, the assumption $\mathbf{a}\cdot \mathbf{\omega }=0$ givesand henceUnitary evolution rotates the Bloch vector and preserves its length. Since $|\mathbf{a}|<1$, the state remains mixed, whereas $|{\uparrow }_x⟩$ has the unit Bloch vector $(1,0,0)$. The system is therefore never definitely in $|{\uparrow }_x⟩$.

Conjugation by ${\sigma }_z$ maps $(r_x,r_y,r_z)$ to $(-r_x,-r_y,r_z)$. The phase-flip channel therefore acts on the Bloch vector as

Take $|L⟩=|0⟩$ and $|R⟩=|1⟩$. Initially the two masses are in the product stateThe branch-dependent Newtonian gravitational potential energy is $-G{m}^2/d_{ab}$. Under the stated approximation, only the $|RL⟩$ branch acquires an appreciable relative phase, so after time $t$,The determinant of its two-by-two coefficient matrix is $(1-e^{i\varphi })/4$, which is nonzero unless $\varphi$ is a multiple of $2\pi$. Thus the state generally has Schmidt rank two: the branch-dependent gravitational phase creates gravitationally induced entanglement.

An entanglement witness has a bound obeyed by every separable quantum state and violated by at least one entangled state. For a product state with Bloch vectors $\mathbf{r}$ and $\mathbf{s}$,by the Cauchy-Schwarz inequality. Convexity gives the same bound for every separable mixed state. Consequently $W>1$ certifies entanglement; in conventional operator form, one of $I\mp (X\otimes Z+Y\otimes Y)$ has negative expectation whenever the absolute-value criterion is violated.

For the state above, direct use of the Pauli matrices givesWith the supplied values,which is close to $\pi$. Hence $W\simeq 2.00$, and to the nearest integerThis is the operating principle of the Bose--Marletto--Vedral experiment.

Write the spin-one-half Heisenberg antiferromagnet ason a bipartite lattice of coordination number $z$. A one-spin density operator is $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, where $\mathbf{r}$ is its Bloch vector and $|\mathbf{r}|\le 1$. In a two-sublattice product state,The minimum is $-1/4$, attained by pure antiparallel Bloch vectors, so the minimizing product state is a Néel state. In the $z\to \infty$ limit this mean-field approximation becomes exact and the ground-state energy per bond is therefore

The phrase “energy density” requires a coupling convention. For the unscaled Hamiltonian above, every site belongs to $z/2$ bonds andwhich diverges as $z\to \infty$. With the standard Kac normalizationthe finite energy density isIf the convention divides by the spatial dimension $d=z/2$ instead, the answer is $-J/4$ per site. A Hamiltonian written with ${\mathbf{\sigma }}_i\cdot {\mathbf{\sigma }}_j$ rather than ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ multiplies all these energies by four.

For a mixed spin-half state, the Bloch vector formula givesPhysical consistency is equivalent to the resulting Bloch vector having length at most one.

Three linearly independent Hermitian operators do not always suffice. The operatorsare linearly independent, but the first expectation is always one and none detects the $y$ component. States with Bloch vectors $(0,c,0)$ and $(0,-c,0)$ have identical expectations for all three. The correct criterion for informationally complete three-observable qubit tomography is that the three traceless parts span the Pauli space.

If the identity is applied with probability $p$ and the three nonidentity Pauli operators each with probability $(1-p)/3$, the Bloch vector retention factor is $(4p-1)/3$. Thus complete depolarization occurs at $p=1/4$, not at zero. The Holevo capacity of a qubit depolarizing channel becomes $1-h_2((2p+1)/3)$ in this identity-probability parametrization.

A qubit depolarizing channel shrinks every Bloch vector by the same factor: ${\Lambda }_p(\rho )=p\rho +(1-p)I/2$. In this retention-parameter convention it is completely positive for $-1/3\le p\le 1$; the usual mixture of unchanged and fully randomized states uses $0\le p\le 1$.

Under repeated identical amplitude damping channels, all excited population eventually relaxes to the ground state for any fixed $0<p\le 1$. The Bloch vector approaches $(0,0,1)$. For $p=1$ the limit is reached in one action, while $p=0$ is the identity channel and preserves every input. Keeping the zero-damping endpoint separate prevents an incorrect universal relaxation claim.

A one-qubit rotation gate is $R_{\alpha }(\theta )=e^{-i\theta {\sigma }_{\alpha }/2}$, where ${\sigma }_{\alpha }$ is a Pauli matrix. It rotates the Bloch vector by angle $\theta$ about the corresponding axis.