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Suppose one equally likely action returns a known normalized pure state $|s⟩$ unchanged, while another returns the maximally mixed state $I/d$. A basic measurement in any orthonormal basis containing $|s⟩$ distinguishes them with success $1-1/(2d)$: guess the pure-state action on its basis label and the reset action on the remaining $d-1$ labels. The first action succeeds with probability one and the second with probability $(d-1)/d$.

The Holevo–Helstrom theorem also proves optimality for these returned states. The Hermitian difference $|s⟩⟨s|-I/d$ has eigenvalues $1-1/d$ and $-1/d$ with multiplicity $d-1$, hence trace norm $2(1-1/d)$. Its optimal equal-prior discrimination success is $1/2+\Vert {\rho }_0-{\rho }_1{\Vert }_1/4$. Random computational-basis permutations fix the uniform superposition, so they supply an example of the invariant action. This compares the returned states in the stated unassisted protocol; it makes no assertion about a more general channel-discrimination problem with an entangled ancillary probe.^[ref]

Measurement backaction is the effect of a measurement in quantum mechanics on later predictions for the measured system. A quantum instrument with one Kraus operator $K_m$ per recorded outcome hasThe conditional density operator ${\rho }_m$ exists only when $p_m>0$. Its nonlinear normalization expresses conditioning; the unnormalized map $\rho \mapsto K_m\rho K_m^{†}$ is linear. More generally, an unresolved outcome can contain several Kraus operators: replace the numerator by ${\sum }_a{K}_{ma}\rho K_{ma}^{†}$ and take its trace for $p_m$. ^[ref].

Keeping the outcome gives a selective quantum measurement. Discarding it gives a nonselective quantum measurement with quantum channel $\rho \mapsto {\sum }_m{p}_m{\rho }_m$. The latter describes the whole measured population without its labels, whereas each ${\rho }_m$ describes a selected subpopulation. Neither description requires the measured quantity to have possessed its eventual value before the interaction.

For a concrete test, prepare the Pauli X eigenstate $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. Its $X=+1$ probability is one. An ideal $Z$ projective measurement gives zero or one with equal probability, leaving $|0⟩$ or $|1⟩$ under the Lüders rule. Discarding the result leaves the maximally mixed state $I/2$. A subsequent $X$ measurement now gives either sign with probability one half, even if the $Z$ result was kept. This observable change in statistics is backaction. Measuring $X$ on the original $|+⟩$ instead leaves it unchanged: disturbance depends on both the preparation and the instrument. Measurement-based quantum feedback uses the recorded outcome to choose a subsequent control operation, so both parts of the instrument matter.

Let $P$ be a rank-$k$ projector with $0<k<N$. No unitary operator can map every pure state whose $P$ probability equals $k/N$ to a state with the same fixed smaller probability. Average $\sqrt{k/N}|g⟩\pm \sqrt{1-k/N}|b⟩$ over good and bad basis vectors and both signs. The average density operator is the maximally mixed state, so every unitary preserves its average $P$ probability. This contradicts a strictly smaller probability for all states in the ensemble. A preparation tailored to one known starting state is a different resource and is used in exact amplitude amplification.

Use the preparation and measurement from part (i). RandomReset returns the density matrix ${\rho }_R=(|0⟩⟨0|+|1⟩⟨1|)/2=I/2$, a maximally mixed state. In the $|+⟩,|-⟩$ basis its two outcomes each have probability $1/2$. In contrast, RandomFlip always gives outcome 0.

Guess RandomFlip after 0 and RandomReset after 1. With the supplied equal priors,An outcome 1 identifies the reset action, but outcome 0 does not identify the flip action with certainty. For example, Bayes' theorem gives posterior probability $2/3$ for the flip action after 0. The total success rate already accounts for the reset runs that also produce 0.

Write ${\rho }_0=|\varphi ⟩⟨\varphi |=(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. Conjugating by a Pauli matrix keeps its own Bloch component and reverses the other two; for example,In the average of the four conjugations, the coefficient of every nonidentity Pauli matrix is zero. ConsequentlyThe Pauli operator basis gives the Bloch representation of a qubit density operator.^[ref]

This is the qubit Pauli twirl, the dimension-two case of the Heisenberg-Weyl twirling channel. The argument uses only trace one, so it also works for mixed qubit inputs. The preparation label must be forgotten: conditioning on the label would retain a known pure branch rather than the maximally mixed state obtained by the ensemble average.

Expand the density matrix of the Bell state and take the partial trace over Alice's qubit:The orthonormality of the computational basis removes the off-diagonal terms. Thus Bob's reduced density matrix is the maximally mixed state

The Pauli matrices are Hermitian matrices with zero trace, so a real Bloch vector makes $\varrho$ Hermitian and gives $\mathrm{tr}\varrho =1$. Their Pauli multiplication identity implies $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Consequently the two eigenvalues of $\mathbf{s}\cdot \mathbf{\sigma }$ are $\pm |\mathbf{s}|$, including the zero-vector case, andA density matrix must also be positive semidefinite. The displayed eigenvalues are nonnegative exactly when $\boxed{|\mathbf{s}|\le 1}$. Thus real coordinates alone are insufficient: the allowed vectors fill the Bloch ball. Its boundary consists of pure states, with eigenvalues one and zero; its center is the maximally mixed state $I/2$.

While the sequence is remembered, the conditional quantum state is the known pure endpoint $C|0⟩$, so its Bloch vector is one of the six Pauli vertices. Partial forgetting leaves a probability distribution $p_j$ over possible endpoints, givingHence the vector lies in the stabilizer octahedron. The mixing describes lost classical information, not a different unitary evolution of the physical qubit.

Under complete unbiased uncertainty over the six endpoints, $p_j=1/6$, the opposite vectors cancel andThis is the maximally mixed state at the center. Mere loss of the particular sequence does not mathematically force uniform probabilities: a known biased distribution of endpoint states would leave its corresponding noncentral mixture. The central conclusion uses the unbiased randomization implicit in complete forgetting in this setting.

Assume $|\alpha {|}^2+|\beta {|}^2=1$. Let Alice's first recorded quantum measurement outcome be $a\in \{0,1\}$. Immediately afterward she assigns the conditional pure quantum state $|a⟩$, with probabilities $p_0=|\alpha {|}^2$, $p_1=|\beta {|}^2$. Bob, who has no record, assigns the density operatorThis is a nonselective quantum measurement, which removes the initial computational basis coherence. It is not generally the original pure quantum state.

After the Hadamard gate, Alice assigns $H|a⟩=|+⟩$ for $a=0$, or $|-⟩$ for $a=1$. Bob assignsBoth $|+⟩$ and $|-⟩$ give either final computational basis outcome with probability $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b⟩$, whereas Bob assignsAlice's conditional quantum states are pure; Bob's are averages over her unobserved records. They agree on all probabilities when supplied the same information. Their first descriptions can coincide if the initial quantum state is a computational basis state; after the second quantum measurement, Bob's maximally mixed state differs from each conditional pure quantum state.

For second-qubit inputs $|0⟩$ and $|1⟩$, c gives control expectation values $\mathrm{Re}{U}_{00}$ and $\mathrm{Re}{U}_{11}$. Preparing the maximally mixed state $I/2$ is equivalent to choosing these inputs independently with equal probability and discarding the choice. Linearity of the density operator evolution and quantum measurement probabilities therefore givesHenceThe factor two is essential: the Hadamard test directly estimates the normalized trace. More generally, a target density operator $\rho$ gives $⟨Z⟩=\mathrm{Re}\mathrm{Tr}(\rho U)$. This argument never requires knowledge of the eigenvectors of $U$.

Yes. Replace the target qubit by a register of $k$ qubits and implement the controlled unitary gateThe algebra in c uses only normalization and unitarity, so for an arbitrary pure register quantum state $|\varphi ⟩$ it still gives $⟨Z⟩=\mathrm{Re}⟨\varphi |U|\varphi ⟩$. For a mixed register quantum state $\rho$, linearity gives $\mathrm{Re}\mathrm{Tr}(\rho U)$. With $\rho =I/2^k$,Only the control qubit needs to be pure. A uniformly randomized computational basis preparation supplies the target maximally mixed state. This is the principle of trace estimation by a maximally mixed Hadamard test. The extension is algebraic; efficiently implementing controlled-$U$ still requires a known suitable gate construction. Moreover, estimating the unnormalized trace to fixed additive precision requires increasingly precise estimation of the normalized signal as $k$ grows. There is no automatic efficient reconstruction of all entries of $U$.

For a density matrix $\rho$ on a $d$-dimensional Hilbert space, its Von Neumann entropy, in bits, iswhere the ${\lambda }_a$ are its eigenvalues and $0{\log }_{2}0=0$. Each summand is nonnegative, proving $S\ge 0$, with equality exactly when one eigenvalue is one and the rest zero, namely for a pure state.

We first prove the nonnegativity of quantum relative entropy needed below. Diagonalize $\rho$ with eigenvalues ${\lambda }_a$ and eigenvectors $u_a$, and $\sigma$ with eigenvalues ${\mu }_b$ and eigenvectors $v_b$. Put $B_{ab}=|⟨u_a,v_b⟩{|}^2$ and $q_a=⟨u_a|\sigma |u_a⟩={\sum }_b{B}_{ab}{\mu }_b$. Each row of $B$ sums to one. Concavity of the ordinary logarithm gives ${\sum }_b{B}_{ab}{\log }_2{\mu }_b\le {\log }_2{q}_a$. ConsequentlyThe final Gibbs inequality follows directly from ${\sum }_{{\lambda }_a>0}{\lambda }_a\log (q_a/{\lambda }_a)\le \log {\sum }_{{\lambda }_a>0}{q}_a\le 0$. Support failure gives $D=+\infty$; zero eigenvalues can otherwise be handled by restriction or regularization. Equality requires $q_a={\lambda }_a$ and equality in the logarithm inequality on every positive-weight row. Thus $u_a$ is a $\sigma$ eigenvector of eigenvalue ${\lambda }_a$ wherever ${\lambda }_a>0$. The remaining diagonal entries of $\sigma$ vanish and positivity makes the remaining rows and columns zero. Hence equality holds exactly when $\rho =\sigma$.

Taking $\sigma =I/d$ gives $D(\rho \Vert I/d)={\log }_{2}d-S(\rho )$, provingThe lower equality is attained exactly by pure states and the upper exactly by the maximally mixed state $I/d$.

For $\bar{\rho }={\sum }_i{p}_i{\rho }_i$, each positive-weight state has support contained in that of $\bar{\rho }$. Expanding the quantum relative entropy givesThis proves Concavity of Von Neumann entropy, with equality precisely when every positive-weight ${\rho }_i$ is the same state.

For the upper entropy bound for a quantum mixture, diagonalize each component, ${\rho }_i={\sum }_a{\lambda }_{ia}|v_{ia}⟩⟨v_{ia}|$, and set $q_{ia}=p_i{\lambda }_{ia}$. Form the rectangular matrix $B$ whose columns are $\sqrt{q_{ia}}{v}_{ia}$. Then $B{B}^{†}=\bar{\rho }$, while the Gram matrix $\Gamma =B^{†}B$ has diagonal $q_{ia}$. The two matrices have the same nonzero eigenvalues: $B$ and $B^{†}$ map corresponding positive-eigenvalue eigenspaces isomorphically. Thus $S(\Gamma )=S(\bar{\rho })$.

If there are $L$ columns, let $W=\mathrm{diag}(1,\omega ,\dots ,{\omega }^{L-1})$, with $\omega =e^{2\pi i/L}$. ThenAll the summands have the same Von Neumann entropy. The just-proved concavity therefore implies $S(\bar{\rho })=S(\Gamma )\le S(\mathrm{diag}\Gamma )=H(q)$. Finally,We have proved both requested bounds:For the upper equality, strictness in the concavity argument forces $\Gamma$ to be diagonal. Its entries between two positive-weight component eigenvectors then force those vectors to be orthogonal. Thus upper equality holds exactly when the supports of the positive-weight component states are mutually orthogonal. In that case their classical label is perfectly distinguishable, explaining the extra $H(p)$ bits.

Prepare the auxiliary system in the maximally mixed state $\rho =I/N$. Then $\mathrm{Tr}(\rho U)=\mathrm{Tr}U/N$, so the two measurement bases giveThere is no need to know the eigenvectors of $U$. Choose a uniformly random member of any known orthonormal basis on each run; discarding the classical preparation label produces $I/N$. Alternatively, discard one half of a maximally entangled pair of $N$-level systems. This is trace estimation by a maximally mixed Hadamard test.

The circuit directly estimates the normalized trace. To demand fixed absolute accuracy $\epsilon$ in the unnormalized trace requires normalized accuracy $\epsilon /N$; straightforward independent frequency sampling consequently uses order $N^2/{\epsilon }^2$ runs at fixed confidence. Thus the construction should not be confused with a dimension-independent guarantee for arbitrary absolute trace precision.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

A unital quantum channel preserves the identity operator: $\Phi (I)=I$. In equal input and output dimensions, it therefore preserves the maximally mixed state. For amplitude damping channel,Hence the channel is not unital for $p>0$; it is unital only in the identity case $p=0$. Trace preservation involves the opposite products $K_i^{†}{K}_i$ and does not imply unitality.

Choose an orthonormal eigenbasis of $V$, with eigenphases ${\varphi }_0,{\varphi }_1$, counting multiplicity. The maximally mixed state is basis independent, soLinearity means that the experiment is the equal mixture of the two eigenstate experiments. After the inverse quantum Fourier transform, the joint state is $\frac{1}{2}{\sum }_{j=0}^1|{\chi }_j⟩⟨{\chi }_j|\otimes |v_j⟩⟨v_j|$, where $|{\chi }_j⟩={\sum }_m{a}_m({\varphi }_j)|m⟩$ andThus the exact output distribution for quantum phase estimation with a maximally mixed target isFor a nonzero phase mismatch, the squared probability amplitude is ${\sin }^2[L(\varphi -2\pi m/L)/2]/\{L^2{\sin }^2[(\varphi -2\pi m/L)/2]\}$, with value one at an exact match. If both eigenphases are exactly representable on the $L$-point grid, their two labels occur with probability $1/2$ each; coincident labels have their probabilities added. For general eigenphases there can be nonzero probabilities at other labels, so it would not be exact to list only the two best approximations with probability $1/2$ each. Each eigenphase is sampled with spectral weight $1/2$, and its nearest-label event has conditional probability at least $4/{\pi }^2$. In particular a nearest label for one eigenphase has unconditional probability at least $2/{\pi }^2$, with additional contributions possible from the other phase.

A unital quantum channel preserves the identity operator: $\Phi (I)=I$. For equal input and output dimensions, this is equivalent to preserving the maximally mixed density operator.

For the amplitude damping channel,Thus the channel is not unital for $p>0$; it is unital only at $p=0$, when it is the identity channel. In particular, it biases the maximally mixed state toward the atomic ground state. The trace-preservation condition involves ${\sum }_j{A}_j^{†}{A}_j$, whereas unitality involves the opposite products ${\sum }_j{A}_j{A}_j^{†}$.

Under the usual reversible-circuit interpretation, there is a genuine obstruction to uniform probability lowering by a unitary in the printed premise. The universal assertion cannot hold for arbitrary $p$: it fails already at the search density $p=k/N$.

An explicit counterexample uses $N=4$, one good basis state $|0⟩$, $p=1/4$ and $p^{\prime }=1/8$. The six pure statesall have good probability $1/4$. Their equally weighted density operator is $I_4/4$, a maximally mixed state. Every unitary operator $C$ leaves this mixture unchanged, so its average output good probability remains $1/4$. The printed assertion would instead make all six output probabilities $1/8$, a contradiction. More generally, averaging states $\sqrt{k/N}|g⟩+e^{i\alpha }\sqrt{1-k/N}|b⟩$ uniformly over good and bad basis vectors and two opposite phases gives $I_N/N$, proving the same obstruction at the actual search density $p=k/N$.

The intended conditional construction needs only a supplied reversible preparation that lowers the success probability of the particular starting state, not of every state. Here is the complete argument under that weaker resource assumption. Count $C$ and $C^{†}$ as supplied, query-free operations, as required for the claimed input-oracle bound. Let $\theta =\arcsin \sqrt{k/N}$ and chooseIf equality holds, use the original preparation. Otherwise use the stipulated preparation on $|{\psi }_0⟩$ to obtain $|{\psi }_{\ast }⟩=C|{\psi }_0⟩$ with good probability $p_{\ast }$. Its reflection operator is implementable byThe good and bad components may be nonuniform, but amplitude amplification applies to this same two-dimensional decomposition. After $j$ iterations of $-I_{{\psi }_{\ast }}{I}_f$, its good probability isThus the stated exact-query conclusion follows from accessible preparation of one known-overlap state and its inverse. If the supplied preparation uses oracle queries, those costs cannot be omitted.

A physically realizable known-state success dilution for exact amplitude amplification is available in the standard enlarged marking-oracle model. Append a flag in $\sqrt{1-q}|0⟩+\sqrt{q}|1⟩$, where $q=p_{\ast }/p$, and call a joint state good only when $f(x)=1$ and the flag is one. This prepared state's success probability is exactly $pq=p_{\ast }$. Reflect about this known product preparation and mark the joint good subspace; $j$ ordinary iterations then succeed with certainty. A supplied controlled phase oracle gives one marking query per iteration, or a Boolean bit oracle computes $f$, applies the joint phase and uncomputes $f$ in two queries. Measuring the data therefore returns a good $x$ with certainty in $O(\sqrt{N/k})$ queries. This changes the state space and marking test, and does not assert the impossible universal $n$-qubit circuit. Controlled access is an additional resource in a bare phase-oracle model, so it is made explicit rather than silently assumed.

Specify the magnetic coupling before fixing its sign: let the magnetic moment be $\mathbf{\mu }=\gamma \mathbf{S}$, where $\gamma$ is a signed gyromagnetic ratio. The Zeeman effect gives $H_Z=-\gamma \mathbf{B}\cdot \mathbf{S}$. For spin one-half, $\mathbf{S}=\hslash \mathbf{\sigma }/2$, so with $B=|\mathbf{B}|$ and $u=\beta \gamma \hslash B/2$,This is the quantum Gibbs state. Its eigenvalues are $(1\pm \tanh u)/2$, so its purity is $(1+{\tanh }^{2}u)/2$. For nonzero level splitting it approaches a pure state as $|u|\to \infty$: positive inverse temperature selects the lower energy, while negative inverse temperature selects the upper one. It is a maximally mixed state at $u=0$, including infinite temperature or zero magnetic splitting. At zero field the Gibbs state remains $I/2$; lowering temperature alone does not select a unique pure state from that degeneracy.

Negative temperature is meaningful for an internally equilibrated spin sector with a bounded energy spectrum and population inversion. It must be sufficiently isolated from unbounded motional degrees of freedom: the oscillator's unbounded spectrum, for instance, makes its partition sum diverge at negative inverse temperature. The two-level spin partition sum itself stays finite for either sign.

An ensemble with normalized quantum states $|{\psi }_a⟩$ and probabilities $p_a\ge 0$, ${\sum }_a{p}_a=1$, has density matrixEach rank-one projector is Hermitian, so ${\rho }^{†}=\rho$. Taking the trace in any orthonormal basis givesIt is also positive: $⟨v|\rho |v⟩={\sum }_a{p}_a|⟨{\psi }_a|v⟩{|}^2\ge 0$. Its eigenvalues ${\lambda }_a$ are therefore nonnegative and sum to one. Hencewith equality exactly when one eigenvalue is one and the others are zero. Thus the purity of a density operator distinguishesEquivalently, a pure state has ${\rho }^2=\rho$ and rank one. In a $d$-dimensional Hilbert space, purity is at least $1/d$, attained by the maximally mixed state $I/d$.