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The absolute value of a linear operator $L$ is the positive square root of an operator $L^{†}L$. Its eigenvalues are the singular values of $L$, its trace is the trace norm of $L$, and it appears in the polar decomposition of a bounded operator $L=U|L|$.

For density operators $\rho ,\sigma$ on the same finite-dimensional Hilbert space, quantum fidelity measures how close their quantum states are. This article uses the unsquared convention:Each square root is the positive square root of an operator, and $\Vert \cdot {\Vert }_1$ is the trace norm. Quantum fidelity is symmetric in its inputs; it is one exactly when $\rho =\sigma$, and zero exactly when $\rho \sigma =0$, meaning that the support of a positive operator for one state is orthogonal to that for the other. Larger quantum fidelity means greater similarity. The separate squared quantum fidelity is $F_{\mathrm{sq}}=F^2$. ^[ref]

For two normalized pure states, quantum fidelity is the magnitude of their inner product, $F=|⟨\psi |\phi ⟩|$. Thus $|0⟩$ compared with itself has $F=1$, while $|0⟩$ and $|1⟩$ have $F=0$. For $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, comparison with $|0⟩$ gives $F=1/\sqrt{2}$, $F_{\mathrm{sq}}=1/2$, and trace distance $D=1/\sqrt{2}$ by the pure-state trace distance and fidelity identity.

Why compare density operators rather than their entries? A change of coordinates should not make the same physical preparations look more or less alike. Unitary invariance of trace distance and fidelity makes quantum fidelity independent of a common basis choice. Multiplying a pure state vector by a global phase leaves its density operator unchanged, so its quantum fidelity with the original state is one. These features make quantum fidelity useful for checking whether quantum channels, state preparations, and Schumacher compression preserve a target quantum state.

The connection to Uhlmann's theorem supplies another beautiful interpretation: quantum fidelity is the largest absolute inner product between pure vectors obtained from a purification of a density operator for each state, using a sufficiently large common reference space. Uhlmann's 1976 paper extended transition probability to algebraic states, including those of a C-star algebra; for density operators its transition probability corresponds to our squared quantum fidelity. The measurement characterization of quantum fidelity gives a complementary description directly in terms of observable outcome statistics. ^[ref]

Subtract the two density operators. The difference is Hermitian with zero diagonal and off-diagonal entry $\alpha \overline{\beta }(1-\overline{c})$. A matrix $\left(\begin{array}{cc}0& a\\ \overline{a}& 0\end{array}\right)$ has eigenvalues $\pm |a|$, because its square is $|a{|}^{2}I$. Its two singular values are therefore both $|a|$. By the definition of the trace norm,For the intended real-overlap phase-flip channel, $1-c=2p$, givingThis vanishes for a computational basis state, since a phase flip then changes only an unobservable global phase. Its largest value at fixed $p$ is $p$, attained when $|\alpha |=|\beta |=1/\sqrt{2}$. The general formula remains correct even when the probabilistic premise of part 1 fails.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

Give $G(H)$ its natural trace-class topology: differentiability means $F(t+h)=F(t)+h\dot{F}(t)+o_1(h)$, with remainder small in trace norm. The Fredholm determinant is the continuous extension of finite-rank determinants; it is multiplicative by finite-rank approximation and satisfiesnear zero. For example its exterior-power series bounds the remainder by $e^{\Vert K{\Vert }_1}-1-\Vert K{\Vert }_1$, proving that estimate directly.

Factor $F(t+h)=F(t)[I+hF(t{)}^{-1}\dot{F}(t)+o_1(h)]$. Multiplicativity and the preceding expansion giveThe determinant is nonzero for invertible $F$, so division yields the required logarithmic derivative. Differentiability only in the bounded-operator norm would not justify this trace formula; the specified topology is essential.

Split $f=f_{-}+a_0+f_{+}$ into strictly negative and strictly positive Fourier parts, and let $A=T(f_{-})$, $B=T(f_{+})$. The Toeplitz multiplication rule is $T(g)T(h)=T(gh)$ when $g$ is coanalytic or $h$ is analytic. It follows directly because the discarded negative Fourier part cannot return to the Hardy space in those cases. Thus exponentiation within either subalgebra is exact, andThe constant terms cancel. If $S=T(z)$ is the unilateral shift, $A={\sum }_{m>0}{a}_{-m}(S^{\ast }{)}^m$ and $B={\sum }_{n>0}{a}_n{S}^n$. The commutator $[(S^{\ast }{)}^m,S^n]$ is finite rank, with trace $n$ when $m=n$ and zero otherwise. For equal powers it is the projection onto the first $n$ basis vectors; for unequal powers its matrix has no diagonal entries. Its trace norm is bounded by $2\min (m,n)$. Smoothness of $f$ makes the weighted double coefficient sum finite, so the commutator series converges in trace norm andApplying part (c) proves the Toeplitz exponential determinant identityIn particular the product is an invertible identity-plus-trace-class operator, so the determinant in the formula is well-defined.

For a finite-rank operator $T$, define $|T|=(T^{\ast }T{)}^{1/2}$ and its trace norm bywhere $s_j>0$ are its nonzero singular values, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T={\sum }_j{s}_j{R}_{u_j,v_j}$, where $R_{x,y}z=⟨z,y⟩x$.

For every bounded operator $B$,Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their orthogonal complement. It attains equality, soThis formula proves the triangle inequality and absolute homogeneity. Also $\Vert T\Vert \le \Vert T{\Vert }_1$, so vanishing of the trace norm forces $T=0$. Thus it is a norm on the vector space of finite-rank operators.

The map $B\mapsto F_B$, where $F_B(T)=\mathrm{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its norm is at most $\Vert B\Vert$. Rank-one tests give the reverse inequality: $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, so taking unit vectors $x,y$ gives $\Vert F_B\Vert =\Vert B\Vert$.

Conversely, let $F$ be a bounded linear functional for the trace norm. The sesquilinear form $\beta (x,y)=F(R_{x,y})$ satisfiesThe Riesz representation theorem gives a unique bounded operator $B$ with $\beta (x,y)=⟨Bx,y⟩$, and $\Vert B\Vert \le \Vert F\Vert$. Since every finite-rank operator is a finite sum of rank-one operators, linearity implies $F(T)=\mathrm{Tr}(BT)$ for every $T$. HenceThe same identification extends to the completion, the trace-class operators. This is trace-class duality; completeness of the original finite-rank space is not required to identify its bounded dual.

Use the unsquared quantum fidelity convention in the question. For any matrix $X$, its trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}=\mathrm{Tr}\sqrt{X{X}^{†}}$: the two positive matrices have the same nonzero eigenvalues, namely the squared singular values. This follows directly from the singular value decomposition.

Set $X=\sqrt{\rho }\sqrt{\sigma }$. Then $X{X}^{†}=\sqrt{\rho }\sigma \sqrt{\rho }$ and $X^{†}X=\sqrt{\sigma }\rho \sqrt{\sigma }$. Their positive square roots therefore have the same trace, provingThe argument also covers singular density operators, since zero singular values contribute zero.

Use the unsquared convention for quantum fidelity. For density operators on a common finite-dimensional Hilbert space,Here $\Vert A{\Vert }_1=\mathrm{Tr}\sqrt{A^{†}A}$ is the trace norm, and all square roots are the positive operator square roots. The two displayed expressions agree because $\sqrt{\rho }\sqrt{\sigma }$ and its adjoint have the same singular values. This convention has $0\le F\le 1$; some literature squares this quantity, but that convention is not used here.

For normalized pure states, the rank-one operator $\sqrt{\rho }\sqrt{\sigma }=⟨\phi |\psi ⟩|\phi ⟩⟨\psi |$ has a single nonzero singular value. Therefore

For any linear operator, the trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$, the sum of its singular values. For a Hermitian operator,The Holevo–Helstrom theorem concerns minimum-error discrimination of two density operators $\rho ,\sigma$. If their prior probabilities are $p$ and $1-p$, the maximum success probability over all quantum measurements isAn optimal binary POVM declares $\rho$ on the positive spectral subspace of $p\rho -(1-p)\sigma$ and declares $\sigma$ on its negative spectral subspace; zero-eigenvalue vectors can be assigned either way. For equal priors the formula becomes $P_{\mathrm{succ}}^{\ast }=\frac{1}{2}+\frac{1}{4}\Vert \rho -\sigma {\Vert }_1$.

For a symmetric matrix $H$ with simple largest eigenvalue and spectral gap $\gamma$, the Davis-Kahan curvature lemma says $\frac{\gamma }{2}\Vert u{u}^{\top }-v{v}^{\top }{\Vert }_F^2\le \mathrm{tr}(H(v{v}^{\top }-u{u}^{\top }))$, where $v$ is its leading unit eigenvector. Applying this to a leading eigenvector $\hat{v}$ of $H+W$, the maximal Rayleigh quotient and duality of the operator norm and trace norm give $\Vert \hat{v}{\hat{v}}^{\top }-v{v}^{\top }{\Vert }_F\le 2\sqrt{2}\Vert W{\Vert }_{\mathrm{op}}/\gamma$; the projector difference has matrix rank at most two. This is a rank-one eigenprojector perturbation bound.

Here $W=M-M_0=A-A_0$ has independent centered bounded upper-triangular entries, apart from symmetry. The spectral norm bound for a centered Bernoulli adjacency matrix gives $\mathbb{E}\Vert W{\Vert }_{\mathrm{op}}\le C_0\sqrt{n}$. Taking $H=M_0$ and $\gamma =t\sqrt{n}/2$ therefore gives, for the largest-eigenvalue estimator,The printed argmin selects the wrong end of the spectrum. For a minimizing unit eigenvector $u$, compare its Rayleigh quotient with any unit vector orthogonal to $v$. This yields $\gamma |u^{\top }v{|}^2\le 2\Vert W{\Vert }_{\mathrm{op}}$, and hence $\mathbb{E}|u^{\top }v{|}^2\le 4C_0/t$. Its mean projector error is at least $\sqrt{2}(1-4C_0/t)$, using $\sqrt{1-x}\ge 1-x$. Along $t=n^{1/4}$ with sufficiently large even $n$, the printed parameter restriction holds and this lower bound tends to $\sqrt{2}$, contradicting an error of order $1/t$. Replace argmin by argmax, or negate the entire matrix before minimizing.

The absolute value of an operator, trace norm, and operator norm are respectivelyThe positive square root of an operator makes $|L|$ well defined. Its eigenvalues $s_j\ge 0$ are the singular values of $L$, so $\Vert L{\Vert }_1={\sum }_j{s}_j$ and $\Vert L{\Vert }_{\mathrm{op}}={\max }_j{s}_j$.

Use the polar decomposition of a bounded operator $L=U|L|$, with the given unitary operator $U$. In an orthonormal eigenbasis $e_j$ for $|L|$,The operator norm assumption and unitarity imply $|⟨e_j|ZU|e_j⟩|\le \Vert Z{\Vert }_{\mathrm{op}}\le 1$. The triangle inequality therefore yields the required trace estimate:Choosing $Z=U^{†}$ attains equality, which also gives the finite-dimensional trace duality formula $\Vert L{\Vert }_1={\max }_{\Vert Z{\Vert }_{\mathrm{op}}\le 1}|\mathrm{Tr}(ZL)|$. A singular $L$ causes no difficulty: in a finite-dimensional square space the polar factor can be extended to a unitary operator on the complementary subspaces.

Write $|\zeta ⟩={\sum }_{i,j}{M}_{ij}|i{⟩}_A|j{⟩}_B$. Normalization says $\Vert M{\Vert }_{\mathrm{HS}}=1$, and the Schmidt rank equals the matrix rank of $M$. Its nonzero singular values are therefore $s_1,\dots ,s_r$ with ${\sum }_j{s}_j^2=1$.

The overlap with the maximally entangled state is $⟨{\varphi }^{+}|\zeta ⟩=\mathrm{Tr}(M)/\sqrt{d}$. The trace norm estimate of Q3(i), with test linear operator $I$, followed by the Cauchy-Schwarz inequality, givesFor two pure states, the unsquared quantum fidelity is the absolute value of their overlap. Hence the maximally entangled overlap bound from Schmidt rank isThe bound is attained by $|\zeta ⟩=r^{-1/2}{\sum }_{j=0}^{r-1}|jj⟩$. Thus uniform Schmidt coefficients on $r$ aligned basis pairs give the greatest possible overlap for that Schmidt rank.

For factorizations $C_i=L_i{L}_i^{\ast }$ and $C_j=L_j{L}_j^{\ast }$, the Procrustes distance between covariance operators is the following infimum over unitary operators $R$:Unitary invariance of the Hilbert-Schmidt norm givesThe polar decomposition of a bounded operator and trace duality implywhere the last equality expresses the trace norm as the sum of the singular values. Taking the infimum proves

The Quantum de Finetti theorem says that the fixed-$k$ reduced density matrix of an exchangeable $N$-particle state approachesas $N\to \infty$. Finite versions bound the trace norm error by a constant of order $d_{\mathrm{loc}}^{2}k/N$. On a bipartite lattice, the corresponding two-sublattice form is a mixtureThis is the mean-field ansatz from the quantum de Finetti theorem.

The bond energy is a linear functional of ${\rho }_{AB}$. A convex combination cannot have energy below its lowest product component, so it remains only to minimizeThe Cauchy-Schwarz inequality gives ${\mathbf{r}}_A\cdot {\mathbf{r}}_B\ge -|{\mathbf{r}}_A||{\mathbf{r}}_B|\ge -1$, with equality for pure antiparallel vectors. This reproduces the Néel state and $-J/4$ per bond found in part (a).

Yes. The limiting state saturates the de Finetti mean-field lower bound: the minimizing product state belongs to the allowed de Finetti mixture, and the finite-de-Finetti error tends to zero as $z\to \infty$. At finite $z$ the theorem gives only an approximation; entanglement and correlated fluctuations can lower the energy below the product-state value by corrections that vanish in the infinite-coordination limit.

For $p=(d+1)/(2d)$, the weighted difference of the two Werner–Holevo channels isThus the map is the transposition map divided by $d$. The diamond norm of the transposition map and invariance of the trace norm under matrix transpose giveSubstitution into the optimal-error formulas shows that an entangled quantum ancilla permits perfect discrimination, whereas every ancilla-free strategy has

For normalized pure states $\rho =|\psi ⟩⟨\psi |$ and $\sigma =|\phi ⟩⟨\phi |$, the trace distance and quantum fidelity satisfyThe last two expressions distinguish the unsquared quantum fidelity from the squared quantum fidelity. Identical pure states give $D=0$, and orthogonal pure states give $D=1$. The equality need not hold for mixed states; the Fuchs-van de Graaf inequalities give bounds in that case.

Brief proof: for two distinct pure states, their two-dimensional span contains the nonzero part of $\rho -\sigma$. This restriction has trace zero and determinant $-(1-F^2)$, hence eigenvalues $\pm \sqrt{1-F^2}$. Its trace norm is twice the positive eigenvalue, giving trace distance $D=\sqrt{1-F^2}$. For identical pure states, $\rho -\sigma =0$ directly.