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Here is a finite-dimensional form of the Kitaev geometrical lemma. Let $A,B$ be positive operators, with nullspaces $S_A,S_B$ and all positive eigenvalues at least $\gamma >0$. First assume $S_A\cap S_B=\{0\}$. Let $P,Q$ be the orthogonal projections onto those nullspaces, and define their smallest angle between two subspaces byIf either nullspace is zero, set $\cos \vartheta =0$. Then

To prove it, spectral decomposition gives $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. Write $C=P+Q$ and $c=\cos \vartheta$. For every vector $z$,Thus $C^2\le (1+c)C$. Since $C$ is positive semidefinite, each of its eigenvalues lies in $[0,1+c]$, so $2I-C\ge (1-c)I$. Consequentlyas claimed.

If there is a common nullspace $S=S_A\cap S_B$, restrict to $S^{\perp }$ and define the angle between $S_A\cap S^{\perp }$ and $S_B\cap S^{\perp }$ there. The same proof bounds the smallest positive eigenvalue of $A+B$ by $2\gamma {\sin }^2(\vartheta /2)$, while $S$ is its zero-energy subspace. This version gives a spectral gap above a possibly degenerate ground state space.

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

Let $\Gamma =T_B$ denote the partial transpose. The partial transpose of a maximally entangled projector iswhere $S$ is the swap operator, satisfying $S^{†}S=I$ and $\Vert S{\Vert }_{\mathrm{op}}=1$. If $\sigma$ has positive partial transpose, then ${\sigma }^{\Gamma }$ is a positive operator of trace one, hence $\Vert {\sigma }^{\Gamma }{\Vert }_1=1$.

For a pure state ${\varphi }^{+}=|\Phi ⟩⟨\Phi |$, the unsquared quantum fidelity obeys $F({\varphi }^{+},\sigma {)}^2=⟨\Phi |\sigma |\Phi ⟩$. Using the trace-adjoint identity for partial transpose and part (i),The overlap is nonnegative, so taking its square root proves the PPT maximally entangled overlap bound:The product state $|00⟩⟨00|$ attains the bound, so it is sharp.

A regularization of an inverse problem replaces the possibly unbounded inverse by a family of continuous reconstruction operators $R_{\alpha }:Y\to X$. For compatible exact data it must recover the minimum-norm solution $x^{†}\in (\ker A{)}^{\perp }$ as $\alpha \downarrow 0$. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, a parameter rule $\alpha =\alpha (\delta )$ must make both bias and amplified noise tend to zero. The target is the minimum-norm solution because data cannot determine a component in the kernel of a linear map.

Tikhonov regularization minimizesThe quadratic norm penalty makes this functional strictly convex and a coercive function. Its first variation, applied to arbitrary real and imaginary perturbations, givesThe positive operator $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$ and has a bounded inverse. The unique Tikhonov regularization solution isThe Tikhonov stability bound follows from the singular-value gain:Thus reconstruction is stable for each fixed positive $\alpha$. Stability is not uniform as $\alpha \downarrow 0$: that limit restores the unbounded inverse. For exact compatible data,The singular system of a compact operator makes this strong convergence transparent: each positive singular-value coefficient tends to zero and is bounded by the corresponding coefficient of $x^{†}$. Consequentlywhenever $\alpha (\delta )\downarrow 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$, for example $\alpha (\delta )=\delta$. This is a convergent regularization of an inverse problem, not merely a bounded formula.

Write the residual of an inverse problem as $r_k=K{u}_{\delta }^{(k)}-f^{\delta }$. The iteration can be rearranged asApplying $K$ gives $(I+\tau K{K}^{\ast })r_{k+1}=r_k$. Since $K{K}^{\ast }$ is a positive operator, its resolvent operator $(I+\tau K{K}^{\ast }{)}^{-1}$ has operator norm at most one. HenceMore explicitly, taking the inner product of the residual equation with $r_{k+1}$ givesThis directly proves the same monotonicity by the Cauchy-Schwarz inequality, including the case $r_{k+1}=0$. An unresolvable data component in $\mathcal{R}(K{)}^{\perp }$ remains unchanged.

Let $H$ be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator $T=H^{\ast }H$. For a normalized emitted signal $h$,The maximizing signal is a right singular vector, equivalently an eigenvector of $T$ with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.

Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere $b_j$ is the receiver response to scatterer $j$, $c_j^{\ast }h$ is its illumination by the source array, and ${\alpha }_j$ is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors $b_j$ and $c_j$. In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector $c_j/\Vert c_j\Vert$ gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest $|{\alpha }_j{|}^2$. Geometric size by itself is not the quantity being ranked.

Repeated adjoint time reversal acoustics is power iteration on $T$. If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if $H$ includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

This symmetric finite-rank integral operator is not positive. For $f(t)=1-\frac{3}{2}t$, set $A_j={\int }_0^1{t}^{j}f(t)dt$. Directly,The associated quadratic form isEvery covariance operator is a positive operator, so this operator is not a covariance operator.

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

A positive operator-valued measure is a family of positive operators $E_i$ satisfying ${\sum }_i{E}_i=I$. Each $E_i$ is called an effect and determines a probability $p_i=\mathrm{Tr}(E_i\rho )$ for input density operator $\rho$. The effects need not be orthogonal projections or commute. This permits measurements with more outcomes than the dimension of the Hilbert space, such as the trine qubit POVM. Coupling to a quantum ancilla and measuring the larger system supplies a physical route to generalized measurements. ^[ref].

For an elementary unsharp $Z$ example on a qubit, takeOn the eigenstate $|0⟩$, $p_{\pm }=(1\pm \eta )/2$; at $\eta =1/2$ they are $3/4$ and $1/4$. At $\eta =0$ the outcome is a fair coin independent of the input, and at $\eta =1$ these effects specify a sharp $Z$ projective measurement. Choosing $K_{\pm }=\sqrt{E_{\pm }}$ specifies one quantum instrument, but $U_{\pm }\sqrt{E_{\pm }}$ for outcome-dependent unitary operators has the same effects and different state updates. Thus POVM does not determine the post-measurement state.

For a Hermitian operator with spectral decomposition $X={\sum }_i{\lambda }_i{P}_i$, its positive part is $X_{+}={\sum }_i\max ({\lambda }_i,0)P_i=(|X|+X)/2$. It is a positive operator supported on the positive spectral subspace. Together with the negative part of a Hermitian operator it satisfies $X=X_{+}-X_{-}$ and $|X|=X_{+}+X_{-}$.

Every bounded positive operator $T$ has a unique positive operator $T^{1/2}$ satisfying $(T^{1/2}{)}^2=T$. The spectral theorem for normal operators on a separable Hilbert space constructs it by applying the scalar square-root function to the spectrum.

A measurement in quantum mechanics turns a quantum state preparation into a classical outcome, such as a detector label. Two questions must be distinguished: how likely is each outcome, and what quantum state remains conditional on that outcome? A positive operator-valued measure answers the first; a quantum instrument answers both. This distinction is useful whenever a measured system participates in a later experiment, including quantum state tomography and measurement-based quantum feedback.

For a density operator $\rho$, the Born rule gives $p_m=\mathrm{Tr}(E_m\rho )$, where the $E_m$ are positive operators satisfying ${\sum }_m{E}_m=I$. Here $I$ is the identity operator, and $\mathrm{Tr}$ denotes the trace. Positivity makes every $p_m$ nonnegative; completeness makes their sum one. For a pure state, $\rho =|\psi ⟩⟨\psi |$ and $p_m=⟨\psi |E_m|\psi ⟩$. A projective measurement is the special case in which the $E_m$ are orthogonal projections onto mutually orthogonal subspaces.

For a qubit, choose the computational basis and write a normalized pure state, up to global phase, asThe angle $\varphi$ is a relative quantum phase; at the poles it does not distinguish states. Measuring the Pauli matrix $Z$ reports $+1$ for label zero and $-1$ for label one, withThese probabilities describe repeated identical preparations. They do not assign a hidden, pre-existing value to every possible measurement basis.

The Born rule grew out of Max Born's 1926 work on quantum scattering: squared probability amplitudes determine the relative frequencies of scattering outcomes. Born explained this statistical interpretation again in his 1954 Nobel lecture. This probability law does not by itself select a post-measurement state. ^[ref]; ^[ref].

A separable positive operator is a finite sum ${\eta }_{AB}={\sum }_y{\alpha }_y\otimes {\beta }_y$ with both factors positive operators. If its trace is one, normalizing the nonzero factors gives a separable quantum state. Conversely, a nonnegative scalar multiple of a separable quantum state is a separable positive operator.

For two finite-dimensional subspaces with projections $P,Q$, define $\cos \vartheta ={\sup }_{\Vert u\Vert =\Vert v\Vert =1}|⟨u,v⟩|=\Vert PQ\Vert$, with vectors drawn from the respective subspaces. The angle is zero when they intersect nontrivially. To measure separation after removing a common intersection, restrict both subspaces to its orthogonal complement. The resulting angle controls sums of positive operators in the Kitaev geometrical lemma.

The support of a positive operator is the orthogonal complement of its kernel. In finite dimension it is the span of eigenvectors with positive eigenvalues, equivalently the operator's image. Quantum relative entropy is finite only when the first state's support lies in the reference state's support.