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An antiunitary operator between complex Hilbert spaces is a surjective antilinear map satisfying the displayed inner product identity. It preserves norms but conjugates scalars. In a chosen orthonormal basis it is $UK$, with $U$ a unitary operator and $K$ coefficientwise complex conjugation. Conversely any such $UK$ is antiunitary. Antilinearity by itself does not preserve norms.

For a Dirac field, the charge-conjugation matrix implements ${\psi }^c=C{\bar{\psi }}^T$. In the standard gamma matrix representation one can take $C=i{\gamma }^2{\gamma }^0$. An overall phase is conventional. The transpose intertwining relation is different from the entrywise complex conjugation involved in an antiunitary operator.

The conjugate Dolbeault operator is the type-$(1,0)$ component of the exterior derivative: $\partial :{\mathcal{A}}^{p,q}\to {\mathcal{A}}^{p+1,q}$. Complex conjugation interchanges $\partial$ and $\bar{\partial }$. On a complex manifold, $d=\partial +\bar{\partial }$, ${\partial }^2={\bar{\partial }}^2=0$, and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

For real smooth functions of compact support in a planar domain $D$, use the Dirichlet inner productChanging the positive normalization factor does not change orthogonality. For complex functions, insert complex conjugation in the second factor to obtain the corresponding Hermitian form.

Let $f:D\to \tilde{D}$ be a conformal bijection, and let $\tilde{u},\tilde{v}$ be smooth functions of compact support on $\tilde{D}$. Its real Jacobian matrix is $Df=|f^{\prime }|R$, where $R$ is a rotation. By the chain rule,The change of variables formula has Jacobian determinant $|f^{\prime }{|}^2$, so this factor cancels:This proves conformal invariance of the planar Dirichlet inner product. By completion it is also an isometry between the corresponding Dirichlet energy spaces. It asserts invariance of the energy form, not of the inhomogeneous Sobolev norm, whose $L^2$ term has a different transformation rule.

Set $t_0=1$ and $t_i=-1$ for spatial indices. The complex conjugation of the scalar $i$ supplies a minus sign, while conjugating each of the three spatial gamma matrices supplies three more. In the original order,Consequently the chirality matrix obeysNo rearrangement of the gamma matrices is needed, so no additional anticommutator sign occurs. The star is entrywise complex conjugation, not matrix transpose or Hermitian conjugation.

Allowed through an s-channel W boson. The initial up quark and down antiquark annihilate into a virtual $W^{+}$, which produces the final charm quark and strange antiquark. The two weak charged current vertices contain the nonzero CKM matrix elements $V_{ud}$ and $V_{cs}$, with complex conjugations determined by fermion-flow conventions.

The left panel shows the unique physical-particle tree-level Feynman diagram. A neutral exchanged gauge boson cannot connect these charged annihilation currents, and flavour-diagonal neutral vertices cannot turn an up quark into a charm quark. There is no additional elementary charged scalar in the minimal Standard Model.

The real Stone-Weierstrass theorem says that a real subalgebra $A\subset C(K,\mathbb{R})$ which contains the constants and separates points is dense in $C(K,\mathbb{R})$ for the uniform norm, when $K$ is compact Hausdorff space. If $A$ is also closed, $A=C(K,\mathbb{R})$. The complex version needs closure under complex conjugation; that extra condition is unnecessary here.

The cyclotomic polynomial ${\Phi }_7=1+X+\cdots +X^6$ is irreducible: replacing $X$ by $X+1$ gives an Eisenstein polynomial at seven. Hence the extension has degree six, andThere is one subgroup of each order $1,2,3,6$, and therefore precisely four intermediate fields. The trivial subgroup fixes $\mathbb{Q}(\eta )$; the whole group fixes $\mathbb{Q}$. The order-two subgroup $\{1,6\}$ is complex conjugation and fixes $\mathbb{Q}(\eta +{\eta }^{-1})$, of degree three: $\eta$ satisfies $X^2-(\eta +{\eta }^{-1})X+1$ and is not real.

The order-three subgroup $\{1,2,4\}$ fixes $u=\eta +{\eta }^2+{\eta }^4$. Its other conjugate is $v={\eta }^3+{\eta }^5+{\eta }^6$. Expanding products and using ${\sum }_{j=1}^6{\eta }^j=-1$ gives $u+v=-1$ and $uv=2$. Thus $u^2+u+2=0$, and this fixed field is $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{-7})$, of degree two. These generators have the required degrees and are fixed by the indicated subgroups, so the Galois correspondence proves they are the full fixed fields.

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.

Complex conjugation interchanges the Dolbeault operator and conjugate Dolbeault operator. Thus $\partial \alpha =0$ implies $\bar{\partial }\bar{\alpha }=0$, and $\bar{\alpha }$ has type $(q,p)$. Its antiholomorphic degree is $p\ge 1$, so the allowed Dolbeault-Poincaré lemma on the polydisc gives a form $\gamma \in {\mathcal{A}}^{q,p-1}$ satisfying $\bar{\partial }\gamma =\bar{\alpha }$.

Conjugate this identity and put $\beta =\bar{\gamma }$. This proves the conjugate Dolbeault-Poincaré lemma:

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore

Let $\sigma (z)=\overline{z}$, componentwise complex conjugation. Interpret the printed $|z^2|$ in the potential as $|z{|}^2={\sum }_j|z_j{|}^2$. With $D=1+|z{|}^2$, the local Fubini-Study form isIts pullback of a differential form under $\sigma$ iswhere the last equality swaps $j,k$ and uses antisymmetry of the wedge product of differential forms. Since ${\sigma }^2=I$, it is an anti-symplectic involution, with fixed-point set ${\mathbb{R}}^n$. Part (c) gives

As usual for Complex projective space and Real projective space, homogeneous coordinates are represented by nonzero vectors. Componentwise complex conjugation descends to the diffeomorphismIt is well defined because $\overline{\lambda z}=\overline{\lambda }\overline{z}$, and ${\tau }^2=I$. Every standard manifold chart is invariant under $\tau$, with ${\phi }_i\circ \tau =\sigma \circ {\phi }_i$. Part (d) and the gluing of the Fubini-Study form therefore imply ${\tau }^{\ast }{\omega }_{FS}=-{\omega }_{FS}$.

A point $[z]$ in the fixed-point set satisfies $\overline{z}=\lambda z$. Applying complex conjugation again gives $|\lambda |=1$. Choose $a\ne 0$ with $a/\overline{a}=\lambda$; then $\overline{az}=az$, so $[z]$ has a real representative. Conversely every real representative determines a point in the fixed-point set. Hence $\mathrm{Fix}(\tau )={\mathbb{RP}}^n$, and part (c) gives

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ and $D_j=-i{\partial }_{x_j}$. Thus $D_j$ has Fourier symbol ${\xi }_j$. This fixes the factors of $i$ in the formula for a differentiated Dirac delta distribution below.

With the Japanese bracket $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$ consists of smooth functions $a$ such that, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,Here $N$ is one fixed finite real order; the constants may depend on $K$. Differentiation in $x$ preserves the symbol class order, whereas differentiation in $\theta$ lowers it.

A phase function is real-valued and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is a positively homogeneous function of degree one in $\theta$, and has nonzero total differential:The nonvanishing condition concerns both sets of variables, not just ${\nabla }_{\theta }\Phi$. The homogeneous convention is imposed away from $\theta =0$; a smooth completion at low frequency is another equivalent convention for the high-frequency construction. Low-frequency changes contribute a smooth function of $x$.

Choose a cutoff function $\rho (\theta )$ equal to one near zero. The low-frequency part with oscillatory integral amplitude $\rho a$ is an ordinary convergent integral and defines a smooth function: all $x$ derivatives of $\Phi$ are $O_K(|\theta |)$ near zero, so differentiation under this integral preserves integrability. For the remaining part define, when $\theta \ne 0$,Then $L{e}^{i\Phi }=e^{i\Phi }$. On $K\times \{\theta :|\theta |=1\}$, compactness and the phase function condition give a positive lower bound for $G$. Homogeneity consequently gives $G\ge c_K|\theta {|}^2$. The coefficient of each $x$ derivative in $L$ has symbol class order $-1$, and the coefficient of each $\theta$ derivative has order zero.

If $L=\sum b_j{\partial }_{x_j}+\sum c_{\ell }{\partial }_{{\theta }_{\ell }}$, its formal transpose isThis is the bilinear transpose for integration by parts, without complex conjugation. It lowers the symbol class order by one. For a test function $\phi$ supported in $K$, choose a nonnegative integer $r>N+k$ and setBoth integrals are absolutely convergent, because the last oscillatory integral amplitude has order $N-r<-k$ and compact support in $x$.

To verify that this defines the intended oscillatory integral, take any $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and insert $\chi (\epsilon \theta )$ in the original integral. Repeated integration by parts gives the preceding expression with $(L^t{)}^r$ applied also to this cutoff function. Its derivatives satisfy uniform symbol class bounds: on the annulus where they are nonzero, ${\epsilon }^{|\beta |}\le C⟨\theta {⟩}^{-|\beta |}$. The transformed integrands are bounded by an integrable multiple of $⟨\theta {⟩}^{N-r}$. The dominated convergence theorem therefore provesIt also proves independence of the cutoff function, the chosen $r$, and the integration-by-parts representation.

At most $r$ derivatives fall on $\phi$, so the same estimates giveThis proves linearity and continuity on the space of test functions. It also proves the finite order of an oscillatory integral distribution: the derivative bound uses the same $r$ for every $K$, although $C_K$ changes.

For $x_0\in X$, take $k=n$, $\Phi =(x-x_0)\cdot \theta$, and $a=(2\pi {)}^{-n}{\theta }^{\alpha }$. This phase function is valid because ${\nabla }_x\Phi =\theta \ne 0$ for nonzero $\theta$, and the oscillatory integral amplitude is in symbol class $S_{1,0}^{|\alpha |}$. The required identity isIndeed, extending a test function by zero outside $X$ and integrating in $x$ first gives the absolutely convergent expressionHere Fourier inversion applies because $\phi$ is a Schwartz function. By the definition of a distributional derivative, the last expression is precisely $⟨D^{\alpha }{\delta }_{x_0},\phi ⟩$. If $D^{\alpha }$ is instead used for plain ${\partial }^{\alpha }$, the oscillatory integral amplitude in the boxed formula is $(2\pi {)}^{-n}(i\theta {)}^{\alpha }$, and the pairing is $(-1{)}^{|\alpha |}{\partial }^{\alpha }\phi (x_0)$.

Not every distribution is one oscillatory integral of the stated class. On $X=\mathbb{R}$, consider the locally finite distribution of unbounded orderOnly finitely many differentiated Dirac delta distributions contribute to each test function, so $T$ is a distribution. Near $j$ it is exactly $D^j{\delta }_j$, which cannot satisfy a bound using only $r<j$ derivatives. Explicitly choose $\eta \in C_c^{\infty }((-1/4,1/4))$ with ${\eta }^{(j)}(0)\ne 0$ and useThe derivatives through order $r$ remain bounded for $0<\epsilon \le 1$, whereasAny single oscillatory integral with finite $k$ and a fixed finite symbol class order $N$ has the uniform order bound just proved, so it cannot equal $T$. The same construction works on any nonempty open set in positive dimension by choosing points that leave every compact set and taking locally finite differentiated Dirac delta distributions there. This obstruction concerns a single fixed-order oscillatory integral, rather than local representations or an infinite sum of them.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

Expanding the normalized maximally entangled state in the definition of the Choi state givesHere the bar denotes componentwise complex conjugation in the basis defining the Choi state. Every term is a positive tensor-product operator, and the whole operator has trace one because $\tilde{\Lambda }$ is assumed to be a quantum channel. It is therefore a separable quantum state, proving entanglement breaking by the separable Choi-state criterion for entanglement breaking.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requiresthe probability-distribution condition alone would not guarantee this. Equivalently, the Kraus operators are $K_j=\sqrt{d{p}_j}|a_j⟩⟨b_j|$, exhibiting the rank-one Kraus representation of an entanglement-breaking channel.

For a linear operator $H$ mapping transmitted signals to recorded signals, the positive operator $H^{\ast }H$ describes adjoint back-propagation after recording. In a reciprocal array, physical time reversal acoustics implements this operation up to a final complex conjugation and the chosen signal conventions.