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arxiv: 2601.09817 · v2 · pith:XJMPPAN5new · submitted 2026-01-14 · 🪐 quant-ph · math-ph· math.MP

Localization of quantum states within subspaces

L. L. Salcedo This is my paper

Pith reviewed 2026-05-21 15:36 UTC · model grok-4.3

classification 🪐 quant-ph math-phmath.MP
keywords localization probabilityquantum subspacesSchur complementoperator decompositionconcavitysuper-additivityquantum informationcryptographic masking
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The pith

Quantum states get a probability of complete localization inside a subspace that is stricter than the usual overlap and obeys concavity plus super-additivity.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper defines a localization probability for a quantum state inside a chosen subspace by decomposing any non-negative operator into a maximal part fully supported inside the subspace and a remaining part with support outside it. This decomposition produces a probability lambda for density operators that is strictly smaller than the conventional trace overlap with the subspace projector. The new probability is concave and super-additive, properties that make it suitable for analyzing composite systems and information-theoretic tasks. Readers care because the construction supplies a rigorous, operationally motivated way to quantify full containment rather than partial overlap, opening direct uses in entropy calculations and cryptographic protocols.

Core claim

Any non-negative operator A admits a unique decomposition A = B + C in which B is the maximal positive operator supported inside a prescribed subspace and C has support disjoint from that subspace. The component B is constructed explicitly through the Schur complement and characterized by an A-dependent inner product together with trace inequalities. When rho is a quantum state, the scalar lambda = Tr(B) equals the probability that rho lies entirely inside the subspace. This lambda is always at most Tr(P rho) and inherits concavity and super-additivity from the decomposition.

What carries the argument

The unique decomposition A = B + C, where B is the maximal positive operator supported inside the subspace and C has disjoint support, obtained via the Schur complement; it isolates the fully localized component of the operator.

If this is right

  • The localization probability is concave, so it is preserved under convex mixtures of states.
  • Super-additivity supplies bounds on the localization of joint states in product subspaces.
  • Natural entropic quantities can be defined from the new probability for use in quantum information.
  • The uniqueness of the decomposition directly yields a simple cryptographic masking scheme.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The same decomposition could be applied to time-evolved states to track how localization changes under open-system dynamics.
  • In quantum error correction the measure might quantify how much a state remains inside a protected code subspace.
  • Derived entropic functionals may connect to existing coherence or correlation measures when the subspace is chosen appropriately.

Load-bearing premise

Every non-negative operator on the Hilbert space admits a unique decomposition into a maximal subspace-supported positive part and a disjoint remainder via the Schur complement.

What would settle it

An explicit non-negative operator that possesses two distinct maximal positive components both supported inside the same subspace, or a concrete state for which the derived localization probability fails to be concave, would refute the uniqueness and the claimed properties.

Figures

Figures reproduced from arXiv: 2601.09817 by L. L. Salcedo.

Figure 1
Figure 1. Figure 1: FIG. 1: Geometric decomposition of a qubit state [PITH_FULL_IMAGE:figures/full_fig_p002_1.png] view at source ↗
Figure 2
Figure 2. Figure 2: FIG. 2: Schematic representation of the decompositions [PITH_FULL_IMAGE:figures/full_fig_p011_2.png] view at source ↗
Figure 3
Figure 3. Figure 3: FIG. 3: Illustration of the decompositions of a qubit state [PITH_FULL_IMAGE:figures/full_fig_p011_3.png] view at source ↗
Figure 4
Figure 4. Figure 4: FIG. 4: Illustration of the concavity property for a qubit in the Bloch [PITH_FULL_IMAGE:figures/full_fig_p014_4.png] view at source ↗
read the original abstract

This work introduces a rigorous notion of localization probability of a quantum state within a given subspace of its Hilbert space. A non-negative operator A is uniquely decomposed as A=B+C, where B is the maximal positive operator supported inside the chosen subspace and C has support disjoint from it. The localized component B can be expressed via the Schur complement and characterized through an A-dependent inner product and suitable trace inequalities. For quantum states, this yields a probability lambda that a state rho be completely contained in a subspace, which is strictly more restrictive than the usual overlap probability Tr(P rho) and enjoys concavity and super-additivity properties. The resulting framework admits natural interpretations in quantum information, including entropic aspects and a simple cryptographic masking scheme based on the uniqueness of the decomposition.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

2 major / 2 minor

Summary. The manuscript introduces a rigorous notion of localization probability λ for a quantum state ρ within a chosen subspace of its Hilbert space. It defines this via a unique decomposition of a non-negative operator A into A = B + C, where B is the maximal positive operator supported inside the subspace (obtained from the Schur complement) and C has support disjoint from the subspace. For quantum states this yields a probability λ that is strictly more restrictive than the standard overlap Tr(Pρ) and satisfies concavity and super-additivity; the framework is illustrated with interpretations in quantum information, entropic quantities, and a cryptographic masking scheme based on decomposition uniqueness.

Significance. If the central mathematical claims hold, the work supplies a new, strictly stronger measure of subspace localization for quantum states together with useful functional properties (concavity, super-additivity) and concrete applications. The parameter-free character of the construction and the explicit link to the Schur complement are strengths that could make the notion a practical addition to the quantum-information toolkit.

major comments (2)
  1. [Definition of the decomposition (likely §2 or §3)] The uniqueness and maximality of the decomposition A = B + C (with B maximal in the Loewner order among positive operators supported inside the subspace) is asserted but not derived in the provided abstract. When the block of A on the orthogonal complement is singular, the generalized Schur complement employs a pseudo-inverse; it is not immediate that the resulting B remains maximal. An explicit proof or counter-example check is required, as this property is load-bearing for the subsequent claims that λ is strictly more restrictive than Tr(Pρ) and inherits concavity and super-additivity.
  2. [Definition of λ for quantum states] The manuscript states that λ is obtained from the trace of the localized component B, yet no explicit formula or normalization step is shown in the abstract. The relation λ = Tr(B) / Tr(A) (or equivalent) must be stated and verified to ensure it is indeed a probability (0 ≤ λ ≤ 1) and that the claimed strict inequality λ ≤ Tr(Pρ) holds with equality only in trivial cases.
minor comments (2)
  1. Notation for the subspace projector P and the inner product induced by A should be introduced once and used consistently; the abstract refers to an “A-dependent inner product” without defining it.
  2. The cryptographic masking scheme is mentioned only in the abstract; a short concrete example (even a 2-qubit illustration) would clarify how uniqueness of the decomposition is exploited.

Simulated Author's Rebuttal

2 responses · 0 unresolved

We thank the referee for the careful reading and constructive comments. The concerns can be addressed by clarifying derivations already present in the full manuscript and by modest revisions to the abstract and introduction for improved accessibility. We respond to each major comment below.

read point-by-point responses

  1. Referee: [Definition of the decomposition (likely §2 or §3)] The uniqueness and maximality of the decomposition A = B + C (with B maximal in the Loewner order among positive operators supported inside the subspace) is asserted but not derived in the provided abstract. When the block of A on the orthogonal complement is singular, the generalized Schur complement employs a pseudo-inverse; it is not immediate that the resulting B remains maximal. An explicit proof or counter-example check is required, as this property is load-bearing for the subsequent claims that λ is strictly more restrictive than Tr(Pρ) and inherits concavity and super-additivity.

    Authors: The full manuscript (Section 2) derives uniqueness and maximality via the Schur complement: Theorem 2.1 shows that B is the unique maximal element in the Loewner order among positive operators supported inside the subspace. For singular blocks we employ the Moore-Penrose pseudoinverse in the generalized Schur complement; the accompanying proof establishes maximality by contradiction—if a strictly larger B′ existed, then A − B′ would fail to be positive semidefinite or the complementary operator C would acquire support inside the subspace. We will add a concise summary of this argument to the introduction and a remark on the singular case to make the reasoning self-contained for readers of the abstract. revision: yes

  2. Referee: [Definition of λ for quantum states] The manuscript states that λ is obtained from the trace of the localized component B, yet no explicit formula or normalization step is shown in the abstract. The relation λ = Tr(B) / Tr(A) (or equivalent) must be stated and verified to ensure it is indeed a probability (0 ≤ λ ≤ 1) and that the claimed strict inequality λ ≤ Tr(Pρ) holds with equality only in trivial cases.

    Authors: Section 3 defines the localization probability for a normalized state ρ (Tr(ρ)=1) by λ = Tr(B), where B is the maximal localized component obtained from the decomposition of A=ρ. By construction 0 ≼ B ≼ ρ, so 0 ≤ λ ≤ 1. Theorem 3.2 proves the strict inequality λ ≤ Tr(Pρ), with equality if and only if C=0 (i.e., the state is fully supported inside the subspace). We will revise the abstract to state the explicit formula λ = Tr(B) for states and to note the normalization and the equality condition. revision: yes

Circularity Check

0 steps flagged

No circularity: new localization probability derived from standard Schur complement decomposition

full rationale

The paper defines a localization probability lambda via the unique decomposition A = B + C of a non-negative operator, with B obtained as the maximal positive part supported in the subspace using the Schur complement. This is a direct mathematical construction resting on standard operator theory and trace inequalities, without any reduction of the central claims to fitted parameters, self-citations, or prior results by the same authors. Properties such as concavity, super-additivity, and strict restrictiveness relative to Tr(P rho) follow from the definition and inequalities rather than being presupposed. The derivation chain is self-contained against external benchmarks in linear algebra.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The central construction rests on standard properties of operators on Hilbert spaces and the existence of the Schur complement; no free parameters, new physical entities, or ad-hoc axioms are introduced in the abstract.

axioms (1)
  • domain assumption Non-negative operators on a Hilbert space admit a unique decomposition into a maximal positive part supported inside a given subspace and a complementary part with disjoint support.
    Invoked when defining B and C from A; this is presented as a rigorous fact enabling the localization probability.

pith-pipeline@v0.9.0 · 5648 in / 1359 out tokens · 50187 ms · 2026-05-21T15:36:18.654645+00:00 · methodology

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Reference graph

Works this paper leans on

41 extracted references · 41 canonical work pages · 1 internal anchor

  1. [1]

    Alternative form from projection 6

    work page

  2. [2]

    Localization of quantum states within subspaces

    Alternative form fromA −1 6 D. Trace inequalities 7 III. Quantum information9 A. Support of a quantum state 9 B. Decomposition of a quantum state along a subspace 9 C. Interpretation of the decomposition as a probability 10 D. Entropic properties 13 E. Measurement of localization 13 F. Cryptographic application 14 IV . Summary and conclusions14 A. Regular...

    work page internal anchor Pith review Pith/arXiv arXiv 2026

  3. [3]

    When X+Y=I, the set{X,Y}is a PVM inH

    work page

  4. [4]

    In addition, when X+Y>0,ran(X) ˙+ran(Y) =H

    If X+Y≥0then X≥0and Y≥0. In addition, when X+Y>0,ran(X) ˙+ran(Y) =H. Proof

    work page

  5. [5]

    Let us show that(ran(X) +ran(Y))⊖ran(X+Y) =0: SinceX+Y is a normal operator,ψ⊥ran(X+Y)iffψ∈ker(X+ Y)

    Clearly ran(X+Y)⊆ran(X) +ran(Y). Let us show that(ran(X) +ran(Y))⊖ran(X+Y) =0: SinceX+Y is a normal operator,ψ⊥ran(X+Y)iffψ∈ker(X+ Y). In that case,(X+Y)ψ=0=⇒Xψ=−Yψ= 0 due to the following: ran(X)∩ran(Y) =0, andψ∈ ker(X)∩ker(Y). Thenψ⊥ran(X) +ran(Y); it follows that ran(X+Y) =ran(X) +ran(Y)

    work page

  6. [6]

    The condition ran(X)∩ran(Y) =0 requires that∀k x kyk =0

    Since the Hermitian operatorsXandYcommute, they share a common spectral decomposition X= n ∑ k=1 xkPk,Y= n ∑ k=1 ykPk,(2.1) where{P k}n k=1 is a PVM andx k,y k ∈R. The condition ran(X)∩ran(Y) =0 requires that∀k x kyk =0. The conditionX+Y=Iimplies that∀k x k +y k =1. Then, for eachkeitherx k =1 andy k =0 orx k =0 andy k =1, hence the set{X,Y}is a PVM. 1 Re...

    work page

  7. [7]

    IfM=0 it follows from clause 1 thatX=Y=0

    LetM:=X+Y. IfM=0 it follows from clause 1 thatX=Y=0. Otherwise we can restrict ourselves to the non-zero subspaceR:=ran(M). WithinRthe operatorMis positive. Then I=M −1/2XM −1/2 +M −1/2Y M−1/2 :=X ′ +Y ′.(2.2) ClearlyX ′ andY ′ are Hermitian and disjoint; hence, they form a PVM inR. ThenXandYare positive operators inRand non-negative inH. WhenM>0 in H,H=r...

    work page

  8. [8]

    2)rank(B) +rank(C) =rank(A)

    A=B+C is also the decomposition of A along ˜Vas a subspace ofran(A). 2)rank(B) +rank(C) =rank(A). 3)ran(B) = ˜V. Proof

    work page

  9. [9]

    The statement then follows from ran(B)⊆ ˜Vand ran(C)∩ ˜V=0, plus uniqueness of the decomposition

    Because the Corollary 2.5BandCvanish on ker(A), thusA,B, andCcan be regarded as operators on ran(A). The statement then follows from ran(B)⊆ ˜Vand ran(C)∩ ˜V=0, plus uniqueness of the decomposition

    work page

  10. [10]

    The statement follows from the Corollary 2.5

    work page

  11. [11]

    The relations ran(B)⊆ ˜V, and ran(C)∩ ˜V=0 im- ply that rank(B)≤dim ˜Vand rank(C) +dim ˜V≤ rank(A). Hence rank(B) +rank(C)≤rank(A).(2.15) Reaching the equal sign requires rank(B) =dim ˜Vand in turn ran(B) = ˜V.□ The Theorem implies that the decomposition effectively oc- curs within ran(A)and it is unchanged ifVis replaced by V∩ran(A). 2 When the decomposi...

    work page

  12. [12]

    A, B, and C are extended as operators onH ′ which vanish onH ′ ⊖H

    IfH⊆H ′, A=B+C is also the decomposition of A alongVas a subspace ofH ′. A, B, and C are extended as operators onH ′ which vanish onH ′ ⊖H

    work page

  13. [13]

    LetW:=ran(C), then A=C+B is the decomposition of A alongWas a subspace ofH

    work page

  14. [14]

    WhenVandV ⊥ are invariant subspaces of A, B(A|V) =P(A|V),C(A|V) =P(A|V ⊥).(2.16)

    work page

  15. [15]

    Forµ≥0, B(µA|V) =µB(A|V), C(µA|V) =µC(A|V). (2.17)

    work page

  16. [16]

    Let B ′ be a Hermitian operator such that A+B ′ ≥0 andran(B ′)⊆V, then B(A+B ′|V) =B(A|V) +B ′ , C(A+B ′|V) =C(A|V). (2.18)

    work page

  17. [17]

    Let U be a unitary operator inH, then B(UAU†|UV) =UB(A|V)U †.(2.19) Proof

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  18. [18]

    and 2) follow fromA=A+0 andA=0+A, being respectively valid decompositions in each case

    work page

  19. [20]

    The exchange of roles between andBandCfollows from the uniqueness of the decomposition sinceBandCare Her- mitian, ran(C)⊆Wand ran(B)∩W=0

    work page

  20. [21]

    Such decomposition fulfills the conditions of the Theorem 2.3 and the statement follows from uniqueness

    WhenVis an invariant subspaceA=PAP+P ⊥AP⊥. Such decomposition fulfills the conditions of the Theorem 2.3 and the statement follows from uniqueness

    work page

  21. [22]

    follows from the uniqueness of the decomposition

    work page

  22. [23]

    The statement is also immediate from the structure of the Schur complement, namely,B=a−b †c−1b

    It follows from uniqueness, sinceA+B ′ = (B+B ′) +C, fulfills the conditions of the unique decomposition ofA+B ′ alongV. The statement is also immediate from the structure of the Schur complement, namely,B=a−b †c−1b. 5

    work page

  23. [24]

    □ It can be noted that in the clause 7 nothing is assumed about the positivity ofB ′, still whenever ran(B ′)⊆V,A+B ′ ≥0 impliesB+B ′ ≥0

    The decompositionUAU † =UBU † +UCU † fulfills the conditions of the Theorem 2.3 and the statement follows from uniqueness. □ It can be noted that in the clause 7 nothing is assumed about the positivity ofB ′, still whenever ran(B ′)⊆V,A+B ′ ≥0 impliesB+B ′ ≥0. This corollary is equivalent to the Lemma 2.10 below. B. Concavity ofB(A|V) Lemma 2.9Let A and B...

    work page

  24. [25]

    IfV⊆V ′ and A=B ′ +C′ is the decomposition along V ′, then B≤B ′

    work page

  25. [26]

    If A≤A ′ and A ′ =B ′ +C ′ is the decomposition of A′ alongV, then B≤B ′

    work page

  26. [27]

    LetV 1 ⊆V, and let A=B 1 +C1 and B=B ′ 1 +C ′ 1 be the decompositions alongV 1, then B ′ 1 ≤B 1 ≤B. ProofThe statements are a straightforward consequence of the Lemma 2.10.□ Proposition 2.12The map A→B(A|V)is super- additive, while A→C(A|V)is sub-additive, that is, B(A1 +A 2|V)≥B(A 1|V) +B(A 2|V), C(A1 +A 2|V)≤C(A 1|V) +C(A 2|V). (2.23) ProofIfA 1 =B 1 +C...

    work page

  27. [28]

    Let Q be the orthogonal projector operator onto the subspace A −1/2 ˜Vwhere ˜V:= ran(A)∩V

    Alternative form from projection Proposition 2.15Let A=B+C be the decomposition of A≥0along the subspaceV⊆H. Let Q be the orthogonal projector operator onto the subspace A −1/2 ˜Vwhere ˜V:= ran(A)∩V. Then 1)B=A 1/2QA1/2,C=A 1/2(I−Q)A 1/2. 2)Tr(B) =Tr(QA). 3)BA −1B=B,BA −1C=0. (2.25) Proof

    work page

  28. [29]

    Also ran(B ′) = ˜V⊆V

    Let B′ :=A 1/2QA1/2,C ′ :=A 1/2(I−Q)A 1/2.(2.26) By constructionB ′ andC ′ are Hermitian andA=B ′ + C′. Also ran(B ′) = ˜V⊆V. On the other hand, since ran(C′)⊆ran(A), ran(C ′)∩V⊆ ˜V, butC ′ is disjoint fromB ′, thus ran(C ′)∩V=0. ThenB ′ =Band C′ =Cby uniqueness of the decomposition

    work page

  29. [30]

    It is a consequence of the previous clause

    work page

  30. [31]

    □ Regarding the second clause, it should be noted that the operatorQdepends onA, as well as onV

    The relations follow from the propertyQ 2 =Q. □ Regarding the second clause, it should be noted that the operatorQdepends onA, as well as onV. ForA>0, the spacesV=ran(B)andW:=ran(C)are not orthogonal, in general. Nevertheless, they can be regarded as orthogonal by modifying the scalar product, namely: ifψ,φ∈ Handψ †φdenotes the (standard) scalar product d...

    work page

  31. [32]

    ProofDue to the Corollary 2.7, the operatorBis the com- ponent ofAalong ˜V:=ran(A)∩Vas a subspace of ran(A)

    Alternative form from A −1 Proposition 2.16Let A≥0onHand A=B+C be its decomposition alongV⊆H, then B−1 = ˜PA−1 ˜P,(2.35) where ˜P is the orthogonal projector operator onto the sub- spaceran(A)∩V. ProofDue to the Corollary 2.7, the operatorBis the com- ponent ofAalong ˜V:=ran(A)∩Vas a subspace of ran(A). The latter can be decomposed as ran(A) = ˜V⊕ ˜V ⊥,(2...

    work page

  32. [33]

    2)Tr(P kB)≥Tr(B k), andTr(P kB) =Tr(B k)iff[P k,B] = 0

    PkBPk ≥B k . 2)Tr(P kB)≥Tr(B k), andTr(P kB) =Tr(B k)iff[P k,B] = 0. 3)Tr(B)≥ n ∑ k=1 Tr(Bk), andTr(B) = n ∑ k=1 Tr(Bk)iff ∀k[P k,B] =0. 4)Tr(B −1)≥ n ∑ k=1 Tr(B−1 k ). A sufficient condition for equal- ity to hold isV⊆ran(A). Proof

    work page

  33. [34]

    Then PkBPk ≥P kBkPk =B k.(2.44)

    SinceV k ⊆Vthe first clause of the Proposition 2.11 impliesB≥B k. Then PkBPk ≥P kBkPk =B k.(2.44)

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    The propertyP kBPk ≥B k implies the following equiv- alence:4 Tr(PkB) =Tr(B k)⇐ ⇒P kBPk =B k,(2.45) 4 IfA≥Band Tr(A)≥Tr(B), then Tr(A−B)≥0 forA−B≥0, and this impliesA−B=0

    Tr(PkB)≥Tr(B k)follows from the previous inequality and the cyclic property of the trace. The propertyP kBPk ≥B k implies the following equiv- alence:4 Tr(PkB) =Tr(B k)⇐ ⇒P kBPk =B k,(2.45) 4 IfA≥Band Tr(A)≥Tr(B), then Tr(A−B)≥0 forA−B≥0, and this impliesA−B=0. 8 thus, the second part of the statement is equivalent to PkBPk =B k ⇐ ⇒[Pk,B] =0.(2.46) Let us...

    work page

  35. [36]

    The second part is also an immediate consequence of the previous clause

    The first part follows from Tr(P kB)≥Tr(B k)using ∑k Pk =PandPB=B. The second part is also an immediate consequence of the previous clause

    work page

  36. [37]

    Let ˜V:=ran(A)∩Vand ˜Vk :=ran(A)∩V k, and let ˜Pand ˜Pk be the corresponding orthogonal projector op- erators. From the Proposition 2.16 it follows that B−1 = ˜PA−1 ˜P,B −1 k = ˜PkA−1 ˜Pk.(2.50) Clearly, ˜Vk ⊆ ˜V, thus ˜Pk ˜P= ˜Pk and so B−1 k = ˜PkB−1 ˜Pk.(2.51) In addition, the ˜Vk are orthogonal among them, andL n k=1 ˜Vk ⊆ ˜V⊆V, hence ∑n k=1 ˜Pk ≤ ˜P≤...

    work page

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    Let B be the component of A alongV⊆Hand P the orthogonal projector ontoV, then PAP≥B andTr(PA)≥Tr(B),(2.55) andTr(PA) =Tr(B)iff[P,A] =0

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