REVIEW 2 major objections 3 minor 1 cited by
Thirty-six officers, artisanally entangled
T0 review · 2 major / 3 minor · reviewed 2026-08-16 · deepseek-v4-flash
Pith's one-line read The paper constructs the first hand-checkable order-6 perfect tensors and classifies them, within a quadratic-form ansatz, as exactly two symmetry orbits.
desk verdict The explicit artisanal solutions and Hadamard families are real, but the 'exactly two orbits' classification is unproven because Lemma 16 evaluates a condition at r=s=0 that is not actually imposed there. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing object is the doubly perfect function $\lambda$ on the discrete phase space $V=\mathbb{Z}_d^{2n}$: a unimodular function for which both the standard auto-correlation $\lambda\star\lambda$ and the twisted auto-correlation $\lambda\,\tilde{\star}\,\lambda$ equal $d^{2n}\delta$; this is exactly the condition for the Weyl–Heisenberg-diagonal unitary $U_\lambda=\sum_a\lambda(a)|\Phi_a\rangle\langle\Phi_a|$ to be two-unitary. The order-six construction uses the Chinese-remainder decomposition of each $\mathbb{Z}_6$ component into $\mathbb{Z}_3\times\mathbb{Z}_2$, the embedding of the non-singlet sector into $\mathbb{Z}_3^3$, and a two-quadratic-form ansatz $P$ on $\mathbb{Z}_3^2$ plus $Q$ on $\mathbb{Z}_3^3$. The argument is carried by a sequence of quadratic Gauss sums in the $\mathfrak{so}(4)$ picture of the qubit sector, which forces $Q_0=Q(\cdot,\cdot,0)$ to have rank one and then narrows the remaining forms to the symmetric and sparse representatives. A parallel algebraic mechanism identifies two-unitarity with quasi-orthogonality of the image of the local observable algebra with both left and right algebras.
What would settle it
Enumerate all pairs of quadratic forms $P$ on $\mathbb{Z}_3^2$ and $Q$ on $\mathbb{Z}_3^3$, a finite set, and evaluate the standard and twisted autocorrelations for every shift in $\mathbb{Z}_6^2$; if any pair outside the $\mathrm{GL}(\mathbb{Z}_3^2)$-orbits of $\lambda_{\mathrm{sym}}$ and $\lambda_{\mathrm{sparse}}$ satisfies $\lambda\star\lambda=d^{2n}\delta$ and $\lambda\,\tilde{\star}\,\lambda=d^{2n}\delta$, Theorem 1 is false. Such an enumeration would also directly verify the compressed rank-one lemma that the classification depends on.
Extended reading notes
Core claim
On the paper's own terms, the central claim is Theorem 1: after decomposing the phase space $\mathbb{Z}_6^2$ as a disjoint union of $\mathbb{Z}_3^2$ and $\mathbb{Z}_3^3$ via the Chinese-remainder isomorphism with an extra variable $m=\hat{x}-\hat{y}$, and writing $\lambda(a)=\omega_3^{\varphi(a)}$ with $\varphi=P(k,l)$ on the two-qutrit sector and $\varphi=P(k,l)+Q(k,l,m)$ on the three-qutrit sector, the doubly perfect condition, meaning no standard or twisted autocorrelations, is satisfied by exactly two orbits under $\mathrm{GL}(\mathbb{Z}_3^2)$. The representatives are $\lambda_{\mathrm{sym}}: P=k^2+l^2,\ Q=-(k+l+m)^2$ and $\lambda_{\mathrm{sparse}}: P=k^2+l^2,\ Q=(l+m)^2$; each yields a two-unitary $U_\lambda$ that is a direct sum of Clifford unitaries acting on the two-qutrit and three-qutrit sectors, and the two orbits are unitarily inequivalent. The paper further proves that a function is doubly perfect if and only if two associated complex Hadamard matrices are proportional to two-unitaries, and sketches a formulation in which two-unitaries correspond one-to-one to unital subalgebras isomorphic to $\mathrm{M}_d$ and quasi-orthogonal to both local observable algebras.
Load-bearing premise
The classification of exactly two orbits depends on a compressed technical lemma that turns a Gauss-sum condition into a rank-one condition on a two-variable quadratic form; if that step gives way, additional orbits could exist, though the two explicit solutions would still be valid.
Editorial extensions
If this is right
- The two explicit solutions can be verified by a short string of quadratic Gauss sums, so existence of an order-six perfect tensor no longer depends on trusting a computer search.
- The classification gives exactly 24 doubly perfect functions in each orbit, and the associated two-unitaries are not unitarily equivalent.
- Every doubly perfect function produces two-unitary complex Hadamard matrices, so the two order-six solutions seed infinite families of such matrices.
- The algebraic reformulation reduces the construction problem to finding unital subalgebras isomorphic to $\mathrm{M}_d$ and quasi-orthogonal to both local observable algebras.
- The Fourier transform is a symmetry of the doubly-perfect condition, so it can be used to move between solutions and to generate new ones.
Reading between the lines
- The same Chinese-remainder strategy could plausibly be adapted to other composite orders left open by the paper, such as $d=10,14,22$, provided the qubit sector admits a finite-field embedding that bypasses the obstruction for dimensions $2\bmod 4$.
- If the quasi-orthogonal-algebra reformulation matures into a construction tool, it would replace a search over tensor entries with a search over subalgebras; the paper says this use is not yet realized, so the immediate test is to find such subalgebras in dimension six directly.
- Because $U_\lambda$ is diagonal in a stabilizer basis and built from a handful of Clifford gates, the hand-made solutions may be substantially easier to turn into gate sequences or physical realizations than the computer-found algebraic matrices; this is an inference about implementation, not a claim in the paper.
- A natural follow-up is to decide whether every order-six doubly perfect function taking values in powers of $\omega_3$ is equivalent to one of the two artisanal solutions; the paper's observation that a known computer solution maps to the symmetric one is suggestive but not a proof.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper constructs explicit order-6 two-unitary matrices (perfect tensors) using a two-sector ansatz that separates phase space into Z3^2 and Z3^3 components, following and generalizing Rather's doubly perfect function framework. It presents two 'artisanal' solutions, the symmetric and the sparse function, gives a direct Gauss-sum verification of the sparse solution, and claims a classification into exactly two GL(Z3^2)-orbits. It also derives infinite families of Hadamard two-unitaries from doubly perfect functions and sketches an algebraic characterization of two-unitaries via quasi-orthogonal subalgebras.
Significance. If the classification were fully established, the paper would provide the first human-checkable construction of order-6 perfect tensors and would explain the structure of previously computer-found solutions. The direct verification of the sparse solution in Section IV C, the explicit Clifford circuit implementation, the Hadamard connection in Theorem 2, and the algebraic perspective in Section VII are genuine contributions that do not depend on the disputed classification step. These parts are machine-checkable or hand-checkable and give the paper independent value. The classification claim, however, is load-bearing for Theorem 1 and is not supported by the proof as written.
major comments (2)
- [§VIII B, Lemma 16 and Eq. (54)] The proof of Lemma 16 uses the r=s=0 case of Eq. (54) to conclude that Q0 has rank one, but that case is not enforced by the orthogonality derivation in Lemma 15. For r=s=0 the qutrit Weyl operator in Eq. (55) is the identity, which is excluded from the L0⊕R0 test space as used in Lemma 15; if instead the element 1⊗K_m were included in L0, the condition would read 2Σ_{k,l}ω^{Q0(k,l)}=0, which is satisfied by no homogeneous quadratic form over F3^2 and would contradict the direct verification of the sparse solution. In addition, Eq. (54) is displayed with a plus sign, but the calculation in Lemma 20 uses a minus sign. At r=s=0 the displayed equation would give 2Σω^{Q0}=0, whereas Lemma 16 states the sum is pure imaginary. Thus the rank-one reduction of Q0, on which the rest of the classification depends, is not proven by the manuscript.
- [§VIII B, converse direction of Theorem 1] Because Lemma 16 is the only argument forcing rank(Q0)=1, and Lemmas 20–22, together with the converse part of Theorem 1, assume this rank-one normal form, the conclusion that there are exactly two orbits of doubly perfect functions in the stated class is not established. The explicit sparse and symmetric solutions remain independently verified in Section IV C, so the existence part is unaffected; what is missing is a valid exclusion of rank-two Q0, which could in principle admit additional orbits. The statement 'exactly two orbits' should either be re-proven with a corrected and non-vacuous argument or replaced by a more limited claim.
minor comments (3)
- [§VIII B, Lemma 15] The text states 'r,s,m∈Z3^2' after Eq. (54), but r and s are summation variables over Z3 for a fixed m∈Z3; the domain notation appears to be a typo and should be corrected.
- [§VIII B, Lemma 23] The final sentence of the proof reads 'Since the trace is not real, trUλsym = trUλsparse≠ trUλsparse', which is not coherent; it should presumably state that the two traces are complex conjugates (or otherwise explicitly non-equal).
- [§II A 2] The paper does not explicitly prove the 'only if' direction of Theorem 2 for the matrix G, only indicating that G arises as a partial transpose of H; a short sentence clarifying that the partial transpose of a two-unitary is two-unitary would improve readability.
Circularity Check
No significant circularity: the two artisanal solutions are verified by direct computation, and the classification proof does not fit parameters or assume its conclusion.
full rationale
The paper's central existence claim is self-contained. Two-unitarity is characterized by the standard and twisted auto-correlation conditions in Eq. (7), and the paper proves this characterization in Sec. VII C rather than merely importing it. The sparse solution lambda_sparse is verified directly in Sec. IV C by reducing the auto-correlation conditions to finite character sums and evaluating them case by case; the symmetric solution is obtained from it by the explicit symmetry -lambda_sym o G^t = lambda_sparse given in Lemma 19. No free parameter is fitted to the target, and no quantity is presented as a prediction but read back from the data. The classification in Theorem 1 is an internal proof within the two-quadratic-form ansatz: the converse direction derives restrictions from Gauss-sum conditions and standard orbit classifications of rank-one quadratic forms, while the direct direction is again the explicit verification. The main debts to prior work, including Rather's ansatz, Bruzda-Zyczkowski's Hadamard observation, and the overlap lemma of Gross et al., are either rederived in the paper or are independently published, parameter-free theorems whose assumptions do not include the target result. The skeptical concern about Lemma 16 concerns the validity of a Gauss-sum argument, not circularity: even if the lemma has a gap, that would be a proof error rather than a reduction of the claim to its own inputs, and the explicit examples remain independently verified.
Assumptions & free parameters
assumptions (5)
- standard math Standard quadratic Gauss sums over F3, including Eq. (A1), are correct.
- standard math Classification of nondegenerate quadratic forms over finite fields by rank and discriminant.
- standard math Chinese remainder theorem and factorization of Weyl-Heisenberg operators for d=3*2.
- standard math Existence of trace-orthonormal bases in finite fields.
- ad hoc to paper The paper-specific ansatz restricts doubly perfect functions to phases of the form omega_3^{P+Q} with P,Q quadratic forms on the Z3 sectors.
Cite this review
Pith. "Pith review of Thirty-six officers, artisanally entangled." pith.science (2026). https://pith.science/paper/E5BESONV
@misc{pith2026250415401,
author = {Pith},
title = {Pith review of: Thirty-six officers, artisanally entangled},
year = {2026},
howpublished = {\url{https://pith.science/paper/E5BESONV}},
note = {Machine review of arXiv:2504.15401}
}
abstract
A perfect tensor of order $d$ is a state of four $d$-level systems that is maximally entangled under any bipartition. These objects have attracted considerable attention in quantum information and many-body theory. Perfect tensors generalize the combinatorial notion of orthogonal Latin squares (OLS). Deciding whether OLS of a given order exist has historically been a difficult problem. The case $d=6$ proved particularly thorny, and was popularized by Leonhard Euler in terms of a putative constellation of "36 officers". It took more than a century to show that Euler's puzzle has no solution. After yet another century, its quantum generalization was resolved in the affirmative: 36 entangled officers can be suitably arranged. However, the construction and verification of known instances relies on elaborate computer codes. In this paper, we present the first human-made order-$6$ perfect tensors. We decompose the Hilbert space $(\mathbb{C}^6)^{\otimes 2}$ of two quhexes into the direct sum $(\mathbb{C}^3)^{\otimes 2}\oplus(\mathbb{C}^3)^{\otimes 3}$ comprising superpositions of two-qutrit and three-qutrit states. Perfect tensors arise when certain Clifford unitaries are applied separately to the two sectors. Technically, our construction realizes solutions to the perfect functions ansatz recently proposed by Rather. Generalizing an observation of Bruzda and \.Zyczkowski, we show that any solution of this kind gives rise to a two-unitary complex Hadamard matrix, of which we construct infinite families. Finally, we sketch a formulation of the theory of perfect tensors in terms of quasi-orthogonal decompositions of matrix algebras.
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Cited by 1 Pith paper
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Works this paper leans on
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[1]
With ωd =e i 2π d (or justω, ifdis clear from context), define operators X:|x⟩7→|x+ 1⟩, Z:|x⟩7→ω x d|x⟩ 6 generalizing the qubit Pauli matrices
Weyl-Heisenberg group Given a dimensiond, label the standard basis{|q⟩}ofC d by representativesx∈{0,...,d−1}ofZ d =Z/(dZ). With ωd =e i 2π d (or justω, ifdis clear from context), define operators X:|x⟩7→|x+ 1⟩, Z:|x⟩7→ω x d|x⟩ 6 generalizing the qubit Pauli matrices. We next describe the composition law of the group generated byXandZexplicitly. In order t...
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[2]
The inner automorphisms, i.e
Clifford group TheClifford groupis the set of unitary automorphisms of the WH group. The inner automorphisms, i.e. the action by conjugation of WH operators, act as the multiplication by a character. Indeed, from (12), w(a)w(b)w(a) † =ω [a,b]w(b). It turns out that the quotient of the Clifford group by the inner automorphisms and phase factors is isomorph...
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[3]
For oddd, the formulas agree with the metaplectic representation up to global phases, which we have chosen to simplify the expressions
Concrete symplectic Clifford unitaries Here, we give explicit formulas for some Clifford unitaries that will be used in this paper. For oddd, the formulas agree with the metaplectic representation up to global phases, which we have chosen to simplify the expressions. (In particular, the unitaries listed below only generate aprojectiverepresentation of the...
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[4]
The substitutionτ d7→τ m d for some powerm∈Zthat is co-prime to the order ofτgives rise to a faithful, but unitarily inequivalent representation of the WH group
The extended Clifford group The construction of the WH group depends implicitly on the choice of a phase factorτ d. The substitutionτ d7→τ m d for some powerm∈Zthat is co-prime to the order ofτgives rise to a faithful, but unitarily inequivalent representation of the WH group. It shares all algebraic properties of the defining representation in the follow...
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[5]
vectorization
The Weyl-Heisenberg basis OnC d⊗C d, define the maximally entangled states |Φ⟩= 1√ d d−1X x=0 |xx⟩,|Φ (p,q)⟩= (τ pqw(p,q)⊗1)|Φ⟩.(16) The state|Φ (p,q)⟩is the “vectorization” ofτ pqd−1/2w(a). This implies that the set{|Φ a⟩}a∈Z n d forms an ortho-normal basis, sometimes known as theWeyl-Heisenberg basis. 8 The unitary UWH =:|p,q⟩7→|Φ (p,q)⟩(17) is Clifford...
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[6]
Applied to basis states, this gives rise to a unitary equivalenceR:C d≃C d1⊗C d2
The Chinese Remainder Obstruction Recall that the Chinese remainder theorem says that ifd=d 1d2 is a product of co-prime numbers, thenx7→ (xmodd 1,xmodd 2)implements a ring isomorphism ofZ d→Z d1×Z d2. Applied to basis states, this gives rise to a unitary equivalenceR:C d≃C d1⊗C d2. Because the definition of the WH group reduces to operations in the ringZ...
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[7]
(10) factors out can be used to remove the summation overkfrom the auto-correlation conditions
Removing thek-dependence The fact that thek-dependence ofλas defined in Eq. (10) factors out can be used to remove the summation overkfrom the auto-correlation conditions. For the variablesa,b∈Z 2 6 appearing in the auto-correlations, use the notation a≃ (k,x) (l,y) ,b≃ (r,u) (s,v) . The auto-correlations become λ⋆λ (a) = X r ω (k+r) 2−r 2 X s,u,v ω ϕ(x+u...
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[8]
The case(x, y) = (0,0) For the case(x,y;u,v) = (0,0; 1,1), the functionϕand hence the difference∆vanishes, and we obtain X s ω(∆∓ls) = X s ω(∓ls) = 3δ(l). 12 For the case(x,y;u,v) = (0,0,u,v)with(u,v)̸= (1,1), labelϕbyn= ˆu−ˆvto find ∆ =ϕ(n;s−l)−ϕ(n;s+l) = (n+s−l) 2−(n+s+l) 2 =−l(n+s)(28) so that X s ω(∆−ls) =ω(−nl) X s ω(−ls) = 3δ(l), X s ω(∆ +ls) = 3ω(−...
Show all 115 references
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[9]
Summing oversleads to quadratic Gauss sums, which can be evaluated using Eq
The case(x, y) = (1,1) For(u,v) = (0,0), the first summand in∆vanishes, leaving us with ∆ =ϕ(1,1;s−l)−ϕ(0,0;s+l) =−(s+l) 2 =−s 2 +ls−l 2. Summing oversleads to quadratic Gauss sums, which can be evaluated using Eq. (A1) to X s ω(∆−ls) =ω(−l 2) X s ω(−s2) =−γω−l2 , X s ω(∆ +ls)...
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[10]
For(u,v) = (0,0), get ∆ =ϕ(n;s−l)−ϕ(0,0;s+l) = (n+s−l) 2−(s+l) 2 = 1 +nl−s(l+n), ⇒ X s ω(∆−ls) = 3ω 2δ(n−l), X s ω(∆ +ls) = 0
The cases(x, y) = (1,0),(0,1) Letn= ˆx−ˆy∈{±1}. For(u,v) = (0,0), get ∆ =ϕ(n;s−l)−ϕ(0,0;s+l) = (n+s−l) 2−(s+l) 2 = 1 +nl−s(l+n), ⇒ X s ω(∆−ls) = 3ω 2δ(n−l), X s ω(∆ +ls) = 0. The cases(x,y;u,v) = (n;n), i.e. those where(x+u,y+v) = (0,0), give again the complex conjugate, by Eq...
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[11]
ForS∈Sp(V), there is a symmetryλ(a)7→λ(S −1a)
[Symplectic maps on phase space]. ForS∈Sp(V), there is a symmetryλ(a)7→λ(S −1a). It can be implemented by choosing an elementUof the Clifford group associated withGand conjugatingU λ byU⊗ ¯U: (U⊗ ¯U)U λ (U†⊗U t) = X a λ(a)(U⊗ ¯U)|Φ a⟩⟨Φa|(U†⊗U t) = X a λ(a)(Uw(a)U†⊗1)|Φ⟩⟨Φ|(Uw...
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[12]
time reversal
[Extended symplectic maps on phase space]. LetPT: (p,q)7→(p,−q). Thenλ(a)7→λ(PT −1a)is a symmetry, which can be implemented by conjugating with the flip operator: FU λ F= X a λ(a)F|Φ a⟩⟨Φa|F= X a λ(a)(1⊗w(a))|Φ⟩⟨Φ|(1⊗w(a)) † = X a λ(a)(w(a)t⊗1)|Φ⟩⟨Φ|(w(a) t⊗1) † = X a λ(PT−1a)...
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[13]
For anyb∈V, there is a symmetryλ(a)7→λ(a−b)
[Linear shifts on phase space]. For anyb∈V, there is a symmetryλ(a)7→λ(a−b). It can be implemented by (w(b)⊗1)U λ(w(b)†⊗1) = X a λ(a)(w(b)w(a)⊗1)|Φ⟩⟨Φ|(w(a) †w(b)†⊗1) = X a λ(a)ω[b,a](w(b+a)⊗1)|Φ⟩⟨Φ|(w(b+a) †⊗1)ω [a,b] = X a λ(a−b)|Φ a⟩⟨Φa|= (1⊗w(b) t)Uλ(1⊗¯w(b)). The first th...
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[14]
For anyb∈V, there is a symmetryλ(a)7→ω [b,a]λ(a)
[Multiplication by a character]. For anyb∈V, there is a symmetryλ(a)7→ω [b,a]λ(a). It can be implemented by (w(b)⊗¯w(b))Uλ = X a λ(a)(w(b)w(a)w(b)†⊗1)|Φ⟩⟨Φ a|= X a ω[b,a]λ(a)|Φ a⟩⟨Φa|=U λ(w(b)⊗¯w(b))
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[15]
Complex conjugationλ7→ ¯λis a symmetry
[Field automorphisms]. Complex conjugationλ7→ ¯λis a symmetry. To implement it, first conjugate the unitary to get Uλ = X a ¯λ(a)(w(Ta)⊗1)|Φ⟩⟨Φ|(w(Ta)⊗1) † = X a ¯λ(T−1a)|Φ a⟩⟨Φa|, and then undo the action ofT∈ESp(V). More generally, assume thatλtakes values in the cyclotomic ...
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[16]
Trivially,λ7→cλfor forc∈C,|c|= 1is a symmetry
[Global phases]. Trivially,λ7→cλfor forc∈C,|c|= 1is a symmetry
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[17]
TheZ 2n d -Fourier transform λ7→Fλ, Fλ)(a) = 1 dn X b ω−[a,b] d λ(b) is a symmetry
[Fourier transform]. TheZ 2n d -Fourier transform λ7→Fλ, Fλ)(a) = 1 dn X b ω−[a,b] d λ(b) is a symmetry. We give a short proof in Lem. 4 below, and an alternative argument in Sec. V B
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[18]
Proof.Recall theconvolution theorem, which says (Ff) (Fg) = 1 dnF(f ⋆g)
The Fourier transform as a symmetry It remains to show: Lemma 4.Ifλ:V→Cis doubly perfect, then so isFλ. Proof.Recall theconvolution theorem, which says (Ff) (Fg) = 1 dnF(f ⋆g). It implies that the Fourier transform satisfies the unitarity and dual-unitarity conditions: Fλ 2 =F...
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[19]
[60] can be mapped to the symmetric artisanal one by this sequence:
Computer-found solution under symmetries It turns out that the computer solutionΛ 3 of Ref. [60] can be mapped to the symmetric artisanal one by this sequence:
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Multiply by the character associated with the vectorb= (2,2),
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Apply the linear mapG= 3 5 1 2 ! ∈SL(Z 2 6),
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generating a two-unitary
Shift byb= (3,3). This could indicate that there are only few orbits – possibly only one – of doubly perfect functions of order six that take values in powers ofω 3. B. Auto-correlation conditions and the characteristic function The WH operators form an orthogonal basis with r...
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[23]
The local observable algebrasLandRare quasi-orthogonal
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[24]
superoperators
IfH=C 2, the (commutative) algebraA={1⊗1,Z⊗Z}, generated by the tensor product of Pauli-z-matrices, is quasi-orthogonal to bothLandR. Quasi-orthogonality thus captures the fact that the correlations measured byZ⊗Z are not locally accessible. The characterization of two-unitari...
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[25]
These can be read off directly from Eq
Qutrit algebras We start with the images of the subalgebra of local observablesL (3),R (3). These can be read off directly from Eq. (24): UλL(3)U† λ has linear basisw(p+q,p+q)⊗w(−q,p)⊗V tw(p,0)V † t +w(p−q,p+q)⊗w(q,p)⊗|Φ 11⟩⟨Φ11|, UλR(3)U† λ has linear basisw(−q,p)⊗w(p+q,p+q)⊗...
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[26]
We will find it advantageous to start describing the spaceUλ(L(2) 0 ⊕R (2) 0 )U† λ
Theso(4;C)-picture The images of the qubit algebrasL (2),R (2) have a more complicated structure than the qutrit ones. We will find it advantageous to start describing the spaceUλ(L(2) 0 ⊕R (2) 0 )U† λ. It turns out that this direct sum contains a basis that is easier to work ...
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[27]
controlled-Wm gate
Image of the qubit algebras Using the notions of Sec. IV B, define D2 :=U Q N2, D 3,m := (1⊗1⊗⟨m|)U Q N3(1⊗1⊗|m⟩), W m :=U WH (D3,mD† 2)U† WH. Then we see thatUλ amounts to a “controlled-Wm gate” in the sense that Uλ (1⊗|Φ m⟩⟨Φ∅|)U† λ =W m⊗|Φm⟩⟨Φ∅|.(44) 23 Taking the adjoint a...
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[28]
VIII, we give the conditions forU λ(L(2) 0 ⊕R (2) 0 )U† λ to be orthogonal toL⊕Rin the so(4)-picture
Quasi-orthogonality for the qubit algebra In preparation of the proof in Sec. VIII, we give the conditions forU λ(L(2) 0 ⊕R (2) 0 )U† λ to be orthogonal toL⊕Rin the so(4)-picture. All calculations in this section will be re-done in Sec. VIII in greater generality. The projecti...
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[29]
(44): Uλ w(a)⊗|Φ m⟩⟨Φ∅| U† λ = (UWH⊗⟨m|) D3U† WH (w(a)⊗1)U WH (D† 2⊗1) (U† WH⊗|m⟩)⊗|Φ m⟩⟨Φ∅|
We need a generalization of Eq. (44): Uλ w(a)⊗|Φ m⟩⟨Φ∅| U† λ = (UWH⊗⟨m|) D3U† WH (w(a)⊗1)U WH (D† 2⊗1) (U† WH⊗|m⟩)⊗|Φ m⟩⟨Φ∅|. The factor acting on the qutrit system simplifies to (UWH⊗⟨m|) D3U† WH (w(a)⊗1)U WH (D† 2⊗1) (U† WH⊗|m⟩) =(UWH⊗⟨m|)(D 3D† 2⊗1)(U † WH⊗|m⟩)w(S 2a) =1 3 ...
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[30]
symmetric
= GL(Z 2 6), which proves the first claim. Using the notions of Eq. (8), fora,a ′∈Z 2 6, a′ = ˆGta⇔ k′ l′ ! =G k l ! and x′ y′ ! =1 x y ! . Thusλ( ˆGt a) =ω ϕ(k′,l′;m). But Q(k′,l′) = (k,l)GQGt(k,l)t, P(k ′,l′,m) = (k,l,m)(G⊕1)Q(G⊕1) t(k,l,m) t. Finally, by Sec. V A,λ7→( ˆG−tλ...
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[31]
Then−λ◦G t lies in theO(Z 2 3)-orbit of a functionλ′, withP′ =1andQ ′ =R sparse a
Ifλis defined byP=1and a formQwithQ 0 in theO(Z 2 3)-orbit ofR sym a . Then−λ◦G t lies in theO(Z 2 3)-orbit of a functionλ′, withP′ =1andQ ′ =R sparse a
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[32]
Proof.For the choice G:= 1 2 2 2 ! direct calculations giveG(−1)G t =1, G(−R sym a )Gt =R sparse a ,−λ sym◦Gt =λ sparse
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