Stationary quantum completions and the readout interface

A nonnegative stationary triple array p(i, y, j) has middle slices My(i, j) = p(i, y, j) and adjacent masses
w(i, y) = ∑︁
j p(i, y, j) = ∑︁
k p(k, i, y). The classical deterministic-readout minimum ∑︁
y rank+(My) is
an existing comparison theorem, not claimed as a new result here. The exact classical comparison is

Omit symbols with zero stationary probability. Write rankC
psd(M) for the least d admitting d-dimensional
complex Hermitian PSD factors Mij = Tr(AiBj). Positive-semidefinite rank and its static correlation-
generation interpretation are established[4, 6]; the additional requirement below is one stationary
recurrent source, not a separate factorization for each observation.

The declared source class has a Hilbert space H = ⨁︁
y Hy, block-diagonal memory states, and
current readout y given by its orthogonal sector. A time-homogeneous Markov step is specified by CP
maps
Φj←y : B(Hy) −→B(Hj),
∑︂
j
Φ∗
j←y(I) = I.

Reading the sector projectively does not change a block-diagonal state. The source cost is the total
Hilbert dimension D = ∑︁
y dim Hy, including its visible current-symbol register. An open Markov
channel, with fresh environment degrees of freedom at each step, is allowed explicitly. This parallels the
fresh randomness allowed in a classical stochastic Markov source; it is not a count of all environmental
or archival degrees of freedom in a closed universe.

Proof. For the lower bound let τy be the stationary positive block of an arbitrary admissible source,
with ∑︁
y Tr τy = 1. Put
Ay
i = Φy←i(τi),
By
j = Φ∗
j←y(I).

Both are positive and p(i, y, j) = Tr(Ay
i By
j ). Hence each sector dimension is at least its slice’s complex
PSD rank; summation proves the bound.
For attainment choose minimum PSD factors separately in each sector. They may be normalized
without increasing their dimensions. Specifically, set Sy = ∑︁
j By
j , restrict to its support, and replace

on that support. Cyclicity of trace and support of every By
j preserve all pairings, while ∑︁
j By
j = I.
Thus Tr Ay
i = w(i, y). Define τy = ∑︁
i Ay
i and restrict further to its support, compressing all factors
there. Pairings and the effect sum are again preserved, and τy is now faithful. Since

a zero w(y, j) implies By
j = 0 on this support. The destination matrix Aj
y also vanishes because its
trace is w(y, j).
For positive w(y, j) set

set the map to zero otherwise. These are CP maps with effects By
j , so each input sector’s summed
map is trace preserving. The constructed state is stationary because
∑︂
y
Φj←y(τy) =
∑︂
y
Aj
y = τj.

It is normalized since ∑︁
y Tr τy = ∑︁
i,y w(i, y) = 1. After any positive-probability observed pair (i, y)
the normalized memory state is exactly σy
i . Its next-output law and next state are

Every finite word is positive when 0 < λ < 1; the noiseless model has forbidden events Yt+2 = Yt.
Adjacent pairs are uniform for all these models.
Each triple slice is My = T λ/9. It has ordinary rank three, so its nonnegative rank is exactly three.
The quantum factors Ay
i = ρi/9, By
j = Eλ
j have size two, and PSD rank one would force ordinary rank
one. Hence Theorem 2.1 and the classical baseline give the exact comparison

The explicit channel is Φj←y(X) = Tr(Eλ
j X)ρy with stationary blocks τy = I/6. Thus the reduced
memory is the physical state of one qubit inside each of three orthogonal current-output sectors; no
hidden classical controller remembering the preceding label is permitted in the resource count.

The POVM is normalized because ∑︁
j nj = 0. The transition is symmetric doubly stochastic, has zero
diagonal and strictly positive off-diagonal entries, and has rank three. To check its rank explicitly,
write ci = cos θi, si = sin θi. Then

and the mutually orthogonal vectors 1, c, s have squared norms m, m/2, m/2. The three nonzero
eigenvalues are 1, −1/2, −1/2. The order-two law is m−2 ∏︁
t Tyt,yt+2. Every triple slice has PSD rank
exactly two; its exact quantum sector cost is 2m.

Proposition 2.2 (Classical sector cost grows with angular resolution). Let rm be the smallest integer
r with m ≤
(︁
r
⌊r/2⌋
)︁. Every classical deterministic-instantaneous source for the phase-circle triple law
has at least mrm states.

Proof. In any nonnegative factorization T = AB with r factors let Si = {l: Ail> 0}. If Si ⊆Sj, the
zero Tjj implies Blj = 0 for every l∈Sj, hence Tij = 0. This contradicts Tij > 0 for i ̸= j. Thus the
m supports are an antichain. To recall the counting bound without relying on a citation, a uniformly
random permutation of r elements has at most one initial segment belonging to an antichain. A set S
is such an initial segment with probability
(︁r
|S|
)︁−1. Taking expectations yields ∑︁
i
(︁r
|Si|
)︁−1 ≤1, hence
m ≤
(︁
r
⌊r/2⌋
)︁, the classical Sperner bound[14]. Apply this to every slice of (1).

The partial trace equals the result of sandwiching X by
√︁Ej ⊗I before tracing the first qubit, so this
is CP. Its effect is Ej ⊗I; summing the effects gives identity. On product inputs,

which proves stationarity.
For the lower bound the four-time cut matrix is Hab,cd = TacTbd/m2, of rank nine. Any d-dimensional
instrument has PSD factors Hab,cd = Tr(AabBcd), where Aab is the unnormalized memory state after
emitting a, b and Bcd = Ψ∗
cΨ∗
d(I) is the effect of the next two outputs. All these factors are nonzero,
because every adjacent pair has probability 1/m2. Their real vector space has dimension at most d2;
thus d ≥3.
Suppose d = 3. The diagonal zeros of T imply zero pairing whenever a = c or b = d. Zero trace
pairing of PSD matrices implies orthogonal ranges. Put

The theorem gives a completed dimension lower bound beyond the generic
√

orthogonal current-symbol sectors. Its diagonal-zero hypothesis is load bearing. For the positive trine
family, the same construction with Eλ
j still gives d ≤4 and the rank argument gives d ≥3. The
following sections prove dimension four on a low-noise interval and dimension three on a high-noise
interval; the intervening boundary remains unresolved.
For m = 3 the unrestricted comparison is exactly four quantum dimensions versus exactly nine
classical HMM states. For larger m, marginalizing H over b, d gives T/m. Nonnegative rank cannot
increase under this operation, so every finite classical HMM needs at least max{9, rm} states, whereas
the quantum minimum remains four. Thus a resolution-dependent classical lower bound survives the
broader interface, although it is weaker than the sector bound. No exact classical optimum for m > 3
is claimed.