Finite unitary averaging and a decoupling second-moment bound

Set $\zeta =\exp (2\pi i\mathrm{/}5)$. The three finite complex-exponential premises in [F9] show $\mathrm{\mid }\zeta \mathrm{\mid }=1$ and ${\zeta }^m=1$ exactly when $5$ divides $m$. Let $G$ be the finite group of monomial unitaries whose nonzero entries belong to $\{1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4\}$. Let $R$ be the real rotation by $\pi \mathrm{/}4$ in coordinates 1 and 2, acting identically elsewhere. Use the symmetric finite list $\mathcal{S}=G\cup \{R,R^{\ast }\}$, retaining multiplicities if any. At each walk step choose the identity with probability $1\mathrm{/}2$ and an element of this list uniformly with the remaining probability. For $j=1,2$, the moment operator on matrices is $T_j(X)=\frac{1}{2}X+\frac{1}{2\mathrm{\mid }\mathcal{S}\mathrm{\mid }}{\sum }_{U\in \mathcal{S}}{U}^{\otimes j}X(U^{\ast }{)}^{\otimes j}\mathrm{.}$ These are self-adjoint contractions for the Hilbert–Schmidt inner product, because the list is inverse-symmetric; laziness makes every eigenvalue nonnegative. Moreover $⟨X,(I-T_j)X⟩=\frac{1}{4\mathrm{\mid }\mathcal{S}\mathrm{\mid }}{\sum }_{U\in \mathcal{S}}\mathrm{\parallel }X-U^{\otimes j}X(U^{\ast }{)}^{\otimes j}{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus their fixed spaces are exactly the simultaneous commutants of the displayed finite unitaries.

For $j=1$, the independent diagonal fifth-root phases force a commuting matrix to be diagonal; permutations force all its diagonal entries equal. For $j=2$, a matrix coefficient from $\mathrm{\mid }i,j⟩$ to $\mathrm{\mid }k,l⟩$ can survive the diagonal phases only when the unordered multisets $\{i,j\}$ and $\{k,l\}$ agree. By the primitive-order identity established in step 1.1, each index occurs with multiplicity 0, 1 or 2; equality modulo 5 of every exponent forces equality of these integer multiplicities. Commutation with all permutations then leaves three constants: $c$ on $\mathrm{\mid }i,i⟩$, $a$ on $\mathrm{\mid }i,j⟩⟨i,j\mathrm{\mid }$ for $i\mathrm{\ne }j$, and $b$ on $\mathrm{\mid }i,j⟩⟨j,i\mathrm{\mid }$. Commutation with $R\otimes R$ applied to $\mathrm{\mid }1,1⟩$, whose rotated vector has nonzero coefficients on both $\mathrm{\mid }1,2⟩$ and $\mathrm{\mid }2,1⟩$, forces $c=a+b$. Conversely $I$ and the swap $F$ commute with every tensor-square unitary. The fixed space is therefore precisely $\mathrm{span}\{I,F\}$, also when $D=2$.

Finite-dimensional spectral decomposition now gives orthogonal projections ${\mathcal{P}}_1,{\mathcal{P}}_2$ onto these fixed spaces and a number $0\le \lambda <1$ such that $\mathrm{\parallel }{T}_j^L-{\mathcal{P}}_j{\mathrm{\parallel }}_{2\to 2}\le {\lambda }^L,j=1,2.$ If the orthogonal complement is zero, take its spectral radius as zero. Every $T_j^L$ is the moment of an finite distribution of length-$L$ unitary words; the limit is only a finite-matrix spectral calculation, not an assumed Haar integral. The first projection is ${\mathcal{P}}_1(X)=\mathrm{tr}(X)I\mathrm{/}D$. For any rank-$K$ projection $P_0$, invariance of trace and swap trace gives ${\mathcal{P}}_2(P_0\otimes P_0)=a_{0}I+b_{0}F,a_0=\frac{K(KD-1)}{D(D^2-1)},b_0=\frac{K(D-K)}{D(D^2-1)}\mathrm{.}$ The coefficients solve $a_0{D}^2+b_{0}D=K^2$ and $a_{0}D+b_0{D}^2=K$.

Choose a coordinate isometry $W_0$ and let $W=U{W}_0$, where $U$ is a finite walk word. Complex conjugation preserves the same limiting first and second moment projections and their norm bounds, so they apply to $P=\bar{W}{W}^T$ as well. The swap identity on matrix units gives the following limiting second moment, with $\mathrm{\Omega }=I_R\mathrm{/}K\otimes {\omega }_E$: ${\lim }_{L\to \mathrm{\infty }}\mathbb{E}\mathrm{\parallel }{\omega }^W-\mathrm{\Omega }{\mathrm{\parallel }}_2^2=\frac{D(KD-1)}{K(D^2-1)}\mathrm{tr}({\omega }^2)+\frac{1-DK}{K(D^2-1)}\mathrm{tr}({\omega }_E^2)\le 2\mathrm{tr}({\omega }^2)\mathrm{.}$ In detail, the squared-state term is $(D^2\mathrm{/}{K}^2)\{a_0\mathrm{tr}({\omega }^2)+b_0\mathrm{tr}({\omega }_E^2)\}$, and the limiting mean is $\mathrm{\Omega }$, so subtract $\mathrm{tr}({\omega }_E^2)\mathrm{/}K$.

The finite moment error has a quantitative bound. Put $d_E=\dim E$ and $\epsilon ={\lambda }^L$. The first-moment error in each matrix unit of the compressed reference is at most $\epsilon$ in Hilbert–Schmidt norm; contracting it with $\omega$ shows $\mathrm{\parallel }\mathbb{E}{\omega }^W-\mathrm{\Omega }{\mathrm{\parallel }}_2\le D\epsilon \mathrm{\parallel }\omega {\mathrm{\parallel }}_2\le D\epsilon$. For the squared-state term, the second-moment error on $P_0\otimes P_0$ has Hilbert–Schmidt norm at most $K\epsilon$. In its trace pairing with $F_E(\omega \otimes \omega )$, the partial trace over the two environment copies has Hilbert–Schmidt norm at most $d_E\mathrm{\parallel }\omega {\mathrm{\parallel }}_2^2\le d_E$ (Cauchy–Schwarz for each diagonal block). Consequently that squared-state error is at most $D^2{d}_E\epsilon \mathrm{/}K$. The cross term with $\mathrm{\Omega }$, whose Hilbert–Schmidt norm is at most $1\mathrm{/}\sqrt{K}$, has error at most $2D\epsilon \mathrm{/}\sqrt{K}$. Hence, with $C_D=D^2{d}_E+2D$, $\mathbb{E}\mathrm{\parallel }{\omega }^W-\mathrm{\Omega }{\mathrm{\parallel }}_2^2\le 2\mathrm{tr}({\omega }^2)+C_D\epsilon \mathrm{.}$ Positivity implies $\mathrm{supp}\omega \subseteq A\otimes \mathrm{supp}{\omega }_E$: if $v\in \ker {\omega }_E$, the marginal quadratic form is a sum of nonnegative diagonal quadratic forms of $\omega$; each is zero, and [F6] makes the corresponding product vectors kernel vectors of $\omega$. Self-adjointness then makes the corresponding rows and columns zero. Compression acts only on $A$, so both ${\omega }^W$ and $\mathrm{\Omega }$ are supported in ${\mathbb{C}}^K\otimes \mathrm{supp}{\omega }_E$; each difference therefore has rank at most $Kr$. Cauchy–Schwarz on singular values and Jensen give the trace-norm bound at most $\sqrt{2Kr\mathrm{tr}({\omega }^2)}+\sqrt{Kr{C}_D\epsilon }$.

Finally write a Hermitian $Z$ as $Z_{+}-Z_{-}$ using [F8]. Each compression of $Z_{+}$ or $Z_{-}$ is positive, so its trace norm is its trace by [F7]; positivity and linearity therefore bound the trace norm of the difference by the sum of the two compressed traces. The first-moment bound $\mathrm{\parallel }\mathbb{E}P-(K\mathrm{/}D)I{\mathrm{\parallel }}_{\mathrm{\infty }}\le \sqrt{K}\epsilon$ gives $\mathbb{E}\mathrm{\parallel }{\mathcal{C}}_W(Z){\mathrm{\parallel }}_1\le (1+D\epsilon \mathrm{/}\sqrt{K})\mathrm{\parallel }Z{\mathrm{\parallel }}_1\mathrm{.}$ Choose a finite $L$ with both $Kr{C}_D{\lambda }^L\le {\delta }^2$ and $D{\lambda }^L\mathrm{/}\sqrt{K}\le \delta$. Such an integer exists by $\lambda <1$; the least such integer is a specified choice. If $r=0$, the positive operator is zero and its decoupling bound is automatic. This proves both assertions with an finite ensemble and exact dimension-dependent moment control.