Operator perturbation bound for QSVT

If either $H_R$ or $H_L$ is zero, both maps are zero and the odd transforms agree; when $H_R=0$ the even right transforms are also the same scalar map on the zero space, and when $H_L=0$ the even right transforms are $P(0)I_{H_R}$ for both. The inequality is then immediate. Assume both dimensions are positive. Let $m=\dim H_L$, $n=\dim H_R$, $q=\min (m,n)$, and extend an SVD of any contraction $C:H_R\to H_L$ to orthonormal bases $u_1,\dots ,u_m$ and $v_1,\dots ,v_n$, with ${\sigma }_j=0$ for any missing singular values up to $q$.

On $K=H_L\oplus H_R$, choose orthonormal input bases $a_1,\dots ,a_m$ in the first summand and $v_1,\dots ,v_n$ in the second, and output bases $u_1,\dots ,u_m$ and $b_1,\dots ,b_n$ in the two summands. For $1\le j\le q$, define $V_C{a}_j=\sqrt{1-{\sigma }_j^2}{u}_j-{\sigma }_j{b}_j$ and $V_C{v}_j={\sigma }_j{u}_j+\sqrt{1-{\sigma }_j^2}{b}_j$. These two images are orthonormal by direct expansion; map any remaining $a_j$ to the corresponding unused $u_j$, and any remaining $v_j$ to the corresponding unused $b_j$. The resulting images form an orthonormal basis, so $V_C$ is unitary. If ${\mathrm{\Pi }}_R^0$ projects onto the second summand and ${\mathrm{\Pi }}_L^0$ onto the first, then ${\mathrm{\Pi }}_L^0{V}_C{\mathrm{\Pi }}_R^0{\mathrm{\mid }}_{H_R}=C$ by the SVD formula. Thus every finite contraction has a PUE unitary completion with fixed subspaces of dimensions $n,m$.

Let $\delta >0$ and set $D=(B-A)\mathrm{/}\delta$. Homogeneity of the operator norm gives $\mathrm{\parallel }D{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$, so $D$ is a contraction. By step 1.2, choose unitary completions $V_A,V_D$ of $A,D$ on the same $K$ with the same projectors ${\mathrm{\Pi }}_R^0,{\mathrm{\Pi }}_L^0$. On $K\oplus K$, let $V=V_A\oplus V_D$ and let ${\mathrm{\Pi }}_R,{\mathrm{\Pi }}_L$ select the right and left subspaces, respectively, in the first copy only. Put $c=(1+\delta {)}^{-1\mathrm{/}2}$, $s=\sqrt{\delta \mathrm{/}(1+\delta )}$, and $W=\left(\begin{array}{cc}cI& -sI\\ sI& cI\end{array}\right)$; $c^2+s^2=1$ makes $W$ unitary. For $x$ in the first-copy right subspace, $W(x,0)=(cx,sx)$, $V$ maps this to $(c{V}_{A}x,s{V}_{D}x)$, and $W^{\ast }$ returns first component $c^2{V}_{A}x+s^2{V}_{D}x$. Projection to the first-copy left subspace therefore gives $c^{2}Ax+s^{2}Dx=(A+\delta D)x\mathrm{/}(1+\delta )=Bx\mathrm{/}(1+\delta )$. Since $V$ and $W$ are unitary, [F8]–[F9] show $V^{\mathrm{\prime }}=W^{\ast }VW$ is unitary; it is a PUE completion of $B\mathrm{/}(1+\delta )$ for the same projectors.

Direct multiplication gives $(W-I{)}^{\ast }(W-I)=2(1-c)I$, hence $\mathrm{\parallel }W-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\sqrt{2-2c}$. With $t=\sqrt{1+\delta }\ge 1$, the inequality $2-2\mathrm{/}t\le t^2-1$ follows from $(t-1{)}^2(t+2)\ge 0$, so $\mathrm{\parallel }W-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \sqrt{\delta }$. For every unit $x$, write $Vx-W^{\ast }VWx=(I-W^{\ast })Vx+W^{\ast }V(I-W)x$. By [F10], $V$ and $W^{\ast }$ preserve norms; the two summands therefore have norms at most $\mathrm{\parallel }W-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Using [F5] and [F4] gives $\mathrm{\parallel }V-V^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2\sqrt{\delta }$.

The exact QSVT phase product contains $d$ signal factors, each either $V$ or $V^{\ast }$; its remaining phase factors are identical for $V$ and $V^{\mathrm{\prime }}$. Telescoping the two products one signal factor at a time gives operator-norm difference at most $d\mathrm{\parallel }V-V^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$: the factor differences for $V$ have that norm, and for $V^{\ast }$ the identity $V^{\ast }-V^{\mathrm{\prime }\ast }=V^{\ast }(V^{\mathrm{\prime }}-V)V^{\mathrm{\prime }\ast }$ plus [F10]–[F11] gives the same norm; multiplying by unchanged unitary factors preserves the bound. Compression by orthogonal projections cannot increase the bound, since projections are contractions by [F6]–[F7]. By [F3], the compressed blocks are $P^{\mathrm{S}\mathrm{V}}(A)$ and $P^{\mathrm{S}\mathrm{V}}(B\mathrm{/}(1+\delta ))$. Steps 2.1 and 3.1 therefore give $\mathrm{\parallel }{P}^{\mathrm{S}\mathrm{V}}(A)-P^{\mathrm{S}\mathrm{V}}(B\mathrm{/}(1+\delta )){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2d\sqrt{\delta }$.

For the second comparison, use step 1.2 to choose completions $V_B,V_0$ of $B$ and the zero map on the same $K$, and put $T=V_B\oplus V_0$. With the same projectors and the same $W,c,s$ from step 2.1, $T$ compresses to $B$, while $W^{\ast }TW$ compresses to $c^{2}B+s^{2}0=B\mathrm{/}(1+\delta )$. The calculation of step 3.1 gives $\mathrm{\parallel }T-W^{\ast }TW{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2\sqrt{\delta }$; the telescoping and compression argument of step 4.1 therefore gives $\mathrm{\parallel }{P}^{\mathrm{S}\mathrm{V}}(B)-P^{\mathrm{S}\mathrm{V}}(B\mathrm{/}(1+\delta )){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2d\sqrt{\delta }$. Combining this with step 4.1 by the triangle inequality proves the asserted $4d\sqrt{\delta }$ bound. If $\delta =0$ the two transforms are identical, and if $d=0$ the transform is independent of the input; both cases satisfy the same bound.