Topological Quantum Algorithms and Braid Compilation — Examples

The coefficients have squared moduli $3\mathrm{/}5$ and $2\mathrm{/}5$, whose sum is one. The channel projectors are $P_1=\mathrm{\mid }1⟩⟨1\mathrm{\mid }$ and $P_{\tau }=\mathrm{\mid }\tau ⟩⟨\tau \mathrm{\mid }$, so they send $\mathrm{\mid }\psi ⟩$ to the corresponding coefficient times its basis vector. Thus the probabilities are $3\mathrm{/}5$ and $2\mathrm{/}5$, and after division by the coefficient magnitudes the conditional states are $\mathrm{\mid }1⟩$ and $\mathrm{\mid }\tau ⟩$ by [F1]–[F2].

If the coefficient of one channel is zero, its projected vector is zero and the corresponding probability is zero. The branch-validity theorem says that this case has no normalized conditional state. If the surviving channel is $a$ with coefficient $\alpha$, normalization gives $\mathrm{\mid }\alpha \mathrm{\mid }=1$, so its conditional vector is $\alpha \mathrm{\mid }a⟩$, representing the same pure state as $\mathrm{\mid }a⟩$. This checks both zero-coefficient boundary cases without fixing a global phase.

Both $I_2$ and $-I_2$ send every target vector to the same ray, since the outputs differ by scalar $-1$; hence their standalone outcome probabilities coincide by [F1]. In the ordered control basis $\mathrm{\mid }0⟩,\mathrm{\mid }1⟩$, the controlled block matrix is $\mathrm{diag}(I_2,-I_2)=Z\otimes I_2$.

Applying this matrix to $\mathrm{\mid }+⟩\mathrm{\mid }0⟩=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{\mid }0⟩\mathrm{/}\sqrt{2}$ gives $\mathrm{\mid }-⟩\mathrm{\mid }0⟩$, whereas controlled $I_2$ leaves the input unchanged. The $X$-basis outcome laws are point masses on opposite outcomes, so their total-variation distance is $1$. This is the relative-phase effect in [F2].

Since $\det I_4=1$, both determinant-corrected extensions in [F1] are identity matrices; their determinants are one and their code restrictions are $I_4$.

The empty word acts as the identity in each block and has exponent sum zero, so both operator differences vanish. It fixes every code vector, hence its restriction to the code is exactly $I_4$ and its projection from the code to the orthogonal leakage complement is zero. Thus the logical error and leakage are zero directly, while [F2] gives physical phase $R_1^0=1$.

Compute ${\eta }_1=64(2^{-13}{)}^{3\mathrm{/}2}=2^6{2}^{-39\mathrm{/}2}=2^{-27\mathrm{/}2}>2^{-14}$, while ${\eta }_2=64(2^{-27\mathrm{/}2}{)}^{3\mathrm{/}2}=2^6{2}^{-81\mathrm{/}4}=2^{-57\mathrm{/}4}<2^{-14}$. Therefore $J=2$ is the least level meeting the tolerance, and [F1] gives $L_2\le L_0{5}^2=25L_0$.

Write the QSD factors from [F3] as $G_1,\dots ,G_M$, acting on the full logical register. Let $E$ be the code isometry and $P=E{E}^{\ast }$. For each factor, apply [F5] with tolerance $\delta \mathrm{/}2$ to obtain a unitary braid operator $B_j$ and a unit scalar ${\omega }_j$. Put ${\widehat{G}}_j={\omega }_j(G_j\otimes I_{\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}})$. The code and leakage components of $B_{j}E-E{\widehat{G}}_j$ each have norm at most $\delta \mathrm{/}2$: they are $E(E^{\ast }{B}_{j}E-{\widehat{G}}_j)$ and $(I-P)B_{j}E$. Their sum therefore has norm at most $\delta$.

The telescoping expansion underlying [F4] applies with the isometry inserted: $B_M\cdots B_{1}E-E{\widehat{G}}_M\cdots {\widehat{G}}_1={\sum }_{j=1}^M{B}_M\cdots B_{j+1}(B_{j}E-E{\widehat{G}}_j){\widehat{G}}_{j-1}\cdots {\widehat{G}}_1\mathrm{.}$ Every outer factor is unitary, so step 1.2 and the triangle inequality bound its norm by $M\delta$. This also bounds logical error and leakage, since compression and orthogonal projection do not increase norm; the target product carries the recorded phase ${\prod }_j{\omega }_j$. The minimum defining $\delta$ gives $M\delta \le \epsilon \mathrm{/}2$, including equality of its two candidates. By [F2], SK length, B6 search work, measurement and phase correction remain separate costs. The supplied $L_0$ and the numerical level calculation give no B6 length bound.