Sparse parity-path transfer and norm

The map $V_x$ permutes the finite orthonormal basis and satisfies $V_x^2=I$, so it is unitary. In a term indexed by $i,b$, the endpoints become $\mathrm{\mid }i-1,b\oplus p_{i-1}⟩$ and $\mathrm{\mid }i,b\oplus x_i\oplus p_i⟩$. Since $p_i=p_{i-1}\oplus x_i$, the second bit is $b\oplus p_{i-1}$; the edge is therefore an ordinary adjacent path edge with the spectator bit unchanged. Thus $V_x{H}_x{V}_x$ is two identical path matrices $H_0$, one for each bit $b$, and $V_x$ fixes $\mathrm{\mid }0,0⟩$ while sending $\mathrm{\mid }n,0⟩$ to $\mathrm{\mid }n,p_n⟩$.

Let $\mathrm{\mid }{D}_i^n⟩$ be the normalized uniform superposition of $n$-bit strings of Hamming weight $i$, and set $\mathrm{\mid }{D}_{-1}^n⟩=\mathrm{\mid }{D}_{n+1}^n⟩=0$. Counting the $i$ downward flips and $n-i$ upward flips gives $J\mathrm{\mid }{D}_i^n⟩=\frac{1}{2}\sqrt{i(n+1-i)}\mathrm{\mid }{D}_{i-1}^n⟩+\frac{1}{2}\sqrt{(i+1)(n-i)}\mathrm{\mid }{D}_{i+1}^n⟩$ for every $0\le i\le n$. Thus $(2\mathrm{/}n)J$ has the stated path weights on the symmetric subspace.

Each $X_r$ permutes an orthonormal basis, so for $v={\sum }_y{c}_y\mathrm{\mid }y⟩$ one has $\mathrm{\parallel }{X}_{r}v{\mathrm{\parallel }}^2={\sum }_y\mathrm{\mid }{c}_y{\mathrm{\mid }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2$, and hence $\mathrm{\parallel }{X}_r{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$. The vector triangle inequality gives $\mathrm{\parallel }Jv\mathrm{\parallel }\le \frac{1}{2}{\sum }_r\mathrm{\parallel }{X}_{r}v\mathrm{\parallel }=(n\mathrm{/}2)\mathrm{\parallel }v\mathrm{\parallel }$, which yields $\mathrm{\parallel }J{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le n\mathrm{/}2$, so $\mathrm{\parallel }{H}_0{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$; unitary conjugation in step 1.1 gives $\mathrm{\parallel }{H}_x{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$.

For each sign string $\sigma \in \{-1,1{\}}^n$, put $u_{\sigma }=2^{-n\mathrm{/}2}{\sum }_{y\in \{0,1{\}}^n}({\prod }_r{\sigma }_r^{y_r})\mathrm{\mid }y⟩$. Factorizing the finite inner-product sum gives $⟨u_{\sigma },u_{\tau }⟩=2^{-n}{\prod }_r(1+{\sigma }_r{\tau }_r)$, so these $2^n$ vectors form an orthonormal basis. Flipping bit $r$ gives $X_r{u}_{\sigma }={\sigma }_r{u}_{\sigma }$. By [F1]–[F2], $e^{-i(\pi \mathrm{/}2){\sum }_r{X}_r}$ acts on $u_{\sigma }$ by $\exp (-i\pi {\sum }_r{\sigma }_r\mathrm{/}2)$. The scalar addition law in [F1] factors this as ${\prod }_r\exp (-i\pi {\sigma }_r\mathrm{/}2)$, and Euler's formula [F3] gives each factor as $-i{\sigma }_r$. Hence $e^{-i(\pi \mathrm{/}2){\sum }_r{X}_r}=(-i{)}^n{X}_1\cdots X_n$. This maps $\mathrm{\mid }{D}_0^n⟩=\mathrm{\mid }{0}^n⟩$ to $(-i{)}^n\mathrm{\mid }{D}_n^n⟩=(-i{)}^n\mathrm{\mid }{1}^n⟩$. The path basis identifies $\mathrm{\mid }{D}_0^n⟩,\mathrm{\mid }{D}_n^n⟩$ with $\mathrm{\mid }0⟩,\mathrm{\mid }n⟩$; step 1.1 then sends the endpoint to $\mathrm{\mid }n,p_n⟩$. Since $H_0=(2\mathrm{/}n)J$, evolution for $t=\pi n\mathrm{/}2$ is $e^{-it{H}_0}=e^{-i\pi J}$, which is the transfer established above.