Signed phase decoding has no alias on the promised spectrum

By [F1], choose an orthonormal eigenbasis and let $H\mathrm{\mid }u⟩=\lambda \mathrm{\mid }u⟩$. The root inner product is linear in its first argument and conjugate-linear in the second (Dirac Notation Conversion). Self-adjointness therefore gives $\lambda \mathrm{\parallel }u{\mathrm{\parallel }}^2=⟨Hu,u⟩=⟨u,H^{\ast }u⟩=⟨u,Hu⟩=\bar{\lambda }\mathrm{\parallel }u{\mathrm{\parallel }}^2,$ so $\lambda$ is real. Also $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }Hu\mathrm{\parallel }\le \mathrm{\parallel }H\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }$, so $\lambda \in [-1,1]$.

On this eigenvector, $e^{+iH{t}_0\mathrm{/}T}$ has eigenvalue $e^{i\lambda t_0\mathrm{/}T}=e^{2\pi i\theta }$, where $\theta$ is the unique representative of $\lambda t_0\mathrm{/}(2\pi T)$ modulo $1$ in $[0,1)$. Each controlled power has eigenphase $\lambda 2^j{t}_0\mathrm{/}T$, so phase estimation encodes this same phase modulo one.

Since $T\ge 8(t_0+1)$, $\mathrm{\mid }\lambda t_0\mathrm{/}T\mathrm{\mid }\le t_0\mathrm{/}T<1\mathrm{/}8<\pi$ by [F4]. Thus the centered representative of the phase angle is exactly $\lambda t_0\mathrm{/}T$, with no alias by a multiple of $2\pi$. Centered decoding gives the stated signed integer $k$ and adjacent eigenvalue grid values differ by $2\pi \mathrm{/}{t}_0$.