Generator convention matters

For the two-vertex graph with one edge, in basis $(e_0,e_1)$ the adjacency swaps the basis vectors, $A{e}_0=e_1$ and $A{e}_1=e_0$, and $D=I$. Thus $L_{+}=I-A$, $L_{-}=-L_{+}$, and $A_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=A$. Put $u_{+}=(e_0+e_1)\mathrm{/}\sqrt{2}$, $u_{-}=(e_0-e_1)\mathrm{/}\sqrt{2}$, and $\psi =(e_0+i{e}_1)\mathrm{/}\sqrt{2}$. Direct inner-product calculations using [F9] give $\mathrm{\parallel }{u}_{+}\mathrm{\parallel }=\mathrm{\parallel }{u}_{-}\mathrm{\parallel }=\mathrm{\parallel }\psi \mathrm{\parallel }=1$ and $⟨u_{+},u_{-}⟩=0$; direct application of $A$ gives $A{u}_{+}=u_{+}$ and $A{u}_{-}=-u_{-}$. Thus $(u_{+},u_{-})$ is an orthonormal eigenbasis.

For the three-vertex path with ordered vertices $0,1,2$, the degrees are $(1,2,1)$ and $A{e}_0=e_1$, $A{e}_1=e_0+e_2$, $A{e}_2=e_1$. Thus $A=\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right),D=\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 1\end{array}\right),A_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=D^{-1\mathrm{/}2}A{D}^{-1\mathrm{/}2}=A\mathrm{/}\sqrt{2},L_{+}=D-A,L_{-}=A-D=-L_{+}\mathrm{.}$ All degrees are positive, so [F1] defines $A_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$. Put $u_0=(e_0-e_2)\mathrm{/}\sqrt{2}$, $u_{+}=(e_0+\sqrt{2}{e}_1+e_2)\mathrm{/}2$, and $u_{-}=(e_0-\sqrt{2}{e}_1+e_2)\mathrm{/}2$. By direct coordinate inner products in the standard orthonormal basis [F9], $(u_0,u_{+},u_{-})$ is an orthonormal basis; direct multiplication gives $A{u}_0=0$, $A{u}_{+}=\sqrt{2}{u}_{+}$, and $A{u}_{-}=-\sqrt{2}{u}_{-}$. The vectors $c=(e_0+e_1+e_2)\mathrm{/}\sqrt{3}$, $u_0$, and $v=(e_0-2e_1+e_2)\mathrm{/}\sqrt{6}$ also have unit norm and pairwise zero inner products. Direct multiplication gives $L_{+}c=0$, $L_{+}{u}_0=u_0$, and $L_{+}v=3v$. Hence the four generator spectra are $\sigma (A)=\{0,\pm \sqrt{2}\}$, $\sigma (A_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}})=\{0,\pm 1\}$, $\sigma (L_{+})=\{0,1,3\}$, and $\sigma (L_{-})=\{0,-1,-3\}$. This directly compares adjacency, normalized adjacency, and both Laplacian signs on the irregular graph.

The spectral definition [F2] gives eigenphases $1,e^{-2i}$ for $U_{L_{+}}(1)$ on $(u_{+},u_{-})$ and $1,e^{2i}$ for $U_{L_{-}}(1)$. By [F3]–[F4], these are equivalently $e^{-i}(\cos 1I+i\sin 1A)$ and $e^i(\cos 1I-i\sin 1A)$: on $u_{+}$ and $u_{-}$, respectively, the coefficients are $1,e^{-2i}$ and $1,e^{2i}$. The $e_1$ coordinate after $U_{L_{+}}(1)$ is $e^{-i}i(\cos 1+\sin 1)\mathrm{/}\sqrt{2}$; after $U_{L_{-}}(1)$ it is $e^{i}i(\cos 1-\sin 1)\mathrm{/}\sqrt{2}$. Thus [F10] gives $p_{+}(1)=(\cos 1+\sin 1{)}^2\mathrm{/}2$ and $p_{-}(1)=(\cos 1-\sin 1{)}^2\mathrm{/}2$.

Put $\xi =(u_0+u_{+})\mathrm{/}\sqrt{2}$, which is unit by step 1.2, and fix the target ray $[\zeta ]:=[U_A(\pi \mathrm{/}\sqrt{2})\xi ]=[(u_0-u_{+})\mathrm{/}\sqrt{2}]$, since $e^{-i\pi }=-1$ by [F4], [F7]. For any real $t$, $U_A(t)\xi =(u_0+e^{-i\sqrt{2}t}{u}_{+})\mathrm{/}\sqrt{2}$ and $U_{A_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}(t)\xi =(u_0+e^{-it}{u}_{+})\mathrm{/}\sqrt{2}$ by [F1]–[F2]. Since the $u_0$ coefficient is the same nonzero number in both target and output, equality of rays forces the scalar multiple to be $1$ and is equivalent to equality of the displayed relative phases.

At $1$, the cosine-series magnitudes $b_k=1\mathrm{/}(2k)!$ and sine-series magnitudes $c_k=1\mathrm{/}(2k+1)!$ decrease to zero: each next factorial multiplies the denominator by a factor at least $2$, so both sequences tend to zero. By [F5]–[F6], the cosine sum lies above its two-term partial sum $1-1\mathrm{/}2=1\mathrm{/}2$, and the sine sum lies above $1-1\mathrm{/}6=5\mathrm{/}6$. Therefore $p_{+}(1)-p_{-}(1)=2\sin 1\cos 1\ge 5\mathrm{/}6>0$. The two Laplacian signs give different same-time measurement laws.

If $[U_A(t)\xi ]=[\zeta ]$, then $e^{-i(\sqrt{2}t-\pi )}=1$. By [F3]–[F4], this forces $\sin (\sqrt{2}t-\pi )=0$ and $\cos (\sqrt{2}t-\pi )=1$; [F8] puts the angle at $k\pi$, and [F7] shows $k$ must be even. Thus $t=\pi \mathrm{/}\sqrt{2}+k\sqrt{2}\pi$. For $k\ge 0$ this is at least $\pi \mathrm{/}\sqrt{2}$; for $k\le -1$ it is at most $\pi \mathrm{/}\sqrt{2}-\sqrt{2}\pi <0$. Hence its least nonnegative value is $\pi \mathrm{/}\sqrt{2}$. The same argument for $A_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ gives $t=\pi +2k\pi$, whose least nonnegative value is $\pi$. For the second family, $k=0$ gives $t=\pi$, every $k\ge 1$ gives a larger time, and every $k\le -1$ gives a negative time. Thus the least nonnegative time is $\pi$. These runtimes differ because $\sqrt{2}>1$.