Adiabatic Computation and Circuit Equivalence — Examples

Put $x=s-1\mathrm{/}2$. Direct multiplication gives $H(s{)}^2=(x^2+1\mathrm{/}4)I$, so its eigenvalues are $\pm \sqrt{x^2+1\mathrm{/}4}$. The ground-to-excited gap is therefore $g(s)=2\sqrt{x^2+1\mathrm{/}4}$, whose minimum is $1$ at $s=1\mathrm{/}2$ and whose endpoint values are $\sqrt{2}$. The ground band is uniformly isolated and has rank one.

Differentiating gives $H^{\mathrm{\prime }}(s)=Z$ and $H^{\mathrm{\prime }\mathrm{\prime }}(s)=0$. Since $Z^{\ast }Z=I$, $\mathrm{\parallel }{H}^{\mathrm{\prime }}(s)\mathrm{\parallel }=1$ for every $s$.

Substituting the endpoint values and derivatives from steps 1.1–1.2 into [F2] gives endpoint contribution $1\mathrm{/}2+1\mathrm{/}2=1$. For the integral, with $x=s-1\mathrm{/}2$, ${\displaystyle {\int }_0^1\frac{7}{g(s{)}^3}ds=\frac{7}{8}{\int }_{-1\mathrm{/}2}^{1\mathrm{/}2}\frac{dx}{(x^2+1\mathrm{/}4{)}^{3\mathrm{/}2}}=\frac{7}{\sqrt{2}},}$ using the antiderivative $x\mathrm{/}((1\mathrm{/}4)\sqrt{x^2+1\mathrm{/}4})$. Hence $A_1(1)=(1+7\mathrm{/}\sqrt{2})\mathrm{/}T$.

The JRS conclusion consequently bounds the projector deviation by $(1+7\mathrm{/}\sqrt{2})\mathrm{/}T$ and the transition probability out of the ground band by its square. For example, any $T\ge 4(1+7\mathrm{/}\sqrt{2})\mathrm{/}\epsilon$ makes the projector deviation at most $\epsilon \mathrm{/}4$.

The path is linear in $s$, hence twice continuously differentiable. Since $Z=\mathrm{diag}(1,-1)$, its eigenvalues are $s-1\mathrm{/}2$ and $-(s-1\mathrm{/}2)$; the gap is $g(s)=2\mathrm{\mid }s-1\mathrm{/}2\mathrm{\mid }$, which vanishes at $s=1\mathrm{/}2$. Therefore no positive uniform full gap exists.

At $s=0$, $H(0)=-Z\mathrm{/}2$ has unique ground vector $\mathrm{\mid }0⟩$; at $s=1$, $H(1)=Z\mathrm{/}2$ has unique ground vector $\mathrm{\mid }1⟩$. The Hamiltonians commute for all times and are diagonal, so the solution of $i{\mathrm{\partial }}_s{U}_T=TH(s)U_T$ maps $\mathrm{\mid }0⟩$ to a scalar phase times $\mathrm{\mid }0⟩$. Its final overlap with the final ground vector is zero, and its transition probability out of the final ground space is exactly $1$ for every $T>0$.

The path satisfies the smoothness assumption but violates the strictly positive uniform-gap hypothesis in Finite JRS adiabatic projector bound. Step 1.2 shows directly that the desired ground-state-following conclusion can fail completely when that hypothesis is removed.

At the algebraic macro-step level, the padding count is $K=\lceil 2\cdot 2\mathrm{/}1\rceil =4$, so $N=6$. The prefix states are $V_0\mathrm{\mid }0⟩=\mathrm{\mid }0⟩$, $V_1\mathrm{\mid }0⟩=X\mathrm{\mid }0⟩=\mathrm{\mid }1⟩$, and $V_2\mathrm{\mid }0⟩=ZX\mathrm{\mid }0⟩=Z\mathrm{\mid }1⟩=-\mathrm{\mid }1⟩$. Since $U_3,\dots ,U_6$ are identities, all later prefix states remain $-\mathrm{\mid }1⟩$. Substitution in the history-state formula of [F1] gives the displayed normalized vector.

For $\mathrm{\ell }=1$, the endpoint patterns are $a_1=00,b_1=10$ on clock bits $(1,2)$; for $\mathrm{\ell }=2,3,4,5$, the internal patterns are $a_{\mathrm{\ell }}=100,b_{\mathrm{\ell }}=110$ on bits $(\mathrm{\ell }-1,\mathrm{\ell },\mathrm{\ell }+1)$; for $\mathrm{\ell }=6$, the endpoint patterns are $a_6=10,b_6=11$ on bits $(5,6)$. On every adjacent pair $(c_{\mathrm{\ell }-1},c_{\mathrm{\ell }})$, the data amplitudes in step 1.1 satisfy ${\varphi }_{\mathrm{\ell }}=U_{\mathrm{\ell }}{\varphi }_{\mathrm{\ell }-1}$: the first two transitions apply $X$ and $Z$, and the last four apply $I$.

By [F1], each $h_{\mathrm{\ell }}$ restricted to its pattern pair has quadratic form $\frac{1}{2}\mathrm{\parallel }{\varphi }_{\mathrm{\ell }}-U_{\mathrm{\ell }}{\varphi }_{\mathrm{\ell }-1}{\mathrm{\parallel }}^2$. Step 2.1 makes this zero for each $\mathrm{\ell }=1,\dots ,6$; all other legal clock amplitudes are outside that term's two-pattern support. Thus every propagation term annihilates $\mathrm{\mid }\eta ⟩$, and the sum $H_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}$ does as well.