Two-level avoided crossing

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 $\epsilon \mathrm{/}4$.