First-order error for two Pauli terms

In the standard inner product, direct conjugate transpose gives $X^{\ast }=X$ and $Z^{\ast }=Z$. Matrix multiplication gives $XZ=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=-iY$ and $ZX=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=iY$. Therefore $[X,Z]=XZ-ZX=-2iY$.

The same matrices give $Y^{\ast }Y=I_2$. Hence [F2] and [F3] give $\mathrm{\parallel }Y\mathrm{\parallel }=1$, and scalar homogeneity of the operator norm yields $\mathrm{\parallel }[X,Z]\mathrm{\parallel }=2$.

Substituting $H_1=X$, $H_2=Z$ and $\mathrm{\parallel }[X,Z]\mathrm{\parallel }=2$ into [F1] gives $\mathrm{\parallel }{S}_1(t\mathrm{/}r{)}^r-U_{X+Z}(t)\mathrm{\parallel }\le (t^2\mathrm{/}(2r))2=t^2\mathrm{/}r$. For the exact finite boundary $t=0$, both evolutions are $I_2$ and the error is zero.

For $H_1=\lambda X$ and $H_2=Z-\lambda X$, both terms are Hermitian and their sum is $Z$. Direct multiplication gives $Z^{\ast }Z=I_2$, and [F2]–[F3] give $\mathrm{\parallel }Z\mathrm{\parallel }=1$. Bilinearity and $[X,X]=0$ give $[H_1,H_2]=\lambda [X,Z]$, whose norm is $2\lambda$ by [step 2.1] and $\lambda \ge 0$. Thus [F1] gives the sufficient certificate $\lambda t^2\mathrm{/}r$. To make this particular bound at most $ϵ$ requires $r\ge \lambda t^2\mathrm{/}ϵ$ when $t\mathrm{\ne }0$. For fixed $t\mathrm{\ne }0$ and $ϵ>0$, this threshold is unbounded as $\lambda$ grows while the total matrix and its norm stay fixed; it shows that this error certificate cannot choose a uniform $r$ from $\mathrm{\parallel }H\mathrm{\parallel }$ and $ϵ$ alone. It is not a lower bound on the actual simulation error.