Hamiltonian Simulation and Product Formulas — Examples

Put $A=-i{H}_1$ and $B=-i{H}_2$, so $AB=BA$. Define $Y(s)=e^{sA}{e}^{sB}$. By [F2] and the product rule, $Y^{\mathrm{\prime }}(s)=A{e}^{sA}{e}^{sB}+e^{sA}B{e}^{sB}=(A+B)Y(s)$, where commutation lets $B$ pass through $e^{sA}$. Also $Y(0)=I$.

The unique solution of $Z^{\mathrm{\prime }}=(A+B)Z$, $Z(0)=I$, is $U_{H_1+H_2}(s)$ by [F2]. Hence $Y(h)=U_{H_1+H_2}(h)$ for every real $h$, which proves the one-step identity in [F1].

Taking $h=t\mathrm{/}r$ and composing the exact identity $r$ times gives $S_1(t\mathrm{/}r{)}^r=U_{H_1+H_2}(t\mathrm{/}r{)}^r=U_{H_1+H_2}(t)$ by the group law in [F2] and the product definition [F1].

Fix $N=2^n$, equip ${\mathbb{C}}^N$ with its standard inner product, and let $\mathrm{\mid }x⟩$ denote its standard basis. For any function $f:[N]\to \{-1,+1\}$ define $H_f={\sum }_{x=1}^{N}f(x)\mathrm{\mid }x⟩⟨x\mathrm{\mid }$. This diagonal matrix is Hermitian and each row has exactly one nonzero entry; the row-location oracle returns $x$ at the nonzero slot, and the entry-value oracle returns $f(x)$ on the diagonal query $(x,x)$. Thus the input is $1$-sparse in [F1].

The canonical ordered rank-one list is $H_f={\sum }_{x=1}^{N}f(x)\mathrm{\mid }x⟩⟨x\mathrm{\mid }$, with $N$ terms, and is determined by $(f(1),\dots ,f(N))$. There are $2^N$ such lists but only $2^N-1$ binary strings of length less than $N$, so any lossless binary encoding has a worst-case length of at least $N$ bits. If a deterministic exact recovery procedure makes fewer than $N$ classical row-value queries, some row $x$ is unqueried; changing only $f(x)$ preserves every reply in its transcript but changes the canonical list, so that transcript cannot determine the list for both inputs.

Since $N=2^n$, this canonical list's description and exact classical recovery costs grow exponentially in the index length $n$. The sparse model supplies only the row/value calls by [F1], whereas the term-list model additionally requires the list and term-evolution access by [F2]; those extra promises therefore cannot be inferred from the sparse interface. No lower bound for every possible alternative decomposition, quantum query procedure, or circuit implementation is asserted.

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.