Simulating Hamiltonians using the Lie-Product formula

linear algebramatricesproof-explanation

Suppose that I have two hermitian matrices $H_1,H_2$, of similar dimension that do not commute. In a section of a paper on simulating Hamiltonians for quantum computing I encounter the following equation that I do not understand. Supposing that $max(||H_1||_2, ||H_2||_2) \sim O(poly(n))$, and $t$ is some real number, $m$ a positive integer, then from the series expansion of the exponential of a matrix,

$(e^{-iH_1t/m}e^{-iH_2t/m})^m \\=(I – i(H_1+H_2)t/m + O(\frac{t^2 max (H_1,H_2)^2}{m^2}))^m \\=(e^{-i(H_1+H_2)t/m} + O(\frac{t^2 max (H_1,H_2)^2}{m^2}))^m $.

I am not sure why the remainder of the terms in the series have smaller maximal singular values than $(t^2/m^2)max(H_1,H_2)^2$ (which means $(t^2/m^2)max(||H_1||_2,||H_2||_2)^2$). Maybe I just don't understand the notation used here, but I am not sure how these terms are absorbed in the big $O$ notation.

I also do not understand how the author went from the second equality to the third equality using the exponential of $-i(H_1+H_2)t/m$.

For fuller context the paper is given here, and the equations in question are on page 11.

Best Answer

Note that $m$ isn't a fixed positive integer; the authors have chosen $m$ sufficiently large depending on $n$ (so possibly $m \to \infty$) such that this bound holds. Let's write $h = \max\{||H_1||_2, ||H_2||_2\}$. By definition, $$e^{iH_jt/m} = \sum_{k=0}^\infty \frac{(-it/m)^k}{k!} H_j^k$$ and as long as $m$ is large, say $m \geq th$ (or more generally, $m = \Omega(th)$), we have $th/m = O(1)$, hence $$\left\lVert \sum_{k=2}^\infty \frac{(-it/m)^k}{k!}H_j^k\right\rVert_2 \leq \sum_{k=2}^\infty \frac{1}{k!} \left(\frac{th}{m}\right)^k = O\left(\frac{t^2h^2}{m^2}\right)$$ giving \begin{align*} e^{iH_1t/m} e^{iH_2t/m} &= \left(I - iH_1t/m + O\left(\frac{t^2h^2}{m^2}\right)\right)\left(I - iH_1t/m + O\left(\frac{t^2h^2}{m^2}\right)\right) \\ &= I - i(H_1 + H_2)t/m + O\left(\frac{t^2h^2}{m^2}\right) \\ &= e^{-i(H_1 + H_2)t/m} + O\left(\frac{t^2h^2}{m^2}\right) \end{align*} where in the last step we have applied the same approximation as before, using the fact that $||H_1 + H_2||_2 \leq 2h$. The authors later need a stronger lower bound on $m$ of the form $m = \Omega(t^2 h^2/\varepsilon)$ to maintain the approximation after raising this quantity to the $m$-th power.