Let $A$ and $B$ be finite-dimensional Hermitian matrices, and let $v$ $:=$ $max${$||A||,||B||$}.
Show that $||(e^{-iAt/m}e^{-iBt/m})^m -e^{-i(A+B)t}||\leq\epsilon$ provided $m = \Omega(v^2t^2/\epsilon)$.
[Math] Lie Product Formula
linear algebra
Related Solutions
On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,
$T^\dagger = T^t = T; \tag 1$
since $T$ is symmetric, it may be diagonalized by some orthogonal operator
$O:V \to V, \tag 2$
$OO^t = O^tO = I, \tag 3$
$OTO^t = D = \text{diag} (t_1, t_2, \ldots, t_m), \; t_i \in \Bbb R, \; 1 \le i \le m, \tag 4$
where
$m = \dim_{\Bbb R}V; \tag 5$
since each $t_i \in \Bbb R$, for odd $n \in \Bbb N$ there exists
$\rho_i \in \Bbb R, \; \rho_i^n =t_i; \tag 6$
we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set
$R = \text{diag} (\rho_1, \rho_2, \ldots, \rho_m); \tag 7$
then
$R^n = \text{diag} (\rho_1^n, \rho_2^n, \ldots, \rho_m^n) = \text{diag} (t_1, t_2, \ldots, t_m) = D; \tag 8$
it follows that
$T = O^tDO = O^tR^nO = (O^tRO)^n, \tag 9$
where we have used the general property that matrix conjugation preserves products:
$O^tAOO^TBO = O^tABO, \tag{10}$
in affirming (9). We close by simply setting
$S = O^tRO. \tag{11}$
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.
Best Answer
HINT: Consider the product of power series $\exp(A/m)\exp(B/m) = I + \frac{A}{m} + \frac{B}{m} + O\left(\frac{1}{m^2}\right)$. Then use this to estimate the logarithm for sufficiently large $m$, then exponentiate and raise to the $m$-th power.
This is from the proof in section 2.4 of "Lie Groups, Lie Algebras, and Representations" by Brian C. Hall.