In trying to understand a method outlined here (page 3, subroutine 1). Consider $$R_3 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} .$$ Let $A$ be a square matrix over $\mathbb{C}$. Define $X_3(A) = R_3 \otimes A + R_3^{\dagger} \otimes A^{\dagger}$, a.k.a $$X_3(A) = \begin{bmatrix} 0_n & 0_n & A \\ 0_n & 0_n & 0_n \\ A^{\dagger} & 0_n & 0_n \end{bmatrix}$$ where $0_n$ is the $n$-dimensional zero matrix. $X_3(A)$ is Hermitian, and the author refers to this as embedding $A$ in a Hermitian matrix.

Next the author makes the assumption that for two square matrices $A_1, A_2$ of the same dimension, we have access to unitary operators $e^{iX_3(A_1) \tau},e^{iX_3(A_2)\tau}$ (this is possible since $X_3(A_i)$ is Hermitian), for $t$ some "simulation time", and $n$ a positive integer designated as the number of applications. This is described in "input assumption #1" (with $\tau$ a time-parameter). In order to obtain an estimate of $e^{iX_3(A_1 + A_2)t}$ , the procedure is described as:

Procedure: Sequentially apply $e^{iA_1 t/n}, e^{iA_2 t/n}$ for a total of $n$ consecutive times , defining

$u_{add}(t) = (e^{iX_3(A_1)t/n} e^{iX_3(A_2) t/n})^n$ where the number of applications of the unitaries with $\tau = t/n$ is proportional to $n = O(t^2 / \epsilon)$, where $\epsilon$ is the error term.

From what I understand for two matrices $A,B$, $e^A e^B = e^{A+B}$ is generally true only if $A,B$ commute, so the above expression say for $n=2$ would be $ e^{iX_3(A_1)t/2} e^{iX_3(A_2)t/2} e^{iX_3(A_1)t/2} e^{iX_3(A_2)t/2}$, what I'm imagining the procedure shows is :

$ e^{iX_3(A_1)t/2} e^{iX_3(A_2)t/2} e^{iX_3(A_1)t/2} e^{iX_3(A_2)t/2} = e^{iX_3(A_1)t/2} e^{iX_3(A_1)t/2}e^{iX_3(A_2)t/2} e^{iX_3(A_2)t/2} = e^{iX_3(A_1)t} e^{iX_3(A_2)t}=e^{iX_3(A_1+A_2)t} $,

but this assumes that $X_3(A_1)$ and $X_3(A_2)$ commute? I'm also not sure what the author means by "apply" these operators.


1 Answer 1


Your reasoning is correct if your two Hamiltonians commute. But, as you say, it doesn't work if they don't commute. In that case, the trick is to find something that approximates the the thing you want.

So, what you should really be thinking about is taking terms in the opposite order: $$ e^{iH_1t/2}e^{iH_2t/2}e^{iH_1t/2}e^{iH_2t/2}\approx(e^{i(H_1+H_2)t/2})(e^{i(H_1+H_2)t/2})\approx e^{i(H_1+H_2)t}. $$ To understand this, be more explicit. Take the first term and compare it to the claimed approximation $$ e^{iH_1\delta t}e^{iH_2t\delta t}-e^{i(H_1+H_2)\delta t} $$ in the limit that $\delta t$ is small. You can perform a Taylor expansion. Thus, $$ (1+iH_1\delta t+\frac12H_1^2\delta t^2+\ldots)(1+iH_2\delta t+\frac12H_1^2\delta t^2+\ldots)-(1+i(H_1+H_2)\delta t+\frac12(H_1+H_2)^2\delta t^2+\ldots). $$ The terms of order $\delta t^0$ and $\delta t^1$ all cancel, so this approximation is accurate up to $O(\delta t^2)$. Now, to get a full evolution of time $t$, I need $t/\delta t$ time steps. All those $\delta t^2$ sized errors could add up, of which there are $t/\delta t$, so the overall error is $O(\delta t)$ for the full sequence. Thus, make $\delta t$ small enough and you've got a fairly accurate simulation.

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    $\begingroup$ This theorem goes by many names; Trotter formula, Trotter-Suzuki formula, Lie product theorem, Lie-Trotter theorem... See en.wikipedia.org/wiki/Lie_product_formula for some additional references. $\endgroup$ Commented Jun 6, 2019 at 15:38

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