What's considered a small time step? I was looking at the following identity that's often used in time evolution:
$$ (e^{xA/n}e^{xB/n})^n \approx (e^{x(A+B)/n})^n$$
This holds when $(\frac{1}{2}(x/n)^2[A,B])^n$ is small. I'm wondering what "small" means. If we consider a physical system with $H = K+V$ (a kinetic and potential term). So in this case, we are looking at $(\frac{1}{2}(-it/n)^2[K,V])^n$. What exactly does "small" mean in this context? (Any references to read up on what "small" means for this would be greatly appreciated!)
 A: First, units: don't forget the factor of $\hbar$ in the time-evolution exponential. The correct form is $e^{-itH/\hbar}$, so that the part in the exponential is dimensionless.
Second, size of operators: One usual way of determining whether an operator $A$ is large or not is to consider its operator norm $||A||$, which is defined as the maximum of $\langle\phi|A^† A|\phi\rangle$ over all possible unit vectors $|\phi\rangle$. Hence, we are probably looking for something like $||\left(1/2(-it/\hbar)^2[K,V]\right)^n||< 1$.
In practice it may be enough for the operator $\Delta = \left(1/2(-it/\hbar)^2[K,V]\right)^n$ to be small just on the current state vector, in the sense that, $\langle\psi_{t=0}|\Delta^†\Delta |\psi_{t=0}\rangle < 1$. I'm not familiar with the proofs of convergence so I don't know if this is the case.
A: Baker–Campbell–Hausdorff formula:
\begin{align*}
e^{X}e^{Y}&=e^{X+Y+\frac12[X,Y]+\frac1{12}[X,[X,Y]]-\cdots}\\
\frac12[X,Y]\approx0\Rightarrow e^{X}e^{Y}&\approx e^{X+Y} 
\end{align*}
where $\approx0$ means small.
A: $xA$ and $xB$ are the arguments of exponentials, and so must be dimensionless. Consequently the only natural scale to compare them to is
$$
\left(\frac{1}{2}\left(\frac{x}{n} \right)^2 [A, B] \right)^n \ll 1
$$
In terms of how much less than $1$, in practice we normally mean either "much smaller than we can measure" or "$0$ when we take the limit $n\rightarrow \infty$" depending on the context.
