# 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!)

• Is this for a Fourier split-step method solution to the Schrödinger equation? Jul 11 '20 at 6:20
• I'm not too familiar with what the Fourier split-step is, so likely not? Jul 13 '20 at 6:19

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.
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.
• Physically is there a way to compare this to momentum or $\hbar$? Jul 10 '20 at 20:55
$$xA$$ 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.