The <matrix> exponential is the absolutely convergent <series>
$$
e^X=\sum_{j=0}^\infty\frac{X^j}{j!}.
$$
Multiplying the two expansions gives
$$
e^{kB}e^{kC}
=I+k(B+C)+k^2\left(\frac12B^2+BC+\frac12C^2\right)+O(k^3).
$$
On the other hand,
$$
e^{k(B+C)}
=I+k(B+C)+\frac{k^2}{2}(B^2+BC+CB+C^2)+O(k^3).
$$
Subtracting yields
$$
\boxed{
e^{kB}e^{kC}=e^{k(B+C)}
+\frac{k^2}{2}(BC-CB)+O(k^3).}
$$
This is the leading <Lie-Trotter splitting commutator error>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 41A / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/41a/b/i/solution