Lie product formula

Remarks

Let $X_m = e^{A/m} e^{B/m}$ and $Y_m = e^{(A + B)/m}$. Both equal $I + (A + B)/m + O(1/m^2)$, so $\norm{X_m - Y_m} = O(1/m^2)$. The identity $X^m - Y^m = \sum_{k=0}^{m-1} X^k (X - Y) Y^{m-1-k}$ and the bound $\norm{X_m}, \norm{Y_m} \le e^{(\norm{A} + \norm{B})/m}$ give $\norm{X_m^m - Y_m^m} = O(1/m)$, and $Y_m^m = e^{A + B}$. If $AB = BA$, then $e^{A/m} e^{B/m} = e^{(A + B)/m}$ and the formula holds for every $m$. In general $e^{A + B} \ne e^A e^B$: for $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, $e^A e^B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, while $e^{A + B} = \begin{pmatrix} \cosh 1 & \sinh 1 \\ \sinh 1 & \cosh 1 \end{pmatrix}$.

Going deeper

For a student

The formula is the basis of splitting methods: to solve $x' = (A + B)x$, one alternates the flows of $x' = Ax$ and $x' = Bx$, which may be easier to compute. In a Lie group it shows that the sum in the Lie algebra corresponds to a limit of products in the group. Trotter extended it to certain unbounded operators on Hilbert spaces, where it is called the Trotter product formula. It is a key step in the proof of the Golden–Thompson inequality.