# Approximating an exponent of non-commutative matrices as a product of exponents

The book Quantum Computation and Quantum Information chapter 4.7.1 presents the following equation. $$$$e^{i(A+B)\Delta t} = e^{i A \Delta t}e^{i B \Delta t} + O(\Delta t^2)$$$$ with non-commuting Hermitian matrices $$A$$ and $$B$$.

I would like to prove the equation above. So far my attempt is to approximate all three exponents \begin{align} \tag{1} e^{i(A+B)\Delta t} &= I + i(A + B) \Delta t + O(\Delta t^2),\\ \tag{2} e^{i A \Delta t} &= I + i A \Delta t + O(\Delta t^2),\\ \tag{3}e^{i B \Delta t} &= I + i B \Delta t + O(\Delta t^2),\\ \end{align} Then multiplying (2) and (3) \begin{align} e^{i A \Delta t}e^{i B \Delta t} &= (I + i A \Delta t + O(\Delta t^2))(I + i B \Delta t + O(\Delta t^2))\\ \tag{4}&=I + i(A + B) \Delta t + O(\Delta t^2). \end{align} Then (4) looks exactly like (1) so I naively conclude $$$$e^{i(A+B)\Delta t} = I + i(A + B) \Delta t + O(\Delta t^2) = e^{i A \Delta t}e^{i B \Delta t} + O(\Delta t^2).$$$$ The last step feels a bit strange but I don't know where the problem is. Could anyone point out a mistake or suggest a proper way of doing it?

• Subtract $(4)$ from $(1)$ and you get the last equality. May 26, 2022 at 7:18