# Prove that e^((A+B) Delta t)=e^(A Delta t) e^(B Delta t) e^(-1/2[A,B] Delta t^2)+O( Delta t^3)

Chad Baldwin 2022-09-26 Answered
Prove that
${e}^{\left(A+B\right)\mathrm{\Delta }t}={e}^{A\mathrm{\Delta }t}{e}^{B\mathrm{\Delta }t}{e}^{-1/2\left[A,B\right]\mathrm{\Delta }{t}^{2}}+O\left(\mathrm{\Delta }{t}^{3}\right)$
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## Answers (1)

Chiecrere2f
Answered 2022-09-27 Author has 9 answers
Expand each term to the order 3 then multiply
${e}^{A\mathrm{\Delta }t}{e}^{B\mathrm{\Delta }t}{e}^{\frac{-1}{2}\left[A,B\right]\mathrm{\Delta }{t}^{2}}=\left(I+A\mathrm{\Delta }t+\frac{\left(A\mathrm{\Delta }t{\right)}^{2}}{2}+O\left(\mathrm{\Delta }{t}^{3}\right)\right)\left(I+B\mathrm{\Delta }t+\frac{\left(B\mathrm{\Delta }t{\right)}^{2}}{2}+O\left(\mathrm{\Delta }{t}^{3}\right)\right)\phantom{\rule{0ex}{0ex}}×\left(I-\frac{1}{2}\left[A,B\right]\mathrm{\Delta }{t}^{2}+O\left(\mathrm{\Delta }{t}^{4}\right)\right)\phantom{\rule{0ex}{0ex}}=I+\left(A+B\right)\mathrm{\Delta }t+\left(AB+\frac{{A}^{2}}{2}+\frac{{B}^{2}}{2}-\frac{1}{2}\left[A,B\right]\right)\mathrm{\Delta }{t}^{2}+O\left(\mathrm{\Delta }{t}^{3}\right)\phantom{\rule{0ex}{0ex}}=I+\left(A+B\right)\mathrm{\Delta }t+\frac{\left(2AB+{A}^{2}+{B}^{2}-AB+BA\right)\mathrm{\Delta }{t}^{2}}{2}+O\left(\mathrm{\Delta }{t}^{3}\right)\phantom{\rule{0ex}{0ex}}=I+\left(A+B\right)\mathrm{\Delta }t+\frac{\left({A}^{2}+{B}^{2}+AB+BA\right)\mathrm{\Delta }{t}^{2}}{2}+O\left(\mathrm{\Delta }{t}^{3}\right)\phantom{\rule{0ex}{0ex}}=I+\left(A+B\right)\mathrm{\Delta }t+\frac{\left(A+B{\right)}^{2}\mathrm{\Delta }{t}^{2}}{2}+O\left(\mathrm{\Delta }{t}^{3}\right)$
${e}^{\left(A+B\right)\mathrm{\Delta }t}=I+\left(A+B\right)\mathrm{\Delta }t+\frac{\left(A+B{\right)}^{2}\mathrm{\Delta }{t}^{2}}{2}+\frac{\left(A+B{\right)}^{3}\mathrm{\Delta }{t}^{3}}{3!}+\cdots \phantom{\rule{0ex}{0ex}}=I+\left(A+B\right)\mathrm{\Delta }t+\frac{\left(A+B{\right)}^{2}\mathrm{\Delta }{t}^{2}}{2}+O\left(\mathrm{\Delta }{t}^{3}\right)-O\left(\mathrm{\Delta }{t}^{3}\right)+{O}_{1}\left(\mathrm{\Delta }{t}^{3}\right)\phantom{\rule{0ex}{0ex}}={e}^{A\mathrm{\Delta }t}{e}^{B\mathrm{\Delta }t}{e}^{\frac{-1}{2}\left[A,B\right]\mathrm{\Delta }{t}^{2}}-O\left(\mathrm{\Delta }{t}^{3}\right)+{O}_{1}\left(\mathrm{\Delta }{t}^{3}\right)\phantom{\rule{0ex}{0ex}}={e}^{A\mathrm{\Delta }t}{e}^{B\mathrm{\Delta }t}{e}^{\frac{-1}{2}\left[A,B\right]\mathrm{\Delta }{t}^{2}}+{O}_{2}\left(\mathrm{\Delta }{t}^{3}\right)\phantom{\rule{0ex}{0ex}}$
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