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 ( A + B ) Δ t = e A Δ t e B Δ t e 1 / 2 [ A , B ] Δ t 2 + O ( Δ t 3 )
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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 Δ t e B Δ t e 1 2 [ A , B ] Δ t 2 = ( I + A Δ t + ( A Δ t ) 2 2 + O ( Δ t 3 ) ) ( I + B Δ t + ( B Δ t ) 2 2 + O ( Δ t 3 ) ) × ( I 1 2 [ A , B ] Δ t 2 + O ( Δ t 4 ) ) = I + ( A + B ) Δ t + ( A B + A 2 2 + B 2 2 1 2 [ A , B ] ) Δ t 2 + O ( Δ t 3 ) = I + ( A + B ) Δ t + ( 2 A B + A 2 + B 2 A B + B A ) Δ t 2 2 + O ( Δ t 3 ) = I + ( A + B ) Δ t + ( A 2 + B 2 + A B + B A ) Δ t 2 2 + O ( Δ t 3 ) = I + ( A + B ) Δ t + ( A + B ) 2 Δ t 2 2 + O ( Δ t 3 )
e ( A + B ) Δ t = I + ( A + B ) Δ t + ( A + B ) 2 Δ t 2 2 + ( A + B ) 3 Δ t 3 3 ! + = I + ( A + B ) Δ t + ( A + B ) 2 Δ t 2 2 + O ( Δ t 3 ) O ( Δ t 3 ) + O 1 ( Δ t 3 ) = e A Δ t e B Δ t e 1 2 [ A , B ] Δ t 2 O ( Δ t 3 ) + O 1 ( Δ t 3 ) = e A Δ t e B Δ t e 1 2 [ A , B ] Δ t 2 + O 2 ( Δ t 3 )
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