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 )

Question

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    )

Chiecrere2f · Accepted Answer

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    )

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)

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