Now I consider the harmonic oscillator problem.The ordinal differential equation is q ˙ = p p ˙ = − q In symplectic Euler method, where p ( m + 1 ) − p ( m ) Δ t = − q ( m ) q ( m + 1 ) − q ( m ) Δ t = p ( m + 1 ) the flow operator ψ d , Δ = ( ( 1 − Δ t 2 ) q + Δ t ⋅ p , − Δ t ⋅ q + p ) is symplectic.Here, the textbook suggests that this flow operator stricktly integrates the modified Hamiltonian and this Hamiltonian is invariant. This modified hamiltonian is H ~ = q 2 + p 2 2 − q p 2 Δ t I cannot understand how to derive H ~ .

Question

Now I consider the harmonic oscillator problem.The ordinal differential equation is                                          q            ˙                                              =        p                                                  p            ˙                                              =        −        q            In symplectic Euler method, where                                                                                                                                p                                      (                    m                    +                    1                    )                                                  −                                  p                                      (                    m                    )                                                                                                                                                                Δ                t                                                                        =        −                  q                      (            m            )                                                                                                                                                          q                                      (                    m                    +                    1                    )                                                  −                                  q                                      (                    m                    )                                                                                                                                                                Δ                t                                                                        =                  p                      (            m            +            1            )                              the flow operator       ψ                  d            ,      Δ        =  (  (  1  −  Δ      t    2    )  q  +  Δ  t  ⋅  p  ,  −  Δ  t  ⋅  q  +  p  ) is symplectic.Here, the textbook suggests that this flow operator stricktly integrates the modified Hamiltonian and this Hamiltonian is invariant. This modified hamiltonian is                                          H            ~                          =                                                                                                                    q                  2                                +                                  p                  2                                                                                                                                              2                                                    −                                                                                                  q                p                                                                                                                              2                                                    Δ        t            I cannot understand how to derive             H      ~      .

Pegoxv · Accepted Answer

Compare the energy levels at the intermediate point  (      p          (      m      )            )    2    +  (      q          (      m      )            )    2    =  (      p          (      m      +      1      )        +      q          (      m      )            Δ    t      )    2    +  (      q          (      m      )            )    2      =  (      p          (      m      +      1      )            )    2    +  (      q          (      m      )            )    2    +  2      p          (      m      +      1      )            q          (      m      )            Δ    t  +  (      q          (      m      )            )    2        Δ        t    2  and  (      p          (      m      +      1      )            )    2    +  (      q          (      m      +      1      )            )    2    =  (      p          (      m      +      1      )            )    2    +  (      q          (      m      )        +      p          (      m      +      1      )            Δ    t      )    2      =  (      p          (      m      +      1      )            )    2    +  (      q          (      m      )            )    2    +  2      p          (      m      +      1      )            q          (      m      )            Δ    t  +  (      p          (      m      +      1      )            )    2        Δ        t    2  The terms of first order in       Δ    t are equal and can be compensated for by subtracting   2  p  q      Δ    t, as the additional terms that introduces are of the order       Δ        t    2   and higher.                    (                  p                      (            m            )                                    )          2                +        (                  q                      (            m            )                                    )          2                −        2                  p                      (            m            )                                    q                      (            m            )                                    Δ                t                            =        (                  p                      (            m            )                          −                  q                      (            m            )                                    Δ                t                  )          2                +        (                  q                      (            m            )                                    )          2                −        (                  q                      (            m            )                                    )          2                          Δ                          t          2                                                  =        (                  p                      (            m            +            1            )                                    )          2                +        (                  q                      (            m            )                                    )          2                −        (                  q                      (            m            )                                    )          2                          Δ                          t          2                                    (                  p                      (            m            +            1            )                                    )          2                +        (                  q                      (            m            +            1            )                                    )          2                −        2                  p                      (            m            +            1            )                                    q                      (            m            +            1            )                                    Δ                t                            =        (                  p                      (            m            +            1            )                                    )          2                +        (                  q                      (            m            +            1            )                          −                  p                      (            m            +            1            )                                    Δ                t                  )          2                −        (                  p                      (            m            +            1            )                                    )          2                          Δ                          t          2                                                  =        (                  p                      (            m            +            1            )                                    )          2                +        (                  q                      (            m            )                                    )          2                −        (                  p                      (            m            +            1            )                                    )          2                          Δ                          t          2                    In the more general case with a Hamiltonian   H  =      1    2        p    2    +  V  (  q  ) and thus      p          (      m      +      1      )        =      p          (      m      )        −      V    ′    (      q          (      m      )        )      Δ    tthe same approach gives the modified Hamiltionian            H      ~        =      1    2        p    2    +  V  (  q  )  −  p      V    ′    (  q  )      Δ    t  ,as also   V  (  q  )  −  p      V    ′    (  q  )      Δ    t  =  V  (  q  −  p      Δ    t  )  +  O  (      Δ    t      )    2

Now I consider the harmonic oscillator problem. The ordinal differential equation is dot q=p dot p=−q

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