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We know that $\dot{q} = \frac{\partial H}{\partial p}$ and $\dot{p} = -\frac{\partial H}{\partial q}$, and we also know the values $Q$ and $P$ respectively of $q$ and $p$ at a later time step $\Delta t$. How could we prove that the quantities $$ \begin{align} Q &= q + {\Delta}t\frac{\partial H}{\partial p}(q,p),\\ P &= p - {\Delta}t\frac{\partial H}{\partial q}(q,p) \end{align} $$ are not symplectic, while $$ \begin{align} Q &= q - {\Delta}t\frac{\partial H}{\partial p}(q,p),\\ P &= p + {\Delta}t\frac{\partial H}{\partial Q}(Q,p) \end{align} $$ are symplectic?

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    $\begingroup$ I am not clear what you are asking. How can you say both that $Q,P$ are $q,p$ at a later time step and also that they are given by the equations above? Surely if we wait for some amount of time $\Delta t$, the location of $Q,P$ will usually be a nonlinear function of the time step $\Delta t$. Are you asking that this holds up to higher order terms in $\Delta t$? If so, you can't choose which of the two expressions for $Q,P$ you get to use. $\endgroup$
    – Ben McKay
    Commented Mar 6, 2021 at 17:55
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    $\begingroup$ Assuming that you are asking that $Q,P$ be computed using a linear approximation, linear in $\Delta t$, then clearly both are symplectic, since one is the flow of $H$ and the other the flow of $-H$. $\endgroup$
    – Ben McKay
    Commented Mar 6, 2021 at 17:56
  • $\begingroup$ @BenMcKay I am sorry to confuse you! This is supposed to be a transformation! And I want to know if it is symplectic or not $\endgroup$
    – Jokerp
    Commented Mar 6, 2021 at 19:31
  • $\begingroup$ @BenMcKay also the difference is that the second hamiltonian does not depend on q but Q $\endgroup$
    – Jokerp
    Commented Mar 6, 2021 at 19:47
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    $\begingroup$ @BenMcKay The sets of equations define different numerical integrators: in the first case $(q_{i+1}, p_{i+1})$ directly in terms of $(q_i, p_i)$, and in the second case $q_{i+1}$ in terms of $(q_i,p_i)$, and $p_{i+1}$ in terms of $(q_{i+1},p_i)$. Anyway, the question seems to be more suited for Math.StackExchange. $\endgroup$
    – Ricardo Buring
    Commented Mar 7, 2021 at 14:21

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I take it that your question is about why the symplectic Euler method is symplectic, while the explicit Euler method is not.

The point is that for a Hamiltonian of the form $H(p,q)=T(p)+V(q)$, the symplectic Euler method can be seen as the composition of the two steps \begin{align*} \tilde{q} &= q_i+T'(p_i)\,\Delta t\\ \tilde{p} &= p_i \end{align*} and \begin{align*} q_{i+1} &= \tilde{q}\\ p_{i+1} &= \tilde{p}-V'(\tilde{q})\,\Delta t \end{align*} whose Jacobians are $$ \frac{\partial(\tilde{q},\tilde{p})}{\partial(q_i,p_i)}=\left(\begin{array}{cc}1&T''(p_i)\,\Delta t\\ 0&1\end{array}\right) ~~~~~\text{ and }~~~~~ \frac{\partial(q_{i+1},p_{i+1})}{\partial(\tilde{q},\tilde{p})}=\left(\begin{array}{cc}1&0\\ -V''(\tilde{q})\,\Delta t&1\end{array}\right) $$ respectively. Since both determinants are identically one, it is readily seen that the symplectic Euler step exactly preserves the phase space volume $dq\wedge dp$.

For the explicit Euler method, no such intermediate step exists, and the Jacobian has a determinant that is not exactly equal to one, hence the phasespace volume is not exactly preserved by the numerical integration.

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