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?



Hamilton equations-Symplectic scheme
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.


