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Consider a Hamilton function $$H_0(x,p) = \frac{p^2}{2m}+ V(x).$$

The canonical equations then read $$\dot{x}(t) = p/m$$ and $$\dot{p}(t) = -V'(x)$$

Now imagine, we add an additional term $$\dot{x}(t) = p/m + g_0(x,p)\quad\text{and}\quad\dot{p}(t) = -V'(x) + g_1(x,p)$$

Then, we can see $g_1$ as an external force, but what is the interpretation of $g_0$? Is it meaningful to call it a friction?

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$mg_0$ is apparently the difference between the kinetic momentum $m\dot{x}$ and the canonical/conjugate momentum $p$. However if the perturbation $g_0,g_1$ does not preserve the canonical structure, i.e. descend from an interaction Hamiltonian, then the notion of canonical/conjugate momentum is not well-defined.

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You can't just arbitrarily add anything, Since the evolution equation for $(x,p)$ is derived from $H$.

In this case, $$\dot{x}=\partial_pH=\frac{p}{m}+g_0(x,p)$$ $$\dot{p}=-\partial_xH=-V'(x)+g_1(x,p)$$ Since you must have, $$dH(q,p)=\partial_qHdq+\partial_pHdp$$ $$\rightarrow -\partial_p\dot{p}=\partial_q\dot{q}$$$$\Rightarrow -\partial_p g_1(x,p)=\partial_qg_0(x,p)$$ If so you can write, $$H=\frac{p^2}{2m}+V(x)+f(x,p)$$ with $$g_0(x,p)=\partial_p f(x,p)\ \ \&\ \ \ g_1(x,p)=-\partial_xf(x,p)$$


One can regard the tern $f(x,p)$ as the term that couples particle's velocity and its position. A specific example of such a Hamiltonian can be, Charge particle in a uniform constant EM field. So if $\vec{B}=B\hat{k}$, Then you can choose $\vec{A}=-By\hat{i}$ so that $$H=\frac{(p_x+eBy)^2}{2m}+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}+e\phi(\vec{r})$$ If you open up the first braket you get a term that will couple the velocity and position.

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