[NOTES/CM-04002] Poisson Bracket Formalism

Poisson Bracket 

We introduce Poisson bracket of two dynamical variables by
\begin{equation}\Label{EQ03} \{F,G\}_{PB}\stackrel{\text{def}}{\equiv}\sum_k\Big( \frac{\partial{F}}{\partial{q_k}}\frac{\partial{G}}{\partial{p_k}}-\frac{ \partial{F}}{\partial{p_k}} \frac{\partial{G}}{\partial{q_k}}\Big ) \end{equation}
{Fundamental Poisson brackets}
It is straight forward to verify that the variables $q_k,p_k$ satisfy the Poisson bracket relations \begin{equation}\Label{EQ15} \{q_i,q_j\}_{PB}=0 ;~~~~~~~~~~ \{p_i,p_j\}=0 \end{equation} \begin{equation}\Label{EQ16} \{q_i,p_j\}_{PB}=\delta_{ij} \end{equation}

Equations of motion in Poisson bracket form
The Hamiltonian equations of motion are
\begin{equation}\Label{EQ01}
\boxed{\frac{\partial{H}}{\partial{p_k}}= \dot q_k, \qquad \frac{\partial{H}}{\partial{q_k}}=-\dot p_k}.
\end{equation}
These equations of motion can be written in the Poisson bracket form
\begin{equation}\Label{EQ01A}
\boxed{ \dot q_k=\frac{\partial{H}}{\partial{p_k}}=, \qquad\dot p_k \frac{\partial{H}}{\partial{q_k}}}.
\end{equation}

Time variation of a dynamical variable

Let $F(q,p,t)$ be a dynamical variable. We compute the equation of motion giving the time evolution of \(F(q,p,t)\):
\begin{eqnarray}\nonumber
\frac{d}{dt}F(q,p,t)
&=& \frac{\partial{F}}{\partial{t}}+\sum_k
\frac{\partial{F}}{\partial{q_k}}\dot q_k
+\frac{\partial{F}}{\partial{p_k}}\dot p_k\\
&=& \frac{\partial{F}}{\partial{t}}+\sum_k\Big(
\frac{\partial{F}}{\partial{q_k}}\frac{\partial{H}}{\partial{p_k}}-\frac{
\partial{F}}{\partial{p_k}}\frac{\partial{H}}{\partial{q_k}}\Big)\Label{EQ02}
\end{eqnarray}
\eqref{EQ02} can then be written as
\begin{equation}
\boxed{ \dd[F(q,p,t)]{t} = \frac{\partial{F}}{\partial{t}}+\{F,H\}_{PB}}
\end{equation}

Constant of motion

If a dynamical variable \(F\) does not depend on time explicitly, we have \begin{equation}
\frac{dF}{dt}= \{F,H\}_{PB}.
\end{equation}
Therefore, if the dynamical variable has zero Poisson Bracket with Hamiltonian $H$,
\begin{equation}
\frac{dF}{dt}=0.
\end{equation}
Thus we have an important result that $F(q,p)$ is a constant of motion if its Poisson bracket of \(F(q,p)\) with the Hamiltonian vanishes.