[NOTES/CM-04005] Poisson Bracket Properties

Important properties of Poisson brackets

1. Linearity if $\alpha_1,\alpha _2$ are numbers and $F_1,F_2,G$ are dynamical variables.Then \begin{equation} \{\alpha_1F_1+\alpha_2F_2,G\}_{PB}=\alpha_1\{F_1,G\}_{PB} \end{equation} \begin{equation} \text{also} ~~ \{F,\beta_1G_1+\beta_2G_2\}_{PB}=\beta_1\{F,G_1\}_{PB}+\beta_2 \{F,G_2\}_{PB} \end{equation}
2. Anti symmetry \begin{equation} \{F,G\}_{PB}=-\{G,F\}_{PB} \end{equation}
3. Product Rule \begin{equation} [F,AB]= A[F,B]+[F,A]B \end{equation} \begin{equation} \{AB,C\}_{PB}= A\{B,C\}_{PB}+\{A,C\}_{PB}B \end{equation}
4. Jacobi Identity \begin{equation} \big\{A,\{B,C\}_{PB}\big\}_{PB}+\big\{B,\{C,A\}_{PB}\big\}_{PB}+\big\{C,\{A,B\}_ {PB}\big\}_{PB}=0 \end{equation} 
5. Chain rule Suppose the dynamical variables are not given in terms of $q_k,p_k$ but in terms of some other set of variables $\{\phi_\alpha\}, \alpha=1,2,..$ which are in term functions of $(q_k,p_k)$then \begin{equation} \{F(\phi),G(\phi)\}_{PB}=\sum_{\alpha,\beta}\pp[F]{\phi_\alpha}\{\phi_\alpha,\phi_\beta\}_{PB}\frac{\partial{G} }{\partial{\phi_\beta}} \end{equation} 

Fundamental brackets 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} These properties are very important and will be called fundamental brackets for the phase space variables. The properties \eqref{EQ15}-\eqref{EQ16} are obvious from the definition.
Proofs of the above properties is straight forward and will be skipped here. 

Poisson bracket theorem

If \(F\) and \(G\) are two constants of motion, their Poisson bracket is also a constant of motion.

Case-I :  We first prove a special case when \(F\) and \(G\) do not depend on time explicitly. The proof for this case follows from Jacobi identity for Poisson brackets. The Jacobi identity for \(F,G,H\) is
\begin{equation}\Label{EQ11}
[F,[G, H]] + [G,[H,F]] + [H, [F,G]] =0.
\end{equation}
The first two terms vanish because \(F\) and \(G\) are constants of motion. Thus we get \begin{equation}[H, [F,G]] =0.\end{equation}Thus the Poisson bracket of \([F,G]\) with the Hamiltonian is zero. Hence \([F,G]\) is a constant of motion.

Case-II : Now consider case when \(F\) and \(G\) are constants of motion depending on time explicitly. Then
\begin{eqnarray}\Label{EQ04} \dd[F]{t} = 0\Longrightarrow \pp[F]{t} + [F,H]=0.\end{eqnarray}
with a similar relation for \(G\). We start with
\begin{eqnarray} \dd{t}[F,G] &=& \pp{t}[F,G] + [[F,G], H]\\ &=& \Big[\pp[F]{t}, G\Big] + \Big[F, \pp[G]{t}\Big] + [[F,G], H] \qquad \text{(Verify!)} \\
&=& -[[F,H],G] - [F, [G,H]] + [[F,G],H]\\ & & \qquad \text{Now rearrange and use Jacobi identity \eqref{EQ11}}\\
&=&0. \end{eqnarray}
In fact a more general result given below holds.

Theorem: Let $q_k(t),p_k(t)$ be the time development of a system on phase space. time development is generated by some Hamiltonian $H(q,p,t)$ if and only if every pair of variables $A(q,p,t)$,$B(q,p,t)$ satisfies the relation \begin{equation} \frac{d}{dt}\big[A,B\big]=\Big[\frac{d}{dt}A,B\Big]+\Big[A,\frac{ dB}{dt}\Big] \end{equation}