Poisson Brackets

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In the derivation of Hamiltons Equations of Motion we discussed the change in a function of two variables \begin{align} \delta F(q,p)= \frac{\partial F}{\partial q} \delta q + \frac{\partial F}{\partial p} \delta p \end{align} The time derivative of such a function with more variables is \begin{align} \frac{dF(q,p)}{dt}= \sum_i\frac{\partial F}{\partial q_i} \dot{q}_i + \frac{\partial F}{\partial p_i} \dot{p}_i \end{align} With Hamiltons Equations of Motion \begin{align} \frac{\partial H}{\partial p_i}=\dot{q}_i \end{align} \begin{align} \frac{\partial H}{\partial q_i}=-\dot{p}_i \end{align} this expression can be written as \begin{align} \dot{F}(q,p)= \sum_i \frac{\partial F}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial F}{\partial p_i} \frac{\partial H}{\partial q_i} \end{align}

We introduce the Poisson Brackets $\{\}$ as a short notation of this equation and write \begin{align} \dot{F}(q,p)= \sum_i \frac{\partial F}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial F}{\partial p_i} \frac{\partial H}{\partial q_i}=\{ F,H \} \end{align} The Poisson Bracket gives the change in $F$ under the symmetry operation $H$.

We will see that use of Poisson Brackets allow a more condensed notation of classical mechanics. Remember the time derivative of a Hamiltonian for a dissipative system. With Poisson Bracket notation we now simply write \begin{align} \frac{dH}{dt}=\{H, H\}=-\frac{\partial L}{\partial t} \end{align} We can use Poisson Brackets to calculate $\dot{p}$ and $\dot{q}$, to this end we introduce further important relations

The Fundamental Poisson Brackets are \begin{align} \{ q_i, q_j \}=0 \end{align} \begin{align} \{ p_i, p_j \}=0 \end{align} \begin{align} \{ q_i, p_j \}= \sum_k \underbrace{\frac{\partial q_i}{\partial q_k}}_{\delta_{ik}} \underbrace{\frac{\partial p_j}{\partial p_k}}_{\delta_{jk}} - \underbrace{\frac{\partial q_i}{\partial p_k} \frac{\partial p_j}{\partial q_k}}_{0}= \delta_{ij} \end{align}

As can be verified by plugging into the relation for the Poisson Bracket and rearranging, the following rules hold \begin{align} \{ A,B \}&=-\{ B,A \} \tag{1}\\ \{ \alpha A+ \beta B,C \}&=\alpha \{ A,C \}+ \beta \{ B,C \} \tag{2}\\ \{ A \cdot B,C \}&=\{ A,C \}\cdot B + A \cdot \{ B,C \} \tag{3}\\ \frac{\partial}{\partial t}\{ A ,B\}&=\{ \frac{\partial A}{\partial t},B \} + \{ A, \frac{\partial B}{\partial t} \} \tag{4}\\ \{ \{ A ,B\},C\} +\{ \{ B ,C\} ,A\}+\{ \{ C ,A\} ,B\}&=0 \tag{5}\\ \end{align}


Example


The harmonic oscillator has the Hamiltonian \begin{align} H=\frac{p^2}{2m}+\frac{k}{2}q^2 \end{align} Using Poisson Brackets we can derive \begin{align} \dot{q}=\{q,H \}=\{q,\frac{p^2}{2m}+\frac{k}{2}q^2\}\overbrace{=}^{rule 2}\frac{1}{2m}\{q,p \cdot p\}-\frac{k}{2}\{q,q \cdot q\}\overbrace{=}^{rule 3} \frac{1}{m} \underbrace{\{q,p\}}_{1} \cdot p+k\underbrace{\{q,q\}}_{0} \cdot q =\frac{p}{m} \end{align} in the same way we caluclate \begin{align} \dot{p}=\{p,H \}=\{p,\frac{p^2}{2m}+\frac{k}{2}q^2\}=-\frac{k}{2}\{p,q \cdot q\}=-kq \end{align}