The Properties of Poisson Brackets
The task of proving the properties of Poisson Brackets (PB) is given as an optional homework in the textbook.
Antisymmetry
This follows from the definition of the PB by swapping \(A\) and \(B\).
Linearity
Linearity has two parts:
- homogeneity: \[\{\lambda A, B\} = \lambda \{A, B\}\]
- additivity: \[\{A+B, C\} = \{A, C\} + \{B, C\}\]
We will show each part separately.
Homogeneity
\[\{\lambda A, B\}\] \[=\sum_i \frac{\partial (\lambda A)}{\partial q_i} \frac{\partial B}{\partial p_i} - \frac{\partial (\lambda A)}{\partial p_i} \frac{\partial B}{\partial q_i}\] \[=\sum_i \lambda \frac{\partial A}{\partial q_i} \frac{\partial B}{\partial p_i} - \lambda \frac{\partial A}{\partial p_i} \frac{\partial B}{\partial q_i}\] \[= \sum_i \lambda \left( \frac{\partial A}{\partial q_i} \frac{\partial B}{\partial p_i} - \frac{\partial A}{\partial p_i} \frac{\partial B}{\partial q_i} \right)\] \[= \lambda \{A, B\}\]
Additivity
\[\{A+B, C\} = \] \[=\sum_i \frac{\partial (A+B)}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial (A+B)}{\partial p_i} \frac{\partial C}{\partial q_i}\] \[=\sum_i \left( \frac{\partial A}{\partial q_i} +\frac{\partial B}{\partial q_i} \right) \frac{\partial C}{\partial p_i} - \left( \frac{\partial A}{\partial p_i} +\frac{\partial B}{\partial p_i} \right) \frac{\partial C}{\partial q_i}\] \[ \sum_i \frac{\partial A}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial A}{\partial p_i} \frac{\partial C}{\partial q_i} + \frac{\partial B}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial B}{\partial p_i} \frac{\partial C}{\partial q_i}\] \[=\{A, C\} + \{B, C\}\]
Product rule
\[\{AB, C\}=\sum_i \frac{\partial (AB)}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial (AB)}{\partial p_i} \frac{\partial C}{\partial q_i}\]
The product rule for the partial derivative is as follows: \[\frac{\partial}{\partial q_i} (AB) = \frac{\partial A}{\partial q_i} B + A \frac{\partial B}{\partial q_i}\] Hence:
\[\{AB, C\}=\sum_i \left( \frac{\partial A}{\partial q_i} B + A \frac{\partial B}{\partial q_i} \right)
\frac{\partial C}{\partial p_i} - \left( \frac{\partial A}{\partial p_i} B + A \frac{\partial B}{\partial p_i} \right)
\frac{\partial C}{\partial q_i}\] \[=\sum_i \left( \frac{\partial A}{\partial q_i} \frac{\partial C}{\partial p_i}
-\frac{\partial A}{\partial p_i} \frac{\partial C}{\partial q_i}\right) B + A \left(
\frac{\partial B}{\partial q_i} \frac{\partial C}{\partial p_i}
-\frac{\partial B}{\partial p_i} \frac{\partial C}{\partial q_i} \right)\] \[=\{A, C\} B + A \{B, C\}\]
Fundamental Poisson Brackets
\[\{q_i, q_j\} = \sum_k \left( \frac{\partial q_i}{\partial q_k} \frac{\partial q_j}{\partial p_k} - \frac{\partial q_i}{\partial p_k} \frac{\partial q_j}{\partial q_k} \right)\]
The coordinates of phase space \(q_i\) and \(p_i\) are independent of each other: \[ \frac{\partial q_i}{\partial p_k} = 0 \, \forall \,i, k\]
Hence: \[\{q_i, q_j\} = 0\]
\[\{p_i, p_j\} = 0\] \[\{q_i, p_j\} = \sum_k \left(\frac{\partial q_i}{\partial q_k} \frac{\partial p_j}{\partial p_k} - \frac{\partial q_i}{\partial p_k} \frac{\partial p_j}{\partial q_k} \right)\]\[=\sum_k \delta_{ik} \delta_{jk} - 0\] \[=\delta_{ij}\]