Exercise 3

The definition of Poisson brackets in Volume I is \[ \{F, G\} = \sum_i \left( \frac{ \partial F}{\partial q_i} \frac{\partial G}{\partial p_i} - \frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q_i} \right) \]

where \(F\) and \(G\) are some functions of \(q_i\) and \(p_i\). These functions and their units are not specified, so lets assume just 1 for it.

The units of the generalized coordinates \(q_i\) and the conjugate momentum \(p_i\) are also not specified, so lets assume the basic SI units m (length) for \(q\) and kg m/s (mass times velocity) for \(p\), respectively. Each term in the Poisson bracket has unit \[ \frac{1}{\text{m}} \cdot \frac{\text{s}}{\text{kg m}} = \frac{\text{s$^2$}}{\text{m m kg s}} = \frac{1}{\text{Nm s}} = \frac{1}{\text{Js}} \] since \[ 1\, \text{N} = 1\, \frac{\text{kg m}}{\text{s$^2$}} \] and \[ 1\, \text{J} = 1\, \text{Nm} \] The unit of \(\hbar\) is Js and i has no unit, so if the Poisson bracket is multiplied by \(\mathrm{i}\hbar\), the unit Js cancels and the unit left over is that of \(F\) times \(G\), as we would expect it for the commutator \([F, G]\).