Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 32A / c

Table of contents
- Solution c

Solution

From

z_{j} = q_{j} + i p_{j}

and

\overset{z}{ˉ}_{j} = q_{j} - i p_{j}

, the chain rule gives

\frac{\partial}{\partial q _{j}} = \frac{\partial}{\partial z _{j}} + \frac{\partial}{\partial z ˉ _{j}}, \frac{\partial}{\partial p _{j}} = i \frac{\partial}{\partial z _{j}} - i \frac{\partial}{\partial z ˉ _{j}} .

Substituting these identities into the canonical Poisson bracket and cancelling the

f_{z_{j}} g_{z_{j}}

and

f_{\overset{z}{ˉ}_{j}} g_{\overset{z}{ˉ}_{j}}

terms gives

{f, g} = - 2 i \frac{\partial f}{\partial z _{j}} \frac{\partial g}{\partial z ˉ _{j}} + 2 i \frac{\partial g}{\partial z _{j}} \frac{\partial f}{\partial z ˉ _{j}} .

This is the Poisson bracket in complex canonical coordinates.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version