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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2018

Paper 1

From $z_j=q_j+ip_j$ and $\bar z_j=q_j-ip_j$, the chain rule gives
$$
\frac{\partial}{\partial q_j}
=\frac{\partial}{\partial z_j}
+\frac{\partial}{\partial\bar z_j},
\qquad
\frac{\partial}{\partial p_j}
=i\frac{\partial}{\partial z_j}
-i\frac{\partial}{\partial\bar z_j}.
$$
Substituting these identities into the canonical <Poisson bracket> and cancelling the $f_{z_j}g_{z_j}$ and $f_{\bar z_j}g_{\bar z_j}$ terms gives
$$
\boxed{\
\{f,g\}
=-2i\frac{\partial f}{\partial z_j}
\frac{\partial g}{\partial\bar z_j}
+2i\frac{\partial g}{\partial z_j}
\frac{\partial f}{\partial\bar z_j}.\
}
$$
This is the <Poisson bracket in complex canonical coordinates>.

Solution

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

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

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

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