I have a volume element in phase space:

$$ d\omega = \prod _{i=1}^{N}(dq_{i},dp_{i})$$

Now I should show the invariance of this product under canonical transformations. I think first I would have to write down a general equation for the canonical transformation then compute the total differential and plug this into the equation given. My problem is now I can't figure out how the equation for a general canonical transformation would look like. Does anybody have a hint how to start? The proof should be done without the concept of Symplectomorphisms like it was already shown in this question Which transformations are canonical?


2 Answers 2


For a canonical transformation that is not a point transformation:

In the symplectic formulation of Hamiltonian mechanics we have

$$ \dot{\boldsymbol{\eta}} = J \frac{\partial H}{\partial \boldsymbol{\eta}}, $$

where $H$ is the Hamiltonian of the system,$\boldsymbol{\eta} = (q_1 \dots q_n p_1 \dots p_n)^T$ and the $2n \times 2n$ matrix J is defined by

$$ J = \begin{bmatrix} 0 & I \\ -I & 0 \end{bmatrix}, $$

with $I$ denoting the $n\times n$ identity matrix, and $0$ the $n\times n$ matrix with only $0$ as entries. Consider a coordinate transformation $\boldsymbol{\xi} = \boldsymbol{\xi}(\boldsymbol{\eta})$. Then

$$ \dot{\xi}_i = \frac{\partial \xi_i}{\partial \eta_j} \dot{\eta}_j = \frac{\partial \xi_i}{\partial \eta_j} J_{jk} \frac{\partial H}{\partial \eta_k} = \frac{\partial \xi_i}{\partial \eta_j} J_{jk} \frac{\partial \xi_l}{\partial \eta_k} \frac{\partial H}{\partial \xi_l}. $$

That is,

$$ \dot{\boldsymbol{\xi}} = D J D^T \frac{\partial H}{\partial \boldsymbol{\xi}}, $$

where $D$ is the Jacobian matrix of the transformation. For the coordinate transformation to be a canonical transformation we thus require $DJD^T = J$. It is straight forward to show that this is equivalent to

$$ \lbrace \xi_i, \xi_j \rbrace_{\boldsymbol{\eta}} = J_{ij}, $$ which the preservation of Poisson brackets.

Now consider the determinant of $J$:

$$ \det(J) = \det(DJD^T) = \det(J) (\det{(D)})^2 \Rightarrow \vert \det(D) \vert = 1. $$

This is condition is the same as conservation of the volume of a phase space volume element:

$$ \text{d}\omega ^\prime = \prod_i \text{d}q_i \text{d}p_i = \prod_i | \det(D) | \text{d}Q_i \text{d}P_i = \text{d}\omega $$


Consider a phase space volume $d\omega = \prod dq^idp_i$. Time evolution can be viewed as the unfolding of canonical transformations, hence, we shall evolve this system $t\rightarrow t+dt$. We wish to show invariance of the measure under this canonical transformation $(q,p) \rightarrow (Q,P)$ where: $$ Q^i = q^i + \dot q^id t, \qquad P^i = p_i + \dot p_idt $$ Compute the new volume $d\omega' = \prod dQ^idP_i$:

\begin{align} d\omega ' &= \prod d( q^i + \dot q^id t, p_i + \dot p_id t)\\ & = \prod ( dq^i + \frac{\partial \dot q^i}{\partial q^i}dt, dp_i + \frac{\partial \dot p_i}{\partial p_i}d t) \end{align} Let us now expand this and retain terms linear in $dt$ we obtain $$ d\omega' \approx \sum_i \bigg( \frac{\partial \dot q^i}{\partial q^i} + \frac{\partial \dot p_i}{\partial p_i}\bigg) = \bigg\{\frac{\partial }{\partial q^i}\frac{\partial H}{\partial p_i} - \frac{\partial }{\partial p_i}\frac{\partial H}{\partial q^i}\bigg\} = 0 $$ Therefore, we have shown invariance of the measure under a canonical transformation.

  • $\begingroup$ yeah but what of transformations which are not point transformation and cannot be generated by an evolution? $\endgroup$ May 16, 2020 at 16:27
  • $\begingroup$ I have added an answer showing this for a canonical transformation which need not be a point transformation. $\endgroup$
    – sondre
    Jan 17, 2021 at 20:38

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