25 Jul 2025
Canonical Transformations, Poisson Brackets, and Lagrange Brackets
Legendre maps, generating functions, canonical conditions, Poisson and Lagrange brackets, and invariance.
Legendre transformations
For a differentiable function $f(x)$, define
\[p=\frac{df}{dx}, \qquad g(p)=px-f(x).\]When $p=f’(x)$ can be inverted for $x(p)$,
\[dg=x\,dp.\]Thus the Legendre transformation replaces the independent variable $x$ by its conjugate slope $p$. In mechanics it replaces velocities by momenta:
\[p_i=\frac{\partial L}{\partial\dot q_i}, \qquad \boxed{H(q,p,t)=\sum_i p_i\dot q_i-L(q,\dot q,t)}.\]The map is locally invertible when
\[\det\left(\frac{\partial^2L} {\partial\dot q_i\partial\dot q_j}\right)\neq0.\]Canonical transformations and generating functions
A transformation $(q,p)\mapsto(Q,P)$ is canonical if the new variables satisfy Hamilton’s equations for a transformed Hamiltonian $K(Q,P,t)$. The old and new phase-space one-forms may differ by an exact differential:
\[\sum_i p_i\,dq_i-H\,dt =\sum_i P_i\,dQ_i-K\,dt+dF_1(q,Q,t).\]Comparing the independent differentials gives the type-1 relations
\[\boxed{p_i=\frac{\partial F_1}{\partial q_i}}, \qquad \boxed{P_i=-\frac{\partial F_1}{\partial Q_i}}, \qquad \boxed{K=H+\frac{\partial F_1}{\partial t}}.\]Partial Legendre transformations of $F_1$ produce the other standard types. For
\[F_2(q,P,t)=F_1+\sum_iQ_iP_i,\]the relations are
\[p_i=\frac{\partial F_2}{\partial q_i}, \qquad Q_i=\frac{\partial F_2}{\partial P_i}, \qquad K=H+\frac{\partial F_2}{\partial t}.\]For
\[F_3(p,Q,t)=F_1-\sum_iq_ip_i,\]they are
\[q_i=-\frac{\partial F_3}{\partial p_i}, \qquad P_i=-\frac{\partial F_3}{\partial Q_i}, \qquad K=H+\frac{\partial F_3}{\partial t}.\]Finally,
\[F_4(p,P,t)=F_1+\sum_iQ_iP_i-\sum_iq_ip_i\]gives
\[q_i=-\frac{\partial F_4}{\partial p_i}, \qquad Q_i=\frac{\partial F_4}{\partial P_i}, \qquad K=H+\frac{\partial F_4}{\partial t}.\]For example, with constant $a\neq0$,
\[F_2(q,P)=aqP\]generates $Q=aq$ and $P=p/a$.
Condition for a canonical transformation
The Poisson bracket in the old variables is
\[\{A,B\}_{q,p} =\sum_i\left( \frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i} -\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i} \right).\]A non-singular transformation is canonical if and only if its new variables obey the fundamental brackets
\[\boxed{\{Q_i,Q_j\}_{q,p}=0}, \qquad \boxed{\{P_i,P_j\}_{q,p}=0}, \qquad \boxed{\{Q_i,P_j\}_{q,p}=\delta_{ij}}.\]Equivalently, put
\[z=(q_1,\ldots,q_s,p_1,\ldots,p_s)^T, \qquad Z=(Q_1,\ldots,Q_s,P_1,\ldots,P_s)^T,\]and define
\[M=\frac{\partial Z}{\partial z}, \qquad J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}.\]The canonical condition is
\[\boxed{MJM^T=J}.\]Poisson brackets and time evolution
The bracket is bilinear and antisymmetric, obeys the product rule, and satisfies the Jacobi identity,
\[\{A,\{B,C\}\} +\{B,\{C,A\}\} +\{C,\{A,B\}\}=0.\]Using Hamilton’s equations,
\[\frac{dA}{dt} =\sum_i\left( \frac{\partial A}{\partial q_i}\dot q_i +\frac{\partial A}{\partial p_i}\dot p_i \right) +\frac{\partial A}{\partial t} =\boxed{\{A,H\}+\frac{\partial A}{\partial t}}.\]Therefore a quantity with no explicit time dependence is conserved precisely when its Poisson bracket with $H$ vanishes.
Lagrange brackets
Let $u_1,\ldots,u_{2s}$ be any non-singular set of phase-space coordinates, with $q_i=q_i(u)$ and $p_i=p_i(u)$. The Lagrange bracket is
\[\boxed{ [u_\alpha,u_\beta] =\sum_i\left( \frac{\partial q_i}{\partial u_\alpha} \frac{\partial p_i}{\partial u_\beta} -\frac{\partial q_i}{\partial u_\beta} \frac{\partial p_i}{\partial u_\alpha} \right)}.\]If $C_{\alpha\beta}={u_\alpha,u_\beta}$ and $L_{\alpha\beta}=[u_\alpha,u_\beta]$, then
\[CL=-I,\]or, equivalently,
\[\sum_\gamma \{u_\alpha,u_\gamma\} [u_\beta,u_\gamma] =\delta_{\alpha\beta}.\]For canonical variables this gives
\[\boxed{[Q_i,Q_j]=0}, \qquad \boxed{[P_i,P_j]=0}, \qquad \boxed{[Q_i,P_j]=\delta_{ij}}.\]In particular, the mixed Lagrange bracket has a plus sign with the definition above.
Invariance of the Poisson bracket
For functions $A(Z)$ and $B(Z)$, the chain rule gives
\[\{A,B\}_{q,p} =\sum_{\alpha,\beta} \frac{\partial A}{\partial Z_\alpha} \frac{\partial B}{\partial Z_\beta} \{Z_\alpha,Z_\beta\}_{q,p}.\]If $(Q,P)$ is canonical, its fundamental brackets reduce the right-hand side to
\[\boxed{\{A,B\}_{q,p}=\{A,B\}_{Q,P}}.\]Thus the Poisson-bracket structure, and consequently Hamilton’s equations, is invariant under a canonical transformation.
Discussion