Poisson bracket

In canonical coordinates,

Functions $F_i$ for which $\{F_i,F_j\}=0$ for all pairs. Their Hamiltonian vector fields commute, and each is constant along the flows of the others. When they are integrals of motion of a Hamiltonian, they are first integrals in involution.

Second-class constraints ${\chi }_a$ with invertible bracket matrix can be eliminated with a modified bracket. For an ordinary even constraint set it is $\{F,G{\}}_D=\{F,G\}-\{F,{\chi }_a\}(C^{-1}{)}^{ab}\{{\chi }_b,G\}$, $C_{ab}=\{{\chi }_a,{\chi }_b\}$. Graded constraints require the matching graded sign convention.

A Dirac bracket on a graded phase space eliminates second-class constraints while respecting the graded Poisson bracket. The fermionic kinetic term $i{\psi }^m{\dot{\psi }}_m/2$ gives $\{{\psi }^m,{\psi }^n{\}}_D=-i{\eta }^{mn}$. Quantization then gives $\{{\hat{\psi }}^m,{\hat{\psi }}^n{\}}_{+}={\eta }^{mn}$.

An even Poisson bracket on a space with commuting and anticommuting variables obeys $\{F,G\}=-(-1{)}^{|F||G|}\{G,F\}$ and the graded Leibniz rule. In particular the bracket of two odd functions is symmetric. Quantization turns the odd-odd bracket into an anticommutator.

The Poisson bracket satisfies

If $F$ and $G$ have no explicit time dependence and satisfy $\{F,H\}=\{G,H\}=0$, the Jacobi identity gives $\{\{F,G\},H\}=0$. Their Poisson bracket is therefore also conserved.

The derivative $D_t={\partial }_t+\{\cdot ,H\}$ is a derivation of the Poisson bracket: $D_t\{f,g\}=\{D_{t}f,g\}+\{f,D_{t}g\}$. The time derivative part obeys the product rule, and the Hamiltonian part obeys it by the Jacobi identity. Therefore explicitly time-dependent conserved quantities also have a conserved Poisson bracket.

Along a trajectory generated by the Hamiltonian $H$,Thus a time-independent observable is conserved exactly when its Poisson bracket with $H$ vanishes.

For a time-independent Hamiltonian $H$, the explicit time partial derivative of a smooth time-dependent conserved quantity is again a conserved quantity. Here conservation means the identity $D_{t}f=f_t+\{f,H\}=0$ on phase space, with the standard Poisson bracket. Differentiating this identity at fixed canonical coordinates givesBecause $H_t=0$, this is exactly $D_t(f_t)=0$, proving the assertion. Iteration gives the same result for all higher explicit time partial derivatives for which the differentiations are defined. By Hamiltonian energy balance, requiring $H$ itself to be a conserved quantity on all trajectories is equivalent to $H_t=0$.

Along Hamilton's equations, spatial derivatives of the Hamiltonian function cancel in its total time derivative. Thus $dH/dt={\partial }_{t}H$. An autonomous Hamiltonian function is conserved; explicit time dependence generally prevents conservation even though phase space is still preserved.

For an autonomous Lagrangian $L(x,\dot{x})$, define its action $A=\int Ldt$ and Hamiltonian $H=\dot{x}{L}_{\dot{x}}-L$. The chain rule givesThus the identity holds along forced trajectories as well as solutions of the homogeneous Euler-Lagrange equation. Explicit time dependence adds $-{\partial }_{t}L$; stochastic work requires a consistent regularization or the Stratonovich chain rule.

The function $D=q\cdot p$ generates the canonical scaling $(q,p)\mapsto (e^{t}q,e^{-t}p)$. An observable $F$ is invariant under this scaling exactly when

For $z_j=q_j+i{p}_j$ and ${\bar{z}}_j=q_j-i{p}_j$,This follows from ${\partial }_{q_j}={\partial }_{z_j}+{\partial }_{{\bar{z}}_j}$ and ${\partial }_{p_j}=i{\partial }_{z_j}-i{\partial }_{{\bar{z}}_j}$.

For an anti-Hermitian matrix $A$, the real quadratic HamiltoniansatisfiesThus $A\mapsto -I_A$ realizes the matrix commutator as a Poisson bracket; it is the quadratic moment-map construction for the unitary action.