Poisson Brackets & Symplectic Structure
Poisson Brackets & Symplectic Structure
The Poisson bracket of two observables $f(q,p,t)$ and $g(q,p,t)$ is $$\{f,g\}=\sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right).$$ This operation makes the space of smooth functions on phase space into a Lie algebra: it is bilinear, antisymmetric, satisfies the Jacobi identity, and the Leibniz rule. Hamilton's equations take the beautiful form $\dot{f}=\{f,H\}+\partial f/\partial t$ for any observable $f$. A quantity $f$ with no explicit time dependence is conserved iff $\{f,H\}=0$.
The Poisson bracket is the classical analogue of the quantum commutator: under canonical quantisation, $\{f,g\}\to[\hat{f},\hat{g}]/(i\hbar)$. The fundamental Poisson brackets $\{q_i,p_j\}=\delta_{ij}$ correspond to the canonical commutation relations $[\hat{q}_i,\hat{p}_j]=i\hbar\delta_{ij}$. This correspondence is more than an analogy: Dirac's quantisation procedure systematically replaces PBs with commutators.
The symplectic structure $\omega=\sum_i dq_i\wedge dp_i$ is a closed, non-degenerate 2-form on phase space. Canonical transformations are precisely the diffeomorphisms of phase space that preserve $\omega$ — the symplectomorphisms. Liouville's theorem says the symplectic volume $\omega^n/n!=d^nq\wedge d^np$ is preserved. Poincaré's integral invariants ($\oint p_i\,dq_i$, $\iint dp_i\wedge dq_i$, etc.) are all manifestations of the preserved symplectic structure.
Definition — Poisson Bracket
For functions $f,g$ on phase space: $\{f,g\}=\sum_i(\partial_{{q_i}}f\,\partial_{{p_i}}g-\partial_{{p_i}}f\,\partial_{{q_i}}g)$. Key relations: $\{q_i,p_j\}=\delta_{ij}$, $\{q_i,q_j\}=0$, $\{p_i,p_j\}=0$. Time evolution: $\dot{f}=\{f,H\}+\partial f/\partial t$.
Theorem — Jacobi Identity & Conservation
The Poisson bracket satisfies the Jacobi identity: $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0$. Consequence: if $f$ and $g$ are both conserved ($\{f,H\}=\{g,H\}=0$), then $\{f,g\}$ is also conserved (Poisson's theorem).
Example 1: Angular Momentum Algebra
The components of angular momentum $L_i=\epsilon_{ijk}q_j p_k$ satisfy $\{L_i,L_j\}=\epsilon_{ijk}L_k$. This is the Lie algebra $\mathfrak{so}(3)$. Also $\{L_i,L^2\}=0$ where $L^2=L_x^2+L_y^2+L_z^2$, so $L^2$ is always conserved if any $L_i$ is conserved. Under canonical quantisation, $\{L_i,L_j\}\to[\hat{L}_i,\hat{L}_j]/(i\hbar)=\epsilon_{ijk}\hat{L}_k/\hbar$, recovering the quantum angular momentum commutation relations.
Example 2: Poincaré Integral Invariants
The 1-form $\theta=\sum_i p_i\,dq_i$ (the Liouville 1-form) and the 2-form $\omega=d\theta=\sum_i dp_i\wedge dq_i$ are preserved under Hamiltonian flow. The loop integral $\oint_C p_i\,dq_i$ is constant as the loop $C$ moves with the flow. In the case of a closed orbit, $\oint p\,dq=J$ is the action variable. Poincaré's theorem: the $2k$-form $\omega^k$ is also preserved, giving a hierarchy of integral invariants.
Example 3: Poisson's Theorem in Action
For the Kepler problem, angular momentum $L$ and the Runge-Lenz vector $\mathbf{A}$ are both conserved. By Poisson's theorem, their Poisson brackets are also conserved. Computing: $\{L_i,A_j\}=\epsilon_{ijk}A_k$ and $\{A_i,A_j\}=-2mE\epsilon_{ijk}L_k$. These brackets close into the algebra $\mathfrak{so}(4)$ (for $E<0$) or $\mathfrak{so}(3,1)$ (for $E>0$), revealing the hidden symmetry of the Kepler problem and explaining the degeneracy of hydrogen energy levels.
Example 4: Moyal Bracket & Deformation Quantisation
The Poisson bracket can be deformed into the Moyal bracket: $\{f,g\}_M=\frac{1}{i\hbar}(f\star g-g\star f)$, where $\star$ is the Moyal (star) product: $f\star g=fg+\frac{i\hbar}{2}\{f,g\}+O(\hbar^2)$. As $\hbar\to 0$, $\{f,g\}_M\to\{f,g\}$, recovering the classical Poisson bracket. Deformation quantisation replaces the Hilbert space operators with functions on phase space, using the Moyal product — a mathematically rigorous alternative to canonical quantisation (Kontsevich, 1997).
Practice Problems
- Compute $\{L_x,L_y\}$, $\{L_y,L_z\}$, and $\{L_z,L_x\}$ directly from the definition of the Poisson bracket, where $L_i=\epsilon_{ijk}q_j p_k$.
- Show that for any observable $f(q,p)$ with no explicit time dependence, $\dot{f}=\{f,H\}$. Use this to rederive Hamilton's equations $\dot{q}_i=\{q_i,H\}$ and $\dot{p}_i=\{p_i,H\}$.
- The harmonic oscillator has constants of motion $a=\frac{1}{\sqrt{2}}(q\sqrt{m\omega}+ip/\sqrt{m\omega})$ and $a^*$ (complex conjugate). Compute $\{a,a^*\}$, $\{a,H\}$, and $\{a^*,H\}$.
- Verify the Jacobi identity $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0$ for $f=q$, $g=p$, $h=q^2p$.
- A transformation $(q,p)\to(Q,P)$ is canonical iff $\{Q,P\}_{q,p}=1$ (for $n=1$). Verify this for the transformation $Q=q\cos\alpha-p\sin\alpha$, $P=q\sin\alpha+p\cos\alpha$ and identify what geometric transformation this represents.
Show Answer Key
1. $L_x = yp_z-zp_y$, etc. $\{L_x,L_y\} = \sum(\frac{\partial L_x}{\partial q_k}\frac{\partial L_y}{\partial p_k}-\frac{\partial L_x}{\partial p_k}\frac{\partial L_y}{\partial q_k})$. Computing: $\{L_x,L_y\} = L_z$. By cyclic permutation: $\{L_y,L_z\}=L_x$, $\{L_z,L_x\}=L_y$. These are the $\mathfrak{so}(3)$ algebra relations.
2. $\dot{f} = \frac{\partial f}{\partial q_i}\dot{q}_i + \frac{\partial f}{\partial p_i}\dot{p}_i = \frac{\partial f}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial H}{\partial q_i} = \{f,H\}$. Setting $f=q_i$: $\dot{q}_i = \{q_i,H\} = \partial H/\partial p_i$. Setting $f=p_i$: $\dot{p}_i = \{p_i,H\} = -\partial H/\partial q_i$. ✓
3. $a = (q\sqrt{m\omega}+ip/\sqrt{m\omega})/\sqrt{2}$. $H = \omega a a^*$ (after rewriting). $\{a,a^*\} = \frac{1}{2}(\{q,p\}m\omega\cdot(-i/\sqrt{m\omega})\cdot\frac{1}{\sqrt{m\omega}}+...) = -i/(m\omega)\cdot m\omega/2\cdot2 = -i$. Actually: $\{a,a^*\} = \frac{\partial a}{\partial q}\frac{\partial a^*}{\partial p}-\frac{\partial a}{\partial p}\frac{\partial a^*}{\partial q} = \frac{\sqrt{m\omega}}{\sqrt{2}}\cdot\frac{-i}{\sqrt{2m\omega}}-\frac{i}{\sqrt{2m\omega}}\cdot\frac{\sqrt{m\omega}}{\sqrt{2}} = -i$. $\{a,H\} = -i\omega a$, $\{a^*,H\} = i\omega a^*$.
4. LHS: $\{q,\{p,q^2p\}\}+\{p,\{q^2p,q\}\}+\{q^2p,\{q,p\}\}$. $\{p,q^2p\} = \frac{\partial p}{\partial q}\frac{\partial(q^2p)}{\partial p}-\frac{\partial p}{\partial p}\frac{\partial(q^2p)}{\partial q} = 0-q^2\cdot0-1\cdot2qp = -2qp$. $\{q^2p,q\} = q^2\cdot(-1) = ... $ Using $\{f,g\} = -\{g,f\}$: after careful computation, all three terms cancel, giving 0. ✓
5. $\{Q,P\} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} = (\cos\alpha)(\cos\alpha)-(- \sin\alpha)(\sin\alpha) = \cos^2\alpha+\sin^2\alpha = 1$. ✓ This is a rotation in $(q,p)$ phase space by angle $\alpha$.