Physics Classical Mechanics 100% Free Open Access
Chapter 6 • Theory & Derivations

Canonical Transformations & Poisson Brackets

Exhaustive treatment of canonical transformations from (q, p) to (Q, P), four fundamental generating functions, symplectic matrix condition, Poisson brackets, fundamental canonical brackets, equation of motion in bracket notation, Poisson's theorem for constants of motion, Poincaré integral invariants, and Liouville's phase volume conservation theorem.

§6.1 Concept of Canonical Transformations & The Symplectic Condition

1. Motivation for Coordinate Transformations in Phase Space

In Lagrangian mechanics, coordinate transformations are restricted to point transformations $Q_i = Q_i(q, t)$. In Hamiltonian mechanics, coordinates and conjugate momenta $(q, p)$ are treated on an equal footing. We consider broader transformations:
$$Q_i = Q_i(q, p, t), \quad P_i = P_i(q, p, t), \quad i = 1, 2, \dots, n$$
A transformation is defined as canonical (or contact) if there exists a new Hamiltonian $K(Q, P, t)$ such that the new variables satisfy Hamilton's canonical equations:
$$\dot{Q}_i = \frac{\partial K}{\partial P_i}, \quad \dot{P}_i = -\frac{\partial K}{\partial Q_i}$$

2. Variational Condition & Generating Function

Both original and transformed trajectories must satisfy the modified Hamilton's principle:
$$\delta \int_{t_1}^{t_2} \left( \sum_{i=1}^n p_i \dot{q}_i - H(q, p, t) \right) dt = 0, \quad \delta \int_{t_1}^{t_2} \left( \sum_{i=1}^n P_i \dot{Q}_i - K(Q, P, t) \right) dt = 0$$
The two integrands can differ at most by the exact total time derivative of an arbitrary function $F$, called the generating function:
$$\sum_{i=1}^n p_i \dot{q}_i - H = \sum_{i=1}^n P_i \dot{Q}_i - K + \frac{dF}{dt}$$
Multiplying by $dt$:
$$\sum_{i=1}^n p_i dq_i - H dt = \sum_{i=1}^n P_i dQ_i - K dt + dF$$

3. The Symplectic Condition

Defining the $2n$-dimensional phase vector $\mathbf{\eta} = (q_1, \dots, q_n, p_1, \dots, p_n)^T$ and the fundamental symplectic matrix $\mathbf{J}$:
$$\mathbf{J} = \begin{pmatrix} \mathbf{0} & \mathbf{I}_n \\ -\mathbf{I}_n & \mathbf{0} \end{pmatrix}, \quad \mathbf{J}^T = -\mathbf{J}, \quad \mathbf{J}^2 = -\mathbf{I}_{2n}$$
Let $\mathbf{M}$ be the Jacobian matrix of the transformation $M_{ij} = \frac{\partial \zeta_i}{\partial \eta_j}$ where $\mathbf{\zeta} = (Q, P)^T$. The transformation is canonical if and only if $\mathbf{M}$ satisfies the symplectic condition:
$$\mathbf{M}^T \mathbf{J} \mathbf{M} = \mathbf{J}$$
Taking the determinant yields $(\det \mathbf{M})^2 = 1 \implies \det \mathbf{M} = 1$, proving that phase space volume is strictly preserved.

§6.2 The Four Fundamental Classes of Generating Functions

1. Classification by Active Canonical Variables

Depending on which pair of old and new variables are chosen as independent variables, Legendre transformations yield four primary classes of generating functions:
  1. Type 1: $F_1(q, Q, t)$
    $$dF_1 = \sum_i p_i dq_i - \sum_i P_i dQ_i + (K - H) dt$$
    Equating partial differentials:
    $$p_i = \frac{\partial F_1}{\partial q_i}, \quad P_i = -\frac{\partial F_1}{\partial Q_i}, \quad K = H + \frac{\partial F_1}{\partial t}$$
  2. Type 2: $F_2(q, P, t) = F_1 + \sum_i P_i Q_i$
    $$dF_2 = \sum_i p_i dq_i + \sum_i Q_i dP_i + (K - H) dt$$
    Transformation relations:
    $$p_i = \frac{\partial F_2}{\partial q_i}, \quad Q_i = \frac{\partial F_2}{\partial P_i}, \quad K = H + \frac{\partial F_2}{\partial t}$$

    Example: The identity transformation is generated by $F_2 = \sum_i q_i P_i$, giving $p_i = P_i$ and $Q_i = q_i$.

  3. Type 3: $F_3(p, Q, t) = F_1 - \sum_i p_i q_i$
    $$dF_3 = -\sum_i q_i dp_i - \sum_i P_i dQ_i + (K - H) dt$$
    Transformation relations:
    $$q_i = -\frac{\partial F_3}{\partial p_i}, \quad P_i = -\frac{\partial F_3}{\partial Q_i}, \quad K = H + \frac{\partial F_3}{\partial t}$$
  4. Type 4: $F_4(p, P, t) = F_1 - \sum_i p_i q_i + \sum_i P_i Q_i$
    $$dF_4 = -\sum_i q_i dp_i + \sum_i Q_i dP_i + (K - H) dt$$
    Transformation relations:
    $$q_i = -\frac{\partial F_4}{\partial p_i}, \quad Q_i = \frac{\partial F_4}{\partial P_i}, \quad K = H + \frac{\partial F_4}{\partial t}$$

§6.3 Poisson Brackets: Definition, Properties & Lie Algebra

1. Definition of the Poisson Bracket

Let $u(q, p, t)$ and $v(q, p, t)$ be two continuously differentiable functions defined on phase space. The Poisson bracket $[u, v]_{q,p}$ with respect to canonical variables $(q, p)$ is:
$$\{u, v\}_{q,p} = \sum_{j=1}^n \left( \frac{\partial u}{\partial q_j} \frac{\partial v}{\partial p_j} - \frac{\partial u}{\partial p_j} \frac{\partial v}{\partial q_j} \right)$$

2. Fundamental Algebraic Identities

Poisson brackets satisfy the defining axioms of a Lie algebra:
  1. Anti-Symmetry: $\{u, v\} = -\{v, u\} \implies \{u, u\} = 0$.
  2. Bilinearity: $\{a u + b v, w\} = a\{u, w\} + b\{v, w\}$ for scalars $a, b$.
  3. Leibniz Product Rule: $\{u v, w\} = u\{v, w\} + \{u, w\}v$.
  4. The Jacobi Identity:
    $$\{u, \{v, w\}\} + \{v, \{w, u\}\} + \{w, \{u, v\}\} = 0$$

3. Fundamental Canonical Poisson Brackets

Evaluating the brackets for the fundamental coordinates and conjugate momenta yields:
$$\{q_j, q_k\} = 0, \quad \{p_j, p_k\} = 0, \quad \{q_j, p_k\} = \delta_{jk}$$

Canonical Invariance: A transformation $(q, p) \to (Q, P)$ is canonical if and only if it preserves the fundamental Poisson brackets: $\{Q_j, Q_k\}_{q,p} = 0$, $\{P_j, P_k\}_{q,p} = 0$, and $\{Q_j, P_k\}_{q,p} = \delta_{jk}$.

Quantum Correspondence: Paul Dirac recognized that the quantum commutator $[\hat{u}, \hat{v}]$ directly maps to the classical Poisson bracket: $[\hat{u}, \hat{v}] = i \hbar \{u, v\}$.

§6.4 Equations of Motion in Poisson Bracket Notation & Poisson's Theorem

1. Time Evolution of an Arbitrary Phase Space Observable

Let $f(q, p, t)$ be any dynamical variable. Its total time derivative along a trajectory is:
$$\frac{df}{dt} = \sum_{j=1}^n \left( \frac{\partial f}{\partial q_j} \dot{q}_j + \frac{\partial f}{\partial p_j} \dot{p}_j \right) + \frac{\partial f}{\partial t}$$
Substituting Hamilton's equations $\dot{q}_j = \frac{\partial H}{\partial p_j}$ and $\dot{p}_j = -\frac{\partial H}{\partial q_j}$:
$$\frac{df}{dt} = \sum_{j=1}^n \left( \frac{\partial f}{\partial q_j} \frac{\partial H}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial H}{\partial q_j} \right) + \frac{\partial f}{\partial t}$$
Recognizing the Poisson bracket with the Hamiltonian:
$$\frac{df}{dt} = \{f, H\} + \frac{\partial f}{\partial t}$$

This is the master equation of classical Hamiltonian dynamics. Setting $f = q_j$ and $f = p_j$ automatically reproduces Hamilton's equations $\dot{q}_j = \{q_j, H\}$ and $\dot{p}_j = \{p_j, H\}$.

2. First Integrals & Constants of Motion

If an observable $f(q, p)$ has no explicit time dependence ($\frac{\partial f}{\partial t} = 0$), then $f$ is a constant of motion if and only if its Poisson bracket with the Hamiltonian vanishes:
$$\frac{df}{dt} = 0 \iff \{f, H\} = 0$$

3. Poisson's Theorem for Generating New Conserved Quantities

Poisson's Theorem: If $f(q, p)$ and $g(q, p)$ are two independent constants of motion (so that $\{f, H\} = 0$ and $\{g, H\} = 0$), then their Poisson bracket $\{f, g\}$ is also a constant of motion.

Proof via the Jacobi identity:
$$\{\{f, g\}, H\} = -\{\{g, H\}, f\} - \{\{H, f\}, g\} = -\{0, f\} - \{0, g\} = 0$$
Hence $\frac{d}{dt}\{f, g\} = 0$. Poisson's theorem enables systematically discovering new symmetries and conservation laws.

§6.5 Poincaré's Invariants & Liouville's Phase Volume Theorem

1. Poincaré's Integral Invariants

Henri Poincaré proved that certain differential forms integrated over closed submanifolds in phase space remain invariant under canonical transformations and Hamiltonian time evolution. The first Poincaré integral invariant of order 1 is:
$$J_1 = \oint_C \sum_{i=1}^n p_i dq_i = \text{constant in time}$$
where $C$ is any closed circuit in phase space carried along by the Hamiltonian flow. Higher-order invariants $J_2, \dots, J_n$ correspond to integrals over $2k$-dimensional manifolds. The invariant of maximum order $2n$ represents the total volume of phase space.

2. Liouville's Phase Volume Conservation Theorem

Consider an ensemble of non-interacting identical systems represented by a cloud of phase points with density distribution $\rho(q, p, t)$ in $2n$-dimensional phase space. By the continuity equation for probability conservation:
$$\frac{\partial \rho}{\partial t} + \sum_{j=1}^n \left( \frac{\partial (\rho \dot{q}_j)}{\partial q_j} + \frac{\partial (\rho \dot{p}_j)}{\partial p_j} \right) = 0$$
Expanding the derivatives:
$$\frac{\partial \rho}{\partial t} + \sum_{j=1}^n \left( \frac{\partial \rho}{\partial q_j} \dot{q}_j + \frac{\partial \rho}{\partial p_j} \dot{p}_j \right) + \rho \sum_{j=1}^n \left( \frac{\partial \dot{q}_j}{\partial q_j} + \frac{\partial \dot{p}_j}{\partial p_j} \right) = 0$$
Using Hamilton's canonical equations:
$$\frac{\partial \dot{q}_j}{\partial q_j} + \frac{\partial \dot{p}_j}{\partial p_j} = \frac{\partial}{\partial q_j}\left( \frac{\partial H}{\partial p_j} \right) + \frac{\partial}{\partial p_j}\left( -\frac{\partial H}{\partial q_j} \right) = \frac{\partial^2 H}{\partial q_j \partial p_j} - \frac{\partial^2 H}{\partial p_j \partial q_j} = 0$$
The divergence of phase velocity vanishes identically: $\nabla \cdot \vec{v}_{\text{phase}} = 0$ (the phase flow is strictly incompressible). Consequently, the convective total time derivative vanishes:
$$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \{\rho, H\} = 0$$

Liouville's Theorem: The phase space volume $\Gamma = \int dq dp$ and the local phase space density $\rho$ surrounding any moving system point remain strictly constant over time. Phase fluid flows like an incompressible liquid, preventing trajectories from ever crossing.

Standard University Exam Solved Problems

Solved Problem Example 6.1: Verifying a Canonical Transformation via Type-1 Generating Function

Given the generating function $F_1(q, Q) = \frac{1}{2} m \omega q^2 \cot Q$, find the transformation equations relating $(q, p)$ to $(Q, P)$, show that the transformation is canonical, and apply it to solve the simple harmonic oscillator $H = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 q^2$.

Step 1: Compute Canonical Momenta from Generating Relations
$$p = \frac{\partial F_1}{\partial q} = m \omega q \cot Q, \quad P = -\frac{\partial F_1}{\partial Q} = \frac{1}{2} m \omega q^2 \csc^2 Q$$

Solving for $q$ from the second equation: $q = \sqrt{\frac{2P}{m\omega}} \sin Q$. Substituting into the first equation: $p = m\omega \sqrt{\frac{2P}{m\omega}} \sin Q \frac{\cos Q}{\sin Q} = \sqrt{2m \omega P} \cos Q$.

Step 2: Verify Fundamental Poisson Bracket {Q, P}
$$\{Q, P\}_{q,p} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} = 1$$

Evaluating the Poisson bracket confirms that the transformation preserves canonical invariants.

Step 3: Transform the Harmonic Oscillator Hamiltonian
$$K(Q, P) = H(q(Q,P), p(Q,P)) = \frac{2m \omega P \cos^2 Q}{2m} + \frac{1}{2} m \omega^2 \left(\frac{2P}{m\omega} \sin^2 Q\right) = \omega P$$

Remarkably, $Q$ is completely cyclic in $K = \omega P$. Hamilton's equations in the new variables become trivial: $\dot{P} = -\frac{\partial K}{\partial Q} = 0 \implies P = \text{const}$, and $\dot{Q} = \frac{\partial K}{\partial P} = \omega \implies Q(t) = \omega t + \beta$.

Final Answer & Physical Insight

Transformation: q = √(2P/(mω)) sin Q, p = √(2mωP) cos Q. Transformed Hamiltonian: K = ω P, yielding immediate linear solution Q(t) = ω t + β, P = E/ω.

Solved Problem Example 6.2: Poisson Bracket Lie Algebra of Angular Momentum Components

Using the Cartesian definitions of orbital angular momentum components $L_x = y p_z - z p_y$, $L_y = z p_x - x p_z$, and $L_z = x p_y - y p_x$, compute the Poisson bracket $\{L_x, L_y\}$ and show that $\{L^2, L_z\} = 0$.

Step 1: Compute {Lx, Ly} Using Fundamental Poisson Brackets
$$\{L_x, L_y\} = \{y p_z - z p_y, z p_x - x p_z\} = \{y p_z, z p_x\} + \{z p_y, x p_z\}$$

Cross terms with no shared coordinates vanish identically. Expanding using the Leibniz product rule: $\{y p_z, z p_x\} = y p_x \{p_z, z\} = -y p_x$.

Step 2: Evaluate the Second Term and Combine
$$\{z p_y, x p_z\} = x p_y \{z, p_z\} = +x p_y \implies \{L_x, L_y\} = x p_y - y p_x = L_z$$

Cyclic permutations yield the complete angular momentum Lie algebra: $\{L_i, L_j\} = \epsilon_{ijk} L_k$.

Step 3: Evaluate Poisson Bracket of L² with Lz
$$\{L^2, L_z\} = \{L_x^2 + L_y^2 + L_z^2, L_z\} = 2 L_x \{L_x, L_z\} + 2 L_y \{L_y, L_z\} + 0 = 2 L_x (-L_y) + 2 L_y (L_x) = 0$$

Because $\{L^2, L_z\} = 0$, total angular momentum magnitude squared $L^2$ and any one of its Cartesian components $L_z$ can be simultaneously conserved in spherically symmetric central force fields.

Final Answer & Physical Insight

Lie algebra: {Li, Lj} = ε_ijk L_k. Consequently, {L², L_z} = 0, proving simultaneous conservation.

Solved Problem Example 6.3: Symplectic Test on a Linear Phase Transformation

Determine the conditions on constants $a, b, c, d$ such that the linear transformation $Q = a q + b p$, $P = c q + d p$ is strictly canonical, and verify Liouville's phase area preservation.

Step 1: Compute the Fundamental Poisson Bracket {Q, P}
$$\{Q, P\}_{q,p} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} = (a)(d) - (b)(c) = a d - b c$$

For the transformation to be canonical, the fundamental Poisson bracket must satisfy $\{Q, P\} = 1$. This requires $a d - b c = 1$.

Step 2: Construct the Jacobian Transformation Matrix M
$$\mathbf{M} = \begin{pmatrix} \partial Q/\partial q & \partial Q/\partial p \\ \partial P/\partial q & \partial P/\partial p \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

The condition $a d - b c = 1$ is precisely the requirement that $\det \mathbf{M} = 1$, which belongs to the special linear group $SL(2, \mathbb{R}) \cong Sp(2, \mathbb{R})$.

Step 3: Verify the Symplectic Condition M^T J M = J
$$\mathbf{M}^T \mathbf{J} \mathbf{M} = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 0 & a d - b c \\ -(a d - b c) & 0 \end{pmatrix} = \mathbf{J}$$

Since $\det \mathbf{M} = 1$, the transformation matrix preserves the symplectic 2-form $dq \wedge dp = dQ \wedge dP$, confirming Liouville's area conservation $dQ dP = dq dp$.

Final Answer & Physical Insight

The transformation is canonical if and only if ad - bc = 1 (unit Jacobian determinant), ensuring symplectic invariance and phase volume preservation.