Hamiltonian Mechanics

4.1 Generalised Momentum

The generalised momentum conjugate to $q_j$ is

$p_j = \frac{\partial L}{\partial \dot{q}_j}$

4.2 The Hamiltonian

The Hamiltonian is defined by the Legendre transform:

$H(q_1, \ldots, q_n, p_1, \ldots, p_n, t) = \sum_{j=1}^n p_j \dot{q}_j - L$

When the transformation is regular (i.e., the Hessian $\partial^2 L / \partial \dot{q}_j \partial \dot{q}_k$ Is non-singular), this is well-defined.

If $L$ does not depend explicitly on time and $V$ is velocity-independent, then $H = T + V$ (total Energy).

4.3 Worked Example: Legendre Transform for the Harmonic Oscillator

Problem. A one-dimensional harmonic oscillator has $L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2$. Find the Hamiltonian.

4.4 Worked Example: Legendre Transform for the Simple Pendulum

Problem. Find the Hamiltonian for a simple pendulum of mass $m$ and length $l$.

4.5 Hamilton’s Equations

Theorem 4.1 (Hamilton’s Equations). The equations of motion in Hamiltonian form are

$\dot{q}_j = \frac{\partial H}{\partial p_j}, \quad \dot{p}_j = -\frac{\partial H}{\partial q_j}$

These are $2n$ first-order ODEs (compared to $n$ second-order ODEs in the Lagrangian formulation).

Proof. From $H = \sum p_j \dot{q}_j - L$:

$dH = \sum \dot{q}_j\, dp_j + \sum p_j\, d\dot{q}_j - \sum \frac{\partial L}{\partial q_j}\, dq_j - \sum \frac{\partial L}{\partial \dot{q}_j}\, d\dot{q}_j - \frac{\partial L}{\partial t}\, dt$

Since $p_j = \partial L / \partial \dot{q}_j$The $d\dot{q}_j$ terms cancel:

$dH = \sum \dot{q}_j\, dp_j - \sum \dot{p}_j\, dq_j - \frac{\partial L}{\partial t}\, dt$

Comparing with $dH = \sum \frac{\partial H}{\partial p_j} dp_j + \sum \frac{\partial H}{\partial q_j} dq_j + \frac{\partial H}{\partial t} dt$:

$\dot{q}_j = \frac{\partial H}{\partial p_j}, \quad \dot{p}_j = -\frac{\partial H}{\partial q_j}, \quad \frac{\partial H}{\partial t} = -\frac{\partial L}{\partial t}$

$\blacksquare$

4.6 Phase Space

Hamiltonian mechanics lives in phase space: the $2n$-dimensional space with coordinates $(q_1, \ldots, q_n, p_1, \ldots, p_n)$. Each point in phase space represents a complete state of the System (positions and momenta).

A phase portrait is the collection of trajectories in phase space. For a 1D harmonic oscillator, the trajectories are ellipses in the $(x, p)$ plane.

4.7 Liouville’s Theorem

Theorem 4.2 (Liouville’s Theorem). The flow in phase space is incompressible: the phase space volume is conserved along trajectories. Equivalently, the phase space density $\rho(q, p, t)$ satisfies:

$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \sum_j \left(\frac{\partial \rho}{\partial q_j}\dot{q}_j + \frac{\partial \rho}{\partial p_j}\dot{p}_j\right) = 0$

Proof. Consider a volume $\Omega$ in phase space. The rate of change of the volume is:

$\frac{d}{dt}\int_\Omega \rho\, dq\, dp = \int_\Omega \frac{\partial \rho}{\partial t}\, dq\, dp$

By the continuity equation in $2n$ dimensions:

$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0$

Where $\mathbf{v} = (\dot{q}_1, \ldots, \dot{q}_n, \dot{p}_1, \ldots, \dot{p}_n)$ is the phase space velocity. Using Hamilton’s equations:

$\nabla \cdot \mathbf{v} = \sum_j \frac{\partial \dot{q}_j}{\partial q_j} + \sum_j \frac{\partial \dot{p}_j}{\partial p_j} = \sum_j \frac{\partial^2 H}{\partial q_j \partial p_j} - \sum_j \frac{\partial^2 H}{\partial p_j \partial q_j} = 0$

By equality of mixed partial derivatives. Therefore:

$\frac{\partial \rho}{\partial t} + \rho\,(\nabla \cdot \mathbf{v}) + \mathbf{v} \cdot \nabla\rho = \frac{\partial \rho}{\partial t} + \mathbf{v} \cdot \nabla\rho = \frac{d\rho}{dt} = 0$

$\blacksquare$

Intuition. Liouville’s theorem is the classical analogue of unitarity in quantum mechanics. It tells us that phase space volume is conserved --- like an incompressible fluid flowing through phase space. This underlies the ergodic hypothesis of statistical mechanics.

4.8 Poisson Brackets

Definition. The Poisson bracket of two functions $f(q, p, t)$ and $g(q, p, t)$ is:

$\{f, g\} = \sum_{j=1}^n \left(\frac{\partial f}{\partial q_j}\frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j}\frac{\partial g}{\partial q_j}\right)$

Theorem 4.3 (Equations of Motion via Poisson Brackets). For any function $f(q, p, t)$:

$\frac{df}{dt} = \frac{\partial f}{\partial t} + \{f, H\}$

In particular, Hamilton’s equations become:

$\dot{q}_j = \{q_j, H\}, \quad \dot{p}_j = \{p_j, H\}$

Proof. Using the chain rule:

$\frac{df}{dt} = \frac{\partial f}{\partial t} + \sum_j \left(\frac{\partial f}{\partial q_j}\dot{q}_j + \frac{\partial f}{\partial p_j}\dot{p}_j\right)$

Substituting Hamilton’s equations:

$\frac{df}{dt} = \frac{\partial f}{\partial t} + \sum_j \left(\frac{\partial f}{\partial q_j}\frac{\partial H}{\partial p_j} - \frac{\partial f}{\partial p_j}\frac{\partial H}{\partial q_j}\right) = \frac{\partial f}{\partial t} + \{f, H\}$

$\blacksquare$

Properties of Poisson Brackets.

Theorem 4.4. The Poisson bracket satisfies:

1. Antisymmetry: $\{f, g\} = -\{g, f\}$
2. Linearity: $\{af + bg, h\} = a\{f, h\} + b\{g, h\}$ for constants $a, b$
3. Leibniz rule: $\{fg, h\} = f\{g, h\} + \{f, h\}g$
4. Jacobi identity: $\{f, \{g, h\}\} + \{g, \{h, f\}\} + \{h, \{f, g\}\} = 0$

Proof. Properties (1)—(3) follow directly from the definition. For the Jacobi identity, write out the terms explicitly:

$\{f, \{g, h\}\} = \sum_j \frac{\partial f}{\partial q_j}\frac{\partial}{\partial p_j}\sum_k \left(\frac{\partial g}{\partial q_k}\frac{\partial h}{\partial p_k} - \frac{\partial g}{\partial p_k}\frac{\partial h}{\partial q_k}\right) - \sum_j \frac{\partial f}{\partial p_j}\frac{\partial}{\partial q_j}\sum_k \left(\frac{\partial g}{\partial q_k}\frac{\partial h}{\partial p_k} - \frac{\partial g}{\partial p_k}\frac{\partial h}{\partial q_k}\right)$

Expanding and collecting terms, the second-order mixed partial derivatives cancel in groups of three (by equality of mixed partials), yielding the Jacobi identity. $\blacksquare$

Theorem 4.5. A quantity $f$ is a constant of motion if and only if $\partial f/\partial t + \{f, H\} = 0$. If $f$ does not depend explicitly on time, $f$ is conserved if and only if $\{f, H\} = 0$.

Proof. Immediate from Theorem 4.3 with $df/dt = 0$. $\blacksquare$

Fundamental Poisson Brackets:

$\{q_j, q_k\} = 0, \quad \{p_j, p_k\} = 0, \quad \{q_j, p_k\} = \delta_{jk}$

4.9 The Hamilton-Jacobi Equation

Definition. Hamilton’s principal function $S(q, t)$ is the action evaluated along the classical path from $(q_0, t_0)$ to $(q, t)$.

Theorem 4.6 (Hamilton-Jacobi Equation). The function $S$ satisfies:

$H\left(q_1, \ldots, q_n, \frac{\partial S}{\partial q_1}, \ldots, \frac{\partial S}{\partial q_n}, t\right) + \frac{\partial S}{\partial t} = 0$

This is a first-order nonlinear PDE in $n + 1$ variables.

Proof. The action from $t_0$ to $t$ is $S = \int_{t_0}^{t} L\, dt'$. The total time derivative is:

$\frac{dS}{dt} = L$

But $S = S(q_1(t), \ldots, q_n(t), t)$So by the chain rule:

$\frac{dS}{dt} = \sum_j \frac{\partial S}{\partial q_j}\dot{q}_j + \frac{\partial S}{\partial t} = L$

From the definition of the conjugate momentum, $p_j = \partial L/\partial \dot{q}_j = \partial S/\partial q_j$ (this can be shown rigorously by varying the endpoint). Therefore:

$L = \sum_j p_j \dot{q}_j + \frac{\partial S}{\partial t} = H + \frac{\partial S}{\partial t}$

Since $dS/dt = L$:

$H + \frac{\partial S}{\partial t} = L = \sum_j p_j\dot{q}_j + \frac{\partial S}{\partial t}$

Which gives $H + \partial S/\partial t = 0$. $\blacksquare$

Intuition. The Hamilton-Jacobi equation is the bridge between classical and quantum mechanics. Schrodinger’s equation can be obtained from it via the substitution $S = -i\hbar \ln\psi$ (up to constants), making $S$ the classical limit of the quantum phase.

Separation of Variables. If $H$ does not depend explicitly on $t$Write $S(q, t) = W(q) - Et$. Then the time-independent Hamilton-Jacobi equation is:

$H\left(q_1, \ldots, q_n, \frac{\partial W}{\partial q_1}, \ldots, \frac{\partial W}{\partial q_n}\right) = E$

Where $W$ is Hamilton’s characteristic function and $E$ is the constant energy.

4.10 Worked Example: Hamilton-Jacobi for the Harmonic Oscillator

Problem. Solve the Hamilton-Jacobi equation for a 1D harmonic oscillator with $H = p^2/(2m) + kx^2/2$.

:::caution Common Pitfall The Lagrangian and Hamiltonian formulations are equivalent only when the Legendre transform from $L$ To $H$ is regular. If $\det(\partial^2 L / \partial \dot{q}_i \partial \dot{q}_j) = 0$The system Has constraints and the Hamiltonian formulation requires special treatment (Dirac brackets or Constraint analysis).

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