Hamiltonian Mechanics | Physics

The generalised momentum conjugate to $q_j$ is

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

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).

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

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

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

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

$\blacksquare$

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.

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.

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:

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

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.

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

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Hamiltonian mechanics is like switching from a video recording of motion to a snapshot of all possible states. While the Lagrangian tracks positions and velocities over time, the Hamiltonian lives in phase space where every point represents a complete snapshot of where something is and how fast it is going. Hamilton’s equations are beautifully symmetric: the rate of change of position depends on how energy changes with momentum, and the rate of change of momentum depends on how energy changes with position. Liouville’s theorem tells us that if you paint a region of phase space, its volume stays the same as it flows around like incompressible paint. This is why statistical mechanics works: the phase space volume that encodes all possible states of a system is preserved under time evolution.

• Lagrangian Mechanics: The Lagrangian formulation provides the foundation for deriving the Hamiltonian via Legendre transform.

• Noether’s Theorem: Noether’s theorem connects symmetries to conserved quantities, which are expressed as Poisson brackets in Hamiltonian mechanics.

• Quantum Mechanics: The Hamiltonian becomes an operator in quantum mechanics, and Poisson brackets become commutators.

• Calculus

• Linear Algebra

• Vector Calculus

• Quantum Computing

Mistake 1: Assuming the Hamiltonian always equals $T + V$ The Hamiltonian equals the total energy $H = T + V$ only when the potential is velocity-independent and the coordinate transformation is scleronomic. For systems with velocity-dependent potentials (like charged particles in electromagnetic fields) or rheonomic constraints, $H \neq T + V$ even though $H$ is still conserved when $\partial L/\partial t = 0$.

Mistake 2: Confusing canonical momentum with mechanical momentum Canonical momentum $p_j = \partial L/\partial \dot{q}_j$ is not always equal to mechanical momentum $m\dot{q}_j$. For a charged particle in a magnetic field, the canonical momentum includes the vector potential: $\mathbf{p} = m\mathbf{v} + q\mathbf{A}$. Always derive canonical momentum from the Lagrangian, not from $m\mathbf{v}$.

Mistake 3: Misapplying Poisson brackets The Poisson bracket $\{f, g\}$ is antisymmetric: $\{f, g\} = -\{g, f\}$. A common error is forgetting this sign when computing brackets of position and momentum. The fundamental brackets are $\{q_j, p_k\} = \delta_{jk}$, $\{q_j, q_k\} = 0$, and $\{p_j, p_k\} = 0$. Mixing up the order changes the sign.