The Hamiltonian (Legendre Transform)

Equivalent forms

H = T + V \;\text{(natural systems)}

H = \frac{p^2}{2m} + V(q)

The Legendre transform turns second-order equations into first-order flow on phase space — dynamics becomes geometry.

Symbol	Name	SI unit	Dimension	Range
$H$	Hamiltonianoutput Total energy as a function of (q, p)	J	ML^2T^-2	0 – 50
$p$	Momentum Canonical momentum of the oscillator	kg·m/s	MLT^-1	-5 – 5
$x$	Position Displacement from equilibrium	m	L	-2 – 2
$m$	Mass Oscillator mass	kg	M	0.1 – 5
$k$	Spring constant Stiffness of the potential ½kx²	N/m	MT^-2	1 – 50

Unit systems

Symbol	SI	CGS	Imperial
$H$	J	erg	ft·lbf
$p$	kg·m/s	g·cm/s	lb·ft/s
$x$	m	cm	ft
$m$	kg	g	lb
$k$	N/m	dyn/cm	lbf/ft

Where it holds

Any Lagrangian system whose velocities can be solved for in terms of momenta (non-degenerate Legendre transform); the framework beneath statistical and quantum mechanics.

Dimensional analysis

H \to [ML^{2}T^{-2}]

p\cdot q

̇

- L \to [MLT^{-1}][LT^{-1}] = [ML^{2}T^{-2}]

Joules on both sides.

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Discovery

William Rowan Hamilton · 1833

Hamilton, who had reformulated optics with a single 'characteristic function', did the same for mechanics in 1833. His H seemed like elegant bookkeeping for a century — until Schrödinger and Heisenberg discovered that quantum mechanics speaks Hamiltonian natively: Ĥψ = iħ∂ψ/∂t.

Try this

Trade velocity for momentum and mechanics becomes geometry: every motion is a walk along an energy contour.

Build H = p²/2m + ½kx² for an oscillator and watch the state glide along an ellipse of constant H in phase space.

Research status: stable

Real-world applications

Quantum mechanics: the Schrödinger equation is built on Ĥ
Statistical mechanics: the Boltzmann factor $e^{-H/k_BT}$
Symplectic integrators for solar-system simulation
Accelerator physics: beam dynamics designed in phase space

Common misconceptions

H is always T + V — false in rotating frames or with time-dependent constraints
H and E are synonyms — H is conserved $iff \partial H/\partial t = 0$ , and equals E only for natural systems
The Legendre transform is just substitution — it's a duality: convex functions of ẋ ↔ convex functions of p

Experimental verification

Energy conservation in isolated systems is the everyday test; in quantum mechanics, measured atomic spectra are literally the eigenvalues of Ĥ — hydrogen's levels match to 12 digits with QED corrections.

Derivation

Define

p = \partial L/\partial

ẋ and eliminate ẋ in favor of p: H(q,

p) = p

ẋ

- L

.

For

L = \tfrac{1}{2}m

ẋ

^{2} - V

:

p = m

ẋ, so ẋ

= p/m

and

H = p(p/m) - [\tfrac{1}{2}m(p/m)^{2} - V] = p^{2}/2m + V

.

The transform is involutive — doing it again returns L.

Limiting cases

L = \tfrac{1}{2}m

ẋ

^{2} - V

, time-independent⟶

H = T + V = E

For 'natural' Lagrangians the Hamiltonian is the conserved total energy.

\partial L/\partial t \neq 0

⟶

H \neq

const and

may \neq E

Driven systems: H can differ from the mechanical energy and need not be conserved.

Rotating reference frame⟶

H = E - \omega L_z

(Jacobi integral)H is conserved but is not the energy — the distinction matters.

What if…

What if H has no explicit time dependence?

$dH/dt = \partial H/\partial t = 0$ — energy conservation for free.

What if I plot all states with the same H?

You get the phase-space orbit itself (in 1D) — motion IS the level curve.

What if the Legendre transform is singular?

You have constraints (e.g., gauge theories) — Dirac developed constrained Hamiltonian dynamics for exactly this.

1

Oscillator energy from (x, p)

Given ·

p:: 1
x:: 0.5
m:: 1
k:: 10

Find · H

Steps

$T = p^{2}/2m = 1/(2\times 1) = 0.5\,\mathrm{J}$
$V = \tfrac{1}{2}kx^{2} = 0.5 \times 10 \times 0.25 = 1.25\,\mathrm{J}$
$H = 0.5 + 1.25 = 1.75\,\mathrm{J}$ — constant along the motion

Result ·

H = p^{2}/2m + \tfrac{1}{2}kx^{2} = 0.5 + 1.25 = 1.75\,\mathrm{J}

2

Legendre transform by hand

Given ·

m:: 2
p:: 4

Find · H for a free particle from

L = \tfrac{1}{2}m

ẋ

^{2}

Steps

$p = \partial L/\partial$ ẋ $= m$ ẋ → ẋ $= p/m = 2\,\mathrm{m/s}$
$H = p$ ẋ $- L = 4\times 2 - \tfrac{1}{2}\times 2\times 4 = 8 - 4 = 4\,\mathrm{J}$
Same as $p^{2}/2m$ — the transform checks out

Result ·

H = p^{2}/2m = 16/4 = 4\,\mathrm{J}

Hamilton's Canonical Equations→Euler–Lagrange Equation→Generalized (Conjugate) Momentum→Poisson Bracket & Liouville Flow→Kinetic Energy→