Hamiltonian — definition H = Σpᵢq̇ᵢ − L: Worked examples

Intuition What this page is

The parent note showed you the recipe: find the momentum $p = \partial L / \partial \overset{q}{˙}$ , invert it, and assemble $H = \sum p_{i} \overset{q}{˙}_{i} - L$ . This page throws every kind of system at that recipe so you never meet a case you haven't already solved. We march through a matrix of scenarios — clean energy cases, cases where $H \neq = E$ , cases where $H$ isn't even conserved, multi-coordinate systems, a magnetic (velocity-dependent) potential, and an exam twist.

Before we start: recall the two questions we always ask about a Hamiltonian.

Is $H$ conserved? Yes exactly when $\partial L / \partial t = 0$ (no naked $t$ in the Lagrangian).
Is $H$ the energy $T + V$ ? Yes only when $T$ is a pure quadratic in the velocities (explained in the refresher just below) and $V$ has no velocity in it. These two questions are independent — that is the whole drama of this page.

Recall Two refreshers you'll need (so this page stands alone)

Euler's theorem, in one line. If $T$ is built purely from terms like $\overset{q}{˙}_{j} \overset{q}{˙}_{k}$ (each term has exactly two velocity factors), then differentiating and summing gives back $2 T$ : $\sum_{i} \overset{q}{˙}_{i} \frac{\partial T}{\partial q ˙ _{i}} = 2 T$ . Picture: each of the two velocity factors "gets counted once", so you recover $T$ twice. This is exactly why $\sum p_{i} \overset{q}{˙}_{i} = 2 T$ and hence $H = 2 T - (T - V) = T + V$ — but only when every term in $T$ has exactly two velocity factors. Jacobi integral, in one line. When $L$ has no explicit $t$ but $T$ is not a pure velocity-quadratic (e.g. a moving constraint adds a velocity-free piece), the conserved quantity $H = \sum p_{i} \overset{q}{˙}_{i} - L$ still exists but is not the energy. That conserved-but-not-energy object has a name: the Jacobi integral. You meet it in cell C4.

The scenario matrix

Every row below is a distinct "class" of behaviour. The last two columns are the two independent questions above. C8 is the exam twist — its true answers are $H$ conserved: NO, $H = E$ : YES (worked out in full below; try to predict it before you get there).

Cell	Scenario class	Example that hits it	$H$ conserved?	$H = E$ ?
C1	Textbook 1D, $H = E =$ const	(a) free-fall particle	✓	✓
C2	2 coordinates, one cyclic	(b) 2D central motion, polar	✓	✓
C3	Explicit time in $L$	(c) driven oscillator	✗	✓ (value drifts)
C4	Moving constraint, $H \neq = E$ but conserved	(d) bead on rotating wire	✓	✗
C5	Velocity-dependent potential (magnetic)	(e) charge in magnetic field	✓	✓ (subtle)
C6	Degenerate / zero input	(f) free particle, $V = 0$ , $p = 0$ limit	✓	✓ (trivial)
C7	Sign / direction cases	(g) particle thrown up vs down	✓	✓
C8	Exam twist: time-scaled potential	(h) growing stiffness	✗	✓

Every cell is worked below and labelled. Prerequisites we lean on: Lagrangian Mechanics, Legendre Transform, Hamilton's Canonical Equations, Conservation Laws & Noether's Theorem, and Phase Space for the pictures.

C1 — The textbook case: 1D free-fall

Worked example (a) A particle of mass

$m$ falling under gravity, height $y$ . $L = \frac{1}{2} m \overset{y}{˙}^{2} - m g y$ . Forecast: guess $H$ before reading on. It should be $\frac{p ^{2}}{2 m} + m g y$ , and it should be both conserved and equal to the energy. Why? Because there's no explicit $t$ , and $T = \frac{1}{2} m \overset{y}{˙}^{2}$ is a clean quadratic.

Momentum: $p = \frac{\partial L}{\partial y ˙} = m \overset{y}{˙}$ . Why this step? The Legendre transform swaps the velocity for its "slope" $p = \partial L / \partial \overset{y}{˙}$ ; we must find that slope first.
Invert: $\overset{y}{˙} = p / m$ . Why this step? $H$ must be a function of $(y, p)$ only — every $\overset{y}{˙}$ has to be evicted.
Assemble: $H = p \overset{y}{˙} - L = p \cdot \frac{p}{m} - (\frac{1}{2} m (p / m)^{2} - m g y) = \frac{p ^{2}}{m} - \frac{p ^{2}}{2 m} + m g y$ . Why this step? This is the definition $H = \sum p_{i} \overset{q}{˙}_{i} - L$ with the inverted velocity plugged in. $H = \frac{p ^{2}}{2 m} + m g y = T + V = E .$ Verify: Hamilton's equations should reproduce free-fall. $\overset{y}{˙} = \partial H / \partial p = p / m$ ✓ (matches step 1). $\overset{p}{˙} = - \partial H / \partial y = - m g$ , i.e. $m \overset{y}{¨} = - m g$ — Newton's law. ✓ Numerically for $m = 2 kg$ , $\overset{y}{˙} = 3 m/s$ , $y = 5 m$ , $g = 9.8$ : $p = 6$ , $H = 36/4 + 2 \cdot 9.8 \cdot 5 = 9 + 98 = 107 J$ .

C2 — Two coordinates, one cyclic: 2D central motion

Worked example (b) A particle of mass

$m$ in a central potential $V (r)$ , polar coordinates $(r, θ)$ . $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - V (r)$ . Forecast: two momenta now. $θ$ never appears in $L$ (only $\dot{θ}$ does) — so $θ$ is cyclic and its momentum $p_{θ}$ (angular momentum) is conserved. $H$ should be $T + V$ .

Momenta: $p_{r} = \frac{\partial L}{\partial r ˙} = m \overset{r}{˙}$ , $p_{θ} = \frac{\partial L}{\partial θ ˙} = m r^{2} \dot{θ}$ . Why this step? Each coordinate gets its own conjugate momentum; both slopes are needed.
Invert: $\overset{r}{˙} = p_{r} / m$ , $\dot{θ} = p_{θ} / (m r^{2})$ . Why this step? Two velocities, two evictions.
Assemble: $H = p_{r} \overset{r}{˙} + p_{θ} \dot{θ} - L$ . $H = \frac{p _{r}^{2}}{m} + \frac{p _{θ}^{2}}{m r ^{2}} - (\frac{p _{r}^{2}}{2 m} + \frac{p _{θ}^{2}}{2 m r ^{2}} - V (r)) = \frac{p _{r}^{2}}{2 m} + \frac{p _{θ}^{2}}{2 m r ^{2}} + V (r) .$ Why this step? Same Legendre assembly; the $\dot{θ}$ term folds neatly because $T$ is quadratic. $H = \frac{p _{r}^{2}}{2 m} + \frac{p _{θ}^{2}}{2 m r ^{2}} + V (r) = E .$ Verify: Since $θ$ is absent from $H$ , $\overset{p}{˙}_{θ} = - \partial H / \partial θ = 0$ — angular momentum conserved, as Conservation Laws & Noether's Theorem predicts for rotational symmetry. ✓ Numerically at $r = 2$ , $p_{r} = 1$ , $p_{θ} = 4$ , $m = 1$ , $V = 0$ : $H = 1/2 + 16/ (2 \cdot 4) = 0.5 + 2 = 2.5$ .

The extra $\frac{p _{θ}^{2}}{2 m r ^{2}}$ term is the centrifugal barrier. Read the figure below like this: the blue curve is the raw attractive potential $V (r) = - k / r$ (a well that pulls inward), the pink curve is the centrifugal term $p_{θ}^{2} / (2 m r^{2})$ (a wall that shoots up near $r = 0$ , since dividing by $r^{2}$ blows up), and the yellow curve is their sum $V_{eff}$ . Notice the yellow curve dips to a minimum at a finite radius — that dip is a stable circular orbit, the radius where inward pull and outward barrier balance. The barrier is the whole reason a central-force particle with nonzero $p_{θ}$ never falls into the centre.

Figure — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

C3 — Explicit time in $L$ : driven oscillator

Worked example (c) A 1D oscillator pushed by an external force

$F (t) = F_{0} cos (Ω t)$ . $L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2} + F_{0} cos (Ω t) x$ . Forecast: the term $F_{0} cos (Ω t) x$ contains $t$ explicitly, so $\partial L / \partial t \neq = 0$ — expect $H$ not conserved. But $T$ is still a clean quadratic and $V$ has no velocity, so $H$ should still numerically equal $T + V$ at each instant.

Momentum: $p = m \overset{x}{˙}$ , invert $\overset{x}{˙} = p / m$ . Why this step? The driving term has no $\overset{x}{˙}$ , so it doesn't touch the momentum.
Assemble: $H = p \overset{x}{˙} - L = \frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} - F_{0} cos (Ω t) x$ . Why this step? Standard assembly; note the drive survives with a minus sign (it was $+ F_{0} cos (Ω t) x$ in $L$ , so $- L$ flips it to $- F_{0} cos (Ω t) x$ ). $H = \frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} - F_{0} cos (Ω t) x .$
Time behaviour: $\frac{d H}{d t} = \frac{\partial H}{\partial t} = + F_{0} Ω sin (Ω t) x \neq = 0$ . Why this step? The total time-derivative of $H$ along motion equals $\partial H / \partial t$ (from Hamilton's equations the $q, p$ parts cancel), and here that partial is nonzero. Verify: Is $H = E$ instant-by-instant? $E = T + V = \frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} - F_{0} cos (Ω t) x$ — yes, identical, because $V$ now includes the drive term. So $H = E$ but neither is conserved: energy pumps in and out. Numeric snapshot: $m = 1, k = 4, F_{0} = 3, Ω = 2$ , at $t = 0$ so $cos = 1, sin = 0$ , $x = 1, p = 2$ : $H = 2 + 2 - 3 = 1$ ; and $d H / d t = 3 \cdot 2 \cdot 0 \cdot 1 = 0$ at this instant. ✓

C4 — Moving constraint: $H \neq = E$ but still conserved

Worked example (d) Bead of mass

$m$ on a wire spun at fixed angular rate $ω$ ; radial coord $r$ . The constraint $ϕ = ω t$ is time-dependent. $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})$ , $V = 0$ : $L = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2} .$ Forecast: danger. $T$ has a piece $\frac{1}{2} m r^{2} ω^{2}$ that carries no $\overset{r}{˙}$ — it's not a pure quadratic in the velocity. So Euler's theorem's " $\sum p \overset{q}{˙} = 2 T$ " fails and we should get $H \neq = E$ . But $L$ has no explicit $t$ (the $ω t$ was substituted away), so $H$ is still conserved.

Momentum: $p_{r} = \partial L / \partial \overset{r}{˙} = m \overset{r}{˙}$ , invert $\overset{r}{˙} = p_{r} / m$ . Why this step? Only $\overset{r}{˙}$ appears; the $ω$ term is velocity-free.
Assemble: $H = p_{r} \overset{r}{˙} - L = \frac{p _{r}^{2}}{m} - (\frac{p _{r}^{2}}{2 m} + \frac{1}{2} m r^{2} ω^{2}) = \frac{p _{r}^{2}}{2 m} - \frac{1}{2} m r^{2} ω^{2}$ . Why this step? The velocity-free term is untouched by the Legendre swap, so it survives into $H$ with the sign of $- L$ : minus. $H = \frac{p _{r}^{2}}{2 m} - \frac{1}{2} m r^{2} ω^{2} .$
Compare to energy: $E = T + V = \frac{p _{r}^{2}}{2 m} + \frac{1}{2} m r^{2} ω^{2}$ . So $H = E - m r^{2} ω^{2} \neq = E$ . Why this step? The rotating wire feeds energy through the constraint; $H$ is a different conserved bookkeeping quantity — the Jacobi integral from the refresher above. Verify: conserved? $\partial L / \partial t = 0$ ✓, so $d H / d t = 0$ . Not the energy? at $r = 1, ω = 2, m = 1, p_{r} = 3$ : $H = 9/2 - 2 = 2.5$ , while $E = 9/2 + 2 = 6.5$ . Different. ✓

C5 — Velocity-dependent potential: charge in a magnetic field

Worked example (e) Charge

$e$ , mass $m$ , in a uniform field $B = B \overset{z}{^}$ , using vector potential $A = \frac{1}{2} B \times r$ . The magnetic Lagrangian is $L = \frac{1}{2} m ∣ \dot{r} ∣^{2} + e \dot{r} \cdot A$ . We keep the full vector throughout — the vectors just carry along for the ride, and to sanity-check numbers at the end we'll drop to one scalar component. Forecast: here the "potential" term $e \dot{r} \cdot A$ contains velocity, so the naive " $H = T + V$ " argument is void. The surprise: because that term is linear in $\dot{r}$ , it cancels out of $H$ and we still get $H =$ kinetic energy.

Momentum: $p = \frac{\partial L}{\partial r ˙} = m \dot{r} + e A$ . This is the canonical momentum, not $m v$ . Why this step? The linear-in-velocity term shifts the slope; forgetting the $e A$ is the classic magnetic mistake.
Invert: $\dot{r} = (p - e A) / m$ . Why this step? Evict the velocity; note the combination $p - e A = m v$ is the kinetic momentum.
Assemble: $H = p \cdot \dot{r} - L$ . Substituting and simplifying (the linear terms cancel): $H = \frac{( p - e A ) ^{2}}{2 m} .$ Why this step? $p \cdot \dot{r} = \frac{1}{m} p \cdot (p - e A)$ and $L = \frac{( p - e A ) ^{2}}{2 m} + \frac{e}{m} A \cdot (p - e A)$ ; subtracting kills the cross terms. Verify: $H = \frac{1}{2} m ∣ v ∣^{2} = T$ — pure kinetic energy, since a magnetic force does no work. So $H = E$ is true, but only after correctly using canonical momentum. Numeric scalar check (one component, $A$ meaning the relevant component of $A$ ): let $m = 1, e = 1, A = 2, p = 5$ : kinetic momentum $= 5 - 2 = 3$ , $H = 3^{2} /2 = 4.5$ ; and $T = \frac{1}{2} m v^{2}$ with $v = 3$ gives $4.5$ . ✓ Same. $L$ is time-independent, so $H$ is conserved too. ✓

C6 — Degenerate / zero input: the free particle

Worked example (f) Free particle,

$V = 0$ , and the limit $p \to 0$ . $L = \frac{1}{2} m \overset{x}{˙}^{2}$ . Forecast: with no potential, $H$ = pure kinetic. At the degenerate point $p = 0$ (particle at rest) $H = 0$ . Nothing should blow up.

Momentum: $p = m \overset{x}{˙}$ , invert $\overset{x}{˙} = p / m$ . Why this step? Standard; no potential to worry about.
Assemble: $H = p \overset{x}{˙} - L = \frac{p ^{2}}{m} - \frac{p ^{2}}{2 m} = \frac{p ^{2}}{2 m}$ . Why this step? Plug the inverted velocity into the definition $H = p \overset{x}{˙} - L$ ; with no potential only the kinetic piece survives.
Degenerate check $p = 0$ : $H = 0$ , $\overset{x}{˙} = \partial H / \partial p = 0$ , $\overset{p}{˙} = - \partial H / \partial x = 0$ . The particle sits still forever — a fixed point in Phase Space. Why this step? Zero input must give a sensible (non-singular) answer; it does. $H = \frac{p ^{2}}{2 m} .$ Verify: $\overset{p}{˙} = 0$ everywhere (no force), so $p$ is conserved — momentum conservation from translational symmetry. At $p = 0$ : $H = 0$ . ✓ At $p = 4, m = 2$ : $H = 16/4 = 4$ . ✓

C7 — Sign / direction cases: thrown up vs down

Worked example (g) The C1 particle again, but examine both signs of velocity: thrown

up ( $\overset{y}{˙} > 0$ , so $p > 0$ ) and thrown down ( $\overset{y}{˙} < 0$ , so $p < 0$ ). $H = \frac{p ^{2}}{2 m} + m g y$ . Forecast: $H$ depends on $p^{2}$ , so both signs give the same energy at the same height — up and down at the same speed are indistinguishable to $H$ . The sign of $p$ only sets the direction of motion in phase space.

Up case: $\overset{y}{˙} = + 3 m/s$ , $m = 2$ , so $p = + 6$ , at $y = 5$ : $H = 36/4 + 2 \cdot 9.8 \cdot 5 = 9 + 98 = 107$ . Why this step? Positive $p$ means $\overset{y}{˙} = \partial H / \partial p = p / m > 0$ , moving up.
Down case: $\overset{y}{˙} = - 3 m/s$ , so $p = - 6$ , same $y = 5$ : $H = (- 6)^{2} /4 + 98 = 9 + 98 = 107$ . Why this step? Negative $p$ means moving down, but $p^{2}$ erases the sign — same energy. Verify: both give $107 J$ ; the phase-space trajectory is a parabola symmetric under $p \to - p$ . ✓ This mirror symmetry is exactly the up/down flight of a thrown ball.

Read the figure below like this: the horizontal axis is momentum $p$ , the vertical axis is height $y$ , and the yellow curve is the set of all $(p, y)$ with the same $H = 107 J$ . The blue dot (right) is "thrown up, $p = + 6$ " and the pink dot (left) is "thrown down, $p = - 6$ " — both sit at the same height $y = 5$ , mirror images across the vertical $p = 0$ axis. That left–right mirror symmetry is the fact that $H$ only sees $p^{2}$ : swapping the direction of travel leaves the energy untouched.

C8 — Exam twist: does explicit time break conservation and energy?

Worked example (h) A particle in a potential whose

strength grows in time: $V (x, t) = \frac{1}{2} k (t) x^{2}$ with $k (t) = k_{0} e^{t / τ}$ . $L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k_{0} e^{t / τ} x^{2}$ . Forecast (guess before solving!): (i) Is $H = E$ ? $T$ is a clean quadratic, $V$ has no velocity — so yes, $H = T + V$ . (ii) Is $H$ conserved? $L$ has explicit $t$ (the $e^{t / τ}$ ) — so no. The trap is to assume " $H = E$ " ⟹ " $H$ conserved". They are separate!

Momentum: $p = m \overset{x}{˙}$ , invert $\overset{x}{˙} = p / m$ . Why this step? The time-dependent stiffness sits in $V$ , not touching $p$ .
Assemble: $H = p \overset{x}{˙} - L = \frac{p ^{2}}{2 m} + \frac{1}{2} k_{0} e^{t / τ} x^{2} = T + V = E$ . ✓ (so $H = E$ ) Why this step? Legendre on a clean quadratic gives $T + V$ exactly; the whole time-dependence rides along untouched inside $V$ . $H = \frac{p ^{2}}{2 m} + \frac{1}{2} k_{0} e^{t / τ} x^{2} .$
Conservation: $\frac{d H}{d t} = \frac{\partial H}{\partial t} = \frac{1}{2} k_{0} \cdot \frac{1}{τ} e^{t / τ} x^{2} \neq = 0$ . Why this step? Along the motion, $d H / d t = \partial H / \partial t$ ; the explicit $t$ in the stiffness leaks energy in. Not conserved. Verify: at $t = 0$ , $τ = 1$ , $k_{0} = 4$ , $m = 1$ , $x = 1$ , $p = 2$ : $H = 2 + 2 = 4$ (energy). $d H / d t = \frac{1}{2} \cdot 4 \cdot 1 \cdot 1 \cdot 1 = 2 \neq = 0$ . So $H = E = 4 J$ right now but rising. ✓ Answer to the twist: $H = E$ YES, $H$ conserved NO.

Recap: which question does each cell answer?

Recall The two independent switches

Conserved? ::: Yes iff $\partial L / \partial t = 0$ (no naked $t$ in $L$ ). Equal to energy? ::: Yes iff $T$ is a pure quadratic in velocities AND $V$ is velocity-free. Cell where conserved but NOT energy ::: C4, bead on rotating wire (the Jacobi integral). Cell where energy but NOT conserved ::: C3 and C8, explicit time in $L$ . Cell where BOTH hold ::: C1, C2, C6, C7 (and C5 after using canonical momentum).

Putting it all together: the eight cells are just the four combinations of the two switches, each illustrated by a concrete system. C1/C2/C6/C7 sit in the happy corner (conserved and energy). C3 and C8 show "energy but drifting" — turn on explicit time and the value of $H$ ceases to be constant, even though at each frozen instant $H$ still equals $T + V$ . C4 shows the opposite mismatch — "conserved but not energy" — where a moving constraint gives you the Jacobi integral instead of $E$ . C5 is the trap disguised as the happy corner: it looks like it should fail (velocity in the potential) but the linearity rescues it, provided you use the canonical momentum $p = m v + e A$ rather than $m v$ . Master these four corners and no exam scenario can surprise you.