# Hamiltonian Mechanics > One more reformulation, and it is the one quantum mechanics actually quantizes. Trading velocity > for momentum turns $n$ second-order equations into $2n$ first-order ones, makes "the state of the > system" a single point in phase space — and equips observables with an algebraic operation, the > Poisson bracket, which is precisely the commutator of Term 1 wearing classical fur. ## Learning Objectives After this lesson you will be able to: 1. Compute **Legendre transforms** and construct the **Hamiltonian** $H(q,p)$ from a Lagrangian. 2. Derive and apply **Hamilton's equations**, and judge when $H = T + V$ and when $H$ equals the energy. 3. Analyze motion in **phase space**: flow, non-crossing trajectories, oscillator ellipses, and the pendulum **separatrix**. 4. Compute **Poisson brackets**, exploit their algebraic properties, and evaluate the **fundamental brackets**. 5. Write time evolution as $df/dt = \{f, H\} + \partial f/\partial t$ and detect conserved quantities via $\{f, H\} = 0$. 6. State **Dirac's canonical quantization rule** and map $\{x, p\} = 1$ to $[\hat x, \hat p] = i\hbar$. --- ## Intuition The Lagrangian lives on positions and velocities; to know where the system goes you still must solve second-order equations. Hamilton's move is to promote the **momentum** to an independent variable. The payoff is conceptual as much as computational: the complete state of the system becomes a single *point* $(q, p)$ in **phase space**, and dynamics becomes a *flow* — every point knows exactly where it is going next. On top of this geometry sits an algebra: the Poisson bracket combines any two observables into a third, and time evolution, conservation laws, and symmetries are all statements about brackets. Quantum mechanics keeps the entire structure and changes one thing: the bracket becomes a commutator, with $\hbar$ setting the exchange rate. --- ## Theory ### The Legendre transform For a convex function $f(x)$ (i.e. $f'' > 0$), the **Legendre transform** is $$ g(p) = p\,x(p) - f\big(x(p)\big), \qquad \text{where } x(p) \text{ solves } p = f'(x). $$ Since $f'$ is strictly increasing, $x(p)$ is well defined. *Geometric meaning:* the tangent to $f$ at $x$ has slope $p$ and $y$-intercept $f(x) - px = -g(p)$, so $g$ re-encodes the curve by its family of tangent lines — slope in, intercept out, no information lost. It is involutive: $dg/dp = x + p\,x'(p) - f'(x)\,x'(p) = x$, so transforming $g$ returns $f$. *Worked scalar example:* $f(v) = \tfrac12 mv^2$ gives $p = mv$, $v = p/m$, and $g(p) = p(p/m) - \tfrac12 m(p/m)^2 = p^2/2m$ — the shape of every kinetic term to come. ### Canonical momentum and the Hamiltonian Define the **canonical momentum** conjugate to $q_i$ by $p_i = \partial L/\partial\dot q_i$ (for $L = \tfrac12 m\dot x^2 - V$ this is $m\dot x$, but *not* always — recall the magnetic Lagrangian from [P.1.2](03-lagrangian-mechanics.md)). The **Hamiltonian** is the Legendre transform of $L$ in its velocity slots: $$ H(q, p, t) = \sum_i p_i\,\dot q_i - L(q, \dot q, t), $$ with every $\dot q_i$ eliminated in favor of $(q, p)$ by inverting $p_i = \partial L/\partial\dot q_i$. ### Hamilton's equations Take the differential of $H$ as defined above: $$ dH = \sum_i\Big(p_i\,d\dot q_i + \dot q_i\,dp_i - \frac{\partial L}{\partial q_i}\,dq_i - \frac{\partial L}{\partial\dot q_i}\,d\dot q_i\Big) - \frac{\partial L}{\partial t}\,dt . $$ The $d\dot q_i$ terms cancel *by the definition* of $p_i$ — this is what the Legendre transform is for. Along an actual motion, Euler–Lagrange gives $\partial L/\partial q_i = \dot p_i$, so $dH = \sum_i(\dot q_i\,dp_i - \dot p_i\,dq_i) - (\partial L/\partial t)\,dt$. Comparing coefficients with $dH = \sum_i\big(\tfrac{\partial H}{\partial q_i}dq_i + \tfrac{\partial H}{\partial p_i}dp_i\big) + \tfrac{\partial H}{\partial t}dt$: $$ \boxed{\;\dot q_i = \frac{\partial H}{\partial p_i}, \qquad \dot p_i = -\frac{\partial H}{\partial q_i}\;} \qquad\text{and}\qquad \frac{\partial H}{\partial t} = -\frac{\partial L}{\partial t}. $$ **Hamilton's equations**: $2n$ first-order equations replacing $n$ second-order ones. Oscillator check: $H = p^2/2m + \tfrac12 kx^2$ gives $\dot x = p/m$, $\dot p = -kx$ — Newton again. ### When is $H = T + V$? When is $H$ the energy? Two *independent* facts. (i) If the map from Cartesian to generalized coordinates has no explicit time dependence, $T$ is a homogeneous quadratic $\tfrac12\sum a_{jk}(q)\dot q_j\dot q_k$; Euler's theorem for homogeneous functions then gives $\sum_i\dot q_i\,\partial T/\partial\dot q_i = 2T$. With $V$ velocity-independent, $p_i = \partial T/\partial\dot q_i$, so $\sum p_i\dot q_i = 2T$ and $H = 2T - (T - V) = T + V = E$. (ii) Independently, $dH/dt = \partial H/\partial t$ (proved below), so $H$ is **conserved** exactly when it has no explicit time dependence. All four combinations occur: the bead on a rotating wire (Worked Example 2) has $H$ conserved yet $H \neq E$, because the rotating constraint makes the coordinate map time-dependent. > **Caution.** $H \neq E$ whenever constraints (or the coordinates) are time-dependent — do not > read "the Hamiltonian" reflexively as "the energy". And beware a notational ambush ahead: in > this program $H$ is the Hamiltonian *here*, but the **Hadamard gate** in quantum-circuit > contexts — [Appendix C](../../appendix/C-notation-glossary.md) flags the clash. Context decides. ### Phase space The state of the system is the point $(q, p) \in \mathbb{R}^{2n}$ — **phase space**. Hamilton's equations attach a velocity vector to every point, defining a **flow**; by uniqueness of ODE solutions, exactly one trajectory passes through each point, so **trajectories never cross**. For the oscillator, $H = p^2/2m + \tfrac12 m\omega^2q^2 = E$ traces an **ellipse** with semi-axes $q_{\max} = \sqrt{2E/m\omega^2}$ and $p_{\max} = \sqrt{2mE}$, of area $$ \pi\,q_{\max}\,p_{\max} = \frac{2\pi E}{\omega}. $$ Phase-space area carries units of *action* (J·s). The old quantum theory (P.2.3) will quantize exactly this area in units of Planck's constant — nature slices phase space into cells of size $2\pi\hbar$. ### Poisson brackets For two observables $f(q, p)$, $g(q, p)$ define the **Poisson bracket** $$ \{f, g\} = \sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right). $$ Its algebra: **antisymmetry** $\{f, g\} = -\{g, f\}$ (swap the two terms) and **bilinearity** in each slot (derivatives are linear) are immediate. The **Leibniz/product rule** $\{fg, h\} = f\{g, h\} + g\{f, h\}$ follows by direct computation: since $\partial(fg)/\partial q_i = f\,\partial g/\partial q_i + g\,\partial f/\partial q_i$ (same for $p_i$), $$ \{fg, h\} = \sum_i\big[(f g_{q_i} + g f_{q_i})h_{p_i} - (f g_{p_i} + g f_{p_i})h_{q_i}\big] = f\{g, h\} + g\{f, h\}. \checkmark $$ Finally the **Jacobi identity** $\{f, \{g, h\}\} + \{g, \{h, f\}\} + \{h, \{f, g\}\} = 0$ holds (a patient expansion of second derivatives; you verify an instance in E4). The **fundamental brackets** follow from $\partial q_i/\partial q_k = \delta_{ik}$ etc.: $$ \{q_i, q_j\} = 0, \qquad \{p_i, p_j\} = 0, \qquad \{q_i, p_j\} = \sum_k \delta_{ik}\delta_{jk} = \delta_{ij}. $$ ### Time evolution is a bracket Let $f(q, p, t)$ be any observable, evaluated along a trajectory. Chain rule plus Hamilton's equations: $$ \frac{df}{dt} = \sum_i\Big(\frac{\partial f}{\partial q_i}\dot q_i + \frac{\partial f}{\partial p_i}\dot p_i\Big) + \frac{\partial f}{\partial t} = \sum_i\Big(\frac{\partial f}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial H}{\partial q_i}\Big) + \frac{\partial f}{\partial t} = \{f, H\} + \frac{\partial f}{\partial t}. $$ The Hamiltonian *generates* time evolution. Special cases: $\dot q = \{q, H\}$ and $\dot p = \{p, H\}$ reproduce Hamilton's equations; $dH/dt = \{H, H\} + \partial H/\partial t = \partial H/\partial t$ (antisymmetry kills $\{H,H\}$), proving the conservation claim above. And for any time-independent observable, $$ f \text{ is conserved} \iff \{f, H\} = 0 . $$ ### The angular momentum algebra With $\mathbf L = \mathbf r \times \mathbf p$: $L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, $L_z = xp_y - yp_x$. Compute $\{L_x, L_y\}$ by summing over $(x, y, z)$; the only nonvanishing partials of $L_x$ are $\partial_y L_x = p_z$, $\partial_z L_x = -p_y$, $\partial_{p_y}L_x = -z$, $\partial_{p_z}L_x = y$, and of $L_y$: $\partial_x L_y = -p_z$, $\partial_z L_y = p_x$, $\partial_{p_x}L_y = z$, $\partial_{p_z}L_y = -x$. Term by term ($x$-, $y$-, $z$-slots): $$ \{L_x, L_y\} = \underbrace{(0)(z) - (0)(-p_z)}_{x} + \underbrace{(p_z)(0) - (-z)(0)}_{y} + \underbrace{(-p_y)(-x) - (y)(p_x)}_{z} = xp_y - yp_x = L_z . $$ Cyclically, $\{L_y, L_z\} = L_x$ and $\{L_z, L_x\} = L_y$ (E5). This closed algebra — computed here with nothing but calculus — will single-handedly determine the quantum theory of angular momentum in [P.6.1](../c06-angular-momentum/01-central-potentials-angular-momentum.md). ### Dirac's rule: the classical shadow of the commutator Here is the headline. In 1925 Dirac observed that the Poisson bracket's algebra — antisymmetry, bilinearity, Leibniz rule, Jacobi identity — is *exactly* the algebra of the operator **commutator** $[\hat A, \hat B] = \hat A\hat B - \hat B\hat A$. **Canonical quantization** promotes phase-space functions to operators and postulates the correspondence $$ \{\,\cdot\,,\,\cdot\,\} \;\longrightarrow\; \frac{1}{i\hbar}\,[\,\cdot\,,\,\cdot\,] . $$ The consequences land one for one. The fundamental bracket $\{x, p\} = 1$ becomes the canonical commutation relation $[\hat x, \hat p] = i\hbar$ — the single equation from which position–momentum uncertainty follows via the Robertson relation ([1.3.2](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md)). The evolution law $df/dt = \{f, H\}$ becomes $d\hat A/dt = \tfrac{1}{i\hbar}[\hat A, \hat H]$, equivalently $i\hbar\,\tfrac{d}{dt}\lvert\psi\rangle = \hat H\lvert\psi\rangle$ — the Evolution Postulate ([1.1.3](../../term-1-quantum-mechanics/c01-postulates/03-evolution-postulate.md)), with the Hamiltonian still the generator of time translation. And $\{L_x, L_y\} = L_z$ becomes $[\hat L_x, \hat L_y] = i\hbar\hat L_z$, fixing the entire quantum angular momentum spectrum before any differential equation is solved. The Poisson bracket is the **classical shadow of the commutator**: everything you computed in this lesson survives quantization with $\{\cdot,\cdot\} \mapsto [\cdot,\cdot]/i\hbar$, and classical mechanics is recovered as the $\hbar \to 0$ limit of the commutator algebra. --- ## Worked Examples ### Example 1 — The oscillator in phase space, with numbers A mass $m = 0.500$ kg on a spring $k = 200$ N/m: $\omega = \sqrt{k/m} = 20.0$ rad/s. Legendre: $p = m\dot x$, so $H = p^2/2m + \tfrac12 kx^2$ (here $H = T + V = E$: time-independent everything). At $E = 1.00$ J the trajectory is the ellipse with $x_{\max} = \sqrt{2E/k} = 0.100$ m and $p_{\max} = \sqrt{2mE} = 1.00$ kg·m/s, traversed clockwise once per period $2\pi/\omega = 0.314$ s. Enclosed area: $\pi x_{\max}p_{\max} = 0.314$ J·s $= 2\pi E/\omega$ ✓ — energy and enclosed action are proportional, with $\omega$ the conversion factor. ### Example 2 — $H$ conserved, yet not the energy The bead on a wire rotating at constant $\Omega$ ([P.1.2](03-lagrangian-mechanics.md), Example 2): $L = \tfrac12 m(\dot r^2 + \Omega^2r^2)$, so $p_r = m\dot r$ and $$ H = p_r\dot r - L = \tfrac12 m\dot r^2 - \tfrac12 m\Omega^2r^2 = \frac{p_r^2}{2m} - \tfrac12 m\Omega^2r^2 . $$ No explicit $t$, so $H$ is **conserved**. But the energy is $E = T = \tfrac12 m(\dot r^2 + \Omega^2r^2) = H + m\Omega^2r^2$, which *grows* as the bead flies outward — the motor does work. The culprit: the time-dependent constraint makes $T$ non-homogeneous in $\dot r$ (it has a velocity-independent piece $\tfrac12 m\Omega^2 r^2$), so the Euler's-theorem step fails and $H \neq T + V$. Conserved Hamiltonian, non-conserved energy — the two notions genuinely split. --- ## Hands-on (Python) ```python import numpy as np import matplotlib.pyplot as plt # --- Phase portraits: oscillator ellipses and the pendulum separatrix --- m, ell, g = 1.0, 1.0, 9.81 w = np.sqrt(g/ell) fig, ax = plt.subplots(1, 2, figsize=(11, 4)) q, p = np.meshgrid(np.linspace(-2, 2, 300), np.linspace(-2, 2, 300)) ax[0].contour(q, p, p**2/(2*m) + 0.5*m*w**2*q**2, levels=10) ax[0].set(title="Oscillator: H contours are ellipses", xlabel="q", ylabel="p") th, pth = np.meshgrid(np.linspace(-2*np.pi, 2*np.pi, 600), np.linspace(-8, 8, 400)) H_pen = pth**2/(2*m*ell**2) - m*g*ell*np.cos(th) ax[1].contour(th, pth, H_pen, levels=20) ax[1].contour(th, pth, H_pen, levels=[m*g*ell], colors="r", linewidths=2) ax[1].set(title="Pendulum: red = separatrix (E = mgl)", xlabel="theta", ylabel="p") plt.tight_layout(); plt.show() # Expected: nested ellipses (left). Right: closed curves (libration) inside the # red separatrix through (+-pi, 0), open wavy curves (full rotation) outside -- # the separatrix is the trajectory that takes infinite time to reach the top. ``` ```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import solve_ivp # --- Energy drift over 100 periods: explicit vs symplectic Euler vs RK45 --- # Oscillator with m = w = 1, (q0, p0) = (1, 0), E0 = 1/2. dt = 0.05 steps = int(100*2*np.pi/dt) E = lambda q, p: 0.5*(p**2 + q**2) q_e = np.empty(steps + 1); p_e = q_e.copy(); q_e[0], p_e[0] = 1.0, 0.0 q_s = q_e.copy(); p_s = p_e.copy() for i in range(steps): # explicit Euler: both updates from the OLD state q_e[i+1] = q_e[i] + dt*p_e[i]; p_e[i+1] = p_e[i] - dt*q_e[i] # symplectic Euler: momentum first, then position with the NEW momentum p_s[i+1] = p_s[i] - dt*q_s[i]; q_s[i+1] = q_s[i] + dt*p_s[i+1] t = np.arange(steps + 1)*dt sol = solve_ivp(lambda t, y: [y[1], -y[0]], (0, t[-1]), [1.0, 0.0], t_eval=t, rtol=1e-8, atol=1e-10) # RK45 by default for qq, pp, lab in [(q_e, p_e, "explicit Euler"), (q_s, p_s, "symplectic Euler"), (sol.y[0], sol.y[1], "RK45")]: plt.semilogy(t, np.abs(E(qq, pp)/0.5 - 1) + 1e-16, label=lab) plt.xlabel("t"); plt.ylabel("|E/E0 - 1|"); plt.legend(); plt.show() # Expected: explicit Euler's energy error grows EXPONENTIALLY (each step # multiplies E by 1 + dt^2); symplectic Euler's stays bounded ~ dt forever; # RK45 is tiny but drifts secularly. The symplectic update respects the # phase-space (bracket) structure -- geometry beats raw accuracy. ``` --- ## Exercises **E1 (easy).** Compute the Legendre transform of $f(v) = av^4$ ($a > 0$, $v > 0$).

Solution

$p = f'(v) = 4av^3 \Rightarrow v = (p/4a)^{1/3}$. Then $g(p) = pv - f = p(p/4a)^{1/3} - a(p/4a)^{4/3} = \big(p - \tfrac{p}{4}\big)(p/4a)^{1/3} = \tfrac{3p}{4}\big(\tfrac{p}{4a}\big)^{1/3} \propto p^{4/3}$. Check: $g'(p) = (p/4a)^{1/3} = v$ ✓ (the involution property).

**E2 (easy).** Construct $H$ for a projectile, $L = \tfrac12 m(\dot x^2 + \dot y^2) - mgy$, write Hamilton's equations, and identify the conserved momentum.

Solution

$p_x = m\dot x$, $p_y = m\dot y$, so $H = (p_x^2 + p_y^2)/2m + mgy$. Equations: $\dot x = p_x/m$, $\dot y = p_y/m$, $\dot p_x = 0$, $\dot p_y = -mg$. Since $x$ is absent from $H$, $p_x$ is conserved — equivalently $\{p_x, H\} = 0$.

**E3 (medium).** For a central potential, $H = \dfrac{p_r^2}{2m} + \dfrac{p_\theta^2}{2mr^2} + V(r)$. Show $\{p_\theta, H\} = 0$ and interpret.

Solution

$\{p_\theta, H\} = \sum_{i\in\{r,\theta\}}\big(\partial_{q_i}p_\theta\,\partial_{p_i}H - \partial_{p_i}p_\theta\,\partial_{q_i}H\big)$. All partials of $p_\theta$ vanish except $\partial p_\theta/\partial p_\theta = 1$, leaving $\{p_\theta, H\} = -\partial H/\partial\theta = 0$ since $\theta$ is absent from $H$. Angular momentum is conserved because the Hamiltonian is rotationally symmetric — the bracket version of the cyclic-coordinate argument, and the classical seed of $[\hat L_z, \hat H] = 0$ in [P.6.1](../c06-angular-momentum/01-central-potentials-angular-momentum.md).

**E4 (medium).** Verify the Jacobi identity for $f = x^2$, $g = p^2$, $h = xp$ (one particle, 1D).

Solution

Inner brackets: $\{p^2, xp\} = 0\cdot x - 2p\cdot p = -2p^2$; $\{xp, x^2\} = p\cdot0 - x\cdot2x = -2x^2$; $\{x^2, p^2\} = 2x\cdot2p - 0 = 4xp$. Then $\{x^2, -2p^2\} = -2(4xp) = -8xp$; $\{p^2, -2x^2\} = -2\{p^2, x^2\} = -2(-4xp) = +8xp$; $\{xp, 4xp\} = 0$. Sum: $-8xp + 8xp + 0 = 0$ ✓.

**E5 (hard).** Show $\{L_y, L_z\} = L_x$ and $\{L_z, L_x\} = L_y$, then prove $\{L^2, L_z\} = 0$ for $L^2 = L_x^2 + L_y^2 + L_z^2$. What does this mean physically?

Solution

The theory computation gave $\{L_x, L_y\} = L_z$; the relabeling $x \to y \to z \to x$ maps $\mathbf L$'s components into each other and leaves the bracket's structure invariant, so $\{L_y, L_z\} = L_x$ and $\{L_z, L_x\} = L_y$ (or repeat the six-term computation). Compactly, $\{L_i, L_j\} = \epsilon_{ijk}L_k$. By Leibniz and antisymmetry: $\{L_x^2, L_z\} = 2L_x\{L_x, L_z\} = -2L_xL_y$, $\{L_y^2, L_z\} = 2L_y\{L_y, L_z\} = +2L_yL_x$, and $\{L_z^2, L_z\} = 0$; the sum vanishes. So $L^2$ is invariant under the rotations any $L_i$ generates: the *magnitude* of angular momentum is compatible with any one component. Quantum translation: $[\hat L^2, \hat L_z] = 0$, which is why states carry the simultaneous quantum numbers $\ell$ and $m$ ([P.6.1](../c06-angular-momentum/01-central-potentials-angular-momentum.md)).

--- ## Checkpoint 1. Define the Legendre transform and give its geometric meaning. 2. Derive Hamilton's equations from $dH$: which terms cancel by construction, and where is Euler–Lagrange used? 3. State the separate conditions for $H = T + V$ and for $H$ conserved, with an example where they come apart. 4. List the four algebraic properties of the Poisson bracket and the fundamental brackets. 5. State Dirac's quantization rule. What do $\{x, p\} = 1$ and $df/dt = \{f, H\}$ become?

Answers

1. $g(p) = px(p) - f(x(p))$ with $p = f'(x)$; it describes the convex curve by its tangent lines (slope $\to$ intercept), losing no information, and is involutive. 2. In $dH = \sum(p\,d\dot q + \dot q\,dp - L_q\,dq - L_{\dot q}\,d\dot q) - L_t\,dt$ the $d\dot q$ terms cancel because $p \equiv \partial L/\partial\dot q$; Euler–Lagrange converts $L_q$ to $\dot p$; matching coefficients of $dq, dp$ gives $\dot q = H_p$, $\dot p = -H_q$. 3. $H = T + V$ needs a time-independent coordinate map (so $T$ is homogeneous quadratic) and velocity-independent $V$; $H$ is conserved iff $\partial H/\partial t = 0$. Bead on a rotating wire: $H$ conserved but $H = E - m\Omega^2r^2 \neq E$. 4. Antisymmetry, bilinearity, Leibniz product rule, Jacobi identity; $\{q_i, q_j\} = \{p_i, p_j\} = 0$, $\{q_i, p_j\} = \delta_{ij}$. 5. $\{\cdot,\cdot\} \to [\cdot,\cdot]/i\hbar$. Then $\{x,p\} = 1 \to [\hat x, \hat p] = i\hbar$, and $df/dt = \{f, H\} \to d\hat A/dt = [\hat A, \hat H]/i\hbar$ — quantum time evolution, equivalently $i\hbar\,d\lvert\psi\rangle/dt = \hat H\lvert\psi\rangle$ ([1.1.3](../../term-1-quantum-mechanics/c01-postulates/03-evolution-postulate.md)).

--- ## Further Reading - **[Gold]** Goldstein, Poole & Safko, Ch. 8 and §9.5 — Hamilton's equations; Poisson brackets and canonical invariants. - **[Sha]** Shankar, Ch. 2 — the Hamiltonian formulation, compact and aimed squarely at quantization. - **[Sak]** Sakurai & Napolitano, §1.6 — canonical commutation relations and Dirac's analogy, up close. --- ← Prev: [Lagrangian Mechanics](03-lagrangian-mechanics.md) · Up: [Pre-Term](../README.md) · Next: [Blackbody Radiation & Planck's Law](../c02-radiation-old-quantum/01-blackbody-radiation.md) →