Exercises — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

(b). The conjugate (generalized) momentum is defined as the derivative of $L$ with respect to the velocity ${\dot{q}}_i$, not the coordinate.

• (a) is only true for a free particle with $L=\frac{1}{2}m{\dot{q}}^2$; it is a special case, not the definition.
• (c) is ${\dot{p}}_i$ by the Euler–Lagrange equation, not $p_i$.
• (d) is $-{\dot{p}}_i$ (Hamilton's equation), not $p_i$.

No. $H$ must be a function of $(q,p,t)$ only. This expression still contains $\dot{x}$. You must invert $p=m\dot{x}\Rightarrow \dot{x}=p/m$ and substitute, giving $H=\frac{p^2}{2m}+V(x)$.

$2n$ first-order ODEs: ${\dot{q}}_i=\partial H/\partial p_i$ and ${\dot{p}}_i=-\partial H/\partial q_i$. This replaces the $n$ second-order Euler–Lagrange equations of Lagrangian Mechanics.

Momentum: $p=\partial L/\partial \dot{x}=m\dot{x}$. Invert: $\dot{x}=p/m$. Assemble & substitute: $H=p\dot{x}-L=p\cdot \frac{p}{m}-\left(\frac{1}{2}m\frac{p^2}{m^2}-\frac{1}{2}k{x}^2\right)=\frac{p^2}{m}-\frac{p^2}{2m}+\frac{1}{2}k{x}^2.$ $\boxed{H=\frac{p^2}{2m}+\frac{1}{2}k{x}^2}=T+V.$

Momenta: $p_x=m\dot{x}$, $p_y=m\dot{y}$, so $\dot{x}=p_x/m$, $\dot{y}=p_y/m$. Assemble: $H=p_x\dot{x}+p_y\dot{y}-L=\frac{p_x^2+p_y^2}{2m}$. Hamilton's equations: $\dot{x}=\frac{\partial H}{\partial p_x}=\frac{p_x}{m},\dot{y}=\frac{p_y}{m},{\dot{p}}_x=-\frac{\partial H}{\partial x}=0,{\dot{p}}_y=0.$ Momenta are constant → straight-line motion, as expected for a free particle.

Momentum: $p=\partial L/\partial \dot{q}=m\dot{q}$. Invert: $\dot{q}=p/m$. (We will now use this to remove every $\dot{q}$.) Assemble (before substituting, keep $\dot{q}$ visible): $H=p\dot{q}-L=p\dot{q}-\left(\frac{1}{2}m{\dot{q}}^2-mgq\right).$ Substitute $\dot{q}=p/m$ everywhere it appears — the first $\dot{q}$ becomes $p/m$, and the $\frac{1}{2}m{\dot{q}}^2$ becomes $\frac{1}{2}m(p/m{)}^2=p^2/(2m)$: $H=p\cdot \frac{p}{m}-\frac{1}{2}m{\left(\frac{p}{m}\right)}^2+mgq=\frac{p^2}{m}-\frac{p^2}{2m}+mgq=\frac{p^2}{2m}+mgq.$ This substitution is legitimate because $\dot{q}=p/m$ holds identically once we define $p$; there are no leftover velocities, so $H=H(q,p)$ as required. Here $T=p^2/2m$ (quadratic in $\dot{q}$) and $V=mgq$ (velocity-independent), so $H=T+V=E$. ✓

$\dot{\theta }=\frac{\partial H}{\partial p_{\theta }}=\frac{p_{\theta }}{m{\ell }^2},{\dot{p}}_{\theta }=-\frac{\partial H}{\partial \theta }=-mg\ell \sin \theta .$ Differentiate the first: $p_{\theta }=m{\ell }^2\dot{\theta }\Rightarrow {\dot{p}}_{\theta }=m{\ell }^2\ddot{\theta }$. Set equal to the second: $m{\ell }^2\ddot{\theta }=-mg\ell \sin \theta .$ This is exactly the Lagrangian equation of motion. ✓ The two first-order Hamilton equations together are equivalent to the one second-order Euler–Lagrange equation.

Total time derivative of $H(q,p,t)$: $\frac{dH}{dt}={\sum }_i\frac{\partial H}{\partial q_i}{\dot{q}}_i+{\sum }_i\frac{\partial H}{\partial p_i}{\dot{p}}_i+\frac{\partial H}{\partial t}.$ Insert Hamilton's equations ${\dot{q}}_i=\partial H/\partial p_i$ and ${\dot{p}}_i=-\partial H/\partial q_i$: $\frac{dH}{dt}={\sum }_i\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i}-{\sum }_i\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i}+\frac{\partial H}{\partial t}=\frac{\partial H}{\partial t}.$ The two sums cancel term by term. So if $\partial H/\partial t=0$, then $dH/dt=0$ — $H$ is conserved. This is the Hamiltonian face of Conservation Laws & Noether's Theorem.

Momentum: $p=\frac{\partial L}{\partial \dot{x}}=-m{c}^2\cdot \frac{-\dot{x}/c^2}{\sqrt{1-{\dot{x}}^2/c^2}}=\frac{m\dot{x}}{\sqrt{1-{\dot{x}}^2/c^2}}.$ Assemble (before inverting, to see the structure): $H=p\dot{x}-L=\frac{m{\dot{x}}^2}{\sqrt{1-{\dot{x}}^2/c^2}}+m{c}^2\sqrt{1-{\dot{x}}^2/c^2}.$ Put over a common denominator $\sqrt{1-{\dot{x}}^2/c^2}$: $H=\frac{m{\dot{x}}^2+m{c}^2(1-{\dot{x}}^2/c^2)}{\sqrt{1-{\dot{x}}^2/c^2}}=\frac{m{\dot{x}}^2+m{c}^2-m{\dot{x}}^2}{\sqrt{1-{\dot{x}}^2/c^2}}=\frac{m{c}^2}{\sqrt{1-{\dot{x}}^2/c^2}}.$ In terms of $p$: from $p\sqrt{1-{\dot{x}}^2/c^2}=m\dot{x}$, algebra gives ${\dot{x}}^2=\frac{p^2{c}^2}{p^2+m^2{c}^2}$, hence $\sqrt{1-{\dot{x}}^2/c^2}=\frac{mc}{\sqrt{p^2+m^2{c}^2}}$ and $\boxed{H=\sqrt{p^2{c}^2+m^2{c}^4}}$ — the famous relativistic energy. This is why Legendre-transforming $L$ (a function of velocity) into $H$ (a function of momentum) is so powerful: momentum is the natural variable of energy.

Momentum: $p=\partial L/\partial \dot{x}=m\dot{x}+eA$. Notice: the momentum is not $m\dot{x}$! It has the extra $eA$ term. Invert: $\dot{x}=\frac{p-eA}{m}$. Assemble: $H=p\dot{x}-L=p\dot{x}-\frac{1}{2}m{\dot{x}}^2-eA\dot{x}+e\phi =(p-eA)\dot{x}-\frac{1}{2}m{\dot{x}}^2+e\phi .$ But $(p-eA)=m\dot{x}$, so $(p-eA)\dot{x}=m{\dot{x}}^2$: $H=m{\dot{x}}^2-\frac{1}{2}m{\dot{x}}^2+e\phi =\frac{1}{2}m{\dot{x}}^2+e\phi =\frac{(p-eA{)}^2}{2m}+e\phi .$ Is $H=T+V$? Numerically $H=\frac{1}{2}m{\dot{x}}^2+e\phi =T+e\phi$. The kinetic energy came out right, but the canonical momentum $p=m\dot{x}+eA$ differs from the kinetic momentum $m\dot{x}$. So while $H$ has the energy value, expressed in $p$ it carries the tell-tale $(p-eA{)}^2$ structure. This is the seed of electromagnetic Hamiltonians.

$\{x,H\}=\frac{\partial x}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial x}{\partial p}\frac{\partial H}{\partial x}=1\cdot \frac{p}{m}-0=\frac{p}{m}=\dot{x}.✓$ $\{p,H\}=\frac{\partial p}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial p}{\partial p}\frac{\partial H}{\partial x}=0-1\cdot kx=-kx=\dot{p}.✓$ Both match $\dot{f}=\{f,H\}$, the Poisson-bracket form of the equations of motion in Phase Space.

(i) Momentum: $p_r=\partial L/\partial \dot{r}=m\dot{r}\Rightarrow \dot{r}=p_r/m$. Assemble and substitute $\dot{r}=p_r/m$: $H=p_r\dot{r}-L=\frac{p_r^2}{m}-\left(\frac{p_r^2}{2m}+\frac{1}{2}m{r}^2{\omega }^2\right)=\frac{p_r^2}{2m}-\frac{1}{2}m{r}^2{\omega }^2.$ (ii) Energy: $E=T+V=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\omega }^2=\frac{p_r^2}{2m}+\frac{1}{2}m{r}^2{\omega }^2$ (using $\dot{r}=p_r/m$; recall $V=0$). (iii) The gap: $H-E=\left(-\frac{1}{2}m{r}^2{\omega }^2\right)-\left(+\frac{1}{2}m{r}^2{\omega }^2\right)=-m{r}^2{\omega }^2\ne 0.$ So $H=E-m{r}^2{\omega }^2$. This is exactly the vertical red gap in the right panel of the figure. Why it fails: the constraint $\varphi =\omega t$ is time-dependent, so $T$ is not a pure homogeneous quadratic in $\dot{r}$ — it has the $\dot{r}$-independent "centrifugal" piece $\frac{1}{2}m{r}^2{\omega }^2$. Euler's theorem (which would give $\sum p_i{\dot{q}}_i=2T$) no longer applies to all of $T$. (iv) Conserved? $L$ has no explicit $t$ ($\omega$ is a constant, $r$ is the variable), so $\partial L/\partial t=0\Rightarrow \partial H/\partial t=0\Rightarrow H$ is conserved. A conserved quantity that is not the energy.

Momentum: $p=\partial L/\partial \dot{x}=m\dot{x}+bx\Rightarrow \dot{x}=\frac{p-bx}{m}$. Assemble: $H=p\dot{x}-L=p\dot{x}-\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2-bx\dot{x}=(p-bx)\dot{x}-\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2.$ Since $(p-bx)=m\dot{x}$, we get $(p-bx)\dot{x}=m{\dot{x}}^2$, so $H=m{\dot{x}}^2-\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=\frac{(p-bx{)}^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2.$ Conserved? $\partial L/\partial t=0$ (no explicit $t$), so $H$ is conserved. ✓ $H=T+V$? The value is $\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=T+V$ — yes, numerically $H=E$ here. The $bx\dot{x}$ term is a total time derivative ($\frac{d}{dt}(\frac{1}{2}b{x}^2)$), so it shifts the canonical momentum by $bx$ but does not change the energy. A subtle case: the momentum is disguised, yet $H=E$ still holds.

(i) Hessian: $\frac{\partial L}{\partial \dot{x}}=bx$, which does not contain $\dot{x}$. So $W=\frac{{\partial }^{2}L}{\partial {\dot{x}}^2}=0.$ Since $\det W=0$, the non-degeneracy condition from the recipe's warning fails. This is a singular (degenerate) Lagrangian. (ii) What goes wrong: the momentum is $p=bx$. Notice it depends only on $x$, not on $\dot{x}$ — so we cannot solve it for $\dot{x}$. Every value of $\dot{x}$ gives the same momentum $bx$. Step 2 of the recipe (invert to get $\dot{x}(x,p)$) is impossible. Instead we get a constraint: $p-bx=0$ must hold identically. The velocity is left undetermined. (iii) The Hamiltonian: $H=p\dot{x}-L=p\dot{x}-bx\dot{x}+\frac{1}{2}k{x}^2=(p-bx)\dot{x}+\frac{1}{2}k{x}^2.$ On the constraint surface $p=bx$, the bracket $(p-bx)$ vanishes, so $H=\frac{1}{2}k{x}^2.$ The velocity $\dot{x}$ has completely disappeared — but not because we substituted it away; it dropped out because its coefficient $(p-bx)$ is forced to zero. This is the fingerprint of a singular system: $\dot{x}$ is never expressible in terms of $p$, and the dynamics is governed by constraints rather than by a clean invertible $H(x,p)$. (Systematically handling such systems is Dirac's constraint theory.) Contrast this with every other problem on the page, where $W=m\ne 0$ and the inversion always succeeded.