Exercises — Canonical transformations — generating functions

$F_3=F_3(p,Q,t)$: the old momentum $p$ and new coordinate $Q$. $q_i=-\frac{\partial F_3}{\partial p_i},P_i=-\frac{\partial F_3}{\partial Q_i},K=H+\frac{\partial F_3}{\partial t}.$ What we did: read off the table row for Type 3. Why the two minus signs: $F_3$ is reached from the master relation by a Legendre swap in $q$ (turning $pdq$ into $-qdp$), which flips the sign in front of $q$; and $Q$ sits on the "$F$-only" side, so its conjugate $P$ also carries a minus (same reason as Type 1).

Yes — it is the identity, trivially preserving Hamilton's equations. It comes from $F_2={\sum }_i{q}_i{P}_i$: then $p_i=\partial F_2/\partial q_i=P_i$ and $Q_i=\partial F_2/\partial P_i=q_i$. Why $F_2$ and not $F_1$: $F_1(q,Q)$ cannot even describe the identity, because for the identity $q$ and $Q$ are the same variable — they are not independent, so $F_1$ is ill-defined. The identity forces a type that mixes $q$ with the new momentum $P$.

True. $K=H+\partial F/\partial t$. If $F=F(q,Q)$ (no explicit $t$), then $\partial F/\partial t=0$, so $K=H$. What this means physically: a time-independent relabelling of phase space cannot inject or remove energy; it only renames points, so the numerical value of the Hamiltonian at a point is unchanged.

Type 1: $p=\partial F_1/\partial q=Q$ and $P=-\partial F_1/\partial Q=-q$. Therefore $\boxed{Q=p,P=-q}$. What it looks like: points in the $(q,p)$ plane rotate by $90^{\circ }$ (see figure below): the old momentum axis becomes the new coordinate axis. Position and momentum are interchangeable.

Type 2 relations:

$p=\partial F_1/\partial q=m\omega q\cot Q$. $P=-\partial F_1/\partial Q=-\frac{1}{2}m\omega q^2\cdot (-{\csc }^{2}Q)=\frac{m\omega q^2}{2{\sin }^{2}Q}$, using $\frac{d}{dQ}\cot Q=-{\csc }^{2}Q=-1/{\sin }^{2}Q$. Solve for $q^2$: $q^2=\frac{2P{\sin }^{2}Q}{m\omega }$, so $q=\sqrt{\frac{2P}{m\omega }}\sin Q$. ✓ Why $\cot Q$ and not something else: the oscillator's phase-space orbits are ellipses. $\cot Q$ is precisely the function whose Type-1 partials turn $(q,p)$ into the polar-like $(Q,P)$ that straightens those ellipses into circles of radius set by $P$.

The Poisson bracket of two functions of $(q,p)$ is $\{A,B\}=\frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial A}{\partial p}\frac{\partial B}{\partial q}.$ Here $\partial Q/\partial q=\cos \alpha$, $\partial Q/\partial p=-\sin \alpha$, $\partial P/\partial q=\sin \alpha$, $\partial P/\partial p=\cos \alpha$. So $\{Q,P\}=\cos \alpha \cdot \cos \alpha -(-\sin \alpha )\cdot \sin \alpha ={\cos }^2\alpha +{\sin }^2\alpha =1.$ Since $\{Q,P\}=1$ (and trivially $\{Q,Q\}=\{P,P\}=0$), the map is canonical for every $\alpha$. What it looks like: rotating the whole $(q,p)$ plane preserves areas — exactly what Liouville's theorem demands. The $90^{\circ }$ case $\alpha =\pi /2$ reproduces Ex. 2.1 ($Q=-p$… up to the sign convention chosen there).

For $Q=\lambda q,P=p$: $\{Q,P\}=\frac{\partial (\lambda q)}{\partial q}\frac{\partial p}{\partial p}-\frac{\partial (\lambda q)}{\partial p}\frac{\partial p}{\partial q}=\lambda \cdot 1-0=\lambda \ne 1$ unless $\lambda =1$. Not canonical. For $Q=\lambda q,P=p/\lambda$: $\{Q,P\}=\lambda \cdot (1/\lambda )-0=1$. Canonical. Generating function (Type 2): we need $Q=\partial F_2/\partial P=\lambda q$ and $p=\partial F_2/\partial q=\lambda P$. Try $F_2=\lambda qP$: then $\partial F_2/\partial P=\lambda q=Q$ ✓ and $\partial F_2/\partial q=\lambda P=p$ ⟹ $P=p/\lambda$ ✓. So $\boxed{F_2=\lambda qP}$. Why the momentum must shrink: stretching positions by $\lambda$ stretches phase-space area by $\lambda$; to keep area fixed (canonicity) the momenta must contract by the same factor.

Type 3: $q=-\partial F_3/\partial p=-(-Q)=Q$ and $P=-\partial F_3/\partial Q=-(-p)=p$. So $Q=q,P=p$: another route to the identity. Analysis: the identity is reachable through different generator types ($F_2=qP$ here vs $F_3=-pQ$) because the choice of which variables to hold fixed is ours; the physical map is the same. This is the Legendre-transform freedom in disguise — see Legendre transformation.

The free particle already has $H=p^2/2m$ independent of $q$, so $q$ is cyclic and $p$ is conserved. The cleanest synthesis is the Hamilton–Jacobi choice: pick $F_2=S(q,P)$ solving $\frac{1}{2m}{\left(\frac{\partial S}{\partial q}\right)}^2=P,$ so that $K=H+\partial F_2/\partial t=P$ (with $P$ the conserved energy). This gives $\partial S/\partial q=\sqrt{2mP}$, i.e. $S=\sqrt{2mP}q$. Then $p=\frac{\partial S}{\partial q}=\sqrt{2mP},Q=\frac{\partial S}{\partial P}=\frac{m}{\sqrt{2mP}}q=\sqrt{\frac{m}{2P}}q.$ Now $K=P$ is independent of $Q$ ⟹ $Q$ is cyclic, $\dot{P}=0$ (energy conserved), and $\dot{Q}=\partial K/\partial P=1$ ⟹ $Q=t+\text{const}$. Interpretation: $Q$ literally became time, and $P$ the energy — the essence of the Hamilton–Jacobi equation and Action–angle variables.

We need $\partial F_2/\partial P=q^2$, so $F_2=q^{2}P$ (plus a $P$-independent piece, take it zero). Then $p=\partial F_2/\partial q=2qP$, giving $\boxed{P=\frac{p}{2q}}$ (for $q\ne 0$). Check $\{Q,P\}$: $\partial Q/\partial q=2q,\partial Q/\partial p=0,\partial P/\partial q=-p/(2q^2),\partial P/\partial p=1/(2q)$, so $\{Q,P\}=2q\cdot \frac{1}{2q}-0\cdot \left(-\frac{p}{2q^2}\right)=1.✓$ Degenerate case: at $q=0$ the map $Q=q^2$ is not invertible ($Q\ge 0$, and $\pm q$ both give the same $Q$) and $P=p/2q$ blows up. So the transformation is canonical only on each half-line $q>0$ or $q<0$ separately — a reminder that "canonical" is often a local statement.

$A$: $Q=p,P=-q$. $B$: $\tilde{Q}=P=-q,\tilde{P}=-Q=-p$. So the net map is $\boxed{\tilde{Q}=-q,\tilde{P}=-p}$ — the point reflection through the origin (rotation by $180^{\circ }$). Why: each $F_1=qQ$ step is a $90^{\circ }$ rotation of phase space; two of them make $180^{\circ }$. Canonical transformations form a group — the composition of canonical maps is canonical, so we never need a new generator to justify the result.

The bracket of the old variables computed with respect to the new ones must equal $1$ for a canonical map: $\{q,p{\}}_{Q,P}=\frac{\partial q}{\partial Q}\frac{\partial p}{\partial P}-\frac{\partial q}{\partial P}\frac{\partial p}{\partial Q}.$ With $q=\sqrt{2P/(m\omega )}\sin Q$ and $p=\sqrt{2Pm\omega }\cos Q$:

From $K=\omega P$ and $\dot{P}=0$, $P$ is constant; and $K=H=E$ on-shell means $E=\omega P$, so $\boxed{P=E/\omega }$. Over one period $Q$ runs $0\to 2\pi$ while $P$ is fixed, so $\oint pdq=\oint PdQ=P{\int }_0^{2\pi }dQ=2\pi P=\frac{2\pi E}{\omega }.$ (The cross-terms integrate to a total differential of $F_1$ around a closed loop and vanish.) What this is: $\oint pdq=2\pi P$ identifies $P$ as the action variable $I=E/\omega$, and $Q$ as its conjugate angle — the full content of Action–angle variables. The oscillator's phase-space ellipse encloses area $2\pi P=2\pi E/\omega$; see the figure.

The Type-1 relations $p=\partial F/\partial q$ and $P=-\partial F/\partial Q$ involve derivatives with respect to $q$ and $Q$. Since $h(t)$ depends only on $t$, $\partial h/\partial q=\partial h/\partial Q=0$, so $p$ and $P$ (hence the whole map) are unchanged. Only $K=H+\partial F/\partial t$ picks up $+h^{\prime }(t)$: $K\to K+h^{\prime }(t)$. Mastery point: a purely time-dependent shift of the generator is a gauge freedom — it rescales the "energy zero" of $K$ without touching physics, mirroring how Hamilton's equations only care about derivatives of $H$. This is why the master relation tolerates "$+dF/dt$": total derivatives are physically invisible.