Canonical transformations — generating functions: Worked examples

The scenario matrix

Every generating-function problem falls into one of these cells. Our job below is to hit every one.

Cell	What makes it distinct	Covered by
A. Direct forward	given $F$ , read off $(Q, P)$ by differentiating	Ex 1
B. Implicit inversion	relations tangle $q, Q$ together; must solve for them	Ex 2
C. Sign / branch care	square roots & inverse trig → quadrant + sign ambiguity	Ex 3
D. Degenerate input	a variable is zero; the origin $q = p = 0$ ( $P = 0$ ) is the one true coordinate singularity of the oscillator action–angle map	Ex 4
E. Time-dependent $F$	$\partial F / \partial t \neq = 0$ , so $K \neq = H$	Ex 5
F. Test-only (no $F$ given)	prove a map is canonical via Poisson brackets	Ex 6
G. Word problem	physical oscillator, real numbers & units	Ex 7
H. Exam twist	a "trap": choose the right type, or catch a non-canonical map	Ex 8

We proceed A→H. Three figures carry the geometry: Figure 1 = the phase-space picture (Ex 3), Figure 2 = the sign-quadrant schematic (Ex 3), Figure 3 = the non-canonical bracket plot (Ex 8).

Example 1 — Cell A: read it straight off

Worked example

$F_{2} = q P + \frac{1}{2} P^{2}$ (single degree of freedom) Find the transformation $(q, p) \to (Q, P)$ .

Forecast: the $q P$ part is the identity (from the parent). The extra $\frac{1}{2} P^{2}$ should nudge $Q$ by something linear in $P$ . Guess $Q = q + P$ , $p = P$ .

Steps.

This is a Type-2 generator, so use $p = \partial F_{2} / \partial q$ and $Q = \partial F_{2} / \partial P$ . Why this step? $F_{2}$ depends on old $q$ and new $P$ — that pairing is the definition of Type 2.
$p = \frac{\partial}{\partial q} (q P + \frac{1}{2} P^{2}) = P$ . Why? only the $q P$ term has a $q$ ; its $q$ -derivative pulls down $P$ .
$Q = \frac{\partial}{\partial P} (q P + \frac{1}{2} P^{2}) = q + P$ . Why? differentiate each term in $P$ : $q P \to q$ , $\frac{1}{2} P^{2} \to P$ .
Invert to express new-in-old: step 2 says $P = p$ ; substitute that into step 3 to get $Q = q + p$ . Why this step? steps 2–3 give the answer partly in terms of the new variable $P$ ; the transformation must read new-in-old, so we use the already-solved relation $P = p$ to eliminate $P$ .

Verify: compute the Poisson bracket ${Q, P} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} = (1) (1) - (1) (0) = 1$ . ✓ Canonical. Forecast was right.

Example 2 — Cell B: relations that tangle, must be solved

Worked example

$F_{1} = \frac{q ^{2}}{2 Q}$ Find $(Q, P)$ in terms of $(q, p)$ .

Forecast: $F_{1}$ mixes $q$ and $Q$ nonlinearly, so we'll get $p$ and $P$ each as ratios; we'll have to algebraically untangle to get $Q (q, p)$ . Expect $Q \propto q / p$ .

Steps.

Type-1 rules: $p = \partial F_{1} / \partial q$ , $P = - \partial F_{1} / \partial Q$ . Why? $F_{1} (q, Q)$ = old $q$ , new $Q$ .
$p = \frac{\partial}{\partial q} \frac{q ^{2}}{2 Q} = \frac{q}{Q}$ . Why? $Q$ is held fixed; $\frac{d}{d q} q^{2} = 2 q$ .
$P = - \frac{\partial}{\partial Q} \frac{q ^{2}}{2 Q} = - (- \frac{q ^{2}}{2 Q ^{2}}) = \frac{q ^{2}}{2 Q ^{2}}$ . Why? $\frac{d}{d Q} Q^{- 1} = - Q^{- 2}$ , and the leading minus flips it positive.
Untangle. From step 2, $Q = q / p$ . Substitute into step 3: $P = \frac{q ^{2}}{2 ( q / p ) ^{2}} = \frac{q ^{2} p ^{2}}{2 q ^{2}} = \frac{p ^{2}}{2}$ . Why this step? the problem asks for new variables as functions of old ones — step 2/3 gave them in terms of a leftover $Q$ , which we eliminate.

So $Q = \frac{q}{p}, P = \frac{p ^{2}}{2} .$

Verify: ${Q, P} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q}$ . Compute: $\frac{\partial Q}{\partial q} = 1/ p, \frac{\partial P}{\partial p} = p, \frac{\partial Q}{\partial p} = - q / p^{2}, \frac{\partial P}{\partial q} = 0$ . So ${Q, P} = (1/ p) (p) - (- q / p^{2}) (0) = 1$ . ✓

Example 3 — Cell C: signs, branches & all four quadrants

Here geometry matters, so we draw the phase plane. The phase point is $(q, p)$ ; its distance from the origin and its angle both carry meaning. Figure 1 shows the phase plane with one worked point in each quadrant.

Figure — Canonical transformations — generating functions

Worked example The oscillator generator, done carefully:

$F_{1} = \frac{1}{2} mω q^{2} cot Q$ with $m = ω = 1$ We found in the parent note $q = 2 P sin Q$ , $p = 2 P cos Q$ (with $m = ω = 1$ ). Given a phase point $(q, p) = (- 1, - 1)$ , find $(Q, P)$ — and get the quadrant right.

Forecast: $q, p$ both negative sits in the third quadrant of the phase plane (see Figure 1). Its radius is $2$ . So expect $P = 1$ and $Q$ an angle in the third quadrant, near $22 5^{\circ}$ .

Steps.

Square-and-add to kill the angle: $q^{2} + p^{2} = 2 P (sin^{2} Q + cos^{2} Q) = 2 P$ . Why this step? $sin^{2} + cos^{2} = 1$ isolates $P$ with no sign ambiguity — squares are always $\geq 0$ .
$P = \frac{q ^{2} + p ^{2}}{2} = \frac{1 + 1}{2} = 1$ . Why? just plug in; $P$ is the (radius $^{2}$ )/2.
Divide to get the angle: $\frac{q}{p} = \frac{sin Q}{cos Q} = tan Q$ , so naively $Q = arctan (q / p) = arctan (1) = 4 5^{\circ}$ . Why the caution? $tan$ repeats every $18 0^{\circ}$ , so $arctan$ alone cannot tell $4 5^{\circ}$ from $22 5^{\circ}$ .
Fix the branch using the signs of $q$ and $p$ (look at the orange point in Figure 1): $q < 0$ means $sin Q < 0$ , $p < 0$ means $cos Q < 0$ — both negative ⟹ third quadrant ⟹ add $18 0^{\circ}$ : $Q = 4 5^{\circ} + 18 0^{\circ} = 22 5^{\circ} = \frac{5 π}{4}$ . Why this step? this is exactly the arctan-quadrant fix: the ratio loses sign information, the individual signs restore it.

All four cases (the four coloured points in Figure 1; the quadrant logic is schematised in Figure 2), with $r = q^{2} + p^{2}$ :

$(q, p)$	quadrant	$sin Q$	$cos Q$	$Q$	why the shift?
$(+, +)$	I	$+$	$+$	$arctan (q / p)$	ratio $> 0$ , $arctan$ already lands in $(0, 9 0^{\circ})$
$(+, -)$	II	$+$	$-$	$arctan (q / p) + 18 0^{\circ}$	$q / p < 0$ so $arctan (q / p) \in (- 9 0^{\circ}, 0)$ — negative, off by $18 0^{\circ}$ ; add it
$(-, -)$	III	$-$	$-$	$arctan (q / p) + 18 0^{\circ}$	$q / p > 0$ so $arctan$ gives $(0, 9 0^{\circ})$ but true angle is $18 0^{\circ}$ ahead; add it
$(-, +)$	IV	$-$	$+$	$arctan (q / p) + 36 0^{\circ}$ (or $-	...

Boundary / axis cases (where the ratio $q / p$ is ill-posed):

point	which is $0$	$tan Q = q / p$ ?	correct $Q$	reason
$q = 0, p > 0$	$sin Q = 0$	$0/ p = 0$ , OK	$Q = 0^{\circ}$	on $+ p$ axis; $cos Q > 0$
$q = 0, p < 0$	$sin Q = 0$	$0/ p = 0$ , OK	$Q = 18 0^{\circ}$	on $- p$ axis; $cos Q < 0$
$q > 0, p = 0$	$cos Q = 0$	$q /0 = \infty$ undefined	$Q = 9 0^{\circ}$	division fails; read from $sin Q > 0$
$q < 0, p = 0$	$cos Q = 0$	$q /0 = \infty$ undefined	$Q = 27 0^{\circ}$	division fails; read from $sin Q < 0$
$q = 0, p = 0$	both	$0/0$ indeterminate	$Q$ undefined	origin: $P = 0$ , angle meaningless (see Ex 4)

Why the axis cases matter: the branch-fix procedure of step 4 divides by $p$ ; on the $p = 0$ axis that division blows up, so you must read the angle straight off the surviving sign of $sin Q$ (i.e. of $q$ ) instead of trusting the ratio.

Verify: reconstruct — $q = 2 \cdot 1 sin 22 5^{\circ} = 2 \cdot (- \frac{1}{2}) = - 1$ ✓, $p = 2 cos 22 5^{\circ} = - 1$ ✓.

Figure 2 turns those sign rules into a schematic map of the four quadrants plus the four axis cases.

Example 4 — Cell D: degenerate / zero input

Worked example What happens at

$Q = 0$ in the oscillator generator? Using $F_{1} = \frac{1}{2} mω q^{2} cot Q$ , examine the point where $Q \to 0$ .

Forecast: $cot Q = cos Q / sin Q \to \infty$ as $Q \to 0$ . Something blows up. But the physical answer ( $q, p$ ) should stay finite — this is a coordinate artefact, like the North Pole on a map of Earth.

Steps.

Recall $q = 2 P / (mω) sin Q$ . As $Q \to 0$ , $sin Q \to 0$ , so $q \to 0$ . Why? the sine factor forces the position to zero regardless of $P$ .
And $p = 2 P mω cos Q \to 2 P mω$ (maximal). Why? $cos 0 = 1$ ; the particle is at the origin moving fastest — exactly the oscillator passing through $q = 0$ .
But the generator itself: $F_{1} \propto q^{2} cot Q$ . With $q = 2 P / (mω) sin Q$ , $q^{2} cot Q = \frac{2 P}{mω} sin^{2} Q \cdot \frac{cos Q}{sin Q} = \frac{2 P}{mω} sin Q cos Q \to 0$ . Why this step? the apparent blow-up cancels: $sin^{2} Q$ beats the single $1/ sin Q$ . The transformation is perfectly regular at $Q = 0$ ; only the formula's naive form looks singular.

Verify (limit): $Q \to 0 lim q^{2} cot Q$ with $q = 2 P sin Q$ ( $m = ω = 1$ ) $= lim_{Q \to 0} 2 P sin Q cos Q = 0$ . ✓ No true singularity.

Where it does break: the Jacobian $\partial (Q, P) / \partial (q, p)$ must be nonzero for the map to be locally invertible. At $q = 0, p = 0$ (the origin, $P = 0$ ) the angle $Q$ is undefined — that single point is the genuine coordinate singularity, just like longitude at a pole.

Example 5 — Cell E: time-dependent generator, so $K \neq = H$

Worked example A uniformly accelerating frame via

$F_{2} = (q - \frac{1}{2} a t^{2}) P + ma t q$ A free particle has $H = \frac{p ^{2}}{2 m}$ . Use the above $F_{2}$ (with constant acceleration $a$ , and $t$ the independent time coordinate every variable evolves in) and find the new Hamiltonian $K$ . Here $Q$ is the position in a frame accelerating at $a$ .

Forecast: because $F_{2}$ depends explicitly on $t$ , the parent's rule $K = H + \partial F / \partial t$ means $K \neq = H$ . The accelerating frame should introduce a fictitious force term (something linear in $Q$ ).

Steps.

Type-2 relations: $p = \partial F_{2} / \partial q$ , $Q = \partial F_{2} / \partial P$ , $K = H + \partial F_{2} / \partial t$ . Why? $F_{2} (q, P, t)$ ; the last term appears because $F_{2}$ depends on the time coordinate $t$ explicitly.
$p = \frac{\partial F _{2}}{\partial q} = P + ma t$ . Why? $\frac{\partial}{\partial q} [(q - \frac{1}{2} a t^{2}) P] = P$ and $\frac{\partial}{\partial q} [ma tq] = ma t$ . So $P = p - ma t$ (momentum measured relative to the moving frame).
$Q = \frac{\partial F _{2}}{\partial P} = q - \frac{1}{2} a t^{2}$ . Why? only the first bracket carries $P$ . So $Q$ is the position minus the frame's displacement $\frac{1}{2} a t^{2}$ — exactly "position in the accelerating frame."
$\frac{\partial F _{2}}{\partial t} = (- a t) P + ma q = - a tP + ma q$ . Why? differentiate the explicit $t$ 's (holding $q, P$ fixed as the partial demands): $\frac{\partial}{\partial t} (- \frac{1}{2} a t^{2}) P = - a tP$ , and $\frac{\partial}{\partial t} (ma tq) = ma q$ .
Assemble $K = H + \partial_{t} F_{2} = \frac{p ^{2}}{2 m} - a tP + ma q$ . Now rewrite in new variables ( $p = P + ma t$ , $q = Q + \frac{1}{2} a t^{2}$ ): $K = \frac{( P + ma t ) ^{2}}{2 m} - a tP + ma (Q + \frac{1}{2} a t^{2}) .$ Expand $\frac{( P + ma t ) ^{2}}{2 m} = \frac{P ^{2}}{2 m} + a tP + \frac{m a ^{2} t ^{2}}{2}$ . The $+ a tP$ cancels the $- a tP$ : $K = \frac{P ^{2}}{2 m} + ma Q + m a^{2} t^{2} (\frac{1}{2} m a^{2} t^{2} + \frac{1}{2} m a^{2} t^{2}) .$ Why this step? we must express $K$ in its own variables to read off dynamics; the leftover pure-time term $m a^{2} t^{2}$ adds no force (its $Q, P$ derivatives vanish) and can be dropped.
Effective Hamiltonian: $K = \frac{P ^{2}}{2 m} + ma Q$ . The term $ma Q$ is a linear potential = fictitious force $- ma$ — precisely what you feel in an accelerating car. Why dropping $m a^{2} t^{2}$ is legitimate: a term depending on $t$ alone is the total time derivative of the function $\int m a^{2} t^{2} d t$ , and adding a total time derivative to a Hamiltonian is a canonical gauge change — it enters the action only as an endpoint term, so it leaves $\dot{Q} = \partial K / \partial P$ and $\dot{P} = - \partial K / \partial Q$ completely unchanged (its $Q$ - and $P$ -derivatives are zero). No dynamics is lost.

Verify: Hamilton in new vars: $\dot{Q} = \partial K / \partial P = P / m$ , $\dot{P} = - \partial K / \partial Q = - ma$ . So in the frame the particle accelerates at $- a$ — the correct pseudo-force. ✓ (See Hamilton–Jacobi equation for the extreme case that drives $K \to 0$ .)

Example 6 — Cell F: no generator given, prove canonicity by Poisson brackets

Worked example Is

$Q = ln (\frac{s i n p}{q}), P = q cot p$ canonical? No $F$ supplied. Test directly.

Forecast: these look engineered to work out — bet on ${Q, P} = 1$ .

Domain first. For $Q = ln (sin p / q)$ to be a real number we need the argument of the log positive: $\frac{sin p}{q} > 0$ . Restrict to the natural patch $q > 0$ and $0 < p < π$ (so $sin p > 0$ ); then $cot p$ in $P$ is finite except at $p = 0, π$ (where $sin p = 0$ and both $Q$ and $P$ blow up). Why this matters: the map is only a valid canonical transformation on the open region where its formulas are smooth and invertible — the log and the $cot$ each carve out that region. On the boundary $p \to 0^{+}$ or $p \to π^{-}$ , $Q \to + \infty$ : an edge, not part of the chart.

Steps.

The only algebraic test of canonicity is ${Q, P} = 1$ (one degree of freedom). Why this step? preserving the fundamental Poisson bracket ${Q, P} = 1$ is equivalent to preserving Hamilton's equations — see Poisson brackets.
Write $Q = ln sin p - ln q$ (valid because both factors are positive on our patch). Partials: $\frac{\partial Q}{\partial q} = - \frac{1}{q}, \frac{\partial Q}{\partial p} = cot p$ . Why? chain rule on each log.
$P = q cot p$ : $\frac{\partial P}{\partial q} = cot p, \frac{\partial P}{\partial p} = - q csc^{2} p$ . Why? $\frac{d}{d p} cot p = - csc^{2} p$ .
${Q, P} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} = (- \frac{1}{q}) (- q csc^{2} p) - (cot p) (cot p) = csc^{2} p - cot^{2} p = 1.$ Why the last equality? the identity $csc^{2} p - cot^{2} p = 1$ (Pythagoras divided by $sin^{2}$ ).

Verify: ${Q, P} = 1$ on the whole patch $q > 0, 0 < p < π$ . ✓ Canonical there.

Example 7 — Cell G: real-world word problem with numbers & units

Worked example A mass on a spring, solved by generator

A block of mass $m = 2 kg$ on a spring of constant $k = 8 N/m$ starts at $q_{0} = 0.10 m$ , at rest. Use the oscillator generator to find the amplitude in action–angle form and the period.

Forecast: angular frequency $ω = k / m = 2 rad/s$ , so period $\approx 3.14 s$ . Energy is all potential at the start: $E = \frac{1}{2} k q_{0}^{2}$ .

Steps.

$ω = k / m = 8/2 = 2 rad/s$ . Why? the harmonic Hamiltonian $H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} q^{2}$ demands $m ω^{2} = k$ , so $ω = k / m$ .
Total energy: at rest at $q_{0}$ , $E = \frac{1}{2} k q_{0}^{2} = \frac{1}{2} (8) (0.10)^{2} = 0.04 J$ . Why? $p = 0$ initially, so $H$ = spring potential energy only.
From the parent, $K = ω P$ and $K = E$ (energy conserved, $F$ time-independent so $K = H = E$ ). Hence the action $P = E / ω = 0.04/2 = 0.02 J\cdots$ . Why this step? $Q$ is cyclic ⟹ $P$ is a conserved constant equal to $E / ω$ .
Angle evolves as $Q = ω t + ϕ$ ( $t$ = the independent time coordinate). Period $T = 2 π / ω = 2 π /2 = π \approx 3.1416 s$ . Why? $Q$ advances by $2 π$ each cycle at constant rate $\dot{Q} = ω$ .
Amplitude check via $q = 2 P / (mω) sin Q$ : max $q_{m a x} = 2 (0.02) / (2 \cdot 2) = 0.01 = 0.10 m = q_{0}$ . Why? started at rest at the turning point, so $q_{0}$ is the amplitude.

Verify (units & numbers): $[P] = J\cdots$ (action, correct dimension); $ω = 2 rad/s$ , $E = 0.04 J$ , $P = 0.02 J\cdots$ , $T = π s \approx 3.14 s$ , amplitude $= 0.10 m$ . ✓ See Action–angle variables.

Example 8 — Cell H: the exam twist (spot the non-canonical map)

Worked example Trap: is

$Q = q^{2}, P = p$ canonical? If not, fix the momentum. Forecast: doubling-type maps usually break the bracket. Bet it's not canonical, and the repair scales $P$ by $1/ (2 q)$ .

Steps.

Test ${Q, P}$ . $\frac{\partial Q}{\partial q} = 2 q, \frac{\partial Q}{\partial p} = 0, \frac{\partial P}{\partial q} = 0, \frac{\partial P}{\partial p} = 1$ . Why? these are the only nonzero partial derivatives of $Q = q^{2}$ and $P = p$ .
${Q, P} = (2 q) (1) - (0) (0) = 2 q \neq = 1$ . Why this matters? a canonical map must give exactly $1$ ; $2 q$ fails everywhere except $q = \frac{1}{2}$ . So $Q = q^{2}, P = p$ is not canonical.
Repair — fix the momentum. Keep $Q = q^{2}$ but seek $P = f (q) p$ with ${Q, P} = 1$ . Then ${q^{2}, f (q) p} = \frac{\partial ( q ^{2} )}{\partial q} \frac{\partial ( f ( q ) p )}{\partial p} = (2 q) f (q) = 1 \Rightarrow f (q) = \frac{1}{2 q}$ . Why this step? we solve for the multiplier that forces the bracket to $1$ ; only a $q$ -dependent scaling of $p$ can cancel the offending factor $2 q$ .
So the canonical version is $Q = q^{2}, P = \frac{p}{2 q}$ . Why this is the fix promised in the statement: the coordinate change $Q = q^{2}$ stays, and only the momentum is rescaled by $1/ (2 q)$ to restore ${Q, P} = 1$ .

Verify: ${q^{2}, \frac{p}{2 q}} = \frac{\partial ( q ^{2} )}{\partial q} \frac{\partial}{\partial p} (\frac{p}{2 q}) - 0 = (2 q) (\frac{1}{2 q}) = 1$ . ✓

Sanity via a generator: this is exactly Type-2 $F_{2} = q^{2} P$ . Check: $Q = \partial_{P} F_{2} = q^{2}$ ✓ and $p = \partial_{q} F_{2} = 2 q P \Rightarrow P = p / (2 q)$ ✓. Consistent. Figure 3 plots ${Q, P} = 2 q$ against the required value $1$ , showing the single accidental crossing at $q = \frac{1}{2}$ .

Active recall

Recall Test yourself (hide answers)

In Ex 3, why can't $arctan (q / p)$ alone give $Q$ , and what happens on the $p = 0$ axis?
In Ex 4, does $F_{1}$ blowing up at $Q = 0$ mean the transformation is singular there? Where is the one true singularity?
In Ex 5, which term of $K$ is the fictitious force, and why can we drop the pure-time term?
In Ex 6, what domain makes $Q = ln (sin p / q)$ well-defined?
In Ex 8, what multiplier repairs $Q = q^{2}$ ?

For which point does

Q = q^{2}, P = p

accidentally satisfy the bracket?

Only

q = \frac{1}{2}

, where

{Q, P} = 2 q = 1

Action of the

m = 2, k = 8

block?

P = E / ω = 0.04/2 = 0.02 J\cdots

Period of that block?

T = 2 π / ω = π \approx 3.14 s

Effective Hamiltonian in Ex 5?

K = \frac{P ^{2}}{2 m} + ma Q

, giving pseudo-force

- ma

Why can a pure-time term be dropped from

K

It is a total time derivative — a canonical gauge change — so its

Q, P

derivatives vanish and dynamics is unchanged.

On the

p = 0

axis, why does the arctan branch-fix fail?

It divides by

p

; instead read

Q

straight from the sign of

sin Q

(i.e. of

q

q > 0 \Rightarrow 9 0^{\circ}

q < 0 \Rightarrow 27 0^{\circ}

Where is the one true coordinate singularity of the oscillator action–angle map?

The origin

q = p = 0

(

P = 0

), where the angle

Q

is undefined.

Domain that makes Ex 6's map well-defined?

q > 0

and

0 < p < π

(so

sin p / q > 0

for the log and

cot p

finite).

Connections

Poisson brackets — the ${Q, P} = 1$ test used in Ex 6 and Ex 8.
Hamilton's equations — the form each example preserves.
Action–angle variables — where Ex 3 and Ex 7 lead.
Hamilton–Jacobi equation — the ultimate time-dependent generator (Ex 5's cousin).
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