Canonical transformations — generating functions: Question bank

Before the traps, we rebuild the three ideas the whole page leans on — the master relation, the four generator types, and the phase-space picture — so no symbol is ever used before it is earned.

Formula The four generator types (what "

$F_{1}, \dots, F_{4}$ " mean) $F$ must depend on one old and one new variable per degree of freedom, giving exactly four choices. Multiply the box by $d t$ and match coefficients of the independent differentials:

F_{1} (q, Q) : F_{2} (q, P) : F_{3} (p, Q) : F_{4} (p, P) : p_{i} = \frac{\partial F _{1}}{\partial q _{i}}, p_{i} = \frac{\partial F _{2}}{\partial q _{i}}, q_{i} = - \frac{\partial F _{3}}{\partial p _{i}}, q_{i} = - \frac{\partial F _{4}}{\partial p _{i}}, P_{i} = - \frac{\partial F _{1}}{\partial Q _{i}}, Q_{i} = + \frac{\partial F _{2}}{\partial P _{i}}, P_{i} = - \frac{\partial F _{3}}{\partial Q _{i}}, Q_{i} = + \frac{\partial F _{4}}{\partial P _{i}}, K = H + \frac{\partial F _{1}}{\partial t} K = H + \frac{\partial F _{2}}{\partial t} K = H + \frac{\partial F _{3}}{\partial t} K = H + \frac{\partial F _{4}}{\partial t}

$F_{2}, F_{3}, F_{4}$ are reached from $F_{1}$ by a Legendre transformation (swapping a variable for its conjugate, e.g. $P d Q = d (P Q) - Q d P$ ). Notice the pattern: the variable appearing only inside $F$ (never as the "free" argument) carries the minus in Types 1 and 3.

The picture below is the arena for every question that follows: phase space, the flat sheet whose two axes are $q$ (horizontal) and $p$ (vertical). A state of the system is one dot; as time runs the dot traces a curve. A CT redraws the grid on this sheet without tearing it.

Figure — Canonical transformations — generating functions

The one geometric fact that makes CTs special is that they preserve oriented area on this sheet. The next figure shows a little patch of states being carried by a CT: it can stretch and shear, but its area is unchanged — this is the visual meaning of the Poisson bracket test ${Q, P} = 1$ and of Liouville's theorem.

True or false — justify

TF1. "Every invertible smooth map

(q, p) \to (Q, P)

is canonical."

False — invertibility is not enough; the map must preserve the symplectic (area) structure, equivalently

{Q, P} = 1

. Most random smooth maps stretch area unevenly and Hamilton's equations break.

TF2. "If the generating function

F

has no explicit time dependence, then

K = H

as functions on phase space."

True — since

K = H + \partial F / \partial t

and

\partial F / \partial t = 0

, the value of the new Hamiltonian equals the old one at each phase point (just re-expressed in

Q, P

TF3. "A canonical transformation cannot change the numerical value of the energy at a point."

False in general — when

F

depends on

t

K \neq = H

, so the new Hamiltonian's value differs; energy conservation as a concept survives but the number attached to it can change.

TF4. "The four generator types

F_{1}

–

F_{4}

describe four different kinds of canonical transformation."

False — they describe the same CTs written with different independent variables; the type is a bookkeeping choice about which old/new pair is treated as free.

TF5. "

F_{2} = \sum_{i} q_{i} P_{i}

generates the identity transformation."

True — the Type-2 rules give

p_{i} = \partial F_{2} / \partial q_{i} = P_{i}

and

Q_{i} = \partial F_{2} / \partial P_{i} = q_{i}

, i.e. nothing changes; this "do-nothing" generator is the base point from which infinitesimal CTs grow.

TF6. "Because

Q = p

P = - q

is canonical, position and momentum play interchangeable roles."

True — with

F_{1} = \sum q_{i} Q_{i}

the Type-1 rules give

p_{i} = \partial F_{1} / \partial q_{i} = Q_{i}

(so

Q_{i} = p_{i}

) and

P_{i} = - \partial F_{1} / \partial Q_{i} = - q_{i}

; the minus is forced by the boxed relation's right-hand

P d Q

, and geometrically it is the

9 0^{\circ}

rotation of the

q

–

p

sheet (rotations preserve area, so it is canonical), showing phase space has no built-in "position" (see Symplectic geometry).

TF7. "The generating-function method might accidentally produce a non-canonical map."

False — any transformation derived from a valid generator is canonical by construction, because it descends from the shared variational principle (the boxed master relation) that fixes Hamilton's form.

TF8. "Canonical transformations preserve phase-space volume."

True — they preserve the symplectic area of figure s02, hence volume in higher dimensions (this is Liouville's theorem viewed through the CT lens).

TF9. "A time-dependent CT that makes

K = 0

leaves the system with no dynamics."

False —

K = 0

means the new variables are all constants of motion; the dynamics is fully encoded in the transformation itself, which is exactly the Hamilton–Jacobi equation strategy.

Spot the error

SE1. "Type-1 rule:

p_{i} = \partial F_{1} / \partial q_{i}

and

P_{i} = + \partial F_{1} / \partial Q_{i}

The sign is wrong:

P_{i} = - \partial F_{1} / \partial Q_{i}

. The minus arises because

P_{i} d Q_{i}

sits on the right of the master relation, so moving it across flips its sign.

SE2. "For

F_{2} (q, P)

we get

Q_{i} = - \partial F_{2} / \partial P_{i}

Wrong sign again:

Q_{i} = + \partial F_{2} / \partial P_{i}

. Type 2 was built by a Legendre swap (

P d Q = d (P Q) - Q d P

) precisely to make this term come out positive — as listed in the four-types box.

SE3. "We can freely choose

F_{1} = F_{1} (q, p, t)

, mixing an old coordinate with its own old momentum."

A generator must depend on one old and one new variable per degree of freedom;

F_{1} (q, p)

uses two old variables and cannot close the transformation equations.

SE4. "Since

K = H + \partial F / \partial t

, adding a constant to

F

changes

K

A constant has zero time-derivative and zero spatial derivatives, so it changes neither

K

nor the transformation; only genuine dependence on

q, Q, t

matters.

SE5. "In the oscillator example

F_{1} = \frac{1}{2} mω q^{2} cot Q

, we solved

H

by integrating a differential equation."

No differential equation was solved — the whole point is that the generator makes

Q

cyclic (

K = ω P

), so the motion

Q = ω t + ϕ

P = const

follows purely algebraically.

SE6. "A point transformation

Q = Q (q)

is only canonical for linear

Q (q)

Any invertible point transformation is canonical once the momenta transform correctly (

P = p \partial q / \partial Q

); nonlinearity is fine, it is generated by

F_{2} = \sum P_{i} Q_{i} (q)

SE7. "Because Poisson brackets are invariant under CTs, computing

{Q, P}

can never test canonicity."

Invariance is why the bracket is a test: a proposed map is canonical iff

{Q, P} = 1

(and

{Q, Q} = {P, P} = 0

) evaluated in the old variables using the definition above.

Why questions

WHY1. Why must the two integrands differ by a total time derivative

d F / d t

, not merely a constant?

A total time derivative integrates to a boundary term whose variation vanishes at fixed endpoints, so it leaves the equations of motion untouched; a mere constant is a special (trivial) case that cannot generate nontrivial variable changes.

WHY2. Why is the generating function a single scalar function rather than a set of formulas for each new variable?

All the transformation equations are partial derivatives of the one scalar

F

, which automatically guarantees the integrability/consistency (mixed partials match) that keeps the map canonical.

WHY3. Why does the Legendre transformation appear when passing from

F_{1}

F_{2}

Switching the free variable from

Q

to its conjugate

P

is exactly a change of "natural variable," which is what a Legendre transformation does; we subtract

\sum Q_{i} P_{i}

to trade

d Q

for

d P

WHY4. Why can a clever CT make a Hamiltonian's coordinate cyclic, and why is that a win?

Q

does not appear in

K

then

\dot{P} = - \partial K / \partial Q = 0

, so

P

is conserved and

\dot{Q}

is constant — the motion becomes trivial straight-line drift, the seed of Action–angle variables.

WHY5. Why are

q

and

p

treated as independent differentials when matching coefficients?

In the master relation

q, Q, t

(Type 1) are chosen as independent free variables, so their differentials can be varied separately; only then must each coefficient match on both sides.

WHY6. Why does the generating-function trick fail if we insist

F

depend on both members of a conjugate pair simultaneously?

Then the transformation equations become degenerate — you cannot solve for all new variables in terms of old ones — because the pair is not independent under the symplectic constraint.

Edge cases

EC1. What happens to the Type-1 relations if

F_{1}

is independent of

Q

Then

P_{i} = - \partial F_{1} / \partial Q_{i} = 0

, forcing all new momenta to vanish — a degenerate, non-invertible map that is not a valid CT.

EC2. Is the swap

Q = p

P = - q

its own inverse?

Applying it twice gives

Q \to - q

P \to - p

, i.e. a

18 0^{\circ}

rotation in phase space, not the identity — so it is order-four, a nice reminder that CTs form a group.

EC3. What is the "smallest" nontrivial CT near the identity?

An infinitesimal one,

F_{2} = \sum q_{i} P_{i} + ϵ G (q, P)

; to first order it shifts variables by

{\cdot, G}

, generating the Hamiltonian flow of

G

itself.

EC4. If

H

is already zero, is any transformation canonical?

The dynamics is trivial (

\overset{q}{˙} = \overset{p}{˙} = 0

), but a map is still canonical only if it preserves the symplectic form; canonicity is a property of the map, independent of whichever

H

you carry.

EC5. Does a time-dependent CT preserve energy conservation?

Not necessarily — with

\partial F / \partial t \neq = 0

K \neq = H

and

K

may itself depend on time, so the conserved quantity (if any) is

K

, not the original

H

EC6. What is the limiting behaviour of the oscillator generator

F_{1} = \frac{1}{2} mω q^{2} cot Q

Q \to 0

Using

q = 2 P / mω sin Q

, the product

q cot Q = 2 P / mω cos Q

stays finite, so

p = mω q cot Q \to 2 mω P

(its maximum) while

q \to 0

— the turning point where all energy is kinetic, with

P

finite throughout.

Recall One-line self-audit

Can I state the master relation from memory and locate every sign? ::: $\sum p \overset{q}{˙} - H = \sum P \dot{Q} - K + d F / d t$ ; the minus on $P$ (Types 1,3) comes from $P d Q$ living on the right.
Do I know the one algebraic test of canonicity? ::: Poisson brackets ${Q, P} = 1$ , ${Q, Q} = {P, P} = 0$ .
Can I explain why cyclic $Q$ is the goal? ::: $\dot{P} = 0$ then $Q$ drifts linearly — the whole trick of Hamilton–Jacobi and action–angle variables.

Connections

Hamilton's equations — the form CTs preserve.
Poisson brackets — the algebraic canonicity test.
Legendre transformation — the swap behind Types 2–4.
Action–angle variables — where cyclic- $Q$ generators lead.
Hamilton–Jacobi equation — the $K = 0$ time-dependent CT.
Liouville's theorem — volume preservation under CTs.
Symplectic geometry — why $q$ and $p$ are on equal footing.