In Hamiltonian mechanics, canonical transformations are important because they preserve the form of Hamilton’s equations. If the old canonical variables $(q_i,p_i)$ are replaced by new variables $(Q_i,P_i)$, the transformation is canonical only when the new variables also satisfy Hamilton’s equations in the same structural form. This allows us to change variables without changing the basic geometry of mechanics.
The deeper use of canonical transformations appears when the change is very small. Infinitesimal canonical transformations reveal how symmetries act in phase space. When such infinitesimal transformations are repeated, they generate finite Lie transformations. When the generator of the transformation leaves the Hamiltonian invariant, the same generator becomes a conserved quantity. This is the Hamiltonian form of the symmetry-conservation connection.
Thus the flow of ideas is:
- canonical transformations preserve Hamiltonian structure,
- infinitesimal canonical transformations are generated by phase-space functions,
- Lie transformations are finite transformations built from infinitesimal ones,
- the Hamiltonian generates time translation,
- symmetry generators give conserved quantities,
- action-angle variables simplify bounded periodic motion.
Consider a type-$2$ generating function close to the identity transformation:
\[F_2(q_i,P_i,t)=\sum_i q_iP_i+\varepsilon G(q_i,P_i,t)\]
where $\varepsilon$ is a small parameter and $G$ is the infinitesimal generator. The transformation equations are
\[p_i=\frac{\partial F_2}{\partial q_i},
\qquad
Q_i=\frac{\partial F_2}{\partial P_i}.\]
Using the given $F_2$,
\[p_i=P_i+\varepsilon\frac{\partial G}{\partial q_i}\]
and
\[Q_i=q_i+\varepsilon\frac{\partial G}{\partial P_i}.\]
Since the transformation is infinitesimal, we may replace $P_i$ by $p_i$ in the first-order correction. Hence
\[P_i=p_i-\varepsilon\frac{\partial G}{\partial q_i}\]
and
\[Q_i=q_i+\varepsilon\frac{\partial G}{\partial p_i}.\]
Therefore the infinitesimal changes are
\[\delta q_i=Q_i-q_i=\varepsilon\frac{\partial G}{\partial p_i},
\qquad
\delta p_i=P_i-p_i=-\varepsilon\frac{\partial G}{\partial q_i}.\]
These are exactly the canonical transformation rules generated by $G$. They show that a single function $G(q,p,t)$ determines how every coordinate and momentum changes.
For any phase-space function $f(q_i,p_i,t)$, the infinitesimal change is
\[\delta f
=
\sum_i\left(
\frac{\partial f}{\partial q_i}\delta q_i
+
\frac{\partial f}{\partial p_i}\delta p_i
\right).\]
Substituting $\delta q_i$ and $\delta p_i$,
\[\delta f
=
\varepsilon
\sum_i
\left(
\frac{\partial f}{\partial q_i}\frac{\partial G}{\partial p_i}
-
\frac{\partial f}{\partial p_i}\frac{\partial G}{\partial q_i}
\right).\]
The expression in brackets is the Poisson bracket ${f,G}$. Therefore
\[\boxed{
\delta f=\varepsilon\{f,G\}
}\]
This is the central formula of infinitesimal canonical transformations. It says that the generator $G$ gives the direction of motion of every phase-space function through the Poisson bracket.
Example: coordinate translation
Let the generator be
$$
G=p.
$$
Then for one degree of freedom,
$$
\delta q=\varepsilon\{q,p\}=\varepsilon,
\qquad
\delta p=\varepsilon\{p,p\}=0.
$$
Hence
$$
q'=q+\varepsilon,
\qquad
p'=p.
$$
Thus the generator $p$ produces a translation in the coordinate $q$. This is why momentum is called the generator of spatial translation.
An infinitesimal transformation gives only a very small change. But if the same infinitesimal transformation is applied repeatedly, it builds a finite transformation. This finite transformation is called a Lie transformation.
Define the Lie operator $L_G$ by
\[L_G f=\{f,G\}.\]
Then one infinitesimal transformation gives
\[f\longrightarrow f+\varepsilon L_G f.\]
Repeated application gives the exponential form
\[f'=\exp(\varepsilon L_G)f.\]
Expanding the exponential,
\[f'
=
f+\varepsilon L_Gf+\frac{\varepsilon^2}{2!}L_G^2f+\frac{\varepsilon^3}{3!}L_G^3f+\cdots.\]
Since $L_Gf=\{f,G\}$, this becomes
\[f'
=
f+\varepsilon\{f,G\}
+\frac{\varepsilon^2}{2!}\{\{f,G\},G\}
+\frac{\varepsilon^3}{3!}\{\{\{f,G\},G\},G\}
+\cdots.\]
Thus a Lie transformation is not a separate kind of transformation. It is the finite transformation obtained by integrating or repeating the infinitesimal canonical transformation generated by $G$.
This is especially useful in perturbation theory. Instead of trying to find one large exact transformation, one chooses a generator that removes or simplifies undesirable terms order by order. In this way, Lie transformations provide a systematic method for simplifying Hamiltonians.
Example: finite coordinate translation
Again take
$$
G=p.
$$
For the coordinate $q$,
$$
L_Gq=\{q,p\}=1.
$$
Applying the operator again,
$$
L_G^2q=L_G(1)=\{1,p\}=0.
$$
Therefore all higher terms vanish, and
$$
q'=\exp(\varepsilon L_G)q=q+\varepsilon.
$$
For the momentum $p$,
$$
L_Gp=\{p,p\}=0.
$$
Therefore
$$
p'=\exp(\varepsilon L_G)p=p.
$$
Hence the finite Lie transformation generated by $G=p$ is
$$
\boxed{
q'=q+\varepsilon,\qquad p'=p.
}
$$
This confirms that the generator $p$ produces a finite translation of the coordinate.
The Hamiltonian as the generator of time translation
The Hamiltonian has a special role because it generates time evolution. Take the generator to be
\[G=H\]
and choose the small parameter to be
\[\varepsilon=dt.\]
Then
\[\delta q_i=dt\{q_i,H\}\]
and
\[\delta p_i=dt\{p_i,H\}.\]
Using the definition of the Poisson bracket,
\[\{q_i,H\}=\frac{\partial H}{\partial p_i}\]
and
\[\{p_i,H\}=-\frac{\partial H}{\partial q_i}.\]
Therefore
\[\delta q_i=dt\frac{\partial H}{\partial p_i},
\qquad
\delta p_i=-dt\frac{\partial H}{\partial q_i}.\]
Dividing by $dt$,
\[\boxed{
\dot q_i=\frac{\partial H}{\partial p_i},
\qquad
\dot p_i=-\frac{\partial H}{\partial q_i}.
}\]
These are Hamilton’s equations. Therefore the Hamiltonian is the generator of time translation in phase space.
This result gives a powerful interpretation of dynamics: time evolution itself is a canonical transformation generated by the Hamiltonian.
For any function $f(q,p,t)$,
\[\frac{df}{dt}
=
\{f,H\}+\frac{\partial f}{\partial t}.\]
If $f$ has no explicit time dependence, then
\[\frac{df}{dt}=\{f,H\}.\]
So the Hamiltonian flow tells how every observable changes with time.
A symmetry is a transformation that leaves the physical structure of the system unchanged. In Hamiltonian mechanics, a continuous symmetry is represented by a generator $G(q,p,t)$. The infinitesimal change of the Hamiltonian under this transformation is
\[\delta H=\varepsilon\{H,G\}.\]
If the Hamiltonian is invariant under the transformation, then
\[\delta H=0.\]
Therefore
\[\{H,G\}=0.\]
Since the Poisson bracket is antisymmetric,
\[\{H,G\}=-\{G,H\}.\]
So the invariance condition may also be written as
\[\{G,H\}=0.\]
Now the total time derivative of $G$ is
\[\frac{dG}{dt}
=
\{G,H\}+\frac{\partial G}{\partial t}.\]
If $G$ has no explicit time dependence and the Hamiltonian is invariant under the transformation generated by $G$, then
\[\frac{dG}{dt}=0.\]
Therefore
\[\boxed{
G=\text{constant}.
}\]
This is the Hamiltonian form of Noether’s theorem:
\[\boxed{
\text{continuous symmetry}
\quad\Longrightarrow\quad
\text{conserved generator}.
}\]
The important point is that a conserved quantity is not merely a number that accidentally remains constant. It is the generator of a continuous canonical symmetry.
The logic is:
- $G$ generates an infinitesimal canonical transformation,
- the change of any phase-space function is $\delta f=\varepsilon{f,G}$,
- if the Hamiltonian is invariant under this transformation, then ${G,H}=0$,
- if $G$ has no explicit time dependence, then $\frac{dG}{dt}=0$,
- therefore $G$ is conserved.
Examples of symmetry generators
1. Spatial translation
If
$$
G=p,
$$
then
$$
\delta q=\varepsilon,
\qquad
\delta p=0.
$$
So $p$ generates spatial translation. If the Hamiltonian does not change under spatial translation, then
$$
\{p,H\}=0
$$
and therefore
$$
p=\text{constant}.
$$
Thus translational symmetry gives conservation of momentum.
2. Rotation about the $z$-axis
If
$$
G=L_z=xp_y-yp_x,
$$
then $G$ generates rotation about the $z$-axis. If the Hamiltonian is invariant under this rotation, then
$$
\{L_z,H\}=0
$$
and therefore
$$
L_z=\text{constant}.
$$
Thus rotational symmetry gives conservation of angular momentum.
3. Time translation
If the Hamiltonian has no explicit time dependence, then
$$
\frac{\partial H}{\partial t}=0.
$$
Using
$$
\frac{dH}{dt}
=
\{H,H\}+\frac{\partial H}{\partial t},
$$
and since
$$
\{H,H\}=0,
$$
we get
$$
\frac{dH}{dt}=0.
$$
Thus invariance under time translation gives conservation of energy.
Action-angle variables
Action-angle variables are special canonical variables designed for bounded periodic motion. They are useful because they convert complicated periodic motion into uniform motion.
The canonical transformation is
\[(q_i,p_i)\longrightarrow(\theta_i,J_i),\]
where $J_i$ are the action variables and $\theta_i$ are the corresponding angle variables.
The action variable tells which orbit the system is moving on. The angle variable tells where the system is on that orbit.
For one degree of freedom, periodic motion means
\[q(t+T)=q(t),
\qquad
p(t+T)=p(t).\]
Therefore the phase-space trajectory is a closed curve in the $(q,p)$ plane. Since the orbit is closed, we define the action variable by the closed integral
\[\boxed{
J=\frac{1}{2\pi}\oint p\,dq.
}\]
The integral $\oint p\,dq$ measures the phase-space area enclosed by the closed orbit. Since it is taken over the whole loop, it does not depend on the instantaneous position of the particle. It depends only on the orbit itself.
The factor $1/2\pi$ is introduced because the conjugate coordinate is an angle variable, and an angle increases by $2\pi$ in one complete cycle.
Thus:
- $J$ labels the closed orbit,
- $\theta$ labels the position on that orbit,
- $J$ behaves like a momentum variable,
- $\theta$ behaves like a coordinate variable.
For action-angle variables, the new Hamiltonian is chosen to depend only on $J$:
\[K=K(J).\]
Then Hamilton’s equations become
\[\dot J=-\frac{\partial K}{\partial \theta}=0\]
and
\[\dot\theta=\frac{\partial K}{\partial J}=\omega(J).\]
Therefore
\[\boxed{
J=\text{constant},
\qquad
\theta(t)=\omega(J)t+\theta_0.
}\]
This is the main advantage of action-angle variables. The original motion may be complicated in $(q,p)$, but in $(\theta,J)$ it becomes uniform motion with constant angular speed.
Why the action variable is introduced
The formula
\[J=\frac{1}{2\pi}\oint p\,dq\]
should not be introduced as an isolated formula. It is meaningful only for systems with bounded periodic motion.
Such systems have the following features:
- the motion is bounded,
- the motion repeats after a period,
- the phase trajectory is closed,
- different energies correspond to different closed phase-space loops,
- we want one variable that labels the whole orbit.
For one degree of freedom, the instantaneous state is represented by $(q,p)$. During periodic motion, the phase point moves around a closed curve. Instead of asking where the particle is at a particular instant, we may ask which closed curve it is moving on. The action variable answers this second question.
The closed integral
\[\oint p\,dq\]
is natural because it is taken around the complete orbit. It measures the whole loop, not one point on the loop. Therefore it remains the same throughout the motion.
So the conceptual shift is:
\[(q,p)
\quad\text{describes instantaneous state,}\]
whereas
\[J
\quad\text{describes the whole periodic orbit.}\]
The conjugate angle variable $\theta$ then describes the running phase along the orbit.
Simple harmonic oscillator
Consider the one-dimensional harmonic oscillator with Hamiltonian
\[H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2.\]
For a fixed energy $E$,
\[E=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2.\]
Solving for $p$,
\[p=\pm\sqrt{2mE-m^2\omega^2q^2}.\]
The phase-space orbit is an ellipse. Its maximum coordinate is
\[q_{\max}=A=\sqrt{\frac{2E}{m\omega^2}},\]
and its maximum momentum is
\[p_{\max}=m\omega A.\]
The area enclosed by the ellipse is
\[\oint p\,dq
=
\pi q_{\max}p_{\max}.\]
Therefore
\[\oint p\,dq
=
\pi A(m\omega A)
=
\pi m\omega A^2.\]
The action variable is
\[J=\frac{1}{2\pi}\oint p\,dq.\]
Thus
\[J
=
\frac{1}{2\pi}\pi m\omega A^2
=
\frac{1}{2}m\omega A^2.\]
But for the harmonic oscillator,
\[E=\frac{1}{2}m\omega^2A^2.\]
Therefore
\[\boxed{
J=\frac{E}{\omega}.
}\]
Hence
\[E=\omega J.\]
So the transformed Hamiltonian is
\[K(J)=\omega J.\]
Hamilton’s equations are
\[\dot J=-\frac{\partial K}{\partial \theta}=0\]
and
\[\dot\theta=\frac{\partial K}{\partial J}=\omega.\]
Therefore
\[\boxed{
J=\text{constant},
\qquad
\theta=\omega t+\theta_0.
}\]
The harmonic oscillator, which appears as back-and-forth motion in $q$, becomes uniform phase motion in $\theta$.
General one-dimensional potential well
Consider a particle moving in a smooth potential $V(q)$ with Hamiltonian
\[H=\frac{p^2}{2m}+V(q).\]
Suppose the particle is trapped between two turning points $q_{\min}$ and $q_{\max}$, where
\[V(q_{\min})=E,
\qquad
V(q_{\max})=E.\]
From the energy equation,
\[E=\frac{p^2}{2m}+V(q),\]
we get
\[p=\pm\sqrt{2m(E-V(q))}.\]
The action variable is
\[J=\frac{1}{2\pi}\oint p\,dq.\]
For back-and-forth motion between the two turning points, the upper and lower halves of the closed phase curve contribute equally. Hence
\[\boxed{
J(E)=
\frac{1}{\pi}
\int_{q_{\min}}^{q_{\max}}
\sqrt{2m(E-V(q))}\,dq.
}\]
This expression is important because it works even when the exact solution $q(t)$ is difficult to find. The action variable can still be obtained from the geometry of the phase-space orbit.
Now we show that the angle variable grows with the physical frequency. Since the transformed Hamiltonian depends only on $J$,
\[K=K(J)=E(J).\]
Therefore
\[\dot J=-\frac{\partial K}{\partial \theta}=0\]
and
\[\dot\theta=\frac{\partial K}{\partial J}=\frac{dE}{dJ}.\]
Using the chain rule,
\[\frac{dE}{dJ}
=
\left(\frac{dJ}{dE}\right)^{-1}.\]
Now differentiate
\[J(E)=
\frac{1}{\pi}
\int_{q_{\min}}^{q_{\max}}
\sqrt{2m(E-V(q))}\,dq.\]
The endpoint terms vanish because the integrand becomes zero at the turning points. Therefore
\[\frac{dJ}{dE}
=
\frac{1}{\pi}
\int_{q_{\min}}^{q_{\max}}
\frac{m}{\sqrt{2m(E-V(q))}}\,dq.\]
Since
\[p=\sqrt{2m(E-V(q))}\]
and
\[\dot q=\frac{p}{m},\]
we have
\[\frac{m}{\sqrt{2m(E-V(q))}}
=
\frac{m}{p}
=
\frac{1}{\dot q}.\]
Therefore
\[\frac{dJ}{dE}
=
\frac{1}{\pi}
\int_{q_{\min}}^{q_{\max}}
\frac{dq}{\dot q}.\]
But
\[\int_{q_{\min}}^{q_{\max}}
\frac{dq}{\dot q}\]
is the time taken to move from one turning point to the other. This is half the period:
\[\int_{q_{\min}}^{q_{\max}}
\frac{dq}{\dot q}
=
\frac{T}{2}.\]
Hence
\[\frac{dJ}{dE}
=
\frac{1}{\pi}\frac{T}{2}
=
\frac{T}{2\pi}.\]
Therefore
\[\frac{dE}{dJ}
=
\frac{2\pi}{T}.\]
Thus
\[\boxed{
\dot\theta=\omega=\frac{2\pi}{T}.
}\]
So for any one-dimensional bounded periodic system,
\[\boxed{
\dot J=0,
\qquad
\dot\theta=\omega.
}\]
This proves that action-angle variables convert general periodic motion into uniform angular motion.
Small-angle pendulum
For a pendulum of length $l$ and mass $m$, the exact Hamiltonian is
\[H=\frac{p_\theta^2}{2ml^2}+mgl(1-\cos\theta).\]
For small oscillations,
\[\cos\theta\simeq 1-\frac{\theta^2}{2}.\]
Therefore
\[mgl(1-\cos\theta)\simeq \frac{1}{2}mgl\theta^2.\]
The Hamiltonian becomes
\[H=
\frac{p_\theta^2}{2ml^2}
+
\frac{1}{2}mgl\theta^2.\]
This is mathematically equivalent to a harmonic oscillator with angular frequency
\[\omega=\sqrt{\frac{g}{l}}.\]
Therefore the action variable is
\[J=\frac{1}{2\pi}\oint p_\theta\,d\theta.\]
Using the harmonic oscillator result,
\[\boxed{
J=\frac{E}{\omega}
=
E\sqrt{\frac{l}{g}}.
}\]
The transformed Hamiltonian is
\[K(J)=\omega J,
\qquad
\omega=\sqrt{\frac{g}{l}}.\]
Therefore
\[\dot J=-\frac{\partial K}{\partial \theta}=0\]
and
\[\dot\theta=\frac{\partial K}{\partial J}=\omega.\]
Hence
\[\boxed{
\dot J=0,
\qquad
\dot\theta=\sqrt{\frac{g}{l}}.
}\]
Thus the small-angle pendulum becomes uniform phase motion in action-angle variables.
Practice questions
Q1. Starting from
\[F_2(q_i,P_i,t)=\sum_i q_iP_i+\varepsilon G(q_i,P_i,t),\]
derive the infinitesimal canonical transformation formulas.
Q2. Prove that for any phase-space function $f(q_i,p_i,t)$,
\[\delta f=\varepsilon\{f,G\}.\]
Q3. Show explicitly that choosing $G=H$ and $\varepsilon=dt$ reproduces Hamilton’s equations.
Q4. Explain why the Hamiltonian is called the generator of time translation.
Q5. If $G$ has no explicit time dependence and satisfies
\[\{G,H\}=0,\]
show that $G$ is conserved.
Q6. Show that momentum generates spatial translation.
Q7. Show that $L_z=xp_y-yp_x$ generates rotation about the $z$-axis.
Q8. Explain why translational symmetry gives conservation of momentum.
Q9. Explain why rotational symmetry gives conservation of angular momentum.
Q10. For the harmonic oscillator, show that
\[J=\frac{E}{\omega}.\]
Q11. For a general one-dimensional potential well, prove that
\[\frac{dJ}{dE}=\frac{T}{2\pi}.\]
Q12. Hence prove that
\[\dot\theta=\frac{dE}{dJ}=\frac{2\pi}{T}.\]
Q13. For the small-angle pendulum, show that
\[J=E\sqrt{\frac{l}{g}}.\]
Q14. Explain why action-angle variables convert bounded periodic motion into uniform phase motion.
Q15. Explain the statement: conserved quantities are generators of continuous canonical symmetries.
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