Canonical Transformations

A canonical transformation is a change of phase-space variables that preserves the Hamiltonian structure. It sends canonical coordinates $(q^i,p_i)$ to new variables $(Q^i,P_i)$ in such a way that Poisson brackets keep their canonical form:

$$\{Q^i,Q^j\}=0, \qquad \{P_i,P_j\}=0, \qquad \{Q^i,P_j\}=\delta^i_j.$$

Equivalently, Hamilton’s equations keep the same shape in the new variables after the Hamiltonian is transformed appropriately. Canonical transformations are the classical mechanics analogue of structure-preserving changes of quantum operators, especially unitary transformations.

Helpful background. Phase Space introduces $q,p$, and Poisson Brackets supplies the test for a canonical change of variables.

Preserving conjugate pairs

Changing variables can simplify a Hamiltonian, but arbitrary variables need not satisfy Hamilton’s equations in their usual form. Canonical transformations preserve the pairing between positions and momenta. Computing the brackets of the new variables is a practical test.

Definition by Poisson Brackets

Let

$$Q^i=Q^i(q,p,t), \qquad P_i=P_i(q,p,t)$$

be new phase-space variables. The transformation is canonical when the new variables satisfy the fundamental Poisson brackets:

$$\{Q^i,Q^j\}=0,$$ $$\{P_i,P_j\}=0,$$

and

$$\{Q^i,P_j\}=\delta^i_j.$$

These brackets are computed using the old canonical variables $(q,p)$. If the conditions hold, then any two functions $F(Q,P)$ and $G(Q,P)$ have the same bracket whether computed in old coordinates or new coordinates:

$$\{F,G\}_{q,p} = \{F,G\}_{Q,P}.$$

This bracket-preservation criterion is often the most direct test.

Definition by Symplectic Form

In canonical coordinates, the standard symplectic form is

$$\omega = \sum_i dq^i\wedge dp_i.$$

A transformation is canonical when it preserves this form. For a time-dependent change of variables, the differentials here are taken on phase space at fixed $t$:

$$\sum_i dQ^i\wedge dP_i = \sum_i dq^i\wedge dp_i.$$

This is the geometric version of preserving Poisson brackets. Symplectic Vector Spaces develops the finite-dimensional linear-algebra viewpoint; in one degree of freedom this means preserving oriented phase-space area. In several degrees of freedom the condition preserves the full symplectic form, not the projected area of each coordinate plane separately.

Linear Test

For linear transformations it is useful to package phase-space coordinates into

$$z= \begin{pmatrix} q\\ p \end{pmatrix}.$$

For one degree of freedom, the Poisson bracket can be written using

$$J= \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}.$$

A linear transformation $z'=Mz$ is canonical precisely when

$$MJM^{T}=J.$$

For the same matrix $M$, this is equivalent to

$$M^{T}JM=J.$$

The first identity preserves the Poisson brackets of coordinate functions; the second preserves the symplectic pairing of tangent vectors. They are algebraically equivalent because $J^{-1}=-J$, not different choices for $M$. For $n$ degrees of freedom use the block matrix $J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}$.

For one degree of freedom, this condition implies $\det M=1$. In higher dimensions, determinant $1$ is not enough; the full symplectic condition is stronger.

Examples

Scaling a Conjugate Pair

Let

$$Q=aq, \qquad P=\frac{p}{a},$$

with $a\ne0$. Then

$$\{Q,P\} = \left\{aq,\frac{p}{a}\right\} = 1.$$

Also $\{Q,Q\}=0$ and $\{P,P\}=0$, so the transformation is canonical.

By contrast,

$$Q=aq, \qquad P=ap$$

gives

$$\{Q,P\}=a^2.$$

This is canonical only when $a^2=1$. It may be invertible, but it does not preserve the canonical bracket for general $a$.

Quarter-Turn in Phase Space

The transformation

$$Q=p, \qquad P=-q$$

is canonical because

$$\{Q,P\} = \{p,-q\} = 1.$$

Geometrically this is a $90^\circ$ rotation in the $(q,p)$ plane, up to units. In quantum mechanics, exchanging position and momentum is closely related to Fourier transformation, though the quantum statement requires normalization conventions and operators.

Point Transformations

A change of configuration coordinate $Q=Q(q)$ can be canonical if the momentum transforms as a covector. In one degree of freedom,

$$P = p\frac{dq}{dQ}.$$

Then

$$\{Q,P\} = \frac{dQ}{dq} \frac{\partial P}{\partial p} = \frac{dQ}{dq} \frac{dq}{dQ} = 1.$$

This is why canonical momentum is not simply a velocity component in arbitrary coordinates. Momentum transforms according to the cotangent-space structure of phase space.

Hamiltonian Flow Is Canonical

Time evolution generated by a Hamiltonian is itself a canonical transformation. If a system flows from initial variables $(q(0),p(0))$ to variables $(q(t),p(t))$, the map from initial phase space to time-$t$ phase space preserves Poisson brackets and phase-space volume.

One way to see bracket preservation is through the Jacobi identity. If observables evolve by

$$\frac{df}{dt}=\{f,H\},$$

then the bracket evolves compatibly:

$$\frac{d}{dt}\{f,g\} = \{\{f,g\},H\}.$$

So the bracket of two evolved observables is the evolved bracket. This is the classical analogue of unitary time evolution preserving commutators.

Infinitesimal Canonical Transformations

A phase-space function $G(q,p)$ generates an infinitesimal canonical transformation by

$$\delta f = \epsilon\{f,G\},$$

where $\epsilon$ is a small parameter. In particular,

$$\delta q^i = \epsilon\frac{\partial G}{\partial p_i}, \qquad \delta p_i = - \epsilon\frac{\partial G}{\partial q^i}.$$

This is Hamiltonian flow with $G$ playing the role of a generator. Taking $G=H$ generates time translations; taking $G=p$ generates translations in $q$; taking angular momentum generates rotations.

For $U(\epsilon)=e^{-i\epsilon\widehat G/\hbar}$ and the convention $A\mapsto U^\dagger A U$, the quantum counterpart is

$$\delta A=\frac{\epsilon}{i\hbar}[A,\widehat G]+O(\epsilon^2).$$

Conjugating in the opposite order reverses the first-order sign. The shared structure is the generator acting by a bracket.

Generating Functions

Canonical transformations can often be built from generating functions. A common type uses a function

$$F_2(q,P,t).$$

The transformation is defined by

$$p_i = \frac{\partial F_2}{\partial q^i}, \qquad Q^i = \frac{\partial F_2}{\partial P_i}.$$

These equations determine a local change of variables when the mixed Hessian $\partial^2F_2/\partial q^i\partial P_j$ is invertible. If the generating function depends explicitly on time, the new Hamiltonian is

$$K(Q,P,t) = H(q,p,t) + \frac{\partial F_2}{\partial t},$$

with $q$ and $p$ rewritten in terms of $Q$, $P$, and $t$.

The identity transformation is generated by

$$F_2(q,P)=\sum_i q^iP_i.$$

Indeed,

$$p_i = P_i, \qquad Q^i = q^i.$$

For a translation by a constant $a$ in one degree of freedom, the choice

$$F_2(q,P)=(q+a)P$$

gives

$$p=P, \qquad Q=q+a.$$

Generating functions are useful because they build canonical transformations automatically, rather than requiring one to check the bracket conditions afterward. Hamilton–Jacobi Theory turns this idea into a method: a suitable action function can generate a canonical transformation to constants of motion.

Hamiltonian in New Variables

Suppose a time-independent canonical transformation rewrites the old variables in terms of new variables:

$$q=q(Q,P), \qquad p=p(Q,P).$$

The new Hamiltonian is the old Hamiltonian expressed in the new variables:

$$K(Q,P)=H(q(Q,P),p(Q,P)).$$

Hamilton’s equations keep their canonical form:

$$\dot Q^i = \frac{\partial K}{\partial P_i}, \qquad \dot P_i = - \frac{\partial K}{\partial Q^i}.$$

If the transformation is time dependent, the extra $\partial F/\partial t$ term in the generating function changes the Hamiltonian. This is the classical version of the familiar quantum fact that a time-dependent unitary transformation adds an extra generator term to the transformed Hamiltonian.

Worked example: a moving coordinate origin

Consider a free particle with $H=p^2/(2m)$ and $m>0$. Let the coordinate origin follow a prescribed twice-differentiable trajectory $a(t)$. This is a passive change of description, generated by

$$F_2(q,P,t)=[q-a(t)]P, \qquad Q=q-a(t),\qquad P=p.$$

The type-$2$ rule, including its time derivative, gives

$$K(Q,P,t)=\frac{P^2}{2m}-\dot a(t)P, \qquad \dot Q=\frac{P}{m}-\dot a(t),\qquad \dot P=0.$$

Hence $\ddot Q=-\ddot a$: a freely moving particle accelerates relative to an accelerating origin. Keeping only $H$ expressed in the new variables would miss this effect. The same result follows from the transformed Lagrangian $L'=m(\dot Q+\dot a)^2/2$: its conjugate momentum is $P=m(\dot Q+\dot a)$, and its Legendre transform is $K$. In this choice of variables, $P$ remains the laboratory momentum, rather than $m\dot Q$.

The generator $K$ is also different from the laboratory energy $E=P^2/(2m)$. Along the motion,

$$\frac{dE}{dt}=0, \qquad \frac{dK}{dt}=\frac{\partial K}{\partial t}=-\ddot a(t)P.$$

For constant origin velocity, both quantities are conserved but differ by $-\dot aP$. For an accelerating origin, $K$ generally changes even though the laboratory particle remains free. Picture transformations makes the corresponding distinction between a quantum frame’s state generator and its energy observable. The general time-dependent generating-function rules are derived in Sussman and Wisdom, §5.4.3.

Quantum Unitary Analogy

Canonical transformations preserve Poisson brackets. Unitary transformations preserve commutators:

$$[UAU^\dagger,UBU^\dagger] = U[A,B]U^\dagger.$$

This is why the analogy is natural:

Classical mechanics	Quantum mechanics
phase-space functions $f(q,p)$	operators $A$
Poisson bracket $\{f,g\}$	commutator $[A,B]/(i\hbar)$
canonical transformation	unitary transformation
generator $G$ acts by $\{\,\cdot\,,G\}$	generator $G$ acts by commutator
Hamiltonian flow	unitary time evolution

The analogy is not an identity. Not every classical canonical transformation has a globally simple quantum unitary implementation. Ordering, domains, topology, and the double-cover behavior of some linear symplectic transformations can matter. The safe statement is that canonical transformations preserve the classical bracket structure that quantum unitary transformations preserve after quantization.

Active and Passive Viewpoints

There are two common readings of a canonical transformation:

• passive: a change of phase-space coordinates used to describe the same physical point;
• active: a map that moves points or observables through phase space.

Both viewpoints are useful, but mixing them causes sign errors. The same warning appears in quantum mechanics when comparing $A\mapsto UAU^\dagger$ with $A\mapsto U^\dagger A U$.

When checking formulas, state which variables are old, which variables are new, and whether the transformation acts on coordinates, states, or observables.

Common Mistakes

• Assuming every invertible change of variables is canonical.
• Checking only phase-space volume preservation instead of the full Poisson-bracket conditions.
• Forgetting that momenta transform as covectors under coordinate changes.
• Ignoring the extra Hamiltonian term for time-dependent generating functions.
• Treating canonical transformations and quantum unitary transformations as exactly the same object.
• Mixing active and passive sign conventions.
• Believing that determinant $1$ is sufficient for a canonical transformation in more than one degree of freedom.

Exercises

1. Decide whether $Q=2q$ and $P=p/2$ is canonical.

Solution

Compute

$$\{Q,P\} = \left\{2q,\frac{p}{2}\right\} = 1.$$

Also $\{Q,Q\}=0$ and $\{P,P\}=0$. Therefore the transformation is canonical.

2. Decide whether $Q=2q$ and $P=2p$ is canonical.

Solution

Compute

$$\{Q,P\} = \{2q,2p\} = 4.$$

The canonical condition would require $\{Q,P\}=1$. The transformation is invertible, but it is not canonical.

3. Use $F_2(q,P)=qP+aP$, with constant $a$, to find the canonical transformation it generates. Then describe a free particle using $F_2(q,P,t)=(q-\tfrac12bt^2)P$, where $b$ is a constant acceleration. Find $K$ and $Q(t)$ for initial laboratory data $q(0)=q_0$, $p(0)=p_0$.

Solution

For a type-$2$ generating function,

$$p=\frac{\partial F_2}{\partial q}=P, \qquad Q=\frac{\partial F_2}{\partial P}=q+a.$$

Thus the transformation translates the coordinate by $a$ and leaves the momentum equal to the new momentum.

For the accelerating origin, $Q=q-bt^2/2$ and $P=p$, so

$$K=\frac{P^2}{2m}-btP, \qquad Q(t)=q_0+\frac{p_0}{m}t-\frac12bt^2, \qquad P(t)=p_0.$$

The trajectory follows either from Hamilton’s equations for $K$ or by transforming the laboratory free trajectory. It has $\ddot Q=-b$. Meanwhile $dK/dt=-bp_0$ and the laboratory energy $p_0^2/(2m)$ remains constant.

4. Let $G=p$ in one degree of freedom. Use $\delta f=\epsilon\{f,G\}$ to find the infinitesimal transformations of $q$ and $p$.

Solution

For the coordinate,

$$\delta q = \epsilon\{q,p\} = \epsilon.$$

For the momentum,

$$\delta p = \epsilon\{p,p\} = 0.$$

Thus $p$ generates translations of $q$.

5. Show that a unitary transformation preserves commutators.

Solution

Use $U^\dagger U=I$:

$$[UAU^\dagger,UBU^\dagger] = UAU^\dagger UBU^\dagger - UBU^\dagger UAU^\dagger.$$

Therefore

$$[UAU^\dagger,UBU^\dagger] = UABU^\dagger - UBAU^\dagger = U[A,B]U^\dagger.$$

References

• V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd ed., Springer, 1989.
• H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.
• G. J. Sussman and J. Wisdom, Structure and Interpretation of Classical Mechanics, 2nd ed., MIT Press, 2015, §5.4.3 — time-dependent generating functions and the transformed Hamiltonian.