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.

Why Quantum Mechanics Needs This

Canonical transformations matter because they are the classical version of preserving the algebra of observables. Quantum mechanics repeatedly uses the corresponding idea:

• position and momentum can be represented in different but equivalent ways;
• unitary transformations preserve commutators and transition probabilities;
• symmetry transformations act on observables without changing the physical algebra;
• canonical quantization starts from Poisson brackets and asks for operators with matching commutators;
• semiclassical methods often choose better phase-space coordinates before approximating an integral.

The key phrase is “preserve structure.” A canonical transformation is not merely an invertible change of coordinates. It preserves the pairing between coordinates and momenta, the Poisson bracket, and the form of Hamiltonian flow.

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:

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

This is the geometric version of preserving Poisson brackets. Symplectic Vector Spaces develops the finite-dimensional linear-algebra viewpoint; here the practical message is that canonical transformations preserve phase-space area in each conjugate pair, and in many degrees of freedom preserve the full symplectic structure.

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

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

Equivalently, with a transpose convention that uses column gradients, one often sees

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

The difference is a bookkeeping convention about whether $M$ acts on coordinates or on gradients. The invariant content is the same: the transformation must preserve the antisymmetric symplectic matrix.

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.

The quantum parallel is

$$\delta A \sim \frac{1}{i\hbar}[A,G],$$

with sign depending on active or passive convention. 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

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

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.

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.

Cross-Links

• Phase Space
• Poisson Brackets
• Hamiltonian Mechanics Review
• Symplectic Vector Spaces
• Symplectic Manifolds, First Look
• Hamilton–Jacobi Theory
• Canonical Commutation Relations
• Classical–Quantum Correspondence
• Quantization vs Classical Limit
• Generators
• Unitary Symmetries
• Heisenberg Group
• Position and Momentum Representations

References

• H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.
• V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd ed., Springer, 1989.
• L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Butterworth-Heinemann, 1976.
• R. Abraham and J. E. Marsden, Foundations of Mechanics, 2nd ed., AMS Chelsea, 2008.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• M. de Gosson, Symplectic Geometry and Quantum Mechanics, Birkhäuser, 2006.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

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 $F_2(q,P)=qP+aP$ to find the canonical transformation it generates.

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.

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

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