5Canonical Transformations

We have done considerable mountain climbing. Now we are in the rarefied atmosphere of theories of excessive beauty and we are nearing a high plateau on which geometry, optics, mechanics, and wave mechanics meet on common ground. Only concentrated thinking, and a considerable amount of re–creation, will reveal the beauty of our subject in which the last word has not been spoken.

Cornelius Lanczos, The Variational Principles of Mechanics [29], p. 229

One way to simplify the analysis of a problem is to express it in a form in which the solution has a simple representation. However, it may not be easy to formulate the problem in such a way initially. It is often useful to start by formulating the problem in one way, and then transform it. For example, the formulation of the problem of the motion of a number of gravitating bodies is simple in rectangular coordinates, but it is easier to understand aspects of the motion in terms of orbital elements, such as the semimajor axes, eccentricities, and inclinations of the orbits. The semimajor axis and eccentricity of an orbit depend on both the configuration and the velocity of the body. Such transformations are more general than those that express changes in configuration coordinates. Here we investigate transformations of phase-space coordinates that involve both the generalized coordinates and the generalized momenta.

Suppose we have two different Hamiltonian systems, and suppose the trajectories of the two systems are in one-to-one correspondence. In this case both Hamiltonian systems can be mathematical models of the same physical system. Some questions about the physical system may be easier to answer by reference to one model and others may be easier to answer in the other model. For example, it may be easier to formulate the physical system in one model and to discover a conserved quantity in the other. Canonical transformations are maps between Hamiltonian systems that preserve the dynamics.

A canonical transformation is a phase-space coordinate transformation and an associated transformation of the Hamiltonian such that the dynamics given by Hamilton's equations in the two representations describe the same evolution of the system.

5.1   Point Transformations

A point transformation is a canonical transformation that extends a possibly time-dependent transformation of the configuration coordinates to a phase-space transformation. For example, one might want to reexpress motion in terms of polar coordinates, given a description in terms of rectangular coordinates. In order to extend a transformation of the configuration coordinates to a phase-space transformation we must specify how the momenta and Hamiltonian are transformed.

We have already seen how coordinate transformations can be carried out in the Lagrangian formulation (see section 1.6.1). In that case, we found that if the Lagrangian transforms by composition with the coordinate transformation, then the Lagrange equations are equivalent.

Lagrangians that differ by the addition of a total time derivative have the same Lagrange equations, but may have different momenta conjugate to the generalized coordinates. So there is more than one way to make a canonical extension of a coordinate transformation.

Here, we find the particular canonical extension of a coordinate transformation for which the Lagrangians transform by composition with the transformation, with no extra total time derivative terms added to the Lagrangian.

Let L be a Lagrangian for a system. Consider the coordinate transformation q = F (t, q′). The velocities transform by

$$\begin{array}{cc}v={\partial }_{0}F\left(t,q\prime \right)+{\partial }_{1}F\left(t,q\prime \right)v\prime .& (5.1)\end{array}$$

We obtain a Lagrangian L′ in the transformed coordinates by composition of L with the coordinate transformation. We require that L′(t, q′, v′) = L(t, q, v), so:

$$\begin{array}{c}\begin{array}{cc}L\prime \left(t,q\prime ,v\prime \right)=L\left(t,F\left(t,q\prime \right),{\partial }_{0}F\left(t,q\prime \right)+{\partial }_{1}F\left(t,q\prime \right)v\prime \right).& (5.2)\end{array}\end{array}$$

The momentum conjugate to q′ is

$$\begin{array}{lll}p\prime \hfill & ={\partial }_{2}L\prime \left(t,q\prime ,v\prime \right)\hfill & \hfill \\ \hfill & ={\partial }_{2}L\left(t,F\left(t,q\prime \right),{\partial }_{0}F\left(t,q\prime \right)+{\partial }_{1}F\left(t,q\prime \right)v\prime \right){\partial }_{1}F\left(t,q\prime \right)\hfill & \hfill \\ \hfill & =p{\partial }_{1}F\left(t,q\prime \right),\hfill & (5.3)\hfill \end{array}$$

where we have used

$$\begin{array}{lll}p\hfill & ={\partial }_{2}L\left(t,q,v\right)\hfill & \hfill \\ \hfill & ={\partial }_{2}L\left(t,F\left(t,q\prime \right),{\partial }_{0}F\left(t,q\prime \right)+{\partial }_{1}F\left(t,q\prime \right)v\prime \right).\hfill & (5.4)\hfill \end{array}$$

So, from equation (5.3),¹

$$\begin{array}{ll}p=p\prime {\left({\partial }_{1}F\left(t,q\prime \right)\right)}^{-1}.\hfill & (5.5)\hfill \end{array}$$

We can collect these results to define a canonical phase-space transformation C_H:²

$$\begin{array}{lll}\left(t,q,p\right)\hfill & =C_{\text{H}}\left(t,q\prime ,p\prime \right)\hfill & \hfill \\ \hfill & =(t,F(t,q\prime ),p\prime {({\partial }_{1}F(t,q\prime ))}^{-1}).\hfill & (5.6)\hfill \end{array}$$

The Hamiltonian is obtained by the Legendre transform

$$\begin{array}{ll}H\prime \left(t,q\prime ,p\prime \right)\hfill & \hfill \\ =p\prime v\prime -L\prime \left(t,q\prime ,v\prime \right)\hfill & \hfill \\ =(p{\partial }_{1}F(t,q\prime ))(({\partial }_{1}F{(t,q\prime )}^{-1}(v-{\partial }_{0}F(t,q\prime ))))-L(t,q,v)\hfill & \hfill \\ =pv-L(t,q,v)-p{\partial }_{0}F\left(t,q\prime \right)\hfill & \hfill \\ =H(t,q,p)-p{\partial }_{0}F\left(t,q\prime \right),\hfill & (5.7)\hfill \end{array}$$

using relations (5.1) and (5.3) in the second step. Fully expressed in terms of the transformed coordinates and momenta, the transformed Hamiltonian is

$$\begin{array}{lll}H\prime \left(t,q\prime ,p\prime \right)=\hfill & H(t,F(t,q\prime ),p\prime {({\partial }_{1}F(t,q\prime ))}^{-1})\hfill & \hfill \\ \hfill & -(p\prime {({\partial }_{1}F(t,q\prime ))}^{-1}){\partial }_{0}F\left(t,q\prime \right).\hfill & (5.8)\hfill \end{array}$$

The Hamiltonians H′ and H are equivalent because L and L′ have the same value for a given dynamical state and so have the same paths of stationary action. In general H and H′ do not have the same values for a given dynamical state, but differ by a term that depends on the coordinate transformation.

For time-independent transformations, ∂₀F = 0, there are a number of simplifications. The relationship of the velocities (5.1) becomes

$$\begin{array}{ll}v={\partial }_{1}F\left(t,q\prime \right)v\prime .\hfill & (5.9)\hfill \end{array}$$

Comparing this to the relation (5.5) between the momenta, we see that in this case the momenta transform “oppositely” to the velocities³

$$\begin{array}{ll}pv=p\prime {\left({\partial }_{1}F\left(t,q\prime \right)\right)}^{-1}{\partial }_{1}F\left(t,q\prime \right)v\prime =p\prime v\prime ,\hfill & (5.10)\hfill \end{array}$$

so the product of the momenta and the velocities is not changed by the transformation. This, combined with the fact that by construction L(t, q, v) = L′(t, q′, v′), shows that

$$\begin{array}{lll}H(t,q,p)\hfill & =pv-L(t,q,v)\hfill & \hfill \\ \hfill & =p\prime v\prime -L\prime \left(t,q\prime ,v\prime \right)\hfill & \hfill \\ \hfill & =H\prime \left(t,q\prime ,p\prime \right).\hfill & (5.11)\hfill \end{array}$$

For time-independent coordinate transformations the Hamiltonian transforms by composition with the associated phase-space transformation. We can also see this from the general relationship (5.7) between the Hamiltonians.

Implementing point transformations

The procedure F->CH takes a procedure F implementing a transformation of configuration coordinates and returns a procedure implementing a transformation of phase-space coordinates:⁴

(define ((F->CH F) state)
  (up (time state)
      (F state)
      (solve-linear-right
        (momentum state)
        (((partial 1) F) state))))

Consider a particle moving in a central field. In rectangular coordinates a Hamiltonian is

(define ((H-central m V) state)
  (let ((x (coordinate state))
        (p (momentum state)))
    (+ (/ (square p) (* 2 m))
       (V (sqrt (square x))))))

Let's look at this Hamiltonian in polar coordinates. The phase-space transformation is obtained by applying F->CH to the procedure p->r that takes a time and a polar tuple and returns a tuple of rectangular coordinates (see section 1.6.1). The transformation is time independent so the Hamiltonian transforms by composition. In polar coordinates the Hamiltonian is

(show-expression
  ((compose
     (H-central 'm (literal-function 'V))
     (F->CH p->r))
   (up 't (up 'r 'phi) (down 'p_r 'p_phi))))

There are three terms. There is the potential energy, which depends on the radius, there is the kinetic energy due to radial motion, and there is the kinetic energy due to tangential motion. As expected, the angle φ does not appear and thus the angular momentum is a conserved quantity. By going to polar coordinates we have decoupled one of the two degrees of freedom in the problem.

If the transformation is time varying the Hamiltonian must be adjusted by adding a correction to the composition of the Hamiltonian and the transformation (see equation 5.8):

$$\begin{array}{ll}H\prime =H\circ C_{\text{H}}+K\hfill & (5.12)\hfill \end{array}$$

The correction is computed by

(define ((F->K F) state)
  (- (* (solve-linear-right
          (momentum state)
          (((partial 1) F) state))
        (((partial 0) F) state))))

For example, consider a transformation to coordinates translating with velocity v:

(define ((translating v) state)
  (+ (coordinates state) (* v (time state))))

We compute the additive adjustment required for the Hamiltonian:

((F->K (translating (up 'v^x 'v^y 'v^z)))
 (up 't (up 'x 'y 'z) (down 'p_x 'p_y 'p_z)))

(+ (* -1 p_x v^x) (* -1 p_y v^y) (* -1 p_z v^z))

Notice that this is the negation of the inner product of the momentum and the velocity of the coordinate system.

Let's see how a simple free-particle Hamiltonian is transformed:

(define ((H-free m) s)
(/ (square (momentum s)) (* 2 m)))

The transformed Hamiltonian is:

(define H-prime
  (+ (compose (H-free 'm)
              (F->CH (translating (up 'v^x 'v^y 'v^z))))
     (F->K (translating (up 'v^x 'v^y 'v^z)))))

(H-prime
  (up 't
      (up 'xprime 'yprime 'zprime)
      (down 'pprime_x 'pprime_y 'pprime_z)))
(+ (* -1 pprime_x v^x)
   (* -1 pprime_y v^y)
   (* -1 pprime_z v^z)
   (/ (* 1/2 (expt pprime_x 2)) m)
   (/ (* 1/2 (expt pprime_y 2)) m)
   (/ (* 1/2 (expt pprime_z 2)) m))

Exercise 5.1: Galilean invariance

Is this result what you expected? Let's investigate.

Recall that in exercise 1.29 we showed that if the kinetic energy is $\frac{1}{2}m{v}^2$ then the translation to a uniformly moving coordinate system introduces extra terms that can be identified as a total time derivative. Since these terms do not affect the Lagrange equations, we can take the kinetic energy in the transformed coordinates to also be $\frac{1}{2}m{(v\prime )}^2$.

Let C_H be the phase space extension of the translation transformation, and C be the local tuple extension. The transformed Hamiltonian is H′ = H ∘ C_H + K; the transformed Lagrangian is L′ = L ∘ C.

a. Derive the relationship between p and p′ both from C_H and from the Lagrangians. Are they the same? Derive the relationship between v and v′ by taking the derivative of the Hamiltonians with respect to the momenta (Hamilton's equation). Show that the Legendre transform of L′ gives the same H′.

b. We have shown that L and L′ differ by a total time derivative. So for any uniformly moving coordinate system we can write the Lagrangian as $\frac{1}{2}m{v}^2$. Similarly, we would expect to always be able to write the Hamiltonian as p²/(2m). Show that this differs from H′ by a total time derivative in the corresponding Lagrangians.

Exercise 5.2: Rotations

Let q and q′ be rectangular coordinates that are related by a rotation R: q = Rq′. The Lagrangian for the system is $L(t,q,v)=\frac{1}{2}m{v}^2-V(q)$. Find the corresponding phase-space transformation C_H. Compare the transformation equations for the rectangular components of the momenta to those for the rectangular components of the velocities. Are you surprised, considering equation (5.10)?

5.2   General Canonical Transformations

Although we have shown how to extend any coordinate transformation of the configuration space to a canonical transformation, there are other ways to construct canonical transformations. How do we know if we have a canonical transformation? To test if a transformation is canonical we may use the fact that if the transformation is canonical, then Hamilton's equations of motion for the transformed system and the original system will be equivalent.

Consider a Hamiltonian H and a phase-space transformation C_H. Let D_s be the function that takes a Hamiltonian and gives the Hamiltonian state-space derivative:⁵

$$\begin{array}{ll}{D}_{s}H(t,q,p)=(1,{\partial }_{2}H(t,q,p),-{\partial }_{1}H(t,q,p)).\hfill & (5.13)\hfill \end{array}$$

Hamilton's equations are

$$\begin{array}{ll}D\sigma =D_{s}H\circ \sigma ,\hfill & (5.14)\hfill \end{array}$$

for any realizable phase-space path σ.

The transformation C_H transforms the phase-space path σ′ (t) = (t, q′ (t), p′ (t)) into σ(t) = (t, q(t), p(t)):

$$\begin{array}{ll}\sigma =C_{\text{H}}\circ \sigma \prime .\hfill & (5.15)\hfill \end{array}$$

The rates of change of the phase-space coordinates are transformed by the derivative of the transformation

$$\begin{array}{ll}D\sigma =D\left(C_{\text{H}}\circ \sigma \prime \right)=\left(D{C}_{\text{H}}\circ \sigma \prime \right)D\sigma \prime .\hfill & (5.16)\hfill \end{array}$$

The transformation is canonical if the equations of motion obtained from the new Hamiltonian are the same as those that could be obtained by transforming the equations of motion derived from the original Hamiltonian to the new coordinates:

$$\begin{array}{ll}D\sigma =\left(D{C}_{\text{H}}\circ \sigma \prime \right)D\sigma \prime =\left(D{C}_{\text{H}}\circ \sigma \prime \right)\left(D_{s}H\prime \circ \sigma \prime \right).\hfill & (5.17)\hfill \end{array}$$

Using equation (5.14), we see that

$$\begin{array}{ll}{D}_{s}H\circ \sigma =\left(D{C}_{\text{H}}\circ \sigma \prime \right)\left(D_{s}H\prime \circ \sigma \prime \right).\hfill & (5.18)\hfill \end{array}$$

With σ = C_H ∘ σ′, we find

$$\begin{array}{ll}{D}_{s}H\circ C_{\text{H}}\circ \sigma \prime =\left(D{C}_{\text{H}}\circ \sigma \prime \right)\left(D_{s}H\prime \circ \sigma \prime \right).\hfill & (5.19)\hfill \end{array}$$

This condition must hold for any realizable phase-space path σ′. Certainly this is true if the following condition holds for every phase-space point:⁶

$$\begin{array}{ll}{D}_{s}H\circ C_{\text{H}}=D{C}_{\text{H}}·D_{s}H\prime .\hfill & (5.20)\hfill \end{array}$$

Any transformation that satisfies equation (5.20) is a canonical transformation among phase-space representations of a dynamical system. In one phase-space representation the system's dynamics is characterized by the Hamiltonian H′ and in the other by H. The idea behind this equation is illustrated in figure 5.1.

We can formalize this test as a program:

(define (canonical? C H Hprime)
  (- (compose (Hamiltonian->state-derivative H) C)
     (* (D C) (Hamiltonian->state-derivative Hprime))))

where Hamiltonian->state-derivative, which was introduced in chapter 3, implements D_s. The transformation is canonical if these residuals are zero.

For time-independent point transformations an appropriate Hamiltonian can be formed by composition with the corresponding phase-space transformation. For more general canonical transformations, we will see that if a transformation is independent of time, a suitable Hamiltonian for the transformed system can be obtained by composing the Hamiltonian with the phase-space transformation. In this case we obtain a more specific formula:

$$\begin{array}{ll}{D}_{s}H\circ C_{\text{H}}=D{C}_{\text{H}}·D_s(H\circ C_{\text{H}}).\hfill & (5.21)\hfill \end{array}$$

Polar-canonical transformation

The analysis of the harmonic oscillator illustrates the use of a general canonical transformation in the solution of a problem. The harmonic oscillator is a mathematical model of a simple spring-mass system. A Hamiltonian for a spring-mass system with mass m and spring constant k is

$$\begin{array}{ll}H(t,x,p_x)=\frac{p_x^2}{2m}+\frac{1}{2}k{x}^2.\hfill & (5.22)\hfill \end{array}$$

Hamilton's equations of motion are

$$\begin{array}{ll}Dx=p_x/m\hfill & \hfill \\ D{p}_x=-kx,\hfill & (5.23)\hfill \end{array}$$

giving the second-order system

$$\begin{array}{ll}m{D}^{2}x+kx=0.\hfill & (5.24)\hfill \end{array}$$

The solution is

$$\begin{array}{ll}x(t)=A\sin (\omega t+\phi ),\hfill & (5.25)\hfill \end{array}$$

where

$$\begin{array}{ll}\omega =\sqrt{k/m}\hfill & (5.26)\hfill \end{array}$$

and where A and φ are determined by initial conditions.

We use the polar-canonical transformation:

$$\begin{array}{ll}(t,x,p_x)=C_{\alpha }(t,\theta ,I)\hfill & (5.27)\hfill \end{array}$$

where

$$\begin{array}{ll}x=\sqrt{\frac{2I}{\alpha }}\sin \theta \hfill & (5.28)\hfill \end{array}$$

$$\begin{array}{ll}{p}_x=\sqrt{2\alpha I}\cos \theta .\hfill & (5.29)\hfill \end{array}$$

Here α is an arbitrary parameter. We define:

(define ((polar-canonical alpha) state)
  (let ((t (time state))
        (theta (coordinate state))
        (I (momentum state)))
    (let ((x (* (sqrt (/ (* 2 I) alpha)) (sin theta)))
          (p_x (* (sqrt (* 2 alpha I)) (cos theta))))
      (up t x p_x))))

And now we just run our test:

(define ((H-harmonic m k) s)
  (+ (/ (square (momentum s)) (* 2 m))
     (* 1/2 k (square (coordinate s)))))

((canonical?
   (polar-canonical 'alpha)
   (H-harmonic 'm 'k)
   (compose (H-harmonic 'm 'k)
            (polar-canonical 'alpha)))
 (up 't 'theta 'I))

(up 0 0 0)

So the transformation is canonical for the harmonic oscillator.⁷

Let's use our polar-canonical transformation C_α to help us solve the harmonic oscillator. We substitute expressions (5.28) and (5.29) for x and p_x in the Hamiltonian, getting our new Hamiltonian:

$$\begin{array}{ll}H\prime (t,\theta ,I)=\frac{\alpha I}{m}{(\cos \theta )}^2+\frac{kI}{\alpha }{(\sin \theta )}^2.\hfill & (5.30)\hfill \end{array}$$

If we choose $\alpha =\sqrt{km}$ then we obtain

$$\begin{array}{ll}H\prime (t,\theta ,I)=\sqrt{\frac{k}{m}}I=\omega I,\hfill & (5.31)\hfill \end{array}$$

and the new Hamiltonian no longer depends on the coordinate. Hamilton's equation for I is

$$\begin{array}{ll}DI(t)=-{\partial }_{1}H\prime (t,\theta (t),I(t))=0,\hfill & (5.32)\hfill \end{array}$$

so I is constant. The equation for θ is

$$\begin{array}{ll}D\theta (t)={\partial }_{2}H\prime (t,\theta (t),I(t))=\omega ,\hfill & (5.33)\hfill \end{array}$$

so

$$\begin{array}{ll}\theta (t)=\omega t+\phi .\hfill & (5.34)\hfill \end{array}$$

In the original variables,

$$\begin{array}{lll}x(t)\hfill & =\sqrt{2I(t)/\alpha }\sin \theta (t)\hfill & \hfill \\ \hfill & =A\sin (\omega t+\phi ),\hfill & (5.35)\hfill \end{array}$$

with the constant $A=\sqrt{2I(t)/\alpha }$. So we have found the solution to the problem by making a canonical transformation to new phase-space variables for which the solution is easy and then transforming the solutions back to the original variables.

Exercise 5.3: Trouble in Lagrangian world

Is there a Lagrangian L′ that corresponds to the harmonic oscillator Hamiltonian H′(t, θ, I) = ωI? What could this possibly mean?

Exercise 5.4: Group properties

If we say that C_H is canonical with respect to Hamiltonians H and H′ if and only if D_sH ∘ C_H = DC_H · D_sH′, then:

a. Show that the composition of canonical transformations is canonical.

b. Show that composition of canonical transformations is associative.

c. Show that the identity transformation is canonical.

d. Show that there is an inverse for a canonical transformation and the inverse is canonical.

5.2.1 Time-Dependent Transformations

We have seen that for time-dependent point transformations the Hamiltonian appropriate for the transformed system is the original Hamiltonian composed with the transformation and augmented with an additive correction. Here we find a similar decomposition for general time-dependent canonical transformations.

The key to this decomposition is to separate the time part and the phase-space part of the Hamiltonian state derivative:⁸

$$\begin{array}{lll}{D}_{s}H(s)\hfill & =(1,+{\partial }_{2}H(s),-{\partial }_{1}H(s))\hfill & \hfill \\ \hfill & =T(s)+\mathcal{D}H(s)\hfill & (5.36)\hfill \end{array}$$

where

$$\begin{array}{ll}T(s)=(1,0,0),\hfill & (5.37)\hfill \end{array}$$

$$\begin{array}{ll}\mathcal{D}H(s)=(0,+{\partial }_{2}H(s),-{\partial }_{1}H(s)),\hfill & (5.38)\hfill \end{array}$$

as code:⁹

(define (T-func s)
  (up 1
      (zero-like (coordinates s))
      (zero-like (momenta s))))

(define ((D-phase-space H) s)
  (up 0 (((partial 2) H) s) (- (((partial 1) H) s))))

If we assume that H′ = H ∘ C_H + K, then the canonical condition (5.20) becomes

$$\begin{array}{ll}{D}_{s}H\circ C_{\text{H}}=D{C}_{\text{H}}·D_s(H\circ C_{\text{H}}+K).\hfill & (5.39)\hfill \end{array}$$

Expanding the state derivative, the canonical condition is

$$\begin{array}{ll}(T+\mathcal{D}H)\circ C_{\text{H}}=D{C}_{\text{H}}·(T+\mathcal{D}(H\circ C_{\text{H}}+K)).\hfill & (5.40)\hfill \end{array}$$

Equation (5.40) is satisfied if the following conditions are met:

$$\begin{array}{ll}\mathcal{D}H\circ C_{\text{H}}=D{C}_{\text{H}}·\mathcal{D}(H\circ C_{\text{H}})\hfill & (5.41)\hfill \end{array}$$

$$\begin{array}{ll}T\circ C_{\text{H}}=D{C}_{\text{H}}·(T+\mathcal{D}K).\hfill & (5.42)\hfill \end{array}$$

The value of T ∘ C_H does not depend on C_H, so this term is really very simple. Notice that equation (5.41) does not depend upon K and that equation (5.42) does not depend upon H.

These can be implemented as follows:

(define (canonical-H? C H)
  (- (compose (D-phase-space H) C)
     (* (D C)
        (D-phase-space (compose H C)))))

(define (canonical-K? C K)
  (- (compose T-func C)
     (* (D C)
        (+ T-func (D-phase-space K)))))

Rotating coordinates

Consider a time-dependent transformation to uniformly rotating coordinates:¹⁰

$$\begin{array}{ll}q=R(\Omega )\left(t,q\prime \right),\hfill & (5.43)\hfill \end{array}$$

with components

$$\begin{array}{ll}x=x\prime \cos (\Omega t)-y\prime \sin (\Omega t)\hfill & \hfill \\ y=x\prime \sin (\Omega t)+y\prime \cos (\Omega t).\hfill & (5.44)\hfill \end{array}$$

As a program this is

(define ((rotating Omega) state)
  (let ((t (time state)) (qp (coordinate state)))
    (let ((xp (ref qp 0)) (yp (ref qp 1)) (zp (ref qp 2)))
      (up (- (* (cos (* Omega t)) xp)
             (* (sin (* Omega t)) yp))
          (+ (* (sin (* Omega t)) xp)
             (* (cos (* Omega t)) yp))
          zp))))

The extension of this transformation to a phase-space transformation is

(define (C-rotating Omega) (F->CH (rotating Omega)))

We first verify that this time-dependent transformation satisfies equation (5.41). We will try it for an arbitrary Hamiltonian with three degrees of freedom:

(define H-arbitrary
  (literal-function 'H
                    (-> (UP Real (UP Real Real Real) (DOWN Real Real Real))
                        Real)))

((canonical-H? (C-rotating 'Omega) H-arbitrary)
 (up 't (up 'xp 'yp 'zp) (down 'pp_x 'pp_y 'pp_z)))
(up 0 (up 0 0 0) (down 0 0 0))

And it works. Note that this result did not depend on any details of the Hamiltonian, suggesting that we might be able to make a test that does not require a Hamiltonian. We will see that shortly.

Since we have a point transformation, we can compute the required adjustment to the Hamiltonian:

((F->K (rotating 'Omega))
 (up 't (up 'xp 'yp 'zp) (down 'pp_x 'pp_y 'pp_z)))
(+ (* Omega pp_x yp) (* -1 Omega pp_y xp))

So, for this transformation an appropriate correction to the Hamiltonian is

$$\begin{array}{ll}K(\Omega )\left(t;x\prime ,y\prime ,z\prime ;p_x^{\prime },p_y^{\prime },p_z^{\prime }\right)=-\Omega \left(x\prime p_y^{\prime }-y\prime p_x^{\prime }\right),\hfill & (5.45)\hfill \end{array}$$

which is minus the rate of rotation of the coordinate system multiplied by the angular momentum. We implement K as a procedure

(define ((K Omega) s)
  (let ((qp (coordinate s)) (pp (momentum s)))
    (let ((xp (ref qp 0)) (yp (ref qp 1))
                          (ppx (ref pp 0)) (ppy (ref pp 1)))
      (* -1 Omega (- (* xp ppy) (* yp ppx))))))

and apply the test. We find:

((canonical-K? (C-rotating 'Omega) (K 'Omega))
 (up 't (up 'xp 'yp 'zp) (down 'pp_x 'pp_y 'pp_z)))
(up 0 (up 0 0 0) (down 0 0 0))

The residuals are zero so this K correctly completes the canonical transformation.

5.2.2 Abstracting the Canonical Condition

We just saw that for the case of rotating coordinates the truth of equation (5.41) did not depend on the details of the Hamiltonian. If C_H satisfies equation (5.41) for any H then we can derive a condition on C_H that is independent of H.

Let's start with an expanded version of equation (5.41):

$$\begin{array}{ll}\mathcal{D}H\circ C_{\text{H}}=D{C}_{\text{H}}·((\mathcal{D}H\circ C_{\text{H}})·D{C}_{\text{H}}),\hfill & (5.46)\hfill \end{array}$$

using the chain rule.

We introduce a shuffle function:

$$\begin{array}{ll}\tilde{J}([a,b,c])=(0,c,-b).\hfill & (5.47)\hfill \end{array}$$

The argument to $\tilde{J}$ is a down tuple of components of the derivative of a Hamiltonian-like function. The shuffle function is linear. Using $\tilde{J}$ we can write $\mathcal{D}H=\tilde{J}\circ DH$.

Let J be the multiplier corresponding to the constant linear function $\tilde{J}$:

$$\begin{array}{ll}J=(D\tilde{J})(s^{\star }),\hfill & (5.48)\hfill \end{array}$$

where s^⋆ is an arbitrary argument, shaped like DH(s), that is compatible for multiplication with s. The value of s^⋆ is irrelevant because D$\tilde{J}$ is a constant function. Then we can rewrite equation (5.46) as

$$\begin{array}{ll}J·DH(C_{\text{H}}(s\prime ))=D{C}_{\text{H}}(s\prime )·J·(DH(C_{\text{H}}(s\prime ))·D{C}_{\text{H}}(s\prime )).\hfill & (5.49)\hfill \end{array}$$

We can move the DC_H(s′) to the left of DH(C_H(s′)) by taking its transpose:¹¹

$$\begin{array}{ll}J·DH(C_{\text{H}}(s\prime ))\hfill & \hfill \\ =D{C}_{\text{H}}(s\prime )·J·({(D{C}_{\text{H}}(s\prime ))}^{\mathcal{T}}·DH(C_{\text{H}}(s\prime ))).\hfill & (5.50)\hfill \end{array}$$

Since ${(D{C}_{\mathrm{H}}(s\prime ))}^{\mathcal{T}}$ is a linear transformation and multiplication is associative for the multipliers of linear transformations, we can write

$$\begin{array}{ll}J·DH(C_{\text{H}}(s\prime ))=D{C}_{\text{H}}(s\prime )·J·{(D{C}_{\text{H}}(s\prime ))}^{\mathcal{T}}·DH(C_{\text{H}}(s\prime )).\hfill & (5.51)\hfill \end{array}$$

This is true for any H if

$$\begin{array}{ll}J=D{C}_{\text{H}}(s\prime )·J·{(D{C}_{\text{H}}(s\prime ))}^{\mathcal{T}}.\hfill & (5.52)\hfill \end{array}$$

As a program, this is^12,13

(define (J-func DHs)
  (up 0 (ref DHs 2) (- (ref DHs 1))))

(define ((canonical-transform? C) s)
  (let ((J ((D J-func) (compatible-shape s)))
        (DCs ((D C) s)))
    (- J (* DCs J (transpose DCs s)))))

This condition, equation (5.52), on C_H, called the canonical condition, does not depend on the details of H. This is a remarkable result: we can decide whether a phase-space transformation preserves the dynamics of Hamilton's equations without further reference to the details of the dynamical system. If the transformation is time dependent we can add a correction to the Hamiltonian to make it canonical.

Examples

The polar-canonical transformation satisfies the canonical condition:

((canonical-transform? (polar-canonical 'alpha))
 (up 't 'theta 'I))
(up (up 0 0 0) (up 0 0 0) (up 0 0 0))

But not every transformation we might try satisfies the canonical condition. For example, we might try x = p sin θ and p_x = p cos θ. The implementation is

(define (a-non-canonical-transform state)
  (let ((t (time state))
        (theta (coordinate state))
        (p (momentum state)))
    (let ((x (* p (sin theta)))
          (p_x (* p (cos theta))))
      (up t x p_x))))

((canonical-transform? a-non-canonical-transform)
 (up 't 'theta 'p))
(up (up 0 0 0) (up 0 0 (+ -1 p)) (up 0 (+ 1 (* -1 p)) 0))

So this transformation does not satisfy the canonical condition.

Canonical condition and Poisson brackets

The canonical condition can be written simply in terms of Poisson brackets.

The Poisson bracket can be written in terms of $\tilde{J}$:

$$\begin{array}{ll}\{f,g\}=(Df)·(\tilde{J}\circ (Dg))=(Df)·J·(Dg),\hfill & (5.53)\hfill \end{array}$$

as can be seen by writing out the components.

We break the transformation C_H into position and momentum parts:

$$\begin{array}{ll}q=A(t,q\prime ,p\prime )\hfill & (5.54)\hfill \end{array}$$

$$\begin{array}{ll}p=B(t,q\prime ,p\prime ).\hfill & (5.55)\hfill \end{array}$$

In terms of the individual component functions, the canonical condition (5.52) is

$$\begin{array}{ll}{\delta }_j^i=\{A^i,B_j\}\hfill & \hfill \\ 0=\{A^i,A^j\}\hfill & \hfill \\ 0=\{B_i,B_j\}\hfill & (5.56)\hfill \end{array}$$

where ${\delta }_j^i$ is 1 if i = j and 0 otherwise. These equations are called the fundamental Poisson brackets. If a transformation satisfies these Poisson bracket relations then it satisfies the canonical condition.

We have found that a transformation is canonical if its position-momentum part satisfies the canonical condition, but for a time-dependent transformation we may have to modify the Hamiltonian by the addition of a suitable K. We can rewrite these conditions in terms of Poisson brackets. If the Hamiltonian is

$$\begin{array}{ll}H\prime (t,q\prime ,p\prime )=H(t,A(t,q\prime ,p\prime ),B(t,q\prime ,p\prime ))+K(t,q\prime ,p\prime ),\hfill & (5.57)\hfill \end{array}$$

the transformation will be canonical if the coordinate-momentum transformation satisfies the fundamental Poisson brackets, and K satisfies:

$$\begin{array}{ll}\{A^i,K\}+{\partial }_0{A}^i=0\hfill & \hfill \\ \{B_j,K\}+{\partial }_0{B}_j=0.\hfill & (5.58)\hfill \end{array}$$

Exercise 5.5: Poisson bracket conditions

Fill in the details to show that the canonical condition (5.52) is equivalent to the fundamental Poisson brackets (5.56) and that the condition on K (5.42) is equivalent to the Poisson bracket condition on K (5.58).

Symplectic matrices

It is convenient to reformulate the canonical condition in terms of matrices. We can obtain a matrix representation of a structure with the utility s->m that takes a structure that represents a multiplier of a linear transformation and returns a matrix representation of the multiplier. The procedure s->m takes three arguments: (s->m ls A rs). The ls and rs specify the shapes of objects that multiply A on the left and right to give a numerical value. These specify the basis. So, the matrix representation of the multiplier corresponding to $\tilde{J}$ is

(let* ((s (up 't (up 'x 'y) (down 'px 'py)))
       (s* (compatible-shape s))
       (J ((D J-func) s*)))
  (s->m s* J s*))

(matrix-by-rows
  (list 0 0 0 0 0)
  (list 0 0 0 1 0)
  (list 0 0 0 0 1)
  (list 0 -1 0 0 0)
  (list 0 0 -1 0 0))

This matrix, J, is useful, so we supply a procedure J-matrix so that (J-matrix n) gives this matrix for an n degree-of-freedom system.

We can now reexpress the canonical condition (5.52) as a matrix equation:

$$\begin{array}{ll}J=D{C}_{\text{H}}(s\prime )·J·{(D{C}_{\text{H}}(s\prime ))}^{\mathcal{T}}.\hfill & (5.59)\hfill \end{array}$$

There is a further simplification available. The elements of the first row and the first column of the matrix representation of $\tilde{J}$ are all zeros. This has simplifying consequences. Consider a general transformation of phase-space states (for two degrees of freedom):

(define C-general
  (literal-function 'C
                    (-> (UP Real (UP Real Real) (DOWN Real Real))
                        (UP Real (UP Real Real) (DOWN Real Real)))))

Consider transformations for which the time does not depend on the coordinates or momenta¹⁴

(define (C-simple-time s)
  (let ((cs (C-general s)))
    (up ((literal-function 'tau) (time s))
        (coordinates cs)
        (momenta cs))))

For this kind of transformation the first row and the first column of the residuals of the canonical-transform? test are identically zero:

(let* ((s (up 't (up 'x 'y) (down 'p_x 'p_y)))
       (s* (compatible-shape s)))
  (m:nth-row
    (s->m s* ((canonical-transform? C-simple-time) s) s*)
    0))

(up 0 0 0 0 0)

(let ((s (up 't (up 'x 'y) (down 'p_x 'p_y)))
      (s* (compatible-shape s)))
  (m:nth-col
    (s->m s* ((canonical-transform? C-simple-time) s) s*)
    0))

(up 0 0 0 0 0)

But for C-general these are not zero. Since the transformations we are considering at most shift time, we need to consider only the submatrix associated with the coordinates and the momenta.

The qp submatrix¹⁵ of dimension 2n × 2n of the matrix J is called the symplectic unit for n degrees of freedom:

$$\begin{array}{ll}{J}_n=\left(\begin{array}{ll}{0}_{n\times n}\hfill & 1_{n\times n}\hfill \\ -1_{n\times n}\hfill & 0_{n\times n}\hfill \end{array}\right).\hfill & (5.60)\hfill \end{array}$$

The matrix J_n satisfies the following identities:

$$\begin{array}{ll}{J}_n^{\mathcal{T}}=J_n^{-1}=-J_n.\hfill & (5.61)\hfill \end{array}$$

A 2n × 2n matrix A that satisfies the relation

$$\begin{array}{ll}{J}_n=A{J}_n{A}^{\mathcal{T}}\hfill & (5.62)\hfill \end{array}$$

is called a symplectic matrix. We can determine whether a matrix is symplectic:

(define (symplectic-matrix? M)
  (let ((2n (m:dimension M)))
    (let ((J (symplectic-unit (quotient 2n 2))))
      (- J (* M J (transpose M))))))

An appropriate symplectic unit matrix of a given size is produced by the procedure symplectic-unit.

If the matrix representation of the derivative of a transformation is a symplectic matrix the transformation is a symplectic transformation. Here is a test for whether a transformation is symplectic:¹⁶

(define ((symplectic-transform? C) s)
  (symplectic-matrix? (qp-submatrix ((D-as-matrix C) s))))

The procedure symplectic-transform? returns a zero matrix if and only if the transformation being tested passes the symplectic matrix test.

For example, the point transformations are symplectic. We can show this for a general possibly time-dependent two-degree-of-freedom point transformation:

(define (F s)
  ((literal-function 'F
                     (-> (X Real (UP Real Real)) (UP Real Real)))
   (time s) (coordinates s)))

((symplectic-transform? (F->CH F))
 (up 't (up 'x 'y) (down 'px 'py)))

(matrix-by-rows
  (list 0 0 0 0)
  (list 0 0 0 0)
  (list 0 0 0 0)
  (list 0 0 0 0))

More generally, the phase-space part of the canonical condition is equivalent to the symplectic condition (for two degrees of freedom) even in the case of an unrestricted phase-space transformation.

(let* ((s (up 't (up 'x 'y) (down 'p_x 'p_y)))
       (s* (compatible-shape s)))
  (- (qp-submatrix
       (s->m s* ((canonical-transform? C-general) s) s*))
     ((symplectic-transform? C-general) s)))

(matrix-by-rows
  (list 0 0 0 0)
  (list 0 0 0 0)
  (list 0 0 0 0)
  (list 0 0 0 0))

Exercise 5.6: Symplectic matrices

Let A be a symplectic matrix: $J_n=A{J}_n{A}^{\mathcal{T}}$. Show that $A^{\mathcal{T}}$ and A⁻¹ are symplectic.

Exercise 5.7: Polar-canonical transformations

Let x, p and θ, I be two sets of canonically conjugate variables. Consider transformations of the form x = βI^α sin θ and p = βI^α cos θ. Determine all α and β for which this transformation is symplectic.

Exercise 5.8: Standard map

Is the standard map a symplectic transformation? Recall that the standard map is: I′ = I + K sin θ, with θ′ = θ + I′, both modulo 2π.

Exercise 5.9: Whittaker transform

Shew that the transformation q = log ((sin p′)/q′) with p = q′ cot p′ is symplectic.

5.3   Invariants of Canonical Transformations

Canonical transformations allow us to change the phase-space coordinate system that we use to express a problem, preserving the form of Hamilton's equations. If we solve Hamilton's equations in one phase-space coordinate system we can use the transformation to carry the solution to the other coordinate system. What other properties are preserved by a canonical transformation?

Noninvariance of pv

We noted in equation (5.10) that point transformations that are canonical extensions of time-independent coordinate transformations preserve the value of pv. This does not hold for more general canonical transformations. We can illustrate this with the polar-canonical transformation. Along corresponding paths x, p_x and θ, I

$$\begin{array}{lll}\hfill x(t)& =\sqrt{\frac{2I(t)}{\alpha }}\sin \theta (t)\hfill & \hfill \\ \hfill p_x(t)& =\sqrt{2I(t)\alpha }\cos \theta (t),\hfill & (5.63)\hfill \end{array}$$

and so Dx is

$$\begin{array}{ll}Dx(t)=D\theta (t)\sqrt{\frac{2I(t)}{\alpha }}\cos \theta (t)+DI(t)\frac{1}{\sqrt{2I(t)\alpha }}\sin \theta (t).\hfill & (5.64)\hfill \end{array}$$

The difference of pv and the transformed p′v′ is

$$\begin{array}{ll}{P}_x(t)Dx(t)-I(t)D\theta (t)\hfill & \hfill \\ =I(t)D\theta (t)(2{\cos }^2\theta (t)-1)+DI(t)\sin \theta (t)\cos \theta (t).\hfill & (5.65)\hfill \end{array}$$

In general this is not zero. So the product pv is not necessarily invariant under general canonical transformations.

Invariance of Poisson brackets

Here is a remarkable fact: the composition of the Poisson bracket of two phase-space state functions with a canonical transformation is the same as the Poisson bracket of each of the two functions composed with the transformation separately. Loosely speaking, the Poisson bracket is invariant under canonical phase-space transformations.

Let f and g be two phase-space state functions. Using the $\tilde{J}$ representation of the Poisson bracket (see section 5.2.2), we deduce

$$\begin{array}{ll}\{f\circ C_{\text{H}},g\circ C_{\text{H}}\}\hfill & \hfill \\ =(D(f\circ C_{\text{H}}))·(\tilde{J}\circ D(g\circ C_{\text{H}}))\hfill & \hfill \\ =(Df\circ C_{\text{H}})·D{C}_{\text{H}}·(\tilde{J}\circ ((Dg\circ C_{\text{H}})·D{C}_{\text{H}}))\hfill & \hfill \\ =(Df\circ C_{\text{H}})·(\tilde{J}\circ Dg\circ C_{\text{H}})\hfill & \hfill \\ =(Df·(\tilde{J}\circ Dg))\circ C_{\text{H}}\hfill & \hfill \\ =\{f,g\}\circ C_{\text{H}},\hfill & (5.66)\hfill \end{array}$$

where the fact that C_H satisfies equation (5.41) was used in the middle. This is

$$\begin{array}{ll}\{f\circ C_{\text{H}},g\circ C_{\text{H}}\}=\{f,g\}\circ C_{\text{H}}.\hfill & (5.67)\hfill \end{array}$$

Volume preservation

Consider a canonical transformation C_H. Let Ĉ_t be a function with parameter t such that (q, p) = Ĉ_t(q′, p′) if (t, q, p) = C_H(t, q′, p′). The function Ĉ_t maps phase-space coordinates to alternate phase-space coordinates at a given time. Consider regions R in (q, p) and R in (q′, p′) such that R = Ĉ_t(R′). The volume of region R′ is

$$\begin{array}{ll}V(R)={\displaystyle {\int }_R\hat{1}}={\displaystyle {\int }_{R\prime }\det (D{\hat{C}}_t),}\hfill & (5.68)\hfill \end{array}$$

where $\hat{1}$ is the function whose value is one for every input. Now if C_H is symplectic then the determinant of DĈ_t is one (see section 4.2.3), so

$$\begin{array}{ll}V(R)=V(R\prime ).\hfill & (5.69)\hfill \end{array}$$

Thus, phase-space volume is preserved by symplectic transformations.

Liouville's theorem shows that time evolution preserves phase-space volume. Here we see that canonical transformations also preserve phase volumes. Later, we will find that time evolution actually generates a canonical transformation.

The symplectic 2-form

Define

$$\begin{array}{ll}\omega ({\zeta }_1,{\zeta }_2)=P({\zeta }_2)Q({\zeta }_1)-P({\zeta }_1)Q({\zeta }_2),\hfill & (5.70)\hfill \end{array}$$

where Q = I₁ and P = I₂ are the coordinate and momentum selectors, respectively. The arguments ζ₁ and ζ₂ are incremental phase-space states with zero time components.

The ω form can also be written as a sum over degrees of freedom:

$$\begin{array}{ll}\omega ({\zeta }_1,{\zeta }_2)={\displaystyle \sum _i(P_i({\zeta }_2)Q^i({\zeta }_1)-P_i({\zeta }_1)Q^i({\zeta }_2)).}\hfill & (5.71)\hfill \end{array}$$

Notice that the contributions for each i do not mix components from different degrees of freedom.

This bilinear form is closely related to the symplectic 2-form of differential geometry. It differs in that the symplectic 2-form is formally a function of the phase-space point as well as the incremental vectors.

Under a canonical transformation s = C_H(s′), incremental states transform with the derivative

$$\begin{array}{ll}{\zeta }_i=D{C}_{\text{H}}(s\prime ){\zeta }_i^{\prime }.\hfill & (5.72)\hfill \end{array}$$

We will show that the 2-form is invariant under this transformation

$$\begin{array}{ll}\omega ({\zeta }_1,{\zeta }_2)=\omega ({\zeta }_1^{\prime },{\zeta }_2^{\prime }),\hfill & (5.73)\hfill \end{array}$$

if the time components of the ${\zeta }_i^{\prime }$ are both zero.

We have shown that condition (5.41) does not depend on the details of the Hamiltonian H. So if a transformation satisfies the canonical condition we can use condition (5.41) with H replaced by an arbitrary function f of phase-space states:

$$\begin{array}{ll}\mathcal{D}f(C_{\text{H}}(s\prime ))=(D{C}_{\text{H}}(s\prime ))·(\mathcal{D}(f\circ C_{\text{H}})(s\prime )).\hfill & (5.74)\hfill \end{array}$$

In terms of ω, the Poisson bracket is

$$\begin{array}{ll}\{f,g\}(s)=\omega (\mathcal{D}f(s),\mathcal{D}g(s))\hfill & (5.75)\hfill \end{array}$$

as can be seen by writing out the components. We use the fact that Poisson brackets are invariant under canonical transformations:

$$\begin{array}{ll}(\{f,g\}\circ C_{\text{H}})(s\prime )=\{f\circ C_{\text{H}},g\circ C_{\text{H}}\}(s\prime ).\hfill & (5.76)\hfill \end{array}$$

Using the relation (5.74) to expand the left-hand side of equation (5.76) we obtain:

$$\begin{array}{ll}(\{f,g\}\circ C_{\text{H}})(s\prime )\hfill & \hfill \\ =\omega ((\mathcal{D}f\circ C_{\text{H}})(s\prime ),(\mathcal{D}g\circ C_{\text{H}})(s\prime ))\hfill & \hfill \\ =\omega ((D{C}_{\text{H}}(s\prime ))·(\mathcal{D}(f\circ C_{\text{H}})(s\prime )),\hfill & \hfill \\ (D{C}_{\text{H}}(s\prime ))·(\mathcal{D}(g\circ C_{\text{H}})(s\prime ))).\hfill & (5.77)\hfill \end{array}$$

The right-hand side of equation (5.76) is

$$\begin{array}{ll}\{f\circ C_{\text{H}},g\circ C_{\text{H}}\}(s\prime )=\omega (\mathcal{D}(f\circ C_{\text{H}})(s\prime ),\mathcal{D}(g\circ C_{\text{H}})(s\prime )).\hfill & (5.78)\hfill \end{array}$$

Now the left-hand side must equal the right-hand side for any f and g, so the equation must also be true for arbitrary ${\zeta }_i^{\prime }$ of the form

$$\begin{array}{ll}{\zeta }_1^{\prime }=\mathcal{D}(f\circ C_{\text{H}})(s\prime )\hfill & \hfill \\ {\zeta }_2^{\prime }=\mathcal{D}(g\circ C_{\text{H}})(s\prime ).\hfill & (5.79)\hfill \end{array}$$

So the ${\zeta }_i^{\prime }$ are arbitrary incremental states with zero time components.

We have proven that

$$\begin{array}{ll}\omega ({\zeta }_1^{\prime },{\zeta }_2^{\prime })=\omega (D{C}_{\text{H}}(s\prime )·{\zeta }_1^{\prime },D{C}_{\text{H}}(s\prime )·{\zeta }_2^{\prime }).\hfill & (5.80)\hfill \end{array}$$

for canonical C_H and incremental states ${\zeta }_i^{\prime }$ with zero time components. Using equation (5.72), we have

$$\begin{array}{ll}\omega ({\zeta }_1^{\prime },{\zeta }_2^{\prime })=\omega ({\zeta }_1,{\zeta }_2).\hfill & (5.81)\hfill \end{array}$$

Thus the bilinear antisymmetric function ω is invariant under even time-varying canonical transformations if the increments are restricted to have zero time component.

As a program, ω is

(define (omega zeta1 zeta2)
  (- (* (momentum zeta2) (coordinate zeta1))
     (* (momentum zeta1) (coordinate zeta2))))

On page 356 we showed that point transformations are sym-plectic. Here we can see that the 2-form is preserved under these transformations for two degrees of freedom:

(define (F s)
  ((literal-function 'F
                     (-> (X Real (UP Real Real)) (UP Real Real)))
   (time s)
   (coordinates s)))

(let ((s (up 't (up 'x 'y) (down 'p_x 'p_y)))
      (zeta1 (up 0 (up 'dx1 'dy1) (down 'dp1_x 'dp1_y)))
      (zeta2 (up 0 (up 'dx2 'dy2) (down 'dp2_x 'dp2_y))))
  (let ((DCs ((D (F->CH F)) s)))
    (- (omega zeta1 zeta2)
       (omega (* DCs zeta1) (* DCs zeta2)))))
0

Alternatively, let z₁ and z₂ be the matrix representations of the qp parts of ζ₁ and ζ₂. The matrix representation of ω is

$$\begin{array}{ll}\omega ({\zeta }_1,{\zeta }_2)=z_1^{\mathcal{T}}·J_n·z_2.\hfill & (5.82)\hfill \end{array}$$

Let A be the matrix representation of the qp part of DC_H(s′) Then the invariance of ω is equivalent to

$$\begin{array}{ll}{z}_1^{\mathcal{T}}·A^{\mathcal{T}}·{\text{J}}_n·A·z_2=z_1^{\mathcal{T}}·J_n·z_2.\hfill & (5.83)\hfill \end{array}$$

But this is true if

$$\begin{array}{ll}{A}^{\mathcal{T}}·J_n·A=J_n,\hfill & (5.84)\hfill \end{array}$$

which is equivalent to the condition that A is symplectic. (If a matrix is symplectic then its transpose is symplectic. See exercise 5.6).

The symplectic condition is symmetrical in that if A is symplec-tic then $A^{\mathcal{T}}$ is symplectic, because the symplectic unit is invertible. The canonical condition

$$\begin{array}{ll}J=D{C}_{\text{H}}(s\prime )·J·{(D{C}_{\text{H}}(s\prime ))}^{\mathcal{T}},\hfill & (5.85)\hfill \end{array}$$

is satisfied by time-varying canonical transformations, and time-varying canonical transformations are symplectic. But if the transformation is time varying then

$$\begin{array}{ll}J={(D{C}_{\text{H}}(s\prime ))}^{\mathcal{T}}·J·D{C}_{\text{H}}(s\prime ),\hfill & (5.86)\hfill \end{array}$$

is not satisfied because J is not invertible. Equation (5.86) is satisfied, however, for time-independent transformations.

Poincaré integral invariant

The invariance of the symplectic 2-form under canonical transformations has a simple interpretation. Consider how the area of an incremental parallelogram in phase space transforms under canonical transformation. Let (Δq, Δp) and (δq, δp) be small increments in phase space, originating at (q, p). Consider the incremental parallelogram with vertex at (q, p) with these two phase-space increments as edges. The sum of the areas of the canonical projections of this incremental parallelogram can be written

$$\begin{array}{ll}{\displaystyle \sum _i\Delta A_i={\displaystyle \sum _i(\Delta q^i\delta p_i-\Delta p_i\delta q^i).}}\hfill & (5.87)\hfill \end{array}$$

The right-hand side is the sum of the areas on the canonical planes;¹⁷ for each i the area of a parallelogram is computed from the components of the vectors defining its adjacent sides. Let ζ₁ = (0, Δq, Δp) and ζ₂ = (0, δq, δp); then the sum of the areas of the incremental parallelograms is just

$$\begin{array}{ll}{\displaystyle \sum _i\Delta A_i=\omega ({\zeta }_1,{\zeta }_2),}\hfill & (5.88)\hfill \end{array}$$

where ω is the bilinear antisymmetric function introduced in equation (5.70). The function ω is invariant under canonical transformations, so the sum of the areas of the incremental parallelograms is invariant under canonical transformations.

There is an integral version of this differential relation. Consider the oriented area of a region R′ in phase space (see figure 5.2). Suppose we make a canonical transformation from coordinates (q′, p′) to (q, p) taking region R′ to region R. The boundary of the region in the transformed coordinates is just the image under the canonical transformation of the original boundary. Let $R_{q^i,p_i}$ be the projection of the region R onto the qⁱ, p_i plane of coordinate qⁱ and conjugate momentum p_i, and let A_i be its area. Similarly, let $R_{q{\prime }^i,p_i^{\prime }}^{\prime }$ be the projection of R′ onto the $q{\prime }^i,p_i^{\prime }$ plane, and let $A_i^{\prime }$ be its area.

The area of an arbitrary region is just the limit of the sum of the areas of incremental parallelograms that cover the region, so the sum of oriented areas is preserved by canonical transformations:

$$\begin{array}{ll}{\displaystyle \sum _i{A}_i={\displaystyle \sum _i{A}_i^{\prime }.}}\hfill & (5.89)\hfill \end{array}$$

That is, the sum of the projected areas on the canonical planes is preserved by canonical transformations. Another way to say this is

$$\begin{array}{ll}{\displaystyle \sum _i{\displaystyle {\int }_{R_{q^i,p_i}}d{q}^{i}d{p}_i={\displaystyle \sum _i{\displaystyle {\int }_{R_{q{\prime }^i,p_i^{\prime }}^{\prime }}dq{\prime }^{i}d{p}_i^{\prime }.}}}}\hfill & (5.90)\hfill \end{array}$$

The equality-of-areas relation (5.90) can also be written as an equality of line integrals using Stokes's theorem, for simply-connected regions $R_{q^i,p_i}$ and $R_{q{\prime }^i,p_i^{\prime }}^{\prime }$:

$$\begin{array}{ll}{\displaystyle \sum _i{\displaystyle {\oint }_{\partial R_{q^i,p_i}}{p}_{i}d{q}^i={\displaystyle \sum _i{\displaystyle {\oint }_{\partial R_{q{\prime }^i,p_i^{\prime }}^{\prime }}{p}_i^{\prime }dq{\prime }^i.}}}}\hfill & (5.91)\hfill \end{array}$$

The canonical planes are disjoint except at the origin, so the projected areas intersect in at most one point. Thus we may independently accumulate the line integrals around the boundaries of the individual projections of the region onto the canonical planes into a line integral around the unprojected region:

$$\begin{array}{ll}{\displaystyle {\oint }_{\partial R}{\displaystyle \sum _i{p}_{i}d{q}^i={\displaystyle {\oint }_{\partial R\prime }{\displaystyle \sum _i{p}_i^{\prime }dq{\prime }^i.}}}}\hfill & (5.92)\hfill \end{array}$$

Exercise 5.10: Watch out

Consider the canonical transformation C_H:

$$(t,x,p)=C_{\text{H}}(t,\theta ,J)=(t,\sqrt{2(J+a)}\sin \theta ,\sqrt{2(J+a)}\cos \theta ).$$

a. Show that the transformation is symplectic for any a.

b. Show that equation (5.92) is not generally satisfied for the region enclosed by a curve of constant J.

5.4   Generating Functions

We have considered a number of properties of general canonical transformations without having a method for coming up with them. Here we introduce the method of generating functions. The generating function is a real-valued function that compactly specifies a canonical transformation through its partial derivatives, as follows.

Consider a real-valued function F₁(t, q, q′) mapping configurations expressed in two coordinate systems to the reals. We will use F₁ to construct a canonical transformation from one coordinate system to the other. We will show that the following relations among the coordinates, the momenta, and the Hamiltonians specify a canonical transformation:

$$\begin{array}{ll}p={\partial }_1{F}_1(t,q,q\prime )\hfill & (5.93)\hfill \end{array}$$

$$\begin{array}{ll}p\prime =-{\partial }_2{F}_1(t,q,q\prime )\hfill & (5.94)\hfill \end{array}$$

$$\begin{array}{ll}H\prime (t,q\prime ,p\prime )-H(t,q,p)={\partial }_0{F}_1(t,q,q\prime ).\hfill & (5.95)\hfill \end{array}$$