Lagrangian and Hamiltonian Dynamics

Volker Perlick

(Lancaster University)

Lecture 4

Canonical Transformations

(Cockcroft Institute, 15 March 2010) In this lecture we will discuss the group of transformations that leave Hamilton’s equations ∂H ∂H x˙ i = , p˙ = − , i i ∂pi ∂x invariant. We begin by introducing a useful notation. 1 n For any two phase space functions f(x ,...,x , p1,...,pn,t) and 1 n g(x ,...,x , p1,...,pn,t) define the “Poisson bracket” n ∂f ∂g ∂f ∂g {f,g} = − . ∂xi ∂p ∂p ∂xi i i i X=1   We compare this expression with the total time derivative of f along a solution to Hamilton’s equations,

n i df ∂f dx ∂f dpi ∂f = + + = dt ∂xi dt ∂p dt ∂t i i X=1   n ∂f ∂H ∂f ∂H ∂f = − + . ∂xi ∂p ∂p ∂xi ∂t i i i X=1   -2- This shows that df ∂f = {f, H} + . dt ∂t Thus, if f does not depend on time explicitly, its total time deriva- tive along solutions of Hamilton’s equations is given by the Poisson bracket with the Hamiltonian.

In particular: If f does not depend on time explicitly, we have df = 0 ⇐⇒ {f, H} = 0 . dt i.e., f is a constant of motion if and only if its Poisson bracket with H vanishes.

-3- The Poisson bracket

n ∂f ∂g ∂f ∂g {f,g} = − ∂xi ∂p ∂p ∂xi i i i X=1   satisfies the following calculation rules.

• Linearity: {f,ag + bh} = a {f,g} + b {f,h} .

• Antisymmetry: {f,g} = − {g,f} .

• Jacobi identity: {f,g},h + {g,h},f + {h,f},g = 0 .    These are the defining properties of a Lie Algebra.

In addition, the Poisson bracket satisfies the derivation rule

{f,gh = g {f,h} + h {f,g} .

-4- We are now ready to define canonical transformations. Definition: A transformation

0i 0i 1 n x = x (x ,...,x , p1,...,pn) ,

0 0 1 n pi = pi(x ,...,x , p1,...,pn) , is called “canonical” if

0i 0j 0 0 0i 0 i {x , x } = 0 , {pi, pj} = 0 , {x , pj} = δj . In the literature one can find several alternative definitions of canon- ical transformations, all of which turn out to be equivalent to the one chosen here. One of these alternative characterisations is the following.

-5- Claim: A transformation is canonical if and only if it leaves the Pois- son bracket of any two phase space functions invariant,

{f,g} = {f 0,g0}0 .

Here

0 01 0n 0 0 1 n f (x ,...,x , p1,...,pn,t) = f(x ,...,x , p1,...,pn,t) ,

0 01 0n 0 0 1 n g (x ,...,x , p1,...,pn,t) = g(x ,...,x , p1,...,pn,t) ,

n ∂f 0 ∂g0 ∂f 0 ∂g0 {f 0,g0}0 = − . ∂x0i ∂p0 ∂p0 ∂x0i i i i X=1  

-6- Proof: For arbitrary transformations, we find

n ∂f ∂g ∂f ∂g {f,g} = − = ∂xi ∂p ∂p ∂xi i i i X=1   n n n ∂f 0 ∂x0j ∂f 0 ∂p0 ∂g0 ∂x0k ∂g0 ∂p0 + j + k ∂x0j ∂xi ∂p0 ∂xi ∂x0k ∂p ∂p0 ∂p i j k j i k i X=1 X=1 X=1    ∂f 0 ∂x0j ∂f 0 ∂p0 ∂g0 ∂x0k ∂g0 ∂p0 − + j + k = 0j 0 0k i 0 i ∂x ∂pi ∂pj ∂pi ∂x ∂x ∂pk ∂x !    n n ∂f 0 ∂g0 ∂f 0 ∂g0 {x0j, x0k} + {p0 , p0 } ∂x0j ∂x0k ∂p0 ∂p0 j k j k j k X=1 X=1 ∂f 0 ∂g0 ∂f 0 ∂g0 + {x0j, p0 } + {p0 , x0k} . 0j 0 k 0 0k j ∂x ∂pk ∂pj ∂x ! The right-hand side is equal to {f 0,g0}0 for all functions f 0 and g0 if and only if the transformation is canonical. 

-7- Note: Here one has to be careful about the notation. For f = xi, the function f 0 is NOT equal to x0i; it is rather xi expressed in terms of 0i 0 0i i0 the x and the pi , i.e., x =6 x .

Examples: • Every point transformation is a canonical transformation. • 0i 0 i The transformation x = pi , pi = −x is canonical.

pi − lines Our old picture remains



















true only under point


















transformations, not un-


der arbitrary canonical

transformations.
xi − lines

-8- The fact that canonical transformations are of crucial relevance for the Hamiltonian formalism is in the following result.

Claim: A transformation is canonical if and only if it preserves the form of Hamilton’s equations for all Hamiltonian functions, i.e., if and only if ∂H ∂H x˙ i = , p˙ = − , i i ∂pi ∂x implies

∂H0 ∂H0 x˙ 0i = , p˙0 = − , 0 i 0i ∂pi ∂x where

0 01 0n 0 0 1 n H (x ,...,x , p1,...,pn,t) = H(x ,...,x , p1,...,pn,t) .

-9- Proof: Assume that Hamilton’s equations hold. Then

n 0 0 n 0 0 0 ∂pi k ∂pi ∂pi ∂H ∂pi ∂H p˙ = x˙ + p˙k = − = i ∂xk ∂p ∂xk ∂p ∂p ∂xk k k k k k X=1   X=1   n n ∂p0 ∂H0 ∂x0j ∂H0 ∂p0 ∂p0 ∂H0 ∂x0j ∂H0 ∂p0 i + j − i + j . ∂xk ∂x0j ∂p ∂p0 ∂p ∂p ∂x0j ∂xk ∂p0 ∂xk k j k j k k j ! X=1 X=1     After rearranging, we find

n ∂H0 ∂H0 p˙0 = {p0, x0j} + {p0, p0 } . i ∂x0j i ∂p0 i j j j ! X=1 An analogous calculation leads to

n ∂H0 ∂H0 x˙ 0i = {x0i, p0 } + {x0i, x0j} . ∂p0 j ∂x0j j j ! X=1 Thus, the primed Hamilton equations hold, for any H0, if and only if the transformation is canonical. 

-10- There are actually two different aspects to the relation between Hamil- ton’s equations and canonical transformations. First, as we have seen, canonical transformatiosn leave Hamilton’s equations invariant. Second, the phase flow defined by the Hamiltonian can be viewed as a one-parameter family of canonical transformations. This can be formulated in the following way.

1 n Claim: Consider any Hamiltonian H x ,...,x , p1,...,pn,t . Fix a time t and a time interval ε. Then the transformation 0
1 n x (t0),...,x (t0), p1(t0),...,pn(t0)   1 n 7−→ x (t0 + ε),...,x (t0 + ε), p1(t0 + ε),...,pn(t0 + ε)

 i  is canonical, where the x (t) and pj(t) satisfy Hamilton’s equations.

-11- Proof: It suffices to give the proof for infinitesimally small ε because then the general result follows by integration. As

i i i x (t0 + ε) = x (t0) + ε x˙ (t0) =

i ∂H 1 n = x (t0) + ε x (t0),...,x (t0), p1(t0),...,pn(t0),t0 , ∂pi   i pi(t0 + ε) = pi(t0) + ε p˙ (t0) =

∂H 1 n = pi(t ) − ε x (t ),...,x (t ), p (t ),...,pn(t ),t , 0 ∂xi 0 0 1 0 0 0 we have to prove that the transformation 

0i i ∂H 1 n x = x + ε x ,...,x , p1,...,pn,t0 , ∂pi   0 ∂H 1 n p = pi − ε x ,...,x , p ,...,pn,t , i ∂xi 1 0 is canonical.  

-12- This is true since ∂H ∂H {x0i, x0j} = {xi, xj} + ε xi , + ε , xj = ∂pj ∂pi n o n o ∂2H ∂2H = 0+ ε − ε = 0 , ∂pipj ∂pjpi 0 0 0i 0 i  and, analogously, {pi, pj} = 0 and {x , pj} = δj . p

Thus, shifting the points along the solution curves x to Hamilton’s equations always gives a canonical transformation. Recall our example of the damped harmonic oscillator.

-13- We will now relate canonical transformations to what is called the symplectic structure of phase space. To that end, we rewrite the Poisson bracket in matrix notation,

∂g ∂x1 ·  ·  · 0 1  ∂g  ∂f ∂f ∂f ∂f  n  {f,g} = ··· ···  ∂x  . 1 n  ∂g  ∂x ∂x ∂p1 ∂pn     −1 0  ∂p1 
     ·   ·   ·   ∂g   ∂pn    The antisymmetric 2n × 2n matrix  

0 1 J =  −1 0    is called the “symplectic matrix” in 2n dimensions.

-14- The Poisson bracket reflects the “symplectic structure” of phase space.

“Symplectic” in greek has the same meaning as “complex” in latin. The name was introduced by Herrman Weyl in the 1920s. It refers to the fact that the 2n × 2n matrix J is “composed” of four n × n blocks, corresponding to the phase space being coordinatised 1 n by x ,...,x and p1,...,pn.

Canonical coordinates can be characterised by the fact that they pre- serve the symplectic structure in a certain sense.

To work this out, we write the chain rule in matrix form.

-15- ∂f ∂f ∂f ∂f ··· ··· = 1 n ∂x ∂x ∂p1 ∂pn   ∂x0j ∂x0j k ∂p ∂f 0 ∂f 0 ∂f 0 ∂f 0 ∂x k = ··· ···     01 0n 0 0 0 0 ∂x ∂x ∂p1 ∂pn ∂pj ∂pj    k   ∂x ∂pk       S  and

∂g | ∂g0 {z } ∂x1 ∂x01  ·   ·  · j ∂p0 · · ∂x0 j · k k  ∂g  ∂x ∂x  ∂g0   n     ∂x0n   ∂x      .  ∂g  =  ∂g0   ∂p   j ∂p0   ∂p0   1   ∂x0 j   1   ·   ∂pk ∂pk   ·   ·     ·   ·       ·     T     ∂g  S  ∂g0   ∂pn   0     ∂pn    | {z }   -16- The unprimed and primed Poisson brackets are ∂g0 ∂x01  ·  · 0 0 0 0 0  ∂g  ∂f ∂f ∂f ∂f ∂x0n  {f,g} = ··· ··· S J ST   , ∂x01 ∂x0n ∂p0 ∂p0  ∂g0  1 n  ∂p0     1   ·   ·   ·   ∂g0   0   ∂pn    ∂g0 ∂x01  ·  · 0 0 0 0 0  ∂g  ∂f ∂f ∂f ∂f ∂x0n  {f 0,g0}0 = ··· ··· J   . ∂x01 ∂x0n ∂p0 ∂p0  ∂g0  1 n  ∂p0     1   ·   ·   ·   ∂g0   0   ∂pn    -17- Thus, the transformation is canonical if and only if its Jacobi matrix leaves the symplectic matrix invariant.

S J ST = J ; For this reason, canonical transformations are also called “symplectic transformations”.

From the determinant theorem we find for a canonical transformation

det(S) det(J) det(ST ) = det(J) .

As det(ST ) = det(S) and det(J) = 1, we have found that a canonical transformation has to satisfy

det(S) = ±1 .

-18- As the volume element transforms with the Jacobi determinant S,

01 0n 0 0 1 n dx ··· dx dp1 ··· dpn = det(S) dx ··· dx dp1 ··· dpn , we have proven that a canonial transformation leaves the phase space volume invariant (up to sign; a negative sign corresponds to a change of orientation from right-handed to left-handed.)

In particular, we have proven Liouville’s theorem:

The phase flow of any Hamiltonian is volume-preserving, i.e., it be- haves like an incompressible fluid.

-19-