Assignment 8. [η j, η k ] = J jk

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1 Assignment 8 Goldstein 9.8 Prove directly that the transformation is canonical and find a generating function. Q 1 = q 1, P 1 = p 1 p Q = p, P = q 1 q We can establish that the transformation is canonical in at least two ways. One is to confirm that the Poisson brackets hold. Another is to verify the symplectic condition. We ll try both just to see how they work. In the first method with Poisson brackets, recall that we have the fundamental Poisson brackets [η j, η k ] = J jk which must be invariant under a canonical transformation. In the current context, that means that [Q j, Q k ] = 0 = [P j, P k ] and [Q j, P k ] = δ jk. In particular, we have [Q 1, P 1 ] = Q 1 = Q 1 + Q 1 q p p q ( ) [Q 1, P ] = Q 1 P P ( ) 0 ( 1) [Q, P 1 ] = Q Q ( ) [Q, P ] = Q P Q P ( ) 1 ( 1) Some of the other Poisson brackets such as [Q i, Q i ] (no sum on i) and [P i, P i ] are identically zero irrespective of the transformation. This can easily be seen by using the definition of the Poisson brackets. On the other hand, we need to find two others [Q 1, Q ] = Q 1 Q Q [P 1, P ] = P P ( ) ( ) ( 1) 1

2 So we have verified that the fundamental Poisson brackets hold for this transformation. canonical. We can also do this by confirming the symplectic condition: MJM T = J To do this, we must construct the Jacobian of the transformation, Therefore it is M ij = ζ i η j where ζ i = (Q 1, Q, P 1, P ) and η j = (q 1, q, p 1, p ). In particular, we have or M 11 = Q 1, M 1 = Q 1 q, M 13 = Q 1, M 14 = Q 1 p M 1 = Q, M = Q q, M 3 = Q, M 4 = Q p M 31 =, M 3 = q, M 33 =, M 34 = p = M 41 = P =, M 4 = P q = 1, M 43 = P, M 44 = P p M = Now, using Maple or Mathematica, one can perform the matrix multiplication to verify that indeed MJM T does yield J. This establishes, as does the Poisson bracket formulation, that the transformation is canonical. We must now find a generating function for this canonical transformation. While there are four basic canonical transformations and four basic types of generating functions associated with these canonical transformations, it is also the case that one can mix these types of transformations and their corresponding generating functions. The easiest way, I now think, to see this is to note the trivial special cases for each of the transformations as listed in Goldstein Table 9.1. Note the four special cases. In each of these, the new position and momentum are very simply related to the old position and momentum. However, they are always related in pairs, e.g. the new momentum will be related to the old coordinates are the old momentum but not to a mixture of the two. In our current case, we have exactly the latter, namely half of our transformation is an example of a trivial special case but the new coordinates are related to a mix of the old coordinates and momentum through Q 1 = q 1 and Q = p. This suggests that we need to have a hybrid or mixed generating function. Taking our form of the transformation and the trivial special cases as a cue, we guess a form F = F (q 1, p, P 1, P ) Q i P i + q p Putting this into the defining form for the canonical transformation, we get p i q i H = P q Q i K + df dt = P i Q i K + F t + F q 1 + F ṗ + F P i p P Q i P i Q i P i + q p + q ṗ i Equating terms, we find the generating function derivatives K = H + F t p 1 = F, q = F p, Q 1 = F Q = F P

3 We now want to integrate these relations for our particular canonical transformation On doing this we get and the full generating function becomes F = p 1 = P 1 + p F p = q = q 1 + P F = Q 1 = q 1 F P = Q = p F (q 1, p, P 1, P ) = q 1 P 1 + q 1 p + p P F = q 1 P 1 + q 1 p + p P Q 1 P 1 Q P + q p 3

4 Goldstein 9.3 By any method, show that the following transformation is canonical: x ( ) P1 sin Q 1 + P, p x = ( ) P1 cos Q 1 Q y ( ) P1 cos Q 1 + Q, p y = ( ) P1 sin Q 1 P where is some fixed parameter. Apply this transformation to the problem of a particle of charge q moving in a plane that is perpendicular to a constant magnetic field B. Express the Hamiltonian for this problem in the (Q i, P i ) coordinates letting the parameter take the form = q c B. From this Hamiltonian, obtain the motion of the particle as a function of time. Because we re going to have to take a bunch of derivatives, we may as well show this is canonical by verifying the invariance of the symplectic condition MJM T = J where M ij = ζ i η j and in this problem, we are letting ζ = (x, y, p x, p y ) and likewise for η. So we need all the partial derivatives: x P1 cos Q 1 Q 1 x Q x sin Q 1 P 1 x P y = 1 P1 sin Q 1 Q 1 y Q y cos Q 1 P 1 y P p x = P1 sin Q 1 Q 1 p x = Q p x = cos Q 1 4 P 1 p x P 4

5 p y = P1 cos Q 1 Q 1 p y Q p y = cos Q 1 4 P 1 p y = P Using Maple or Mathematica, you can construct M and verify the symplectic condition thereby confirming that the transformation is canonical. For the second part of the problem, we can write the Hamiltonian from Eq (note that Goldstein uses inconsistent units) as H ( p q A m c ) where, as in Problem 8.3, we take A B r B z ( îy + ĵx). This gives H ( p x + qb ) z m c y 1 ( + p y qb ) z m c x where we set = qb z /c. Making the transformation to the new phase space coordinates, we get H [ ( ) P1 cos Q 1 Q + m + 1 m [ m P 1 ( P1 sin Q 1 P ) a remarkably simple Hamiltonian. Hamilton s equations are 1 ( ) ] P1 cos Q 1 + Q ( ) ] P1 sin Q 1 + P Q 1 = H = m, Q = H P, P 1 = H Q 1 P = H Q which, when integrated yield Q 1 (t) = m t + Q 1(0) with Q, P 1 and P being constants. If we now write the solution in the original coordinates, we have x(t) [ ( t ) ] P1 sin m + Q 1(0) + P y(t) [ ( t ) ] P1 cos m + Q 1(0) + Q p x (t) = [ ( t ) ] P1 cos m + Q 1(0) Q p y (t) = [ ( t ) ] P1 sin m + Q 1(0) P where, again, Q 1 (0), Q, P 1 and P are all constants which will be given by the initial conditions. 5

6 Goldstein 9.4 (a) Show that the transformation is canonical and find a generating function. Q = p + iaq, P = p iaq ia (b) Use the transformation to solve the linear harmonic oscillator problem. Since this is a problem in just a two dimensional phase space, we have only one nontrivial Poisson bracket to evaluate in order to determine that the transformation is canonical, namely [Q, P ] q,p Q P q p Q P p q 1 = ia ia 1 1. The other Poisson brackets, [Q, Q] and [P, P ], are identically zero. Thus, the transformation is canonical. A generating function for this can be found in a variety of ways. Let s try to get an F 1 (q, Q) (but we could try others too). In that case, we know p = F 1 q, P = F 1 Q To solve these for F 1, we need to express the left hand sides in terms of the independent variables q and Q. To this end, we get Q iaq = F 1 q, Q iaq = F 1 ia Q Integrating these we get F 1 (q, Q) = qq ia q + g 1 (Q) F 1 (q, Q) = 1 4ia Q + qq + g (q) where g 1 (Q) and g (q) are integration constants that obviously depend on the variable not being integrated. For consistency, it is clear that we must have g 1 = Q /(4ia) and g = iaq / yielding F 1 (q, Q) = 1 4ia Q + qq ia q Our generating function then is F = F 1. Note that we did not have to choose to use an F 1. That we could have tried to find the generating function using a different one of the four basic canonical transformations becomes clear if we try to find an F (q, P ). In this case, we have p = F q, Q = F P where in terms of the indpendent variables, q and P, we have p = iap + iaq, Q = ia(p + q) Integrating the partial differential relations for F yields F (q, P ) = iap q + ia q + h 1 (P ) F (q, P ) = iap + iap q + h (q) 6

7 where, again, the integration constants are h 1 and h and consistency requires F (q, P ) = iap + iap q + ia q Of course, this doesn t look like the earlier generating function, but recall that in order to compare, we have to construct F = F QP and write it in terms of the earlier q and Q (or express the previously found generating function in terms of q and P ). Doing this we get F = F QP = iap + iap q + ia q QP ( Q iaq ) Q iaq = ia + iaq qia ia = 1 4ia Q + qq ia q + ia q Q Q iaq ia which is exactly the generating function we got before. Using this canonical transformation in the harmonic oscillator problem, we first factor the Hamiltonian written in standard coordinates H ( p + m ω q ) m m ( p + imωq )( p imωq ) m Q iap = iω QP where we have made the identification a = mω. Hamilton s equations are particularly simple: Q = H P = iωq P = H Q = iωp which are uncoupled equations with simple exponential solutions: where Q 0 and P 0 are constants. Q(t) = Q 0 e iωt P (t) = P 0 e iωt. 7

8 Goldstein 9.31 Show by the use of Poisson brackets that for a one-dimensional harmonic oscillator there is a constant of the motion u defined as k u(q, p, t) = ln(p + imωq) iωt, ω = m What is the physical significance of this constant of the motion? For the harmonic oscillator, the Hamiltonian in usual coordinates is The equation of motion for the quantity u is given by with the Poisson bracket being given by The explicit time dependence yields H ( p + m ω q ) m du dt [u, H] u H q = [u, H] + u t p u p H q = imω p + imωq p m 1 p + imωq m ω q m = iω u t = iω Combining, it is now obvious that du/dt and u is a conserved quantity. So what to make of this conserved quantity? Rearranging a bit, we have or, on equating real and imaginary parts, p + imωq = e u e iωt p = e u cos ωt mωq = e u sin ωt we see that u is related to the (constant) amplitude of the oscillation of our harmonic oscillator. 8

9 Goldstein 9.34 Obtain the motion in time of a linear harmonic oscillator by means of the formal solution for the Poisson bracket version of the equation of motion as derived from Eq Assume that at time t the initial values are q 0 and p 0. The Hamiltonian, of course, is H = p m + 1 mω q and the formal solution of the problem via Poisson brackets is u(t) = u 0 + t [u, H] 0 + t! [[u, H], H] 0 + t3 3! [[[u, H], H], H] 0 + where u is any dynamical variable such as p or q. To this end we have to calculate a bunch of Poisson brackets defined, of course, in our problem as For u = q, we get while for u = p, we get [u, H] = u H q p u p [q, H] = H p = p m H q [p, H] = H q = mω q It should hopefully be clear that the subsequent nested Poisson brackets in the formal expansion will alternate between these two results (with additional constant factors, of course). In particular, we find [[q, H], H] m [p, H] = ω q and Intuiting the pattern, we have [[[q, H], H], H] = ω [q, H] = ω m p [[[[q, H], H], H], H] = ω m [p, H] = ω4 q [[p, H], H] = mω [q, H] = ω p [[[p, H], H], H] = ω [p, H] = mω 4 q [[[[p, H], H], H], H] = mω 4 [q, H] = ω 6 p q(t) = q 0 + t p 0 m t! ω q 0 t3 p 3! ω 0 m + t4 4! ω4 q 0 + n (ω t)n = q 0 ( 1) + p 0 n (ω t)n+1 ( 1) (n)! mω (n + 1)! n=0 = q 0 cos(ωt) + p 0 mω sin(ωt) n=0 p(t) = p 0 t mω q 0 t! ω p 0 + t3 3! mω4 q 0 + t4 4! ω6 p 0 + n (ω t)n n (ω t)n+1 = p 0 ( 1) mω q 0 ( 1) (n)! (n + 1)! n=0 = p 0 cos(ωt) mω q 0 sin(ωt) n=0 Isn t this beautiful? 9