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Canonical Transformations Canonical transformations November 23, 2014 A Recall that we have defined a symplectic transformation to be any linear transformation M B leaving the symplectic form invariant, AB A B CD Ω ≡ M C M DΩ Coordinate transformations, χA ξB which are symplectic transformations at each point are called canonical. Specifically, those functions χA (ξ) satisfying @χC @χD ΩCD ≡ ΩAB @ξA @ξB are canonical transformations. Canonical transformations preserve Hamilton’s equations. 1 Poisson brackets We may also write Hamilton’s equations in terms of Poisson brackets between dynamical variables. By a dynamical variable, we mean any function f = f ξA of the canonical coordinates used to describe a physical system. We define the Poisson bracket of any two dynamical variables f and g by @f @g ff; gg = ΩAB @ξA @ξB The importance of this product is that it is preserved by canonical transformations. We see this as follows. Let ξA be any set of phase space coordinates in which Hamilton’s equations take the form dξA @H = ΩAB (1) dt @ξB and let f and g be any two dynamical variables. Denote the Poisson bracket of f and g in the coordinates A A ξ be denoted by ff; ggξ. In a different set of coordinates, χ (ξ) ; we have @f @g ff; gg = ΩAB χ @χA @χB @ξC @f @ξD @g = ΩAB @χA @ξC @χB @ξD @ξC @ξD @f @g = ΩAB @χA @χB @ξC @ξD Therefore, if the coordinate transformation is canonical so that @ξC @ξD ΩAB = ΩCD @χA @χB 1 we have @f @g ff; gg = ΩAB = ff; gg χ @ξC @ξD ξ and the Poisson bracket is unchanged. We conclude that canonical transformations preserve all Poisson brackets. Conversely, a transformation which preserves all Poisson brackets satisfies ff; ggχ = ff; ggξ @ξC @ξD @f @g @f @g ΩAB = ΩCD @χA @χB @ξC @ξD @ξC @ξD for all f; g and must therefore be canonical. An important special case of the Poisson bracket occurs when one of the functions is the Hamiltonian. In that case, we have @f @H ff; Hg = ΩAB @ξA @ξB @f @H @f @H = i − i @x @pi @p @xi @f dxi @f dp = − − i @xi dt @pi dt df @f = − @t @t or simply, df @f = ff; Hg + @t @t This shows that as the system evolves classically, the total time rate of change of any dynamical variable is the sum of the Poisson bracket with the Hamiltonian and the partial time derivative. If a dynamical variable @f has no explicit time dependence, @t = 0, then the total time derivative is just the Poisson bracket with the Hamiltonian. i The coordinates provide another important special case. Since neither x nor pi has any explicit time dependence, we have dxi = H; xi dt dp i = fH; p g (2) dt i or simply ξ_A = H; ξA , and we can check this directly that this reproduces Hamilton’s equations, dq i = H; xi dt N X @xi @H @xi @H = − @xj @p @p @xj j=1 j j N X @H = δ ij @p j=1 j @H = @pi 2 and dp i = fH; p g dt i N X @pi @H @pi @H = − @q @p @p @q j=1 j j j j @H = − @qi @qi @pi Notice that since qi; pi and are all independent, and do not depend explicitly on time, = = 0 = @pj @qj @qi @pi @t = @t : We also have the commutator of the Hamiltonian with the Hamiltonian itself, dH @H = fH; Hg + dt @t @H = @t so if the Hamiltonian is not explicitly time-dependent, then it is a constant of the motion. @f More generally, a dynamical variable with no explicit time dependence, @t = 0, is a constant of the motion if and only if it has vanishing Poisson bracket with the Hamiltonian, fH; fg = 0. 2 Canonical transformations i We now define the fundamental Poisson brackets. Suppose x and pj are a set of coordinates on phase space m such that Hamilton’s equations hold. Since they themselves are functions of (x ; pn) they are dynamical A m variables and we may compute their Poisson brackets with one another. With ξ = (x ; pn) we have @xi @xj xi; xj = ΩAB ξ @ξA @ξB N X @xi @xj @xi @xj = − @xm @p @p @xm m=1 m m = 0 for xi with xj; @xi @p xi; p = − p ; xi = ΩAB j j ξ j ξ @ξA @ξB N i i X @x @pj @x @pj = − @xm @p @p @xm m=1 m m N X i m = δmδj m=1 i = δj i for x with pj and finally @p @p fp ; p g = ΩAB i j i j ξ @ξA @ξB N X @pi @pj @pi @pj = − @xm @p @p @xm m=1 m m = 0 3 for pi with pj: The subscript ξ on the bracket indicates that the partial derivatives are taken with respect A i to the coordinates ξ = x ; pj : We summarize these relations as A B AB ξ ; ξ ξ = Ω However, since Poisson brackets are preserved by canonical transformations, this will hold in any canonical A B AB coordinates, ξ ; ξ χ = Ω . We summarize the results of this subsection with a theorem: Let the coordinates ξA be canonical. Then a coordinate transformation χA (ξ) is canonical if and only if it satisfies the fundamental bracket relation A B AB χ ; χ ξ = Ω For proof, note that the bracket on the left is defined by @χA @χB χA; χB = ΩCD ξ @ξC @ξD so in order for χA to satisfy the canonical bracket relation we must have @χA @χB ΩCD = ΩAB (3) @ξC @ξD which is just the condition shown above for the coordinate transformation χA (ξ) to be canonical. Conversely, suppose the transformation χA (ξ) is canonical, so that eq.(3) holds. Then, computing the Poisson bracket @χA @χB χA; χB = ΩCD = ΩAB ξ @ξC @ξD so χA satisfies the fundamental bracked relation. In summary, each of the following statements is equivalent: 1. χA (ξ) is a canonical transformation. 2. χA (ξ) is a coordinate transformation of phase space that preserves Hamilton’s equations. 3. χA (ξ) preserves the symplectic form, according to @ξC @ξD ΩAB = ΩCD @χA @χB 4. χA (ξ) satisfies the fundamental bracket relations A B AB χ ; χ ξ = Ω These bracket relations represent a set of integrability conditions that must be satisfied by any new set of canonical coordinates. When we formulate the problem of canonical transformations in these terms, it is not i j j obvious what functions q x ; pj and πi x ; pj will be allowed. Fortunately there is a simple procedure for generating canonical transformations, which we develop in the next section. We end this section with three examples of canonical transformations. 4 2.1 Example 1: Coordinate transformations i i Let x ; pj be one set of canonical variables. Suppose we define new configuration space variables, q ; be an arbitrary invertible function of the spatial coordinates: qi = qi xj i We seek a set of momentum variables πj such that q ; πj are canonical. For this they must satisfy the fundamental Poisson bracket relations: i j q ; q x;p = 0 i i q ; πj x;p = δj fπi; πjgx;p = 0 Check each: N X @qi @qj @qi @qj qi; qj = − x;p @xm @p @p @xm m=1 m m = 0 j since @q = 0: For the second bracket, @pm i i δj = q ; πj x;p N i i X @q @πj @q @πj = − @xm @p @p @xm m=1 m m N i X @q @πj = @xm @p m=1 m i Since q is independent of pm; we can satisfy this only if m @πj @x = j @pm @q Integrating gives @xn π = p + c (x) j @qj n j i with the cj an arbitrary functions of x . Choosing cj = 0; we compute the final bracket: @πi @πj @πi @πj fπi; πjgx;p = m − m @x @pm @pm @x @ @xn @ @xs @ @xn @ @xs = m i pn j ps − i pn m j ps @x @q @pm @q @pm @q @x @q @xm @ @xn @xm @ @xn = p − p @qj @xm @qi n @qi @xm @qj n @2xn @2xn = p − p @qj@qi n @qi@qj n = 0 @f Exercise: Show that the final bracket, fπi; πjgx;p still vanishes provided ci = @qi for some function f (q). 5 Therefore, the transformations qj = qj(xi) @xn @f π = p + j @qj n @qj is a canonical transformation for any functions qi(x): This means that the symmetry group of Hamilton’s equations is at least as big as the symmetry group of the Euler-Lagrange equations. 2.2 Example 2: Interchange of x and p: The transformation i q = pi i πi = −x is canonical. We easily check the fundamental brackets: i j q ; q x;p = fpi; pjgx;p = 0 i j q ; πj x;p = pi; −x x;p j = x ; pi x;p j = δi i j fπi; πjgx;p = −x ; −x x;p = 0 i Interchange of x and pj; with a sign, is therefore canonical. The use of generalized coordinates in Lagrangian mechanics does not include such a possibility, so Hamiltonian dynamics has a larger symmetry group than Lagrangian dynamics. For our next example, we first show that the composition of two canonical transformations is also canon- ical. Let (χ) and χ (ξ) both be canonical. Defining the composition transformation, (ξ) = (χ (ξ)) ; we compute @ A @ B @ A @χE @ B @χF ΩCD = ΩCD @ξC @ξD @χE @ξC @χF @ξD @χE @χF @ A @ B = ΩCD @ξC @ξD @χE @χF @ A @ B = ΩEF @χE @χF = ΩAB so that (ξ) is canonical.
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