The Liouville Theorem As a Problem of Common Eigenfunctions

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We start by giving the definition of the eigenfunctions of a real-valued function f(qi,pi,t): We shall say that S(qi,t) is an eigenfunction of f(qi,pi,t), with eigenvalue λ, if S is a solution of the first-order partial differential equation ∂S f(qi, ,t)= λ. (5) ∂qi (It may be noticed that if f is a time-independent Hamiltonian, then (5) is the corresponding time- independent HJ equation.) We note that if S(qi,t) is an eigenfunction of f(qi,pi,t) with eigenvalue λ, then so it is S(qi,t)+φ(t), for any function φ(t) of t only, and that the solutions of (5) will depend parametrically on λ. Of course, in order for (5) to be a differential equation, f must depend on one of the pi, at least. For instance, according to this definition, the eigenfunctions of the function

F (q,p,t)= mωq sin ωt + p cos ωt, (6) where m and ω are constants, are the solutions of the diﬀerential equation ∂S mωq sin ωt + cos ωt = λ, ∂q which can be readily integrated giving mω S = λq sec ωt − q2 tan ωt + φ(t), (7) 2 where φ(t) is an arbitrary function of t only. Note that λ may be a function of t. If S(qi,t) is a common eigenfunction of f(qi,pi,t) and g(qi,pi,t), with eigenvalues λ and µ, respectively, that is, S satisﬁes (5) and

∂S g(qi, ,t)= µ, ∂qi then, diﬀerentiating with respect to qi, making use of the chain rule, we obtain

∂f ∂f ∂2S ∂g ∂g ∂2S + = 0 and + =0, ∂qi ∂pj ∂qj ∂qi ∂qi ∂pj ∂qj∂qi hence,

∂f ∂g ∂g ∂f ∂f ∂2S ∂g ∂g ∂2S ∂f {f,g} = − = − + ∂qi ∂pi ∂qi ∂pi ∂pj ∂qj ∂qi ∂pi ∂pj ∂qj ∂qi ∂pi ∂2S ∂2S ∂f ∂g = − =0. ∂q ∂q ∂q ∂q ∂p ∂p j i i j i j

Thus, if f(qi,pi,t) and g(qi,pi,t) possess common eigenfunctions, then {f,g} = 0. In order to see that the converse is also true, we now assume that {f,g} = 0. If f and g are functionally independent, then there exists, locally at least, a set of canonical coordinates, Qi, Pi, such that, P1 = f and P2 = g. Then, the eigenvalue equations for f and g are ∂S/∂Q1 = λ and ∂S/∂Q2 = µ, which have the simultaneous solutions S = λQ1 + µQ2 + φ(Q3,...,Qn,t), where φ is an arbitrary function of n − 1 variables, thus showing that f and g have common eigenfunctions. (Note that this does not mean that every eigenfunction of f is an eigenfunction of g (cf. [7], Sec. 2.9).) The expression for S in terms of the original coordinates, (qi,t), is not given by the simple substitution of the Qi as functions of (qi,pi,t) [8]; what is relevant here is the existence of common eigenfunctions for f and g. In the case where f and g are functionally dependent, the eigenvalue equations for f and g are equivalent to each other and, trivially, possess common solutions.

2 2.1 Alternative formulation of the Liouville theorem

We now assume that F1,...,Fn are n functions satisfying (1)–(3), and we consider a common eigenfunction S(qi,t) of F1,...,Fn, with eigenvalues λ1,...,λn, respectively, then, assuming that the eigenvalues are constant, diﬀerentiating with respect to t both sides of the equation ∂S Fi(qj , ,t)= λi, (8) ∂qj making use of the chain rule, the Hamilton equations and (1), we have

∂Fi ∂Fi d ∂S ∂Fi 0 = q˙j + + ∂qj ∂pj dt ∂qj ∂t ∂F ∂H ∂F ∂2S ∂2S ∂F ∂H ∂F ∂H = i + i + q˙ − i + i ∂q ∂p ∂p ∂t∂q ∂q ∂q k ∂q ∂p ∂p ∂q j j j j k j j j j j ∂F ∂2S ∂2S ∂H ∂H = i + + , i =1, . . . , n. ∂p ∂t∂q ∂q ∂q ∂p ∂q j j k j k j By virtue of (3), the last equations are equivalent to

∂2S ∂2S ∂H ∂H 0 = + + ∂t∂qj ∂qk∂qj ∂pk ∂qj ∂ ∂S ∂S = + H(q , ,t) , j =1, . . . , n, ∂q ∂t k ∂q j k which implies that the expression inside the brackets is a function of t only, ∂S ∂S + H(qk, ,t)= χ(t). ∂t ∂qk Thus, t S˜ = S − χ(u) du Z is a solution of the HJ equation. We can verify that this solution is complete by diﬀerentiating (8) with respect to λj , which gives 2 ∂Fi ∂ S = δij . ∂pk ∂λj ∂qk 2 Taking into account (3), this last equation shows that det(∂ S/∂λj∂qk) 6= 0.

3 Examples

In this section we give some examples of the method presented above.

3.1 One-dimensional harmonic oscillator The function F (q,p,t)= mωq sin ωt + p cos ωt already considered above [see (6)], is a constant of motion if the Hamiltonian is given by

2 2 p mω 2 H = + q , (9) 2m 2 where ω is a constant. According to the results of the preceding section, if λ is a constant, mω S = λq sec ωt − q2 tan ωt + φ(t) (10) 2

3 must be a solution of the HJ equation for the Hamiltonian (9), if the function φ is appropriately chosen. A direct computation yields

2 2 2 1 ∂S mω ∂S λ ′ + q2 + = sec2 ωt + φ (t), 2m ∂q 2 ∂t 2m and, therefore, choosing φ(t)= −λ2 tan ωt/2mω, we obtain the complete solution of the HJ equation

λ2 mω2 tan ωt S = λq sec ωt − + q2 . 2m 2 ω Note that, in this case, H is also a constant of motion and, as pointed out above, the equation that determines the eigenfunctions of H is just the time-independent HJ equation. However, the constant of motion (6) leads to simpler expressions.

3.2 Particle in a time-dependent force ﬁeld As a second example we consider the time-dependent Hamiltonian p2 H = − ktq, 2m where k is a constant. One can readily verify that kt2 F = p − 2 is a constant of motion and that the eigenfunctions of F , i.e., the solutions of ∂S kt2 − = λ, ∂q 2 are given by kt2 S = λq + q + φ(t), (11) 2 where φ(t) is an arbitrary function of t only. Then

2 2 2 2 2 4 1 ∂S kt ∂S λ λkt k t ′ − + = + + + φ (t), 2m ∂q 2 ∂t 2m 2m 8m hence, choosing λ2t λkt3 k2t5 φ(t)= − − − , 2m 6m 40m (11) is a complete solution of the HJ equation (which is not separable).

3.3 Particle in two dimensions As a ﬁnal example we consider the Hamiltonian

2 2 1 eB eB H = p + y + p − x , (12) 2m x 2c y 2c " # which corresponds to a charged particle of mass m and electric charge e in a uniform magnetic ﬁeld B. The functions 1 1 mω mω F1 = (1 + cos ωt)p − p sin ωt + x sin ωt − (1 − cos ωt)y, (13) 2 x 2 y 4 4 1 1 mω mω F2 = (1 + cos ωt)p + p sin ωt + y sin ωt + (1 − cos ωt)x, (14) 2 y 2 x 4 4 where ω ≡ eB/mc, are constants of motion in involution, which correspond to the values of the canonical momenta px and py, respectively, at t = 0.

4 From (13) and (14) one ﬁnds that the common eigenfunctions of F1 and F2, with eigenvalues λ1 and λ2, respectively, are

1 mω 2 2 S = λ1x + λ2y + tan ωt λ2x − λ1y − (x + y ) + φ(t), 2 4 h i where φ(t) is an arbitrary function of t only. Substituting this expression into the HJ equation one ﬁnds that S is a solution of this equation if and only if

2 2 λ1 + λ2 ′ sec2 1 ωt + φ (t)=0, 2m 2 hence, 2 2 1 mω mω tan 2 ωt S = λ1x + λ2y − λ1 + y + λ2 − x 2 2 mω is a complete solution of the HJ equation.

4 Concluding remarks

As pointed out above, the formulation of the Liouville theorem given here makes use of terms analogous to those employed in the standard formalism of quantum mechanics, thus providing another example of the parallelism between both theories. Another advantage of the version of the Liouville theorem given above is that its proof is shorter than those usually presented in the textbooks.

References

[1] Whittaker, E.T.: A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, 4th ed. Cambridge University Press, Cambridge (1993). [2] Vilasi, G.: Hamiltonian Dynamics. World Scientific, Singapore (2001). [3] Babelon, O., Bernard, D., Talon, M.: Introduction to Classical Integrable Systems. Cambridge University Press, Cambridge (2003). [4] Fasano, A., Marmi, S.: Analytical Mechanics. Oxford University Press, Oxford (2006). [5] DiBenedetto, E.: Classical Mechanics. Birkhäuser, New York (2011). [6] Torres del Castillo, G.F.: Applications and extensions of the Liouville theorem on constants of motion, Rev. Mex. F´ıs. 57, 245-249 (2011). [7] Sneddon, I.N.: Elements of Partial Differential Equations. Dover, New York (2006). [8] Torres del Castillo, G.F., Cruz Dom´ınguez, H.H., de Yta Hernández, A., Herrera Flores, J.E., Sierra Mart´ınez, A.: Mapping of solutions of the Hamilton–Jacobi equation by an arbitrary canonical transformation, Rev. Mex. F´ıs. 60, 301-304 (2014).