In a remarkable paper Chandrasekar et al. showed that the (second-order constant-coefficient) classical equation of motion for a damped harmonic oscillator can be derived from a Hamiltonian having one degree of freedom. This paper gives a simple derivation of their result and generalizes it to the case of an nth-order constant-coefficient differential equation.

It seems unlikely that the equation for the damped classical harmonic oscillator

$x ̈ + 2 γ x ̇ + ω 2 x = 0 ( γ > 0 ) ,$
(1)

a dissipative system, could be derived from a Hamiltonian. This is because $E= 1 2 x ̇ 2 + 1 2 ω 2 x 2$, the standard expression for the total energy, is not conserved for γ≠0. Indeed, it satisfies the equation

$d d t 1 2 x ̇ 2 + 1 2 ω 2 x 2 = − 2 γ x ̇ 2 ,$
(2)

showing that it decreases with time.

Nonetheless, Bateman1 made the remarkable observation that if one appends the time-reversed oscillator equation with undamping (gain) instead of damping,

$y ̈ − 2 γ y ̇ + ω 2 y = 0 ( γ > 0 ) ,$
(3)

then even though the two oscillators are independent and noninteracting [the two evolution equations (1) and (3) are not coupled], the equations of motion (1) and (3) can be derived from the time-independent quadratic Hamiltonian

$H = p q + γ ( y q − x p ) + ω 2 − γ 2 x y$
(4)

involving the two pairs of canonical variables (x, p) and (y, q). To derive the original Equation (1) we use the pair of Hamilton’s equations

$x ̇ = ∂ H ∂ p = q − γ x ,$
(5)
$q ̇ = − ∂ H ∂ y = − γ q − ω 2 − γ 2 x .$
(6)

Substituting for q in (6) gives $q ̇ =−γ x ̇ − ω 2 x$, and then differentiating (5) with respect to t gives (1). The partner Equation (2) can be similarly obtained from the remaining Hamilton’s equations

$y ̇ = ∂ H ∂ q = p + γ y ,$
(7)
$p ̇ = − ∂ H ∂ x = γ p − ω 2 − γ 2 y .$
(8)

It is worth noting that the Hamiltonian (4) is $PT$ symmetric2 if we define the action of $P$ as interchanging the two oscillators

$P : x → y , y → x , p → q , q → p ,$
(9)

while $T$ reverses the signs of the momenta,

$T : x → x , y → y , p → − p , q → − q .$
(10)

The PT symmetry of H in (4) and the success of Bateman’s strategy depend crucially on the gain and loss terms in (1) and (3) being exactly balanced. As a consequence of the gain/loss balance, the system possesses a conserved quantity, namely the value of H. However, the energy has the complicated form (4) and is not a simple sum of kinetic and potential energies.

Recently it was shown3 that the equation of motion (1) of the damped oscillator can be derived from a (nonquadratic) time-independent Hamiltonian depending on only a single canonical pair (x, p). This remarkable result was proved by using a modification4 of the Prelle-Singer approach5 to identify integrals of motion of dynamical systems. This paper is a comment on the interesting work in Refs. 3 and 4.

In Ref. 3 different forms of the Hamiltonian were given depending on whether the system was overdamped, underdamped, or critically damped. In particular, for the overdamped case (γ > ω) the Hamiltonian takes the unconventional form

$H = A x p + B p δ ,$
(11)

where A, B, and δ are constants that we will specify later. Different functional forms were given for the other cases in order to have a real Hamiltonian. However, since this is not a concern for us, the functional form of (11) serves for all cases (apart from an obvious modification in the case of critical damping).

In this paper we show in Sec. II how the Hamiltonian (11) can be derived by elementary means and we identify the coefficients therein in terms of the (generally complex) eigenfrequencies of the problem. The conserved quantity, written in terms of $x ̇$ and x, is similarly identified. In Sec. III we go on to apply the procedure to the third-order linear equation with constant coefficients, a particular example of which describes the nonrelativistic self-acceleration of a charged oscillating particle.6 The additional conceptual problems that arise in this case are discussed in detail. In Sec. IV we generalize the procedure to an arbitrary nth-order constant-coefficient equation. Section V discusses the difficult problem of quantizing Hamiltonians of this type. Finally, Sec. VI gives a brief summary.

We begin with (1) and substitute x(t) = eiνt. This gives a quadratic equation for the frequency ν,

$ν 2 + 2 i γ ν − ω 2 = 0 ,$
(12)

which factors as

$ν − ω 1 ν − ω 2 = 0 ,$
(13)

where

$ω 1 + ω 2 = − 2 i γ , ω 1 ω 2 = − ω 2 ,$
(14)

with

$ω 1 , 2 = − i γ ± Ω = − i γ ± ω 2 − γ 2 .$
(15)

The generic form of a Hamiltonian H(x, p) that can generate the evolution equation (1) is given in (11). There are two such Hamiltonians, corresponding to the two eigenfrequencies in (15). The first is

$H 1 = − i ω 1 x p + ω 1 ω 1 − ω 2 p 1 − ω 2 / ω 1 .$
(16)

A second and equally effective Hamiltonian is obtained by interchanging ω1 and ω2,

$H 2 = − i ω 2 x p + ω 2 ω 2 − ω 1 p 1 − ω 1 / ω 2 .$
(17)

As mentioned in Sec. I, Hamiltonians of this form appear in Ref. 3 for the case of over-damping (γ2 > ω2), where they are real since in that case ω1 and ω2 are purely imaginary, but they apply equally well when γ2 < ω2 if we are not concerned with the reality of the Hamiltonian. Indeed, the Hamiltonian is no longer the standard real energy, which is not conserved. Rather, it is a complex quantity which is conserved and from which the equations of motion can be derived in the standard way.

For the Hamiltonian H1, Hamilton’s equations read

$x ̇ = ∂ H 1 ∂ p = − i ω 1 x + p − ω 2 / ω 1 ,$
(18)
$p ̇ = − ∂ H 1 ∂ x = i ω 1 p .$
(19)

We then take a time derivative of (18) and simplify the resulting equation first by using (19) and then by using (18),

$x ̈ + i ω 1 x ̇ = − ω 2 ω 1 p − 1 − ω 2 / ω 1 p ̇ = − i ω 2 p − ω 2 / ω 1 = − i ω 2 x ̇ + i ω 1 x .$
(20)

Thus,

$x ̈ + i ω 1 + ω 2 x ̇ − ω 1 ω 2 x = 0 ,$
(21)

which reduces to (1) upon using (14).

The evolution equation (1) has one conserved (time-independent) quantity, which can be expressed in terms of x(t) only. To find this quantity, we begin with (18) and solve for p,

$p = x ̇ + i ω 1 x − ω 1 / ω 2 .$
(22)

We then use this result to eliminate p from H1. Since H1 is time independent, we conclude that

$C 1 = x ̇ + i ω 2 x ω 2 / x ̇ + i ω 1 x ω 1$
(23)

is conserved. Had we started with the Hamiltonian H2 we would have obtained the conserved quantity

$C 2 = x ̇ + i ω 1 x ω 1 / x ̇ + i ω 2 x ω 2 ,$
(24)

but this is not an independent conserved quantity because C2 = 1/C1. These conserved quantities were also found in Ref. 3 for the case of over-damping.

When γ = 0, these results reduce to the familiar expressions for the simple harmonic oscillator. In that case we let ω = ω1 = − ω2, so that H1 becomes

$H 1 = − i ω x p + 1 2 p 2 ,$
(25)

which is related to the standard Hamiltonian for the simple harmonic oscillator by the change of variable ppiωx. The conserved quantities C2 and C1 become simply $( x ̇ 2 + ω 2 x 2 ) ± ω$, in which we recognize the usual conserved total energy.

In this section we show how to construct a Hamiltonian that gives rise to the general third-order constant-coefficient evolution equation

$( D + i ω 1 ) ( D + i ω 2 ) ( D + i ω 3 ) x = 0 ,$
(26)

where Dd/dt. The Hamiltonian that we will construct has just one degree of freedom.

A physical example of such an equation is the third-order differential equation

$m x ̈ + k x − m τ x ⃛ = 0$
(27)

that describes an oscillating charged particle subject to a radiative back-reaction force.6 Following Bateman’s approach for the damped harmonic oscillator, Englert7 showed that the pair of noninteracting Equation (27) and

$m y ̈ + k y + m τ y ⃛ = 0$
(28)

can be derived from the quadratic Hamiltonian

$H = p s − r q m τ + 2 r s m τ 2 + p z + q w 2 − m z w 2 + k x y .$
(29)

This Hamiltonian contains the four degrees of freedom (x, p), (y, q), (z, r), and (w, s). An interacting version of this model was studied in Ref. 8. In fact, we find that the two equations of motion (27) and (28) can be derived from the simpler quadratic Hamiltonian

$H = p r + q z m τ − r z τ + k x y ,$

which has only the three degrees of freedom (x, p), (y, q), and (z, r). A similar three-degree of freedom Hamiltonian was also found in Ref. 7.

Our objective here is to find a one-degree-of-freedom Hamiltonian that can be used to derive the third-order differential equation (26). Note that the general solution to (26) is

$x = a 1 e − i ω 1 t + a 2 e − i ω 2 t + a 3 e − i ω 3 t ,$
(30)

where ak are arbitrary constants. If we form (D + 2)(D + 3) x, that is, $x ̈ +i ( ω 2 + ω 3 ) x ̇ − ω 2 ω 3 x$, we obtain

$a 1 e − i ω 1 t = − ( D + i ω 2 ) ( D + i ω 3 ) x ( ω 1 − ω 2 ) ( ω 1 − ω 3 )$
(31)

in which the constants a2 and a3 do not appear. Similarly, we have

$a 2 e − i ω 2 t = − ( D + i ω 3 ) ( D + i ω 1 ) x ( ω 2 − ω 3 ) ( ω 2 − ω 1 ) ,$
$a 3 e − i ω 3 t = − ( D + i ω 1 ) ( D + i ω 2 ) x ( ω 3 − ω 1 ) ( ω 3 − ω 2 ) .$
(32)

So, assuming that the frequencies ωk are all different, there are two distinct conserved quantities, namely

$C 2 = [ x ̈ + i ( ω 2 + ω 3 ) x ̇ − ω 2 ω 3 x ] [ x ̈ + i ( ω 1 + ω 2 ) x ̇ − ω 1 ω 2 x ] − ω 1 / ω 3 ,$
$C 3 = [ x ̈ + i ( ω 2 + ω 3 ) x ̇ − ω 2 ω 3 x ] [ x ̈ + i ( ω 1 + ω 3 ) x ̇ − ω 1 ω 3 x ] − ω 1 / ω 2 .$
(33)

These expressions and the equation of motion can be derived from the Hamiltonian

$H = − i ω 1 x p + b 2 ω 1 ω 1 − ω 2 p 1 − ω 2 / ω 1 + b 3 ω 1 ω 1 − ω 3 p 1 − ω 3 / ω 1 ,$
(34)

where b2 and b3 are arbitrary constants. Thus, $p ̇ ≡−∂H/∂x=i ω 1 p$. This means that pe1t, so that 1/p is directly related to the combination in (31).

Then, from Hamilton’s equation $x ̇ ≡∂H/∂p$ and from further differentiation with respect to t, we obtain

$x ̇ = − i ω 1 x + b 2 p − ω 2 / ω 1 + b 3 p − ω 3 / ω 1 ,$
$x ̈ = − i ω 1 x ̇ − i ω 2 b 2 p − ω 2 / ω 1 − i ω 3 b 3 p − ω 3 / ω 1 ,$
(35)
$x ⃛ = − i ω 1 x ̈ − ω 2 2 b 2 p − ω 2 / ω 1 − ω 3 2 b 3 p − ω 3 / ω 1 .$

This further differentiation is crucial because it is required to eliminate the constants b2 and b3, and thus to obtain a differential equation that only contains x(t) and is independent ofb2andb3. Combining these equations and performing some simplifying algebra, we obtain

$x ⃛ + i ( ω 1 + ω 2 + ω 3 ) x ̈ − ( ω 1 ω 2 + ω 2 ω 3 + ω 3 ω 1 ) x ̇ − i ω 1 ω 2 ω 3 x = 0 .$
(36)

We emphasize that the constants b2 and b3 have disappeared in this combination and we have reconstructed the equation of motion (26). These constants are reminiscent of Lagrange multipliers, but they are unlike Lagrange multipliers in that we do not vary the Hamiltonian with respect to them. Rather, we require that the equations of motion be independent of these constants.

Using only derivatives up to the second order, we can find expressions for b2pω2/ω1 and b3pω3/ω1, namely

$b 2 p − ω 2 / ω 1 = i [ x ̈ + i ( ω 1 + ω 3 ) x ̇ − ω 1 ω 3 x ] / ( w 2 − ω 3 ) ,$
$b 3 p − ω 3 / ω 1 = i [ x ̈ + i ( ω 1 + ω 2 ) x ̇ − ω 1 ω 2 x ] / ( ω 3 − ω 2 ) ,$
(37)

in which we recognize two of the quantities in square brackets that appear in (33). The third such quantity, namely $x ̈ +i ( ω 1 + ω 2 ) x ̇ − ω 1 ω 2 x$, is closely related to H,

$x ̈ + i ( ω 2 + ω 3 ) x ̇ − ω 2 ω 3 x = − ( ω 1 − ω 2 ) ( ω 1 − ω 3 ) x + i b 2 ( ω 3 − ω 1 ) p − ω 2 / ω 1 + i b 3 ( ω 2 − ω 1 ) p − w 3 / ω 1$
$= i ( ω 1 − ω 2 ) ( ω 3 − ω 1 ) H / ( ω 1 p ) .$

We conclude that

$C 2 = [ i ( ω 1 − ω 2 ) ( ω 3 − ω 1 ) H / ( ω 1 p ) ] p [ i ( ω 2 − ω 3 ) b 3 ] − ω 1 / ω 3 ∝ H b 3 − ω 1 / ω 3 .$

Similarly, we have

$C 3 ∝ H b 2 − ω 1 / ω 2 .$

Thus C2 and C3 are constants of the motion because they are both proportional to the Hamiltonian, with proportionality constants given by powers of b2 and b3, respectively.

To summarize, by eliminating the parameters b2 and b3 the evolution equation (26) can be derived from the unusual time-independent Hamiltonian (34) containing the single coordinate variable x and its conjugate momentum p. This Hamiltonian is a conserved quantity, which can be expressed as

$H = i ω 1 p x ̈ + i ( ω 2 + ω 3 ) x ̇ − ω 2 ω 3 x ( ω 1 − ω 2 ) ( ω 1 − ω 3 ) .$
(38)

The conserved quantities C2 and C3 are both proportional to H.

Before moving on, we must explain how a Hamiltonian with a single degree of freedom can give rise to a differential equation whose order is greater than two. The problem is as follows. Our Hamiltonian has the generic form

$H = a x p + f ( p ) .$
(39)

Therefore, the equations of motion are simply

$x ̇ = a x + g ( p ) ,$
(40)

where g(p) = f′(p), and

$p ̇ = − a p .$
(41)

First, we solve (41)

$p ( t ) = C e − a t ,$
(42)

where C is an arbitrary constant. Next, we return to (40), which becomes

$x ̇ = a x + g C e − a t$

after we eliminate p by using (42). This is a first-order equation. Thus, its solution has only two arbitrary constants:

$x ( t ) = ϕ ( t , C , D ) .$
(43)

We obtain the higher-order differential equation (36) by the sequence of differentiations in (35) that were required to eliminate the constants b2 and b3. Of course, the solution to an nth-order equation can incorporate n pieces of data such as n initial conditions: x(0), $x ̇ ( 0 )$, $x ̈ ( 0 )$, $x ⃛ ( 0 )$, and so on. How is it possible to incorporate n pieces of data with only two arbitrary constants C and D? There appear to be n − 2 missing arbitrary constants.

The answer is that the n − 2 pieces of initial data determine n − 2 parameters bk multiplying each of the fractional powers of p in H. (One parameter can always be removed by a scaling.) We can incorporate the initial data into the Hamiltonian in the form of these parameters. These parameters specify an ensemble of Hamiltonians, all of which give a unique nth-order field equation that is independent of these parameters and is capable of accepting n pieces of initial data.

For the triple-dot equation we can see from (37) that the ratio $b 2 1 / ω 2 / b 3 1 / ω 3$ is related to the initial conditions. So, for the case of the third-order equation, the three arbitrary constants are C, D, and $b 2 1 / ω 2 / b 3 1 / ω 3$. We emphasize that the Hamiltonian gives the higher-order equations of motion precisely because of the requirement that the parameters bk drop out from the equation of motion. The parameters bk in the Hamiltonian are crucial because they incorporate the initial data and are determined by the initial data. The nonzero parameter b3 forces the evolution equation to be third order. Without b3 the Hamiltonian does not know about the third frequency ω3. Indeed, if b3 = 0, (34) reduces to (16) (with b2 = 1). [This is consistent with (37) because setting b3 = 0 there implies that $x ̈ +i ( ω 1 + ω 2 ) x ̇ − ω 1 ω 2 x=0$.]

Finally, one may ask what would happen if we followed the standard procedure for deriving the equations of motion for x(t) from the Hamiltonian equations of motion $p ̇ =−∂H/∂x$ and $x ̇ =∂H/∂p$. This would mean solving the second equation for p in terms of x and $x ̇$ and then substituting back in the first to obtain a second-order equation for x(t). In our case that would mean solving the first of Equations (35) for p, which is not possible for general values of the parameters. However, it is instructive to see how this procedure works if we choose the parameters so that an explicit solution is possible. For example, if we choose ω1 = 1, ω2 = − 2, ω3 = 4, b2 = 2, and b3 = 1, the equation can be solved to give $p 2 =−1+ x ̇ + i x + 1$. Substituting back into the equation $p ̇ =ip$ gives, after some algebra, the nonlinear second-order equation

$( x ̈ − 3 i x ̇ + 4 x − 4 i ) 2 = − 16 ( x ̇ + i x + 1 ) .$

Further manipulation shows this to be equivalent to the constancy of C2/C3.

So, in the cases where the standard procedure can be followed in practice, the resulting nonlinear second-order equation is equivalent to an equation for a constant of the motion (which of course depends on the parameters b2 and/or b3).

It is straightforward to generalize to the case of an arbitrary nth-order linear constant-coefficient evolution equation

$∏ r = 1 n D + i ω r x ( t ) = 0 ,$
(44)

whose general solution is

$x ( t ) = ∑ r = 1 n a r e − i ω r t .$
(45)

For simplicity, we assume first that the frequencies ωr are all distinct; at the end of this section we explain what happens if some of the frequencies are degenerate.

Corresponding to (32) and (33), we have

$e − i ω s t ∝ ∏ r ≠ s D + i ω r x ( t ) .$
(46)

Thus, the quantity

$Q s ≡ ∏ r ≠ s n D + i ω r x ( t ) 1 / ω s$
(47)

is proportional to eit for all s. Hence, the n − 1 independent ratios Qs/Q1 (s > 1) are all conserved. Any other conserved quantities can be expressed in terms of these ratios.

The equation of motion and the conserved quantities can be derived from the Hamiltonian

$H = − i ω 1 x p + ∑ r ≠ 1 n b r ω 1 p 1 − ω r / ω 1 ω 1 − ω r ,$
(48)

which is the nth order generalization of (34) for the cubic case. In this expression the n − 1 coefficients br are arbitrary, and must be eliminated to give the nth-order equation of motion. Note that in constructing the Hamiltonian H there is nothing special about the subscript “1” and it may be replaced by the subscript “s” (1 < sn).

Until now, we have assumed that the frequencies ωr are all distinct. However, if some of the frequencies are degenerate, there is a simple way to construct the appropriate Hamiltonian: If the frequencies ω1 and ω2 are equal, we make the replacement

$ω 1 ω 1 − ω 2 p 1 − ω 2 / ω 1 ⟶ log ( p ) .$
(49)

(In making this replacement we are shifting the Hamiltonian by an infinite constant.) Thus, for ω1 = ω2 the Hamiltonian H1 in (16) reduces to

$H 1 = − i ω 1 x p + log p .$
(50)

Hamilton’s equations for this Hamiltonian immediately simplify to (21) with ω1 = ω2. Similarly, for the case ω1 = ω2 the Hamiltonian (34) reduces to

$H = − i ω 1 x p + b 2 log p + b 3 ω 1 ω 1 − ω 3 p 1 − ω 3 / ω 1$
(51)

and Hamilton’s equations for this Hamiltonian readily simplify to (36) with ω1 = ω2.

Also, if the frequencies are triply degenerate, ω1 = ω2 = ω3 = ω, the Hamiltonian in (34) is replaced by

$H = − i ω x p + b log p + 1 2 c ( log p ) 2 ,$
(52)

where b and c are two parameters that are determined by the initial data. Once again, Hamilton’s equations for this Hamiltonian combine to give (36) with ω1 = ω2 = ω3 = ω.

The obvious question to be addressed next is whether it is possible to use the Hamiltonians that we have constructed to quantize classical systems that obey linear constant-coefficient evolution equations. Let us begin by discussing the simple case of the quantum harmonic oscillator (QHO), whose Hamiltonian H1 is given in (25).

One possibility is to quantize the Hamiltonian in p-space by setting x = id/dp. Then by shifting E by ω we see that the time-independent Schrödinger eigenvalue equation is

$H 1 ψ ̃ ( p ) = ω p d d p + 1 2 p 2 ψ ̃ ( p ) = E ψ ̃ ( p ) ,$
(53)

whose solution is

$ψ ̃ ( p ) ∝ p E / ω e − p 2 / ( 4 ω ) .$
(54)

In this way of doing things we derive the quantization condition by demanding that $ψ ̃$ be a well defined, nonsingular function, which requires that E = , where n is a non-negative integer.9 However, these “momentum-space” eigenfunctions are problematical because p has no clear physical interpretation as a momentum. The p-space eigenfunctions are certainly not orthonormal in any simple sense because they do not solve a Sturm-Liouville boundary-value problem.10

However, we can calculate the corresponding x-space eigenfunctions by Fourier transform using the formula11

$H n ( z ) = ( − i ) n 2 π e z 2 ∫ − ∞ ∞ d p e i p z p n e − p 2 .$
(55)

We find that

$ψ n ( x ) ∝ e − 1 2 ω 2 x 2 φ n ( x ) ,$
(56)

where φn(x) is the nth eigenfunction of the QHO. This is consistent with our remark above that H1 is related to the standard QHO Hamiltonian by the transformation ppiωx. This transformation is achieved at the operator level by the similarity transformation peω2x2/2peω2x2/2.12 Because of this additional factor, our eigenfunctions are orthonormal with respect to the metric η = eω2x2. As an alternative approach, we can cast (25) in x-space as

$H 1 = − 1 2 d 2 d x 2 − ω 2 1 + x d d x ,$
(57)

from which we can obtain the ψn(x) directly.

To summarize, the quantized version of (25) corresponds to a transformed version of the QHO, where the x-space eigenfunctions are simply related to the standard eigenfunctions and are orthonormal with respect to an additional weight function. The p-space eigenfunctions can be written down but their interpretation is not obvious (the operator p corresponds to the conventional raising operator a) and are not orthogonal in any simple way. Moreover, p, represented as −id/dx, is not Hermitian because the overlap integral between two wave functions has to be calculated with the inclusion of the metric η(x), so integration by parts is no longer simply a matter of a minus sign. Instead, p is pseudo-Hermitian13,14 with respect to η, i.e., p = ηpη−1. In p space the weight function eω2x2 becomes the highly nonlocal operator eω2d2/dp2.

If we generalize to the damped harmonic oscillator, we can still find a solution $ψ ̃ ( p )$ to the time-independent Schrödinger equation, namely3

$ψ ̃ ( p ) ∝ p E / ω 1 exp − ω 1 ( ω 1 − ω 2 ) 2 p 1 − ω 1 / w 2 ,$
(58)

but even if we take E = 1 to make the prefactor nonsingular, we are still left with a nonintegral, and in general complex, power of p in the exponential. (See, also, the comments in Ref. 1.) Thus, in addition to the previously discussed problems with $ψ ̃ ( p )$, we would now have to consider it to be a function in a cut plane. Moreover, there is no simple formula like (55) whereby one could obtain the x-space eigenfunctions. Furthermore, if we cast the equation in x-space we obtain

$H 1 = 1 1 − ω 2 / ω 1 − i d d x − i d d x − ω 2 / ω 1 − i ( ω 1 − ω 2 ) x ,$
(59)

in which the difficulty associated with a fractional derivative is manifest.

Evidently, quantizing Hamiltonians of the form in (39) is nontrivial. The problem of quantizing the cubic equation describing the back-reaction force on a charged particle was solved in Ref. 8. However, the system that was actually quantized was a pair of coupled cubic equations in the unbroken$PT$-symmetric region. Thus, it may be that the most effective approach for quantizing a Hamiltonian of the form (39) is to introduce a large number of additional degrees of freedom.

We have shown that any nth-order linear constant-coefficient evolution equation can be derived from a nonconventional but simple Hamiltonian of the form (39). Remarkably, this Hamiltonian has only one degree of freedom, that is, one pair of dynamical variables (x, p). Furthermore, we have shown that for such a system there are n − 1 independent constants of the motion and we have constructed these conserved quantities in terms of x(t) and its time derivatives. However, we find that it is not easy to formulate a general procedure to quantize the system described by the Hamiltonian, and this remains an extremely interesting but difficult open problem.

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