29 April 2002

Physics Letters A 296 (2002) 312–316

www.elsevier.com/locate/pla

Reply

Reply to “Comments on Bouda and Djama’s‘Quantum Newton’s Law’ ”✩

A. Bouda∗, T. Djama

Laboratoire de Physique Théorique, Université de Béjaïa, Route Targa Ouazemour, 06000 Béjaïa, Algeria

Received 13 August 2001; received in revised form 13 March 2002; accepted 14 March 2002

Communicated by P.R. Holland

Abstract

In this reply, we hope to bring clarifications about the reservations expressed by Floyd in his comments, give furtherexplanations about the choice of the approach and show that our fundamental result can be reproduced by other ways. Wealso establish that Floyd’s trajectories manifest some ambiguities related to the mathematical choice of the couple of solutionsof Schrödinger’s equation. 2002 Elsevier Science B.V. All rights reserved.

PACS: 03.65.Bz; 03.65.Ca

Keywords: Quantum law of motion; Lagrangian; Hamiltonian; Quantum Hamilton–Jacobi equation; Jacobi’s theorem

In Floyd’s comments [1] on our previous Letter [2],after having showed that

(1)2Et = S0,

it is stated that our equation of motion is the quantumreduced action. Firstly, we indicate that the aboverelation, up to an additive constant, is already writtenin our Letter [2, Eq. (39)]. Secondly, relation (1) isvalid only in the particular free particle case and itdoes not work for other potentials. Therefore, wecannot assert that our equation of motion is thequantum reduced action. With regard to our velocity,it is an instantaneous velocity of the particle which

✩ PII of original article: S0375-9601(02)00278-5.* Corresponding author.

E-mail addresses: bouda_a@yahoo.fr (A. Bouda),djama.toufik@caramail.com (T. Djama).

is localized in space at each time. Furthermore, theknowledge of all the integration constants, even thenon-classical onesa andb, determines univocally thetrajectory and the velocity at each time.

Concerning the representation of the Hamilton’sprincipal functionS as an integral of a Lagrangian,Floyd claimed that his finding can be generalized forthe case whereh is not considered close to 0. From thequantum Hamilton principal function, he proposed theLagrangian

L(x, x, x, ˙x) = ∂S0

∂xx − 1

2m

(∂S0

∂x

)2

− V (x)

(2)

+ h2

4m

[3

2

(∂S0

∂x

)−2(∂2S0

∂x2

)2

−(∂S0

∂x

)−1(∂3S0

∂x3

)],

0375-9601/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved.PII: S0375-9601(02)00277-3

A. Bouda, T. Djama / Physics Letters A 296 (2002) 312–316 313

and suggested that one should start from some of therelations in Ref. [3] giving the derivatives ofS0 withrespect tox in terms of temporal derivatives ofx. Headded that the resultingL(x, x, x, ˙x) and the resultingLagrange equation are cumbersome. We would like toindicate that it is not only cumbersome to express theLagrangian but Faraggi–Matone’s relations, as givenin [3], do not allow us to express the Lagrangian (2) asa function of(x, x, x, ˙x). In fact, these relations are

P =m

(1− ∂Q

∂E

)x,

∂P

∂x= −m ∂2Q

∂x∂Ex +m

(1− ∂Q

∂E

)x

x,

∂2P

∂x2= −m

∂3Q

∂x2∂Ex − 2m

∂2Q

∂x∂E

x

x

+m

(1− ∂Q

∂E

)( ˙xx2

− x2

x3

),

in which the quantum potential,Q, must be substitutedby

(3)Q= h2

4m

[1

P

∂2P

∂x2− 3

2

(1

P

∂P

∂x

)2]

and P = ∂S0/∂x is the conjugate momentum. It isclear thatx, x and ˙x are related toP , ∂P/∂x, . . . , ∂4

P/∂x4 and ∂P/∂E, . . . , ∂5P/∂x4∂E. In our pointof view, it is not possible to express(P, ∂P/∂x,∂2P/∂x2), and, therefore, the Lagrangian (2), only interms of(x, x, x, ˙x). In addition, the constantE willappear in (2) and will be redundant. In order to avoidthe above higher derivatives, an alternative is to use thesolution of the quantum stationary Hamilton–Jacobiequation (QSHJE). In compensation, we incorporatefrom the beginning of the formalism hidden parame-ters which are represented by the non-classical integra-tion constants appearing in the reduced action. That’sprecisely what we have done in Ref. [2].

Furthermore, if we would take up the Lagrangiandepending on(x, x, x, ˙x) and keep the definition

(4)S =∫Ldt,

the quantum equation of motion, deduced by appeal-ing to the least action principle, is

(5)d3

dt3

∂L

∂ ˙x − d2

dt2

∂L

∂x+ d

dt

∂L

∂x− ∂L

∂x= 0.

First, we see that if the dependence on˙x of L is notlinear, we will obtain an equation of sixth order. Thisis not compatible with the QSHJE which indicates thatthe fundamental law of motion must be a forth orderequation [2]. In addition, the corresponding Hamil-tonian can be constructed as follows: we calculate thetotal derivative with respect tot of L, and look for theexistence of any constant of motion with the use of (5)in the stationary case. We get

H =(∂L

∂x− d

dt

∂L

∂x+ d2

dt2

∂L

∂ ˙x)x

(6)+(∂L

∂x− d

dt

∂L

∂ ˙x)x + ∂L

∂ ˙x˙x −L,

so that

(7)dH

dt= 0,

if ∂L/∂t = 0. At this stage, many difficulties appearin the search of canonical equations. In fact, if wewrite the Hamiltonian as a function of(x,P ), thislast set of variables will not be sufficient to substitutethe set(x, x, x, ˙x). If we add to the set(x,P ) thederivativesP and P , we lose the symmetry betweenthe canonical variablesx andP . If we write H as afunction of (x,P, x, P ), in the classical limith → 0,P and x will form a redundant subset. It is not easy,may be impossible, to construct an Hamiltonian withcanonical variables from which we use the Hamilton–Jacobi procedure to get to the third order well-knownQSHJE.

Now, let us consider the problem of constantswhich seem forming a redundant subset. In order togo round this problem, let us present the Lagrangianformulation in the following manner. We appeal to thequantum transformation

x → x,

introduced by Faraggi and Matone [3,4], with whichthe quantum equations take the classical forms. Then,we write the quantum Lagrangian in the form

(8)L(x, ˙x)= 1

2m ˙x2 − V (x).

The hidden parameters introduced in [2], the energyand the coordinatex are absorbed inx. In (8), we canconsiderx and ˙x as independent variables and then the

314 A. Bouda, T. Djama / Physics Letters A 296 (2002) 312–316

equation of motion,

(9)d

dt

∂L

∂ ˙x − ∂L

∂x= 0,

resulting from the least action principle leads to

(10)md2x

dt2= −dV

dx.

This relation recalls us the classical Newton’s law. AsV (x)= V (x), integrating Eq. (10) gives

(11)m

(dx

dt

)2

+ V (x)=E,

where the integration constantE is identified to theenergy of the system because, in the classical limith → 0, x reduces tox [3,4] and (11) must reproducethe classical law of the energy conservation. Untilnow, we have no redundant subset. Let us at presentexpress (11) in terms ofx. Again, the relationV (x)=V (x) allows us to write

(12)m

(dx

dt

)2(∂x

∂x

)2

+ V (x)= E.

Taking into account the expression

(13)∂x

∂x= ∂S0/∂x√

2m(E − V (x)),

which we deduce from Eq. (8) in [2] or (56) in [4],Eq. (12) leads to

(14)x∂S0

∂x= 2

[E − V (x)

].

This equation is exactly the same as the one we getin [2] by expressing the Lagrangian in terms of(x, x)

and hidden parameters.We stress that it is also possible to reproduce

Eq. (14) with an Hamiltonian formulation. In fact, asshown by Faraggi and Matone [3,4], the QSHJE canbe written as

(15)E = 1

2m

(∂S0

∂x

)2(∂x

∂x

)2

+ V (x).

SubstitutingE by the HamiltonianH and ∂S0/∂x

by P , we get

(16)H = P 2

2m

(∂x

∂x

)2

+ V (x),

which leads to the canonical equation

(17)x = ∂H

∂P= P

m

(∂x

∂x

)2

,

since x does not depend on the derivative ofx. Byusing this last equation in (15), we reproduce (14).As for the quantum version of Jacobi’s theorem [2],relations (16) and (17) constitute another proof thatwe can reproduce our fundamental result, Eq. (14),without appealing to any Lagrangian formulation.

In the last reservation expressed by Floyd, it isstated that our use of the quantum coordinate im-plies that classical mechanics would be consistent withthe quantum equivalence postulate (QEP). In his rea-soning, he considered two classical systems,AclassicalandBclassical, and their corresponding quantum sys-tems,AquantumandBquantum. According to the QEP,AquantumandBquantumcan be connected by coordinatetransformation. It follows thatAclassicalandBclassicalcan be also connected because the quantum transfor-mation would relateAquantumandAclassicalas well asBquantumandBclassicalconsistent with QEP. We wouldlike to emphasize that we have not assumed that trans-formation (8) in [2] follow from the QEP. This equa-tion is just a step which allow us to reduce the QSHJEto the classical form in order to apply classical laws tothe quantum motion. Of course, this step is differentfrom the maps which we consider when we connectdifferent states.

Now, let us discuss the validity of Floyd’s versionof Jacobi’s theorem. We stress that in classical me-chanics this theorem is a consequence of a particularcanonical transformation which makes the new Hamil-tonian vanish. The resulting Hamilton–Jacobi equa-tion is a first order one. In quantum mechanics, if weuse the coordinatex with which the quantum laws takethe classical forms, we can reproduce the procedure ofthe canonical transformation making the new Hamil-tonian vanish and get to the Jacobi’s theorem so that

(18)(t − t0)1 =[∂S0(x)

∂E

]x=cte

.

The resulting QSHJE, in whichx is the variable,will be a first order one. With regard to the Jacobi’stheorem as written by Floyd [5],

(19)(t − t0)2 =[∂S0(x)

∂E

]x=cte

,

A. Bouda, T. Djama / Physics Letters A 296 (2002) 312–316 315

we observe that there is no procedure which startsfrom an Hamiltonian formulation and leads to thethird order well-known QSHJE. AsS0(x)= S0(x), thedifference between (18) and (19) can be showed in thefollowing relations

(t − t0)2 =[∂

∂ES0(x, a, b,E)

]x=cte

=[∂

∂ES0[x(x, a, b,E),E]

]x=cte

=[∂S0

∂E

]x=cte

+ ∂S0

∂x

[∂x

∂E

]x=cte

(20)= (t − t0)1 + ∂S0

∂x

[∂x

∂E

]x=cte

.

Now, let us consider the argument proposed by Floydto justify the use of its version of Jacobi’s theorem.From the relationS0 = S+Et between the reduced ac-tion S0(x,E,a, b) and the Hamilton’s principle func-tion S(x, t,E,a, b), he first calculated the derivativewith respect tot and got

(21)∂S

∂t= −E.

Then, he calculated the derivative with respect toE

and got

(22)∂S0

∂E= ∂S

∂E+ ∂t

∂E+ t .

In (21), he consideredE andt as independent, whilein (22) t is considered as a function ofE. Furthermore,he substituted in (22)∂S/∂E by (∂S/∂t)(∂t/∂E).Firstly, in our point of view, when we considerS(x, t,E,a, b), all the elements of the set(x, t,E,a, b) must be seen as independent. Secondly, even ifwe suppose thatt = t (E), we cannot substitute∂S/∂Eby (∂S/∂t)(∂t/∂E) because in this case we haveS =S(x, t (E),E,a, b) and we must write

(23)∂S

∂E= ∂S

∂E

∣∣∣∣t=cte

+ ∂S

∂t

∣∣∣∣E=cte

∂t

∂E.

Thirdly, he got∂S0/∂E = t representing the Jacobi’stheorem. We observe in describing the motion for anyinitial condition that there is an integration constantmissing from this equation.

To conclude this discussion about Floyd’s versionof Jacobi’s theorem, let us show that the trajectoriesdepend on the choice of the couple of Schrödinger’s

solutions. The reduced action is [2]

(24)S0 = harctan

(aφ1

φ2+ b

)+ hλ,

(φ1, φ2) being a real set of independent solutions ofSchrödinger’s equation. As an example, we consider afree particle of energyE and we setk = √

2mE/h. Ifwe choose

(25)φ1 = sin(kx), φ2 = cos(kx),

and use Jacobi’s theorem as proposed by Floyd, we get

(26)

t = t0 + ma

hkx[(

1+ b2)cos2(kx)+ a2 sin2(kx)

+ 2ab sin(kx)cos(kx)]−1

.

Another possible choice is

(27)θ1 = sin(kx)+ g(k)cos(kx),

(28)θ2 = cos(kx)+ f (k)sin(kx),

wheref andg are two arbitrary real functions ofksatisfying the conditionfg = 1. We indicate thatFloyd [6] has also used linear combinations of Schrö-dinger’s solutions with coefficients depending onk.For simplicity, we choose in what followsg(k)= 0.Now, let us look for the existence of three real para-metersa, b andt0 with which the reduced action takesthe form

(29)S0 = harctan

(aθ1

θ2+ b

)+ hλ

as in (24), and the resulting Floyd’s trajectory,

t = t0 + ma

hk

[x − df

dksin2(kx)

]

(30)

×[(

1+ b2)cos2(kx)

+ (a2 + b2f 2 + f 2 + 2abf

)sin2(kx)

+ 2(f + ab+ b2f

)sin(kx)cos(kx)

]−1,

reproduces the same quantum equation as (26) forevery f (k). This implies that the right-hand sidesof (26) and (30) must be identical. Forx = 0, thisidentity givest0 = t0, and, therefore, forx = π/2k andx = 3π/2k, we deduce that

(31)aa

(π

2k− df

dk

)= π

2k

(a2 + b2f 2 + f 2 + 2abf

)

316 A. Bouda, T. Djama / Physics Letters A 296 (2002) 312–316

and

aa

(3π

2k− df

dk

)= 3π

2k

(a2 + b2f 2 + f 2 + 2abf

),

(32)

respectively. These two last equations cannot be simul-taneously satisfied unless one hasdf/dk = 0. Sincethe functionf (k) is arbitrary, the identification ofEqs. (31) and (32) leads to a contradiction. Thus, weget to the unsatisfactory conclusion for Floyd’s trajec-tories: the mathematical choices affect the physics re-sults. This is not the case for our formulation for whichwe clearly showed [7] for any potential that the tra-jectories are independent on the choice of the couple(φ1, φ2).

The ambiguity appearing in the definition of thederivative∂S0/∂E when we consider the dependenceon E of the integration constants is pointed outby Faraggi–Matone [3]. In order to allow “a non-ambiguous definition of time parametrization”, theysuggested that all the terms depending onE and whichcan be absorbed in a redefinition of the integrationconstants should not be considered in evaluating∂S0/∂E. With this proposal, we can indeed showthat Floyd’s trajectories are independent on the choiceof the couple(φ1, φ2). In fact, let us consider thetransformation

(33)φ1 → θ1 = µφ1 + νφ2,

(34)φ2 → θ2 = αφ1 + βφ2,

where the real parameters(µ, ν,α,β) are dependingonE and satisfying the conditionµβ = να. If we writefor any potential the new reduced action as in (29),with the same procedure developed in Ref. [7], we canfind a = a(a, b,µ, ν,α,β) andb = b(a, b,µ, ν,α,β)

in such a way as to guarantee that∂S0/∂x = ∂S0/∂x.This equality implies that, up to an additive constant,S0 andS0 are identical. In other words, we can write

(35)

S0 = harctan

[a(a, b,µ, ν,α,β)

φ1

φ2

+ b(a, b,µ, ν,α,β)

],

where we have omitted the additive constant. Accord-ing to Faraggi–Matone’s proposal, it follows that

(36)∂S0

∂E= ∂S0

∂E.

However, this procedure of evaluating∂S0/∂E leadsto some unsatisfactory results. As an example, if wecalculate the time reflection for a semi-infinite rectan-gular barrier [6], we get a vanishing value. Anotherunsatisfactory aspect of this procedure appears whenwe consider the free case withE = 0 as a limit fromarbitraryE. In fact, if we want to reproduce the twoindependent solutionsφ0

1 = x andφ02 = 1 of the free

case [3,8] from the solutions (25) when we considerthe limitE → 0, we must rescaleφ1(x) as follows

(37)φ1 = k−1 sin(kx).

If we want to keep this possibility of reproducing thefree case in the limitE → 0 when we apply Jacobi’stheorem, we must write explicitly the factork−1 in theexpression of the reduced action and this will give riseto a further term in the right-hand side of (26). It isclear that this creates confusion in the definition oftime parametrization.

Acknowledgements

We would like to thank E.R. Floyd for interestingdiscussions and encouragements despite our disagree-ments about the formulation of trajectory representa-tion of quantum mechanics.

References

[1] E.R. Floyd, Comments on Bouda and Djama’s “QuantumNewton’s Law”, submitted to Phys. Lett. A.

[2] A. Bouda, T. Djama, Phys. Lett. A 285 (2001) 27.[3] A.E. Faraggi, M. Matone, Int. J. Mod. Phys. A 15 (2000) 1869.[4] A.E. Faraggi, M. Matone, Phys. Lett. A 249 (1998) 180.[5] E.R. Floyd, Phys. Rev. D 26 (1982) 1339.[6] E.R. Floyd, Found. Phys. Lett. 13 (2000) 235.[7] A. Bouda, T. Djama, quant-ph/0108022.[8] A.E. Faraggi, M. Matone, Phys. Lett. B 445 (1998) 77.