Open access peer-reviewed chapter

# On the Nonuniqueness of the Hamiltonian for Systems with One Degree of Freedom

By Sikarin Yoo-Kong

Submitted: November 6th 2018Reviewed: June 14th 2019Published: December 13th 2019

DOI: 10.5772/intechopen.88069

## Abstract

The alternative Hamiltonians for systems with one degree of freedom are solved directly from the Hamilton’s equations. These new Hamiltonians produce the same equation of motion with the standard one (called the Newtonian Hamiltonian). Furthermore, new Hamiltonians come with an extra-parameter, which can be used to recover the standard Hamiltonian.

### Keywords

• Hamiltonian
• Lagrangian
• nonuniqueness
• variational principle
• inverse problem of calculus of variations

## 1. Introduction

It was well known that the Lagrangian possesses the nonuniqueness property. It means that the constant can be added or multiplied into the Lagrangian: LNẋxαLNẋx+β

Furthermore, the total derivative term can also be added to the Lagrangian without alternating the equation of motion: LNẋxαLNẋx+β+df/dt,

where =fxt. This fact can be seen immediately from the variational principle with the action functional:

Sx=0TdtαLNẋx+β+dfdt=0TdtαLNẋx+β+fTf0E1

Obviously, the last two terms contribute only at the boundary. Then the variationxx+δxon the action and δSx=0, with conditions δx0=δxT=0,give us the same Euler-Lagrange equation:

xLNẋxddtẋLNẋx=0E2

The standard Lagrangian takes the form

LNẋx=TẋVxE3

where Tẋis the kinetic energy and Vxis the potential energy of the system. For a system with one degree of freedom, the kinetic energy is Tẋ=mẋ2/2. The equation of motion associated with the Lagrangian Eq. (1) is

x¨=1/mdVx/dx=defQE4

Recently, it has been found that actually there is an alternative form of the Lagrangian called the multiplicative form [1, 2, 3]: Lẋx=FẋGx, where F and G are to be determined. Putting this new Lagrangian into the Euler–Lagrange Eq. (2), we obtained

Lλẋx=mλ2eExẋmλ2+ẋλ20ẋeExq̇mλ2dq̇E5

where Exẋ=mẋ2/2+Vxis the energy function and mλ2is in the energy unit. We find that under the limit λwhich is very large limλLλẋxmλ2=LNẋx, we recover the standard Lagrangian. The derivation of Eq. (5) can be found in the Appendix. Interestingly, this new Lagrangian can be treated as a generating function producing an infinite hierarchy of the Lagrangian:

Lλẋx=j=01j!1mλ2j1LjẋxE6

where

Ljẋx=k=0jj!TjkVkjk!k!2j2k+1E7

These new Lagrangians Ljẋx, however, produce the same equation of motion. Equations (6) and (7) provide an alternative way to modify the Lagrangian Eq. (3).

The problem studied in [1, 2, 3] that is actually related to the inverse problem of calculus of variations in the one-dimensional case. The well-known result can be dated back to the work of Sonin [4] and Douglas [5].

Theorem (Sonin): For every function Q,there exists a solution gLof the equation:

gQx¨=xLẋxddtẋLẋx.whereg=2ẋ2Lẋx0E8

What we did in [1, 2, 3] is that we went further to show that actually Eq. (8) admits infinite solutions.

In the present chapter, we will construct the Hamiltonian hierarchy for the system with one degree of freedom. In Section 2, the multiplicative Hamiltonian will be solved directly from Hamilton’s equations. In Section 3, the physical meaning of the parameter λwill be discussed. In Section 4, the redundancy of the Hamiltonians and Lagrangians will be explained. In the last section, a summary will be delivered.

## 2. The multiplicative Hamiltonian

To obtain the Hamiltonian, we may use the Legendre transformation:

HNpx=pẋLNẋxE9

where p=L/ẋ=mẋis the momentum variable. The standard form of the Hamiltonian is

HNpx=p22m+VxE10

which is nothing but the total energy of the system. The action is then

Spx=0TdtpẋHNpxE11

With the variations xx+δxand pp+δp, with conditions δx0=δxT=0, the least action principle δSx=0gives us

ẋ=pHNpx,ṗ=xHNpxE12

which are known as a set of Hamilton’s equations.

We now introduce a new Hamiltonian, called the multiplicative Hamiltonian, in a form

Hpx=KpWx,E13

where Kpand Wxare to be determined. Equations (12) and (3) give us a new equation:

0=1mxHNpx+ṗ2p2HNpx+pm2xpHNpxE14

Replacing HNby Hand inserting Eqs. (13) into (14), we obtain

0=d2Kdp2+1mṗWdWdxpdKdp+KE15

Now we define

A=def1mṗWdWdxE16

where Ais a constant to be determined. Equation (16) can be immediately solved and result in

Wx=aemAVxE17

where ais a constant of integration. Substituting Eq. (17) into Eq. (15), we find that the function Kpis in the form

Kp=beAp22E18

where bis another constant. Then the multiplicative Hamiltonian Eq. (13) becomes

Hpx=ceAp22mAVxE19

where c=ab. If we now choose c=mλ2and A=1/m2λ2, the Hamiltonian Eq. (19) becomes

Hλpx=mλ2eHNpxmλ2E20

Inserting Eq. (20) into Eq. (14), we find that

dVxdxp2m2λ2+1=ṗp2m2λ2+1dVxdx=ṗE21

which is the equation of motion of the system. Then this new Hamiltonian Eq. (20) gives us the same equation of motion as Eq. (10).

For the case mλ2HNpx, we find that the multiplicative Hamiltonian

Hλpxmλ2+HNpxE22

gives back the standard Hamiltonian. The constant mλ2does not alter the equation of the motion of the system.

We find that the multiplicative Hamiltonian Eq. (20) can also be directly obtained from the Legendre transformation:

Hλpx=pλẋLλẋxE23

where

pλ=ẋLλẋx=1λ20peζ22m2λ2mE24

Inserting Eqs. (24) and (5) into Eq. (23), we obtain

Hλpx=mλ21λ20peζ22m2λ2mpm
mλ2ep22mλ2+pm2λ20peζ22m2λ2meVxmλ2
=mλ2eHNpxmλ2E25

which is identical to Eq. (20).

Furthermore, we can rewrite the multiplicative Hamiltonian Eq. (20) in terms of the series:

Hλpx=j=01j!1mλ2j1HjpxE26

whereHjpxHNj=p2/2m+Vxj. It is not difficult to see that Hjpxproduces exactly the equation of motion Eq. (21).

From the structure of Eqs. (26) and (6), it must be a hierarchy of the Legendre transformation. To establish such hierarchy, we start to rewrite the momentum Eq. (24) in the form

pλ=1λ20peζ22m2λ2m=j=01j!1mλ2j1pjE27

where

pjpx=j!pj1Vx+p2j1j12j12j1mj1,j1andp=mẋE28

Then the Legendre transformation Eq. (23) becomes

0=j=01j!1mλ2j1Ljẋxpjẋ+HjpxE29

Eq. (29) holds if

Ljẋx=pjẋ+HjpxE30

which are the Legendre transformations for each pair of the Hamiltonian Hjpxand LagrangianLjẋxin the hierarchy.

Next, we consider the total derivative dHjpx=dpjẋdLjẋxresulting in

dxHjx+ṗpjp+dpHjpẋpjp=0E31

Eq. (31) holds if

Hjx=ṗpjp,Hjp=ẋpjpE32

Eq. (32) can be considered as the modified Hamilton’s equations for each Hjpxin the hierarchy. Obviously, for j=1, we retrieve the standard Hamilton’s Eq. (12), since p1=p=mẋ.

From the structure of the multiplicative Hamiltonian Eq. (20), it seems to suggest that the exponential of the function, defined on phase space, is always a solution of the Eq. (14). Then we now introduce an ansatz form of the Hamiltonian as

Ha,bpx=beaZpxE33

where aand bare constants to be determined. Substituting Eq. (33) into Eq. (14), we obtain

0=1mZx+ṗ2Zp2+pm2Zxp+aṗZp2+pmZpZxE34

We find that if we take HNpx=Zpxto be the standard Hamiltonian, the first three terms in Eq. (34) give us back Eq. (14). Then the last bracket must vanish and gives us an extra-relation:

0=ṗHNp2+pmHNpHNxE35

or

0=ṗHNp+pmHNxE36

We immediately see that actually Eq. (36) is a consequence of the conservation of the energy of the system:

0=dHNdt=HNppt+HNxxt=ṗHNp+pmHNxE37

Then what we have here is another equation that can be used to determine for the Hamiltonian subject to the equation of motion Eq. (21). To see this, we may start with the standard form of the Hamiltonian HNpx=Tp+Vx, where Tpis a function of the momentum and to be determined. Inserting the Hamiltonian into Eq. (36), we obtain

0=ṗdTdp+pmdVdxE38

Using Eqs. (21) and (38), it can be rewritten in the form

0=ṗdTdppmE39

Since ṗ0, it means that the term inside the bracket must be zero and

dT=pmdpTp=p22m+CE40

where Cis a constant which can be chosen to be zero. So we successfully solved the standard Hamiltonian.

Next, we put Zpx=KpWxwhich is in the multiplicative form Eq. (13) into Eq. (36), and we obtain

0=WṗWdKdp+pmKdWdxE41

or

mKpdKdp=1WdWdVE42

We see that both sides of Eq. (42) are independent to each other. Then Eq. (42) holds if both sides equal to a constant β. We have now for the left-hand side

mKpKp=β
dKK=βpmdpKp=C1eβTpE43

where C1is a constant to be determined. Next, we consider the right-hand side

1WdWdV=β
dWW=βdVWx=C2eβVxE44

where C2is a constant to be determined. Then finally the function Zpxbecomes

Zpx=C1C2eβHNpxE45

where HNpxis the standard Hamiltonian. If we now choose C1C2=mλ2and β=1/mλ2, the function Zpxis exactly the same with Eq. (20).

We see that with Eq. (36) the Hamiltonian can be easily determined. Here we come with the conclusion that in every function Q=dVdx,there exist infinite Hamiltonians of equation

Qṗ=ṗHp+pmHxE46

The existence of solutions of Eq. (46) implies that actually we can do an inverse problem of the Hamiltonian for the systems with one degree of freedom.

Remark: The perspective on nonuniqueness of Hamiltonian, as well as Lagrangian, here in the present work is quite different from those in Aubry-Mather theory [6, 7] (see also [8]). What they had been investigating is the modification of the Tonelli LagrangianLηLη̂, where mechanical Lagrangian LNẋx=TẋVxis one of Tonelli Lagrangians. Here η̂=<ηx,ẋ>:TMRand ηxis a closed 1-from on the manifold M. This means that dtLηand dtLwill have the same extremals and therefore the same Euler-Lagrange evolution, since δdtη̂=0. Thus for a fixed L, the extreamise of the action will depend only on the de Rham cohomology class c=ηH1MR. Then we have a family of modified Lagrangians, parameterized over H1MR. With the modified Tonelli LagrangianLη, one can easily find the associated Hamiltonian Hηxp=Hxηx+p, where the momentum is altered: pp+ηx. Then we also have a family of modified Hamiltonians, parameterized over H1MR. To make all this more transparent, we better go with a simplest example. Consider the modified LagrangianLϵLN+ϵẋ, where ϵis a constant. We find that a new action differs from an old action by a constant depending on the endpoints, abdtLϵ=abdtLN+ϵxbxa,and they give exactly the same Euler-Lagrange equation (see also Eq. (1)). With this new LagrangianLϵ, we can directly obtain the Hamiltonian Hϵxp=HNxp+ϵ.

## 3. Harmonic oscillator

In this section, we give an explicit example, e.g., the harmonic example, and also give the physical interpretation of the parameter λ. The standard Hamiltonian for the harmonic oscillator reads

Hpx=p22m+kx22E47

Then the multiplicative Hamiltonian for the harmonic oscillator is

Hλpx=mλ2e1mλ2p22m+kx22E48

Now we introduce η=xp,and then we consider

dtλ=JHλ∂ηwhere∂η=/∂x/∂pE49

wheretλis a time variable associated with the multiplicative Hamiltonian and Jis the symplectic matrix given by

J=0110E50

Inserting Eq. (48) into Eq. (49), we obtain

dtλ=k=01k!1mλ2k1JHk∂η=defk=0dtkE51

where

ddtk=Ek1k1!mλ2k1ddtE52

where E=T+Vis the energy function and tis the standard time variable associated with the Hamiltonian Eq. (47). Equation (51) suggests that the λ-flow is comprised of infinite different flows on the same trajectory on the phase space (see Figure 1).

This means that we can choose any Hamiltonian in the hierarchy to work with. The physics of the system remains the same but with a different time scale. Then we may say that the parameter λplays a role of scaling in the Hamiltonian flow on the phase space. From Eq. (52), we see that as mλ2, only the standard flow survives, and of course we retrieve the standard evolution t1=tof the system on phase space.

Next we consider the standard Lagrangian of the harmonic oscillator

LNẋx=mẋ22kx22E53

and the multiplicative Lagrangian is

Lλẋx=mλ2eExẋmλ2+ẋλ20ẋeExq̇mλ2dq̇E54

whereExẋ=mẋ2/2+kx2/2is the energy function. We know that Lagrangian Eq. (54) can be rewritten in the form

Lλẋx=j=01j!1mλ2LjẋxE55

where

Ljẋx=k=0jj!mẋ2/2jkkx2/2kjk!k!2j2k+1E56

The action of the system is given by

Sx=0TdtLλẋx=j=10TdtjLjẋxE57

where

dtj=1j!1mλ2k1dtE58

The variation xx+δxwith conditionsδx0=δxT=0results in

δSx=j=10TdtjLjxddtjLjxjE59

wherexj=dx/dtj. Least action principle δSx=0gives infinite Euler-Lagrange equations

0=LjxddtjLjxj,j=1,2,3E60

which produce the equation of motions

d2xdtj2=kxmE61

associated with different time variables. Again in this case, we have the same structure of equation of motion for each Lagrangian in hierarchy but with a different time scale. From Eq. (58), we see that as mλ2, only the standard flow survives, and of course we retrieve the standard evolution t1=tof the system. Then the parameter λalso plays the role of scaling in the Lagrangian structure.

## 4. Redundancy

From previous sections, we see that there are many forms of the Hamiltonian that you can work with. One may start with the assumption that any new Hamiltonian is written as a function of the standard Hamiltonian HN: H=fHN. Inserting this new Hamiltonian into Hamilton’s equations, we obtain

fEHx=pτ,fEHp=xτE62

wherefE=dfHN/HNwith fixing HN=Eand t=fHNτis the rescaling of time parameter. This result agrees with what we have in Section 3, rescaling the time evolution of the system. However, there are some major different features as follows. The first thing is that our new Hamiltonians contain a parameter λ, since the explicit forms of the Hamiltonian are obtained. With this parameter, it makes our rescaling much more interesting with the fact that the rescaling time variables depend on also the parameter (see Eq. (52)). Then it means that we know how to move from one scale to another scale and of course we know how to obtain the standard time evolution by playing with the limit of the parameter λ. Without explicit form of the new Hamiltonian, which contains a parameter, we cannot see this fine detail of family of rescaling time variables, since there is only a fixed parameter E. The second thing is that actually the new Hamiltonian Eq. (20), which is a function of the standard Hamiltonian, can be obtained from the Lagrangian Eq. (5) by means of Legendre transformation. What we have seen is that Lagrangian Eq. (5) is nontrivial and is not a function of the standard Lagrangian. Again this new Lagrangian contains a parameter λ, the same with the one in the new Hamiltonian. With this parameter, the Lagrangian hierarchy Eq. (7) is obtained. What we have here is a family of nontrivial Lagrangians to work with, producing the same equation of motion, as a consequence of nonuniqueness property. An importance thing is that there is no way you can guess the form of this family of Lagrangian without our mechanism in the appendix. This means that the Hamiltonian in the form H=fHNcannot deliver all these fine details. The explicit form of the Hamiltonian Eq. (20) allows us to study in more detail and is definitely richer than the standard one.

## 5. Summary

We show that actually there exist infinite Hamiltonian functions for the systems with one degree of freedom. We may conclude that there exists the reverse engineering of the calculus of variation on phase space (see Eq. (44)). Furthermore, the solution of Eq. (44) exists not only as one but infinite. Interesting fact here is that these new Hamiltonians come with the extra-parameter called λ. We give the interpretation that the term mλ2involves the time scaling of the system. This means that we can pick any Hamiltonian or Lagrangian to study the system, but the evolution will be in different scales.

In the case of many degrees of freedom, the problem turns out to be very difficult. Even in the case of two degrees of freedom, the problem is already hard to solve from scratch. We may start with an anzast form of the Lagrangian: Lẋẏxy=FẋẏGxy. This difficulty can be seen from the fact that we have to solve a non-separable coupled equation. A mathematical trig or further assumptions might be needed for solving Fẋẏand Gxy. The investigation is now monitored.

Furthermore, promoting the Hamiltonian Eq. (20) to be a quantum operator in the context of Schrodinger’s equation is also an interesting problem. This seems to suggest that an alternative form of the wave function for a considering system is possibly obtained. This can be seen as a result from that fact that with new Hamiltonian operator, we need to solve a different eigenvalue equation, and of course a new appropriate eigenstate is needed. From the Lagrangian point of view, extension to the quantum realm in the context of Feynman path integrals is quite natural to address. However, this problem is not easy to deal with since the Lagrangian multiplication is not in the quadratic form. Then a common procedure for deriving the propagator is no longer applicable. Further study is on our program of investigation.

## Acknowledgments

The author would like to thank Kittikun Sarawuttinack, Saksilpa Srisukson, and Kittapat Ratanaphupha for their interests and involving themselves in the investigation on this topic.

## Notes

The content in this chapter is collected from a series of papers [1, 2, 3].

## Appendix

In this section, we will demonstrate how to solve the multiplicative Lagrangian Eq. (5). We introduce here again the Lagrangian Lẋx=FẋGx, where F and G are to be determined. Inserting the Lagrangian into the action and performing the variationxx+δx, with conditions δx0=δxT=0, we obtain

δSx=0TdtddtGdFdẋ+FdGdxδxE63

The least action principle states that the system will follow the path which δS=0resulting in

ddtGdFdẋ+FdGdx=0E64

Eq. (64) can be rewritten in the form

d2Fdẋ21x¨GdGdxFẋdFdẋ=0E65

Using equation of motion, we observe that the coefficient of the second term depends only xvariable. Then we may set

1x¨GdGdx=defA1GdGdx=AmdVdxE66

We find that it is not difficult to see that the function Gthat satisfies Eq. (66) is

Gx=α1eAVx/mE67

where α1is a constant to be determined. Inserting Eq. (66) into Eq. (65), we obtain

d2Fdẋ2AFẋdFdẋ=0E68

and the solution Fis given by

Fẋ=α2ẋα3eAẋ2/2+ẋA0ẋdveAv2/2E69

whereα2and α3are constants. Then the multiplicative Lagrangian is

Lẋx=k1ẋk2eAẋ2/2+ẋA0ẋdveAv2/2eAVx/mE70

wherek1=α1α2and k2=α1α3are new constants to be determined. We find that if we choose k1=0,A=1/λ2which is a unit of inverse velocity squared and k2=mλ2which is in energy unit, Lagrangian Eq. (70) can be simplified to

limλLλẋxmλ2=mẋ22Vx=LNẋxE71

the standard Lagrangian at the limit λapproaching to infinity. Therefore, the Lagrangian Eq. (70) is now written in the form

Lλẋx=mλ2eẋ2/2λ2+ẋλ20ẋdvev2/2λ2eVx/mλ2E72

which can be considered as the one-parameter extended class of the standard Lagrangian.

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© 2019 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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Sikarin Yoo-Kong (December 13th 2019). On the Nonuniqueness of the Hamiltonian for Systems with One Degree of Freedom, Progress in Relativity, Calin Gheorghe Buzea, Maricel Agop and Leo Butler, IntechOpen, DOI: 10.5772/intechopen.88069. Available from:

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