Quantum Integrable Systems and Quantum

Quantum Integrable Systems and Quantum McMillan Map

A PROJECT REPORT

submitted by

SABYASACHI CHAKRABORTY (PH08C023) for the award of the degree

of

MASTER OF SCIENCE in

PHYSICS

DEPARTMENT OF PHYSICS INDIAN INSTITUTE OF TECHNOLOGY MADRAS APRIL 2010

CERTIFICATE

This is to certify that the thesis titled Quantum Integrable Systems and Quantum McMillan Map, submitted by Sabyasachi Chakraborty, to the Indian Institute of Technology Madras, for the award of the degree of Master of Science, is a bona fide record of the work done by him under my supervision. The contents of this thesis, in full or in parts, have not been submitted to any other Institute or University for the award of any degree or diploma.

Dr. Arul Lakshminarayan Advisor, Associate Professor, Dept. of Physics, IIT-Madras, 600 036

Place: Chennai Date: 19 April 2010

ACKNOWLEDGEMENTS I would like to take this opportunity to express my sincere gratitude to many individuals who have given me a lot of support during my two years masters program at IIT Madras. Without them my project would not have been completed successfully on time. At the outset I am indebted to my advisor Dr. Arul Lakshminarayan for giving me an opportunity to work in the field of dynamical systems. I am also thankful to him for his patience, excellent guidance and continuous motivation throughout my entire project work. Thanks to all my friends during my M.Sc. period for their continuous assistance. But specially I am indebted to Arko and Malay. They have shared their experience and guided me from time to time with invaluable discussions. Thanks to Ankur Guha Roy for helping me out in making Matlab codes and also with LATEX 2ε .Without his constant inspiration and support this wouldnot have been possible. Finally I dedicate this thesis to my parents and my elder brother for their much needed support and encouragement through my entire life.

i

ABSTRACT Painlevé test for the continuous system proves the integrability of the system. But it has its drawbacks. It cannot deal with the singularity worse than the movable singularity. The KdV equation has been proved integrable with the help of this test. The study of discrete Painlevé test proves the fact that the McMillan map is an integrable, area preserving map. The McMillan map is obtained from the δ kicked harmonic oscillator system. Now the δ function represented in the Hamiltonian can be represented as a summation of the cosine function. ∞ ∞ X X t t cos(2nπ ). Now in the later part I have tried to study δ( − n) = T T n→∞ n→∞ the nature of the stroboscopic plots when the δ function is removed and a single cos term is implemented. I have given the plots for different values of ω. The nature of these plots looks chaotic. Apart from that I also studied some Knot theory which determines the statistical integrability by calculating the partition function.

ii

TABLE OF CONTENTS LIST OF TABLES

v

LIST OF FIGURES

1

1 Introduction

2

1.1

Algorithm of The Painlev´ e Property . . . . . . . . . . . .

2

1.2

Scopes and Limitations of The Test . . . . . . . . . . . . .

3

1.2.1

Scopes . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.2.2

Limitations . . . . . . . . . . . . . . . . . . . . . . . .

4

Example:The Korteweg-de Vries Equation . . . . . . . . . . . .

4

1.3

2 The Painlev´ e Test To Detect The Integrability of A Discrete Equation 2.1

Algorithm of The Discrete Painlev´ e Test . . . . . . . . . . 2.1.1

2.2

2.3

Lax Pairs and The Deduction of The Lattice KdV Equation . . . . . . . . . . . . . . . . . . . . . . . . . .

10 11

Singularity Confinement of The Maps . . . . . . . . . . . .

15

2.2.1

15

The Lattice KdV Equation . . . . . . . . . . . . . .

Singularity Confinement of The McMillan Map

. . . . .

16

2.3.1

Considering d = 1 and f = 0 . . . . . . . . . . . . . .

20

2.3.2

Considering d 6= 0,so taking d = 1 . . . . . . . . . . .

21

3 Jacobian Elliptic Function 3.1

10

23

Study of The Jacobian Elliptic Functions . . . . . . . . . .

23

3.1.1

Identities of The Elliptic Functions . . . . . . . . .

24

3.1.2

Addition Formula for The Elliptic Functions . . .

24

4 Constant of Motion For The Classical McMillan Map 4.0.3

The Classical McMillan Map . . . . . . . . . . . . .

iii

26 26

4.0.4

Constant of Motion For The Classical McMillan Map

27

4.0.5

The Constant of Motion is Parametrized in Terms of The Elliptic Functions . . . . . . . . . . . . . . . .

29

5 Constant of Motion For The Quantum McMillan Map 5.0.6

Constant of Motion For The Quantum McMillan Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Quantum McMillan Map And Stroboscopic Plots

33 33 39

6.0.7

1.5 Degree of Freedom System . . . . . . . . . . . .

39

6.0.8

Stroboscopic Plots . . . . . . . . . . . . . . . . . . . .

40

7 Quantum McMillan Map and Some Aspects of The McMillan Map 7.0.9

44

δ Kicked Harmonic Oscillator . . . . . . . . . . . . .

44

7.0.10 Some Aspects of The McMillan Map . . . . . . . .

45

7.0.11 Stable Points of The McMillan Map . . . . . . . . .

46

7.0.12 Stability Criterion of The McMillan Map . . . . .

48

7.0.13 Determination of The Stability of The Orbit(q = 0)

50

LIST OF TABLES

v

LIST OF FIGURES 6.1

Phase portrairs for parametrically forced harmonic oscillator . .

41

6.2

Stroboscopic plots for parametrically forced harmonic oscillator

42

6.3

For parametrically forced pendulum . . . . . . . . . . . . . . . .

43

7.1

Double well potential . . . . . . . . . . . . . . . . . . . . . . . .

53

7.2

stability of the orbit with q = 0 . . . . . . . . . . . . . . . . . .

53

7.3

stroboscopic plot . . . . . . . . . . . . . . . . . . . . . . . . . .

54

7.4


54

7.5


55

7.6


55

1

CHAPTER 1

Introduction

The whole dynamical system can be divided into two classes. First when the system is chaotic there the solutions of the system depends sensitively on the initial conditions. And the next is integrable system where the solutions of the systems can be written down in terms of the elementary functions which is not the case for the chaotic system.

Painlevé criterion which detects the integrability of a continuous system as well as a discrete system is not a fool proof criterion. It has its drawbacks. In this section we will discuss the integrability of a continuous system and also discuss the drawbacks.

Suppose the equation is of the form fxx = K(x, f, fx ). The restrictions on K is that it has to be analytic in x, rational in fx and algebraic in f. The Painlevé property can work if the equations contain no worse singularities than the movable poles so logarithmic Branch points and essential singularities are also taken under consideration.

1.1

Algorithm of The Painlev´ e Property

This algorithm gives us the opportunity to find out weather or not the an ODE or an PDE is integrable or not[1].

For a partial differential equation or PDE case where the function f is repre-

sented as f = f (t, x). Now f is expressed in terms of the Laurent series f =g

α

∞ X

uk g k

(1.1)

n=0

As expressed previously it should have only movable poles. Again u0 (t, x) 6= 0 and α is considered to be a negative integer. The criterion on g is again g(t, x) = 0 with gx (t, x) 6= 0

• Step 1: First we have to determine the negative integer α from the leading ordered term f ∝ u0 g α .

• Step 2: Secondly we have to calculate the non-negative integer powers r, which is known as the resonances, this is done by substituting the term f ∝ u0 g α + ur g r+α into (1.1) and retaining the most singular term.

• Step 3: At resonance conditions i.e for certain values of r ur has to be arbitrary due to its vanishing coefficient.

These three conditions are known as the compatibility condition and if these three are satisfied the equation is said to be integrable. But as we have discussed earlier it can work if the equations contain no worse singularities than the movable poles so this criterion is necessary but not sufficient.

1.2 1.2.1

Scopes and Limitations of The Test Scopes

•The Pianlevé test works for single ODE or PDE.

3

•The parameters and the number of independent variable is unlimited.

1.2.2

Limitations

•It can deal only with movable singularities and not any other kind of singularity.

•Arbitrary parameters of f and arbitrary functions of f and its derivatives are not allowed.

•We have to determine the u1 and u2 respectively because selective determination of u’s will lead to the wrong result.

1.3

Example:The Korteweg-de Vries Equation

The KdV equation is in the form ft + f fx + bfxxx = 0

(1.2)

. Where ft and fx are the time and space derivative of the function respectively. Now to solve for integrability we first put f = u0 g alfa

(1.3)

. Now putting this form in the terms of equation (1.2) we find that ft = u0 (alfa )g (alfa −1)

∂g ∂t

(1.4)

.The derivatives of u is not considered here since it doesn’t contain alfa so minimum value of alfa could’t be obtained from it.

4

∂g ∂x

(1.5)

∂ 2g ∂g + u0 (alfa )(alfa − 1)g alfa −2 2 ∂x ∂x

(1.6)

fx = u0 (alfa )g (alfa −1)

fxx = u0 (alfa )g alfa −1

2

3g

∂ g ∂ alfa −2 ∂g ( ∂x )( ∂x fxxx = u0 (alfa )g (alfa −1) ∂x 3 +u0 (alfa )(alfa −1)g 2 )+u0 (alfa )(alfa − 2

∂ g alfa −3 ∂g 3 1)g alfa −2 ∂x ( ∂x ) 2 + u0 (alfa )(alfa − 1)(alfa − 2)g

Putting these terms in the equation number (1.2) we find that f fx = u20 (alfa )g (2alfa −1)

∂g ∂x

(1.7)

2

3g

∂ g ∂ alfa −2 ∂g ( ∂x )( ∂x bfxxx = bu0 (alfa )g (alfa −1) ∂x 3 +bg 2 )u0 alfa (2alfa −1)+bu0 (alfa )(alfa −

∂g 3 1)(alfa − 2)g alfa −3 ( ∂x )

So the minimum powers of g are [2alfa − 1, alfa − 3]

(1.8)

And the corresponding coefficient of those terms are given by [u20 alfa

∂g ∂g , bu0 alfa (alfa − 1)(alfa − 2)( )3 ] ∂x ∂x

(1.9)

As alfa is a replacement for α and as it is a negative integer as considered

5

previously so to get the value of alfa we equate the two exponents of g and we get alfa = −2

(1.10)

and thus the minimum power of g is -5. Now putting g = −5 and alfa = −2 in (1.2) we get

ft

= −2u0 g −3 (∂t g)

(1.11)

f fx

= −2u20 g −5 (∂x g)

(1.12)

bfxxx = −24u0 bg −5 (∂x g)3

(1.13)

So the most dominant term which is considering the minimum powers of g and using equation (1.2) we have − 2u0 g −5 [12b(∂x g)3 + u0 (∂x g)]

(1.14)

So solving equation (1.14) we have u0 = −12b(∂x g)2

(1.15)

Again to get the resonance condition we put f = u0 g alfa + ur g r+alfa

(1.16)

And the corresponding time and space derivatives are written by ft = (r + alfa )ur g r+alfa −1 ∂t g + u0 (alfa )g alfa −1 (∂x g)

(1.17)

fx = (r + alfa )ur g r+alfa −1 ∂x g + u0 (alfa )g alfa −1 (∂x g)

(1.18)

6

fxx = ur (r + alfa )(r + alfa − 1)g r+alfa −2 (∂x g)2 + (r + alfa )ur g r+alfa −1 (∂xx g)+ u0 (alfa )g (alfa −1) ∂xx g + u0 (alfa )(alfa − 1)g alfa −2 (∂x g)2 fxxx = ur (r + alfa )(r + alfa − 1)(r + alfa − 2)g r+alfa −3 (∂x g)3 + 2ur (r + alfa )(r + alfa − 1)g r+alfa −2 (∂x g)(∂xx g) +ur (r + alfa )(r + alfa − 1)g r+alfa −2 (∂x g)(∂xx g) + ur (r + alfa )g r+alfa −1 (∂xxx g) +u0 (alfa )g alfa −1 ∂xxx g + u0 (alfa )(alfa − 1)g alfa −2 (∂x g)(∂xx g) +2u0 (alfa )(alfa − 1)g alfa −2 ∂x g(∂xx g) + u0 (alfa )(alfa − 1)(alfa − 2)g alfa −3 (∂x g)3 This term consists the minimum powers of g i.e g r−5

fxxx = ur (r + alfa )(r + alfa − 1)(r + alfa − 2)g r+alfa −3 (∂x g)3

(1.19)

f fx = −12b(∂x g)3 ur (r − 4)g r−5

(1.20)

This term also consists the minimum powers of g. So collecting the dominant term i.e the term having g r−5 is collected

7

ur (r − 2)(r − 3)(r − 4)g r−5 (∂x g)3 − 12bur (r − 4)g r−5 (∂x g)3

(1.21)

= b(∂x g)3 (r − 6)(r − 4)(r + 1)ur g r−5

(1.22)

As mentioned previously that r is a positive integer so we take the values r = 6 and r = 4 where the maximum resonance occurs at r = 6

Now putting these results into (1.2) we have

f=

6 X

uk g k−2

(1.23)

k=0

with u0 = −12b(∂x g)2

(1.24)

So the terms present in the (1.2) becomes

fx =

6 X k=0

fxx =

(k − 2)uk g k−3 (∂x g)

6 X [(k − 2)uk g k−3 ∂xx g + (k − 2)(k − 3)uk g k−4 (∂x g)2 ] k=0

fxxx =

6 X k=0

[uk (k − 2)g k−3 ∂xxx g + (k − 2)(k − 3)uk g k−4 ∂x g∂xx g

+2(k − 2)(k − 3)uk g k−4 ∂x g∂xx g + uk (k − 2)(k − 3)(k − 4)g k−5 (∂x g)3 ]

8

(1.25)

(1.26)

f fx =

6 X

uk um g m+k−5 ∂x g

(1.27)

k,m=0

Putting these terms we get the coefficient of g −4 as 6b(∂x g)2 ((−12b(∂x g)2 )x − 36b(∂x g)(∂xx g) + 5u1 gx )

(1.28)

Same as for u2 and u3 i.e the coefficient is non vanishing but for u4 the coefficient of 1/g is 0 so u4 is arbitrary. So the first compatibility condition is satisfied.

The coefficient of g is 0 i.e u6 is arbitrary. So the compatibility condition is satisfied and the equation is integrable. On the other hand the cylindrical KdV equation which is of the form fx + fxxxx + 6f fxx + 6(fx )2 + ftx = 0 2t is not integrable because the compatibility condition is not satisfied.

9

(1.29)

CHAPTER 2

The Painlev´ e Test To Detect The Integrability of A Discrete Equation

To test the integrability of a system several methods are used. For example Knot theory helps to calculate the partition function of a system and thus it helps to find out weather or not a system is statistically integrable or not. For integrable mappings the singularities are confined and they do not propagate as time goes on, Now in this case we are discretizing the spatial and time derivatives to obtain a discrete equation and in order to prove the integrability of that equation one needs to find out the discrete Painlevé test.

Now integrable lattices are of particular importance because their different approximation generates a set of integrable partial differential equations. Until now there has been no way to predict the integrability of mappings. We will show later the stroboscopic maps which helps us to find out the integrability of a system.

Until then the present work gives a new criterion for assessing the integrability of discrete systems and it is based on the movable singularities of a mapping and so it is analogous to the Painlevé criterion for continuous systems. In this section we will mainly deal with the McMillan maps and we will try to see the nature of its singularities i.e weather it is movable or not.

2.1

Algorithm of The Discrete Painlev´ e Test

This algorithm is based on to find out the singularities in a map and see if it is movable or not. If the singularity condition is satisfied then we can ensure the

system to be integrable. Now considering the map equation i−1 i i xj+1 = xi−1 j+1 + 1/xj − 1/xj+1

(2.1)

This equation is known as the lattice KdV equation and the proof of this map equation comes from the Lax pairs. We are going to discuss the Lax pairs in the next section but before doing that the ultimate proof of the discrete Painlevé equations are in fact their effective linearization.

2.1.1

Lax Pairs and The Deduction of The Lattice KdV Equation

In most cases the Lax pairs are of the form ξφn,ξ = Ln φn

(2.2)

φn+1 = Mn φn

(2.3)

. And

. From these two equations we write[2] ξφn+1,ξ

= Ln+1 (ξ)φn+1

(2.4)

= Ln+1 (ξ)Mn (ξ)φn

(2.5)

= Mn,ξ (ξ)φn,ξ

(2.6)

And from (2.3) we write φn+1,ξ

ξφn+1,ξ = Mn,ξ (ξ)Ln (ξ)φn

11

(2.7)

So equating (2.5) and (2.7) we write Ln+1 (ξ)Mn (ξ)φn = Mn,ξ (ξ)Ln (ξ)φn

(2.8)

[Ln+1 (ξ)Mn (ξ) − Mn,ξ (ξ)Ln (ξ)]φn = 0

(2.9)

Equation (2.9) gives the compatibility condition. Now for convenience we write equation (2.9) in the form ′

Lj (k)Mj (k) = Mj+1 (k)Lj (k)

(2.10)

In which Lj and Mj are discrete analogous of Zakharov-Shabat type of systems, k being a spectral parameter and the prime denotes the spectral time shift.

Now the lattice KdV equation is also expressed in the form ′

v2j−1 = v2j

(2.11)

v2j = v2j+1 + ǫδ/v2j − ǫδ/v2j+2

(2.12)

and ′

These two equations are obtained from (2.10) where Lj = V2j V2j−1

(2.13)

where 

M =

uj

1

λ2j 0

12

 

(2.14)

And 

Vi = 



vi 1

(2.15)



λi 0

And also using the equations uj = u2j−1 − ǫδ/v2j

(2.16)

and



′

So Lj (k)M (k) =  

or,  

′

v2j 1 ′

λ2j 0

v2j

′

1

′

0

λ2j



 

(2.17)

λ2j+1 = k 2 − p2

(2.18)

′

1

v2j−1 ′

λ2j−1 0

 

uj v2j−1 + λ2j v2j−1



uj λ2j−1

λ2j−1

′

′

1

λ2j

0





= 

′

v2j−1 λ2j

λ2j {uj v2j−1 + λ2j } 

=

′

′

′

′

′

′

′

′

′

′

λ2j {uj v2j−1 + λ2j }

′

v2j−1 λ2j

u2j+1 1 λ2j+2 0

 

 

v2j

1

λ2j

0

 

λ2j v2j−1

  

So comparing the 4th element we have ′

v2j−1 = v2j

(2.19)

Likewise comparing the 1st element we have

′

′

′

′

′

u2j−1 v2j v2j−1 − ǫδ/v2j v2j v2j−1 + λ2j−1 u2j−1 = u2j+1 v2j v2j−1 − ǫδ/v2j+2 v2j v2j−1 + ′

u2j+1 λ2j−1 + ǫδ/v2j λ2j−1 − ǫδ/v2j+2 λ2j−1 13

v2j−1

1

λ2j−1 0

v2j v2j−1 + λ2j−1 v2j



v2j {uj v2j−1 + λ2j } + uj λ2j−1 v2j v2j−1 + λ2j−1 ′

1

λ2j+2 0

v2j {uj v2j−1 + λ2j } + uj λ2j−1 v2j v2j−1 + λ2j−1 ′



uj+1

= 

′

′

uj



′

′

′

′

λ2j = k 2 − q 2

λ2j

 

 

From the 2nd element we have ′

′

′

v2j v2j−1 + λ2j−1 = uj+1 v2j + λ2j

′

(2.20)

′

v2j v2j + λ2j−1 = uj+1 v2j + λ2j

′

′

′

(2.21)

′

v2j v2j−1 + λ2j−1 = uj+1 v2j−1 + λ2j

′

′

(2.22)

′

v2j v2j−1 = [u2j+1 − ǫδ/v2j+2 ]v2j−1 + (λ2j − λ2j+1 )

(2.23)

= [u2j+1 − ǫδ/v2j+2 ]v2j + (k 2 − q 2 − k 2 + p2 )

(2.24)

= [u2j+1 − ǫδ/v2j+2 ]v2j + ǫδ

(2.25)

So ′

′

′

v2j = [u2j+1 − ǫδ/v2j+2 ]v2j /v2j−1 + ǫδ/v2j−1

′

′

v2j = [u2j+1 − ǫδ/v2j+2 ] + ǫδ/v2j−1

(2.26)

(2.27)

and comparing the 3rd term in the matrix we have ′

′

′

λ2j uj v2j−1 + λ2j λ2j = λ2j+2 λ2j − 1 + λ2j+2 v2j v2j−1

(2.28)

From comparing the 1st term and after simplifying we get ′

[u2j−1 − ǫδ/v2j ]v2j−1 = (λ2j−1 − λ2j ) + v2j v2j−1

(2.29)

u2j−1 v2j − ǫδ = −ǫδ + v2j v2j−1

(2.30)

14

u2j−1 = v2j−1

(2.31)

u2j+1 = v2j+1

(2.32)

v2j = v2j+1 + ǫδ/v2j − ǫδ/v2j+2

(2.33)

So we can write ′

So together the equation (2.34) and the equation (2.19) gives us the form of the lattice KdV equation that we wrote in a different form i.e ′

v2j = v2j+1 + ǫδ/v2j − ǫδ/v2j+2

(2.34)

and ′

v2j−1 = v2j

2.2 2.2.1

(2.35)

Singularity Confinement of The Maps The Lattice KdV Equation

Before going to the McMillan maps we will discuss the singularity confinement of the Lattice KdV map. Now the previous derived equation of the map can be written in the way i i xi+1 = xi−1 j j+1 + 1/xj − 1/xj+1

(2.36)

.

Evolution takes place along the increasing direction of i. Now assuming xij = 0 15

(2.37)

we have xi+1 →∞ j

(2.38)

xi+1 j−1 → ∞

(2.39)

i−1 xi+3 + 1/xij−1 − 1/xi+2 j−1 = xj j

(2.40)

[3] Again

But

i+3 So xi+3 j−1 and xj−2 gives a finite value. Thus the singularity is confined i.e it does n’t

propagate. One can make a loop and can come back to the initial condition. For a non integrable system singularities are present at the boundaries and crossing over is not possible. This principle will help us to find out the integrability of the McMillan maps as we will see in the next section.

2.3

Singularity Confinement of The McMillan Map

The discretized an harmonic oscillator can be written in the following manner [xn+1 − 2xn + xn−1 ]/δ 2 = −αxn − βx2n [xn+1 + xn−1 ]/2

(2.41)

It can also be written in the following manner zn+1 + zn−1 =

16

2µzn 1 + zn2

(2.42)

This is more well known as the McMillan maps. Now let zn = µ=1−

αδ 2 2

So eqn (2.42) becomes =

2zn 2 [1 1+zn

=

−

αδ 2 ] 2

zn (2−αδ 2 ) 2 1+zn

q

βδ 2 xn 2

and

(2.43)

(2.44)

so

r

[zn+1 + zn−1 ][1 + zn2 ] = zn [2 − αδ 2 ]

(2.45)

zn+1 =

q

βδ 2 xn+1 2

(2.46)

zn−1 =

q

βδ 2 xn−1 2

(2.47)

βδ 2 βδ 2 2 [xn+1 + xn−1 ][1 + x ] = xn [2 − αδ 2 ] 2 2 n

2

[xn+1 + xn−1 − 2xn ] = − βδ2 x2n (xn+1 + xn−1 ) − αδ 2 xn

[xn+1 − 2xn + xn−1 ]/δ 2 = −αxn − βx2n [xn+1 + xn−1 ]/2

(2.48)

(2.49)

(2.50)

So we get back equation (2.41)

Now equation (2.41) is a particular case of the equation xn+1 + xn−1 = −

ax2n + bxn + c dx2n + exn + f

(2.51)

Although the singularity confinement was first discovered in a 2-D lattice, the best

17

way to understand how it operates in the case of a 1-D system, Now considering the most classical example we write which is the McMillan maps we write xn+1 + xn−1 =

2µxn 1 − x2n

(2.52)

This map is integrable which we will see after the singularity confinement approach. Now from the map equation it is seen that the map has a singularity at xn = 1. Now let us assume that x0 is finite and x1 blows up at 1. So we consider a small neighborhood of 1 and assume that x1 = 1 + ǫ

(2.53)

[4] So using that and putting n = 1 we have from (2.52) x2 + x0 =

x2 =

=

=

2µx1 1−x21

2µ(1+ǫ) [1−(1+ǫ2 )]

2µ(1+ǫ) −ǫ(ǫ+2)

2µ −2ǫ−ǫ2

+

(2.54)

− x0

− x0

2µǫ −2ǫ−ǫ2

(2.55)

(2.56)

− x0

(2.57)

As we have considered ǫ to be a very small quantity so after binomially expanding it we get = −2µ[ǫ(ǫ + 2)]−1 − 2µ[2 + ǫ]−1 − x0

(2.58)

= − µǫ [1 + 2ǫ ]−1 − µ[1 + 2ǫ ]−1 − x0

(2.59)

= − µǫ +

µ 2

−µ+ 18

µǫ 2

− x0

(2.60)

= − µǫ − (x0 + µ2 ) + O(ǫ)

(2.61)

x3 = −1 + ǫ + O(ǫ2 )

(2.62)

Similarly

And similarly for x4 we iterate the map equation further and get x4 + [

µ −µ − (x0 + )] = ǫ 2

=

2µ(−1+ǫ) 1−(−1+ǫ)2

2µ(−1+ǫ) −(ǫ2 −2ǫ)

(2.63)

(2.64)

= −2µ[2ǫ − ǫ2 ]−1 + 2µǫ[2ǫ − ǫ2 ]−1

(2.65)

= − µǫ [1 − 2ǫ ]−1 + µ[1 − 2ǫ ]−1

(2.66)

= − µǫ [1 + 2ǫ ] + µ[1 + 2ǫ ]

(2.67)

= − µǫ −

µ 2

+µ

(2.68)

So x4 −

µ µ − x0 − = − µǫ + ǫ 2

x4 = x0 + µ + O(ǫ)

µ 2

(2.69)

(2.70)

So not only the singularity is confined but also the map has recovered a memory since the term x4 contains x0 . Using (2.51) where the McMillan map is expressed

19

in a more general way. xn+1 + xn−1 = −

ax2n + bxn + c dx2n + exn + f

(2.71)

Similarly this equation can also be studied from its singularities. Two different cases can be observed. One with d 6= 0 in that case we can consider d = 1 for our convenience and the next step is to consider d = 0. In both the cases we consider f =0

2.3.1

Considering d = 1 and f = 0

The equation becomes xn+1 + xn−1 = −

ax2n + bxn + c xn (xn + e)

(2.72)

Let us assume that xn = 0 so xn+1 + xn−1 → ∞

(2.73)

Now iterating the map equation once more xn+2 = −

=

−ax2n x2n+1 +exn+1

−

= − (1+ a e

xn+1

[ax2n+1 +bxn+1 +c] [x2n+1 +exn+1 ]

− xn−1

bxn+1 x2n+1 +exn+1

−

c x2n+1 +exn+1

b xn+1 +c

−

c x2n+1 +exn+1

)

−

= −a

20

(2.74)

−0

(2.75)

(2.76)

(2.77)

2.3.2

Considering d 6= 0,so taking d = 1

Again the generalized map equation becomes xn+1 + xn−1 = −

ax2n + bxn + c xn (xn + e)

(2.78)

Now considering xn = 0 we have xn+1 + xn−1 → ∞

xn+2 = −

[ax2n+1 +bxn+1 +c] [xn+1 (xn+1 +e)]

(2.79)

− xn

(2.80)

[3] ax2

= − xn+1 (xn+1 n+1 +e

= − 1+ a e

(2.81)

(2.82)

xn+1

= −a

(2.83)

Similarly if we successively iterate the map then to have x4 finite a(a2 − 1)

(2.84)

has to be satisfied.

So this procedure implied that the singularity is confined it does not propagate as time goes on. So the map is completely integrable. In the next chapters we will discuss the constants of motion of the map and the solution of the map is expressed in terms of the Jacobian Elliptic functions.

21

The generalized map equation can be written as xn+1 =

f1 (xn ) − xn−1 f2 (xn ) f2 (xn ) − xn−1 f3 (xn )

22

(2.85)

CHAPTER 3

Jacobian Elliptic Function

The Jacobian Elliptic Functions have a profound role in various aspects of physics. Jacobian Elliptic Functions are a set of basic elliptic functions that has a relevance to some applications for example the pendulum motion.

3.1

Study of The Jacobian Elliptic Functions

The basic elliptic functions are defined as sn(x) = sn(x, k) = sin ψ

(3.1)

cn(x) = cn(x, k) = cos ψ

(3.2)

and dn(x) = dn(x, k) =

q

1 − k 2 sin2 ψ

(3.3)

For k = 0 and k = 1 we recover the circular (trigonometric) functions and the hyperbolic functions. When k = 0 we have x = ψ and so sn(x, 0) = sin(x)

(3.4)

cn(x, 0) = cos(x)

(3.5)

dn(x, 0) = 1

(3.6)

Again putting k = 1 we get

3.1.1

sn(x, 1) = tanh(x)

(3.7)

cn(x, 1) = sech(x)

(3.8)

dn(x, 1) = sech(x)

(3.9)

Identities of The Elliptic Functions

Elliptic functions follow some identities. These are as follows cn2 + sn2 = 1

(3.10)

dn2 + k 2 sn2 = 1

(3.11)

′

(3.12)

sn (x) = cn(x)dn(x)

′

cn (x) = −sn(x)dn(x)

′

dn (x) = −k 2 sn(x)cn(x)

3.1.2

(3.13)

(3.14)

Addition Formula for The Elliptic Functions

Using the previous identities and the definitions of the Elliptic functions we get[5] cn(u + v, 1) = sech (u + v)

24

(3.15)

1 cosh(u+v)

(3.16)

1 cosh u cosh v+sinh u sinh v

(3.17)

sech u sech v 1+tanh(u) tanh(v)

(3.18)

sech u sech v [1−tanh(u) tanh(v)] 1−tanh2 (u) tanh2 (v)

(3.19)

=

=

=

=

cn(u + v, 1) =

cn(u,1)cn(v,1)−sn(u,1)sn(v,1)dn(u,1)dn(v,1) 1−sn2 (u,1)sn2 (v,1)

(3.20)

Similarly cn(u + v) =

cn(u)cn(v)−sn(u)sn(v)dn(u)dn(v) 1−k2 sn2 (u)sn2 (v)

(3.21)

sn(u + v) =

sn(u)sn(v)dn(v)+sn(v)cn(u)dn(u) ∆(u,v)

(3.22)

and

Where ∆ = 1 − k 2 sn2 (u)sn2 (v)

(3.23)

This addition formula will be of much need in the later part when we will discuss the integral of motion of the map and we will see that the integral can be parametrized in terms of the Jacobian Elliptic Functions.

25

CHAPTER 4


As we have seen in the 2nd chapter that due to the singularity confinement method the McMillan map is integrable. Now in this section we will try to find out the integral of motion for the McMillan map in both quantum and Classical picture.

4.0.3

The Classical McMillan Map

The generalized equation for the McMillan map is given in the form[4] xn+1 =

f1 (xn ) − xn−1 f2 (xn ) f4 (xn ) − xn−1 f3 (xn )

(4.1)

With f3 = 0, f2 = f4 = 1 and f1 =

2µxn 1 − x2n

(4.2)

we get back the McMillan map equation which is given by xn+1 + xn−1 =

2µxn 1 − x2n

(4.3)

It may happen that for a certain n the mapping (apparently) loses one degree of freedom. This occurs when xn+1 is independent of xn−1 and this occurs whenever xn+1 =

f1 (xn ) f4 (xn )

f1 (xn ) − xn−1 f2 (xn ) f1 (xn ) = f4 (xn ) f4 (xn ) − xn−1 f3 (xn )

(4.4)

(4.5)

f1 (xn )f4 (xn ) − xn−1 f1 (xn )f3 (xn ) = f1 (xn )f4 (xn ) − xn−1 f4 (xn )f2 (xn )

(4.6)

xn−1 [f1 (xn )f3 (xn ) − f4 (xn )f2 (xn )] = 0

(4.7)

f1 (xn )f3 (xn ) − f4 (xn )f2 (xn ) = 0

(4.8)

f2 (xn ) f1 (xn ) = f4 (xn ) f3 (xn )

(4.9)

This is the criterion for xn+1 to be independent of xn−1 So xn+1 =

f2 (xn ) f1 (xn ) = f4 (xn ) f3 (xn )

(4.10)

This equation holds unless xn−1 was such that both the numerator and the denominator vanish. In that case xn−1 =

f1 (xn ) f4 (xn ) = f2 (xn ) f3 (xn )

(4.11)

So there are two ways by which we can preserve the singularity confinement method. Either the previous relation is satisfied or it is not satisfied. For the later case xn+1 is independent of xn−1 . So one degree of freedom is definitely lost. The main idea is that the singularity confinement is nothing but the loss or appearance of a degree of freedom

4.0.4


We write the map equation in the form[6] ′

x =

f1 (y) − xf2 (y) f2 (y) − xf3 (y)

27

(4.12)

and ′

′

y =

′

g1 (x ) − yg2 (x ) g2 (x′ ) − yg3 (x′ )

(4.13)

′

where x = xn+1 and y = xn with ′

(4.14)

′

(4.15)

f (x) = (A0 (X )) × (A1 (X)) and g(x) = (AT0 (X )) × (AT1 (X))



x2





αi βi γi



        X = x  Ai =  δi ǫi ξi      1 κi λi µi All the mappings considered here are symmetric. i.e f = g thus f (x) can be written as 

  f (x) =  

α0 x2n+1 + β0 xn+1 + γ0





α1 x2n + β1 xn + γ1

    + ǫ0 xn + ξ0  ×  β1 x2n + ǫ1 xn + ξ1   γ0 x2n+1 + ξ0 xn+1 + µ0 γ1 x2n + ξ1 xn + µ1 β0 x2n+1

    

(4.16)

Now the invariant is given by

(α0 + kα1 )x2n x2n+1 + (β0 + kβ1 )(x2n xn+1 + xn x2n+1 ) + (γ0 + kγ1 )(x2n + x2n+1 ) + (ǫ0 + kǫ1 )xn xn+1 + (ξ0 + kξ1 )(xn + xn+1 ) + (µ0 + kµ1 ) = 0

28

4.0.5

The Constant of Motion is Parametrized in Terms of The Elliptic Functions

Now let us take the constant of motion and it can be written into a different form as

ax2 y 2 + b(x2 y + xy 2 ) + c(x2 + y 2 ) + 2dxy + e(x + y) + f = 0 where we write xn+1 = x and xx = y in the previous equation.

Now applying a bilinear transformation of the form[7]

x→

αx+β γx+δ

and y→

αy+β γy+δ

So putting these terms in the previous equation we have

2

a (αx+β) (γx+δ)2 αy+β ] γy+δ

(αy+β)2 (γy+δ)2

2

+ b (αx+β) (γx+δ)2

(αy+β) (γy+δ)

2

(αy+β) (αy+β) + b (αx+β) + 2α (αx+β) + c[ αx+β + (γx+δ) (γy+δ)2 (γx+δ) (γy+δ) γx+δ

+f =0

we have parametrized the above equation in such a manner that all the odd powers goes to zero whereas all the even powers remain. So that implies we are making the term e(x + y) and b[x2 y + y 2 x] going to zero. Now putting the term e(x + y) going to zero we have e(x + y) = e[

αx + β αy + β + ]=0 γx + δ γy + δ

(4.17)

2αγxy + (x + y)(αδ + βγ) + 2βγ) + 2βδ =0 (γx + δ)(γy + δ)

(4.18)

2αγxy + 2βδ = −(x + y)(αδ + βγ)

(4.19)

29

(x + y)(αδ + βγ) =0 2(αγxy + βδ)

(4.20)

So to satisfy this equation we have to fulfill the criterion αδ + βγ = 0

(4.21)

After making the odd powers zero we are left with the canonical form x2 y 2 + 1 + c(x2 + y 2 ) + 2dxy = 0

(4.22)

y 2 [x2 + c] + 2dxy + (cx2 + 1)] = 0

(4.23)

Now solving for y we get y=

−xd ±

y=

−xd ±

√

d2 x2 − cx4 − x2 − c2 x2 − c (x2 + c)

(4.24)

p −c + (d2 − 1 − c2 )x2 − cx4 (x2 + c)

(4.25)

The argument contains a quartic function of x, so it can be written as a perfect square term by transforming x to the variable u. Let 1

x = k 2 sn(u)

(4.26)

x2 = k sn2 (u)

(4.27)

or

or k=

x2 sn2

30

(4.28)

Similarly k −1 =

sn2 x2

(4.29)

Now k + k −1 =

(d2 − 1 − c2 ) c

(4.30)

The argument is given by = −c[1 − (k + k −1 )x2 + x4 ]

(4.31)

1 = [(1 + k 2 sn4 (u)) − k sn2 (u)(k + )] k

(4.32)

= −c[(1 + k 2 sn4 (u)) − sn2 (u) − k 2 sn2 (u)]

(4.33)

= −c[(1 − sn2 (u) + k 2 sn2 (u)(sn2 (u) − 1)]

(4.34)

= −c(1 − k 2 sn2 (u))(1 − sn2 (u))

(4.35)

= −c cn2 (u) dn2 (u)

(4.36)

We define a parameter η in such a way that d=

cn(η) dn(η) k sn2 (η)

(4.37)

and again choosing d in such a manner that d=

cn(η) dn(η) k sn2 (η)

31

(4.38)

so the expression for y becomes y=

−xd ±

p −c cn2 (u) dn2 (u) c + k sn2 (u)

1

=

dn(η) −k − 2 sn(u) cn(η) ± k sn2 (η) 1 −k sn2 (η)

q

1 k sn2 (η)

(4.39)

cn2 (u)dn2 (u)

+ k sn2 (u)

(4.40)

1

1

−k 2 sn(u) cn(η) dn(η) ± k 2 sn(η) cn(u) dn(u) = −1 + k 2 sn2 (η)sn2 (u)

(4.41)

So 1

1

−k 2 sn(u) cn(η) dn(η) ± k 2 sn(η) cn(u) dn(u) y= −1 + k 2 sn2 (η)sn2 (u)

(4.42)

Now using the addition formula of the Jacobian Elliptic Function we find that 1

y = k 2 sn(u ± η)

(4.43)

And 1

x = k 2 sn(u)

(4.44)

So it is proved that the constant of motion is parametrized in terms of the Jacobian Elliptic Function As the classical McMillan map is integrable so it has solutions and they are represented in terms of the Jacobian Elliptic Functions[8] and the solutions are xn = x0 cn(Ωn, k)

(4.45)

and k = x0

dn(Ω) sn(Ω)

32

(4.46)

CHAPTER 5

Constant of Motion For The Quantum McMillan Map

Now as we have deduced the classical constant of motion our next task naturally is to find out the constant of motion for the Quantum McMillan maps. Here we are considering the same map equation but the only change to that of the classical case is that in the quantum case x and y are operators, thus they follow the canonical commutation relation which is as follows [x, y] = i¯ h

(5.1)

. The constant of motion would be same to that of the classical case with some correction factor added to it. We have to determine that correction term and find out the constant of motion for the quantum McMillan maps.

5.0.6

Constant of Motion For The Quantum McMillan Map

Let the map equation is given[9] x′ = −x − g(y)

(5.2)

y ′ = −y − h(x)

(5.3)

and

where g(x) =

δx2 + ǫx + ξ αx2 + βx + γ

(5.4)

h(x) =

βx2 + ǫx + λ αx2 + βx + κ

(5.5)

similarly

where [x, y] = i¯ h

(5.6)

follows the canonical commutation relation because in the quantum case x, y are considered to be operators. In the previous chapter we have discussed that the classical constant of motion is Ic (x, y) = αx2 y 2 + βx2 y + γx2 + δxy 2 + ǫxy + ξx + κy 2 + λy

(5.7)

For the quantum case Iq = Ic + i¯ hf (x, y)

(5.8)

where f(x,y) is the correction term. Now the map can be factorized in terms of the involutions Hq and Gq where

Gq : x′ = −x − g(y)

(5.9)

y′ = y

(5.10)

,

34

and H q : x′ = x

(5.11)

y ′ = −y − h(x)

(5.12)

,

Considering the mapping of the form Iq (x) = Iq (Gq (x))

(5.13)

Iq (x) = Iq (Hq (x))

(5.14)

and

Using these previous equations we can write Ic (x, y) + i¯ hf (x, y) = Ic (x′ , y ′ ) − i¯ hf ′ (x′ , y ′ )

(5.15)

Now Ic (x′ , y ′ ) is given by

Ic (x′ , y ′ ) = αx2 (−y − h(x))2 − βx2 (y + h(x)) + γx2 + δx(−y − h(x))2 − ǫx(y +

h(x)) + ξx + k(y + h(x))2 − λ(y + h(x))

= αx2 y 2 + αx2 h(x)2 + αx2 [yh(x) + h(x)y] − βx2 y − βx2 h(x) + γx2 + δx[y 2 +

h(x)2 + yh(x) + h(x)y] − ǫxy − ǫxh(x) + ξx + ky 2 + kh(x)2 + kyh(x) + kh(x)y − λy − λh(x)

Now putting this is equation (5.15) we have

⇒ αx2 h(x)2 +αx[x, y]h(x)+αx2 h(x)y−βx[x, y]−βx2 h(x)+δxh(x)2 +δ[x, y]h(x)+ 35

δxh(x)y − ǫ[x, y] − ǫxh(x) + kh(x)2 +kyh(x) + kh(x)y − λy − λh(y) − i¯ hf ∗ (x, −y − h(x)) − βx[x, y] − ǫ[x, y] − λy − i¯ hf (x, y) = 0

Using the canonical commutation relation (xy − yx) = i¯ h and putting the imaginary part to zero we get

(xy − yx) = i¯ h

(5.16)

⇒ αx(i¯ h)h(x)+αx(i¯ h)h(x)−βx(i¯ h)+δ(i¯ h)h(x)+δh(x)i¯ h −ǫi¯ h −i¯ hf ∗ (x, −y − h(x)) − βx(i¯ h) − ǫ(i¯ h) − i¯ hf (x, y) = 0

⇒ 2αxh(x) − 2βx + δh(x) − 2ǫ − f ∗ (x, −y − h(x)) − f (x, y) = 0

(5.17)

⇒ f (x, y) + f ∗ (x, −y − h(x)) − 2αxh(x) + 2βx − 2δh(x) + 2ǫ = 0

(5.18)

Again considering the other equation we get ′

′

′

′

Ic (x, y) + i¯ hf (x, y) = Ic (x , y ) − i¯ hf ∗ (x , y )

(5.19)

Now using the map equation ′

x = −x − g(y);

′

y =y

36

(5.20)

(5.21)

Ic (x, y) = αx2 y 2 + βx2 y + γx2 + δxy 2 + ǫxy + ξx + ky 2 + λy

⇒ αx2 y 2 + βx2 y + γx2 + δxy 2 + ǫxy + ξx + ky 2 + λy + i¯ hf (x, y) = α(−x −

g(y))2 y 2 + β(−x − g(y))2 y + γ(−x − g(y))2 −δ(x + g(y))y 2 − ǫ(x + g(y))y − ξ(x +

g(y)) + ky 2 + λy − i¯ hf ∗ (−x − g(y), y)

⇒ αg(y)2 y 2 + αxy 2 g(y) + αg(y)xy 2 + βg(y)2 y + βxyg(y) + βg(y)xy + γg(y)2 +

γxg(y)+γg(y)x− δxy 2 −δg(y)y 2 −ǫxy −ǫg(y)y −ξx−ξg(y)−i¯ hf ∗ (−x−g(y), y)− δxy 2 − ǫxy − ξx − i¯ hf (x, y) = 0

⇒ α[x, y]yg(y) + αg(y)2 y 2 + αg(y)[x, y]y + βg(y)2 y + β[x, y]g(y) + βg(y)[x, y] +

γg(y)2 + γxg(y) + γg(y)x −δ[x, y]y − δg(y)y 2 − ǫ[x, y] − ǫg(y)y − ξx − ξg(y) − i¯ hf ∗ (−x − g(y), y) − δ[x, y] − ǫ[x, y] − ξx − i¯ hf (x, y) = 0

Making the imaginary term equal to zero

⇒ i¯ hαyg(y) + αg(y)i¯ hy + βi¯ hg(y) + βi¯ hg(y) − δi¯ hy − ǫi¯ h − i¯ hf ∗ (−x − g(y), y) − δi¯ hy −ǫi¯ h − i¯ hf (x, y) = 0

⇒ 2αyg(y) + 2βg(y) − 2δy − 2ǫ − f (x, y) − f ∗ (−x − g(y), y) = 0

(5.22)

⇒ f (x, y) + f ∗ (−x − g(y), y) − 2αyg(y) − 2βg(y) + 2δy + 2ǫ = 0

(5.23)

So in short the equation from which f (x, y) the correction factor is to be deduced 37

is given by

f (x, y) + f ∗ (x, −y − h(x)) − 2αxh(x) + 2βx − 2δh(x) + 2ǫ = 0

(5.24)

⇒ f (x, y) + f ∗ (−x − g(y), y) − 2αyg(y) − 2βg(y) + 2δy + 2ǫ = 0

(5.25)

Iq (x, y) = Ic (x, y) + i¯ hf (x, y)

(5.26)

and

So

where f (x, y) is to be determined from the previous two equations

38

CHAPTER 6

Quantum McMillan Map And Stroboscopic Plots

It is not always possible to find out a mathematical expression for any system which will help us to find out weather or not a system is integrable. Now other option is to do numerically or to find out the stroboscopic plots. From there we can tell something about the integrability of the system. In this section we will study about the 1.5 degrees of freedom and also look at the phase portraits of the parametrically forced pendulum and simple harmonic oscillator. We will also see the stroboscopic plots in these two cases. For harmonic oscillator case it will give rise to the well known Mathieu functions but for pendulum the behavior is very much chaotic in nature. All the systems studied here are one degree of freedom system with time part present explicitly i.e the systems are non-autonomous. This is well known as 1.5 degree of freedom system.

6.0.7

1.5 Degree of Freedom System

Considering the Hamiltonian of the form H(q, p, t) = f (p) + g(t)V (q)

(6.1)

[10] where g(t) = g(1 + ǫ(2πt)) and f (p) = p2 /2.

Again for harmonic oscillator V (q) = q 2 /2 and for parametrically forced pendulum V (q) = sin(q(t))

Now the phase portrait of the parametrically harmonic oscillator is given by

solving the Hamilton’s equation of motion which

6.0.8

∂H ∂q = ∂t ∂p

(6.2)

∂p ∂H =− ∂t ∂p

(6.3)

Stroboscopic Plots

Considering the forcing to be periodic we will study the phase space of the parametrically forced system only at times separated by the period of forcing. So a discrete set of points would be obtained and if the points forms a definite pattern and are bound we can infer that the system is integrable, whereas for the case of a system where the points are all scattered it can be inferred that the system is chaotic in nature.

Now for the parametrically forced harmonic oscillator the phase portrait and the stroboscopic plot are given below and after that the phase portrait of the parametrically forced pendulum case is given. We can easily see that in case of the harmonic oscillator the stroboscopic plots form a pattern where in case of the pendulum it gives scattered points which represent the chaotic nature. Now this plots are obtained by solving the differential equation which comes from the Hamilton’s equation. For example in the Harmonic Oscillator case the equation is given by d2 q = −g[1 + ǫsin(2πt)]q dt2

(6.4)

and for the parametrically forced pendulum case it is d2 q = −g[1 + ǫsin(2πt)] cos(q(t)) dt2

40

(6.5)

ε =2 4 3 2

p

1 0 −1 −2 −3 −4 −4

−3

−2

−1

0 q

1

2

3

4

(a) Phase portrait for parametrically forced harmonic oscillator ε =5 6

4

p

2

0

−2

−4

−6 −5

−4

−3

−2

−1

0 q

1

2

3

4

5

(b) b parametrically forced harmonic oscillator with ε=10 15

10

p

5

0

−5

−10

−15 −8

−6

−4

−2

0 q

2

4

6

8

(c) c

Figure 6.1: Caption of subfigures (a), (b) and (c)

41

Stroboscopic plot for harmonic Oscillator 6

4

p

2

0

−2

−4

−6 −2.5

−2

−1.5

−1

−0.5

0 q

0.5

1

1.5

2

2.5

(a) with one initial condition q0 = 3 and p0 = 0 with many initial conditions 8 6 4

p

2 0 −2 −4 −6 −8 −2.5

−2

−1.5

−1

−0.5

0 q

0.5

1

1.5

2

2.5

(b) with many initial conditions


42

phase portrait of parametrically forced pendulum 2 1.5 1 0.5 p

0 −0.5 −1 −1.5 −2 2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

q

(a) Phase portrait for the parametrically forced pendulum for g=5 3

2

1

p

0

−1

−2

−3 0

5

10

15

20

25 q

30

35

40

45

50

(b) with different value of ǫ

stroboscopic plot for parametrically forced pendulum 15

p

10

5

0

−5 0

100

200 q

300

400

(c) stroboscopic plot for parametrically forced pendulum


43

CHAPTER 7

Quantum McMillan Map and Some Aspects of The McMillan Map

In this section we will study the delta kicked harmonic oscillator and see how the Quantum McMillan map arises from it. We will also study some basic structures of the McMillan map and then instead of the delta kick we will use other mathematical functions and we will see how the phase portrait looks like along with the stroboscopic plots.

7.0.9

δ Kicked Harmonic Oscillator

The δ kicked harmonic oscillator is given by the expression ∞ X p2 2 2 2 + [q − ds(a)ns(a)log(q + cs (a))] δ(t − n) H= 2 n=0

(7.1)

[11; 12] Now the Quantum McMillan map is derived from this equation and it is given by qn+1 = pn+1 + qn

(7.2)

and pn+1 = pn − 2qn +

2(ds(a)ns(a) − cs2 (a))qn qn2 + cs2 (a)

(7.3)

7.0.10

Some Aspects of The McMillan Map

Area Preservation of The McMillan Map To deduce the area preservation of the McMillan map we have to calculate the Jacobian Matrix and if the determinant of the matrix comes out to be one then this map is area preserving. 

J =

∂f ∂pn

∂f ∂qn

∂g ∂pn

∂g ∂qn

 

(7.4)

Where f and g are defined as follows

2[ds(a)ns(a) − cs2 (a)] qn qn2 + cs2 (a)

(7.5)

∂f =1 ∂pn

(7.6)

∂f 2[ds(a)ns(a) − cs2 (a)] 2[ds(a)ns(a) − cs2 (a)]qn (−1) + 2qn = −2 + ∂qn (qn2 + cs2 (a)) (qn2 + cs2 (a))2

(7.7)

f = pn − 2qn +

Similarly the other equation can be written as

qn+1 = pn − qn +


(7.8)

Thus g is written in the form

g = p n − qn +

2[ds(a)ns(a) − cs2 (a)] qn qn2 + cs2 (a) 45

(7.9)

So differentiating it with respect to pn and qn we have

∂g =1 ∂pn

(7.10)

[ds(a)ns(a) − cs2 (a)]qn 2[ds(a)ns(a) − cs2 (a)] ∂g −2 2qn = −1 + ∂qn qn2 + cs2 (a) (qn2 + cs2 (a))2

(7.11)

So the determinant of the matrix becomes

2[ds(a)ns(a)−cs2 (a)] 2 +cs2 (a) qn 2 (a)] [ds(a)ns(a)−cs 4qn2 (q2 +cs2 (a))2 n

|J| = −1 +

2

2

(a)] (a)] − 4qn2 [ds(a)ns(a)−cs + 2 − 2 [ds(a)ns(a)−cs + (q 2 +cs2 (a))2 (q 2 +cs2 (a)) n

n

=1

Thus the map is area preserving.

7.0.11

Stable Points of The McMillan Map

Type and staility are determined by the sign and the magnitude of the eigen values of the Jacobian matrix. If the absolute value of the real parts of all the eigenvalues are less than one then the map is a stable one. On the other hand if the real part of the eigen-values becomes greater than one then it is unstable. Now first we have to deduce the stable points of the map.

The map equation is given y qn+1 = pn − qn +


46

(7.12)

And pn+1 = pn − 2qn +

2[ds(a)ns(a) − cs2 (a)] qn [qn2 + cs2 (a)]

(7.13)

From here let us write qn+1 = qn = q ∗

(7.14)

pn+1 = pn = p∗

(7.15)

and

So putting that we get q ∗ = p∗ − q ∗ + 2

2q ∗ =

[ds(a)ns(a) − cs2 (a)] ∗ q q ∗2 + cs2

2[ds(a)ns(a) − cs2 (a)] ∗ q q ∗2 + cs2 (a)

(7.16)

(7.17)

q ∗2 = ds(a)ns(a) − 2cs2 (a)

(7.18)

p q ∗2 = ± ds(a)ns(a) − 2csa

(7.19)

again putting this to the other equation we get [ds(a)ns(a) − cs2 (a)] ∗ q q =p −q +2 q ∗2 + cs2 (a)

(7.20)

p∗ = 0

(7.21)

∗

∗

∗

or,

47

7.0.12

Stability Criterion of The McMillan Map

For Henon Map x∗ = 1 − ax∗2 + y ∗

(7.22)

y ∗ = bx∗

(7.23)

and

2

the stability criterion is given by 3 (b−1) > a Now following the same procedure 4 we have the map equation 2[ds(a)ns(a) − cs2 (a)] qn qn2 + cs2 (a)

(7.24)

2[ds(a)ns(a) − cs2 (a)] = pn − 2qn + qn qn2 + cs2 (a)

(7.25)

qn+1 = pn − qn +

pn+1

p fixed points are (0, ± ds(a)ns(a) − 2cs2 (a)) The Jacobian Matrix is given by 

M =

∂qn+1 ∂qn ∂pn+1 ∂qn

∂qn+1 ∂pn ∂pn+1 ∂pn

 

(7.26)

det (M − λI) = 0

48

(7.27)



det

2 4qn − [ds(a)ns(a)−cs2 (a)] 2 4qn − [ds(a)ns(a)−cs 2 (a)]

1−

⇒ (1 − λ)[(1 − λ) −

⇒ (1 − λ)[(1 − λ) − ⇒ (1 − λ2 ) −

λ

1 1−λ

 

2 4qn ] [ds(a)ns(a)−cs2 (a)]

+

2 4qn [ds(a)ns(a)−cs2 (a)]

=0

4q ∗2 ] [ds(a)ns(a)−cs2 (a)]

+

4q ∗2 [ds(a)ns(a)−cs2 (a)]

=0

4q ∗2 [ds(a)ns(a)−cs2 (a)]

+

⇒ λ2 + 1 − 2λ +

⇒ λ2 + λ[

4q ∗2 λ [ds(a)ns(a)−cs2 (a)]

+

4q ∗2 [ds(a)ns(a)−cs2 (a)]

4q ∗2 λ =0 [ds(a)ns(a) − cs2 (a)]

4[ds(a)ns(a) − 2cs2 (a)] − 2] + 1 = 0 [ds(a)ns(a) − cs2 (a)]

=0

(7.28)

(7.29)

For the stability criterion to be satisfied we have to have |Reλ| < 1

(7.30)

p ⇒ cs(a) 2cs2 (a) − ds(a)ns(a) < ds(a)ns(a) − 2cs2 (a)

(7.31)

⇒ cs2 (a)[2cs2 (a) − ds(a)ns(a)] < ds2 (a)ns2 (a) + 4cs4 (a) − 4ds(a)ns(a)cs2 (a) ⇒ [2cs4 (a) − cs2 (a)ds(a)ns(a)] < [ds2 (a)ns2 (a) + 4cs4 (a) − 4ds(a)ns(a)cs2 (a)] 49

⇒ [2cs4 (a) − 3ds(a)ns(a)cs2 (a) + ds2 (a)ns2 (a)] > 0

(7.32)

⇒ 2cs2 (a)[cs2 (a) − ds(a)ns(a)] − ds(a)ns(a)[cs2 (a) − ds(a)ns(a)] = 0 (7.33)

7.0.13

⇒ [2cs2 (a) − ds(a)ns(a)][cs2 (a) − ds(a)ns(a)] > 0

(7.34)

⇒ [cs2 (a) − ds(a)ns(a)] > 0

(7.35)

[2cs2 (a) − ds(a)ns(a)] > 0

(7.36)

Determination of The Stability of The Orbit(q = 0)

Consider the Hamiltonian of the form given by ∞ X p2 2 2 2 H= + [q − ds(a)ns(a)log(q + cs (a))] δ(t − n) 2 n=0

(7.37)

Now we replace the delta function with that of the cosine function. The stroboscopic plots are given afterwards for different values of the frequency. So the new Hamiltonian becomes

50

H=

p2 + [q 2 − A log(q 2 + B)]cos(ωt) 2

(7.38)

dq =p dt

(7.39)

dp 2Aq = −2qcos(ωt) + 2 cos(ωt) dt q +B

(7.40)

and the potential term is given by

V (q) = [q 2 − Alog(q 2 + B)]

(7.41)

For A=10,B=1 it becmes a perfect double well potential. This equation don’t have a map counterpart. So let us consider that at time t the phase co-ordinates are (q(t), p(t))

(7.42)

t + δt

(7.43)

and after time

the co-ordinates are given by (q + δq, p + δp)

d(δq) = δp dt 51

(7.44)

2Acos(ωt) 4Aq 2 cos(ωt) d(δp) = −2cos(ωt)δq + δq − δq dt (q 2 + B) (q 2 + B)2

(7.45)

u(t) = δq

(7.46)

v(t) = δp

(7.47)

let

So the equation turns out to be

du =v dt

(7.48)

dv 2Acos(ωt) 4Aq 2 cos(ωt) = [−2cos(ωt) + − ]u dt (q 2 + B) (q 2 + B)2

(7.49)

So we have a second order differential equation in u

2Acos(ωt)u 4Aq 2 cos(ωt) d2 u = −2cos(ωt)u + − u dt2 (q 2 + B) (q 2 + B)2 log[u(t)]2 t→∞ t

solving this equation numerically and finding lim

52

(7.50)

If the solution comes out to be zero then the system is stable, Otherwise if it is greater than that the system is unstable.

The double well potential is given by the figure below

Double well for A=10,B=1 60 50 40

V(q)

30 20 10 0 −10 −20 −10

−8

−6

−4

−2

0 q

2

4

6

8

10

Figure 7.1: Double well potential

Now we have iterated the previous differential equation to show weather or not it has a stable orbit about q = 0 or not. This figure proves that the at q = 0 orbit stability

−4

2

x 10

1.5 1 0.5

u(t)

0 −0.5 −1 −1.5 −2 −2.5 0

1

2

3

4

5 time

6

7

8

9

10 4

x 10

Figure 7.2: stability of the orbit with q = 0

is actually an unstable one because the value is greater than zero. Now changing the value of ω we get different stroboscopic plots. The plots are given below

53

ω =2.2 15

10

p

5

0

−5

−10

−15 −100

−80

−60

−40

−20

0 q

20

40

60

80

100

Figure 7.3: stroboscopic plot

plot for ω =2.5 50 40 30 20

p

10 0 −10 −20 −30 −40 −50 −200

−150

−100

−50

0 q

50


54

100

150

200

ω =2.7 10 8 6 4

p

2 0 −2 −4 −6 −8 −10 −40

−30

−20

−10

0 q

10

20

30

40


ω =3.5 15

10

p

5

0

−5

−10

−15 −60

−40

−20

0 q

20


55

40

60

REFERENCES [1] S. A. Willy Hereman, “The painleve test for nonlinear ordinary and partial differential equations,” 2004, pp. 1–15. [2] V. G. P. F. W. Nijhoff, H.W.Capel, “Integrable quantum mappings,” in PHYSICAL REVIEW A, 1992, pp. 2155–2157. [3] V. P. B. Grammaticos, A. Ramani, “Do integrable mappings have the painleve property,” in PHYSICAL REVIEW LETTERS, 1991, pp. 1825–1827. [4] B. Grammaticos and A. Ramani, “Integrability-and how to detect it,” pp. 72–82. [5] W. F. E. J. V. Armitage, Elliptic Functions, 1st ed. Press, 2006.

Cambridge University

[6] G. R. W. Q. et al, “Integrable mappings and soliton equations,” in Physica D. [7] R. J. Baxter, Exactly Solved Models In Statistical Mechanics, 3rd ed. demic Press, London NW1 7DX, 1989.

Aca-

[8] S. R. M. Lakshmanan, Nonlinear Dynamics, 2nd ed. Springer, 2005. [9] F. W. N. G. R. W. Quispel, “Integrable two dimensional quantum mappings,” in Physics Letters A, 419-422, 1992. [10] A. Lakshminarayan, Hamiltonian Chaos, Classical and Quantum Aspects, 2004. [11] Lecture 2 by Arul Lakshminarayan, Hamiltonian Chaos, 2004. [12] G. M. Zaslavsky, The Physics of Chaos in Hamiltonian Systems, 2nd ed. Imperial College Press, 2007.

56