Magnetna cigra Svemira

8/14/2019 Magnetna cigra Svemira

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The magnetic top of Universe as a model of quantum spin

Source file of A.O. Barut, M. Bozic and Z. Maric

Substitution, conversion and transformation by Dusan Stosic

Abstract

The magnetic top is defined by the property that the external magnetic field B coupled to theangular velocity as distinct from the top fhose magnetic moment is independent of angular

velocity. This allows one to

construct a "gauge" theory of the top where the caninical angular momentum of the ooint particle

and the B field plays the role of the gauge potential. Magnetic top has four constants of motion sothat Lagrange equations for Euler angles , , (wich define the orientation of the top) are

solvable, and are solved here. Although the Euk=ler angles have comlicated motion.,the

canonical angular momentum s, interpreted as spin , obeys precisely a simple precession

equation. The Poisson brackets of allow us further to make an unambiguous quantization ofspin , leading to the Pauli spin Hamiltonian. The use of canonical angular momentumalleviates

the ambiguity in the ordering of the variables in the Hamiltonian. A detailed

gauge theory of the asimmetric magnetic top is alsou given.


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Euler angles - The xyz (fixed) system is shown in blue, the XYZ (rotated) system

is shown in red. The line of nodes, labelled N, is shown in green.

Contentspage

3

IntroductionI.

II . Lagrangian and Hamiltonian of the symmetric magnetic top 6

III. Lagrange equation for the magnetic topand their solutions for

constant magnetic field1 10IV. The torque equation and its equivalence with the Lagrange

equations 17

V. Hamilton's equations for the magnetic top 18VI. Quantum magnetic top 21

VII. The states of the quantum magnetic top 26

VIII. The Asymmetric Magnetic top 29Appendix A.Top with magnetic moment fixed in the body frame 36

I. Inroduction

References 41Whereas the coordinates and momenta of quantum particles have a classical origin

or a classical

counterpart,the spin is generally thought to have no classical origin. It is, in Pauli

words,"a calassicay non-explanable two-valuedness"{1} .Thus, the spin and coordinates are not on the same

footing as far as the

picture of the particles is concerned.In atomic physics the role of spin is enormous due to the Pauli-principle and spin

statistics connection,althougt the numerical values of spin orbit terms are small.

In nuclear and particle physics and in very high energy physics, there spin hyperfine

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terms turned out to play an essential role, whose theoretical understandig is still

lacking (2). Even in the interpretation and foundations of quantum theory, the nature

of spin seems to be rather crucial, and a need for a classical model of spin has longbeen felt (3).

Our knowledge about the importance of spin in all these areas comes from the

widespread and succesfull applicability of Pauli and Dirac matrices and spinrepresentation of Galillei and Poincare groups. Although there is no mystery is

actually some mystery in the physical origin and in the visualization of spin.

(It cocerns the spin 1/2 as well as the higher spins). Because of all those reasonsthere has been in the past many attempts to identify internal spin variables and to main

clssical models of spin, both of Pauli (4-12) as well as of Dirac spin (13-18)

But , none ofthe nonrelativistic spin models has been generally accepted, either

because none of the propsed models is without shorthcomings and difficulties or

because the prevailling attitude of physicists towards internal spin variables is, in

Schulman's words: a general unconfortablenes at the mention of internal spin variablesand a reliance on the more formal, but nevertheless completely adequate, spinor

wave functions which are labelled basis vectors for a representation of so*3)

but are endowed with no further properties"(10)In this paper, we shall consider the nonrelativistic Pauli spin, and a minimal

classical model - in the sense of the smallest possible phase space dimension

- underlying the Pauli equation. Our classical model of quantum spin is basedon magnetic top , wich we define as a top whose mafnetic moment is

proportional to the angular velocity(Chapter II) By solving the classical

equation of motion of the magnetic top we shall show that it has, by virtue

of the special coupling to the magnetic field, a unique property that the motionof its magnetic moment is one dimensional (i.e ptecessio around the magnetic field)

whereas the top itself performs a complicated three-dimensional motion

(Chapters III and IV).The motion of the magnetic moment of the magnetic top is different in an essential

way from the motion of the top which carries magnetic moment fixed in the body

frame. Namely, a magnetic moment which is fixed to the top preform athree-dimensional motion (precession with nutation) since it shares the motion

of the body to which it is attached (Apendix A). This distinction is the consecuence

of the differnce in the form of the two Lagrangian. The potential in the Lagrangianof magnetic top (Chapter II) is angular velocity

dependent whereas the potential of the top which carries magnetic moment is velocity

indeoendent (Apendix A). Also, Hamiltonian of the latter top is simple sum of kinetic

and angular velocity independent potential wheras Hamiltonian of magnetic top isnot of this form(Chapter II).

It is necessery to relize those differences in order to understand the difference

between our work and previous works (8,9,10( on the classical models of spinwhich were also based on the top.

In Rosen work, classical model of spin is in fact the top with angular velocity

independent potential (8). In our oppinion this model is unsatisfactory because

for quantum spin there exists the linear relation

s between magnetic

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moment operator and spin angular momentum operators,whereas, such

a relation does not characterize Rosen's classical model in which it is assumed

that Hamiltonian is a sum of kinetic energy

2

and potential energy B

is

independent of angular velocity. But this is possible only if is independentof spin angular momentum.

The Lagrangian of the magnetic top is identical with the Lagrangian of the Bopp

and Haag (9) model of spin. But the procedure of the construction of the Hamiltonian

and subsequent quantization procedures differ in our and in the Bopp and Haagaproach (Chapter VI).

Certain authors have arged in the past that the top is not an appropriate model

of spin, because its configuration space (which is three dimensional ) is largerthan it is necessert. Namely, in Nielsen and Rohrlich words (11) "quantum-mechanical

perticle of definite spin is essentially one-dimension (since it is completelt by

the eigenstates of one coordinate) so Schulman's formulation seems over complicated".It follous from our analysis that this remark is not applicable to the magnetic topbecause although its configuration space is three-dimensional, the magnetic moment

of magnetic top precesses around constant magnetic field (Chapter III). Moreover, in

the light of this result it becomes understadable why Pauli theory of the spin motion ina magnetic field has been so succseful despite the fact that it avoids to answer the

question as to what the internal spin variables are and what the variables conjugate

to spin are. The explanation is simple. It is a satisfactory theory for those phenomenafor which only thr motion of magnetic moment is relevant. But, are there phenomena

determined by the motion of the magnetic top itself. Our answer is positive. One

example is the phase change of spinors in magnetic fields (Chapter VII).

II Lagrangian and Hamiltonian of the symetric magnetic top

As stated in the Inroduction we shell use the word"top" to denote the mechanical

object whose orientation in the reference frame is discribed by Euler angles , , .Magnetic top by definition has a magnetic moment proportional to its angular

momentum

sv . =

sv . cm gm sec

(1)

Mtopsv gsv sv

topsv 6. 05 10stattesla

=

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gsv sv

6. 05 1091erg

stattesla=

The angular momentum itself is proportional to the vector of angular velocity

(2)

sv

Isv sv

Isvi

sv..i ei Isv x.sv i y.sv j+ z.sv k+( )

are unit vectors of the coordinates system attached to the body

and whose orientation iz the Laboratory frame are three Euler angles.

.

..

.

i

1.17110 95 gmcm2

sec-1

Isv x.sv i y.sv j+ z.sv k+( )3

1.17110 95 gm cm2

sec-1

sv . gm cm sec

Isv sv

1 .1 71 1 095 gmcm2 sec-1=

are unit vectors along the axis of the Laboratory reference

frame. The components of in the Laboratory frame are :x cos ( ) ' sin ( ) sin ( ) '+( )

y sin ( ) ' cos ( ) sin ( ) 'z ' cos ( ) '+

The components of in the body-fixsed frame, on the other hand are:1 sin ( ) sin ( ) cos ( ) '+( )

2 sin ( ) cos ( ) ' sin ( ) '

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3 cos ( ) ' '+

The kinetic energy T.sv of the free symmetrical top is a simple function of (or )

a b: =

( )

Tsv

Isv sv 2

2

sv 2

2 Isv1

2'( )

2'( )

2sin ( )

2+ '( ) '( ) cos ( )+

2+

Ha

d

: =

Had: =

Had: =

sv

2 Isv1 .0 31 1 0

77 gmcm2 sec-2=

g m cm2

'( ) 2 '( ) 2 sin ( ) 2+ '( ) '( ) cos ( )+ 2+ 0 gm cm2 sec-2=

g m cm2

2 '( ) 2 '( ) 2 sin ( ) 2+ '( ) '( ) cos ( )+ 2+

sv 2

2 Isv+ 1. 031 1077 gm cm2 sec-2=

1

2 xd

d

2

z

dd

2

sin ( ) 2+zd

d

y

dd

cos ( )+

2

+

sv

2

2 Isv

Tsv

Isv sv

2:=

According to classical electrodynamics the potential energy of the magnetic moment M

in a magnetic field B is:sv .=

( )

Vsv Mtopsv Bsv

Vsv Mtopsv Bsv:=sv . gmcm sec=

Conseqently , the Lagrangian takes the form:sv

Rgsv

2 .062 1 077 gmcm2 sec-2=

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Isv

sv

2Mtopsv Bsv+ 5. 154 10

77 gm cm2 sec-2=

Conseqently , the Lagrangian takes the form:( )

Lsv Tsv VsvIsv sv

2

2Mtopsv Bsv+

sv sv . gm cm sec

Isv sv 2

2Mtopsv Bsv+ 5. 154 10

77 gm cm2 sec-2=

But , for our magnetic top we assume that the relation(1) is valid. By incorporating

this relation into the Lagrangian we get:

LIsv sv

2

2gsv Isv sv

Bsv+

Lsv

Isv sv 2

2gsv Isv sv

Bsv+:=

sv . gm cm sec

Isv sv

2gsv Isv sv

Bsv+ 5. 154 10

77 gm cm2 sec-2=

It is important to realize that this Lagrangian is different, in an essential way , from

the Lagrangian studied in classical electromagnetism, where is a fixed vectorin body frame and

P Ixd

dI B1 B+

b . sec

a b: =

Had: =

p 'L

d

d

:= Isv ' Isv gsv Bs v+

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sv sv sv sv . gm cm sec

P Iy

dd

cos ( )xd

d+

I g1Bz+

P Izd

dcos ( )

zd

d+

I g1Bz+

where( )

( )

B Bx cos ( ) By sin ( )+

R1 sin ( )0

cos ( )

0

0

0

0

0

cos ( )

sin ( )

sin ( )

cos ( )

sin ( )0

cos ( )

0

0

1

:=

sin ( )0

cos ( )

0

0

0

0

0

cos ( )

sin ( )

sin ( )

cos ( )

sin ( )0

cos ( )

0

0

1

.

0.554

0

.

0.6740

.0.489

0 =

Bz1 Bx sin ( ) sin ( ) By cos ( ) sin ( ) Bz cos ( )+

Following general procedures we need now to express , , and

in terms of , and

R1 0.554

0

0.6740

0.4890

=

xd

d

PI

g1 B

y

d

d

P P cos ( ) I g1 Bz cos Bz((

I s in ( ) 2

zd

d

P P cos ( ) I g1 Bz1 cos Bz((

I s in ( ) 2

Bicause the dependence of the Lagrangian on , and is

trough angular velocity

it is usefull ro express angular velocity

through

the cannonical moment , and

sv.x Isv sv.x cos ( ) Psin ( )

sin ( )P+

sin ( ) cos ( )sin ( )

P gsv Isv Bsv.x

sv.y Isv sv.y sin ( ) Pcos ( )

sin ( ) P+cos ( ) cos ( )

sin ( ) P gsv Isv Bsv.y .

sv sv.x . gm cm sec

.

sv sv sv sv

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2

: =

2

: =

.

Isv sv.y sin ( ) Pcos ( )

sin ( )P+

cos ( ) cos ( )

sin ( )P gsv Isv Bsv.y

sin Psin ( )

P+

sin ( )P gsv Isv Bsv.y 0 gm cm sec

-=

. gm cm sec=

x cos Psin ( )

P+

sin ( )P gsv Isv Bsv: =

( )

y sin Psin ( )

P+

sin ( )P gsv Isv Bsv: =

z gm cm sec=

sx cos Psin ( )

P+

sin ( )P: =

x gm cm sec=

y g m c m s ec=

sy

sin

P sin ( )

P+

sin ( )P

: =

x . gm cm sec

sv sv sv . gm cm sec

sv sv.y . gm cm sec

x sv s v sv gm cm sec

z I x P g 1 IBz

We shallnow define a new vector quantity - cannonical angular

momentum s, by

sx cos ( ) Psin ( )

sin ( )P+

sin ( ) cos ( )

sin ( )P

sy sin ( ) Pcos ( )

sin ( )P+

cos ( ) cos ( )

sin ( )P

sz P

It is seen ...take the form

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s g1 I B

The latter relation is analogous to the relation between the

kinetic momentum in the electromagnetic field of the vector potential A

L1

2m q 2 e A q+

1 m q1

P

1 q1

Ld

d

m q p e A p m q e A+

m q exp 1( ) A+Now we are ready to write the Hamiltonian of the magnetic top

according to

2: =

. gm cm sec

H P P+ I

( ) 2

2 g1 I

B( )( )

. gm cm sec:

P

sec

P

sec

+ Isvsv

2

2

gsv Isv sv

Bsv

1. 428 1019

Msv c

2

2

Psec

Psec

+ Isv sv

2

gsv Isv sv

Bsv

1.428 1019

1. 947 1085 gm cm sec-2

5. 293 109 cm=

10


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Isv

sv

2

gsv Isv sv

Bsv

Msv c2

2

3=

sec sec+ 1. 472 1096

gm cm2 sec-2=

. gm cm sec

2

: =

After some algebra we obtainy a0: =

x a0: =

z a0: =

s2:=

2

: =

s

2 Ig1 s B g12 I B2+

s

2 Ig1 s B g12 I B2+

H I

( ) 2

2

( ) 2

2 IM

2

2 I g12

s g1 I B( )2

2 Is

2

2 Ig1 s B g12I B2+

sv . cm gm sec=

s

2 Ig1 s B g1 2 I B2+ =

I

2

: =

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Isv

sv 2

2

sv 2

2 I

sv

Mtopsv2

2 Isv gsv2

s gsv Isv Bsv( )2

2 Isv

s2

2 Isvgsv s Bsv gsv

2Isv Bsv

2+

5

1. 031 1077

1. 031 1077

1. 031 1077

2. 319 1077

1. 289 1077

erg=

topsv

2 Isv gsv2

1 .0 31 1 077 gmcm2 sec-2=

sv . cm gm sec=

s

2 Isvgsv s Bsv gsv

2Isv Bsv

2+

6.251. 031 10

77 gm cm2 sec-2=

Isv

sv

2

sv

2

2 Isv

Mtopsv2

2 Isv gsv2

s gsv Isv Bsv( )2

Isv

4.5( )

s2

2 Isv

gsv s Bsv gsv2

Isv Bsv2+

6.25

1. 031 1077

1. 031 1077

1. 031 1077

1. 031 1077

1. 031 1077

erg=

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.50 Isv sv2

.50

sv2

Isv

.50

M

topsv

2

Isv gsv2( )

.50

s 1. gsv Isv Bsv( )2

Isv

.10s

2

Isv

.20 gsv s Bsv .20 gsv2 Isv Bsv

2+

4

Msv c2

2

1

4 gsv2

Bsv2( )

2 gsv s Bsv 5 Msv c2+ 4 gsv

2 Bsv2 s2 20 gsv s Bsv Msv c

2+ 25 Msv2 c4+( )+

1

4 gsv2

Bsv2( )

2 gsv s Bsv 5 Msv c2 4 gsv

2 Bsv2 s2 20 gsv s Bsv Msv c

2+ 25 Msv2 c4+( )+

sv c

21 .0 31 1 0

77 gmcm2 sec-2=

5.572 1011 2 gm cm2( )

1.194=

sv

Mtopsv

sv

5. 166 104

5.166

10

4

cm1

gm0

sec-1=

s

2 Isvgsv s Bsv gsv

2Isv Bsv

2+

51. 289 10

77 gm cm2 sec-2=

So ,again the form of the Hamiltonian ......

1

2 xd

d

2

z

dd

2

sin ( ) 2+zd

d

y

dd

cos ( )+

2

+

me c

2

2

1

2 mep e A( )2

( )2

2 I

H1

2 mp e A( )2

gm

2 sec2 x

dd

2

zd

d

2

sin ( ) 2+zd

d

y

dd

cos ( )+

2

+

me c

2+ 2. 18 10 11 erg=

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1

2 mep e A( )2

s g1 I B( )2

2I

L1

2me q

2 e A q+

1

me

2 me g h1( )

1

2

c

1me

2me g h1( )

1

2

c

1. 589 109

1.589 10 9

cm0

sec0=

Ovde dodje tekst . gm cm sec

2 gc

2 12

( ) 4 .322 10 gm cmsec-=

III . Lagrange equations for the magnetic top and theirsolutions for constant magnetic fields

We shell now write and solve Lagrange equations of motion for magnetic top in a constantmagnetic field, assumed to be directed along the z-axis of the space-fixed reference frame. This

assumption does not reduce the generality of our solution, since the orientation of the Laboratory

frame may be chosen convenniently. With this assumption the Lagrangian (8) takes the form :18Lsv1 2

'2

'2

+ '2

+ 2 ' ' cos ( )+ gsv Bsv Isv ' ' cos ( )+( )+ sec gsv Bsv Isv+: =

sv1 . gm cm sec=

Because this Lagrangian does not depende on f and c the momenta and are integrals ofmotions :

Ha 'Lsv1

d

d

d

d Lsv1

d

d

Ha

P secd

d

19

Ha 'Lsv1

d

d

d

d Lsv1

d

d

Ha

P secd

d

Hence the corresponding two Lagrange equations reduce to two first order differential equations :20

' ' cos ( )+ gsv Bsv+P

Isv

21

' cos ( ) '+ gsv Bsv cos ( )+PIsv

The third Lagrange equation is a second order differential equation

22

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''2

Ha

d

:=

Ha 'sv

dd sv

d 0 gm cm sec-=

s n+ gsv sv

s n+ gm cm 0 gm cm sec=

In order to solve the latter equation we shall substitute into it the following expressions

23

1'Isv sin ( )

2

gsv Bsv:=

24

Isv sin ( )2

gsv Bsv 0 sec

-1=

1'

Isv sin ( )2

: =

Isv sin ( )2

3 .5 21 1 0

18 sec-1=

obtained from eqs.(20)

Isv sin ( )2

Isv sin ( ) 1. 24 10 35 sec-2=

25

''

Isv sin ( )

2

Isv

sin ( )+ 1. 24 10 35 sec-2=

Now we note the remarkable identities

sin ( )2. 342 10

95 gm cm2 sec-1=

26

'

sin ( )sec

1d

0gmcm2

sec-1=

sin ( )2

2. 342 1095 gm cm2 sec-1=

27

'

sin ( )d0 gm cm

2=

sin ( )2

2. 342 1095 gm cm2 sec-1=

With the aid of those identities we transforme equation (25) to any one of following two forms :

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''

Isv sin ( ) '

Isv sin ( )d: =

Isv sin ( ) '

Isv sin ( )d 0 sec-1=

sec

''

Isv sin ( ) '

Isv sin ( )d: =

Isv sin ( ) '

Isv sin ( )d 0 sec-1=

Now multiplying bots equations with = we find

d '2 cos

Isv sin ( ) :=

d '2 cos

Isv sin ( ) :=

cos

Isv sin ( ) 1.24 10

35 sec-2=

'2 co s

Isv sin ( ) + 1. 24 10 35 sec-2=

A '2P

P

cos ( )

Isv sin ( )

2

+:=

'02 co s

Isv s in ( ) + 1.24 10 35 sec-2=

'02 co s

Isv s in ( ) + 1.24 10 35 sec-2=

1.24 10 sec=

BP P cos ( )

Isv sin ( )

2

:=

So, we found two other integrals of motion. In order to find (t). It is sufficient to use of themd

1

P P cos ( )

A Isv sin ( )

2

A dt

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d

1

P P cos ( )

A Isv sin ( )

2

A dt

or

After some algebraic operations we recognize on the left hand site an integrable function ;x=sin()

32dcos ( )

a b cos c cos ( ) 2+( )+dt

where( )

a '02

P2 P 2+( ) cos 0( )2 2 P( ) P cos 0( )

Isv sin 0( )2

+

a '02

gm cm2

sec-2( )

P2

P2+( ) cos 0( )

2 2 P( ) P cos 0( )

Isv sin 0( )2

+:=

b2 P P

Isv2

Isv2

2 .4 79 1 035 sec-2=

b 2 cos 0( ) '02 '0 gsv Bsv+( )

2+ ' '0 gsv Bsv+( ) 1 cos ( )2+( )+:=

2 cos 0( ) '02 '0 gsv Bsv+( )

2+ ' '0 gsv Bsv+( ) 1 cos ( )2+( )+ 1. 518 10 51 sec-2=

33

c10 gm cm sec 2 cos 0 ++

Isv2

sin 0( )2

:=

. sec

33

'02

sec2 '0 gsv Bsv+( )

2+ '02+ 2 cos ( ) '0 '0 gsv Bsv+( )+ 2.467 sec-2=

b b gm cm:=

The solution readsa c : =

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cos ( )

2 csin c( ) t asin

2 c cos 0( ) b+

+

b

2 c:=

34

cos ( )

2 c

sin c( ) t asin2 c cos 0( ) b+

+

b

2 c

:=

where

35

4 a c1 b2 4 '04 '0

2sin 0( )

2 '02 '0 gsv Bsv+( )

2+ sin 0( )4 '0

2 '0 gsv Bsv+( )2++:=

4 a c : =

Therefore cos( ) oscillates with the period

T0 2

c

between the two values

determined by36

cos 2( ) b

2 c1:=

cos 1( ) b+

2 c1:=

Consequently, oscillates with the same period between the corresponding values and

: depending on the initial condition.

Now we are ready to determine and . By integrating the equation (23) we find :

g

sv

Bsv

t

0

tP P cos ( )

Isv sin ( ) 2

d+

gsv Bsv t

0

P P

Isv sin ( )2

1

td

d

d+ 2=

37

gsv Bsv t

0

A Isv sin ( )

cos ( )

1P P cos ( )A Isv sin ( )

2

P P

Isv sin ( )2

d+

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A Isv sin ( )

1

P P

A Isv sin ( )

2

0=

38 a0 gsv Bsv t asin

A Isv sin ( )cos ( )

asinA Isv sin 0( )

cos 0( )

+

38 b0 gsv Bsv t asin


asinA Isv sin 0( )

cos 0( )

+

In an analogous way we obtain

39 a

0asin


asin

A Isv sin 0( )cos

0( )

39 b 0 asin


asin

A Isv sin 0( )cos 0( )

Not that only (t) depends on the magnetic field

Implicit assumption that sin( )=0 , there exist particular solution of Lagrange equation, which

are characterised by : sin (t) for any value of t. Such solution exist for initial conditios :0 0

or 0 ( ) , '00

,

P0

Isv

P0

Isv

'0

'0

+ gsv

Bsv.

+

Lagrange equation 20-22 are then equvalent to

:

P0

Is v

P0

Is v

'0 '0+ gsv Bs v+ 1=

41

' '+ gsv Bsv+P0Isv

P0Isv

PIsv

PIsv

1=

'( ) 0

The solution of the latter equations are :

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Isv

gsv Bsv

t 0+ 0+ 3.142=

21.571=

+PIsv

gsv Bsv

t 0+ 0+

Isv

1.571=

The fact that for those initial conditions the equations give only the dependence of ( + ) on

t and do not give the dependence on t of each angle separately is understandable. When the z-axis

of the body frame coincides with the z-axis of laboratory frame the rotation described by (t)and (t) are rotations about the same axis (z-axis) and consequently the angles (t) and

(t) do not appear separately but together in a sum.

Having determined the solution of Lagrange equations of motion we may now determine thetime dependence of the most important quantity for our purpouse, i.e. kinetic angular momentum

(2) and cannonical (spin) - (12). By virtue of the equations (23) and (24) we find that is a

constant of motion

z IsvPIsv

gsv Bsv

P gsv Bsv Isv z0

41 sv sv sv gmcm sec=

Isv Isv

gsv Bsv

0gmcm2 sec-1=

Further, taking into account the relation (24) and (30) and introducing the angle such that : ' A cos ( ) ' sin ( ) A s in ( )

asinP P cos ( )

A Isv sin ( )

' 0


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t.

t( ) 0 gsv Bsv t asinP P cos 0( )A Isv sin 0( )

'0 0>,

t( ) 0 gsv Bsv t( ) asinP P cos 0(

A Isv sin 0( )

'0 0


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IV .The Torque equation and its

equivalence with Lagrange equation

We are going to demonstrate the equivalence of the torque equationtopsv sv . gm cm sec=

sv sv sv . gm cm sec

sec

t sv

d

dN Mtopsv Bsv gsv sv

Bsv

( )

wits the Lagrange equation(20,21) and (22) by substituing into the torque equation the

expressions (3)for the components of the vector .sv assuming B=Bk

sec=

46

tcos ( ) ' sin ( ) sin ( ) '+ gsv Bsv sin ( ) ' cos ( ) sin ( ) '( )

d

a . sec=

sec

47

tsin sec( ) ' cos sec( ) sin sec( ) '+ gsv Bsv cos ( ) ' sec sin ( ) sin ' s ec( ) ' sec( )

d

48

t cos +

d

From eq.(48) we obtains immediately one of the integral of motion

49

' cos ( )+( ) ' zIsv

x0Isv

But , this equation is equvalent to the Lagrange equation (20) , the relation between the constantsbeing :

50 z P gsv Bsv Isv

Next , by multiplying the equation (46) and (47) by cos( ) and sin( ) respectively, and

summing the resultant expression, we find the Lagrange (22).The third equvalence between theLagrange and torque equations may be established after the following operations. First, wemultiply (20) with -cos( ) and sum with (21).This given :

51

' s in ( )PIsv

PIsv

co s ( )

Differentiation of the latter equation gives :

22


23/39

52 '' sin ( ) 2 cos ( ) ' '+

Isv

'

On obtains the same equation by multiplaying Eqs. (46) and (47) with cos( ) and -sin( ),

respectively and then summing the resultant expression. Hence the equvalence is proved.

V. Hamilton's equations for the magnetic topFrom (16) we eassily derive Hamilton's equations for the magnetic top

'P

Hd

d

PIsv

gsv Bx cos ( ) By sin ( )+( )

'P

Hd

d

P P cos ( )

Isv sin ( )2

gsv

Bx sin ( ) By cos ( )( )sin ( )

cos ( )+ Bz

'P

Hd

d

P P cos ( )

I

sv

sin ( ) 2gsv

Bx sin ( ) By cos ( )

sin ( )

+

P' H

d

dP

2cos ( ) P

2cos ( )+ P P 1 cos ( )

2+( ) gsv Bxsin ( ) P P cos ( )(

sin ( ) 2

+

gsv Bycos ( ) P cos ( ) P( )

sin ( ) 2+

...

P2

cos ( ) P2

cos ( )+ P P 1 gm cm2 cos ( ) 2+( ) gsv Bx

sin ( ) P P cos ( )( )

sin ( ) 2

+

gsv( ) Bycos ( ) P cos ( ) P( )

sin ( ) 2+

...

P '

Hd

d

53b

P'

Hd

d

By taking B along z axis, we obtain the simpler equations

'P

Isv

P'P P cos ( )( ) P cos ( ) P

Isv sin ( )

2

'P P cos ( )

Isv s in ( )2

gsv Is v Bsv

P const

P ' 0

23


24/39

'P P cos ( )

Isv sin ( )2

P const

P' 0

We see that Hamilton equations for ' and ' are identical with the equations (23) and (24)which were derived form the Lagrange equations (20) and (21). By combining the equations for

' and through

''P'

Isv

we find the Lagrange equations (25).Now we shall show that Hamilton's formalisme for magnetic top leads also to the torque

equation for the motion of spin for this purpose we shall use the Poisson-bracket formalism.

By applying the general dynamical for any quantity u{q.a,p.a) in phase space (q,a,p,a) for theequations of motion of the angular momentum we find :

t 1

d

d i H,( )

q i

d

d

iH

d

d

p i

d

d

q i

d

d

jH

d

d

p 1

d

d jH i j,(

d

d

For the Poisson brackets of spin components we after some calculation x y,( ) z gsv Isv Bsvz+z sv sv sv . gm cm sec

x z,( y gsv Isv Bsvy z y,( ) x gsv Isv Bsv.x

We have also from(16)

56

xH

d

d

xIsv

yH

d

d

yIsv

zH

d

d

zIsv

By supstitution (56) and (57) into (55) we find again the torque equation (45), i.e.

57

t x

d

dgsv y Bx z By( )

t y

d

dgsv z Bx x Bx( )

58

t z

d

dgsv x By y Bx( )

24


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It is well known that it follows from (58) that is a constant of motion

59

t

2d

d0

Before we start to quantiye this system let us note that due to the equalities

q i

d

d qsi gsv Isv Bi( )

d

d qsi

d

d

60

p i

d

d psi gsv Isv Bi( )

d

d psi

d

d

we have the follwing important relations61 i j,( ) si sj,( )

Taking this relation into account we find the Poisson brackest of the components of the canonical

angular momentum or spin vector s.62si sj,( ) i j, k,( ) sk,

as wellas the dynamical equation for s

58 '

ts

d

dgsv s Bsv

VI. Quantum magnetic top

In order to quantze the motion, we shall aply two standard quantization procedures.1) Cannonicalquantization and 2) Schrodinger quantization. The third form of quantization, the path integral

formalism, will be discussed separately.

1) Canonical quantization

It is well known that in the framework of this formalism one passes from the classical to the

quantum case by replacing the classical dynamical variables f(p,q) , g(p,q), etc. by operators F,G,

etc.in some Hilbert space of states, in such a way that the Lie product in the space of classicalfunctions, defined as a Poisson bracket :

f g,( )q

fd

d

p

gd

d

pf

d

d

q

gd

d+

is replaced by the Dirac commutator (quantum Poisson bracket)

F G,( )0 i h( ) 1 F G G F( ) i h( ) 1 F G,( )which now plays the role of the Lie product in the space of operators.The Dirac Lie product

conserves the structure of Lie algebra of classical functions with Poisson bracket as the Lie

product. The equation of motion for a dynamical variable F now reads

tF

d

d

1

i hF H,( ) F H,( )Q

where H is Hamilton operator associeted with the classical Hamiltonian H(p,q).The basic quantity of the magnetic top is cannonical angular momentum s. Taking into account

the Poisson bracket (62 )

25


26/39

of the components of s and the requirement that the quantum Poisson bracket (s.i,s.j)^0 have to

conserve the structure of the classical Lie algebra we may immediatly write the Dirac bracket of

the components s.i of the operator of cannonical angular momentum s.si sj+( ) i j, k,( ) sk

It follows strainghtforwardly that the commutators of the components of s have to be :

si sj+( ) si sj sj si( ) i h i j, k,( ) sk( )One further step leads now to Hamilton operator of the quantum magnetic top. Inthe classical

Hamiltonian (16) canonical angular momentum s has to be substituted by the operator s.

Hs

2

2 Isvgsv s Bsv

gsv2

Isv Bsv2

2+

s

2 Isvgsv s Bsv

gsv sv sv

2+ 2. 319 1077 gm cm2 sec-2=

The components of the well known Pauli spin operatpor

x0

1

1

0

:=

z1

0

0

1:=

y0

1

10

:=

2 x

5. 855 1094

.

0 gm cm

2sec

-1=

sv

2 y

5. 85 5 1094

.

0 gm cm

2sec

-1=

sv

2 z

.

0 5.855 1094 gm cm

2sec

-1=

satisfy the commutation relations (65) and therefore Pauli operators represent one possiblerepresentation of quantum canonical angular momentum operators. But of cource there are many

other bigher dimensional representationss.

In the two-dimensional spin space spanned by two eigenstates of s.z

s1

0

s0

1

the cotribution of the term

s

2 Isvgsv s Bsv

gsv sv sv

2+

to the eigenstates is constant(independent of the state) and we argue that those two terms in the quantum Hamiltonian give a

constant energy shift. In this way we conclude that Pauli Hamiltonian69

HP gsv s Bsv gsv sv

2

Bsv

gsvsv

2 x Bsv 2.062 1077.

0 gm cm

2sec

-2=

26


27/39

is the dynamical part of the Hamiltonan and one of the quantum representation of the magnetic

quantum top

One shorthcoming of this representation is that it does not contain quantum analogues of ,, and p. , p. and p. . But this shorthcoming may be removed by applyng the

Schrodinger quantization (22).

ii^0) Schrodinger quantization

In Schrodinger quantization, with canonical momenta P. ,P. and P. one associates theoperators of canonical momenta P. ,P. ,P.

P i sv d

d

70

P i sv d

d

P i sv d

d

By substituing into (12) the canonical momenta P. ,P. ,P. in terms of the above

operators, we find the differential representation of the s.x,s.y,s.z.

sx cos ( ) i hd

d

sin ( )cos ( )

sin ( )

i hd

d+

sin ( )

sin ( ) i hd

d

71

sy sin ( ) i hd

d

cos ( )cos ( )

sin ( )

i hd

d+

cos ( )

sin ( ) i hd

d

sz

i hdd

It is eqsy to see that commutators of the above differential operators sartisfy the commutation

relations (65) . By squaring the operators (71) and by summing the resultant expression we obtainthe differential representation of the operator s^2.

s2

sx2

sy2+ sz

2+( )2

h2d

d

2

cot ( )

h2d

d

1

sin ( ) 2 2h

2d

d

2

2h

2d

d

2

+

+ 2

cot ( )

sin ( )

2 dh

2d

d

2

+

The differential representation of the Hamilton operator (66) reads :

H1

2 Isv2

h2d

d

2

cot ( )

h2d

d

1

sin ( ) 2 2h

2d

d

2

2h

2d

d

2

+

+ 2

cot ( )

sin ( )

2 dh

2d

d

2

+

gsv Bsv i

hd

d

+

gsv 2 Isv Bsv 22

+

As in the case of Pauli representation, in the subspace spanned by the eigenstates of s^2associated with the eigenvalue s*(s+1), the contribution of the first two terms to energy

eigenvalues is independent of the states. Thr ramaininig term is another possible representation of

the Pauli HamiltonianBsv Bsv k

27


28/39

s1

2

HP gsv i h Bsv d

d

We want to stress here that s is quantum analogue of the canonical angular momentum s and not

of the kinetic angular momentum . In the absence of the field the angular momentumcoincides with the canonical angular momentum s. In the works of Bopp and Haag (9) and Dahl(13) the operators (71) and (72) have been derived starting form the free top and from the angular

momentum =I.sv* .sv expressed trough the momenta P.1 and P.2 of two point particles at

point with radius vectors r.1 and r.2 (with constant mutual angle u).73 Isv sv P1 P2 P2 r2+

Judd (23) associetedthe same differential operators with =I.sv* .sv of the free top using thecorresondence rule (70). Rosen also uses those differential operators (8).

The subsequent procedure of Bopp and Haag in the presence of the field consists in the

following steps : Yhey substituted the expressions (9) for P. ,P. ,P. valid in the presence

of the field into the following relation between angular momentum components .x, .y, .z (denoted in their paper by M= (m.x,M.y,M.z)) and canonical momenta P. ,P. ,P.

Mx x cos ( ) Psin ( )

sin ( )P

sin ( ) cos ( )

sin ( )P

My y sin ( ) Pcos ( )

sin ( )P

cos ( ) cos ( )

sin ( )P+

74Mz z P

But , as it is seen from (11) this latter relation is valid in the absence of the field. In this wayBopp and Haag obtained the relation

75M 'M M'+ Isv sv gsv Isv Bsv+

sv sv sv sv sv . gm cm sec=

which they substitued into H expressed through , , , ', ', ' (H=I.sv* .sv^2/2)

In this way they found76

HM

2

2 Isv

gsv M Bsv

gsv Isv Bsv

2+

In the next step Bopp and Haag claim that the quantum analogue of M is the operator (71)In the above reasoning the justification of the use of the relation (74) in the presence of the

field is missing. Consequently, the theoretical meaning of the relation (75) (the relation (36) inBopp and Haag paper) is missing too. In our reasoning, which strictly follows the standardprocedure for the construction of the Hamiltonian (which has to be considered as a function in

phase space , , ,P. ,P. ,P. and not as a function of , , , ', ', ')

we obtain the relation (11) which takes the place of Bopp and Haag relation (360. But , then wedefine in (12) a new quantity s and we look for the quantum analogue of this quantity. In this way

we make a clear distinction between angular momentum .sv=I.sv* .sv and canonical

angular momentum s and this distinction is theoretically justified in the framework of

28


29/39

Hamiltonian formalism. Moreover, the analogousdistinction betweein the kinetic momentum mv

and canonical momentum p is standard in the gauge theory of point particles. On the other hand,

theoretical status of Bopp and Haag quantity M,M' and'M has not been established.The quantization based on the form (66) of the Hamiltonian has one more advantage. One

discovers this advantage if one tries to quantize on the basis of the Hamiltonian expressed

through phase space variables , , ,P. ',P. ',P. ' .

H1

2 IsvP

2P

2

sin ( ) 2+ P

2+

2 P P

cos ( )

sin ( ) 2 gsv Bsv P

gsv2

Bsv2 Isv

2+

gsv Bsv Pgsv sv sv

2+ 1. 237 1078 gm cm2 sec-2=

2 P Psin ( ) 2

cm 1 .47 6i 10271 gm cm sec-=

1

2 IsvP

2

sin ( ) 2+ P

2+

2. 749 10109 gmcm2 sec-2=

The direct substitution of the phase space variables by operators (70) into the above form of H

leads to the operator which differs from Hamilton operator (66a) by the absence of the terms -

(h^2 /2*I.sv)*cotg. ^

d

This difference is due to the ambiguity in the ordering of , , ,P. ,P. ,P. in (16')It seems that the use of canonical angular momentum implicitly alleviates this ambiguity and

provides the correct ordering

VII The states of the quantum magnetic top

s1

2

With Pauli representation of the spin operators, the associeted quantum states are the spinors

which are linear combinations of two basic states

1

0 and

0

1 , namely the eigenstatesof .z

1

0

0

1+

The two eigenstates of the Pauli Hamiltonian are very often written in terms of the polar ( .B)

and asimuthal ( .B) angle of the vector B.

B B e

I B

2cos

2

1

0 e

I Bsin

2

0

1+

,

78

B B e

i B

2 2

sin

1

0 e

i Bcos

2

0

1+

,

In this way of writting one stresses the fact that the eigenstates of the spin Hamiltonian in a

magnetic field are the eigenstates of the component s.B of the spin operator s.As is well known, the differential operators (71) and (72) can act on larger spaces of states

than the space of Pauli spinors and these spaces are richer in informations than are Pauli states.

29


30/39

The operator s^2 has the eigenvalues s*(s+1) where s takes all integer and half integer values. In

the corresponding subspaces D^s thetwo-valued representations of the Rotation group are

realized (9).In the case of s=2/2, which is of interest to us here, the basic states of D^1/2 are usually

chosen to be the eigenstates of s.z which are the following functions of , , , (9,13)

u1

2 , ,( ) i e

i2

2

+

cos2

2 2

79 a

u 1

2 , ,( ) i e

i2

2

+

sin2

2 2

u1

2

, ,( ) i ei

2

2

+

cos2

2 2

79 b

u1

2 , ,( ) i e

i

2

2

+

cos2

2 2

Therefore ,the use of differential operators (71) instead of Pauli operators (67) implies thedescription of spin states by probabillity amplitudes u.n and their linear combinations insread by

matrices

1

0and

0

1 and their linear combinations.

Is there any advantage of using wave functions U.n( , , ,) to describe spin states rather

than Pauli spinor ?The first advantage is that with u.n( , , ),the spin is no longer a strange and

abstruse quantum-mechanical object fitted into the general quantum-mechanical framework.From this advantage follows the second one. It is telated to the understanding of the law of

transformation of spin states under rotation.

The property of spinors

to change sign under 2* rotation which is a cocequence

of the law of transformation of spinor under rotation for an arbitrary angle

Rz ( ) ,( ) ei

2x

ei

2x

e1 B( ) +

cos B

2

ei B +( )

sin B

2

cos B

2

30


31/39

Rz ( ) ,( ) ei

2

xe

i2

xe

1 B( ) +

cos B

2

ei B +( )

sin B

2

cos

B2

has been the subject of studies (both theoretically and experimentally), discussions andcontroversies (25-31). The source of controversies lies in the difficulties to physically

understandthis property. Namely, if one uses for the states

1

0 and

0

1 the usual physicalpicture of the spin vector alongz-axis, one can hardly understand what is the physical reason for

the phase changes by - and under 2 rotation (32,33).

It seems that these difficuilties are removed if one interpretes the spin property as amodification of the interaction between a magnetic field and magnetic top such that u.n( , ,

) is the probability amplitude of the angles , , under this motion. The effect ofR.z(

) on U.n( , , )

Rz ( )u1

2 , ,( ) i e

1

2 +( )

cos

2

2 2

81

Rz ( )u1

2 , ,( ) i e

1

2 +( )

sin

2

2 2

is seen to be due to the change of the angle by -

In our study of the classical magnetic top we saw that to the simple precession of spin with

frequency -g.sv*B.sv corresponds a more complicated motion of the magnetic top, in which the

angles and a very complicated functions of time. In the absence of precession of spin(when spin is along the z-axis) the body rotates with frequency -g.sv*B.sv+P. /I.sv wich is

different from Larmor frequecy .L=-g.sv*B.sv. Consequently, when the azimuthal angle(t)- (t) of spin vector changes by - (or does not change at all) the orientation angles ,

, ,change by 2 (or zero) , , do not necessarlly leads to the initial

orientation.

We expect that those differeces in the motion of the angular momentum and of the body, inthe classical case, have their counerparts in the quantized motions. They might explain the

strange transformation properties of spinors under rotation. But, the full understanding requires

more detailed study of the quantized motion of the magnetic top.sv sv . sec

gsv BsvIsv

+ 7 .0 42 1 0 18 sec-1=

Isv

3 .5 21 1 018 sec-1=

VIII. The Asymmetric Magnetic TopIt seems worthwhile to generalize the above study to the case of an asymmetric top for which the

31


32/39

simple relation (2) betweein the kinetic angular momentum and the angular velocity is

no longer valid. Istead the following relation holds

82

i

Ii i ei

where e.i are unit vectors along the body fixed frame for which the moment inertia tensor isdiagonal. In addition we shall assume, insread of relation (1), the more general relation between

the kinetic angular momentum and magnetic momentumM

i

gi i eii

gi Ii i ei

83Consequently , the Lagrangian of the magnetic top in a magnetic field B=Bk reads

84

L

i

Isv sv2

2 Mtopsv Bsv+i

Isv sv2

2i

gsv Isv sv Bsv+i

Isv sv2

2i

sv Bsv1+

where B.i are the components of B in body-fixed frame85

B

i

Bi ei( ) Bsin ( ) sin ( )

sin ( ) cos ( )

cos ( )

Bi Ii gi Bi

i

sv sv

2 Mtopsv Bsv+ 5. 154 1077 gm cm2 sec-2=

i

sv sv

2i

gsv Isv sv Bsv+ 5. 154 1077 gm cm2 sec-2=

i

sv sv

2i

sv Bsv1+ 3.217 1089 eV=

Bohr radius by coeficient

i

sv sv

2i

sv Bsv1+

9. 74 1085 gm cm sec-2

5. 292 109 cm=

As in the case of symmetric top, the three coordinates which determine the orientation of the top

in the laboratory frame are :q1

86

32


33/39

q2

q3

Since the components of angular velocity are linear functions, eq.(4), of the time derivative

of the angles, it is appropriate to write the set of relations (4) in matrix form

87

C q( ) q'

C q( )

cos ( )

sin ( )

0

sin ( ) sin ( )

sin ( ) cos ( )

cos ( )

0

0

1

88

q'

'

'

'

Using this notation we write the Lagrangian as

84 a

L1

2

i

Ii

n

Cin q( ) q1n

2

in

Cimq'

n

Bi+

Consequently ,the canonical momenta are :89

Pkk'k

Ld

di

I Cin q( ) q'n Cik Ak+

90

A B

g1 I1 sin ( ) cos ( ) g2 I2 sin ( ) sin ( ) cos ( )

g1 I1 sin ( )2 sin ( ) 2 g2 I2 sin ( )

2 cos ( ) 2+ g3 I3 cos ( )2+

g3 I3 cos ( )

A

A

A

We shall define the quantity :

.k Pk Ak

i

Ii

n

Cin q( ) q 'n Cinin

CinT

Ii Cik q'n

.kni

CniT

Cik q'n

or in matrix form

CT C q'n

whereCij Ii Cij

is the transposed matrix of C

33


34/39

C

I1 cos ( )

I2 sin ( )

0

I1 sin ( ) sin ( )

I2 sin ( ) cos ( )

I3 cos ( )

0

0

I3

It follows from (91) that

q' C 1 CT( )1

C1

cos ( )I1

sin ( )

I1 sin ( )

sin ( ) cos ( )I1 sin ( )

sin ( )I2

cos ( )

I2 sin 0(

cos ( ) cos ( )I2 sin ( )

0

0

1

I3

CT( ) 1

cos ( )

sin ( )

0

sin ( )

sin ( )cos ( )

sin ( )0

sin

sin ( )

cos ( )

cos ( )sin ( )

cos ( )

1

In order to express q'.1^ in terms of '.1^ we need the product C^-1*(C^T)^-1 wich

reads :

g C1

CT( ) 1

cos ( ) 2

I1

sin ( ) 2

I2

+

cos ( ) sin ( )

sin ( )

1

I1

1

I2

cos ( ) sin ( ) cos ( )sin ( )

1

I1

1

I2

cos sin ( )( )

sin ( )1

I1

1

I2

1

sin ( )2

sin ( ) 2

I1

cos ( ) 2

I2

+

cos ( )

sin ( ) 2sin ( ) 2

I1

cos ( )I2

2

+

cos ( ) sin ( ) cos (sin ( )

cos ( )

sin ( )2

sin ( ) 2

I1

cos ( ) 2

sin ( ) 2sin ( ) 2I1 +

94

with the aid of matrix elements g.ik, the relation (92) read

q'k

i

gki .ii

gki Pi Ai( )

Having expressed velocities q'.k in terms of momenta P.i we are now ready in construct the

Hamiltonian of the asymmetric top starting from the general relation

H p q' L p q' A q'( )i

I1

2

C q'( )i[ ]

2

Using (92) the first two terms take the form

p q' A q'( ) q' C 1 CT( )1

.12

i

1

2

k

.k Cki1

j

CiT( )

1 .j 1

2 C 1 CT( )

1

It folows now that

96

34


35/39

H1

2 C 1 CT( )

1

i

Ii

2i( )

2 T 12

p A( ) C 1 CT( )1

p A( ) 12

p A( ) G p A( )

H1

2p A( ) G p A( ) 1

2gik q( ) p A( ) .i p A( ) .k

Eqs. (92a) and (96) suggest to interpret g.ik(q) as the metric tensor in the space of the kinetic

momentaThe more explicit form of H reads ;

H I sin ( ) 2 I1cos ( ) 2

I2

+

P A( )

21

sin ( ) 2P A( )

2+cos ( ) 2

sin ( ) 2P A( )

2+

1

2 I3P A( )

2cos ( ) sin ( )

sin ( )

I2 I1

I1 I2 p A( ) P A( )++

...

cos ( ) sin ( ) cos ( )sin ( )

I2 I1

I1 I2p A( ) P A( )

cos ( )

sin ( )2

sin ( ) 2

I1

cos ( ) 2

I2

+

p A( ) P A( )+

...

+

...

For the symmetric top (I.1=I.2=I.3) and for g.1=g.2=g.3=g, the above Hamiltonian reduces to the

form given in (18').

Hamilton and Lagrange equations follow directly from the above expressions for Hamiltonian

and Lagrangian.Canonical angular momentum

Let us now express the canonical angular momentum s through the canonical momenta P. ,P.

,P. . For this purpose we shall substitute the relations (87) and (92a) into (82).

i

I1 1 e1i

I1

k

Cik q'k e1i

I1.k

k

Cik C1

CT( ) 1

CT

( )1

CT

( )1

P CT

( )1

A97

cos ( )

sin ( )

0

sin ( )

sin ( )

cos ( )

sin ( )0

sin ( )sin ( )

cos ( )

cos ( )sin ( )

cos ( )

0

P A

P A

P A

If we compare the latter relation with the relation (12) and (13) we conclude that in the phasespace expressios of for asymmetric top the first term is the same function of phase space

variables as is the function s defined in (12). So,we shall call the quantity

s CT( ) 1 p

cos ( ) Psin

( )

sin ( ) P+sin

( )

sin ( ) cos ( ) P+

sin ( ) Pcos ( )

sin ( )P+

cos ( )sin ( )

cos ( ) P+

P

i

si ei

the canonical angular momentum, or simply spin of the asymmetric top. The components of s in

the laboratory frame are identical with the ones given in (12). To the quantity .i((C^T)^-

35


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1*A)*e.i we shall give the name

a

i

CT( ) 1 A .i ei

Its components in the body frame are :a1 g1 I1 sin ( ) sin ( )

a2 g2 I2 sin ( ) cos ( )

a3 g3 I3 cos ( )

The relation (97) turns into : s a

Now it is matter of simple algebra to express the Hamiltonian of the asymmetric magnetic top in

terms of its spin100

H1

2C

T C 1

jik

1

2Cji j Cik

1 k1

2

jk

j k

i

Cjk Cik1( )

1

2

jk

j kjkIj

j

j2

2 Ij

A remarkable simplification occurs if we choose the constants g.1 that we introduces in (83) tosatisfy g.1^2 =g^2/I.1. Then the Lagrangian and Hamiltonian become

L1

2 g2

2 B+

101

H1

2g 2

Components of the spin vector in the body frame satisfy the folowing equations of motion :

s'j H s,( )

i i

Hd

d

k pk i

d

d

qksi

d

d pk

sid

d

qk i

d

d

i i

Hd

d

i si,( )

i si,( ) si ai si,( ) ai si,( )n

ijn sn ai sj,( )

ai sj,( )k

qk

sid

d

pk

sjd

d

s'j

i

si ai( )Ii

n

ij sn ai sj,( )

Appendix A: Top with magnetic moment fixed in the body frameA top is fully characterized and specified by its coupling. In this paper we have defined and

studied magnetic top characterized by a velocity dependent magnetic moment. In order to make

more clear our argumentation that the magnetic top is the more appropriate classical model ofspin we shall present here a theory of the top which carries the magnetic moment M attached to

the body. That implies that the magnetic moment M is independent of the angular velocity (for if

M were dependent on angular velocity it could not be constant in the body frame).Consequently, the coupling with magnetic field B is velocity independent.V M B

This potential has the smae form as the potential top (with mass M.sv and center of masscoordinate R) in the gravitational field g.

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V Msv R gsv

M being analogous to M.0*R playing the role of gravitationald field g.We shall deal here with the axialy symmetric top (I.1=I.2) and shall assume that M is along the

body z-axis, i.e. M=M.z. Then, making B along the z-axis of the laboratory frame (B=Bk), wich

does not reduce thr generality of our results,we write the ineraction potential V in the form

A3V M B cos ( )

The Lagrangian is differnce of kinetic and potential energy termsA4

L T VI1

21

2 22+( )

I3

23

2+ M cos ( )+

I1

2 '2 '2 sin ( ) 2+( )

I3

2' ' cos ( )+( ) 2+ M B cos (+

In order to construct the Hamiltonian we folow the usual procedure. Canonocal momenta are the

following functions of , and

P 'L

d

dI1 '

P ' Ld

d I1 sin ( )2

' I3 ' ' cos ( )+( ) cos ( )+

P 'L

d

dI3 ' cos ( ) '+(

Now one express velocities , , and ' in the terms of canonical momenta

'PI.1

'P P cos ( )(

I1 sin ( )2

'P

I3P P cos ( )( ) cos ( )I2 sin ( )

2

By substituting the latter expression into T andp q ' P ' P+ ' P+

one obtain

p q 2 T 2P

2

2 I1

P2

2 I3+

P2

2 I1 sin ( )2

+P

2cos ( ) 2

2 I3 sin ( )2

+ P Pcos ( )

I1 sin ( )2

and consequently

H

, ,P

,P

,P

,( ) ' P

' P

+ ' P

+ , ,P

,P

,P

,( )V

+T V

+

P2

2 I1

P2

2 I3+

P2

2 I1 sin ( )2+

P2

cos ( ) 2

2 I3 sin ( )2

P Pcos (

I1 sin+

...

Therefore , in agreement with the general theory, to the Lagrangian with velocity independent

potential there corresponds a Hamiltonian which is a simple sum of kinetic and potential energy

terms. The Hamiltonian (16) of the magnetic top does not have this property, again in agreement

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with the general theory, since the interaction term in The Lagrangian (7) is dependent on

velocities ', ', and '

Hamilton's equations of motionThe first three Hamilton1s equations are the eqs.(A6) .The remaining three read :

P'

H

d

d

P2 cos ( ) P cos ( ) P P 1 cos ( )

2+( )+

I1 sin ( )2

M B sin ( )+

P '

Hd

d

A9

P' H

d

d

Comparing the equations (A6) and (A9) with Hamilton's equations (54) of the magnetic top wefind that equations for ',P'. and P'. in both sets are the same whereas the equations for

', ', and P'. are different. Since the Hamiltonian of the top with velocity independent

magnetic moment is identical to the gravitational top, the corresponding Hamilton's equation are

to be found in literature (24). Here we shall review the well known qualitative analysis (19)Two immediate first integrals of motion are :

P I3 ' ' cos ( )+ I3 3 P0(

P I1 sin ( )2 ' I3 ' I3 ' ' cos ( )+( )+ P0+

Since the system is conservative the total energy is the third integral o f motion.

E T V+I1

2'2 '2 sin ( ) 2+( )

I3

2

P0

I3

2+ M B cos ( )+

Only three additional quadratures are needed to solve the problrm. From the above three integralsit is possible to express ', ' and ' as functions of ' and constant of motion

'P0 P0 cos ( )

I1 sin ( )2

'P0

I3

cos ( )P0 P0 cos ( )

I1 sin ( )2

A13I1

2'2

P0 P0 cos ( )( )2I1 sin ( )

2+

M B cos ( )+ E

P02

2 I3

E'

The equation (A13) differs from the corresponding equation (31) of the magnetic top by presence

of the term M*B*cos( ). As we are going to see, due to the presence of this term the equation

(A13) leads to an eltptic integral (with cubic polynomial under the integral sign) On the otherhand the equation (30) leads to the equation (32) with square polynomial and therefore is

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integrable. From (A13) it follows :

A14

' I1 sin ( ) sin ( )2

2 I1 E' M B cos ( )( ) P0 P0 cos ( )2( )

1

2

A15

T

COS 0( )( )

COS ( ) 1( )

xdcos ( ) .

2 I1 1 cos ( )2( ) E' M B cos ( )( ) P0 P0 cos ( )

2( )

1

2

d

Since the solution of the equation (A15) cannot be writwn in an analytic form, the sme is valid

for the solution of equation (A11) and (A12). But, the qualitative featrures of the solution (t)of the equation *A13) are known (19). They are pictured on Fig.3 in which the possible shapes

for the locus of the body axis on the unit sphere are indicated. Recalling that M was assumed to

be along e.x. this figure presents also the motion of the magnetic moment M fixed with the body.Therefore, M follows the motion of the body, quite different from M and of the magnetic top

wich move with respect to the body. So, M performs a complicated motion (precession withnutation) whereas and M simpli precess.

One of the authors (M.B) would like to thank Professor Abdus Salam, the International Centre for

Theoretical Physics, Trieste.

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Magnetna cigra Svemira