Student Seminar: Classical and Quantum Integrable Systems

Preprint typeset in JHEP style - HYPER VERSION
Student Seminar:
Classical and Quantum Integrable Systems
Gleb Arutyunov a
a Institute for Theoretical Physics and Spinoza Institute, Utrecht University
3508 TD Utrecht, The Netherlands
Abstract: The students will be guided through the world of classical and quantum
integrable systems. Starting from the famous Liouville theorem and finitedimensional
integrable models, the basic aspects of integrability will be studied including
elements of the modern classical and quantum soliton theory, the Riemann-
Hilbert factorization problem and the Bethe ansatz.
Delivered at Utrecht University, 20 September 2006- 24 January 2007

Contents
1. Liouville Theorem 2
1.1 Dynamical systems of classical mechanics 2
1.2 Harmonic oscillator 5
1.3 The Liouville theorem 7
1.4 Action-angle variables 9
2. Examples of integrable models solved by Liouville theorem 11
2.1 Some general remarks 11
2.2 The Kepler two-body problem 12
2.2.1 Central fields in which all bounded orbits are closed. 15
2.2.2 The Kepler laws 17
2.3 Rigid body 20
2.3.1 Moving coordinate system 20
2.3.2 Rigid bodies 21
2.3.3 Euler’s top 23
2.3.4 On the Jacobi elliptic functions 27
2.3.5 Mathematical pendulum 29
2.4 Systems with closed trajectories 31
3. Lax pairs and classical r-matrix 32
3.1 Lax representation 32
3.2 Lax representation with a spectral parameter 34
3.3 The Zakharov-Shabat construction 36
4. Two-dimensional integrable PDEs 41
4.1 General remarks 42
4.2 Soliton solutions 43
4.2.1 Korteweg-de-Vries cnoidal wave and soliton 43
4.2.2 Sine-Gordon cnoidal wave and soliton 45
4.3 Zero-curvature representation 47
4.4 Local integrals of motion 49
5. Quantum Integrable Systems 55
5.1 Coordinate Bethe Ansatz (CBA) 56
5.2 Algebraic Bethe Ansatz 68
5.3 Nested Bethe Ansatz (to be written) 79
6. Introduction to Lie groups and Lie algebras 80
– 1 –

7. Homework exercises 95
7.1 Seminar 1 95
7.2 Seminar 2 96
7.3 Seminar 3 97
7.4 Seminar 4 98
7.5 Seminar 5 100
7.6 Seminar 6 102
7.7 Seminar 7 105
7.8 Seminar 8 107
1. Liouville Theorem
1.1 Dynamical systems of classical mechanics
To motivate the basic notions of the theory of Hamiltonian dynamical systems consider
a simple example.
Let a point particle with mass m move in a potential U(q), where q = (q 1 , . . . q n )
is a vector of n-dimensional space. The motion of the particle is described by the
Newton equations
m¨q i = − ∂U
∂q i
Introduce the momentum p = (p 1 , . . . , p n ), where p i = m ˙q i and introduce the energy
which is also know as the Hamiltonian of the system
H = 1
2m p2 + U(q) .
Energy is a conserved quantity, i.e. it does not depend on time,
dH
dt = 1 m p iṗ i + ˙q i ∂U
∂q i = 1 m m2 ˙q i¨q i + ˙q i ∂U
∂q i = 0
due to the Newton equations of motion.
Having the Hamiltonian the Newton equations can be rewritten in the form
˙q j = ∂H
∂p j
,
ṗ j = − ∂H
∂q j .
These are the fundamental Hamiltonian equations of motion. Their importance lies
in the fact that they are valid for arbitrary dependence of H ≡ H(p, q) on the
dynamical variables p and q.
– 2 –

The last two equations can be rewritten in terms of the single equation. Introduce
two 2n-dimensional vectors
( ) p
x = , ∇H =
q
(
∂H
)
∂p j
∂H
∂q j
and 2n × 2n matrix J:
J =
( ) 0 −I
I 0
Then the Hamiltonian equations can be written in the form
ẋ = J · ∇H , or J · ẋ = −∇H .
In this form the Hamiltonian equations were written for the first time by Lagrange
in 1808.
Vector x = (x 1 , . . . , x 2n ) defines a state of a system in classical mechanics. The
set of all these vectors form a phase space M = {x} of the system which in the present
case is just the 2n-dimensional Euclidean space with the metric (x, y) = ∑ 2n
i=1 xi y i .
The matrix J serves to define the so-called Poisson brackets on the space F(M)
of differentiable functions on M:
{F, G}(x) = (∇F, J∇G) = J ij ∂ i F ∂ j G =
n∑
j=1
( ∂F
∂p j
∂G
∂q j − ∂F
∂q j ∂G
∂p j
)
.
Problem. Check that the Poisson bracket satisfies the following conditions
{F, G} = −{G, F } ,
{F, {G, H}} + {G, {H, F }} + {H, {F, G}} = 0
for arbitrary functions F, G, H.
Thus, the Poisson bracket introduces on F(M) the structure of an infinitedimensional
Lie algebra. The bracket also satisfies the Leibnitz rule
{F, GH} = {F, G}H + G{F, H}
and, therefore, it is completely determined by its values on the basis elements x i :
{x j , x k } = J jk
– 3 –

which can be written as follows
{q i , q j } = 0 , {p i , p j } = 0 , {p i , q j } = δj i .
The Hamiltonian equations can be now rephrased in the form
ẋ j = {H, x j } ⇔ ẋ = {H, x} = X H .
A Hamiltonian system is characterized by a triple (M, {, }, H): a phase space
M, a Poisson structure {, } and by a Hamiltonian function H. The vector field X H
is called the Hamiltonian vector field corresponding to the Hamiltonian H. For any
function F = F (p, q) on phase space, the evolution equations take the form
dF
dt = {H, F }
Again we conclude from here that the Hamiltonian H is a time-conserved quantity
dH
dt
= {H, H} = 0 .
Thus, the motion of the system takes place on the subvariety of phase space defined
by H = E constant.
In the case under consideration the matrix J is non-degenerate so that there
exist the inverse
J −1 = −J
which defines a skew-symmetric bilinear form ω on phase space
ω(x, y) = (x, J −1 y) .
In the coordinates we consider it can be written in the form
ω = ∑ j
dp j ∧ dq j .
This form is closed, i.e. dω = 0.
A non-degenerate closed two-form is called symplectic and a manifold endowed
with such a form is called a symplectic manifold. Thus, the phase space we consider
is the symplectic manifold.
Imagine we make a change of variables y j = f j (x k ). Then
ẏ j = ∂yj
}{{} ∂x k
A j k
ẋ k = A j k J km ∇ x mH = A j km ∂yp
kJ ∂x m ∇y pH
– 4 –

or in the matrix form
ẏ = AJA t · ∇ y H .
The new equations for y are Hamiltonian if and only if
AJA t = J
and the new Hamiltonian is ˜H(y) = H(x(y)).
Transformation of the phase space which satisfies the condition
AJA t = J
is called canonical. In case A does not depend on x the set of all such matrices form
a Lie group known as the real symplectic group Sp(2n, R) . The term “symplectic
group” was introduced by Herman Weyl. The geometry of the phase space which
is invariant under the action of the symplectic group is called symplectic geometry.
Symplectic (or canonical) transformations do not change the symplectic form ω:
ω(Ax, Ay) = −(Ax, JAy) = −(x, A t JAy) = −(x, Jy) = ω(x, y) .
In the case we considered the phase space was Euclidean: M = R 2n . This is not
always so. The generic situation is that the phase space is a manifold. Consideration
of systems with general phase spaces is very important for understanding the
structure of the Hamiltonian dynamics.
1.2 Harmonic oscillator
Historically it is proved to be difficult to find a dynamical system such that the
Hamiltonian equations could be solved exactly. However, there is a general framework
where the explicit solutions of the Hamiltonian equations can be constructed. This
construction involves
• solving a finite number of algebraic equations
• computing finite number of integrals.
If this is the way to find a solution then one says it is obtained by quadratures.
The dynamical systems which can be solved by quadratures constitute a special
class which is known as the Liouville integrable systems because they satisfy the
requirements of the famous Liouville theorem. The Liouville theorem essentially
states that if for a dynamical system defined on the phase space of dimension 2n one
finds n independent functions F i which Poisson commute with each other: {F i , F j } =
0 then this system van be solved by quadratures.
– 5 –

To get more insight on the Liouville theorem let us consider the simplest example
– harmonic oscillator. The phase space has dimension 2 and the Hamiltonian is
H = 1 2 (p2 + ω 2 q 2 ) ,
while the Poisson bracket is {p, q} = 1. Energy is conserved, therefore, the phase
space is fibred into ellipses H = E.
p
stationary point
H=E=const −−energy levels
q
HARMONIC OSCILLATOR −− PROTOTYPE OF LIOUVILLE INTEBRABLE SYSTEMS
Problem.
system
Rewrite the Poisson bracket {p, q} = 1 and the Hamiltonian in the new coordinate
p = ρ cos(θ) , q = ρ ω sin(θ) .
The answer is
The hamiltonian is
{ρ, θ} = ω ρ .
H = 1 2 ρ2 → ρ = √ 2H .
We see that ρ is an integral of motion. Equation for θ:
˙θ = {H, θ} = ρ{ρ, θ} = ω ⇒ θ(t) = ωt + θ 0 .
This means that the flow takes place on the ellipsis with the fixed value of ρ.
Generalization to n harmonic oscillators is easy:
H =
n∑
i=1
1
2 (p2 i + ω 2 i q 2 i ) .
– 6 –

Commuting integrals
F i = 1 2 (p2 i + ω 2 i q 2 i ) .
Define the common level manifold
M f = {x ∈ M : F i = f i , i = 1, . . . , M}
This manifold is isomorphic to n-dimensional real torus which is a cartesian product
of n topological circles. These tori foliate the phase space and can be parametrized
with n angle variables θ i which evolve linearly in time with frequencies ω i . This
motion is conditionally periodic: if all the periods T i = 2π
ω i
are rationally dependent:
T i
T j
= rational number
the motion is periodic, otherwise the flow is dense on the torus.
1.3 The Liouville theorem
The system is Liouville integrable if it possesses n independent conserved quantities
F i , i = 1, . . . , n, {H, F i } which are in involution
{F i , F j } = 0 .
The Liouville theorem. Suppose that we are given n functions in involution on a
symplectic 2n-dimensional manifold
Consider a level set of the functions F i :
F 1 , . . . , F n , {F i , F j } = 0 .
M f = {x ∈ M : F i = f i , i = 1, . . . , n}
Assume that the n functions F i are independent on M f . In other words, the n-forms
dF i are linearly independent at each point of M f . Then
1. M f is a smooth manifold, invariant under the flow with H = H(F i ).
2. If the manifold M is compact and connected then it is diffeomorphic to the
n-dimensional torus
T n = {(ψ 1 , . . . , ψ n ) mod 2π}
3. The phase flow with the Hamiltonian function H determines a conditionally
periodic motion on M f , i.e. in angular variables
dψ i
dt = ω i , ω i = ω i (F j ) .
– 7 –

4. The equations of motion with Hamiltonian H can be integrated by quadratures.
Let us outline the proof. Consider the level set of the integrals
M f = {x ∈ M : F i = f i , i = 1, . . . , M} .
By assumptions, the n one-forms dF i are linearly independent at each point of M f ;
by the implicit function theorem, M f is an n-dimensional submanifold on the 2ndimensional
phase space M. Moreover, the n linearly-independent vector fields
ξ Fi = {F i , . . .}
are tangent to M f and commute with each other.
Let α = ∑ i p idq i be the canonical 1-form and ω = dα = ∑ i dp i ∧ dq i is the
symplectic form on the phase space M. Consider a canonical transformation
i.e.
(p i , q i ) → (F i , ψ i )
ω = ∑ i
dp i ∧ dq i = ∑ i
dF i ∧ dψ i
such that F i are treated as the new momenta. If we found this transformation then
equations of motion read as
˙ F j = {H, F j } = 0 ,
˙ψ j = {H, ψ j } = ∂H
∂F j
= ω j .
Thus, ω i are constant in time. In these coordinates equations of motion are solved
trivially
F j (t) = F j (0) , ψ j (t) = ψ j (0) + tω j .
Thus, we see that the basic problem is to construct a canonical transformation
(p i , q i ) → (F i , ψ i ). This is usually done with the help of the so-called generating
function S. Consider M f : F i (p, q) = f i and solve for p i : p i = p i (f, q). Consider the
function
We see that
and we further define
S(f, q) =
∫ m
m 0
α =
∫ q
p j = ∂S
∂q j
ψ j = ∂S
∂f j
q 0
∑
i
p i (f, ˜q)d˜q i
– 8 –

Thus, we have
Since d 2 S = 0 we get
i.e. the transformation is canonical.
dS = ∂S
∂q j
dq j + ∂S
∂f j
df j = p j dq j + ψ j df j
∑
dp j ∧ dq j = ∑
j
j
df j ∧ dψ j ,
The next point is to show that S exists, i.e. it does not depend on the path. If
we have a closed path from m 0 to m and from m to m 0 and assume that M f does
not have non-trivial cycles then by the Stokes theorem we get
∫ m0
∫ ∫
∆S = α = dα = ω = 0
m 0
because the form ω vanishes on M f :
ω(ξ Fi , ξ Fj ) = {F i , F j } = 0 .
In case the manifold M f has non-trivial cycles the situation changes and one gets
the change of S given by integral of α over a cycle
∫
∆ cycle S = α
which is a function of F i only! This tells us that in this case the variables ψ j are
multi-valued.
cycle
Mention Darboux
1.4 Action-angle variables
As follows from the Liouville theorem under suitable assumptions of compactness
and connectedness motion of a dynamical system in the 2n-dimensional phase space
happens on a n-dimensional torus T n being a common level of n commuting integrals
of motion. The torus has n fundamental cycles C j which allow to introduce the
“normalized” action variables
I j = 1
2π
∮
C j
p i (q, f)dq i ≡ 1
2π
∮
C j
α ,
where f i define the common level T n of the commuting integrals F i . The variables
I j are functions of f i only and therefore they are constants of motion. The angle
variables are introduced as independent angle coordinates on the cycles
1
2π
∮
C j
dθ i = δ ij .
– 9 –

Let us show that the variables (I i , θ i ) are canonically conjugate. For that we need
to construct a canonical transformation (p i , q i ) → (I i , θ i ). Consider a generating
function depending on I i and q i :
We see that
Let us introduce
S(I, q) =
∫ m
m 0
α =
∫ q
q 0
p i (q ′ , I)dq ′ i .
p j = ∂S
∂q j
=⇒ p = p(q, I).
θ j = ∂S
∂I j
=⇒ θ = θ(q, I).
and show that θ j are indeed coincide with the properly normalized angle variables.
We have
∮
1
dθ i = 1 ∮
d ∂S = ∂ ( ∮ 1
)
dS = ∂ ( ∮ 1 ∂S
dq k +
∂S dI k
2π C j
2π C j
∂I i ∂I i 2π C j
∂I i 2π C j
∂q k ∂I
} {{ k
}
Furthermore,
= ∂ ( ∮ 1
)
α = δ ij .
∂I i 2π C j
=0 on C j
)
( ∂S
) (
dI i ∧ dθ i = −d(θ i dI i ) = −d dI i = −d dS − ∂S )
dq i = d(p i dq i ) = dp i ∧ dq i .
∂I i ∂q i
Problem. Find action-angle variables for the harmonic oscillator.
We have
E = 1 2 (p2 + ω 2 q 2 ) =⇒ p(E, q) = ± √ 2E − ω 2 q 2 .
and, therefore,
I = 1 ∮
dq √ 2E − ω
2π
2 q 2 = 2
E
2π
∫ √ 2E
ω
− √ 2E
ω
The generating function of the canonical transformation reads
while for the angle variables we obtain
S(I, q) = ω
∫ q
dx √ 2I − x 2 ,
dq √ 2E − ω 2 q 2 = E ω .
θ = ∂S
∂I = ω ∫ q
dx
√
2I − x
2 = ω arctan
q
√
2I − q
2
=⇒ q = √ 2I sin θ ω .
Finally, we explicitly check that the transformation to the action-angle variables is canonical
(
dI
dp ∧ dq = ω √ −
2I − q
2
qdq
)
√ ∧ dq =
2I − q
2
ω
√
2I − q
2 dI ∧ d(√ 2I sin θ ω
)
= dI ∧ dθ .
– 10 –

2. Examples of integrable models solved by Liouville theorem
2.1 Some general remarks
Problem. Consider motion in the potential
V (q) =
Solve eoms and find a period of oscillations. One has
t − t 0 =
∫ q
q 0
dq
√
= −
2(E − g2
sin 2 q )
∫ q
q 0
g2
sin 2 q , E > g2 .
d cos q
√ √
2E
(E−g 2 )
E
− cos 2 q
∫ arccos q
= −
arccos q 0
Thus, motion happens on the interval q 0 < q < π − q 0 and taking q 0 = arcsin
We see from here that
Period is
cos √ } {{
2E
}
ω
It does not depend on g 2 !!!
t = − 1 √
2E
(
arcsin
x
√
E−g 2
E
( π
t = cos
2 − arcsin x
)
√ =
1 − g2
E
T = 2π ω =
)
2π √
2E
.
| x=cos q
q
E−g
x=
2
E
x
√
1 − g2
E
dx
√ √ .
2E
(E−g 2 )
E
− x 2
√
g 2
E
= √
1
cos q .
1 − g2
E
one gets
Problem. Consider a one-dimensional harmonic oscillator with the frequency ω and compute the
area surrounded by the phase curve corresponding to the energy E. Show that the period of motion
along this phase curve is given by T = dS
dE .
A curve is an ellipsis
( x
a
) 2
+
( y
b
) 2
= 1
with the area
S = 2b
∫ a
−a
dx √ ∫ π √
1 − x 2 /a 2 2
= 2ba dφ cos φ 1 − sin 2 φ = 2ab
− π 2
∫ π
2
− π 2
dφ cos 2 φ = πab .
We have to identify a = ρ, b = ρ ω
so that
S = πab = π ρ2
ω = 2π ω E .
From here we see that
dS
dE = 2π ω = T ,
where T is a period of motion. The last expression has the same form as the first law of thermodynamics
dE = 1 T
dS provided that 1/T is the temperature (the period ≡ the inverse temperature).
Problem. Let E 0 be the value of the potential at a minimum point ξ. Find the period T 0 =
lim E→E0 T (E) of small oscillations in a neighborhood of the point ξ.
– 11 –

We have
H = p2
2
p2
+ V (x) =
2 + V } {{ (ξ) + V ′ (ξ)(x − ξ) + 1 } } {{ } 2 V ′′ (ξ)(x − ξ) 2 + · · ·
const =0
Effectively we have motion described by the harmonic oscillator with the Hamiltonian
H eff = p2
2 + 1 2 V ′′ (ξ)q 2
whose frequency is ω = √ V ′′ (ξ). Therefore the period of small oscillations is
T 0 =
2π
√
V
′′
(ξ) .
2.2 The Kepler two-body problem
Here we consider one of the historically first examples of integrable systems solved
by the Liouville theorem: The Kepler two-body problem of planetary motion.
In the center of mass frame eoms are
d 2 x i
dt 2
(r)
= −∂V , r =
∂x i
√
x 2 1 + x 2 2 + x 2 3
In the original Kepler problem V (r) = − k , k > 0. The Hamiltonian
r
H = 1 2
3∑
p 2 i + V (r)
i=1
and the bracket {p i , x j } = δ ij .
Problem . Show that the angular momentum
⃗J = (J 1 , J 2 , J 3 ) ,
J ij = x i p j − x j p i = ɛ ijk J k
is conserved.
J˙
ij = ẋ i p j − x i ṗ j − (i ↔ j) = p i p j + ∂V
∂r x ∂r
i − (i ↔ j) = ∂V
∂x j ∂r
Note that this is a consequence of the central symmetry.
(
x i
∂r
∂x j
− x j
∂r
∂x i
)
= 0
Problem. Compute the Poisson brackets
Show that there are three commuting quantities
{J i , J j } = −ɛ ijk J k
H, J 3 , J 2 = J 2 1 + J 2 2 + J 2 3
Rewrite the canonical one form in the polar coordinates
x 1 = r sin θ cos φ, x 2 = r sin θ sin φ, x 3 = r cos θ
– 12 –

We find
α = ∑ i
p i dx i = p r dr + p θ dθ + p φ dφ ,
where the original momenta are expressed as
p 1 = 1 (
sin φ
)
rp r cos φ sin θ + p θ cos θ cos φ − p φ ,
r
sin θ
p 2 = 1 (
cos φ
)
rp r sin φ sin θ + p θ cos θ sin φ + p φ ,
r
sin θ
p 3 = p r cos θ − 1 r p θ sin θ .
Conserved quantities
(
p 2 r + 1 r 2 p2 θ +
H = 1 2
J 2 = p 2 θ + 1
sin 2 θ p2 φ
J 3 = p φ
1
)
r 2 sin 2 θ p2 φ + V (r)
To better understand the physics we note that the motion happens in the plane
orthogonal to the vector J. ⃗ Without loss of generality we can rotate our coordinate
system such that in a new system J ⃗ has only the third component: J ⃗ = (0, 0, J3 ).
This simply accounts in putting in our previous formulae θ = π . Then we note that
2
p 2 φ
˙φ = {H, φ} = {
2r 2 sin 2 θ , φ} =
that for θ = π 2 expresses the integral of motion p φ as
p φ = r 2 ˙φ.
p φ
r 2 sin 2 θ
This is the conservation law of angular momentum discovered by Kepler through
observations of the motion of Mars. The quantity p φ = J has a simple geometric
meaning. Kepler introduced the sectorial velocity C:
C = dS
dt ,
where ∆S is an area of the infinitezimal sector swept by the radius-vector ⃗r for time
∆t:
∆S = 1 2 r · r ˙φ∆t + O(∆t 2 ) ≈ 1 2 r2 ˙φ∆t .
This is the (second) law discovered by Kepler: in equal times the radius vector sweeps
out equal areas, so the sectorial velocity is constant. This is one of the formulations
of the conservation law of angular momentum. 1
1 Some satellites have very elongated orbits. According to Kepler’s law such a satellite spends
most of its time in the distant part of the orbit where the velocity ˙φ is small.
– 13 –

We can now see how the solution can be found by using the general approach
based on the Liouville theorem. The expressions for the momenta on the surface of
constant energy and J = J 3 are
√
p r = 2(H − V ) − J 2
r , p 2 φ = J 3 = J .
We can thus construct the generating function of the canonical transformation from
from the Liouville theorem
∫ r
S =
√2(H − V ) − J ∫ 2 φ
r + Jdφ
2
and the associated angle variables
We have eoms
Integrating the first one we obtain
ψ H = ∂S
∂H ,
ψ J = ∂S
∂J
˙ψ H = 1 , ˙ψ J = 0 .
and, therefore,
The equation for ψ J gives
ψ J = −
t − t 0 =
∫ r
ψ H = t − t 0
∫ r
dr
√
.
2(H − V ) − J2
r 2
Jdr
√
+ φ = 0 ,
r 2 2(H − V ) − J2
r 2
so that
φ =
∫ r
Jdr
√
r 2 2 ( ) .
E − V (r) − J 2
2r 2
Generically, equation which defines the values of r at which ṙ = 0:
E − V (r) − J 2
2r 2 = 0
has two solutions: r min and r max , they are called pericentum and apocentrum respectively
2 . When ṙ = 0, ˙φ ≠ 0. The r oscillates monotonically between rmin and
2 If the earth is the center then r min and r max are called perigee and apogee, if the sun – perihelion
and apohelion, if the moon – perilune and apolune.
– 14 –

max while φ changes monotonically. The angle between neighboring apocenter and
pericenter is given by
∆φ =
∫ rmax
r min
Jdr
√
r 2 2 ( ) .
E − V (r) − J2
2r 2
Generic orbit is not closed! It is closed only if ∆φ = 2π m , m, n ∈ Z, otherwise it is
n
everywhere dense in the annulus. The annulus might degenerate into a circle.
2.2.1 Central fields in which all bounded orbits are closed.
Determination of a central potential for which all bounded orbits are closed is called
the I.L.F. Bertrand problem.
There are only two cases for which bounded orbits are closed
V (r) = ar 2 , a ≥ 0 ,
V (r) = − k r , k ≥ 0 .
To show this we have to solve several problems.
Problem. Show that the angle φ between the pericenter and apocenter is equal to the half-period
of an oscillation in the one dimensional system with potential energy W (x) = V (J/x) + x2
2 .
Substitution r = J r gives ∆φ =
∫ xmax
x min
dx
√
2(E − W (x))
.
Problem. Find the angle φ for an orbit close to the circle of radius r.
Effectively the angle φ is described by half-period of oscillation
∆φ =
∫ xmax
x min
dx
√
2(E − W (x))
.
We have
∆φ = π ω , ω = √ W ′′ (x) ,
where x = J r . We find W ′ (x) = ∂ x V (J/x) + x = − J x 2 V ′ (J/x) + x ,
We have to take
W ′′ (x) = 2 J x 3 V ′ (J/x) + J 2
x 4 V ′′ (J/x) + 1 .
− J x 2 V ′ (J/x) + x = 0 =⇒ x3
J = V ′ (J/x) =⇒ J
r 3/2 = √ V ′ (r) .
– 15 –

Thus,
and, therefore, the half-period is
W ′′ (x) = J )
(3V ′
x 3 (r) + rV ′′ (r) = 3V ′ (r) + rV ′′ (r)
V ′ (r)
√
V
∆φ circ = π
′ (r)
3V ′ (r) + rV ′′ (r) .
Problem. Find the potentials V for which the magnitude of ∆φ circ does not depend on the radius.
We have to require
( 3V ′ (r) + rV ′′ (r)
V ′ (r)
) ′ ( rV ′′ (r)
) ′ (
= =
V ′ r ( log V ′ (r) ) ) ′
′
= 0 ,
(r)
i.e.
log V ′ (r) = const
∫ 1
r
= s log r + m , s, m = const .
Further,
V ′ (r) = const r s , =⇒ V (r) = ar α ,
or V (r) = b log r if s = −1. Finally, the expression
V (r) = ar α we will get
V ′ (r)
3V ′ (r)+rV ′′ (r)
should be positive. If we take
V ′ (r)
3V ′ (r) + rV ′′ (r) = α
3α + α(α − 1) = 1 > 0 =⇒ α > −2 .
2 + α
Finally, we also have
∆φ circ =
π
√ 2 + α
.
Here the logarithmic case correspond to α = 0. Particular cases are α = 2 which gives ∆φ circ = π 2
and α = −1 which gives ∆φ circ = π.
Problem. Let V (r) → ∞ as r → ∞. Find
lim ∆φ circ(E, J)
E→∞
Let us make a substitution x = yx max , we get
∫ 1
dy
∆φ circ = √
y min 2(Q(1) − Q(y))
(
where Q(y) = y2
2 + 1 J
x
V 2
max yx max
). As E → ∞ we have x max → ∞ and y min → 0 and the second
term in Q can be discarded. Thus, we get
∆φ circ =
∫ 1
0
dy
√
1 − y
2 = π 2 .
Problem. Let V (r) = −kr −β , where 0 < β < 2. Find
lim ∆φ circ(E, J)
E→−0
– 16 –

One has
∆φ circ =
∫ xmax
x min
dx
√
2E + 2k x
J β − x 2
β
E→0
→
∫ xmax
x min
dx
√
2k
x
J β − x 2
β
Rescale x = αy with α satisfying the relation 2k
J β α β = α 2 , then we get
∆φ circ =
∫ 1
We note that the result does not depend on J.
0
dy
√
yβ − y = π
2 2 − β .
Now we are ready to find the potentials for which all bounded orbits are closed. If
all bounded orbits are closed, then, in particular, ∆φ circ = 2π m = const. That means
n
that ∆φ circ should not depend on the radius, which is the case for the potentials
V (r) = ar α , α > −2 and V (r) = b log r .
In both cases ∆φ circ = √ π
2+α
. If α > 0 then lim E→∞ ∆φ circ (E, J) = π and therefore
2
α = 2. If α < 0 then lim E→0 ∆φ circ (E, J) = π . Then we have an equality
2+α
π
= √ π
2+α 2+α
which gives α = −1. In the case α = 0 we find ∆φ circ = √ π 2
which
is not commensurable with 2π. Therefore all bounded orbits are closed only for
V = ar 2 and U = − k .
r 2
2.2.2 The Kepler laws
For the original Kepler problem we have
and
Integrating we get
∫
φ =
V (r) = − k r + J 2
2r 2 .
Jdr
√
r 2 2(E + k − J 2
)
r 2r 2
φ = arccos
J
√ − k r J
.
2E + k2
J 2
An integration constant is chosen to be zero which corresponds to the choice of an
origin of reference for the angle φ at the pericenter. Introduce the notation
√
J 2
k = p , 1 + 2EJ 2
= e ,
k 2
This leads to
r =
p
1 + e cos φ
– 17 –

This is the so-called focal equation of a conic section. When e < 1, i.e. E < 0, the
conic section is an ellipse. The number p is called a parameter of the ellipse and e
the essentricity. The motion is bounded for E < 0.
The semi-axis a is determined as
2a =
p
1 − e + p
1 + e = 2p
1 − e . 2
We also have
Thus,
c = a −
p
1 + e = 1 ( p
2 1 − e − p )
= ep
1 + e 1 − e . 2
c
a = e .
b a p
c O
p
p
1−e 1+e
Keplerian ellipse
Obviously, we have three distinguished points
φ = 0 : r = p
1 + e ,
φ = π 2 : r = p ,
We can now formulate the Kepler laws:
φ = π : r = p
1 − e .
1. The first law: Planets describe ellipses with the Sun at one focus.
2. The second law: The sectorial velocity is constant.
3. The third law: The period of revolution around an elliptical orbit depends only
on the size of the major semi-axes. The squares of the revolution periods of
two planets on different elliptical orbits have the same ratio as the cubes of
their major semi-axes.
Let us prove the third law. Let T be a revolutionary period and S be the area swept
out by the radius vector over the period. We have
S = πab = πa 2√ √
1 − e 2 = π 1 − e2 = π
(1 − e 2 ) 2 (1 − e 2 ) 3 2
p 2
p 2
= πkJ
( √ 2|E|) 3 ,
– 18 –

while
i.e.
a =
p
1 − e = k
2 2|E| .
On the other hand, since the sectorial velocity C is constant we have
∫ T
0
C =
∫ T
0
dt dS
dt = S , =⇒ CT = J 2 T = S ,
T = 2S J =
2πk
( √ 2|E|) = √ 2π a 3 2 .
3 k
It is interesting to note that the total energy depends only on the major semi-axis
a and it is the same for the whole set of elliptical orbits from a circle of radius a
to a line segment of length 2a. The value of the second semi-axis do depend on the
angular momentum.
The Runge-Lenz vector and the Liouville torus. The phase space of the motion in the
central field is T ∗ R 3 , i.e. it is six-dimensional. There are four conserved integrals:
three components of the angular momentum J i and the energy E. This shows that
the motion happens on the two-dimensional manifold. In case of the bounded motion
it is the two-dimensional Liouville torus. Thus, there are two frequencies associated
and when they are not rationally commensurable the orbits are not closed but rather
dense on the torus. For the specific Kepler motion (with any sign of k) there is one
more non-trivial conserved quantity appears which is absent for a generic central
potential: The Runge-Lenz vector (for definiteness we assume that k > 0):
⃗R = ⃗v × ⃗ J − k ⃗r r .
Problem. Show that the Runge-Lenz vector is conserved.
Indeed, we have
˙⃗R = ˙⃗v ×
}{{}
J ⃗ −k ⃗v r + k⃗r(⃗v⃗r) r 3
m ⃗r×⃗v
= m ˙⃗v × (⃗r × ⃗v) − k ⃗v r + k⃗r(⃗v⃗r) r 3 .
On the other hand,
m ˙⃗v = − ∂U ⃗r
∂r r = −k ⃗r r 3
and, therefore,
˙⃗R = −k 1 r 3 ⃗r × (⃗r × ⃗v) − k⃗v r + k⃗r(⃗v⃗r) r 3
Further one has to use the formula
⃗r × (⃗r × ⃗v) = (⃗v⃗r)⃗r − r 2 ⃗v
– 19 –

to show that ˙⃗ R = 0. The last formula can be proved by noting that the vector
⃗r × (⃗r × ⃗v) = α⃗r + β⃗v
is orthogonal to ⃗r. Thus, multiplying both sides by ⃗r we get
0 = αr 2 + β(⃗v⃗r) .
On the other hand, multiplying both sides by ⃗v we get
(⃗v, ⃗r × (⃗r × ⃗v)) = α(⃗v⃗r) + βv 2
which gives
(⃗v, ⃗r × (⃗r × ⃗v)) = −(⃗r × ⃗v, ⃗r × ⃗v = −r 2 v 2 sin φ = −r 2 v 2 (1 − cos 2 φ)
= −r 2 v 2 + (⃗v⃗r) 2 = α(⃗v⃗r) + βv 2 .
These two equations allows one to find
α = (⃗v⃗r) , β = −r 2 .
2.3 Rigid body
2.3.1 Moving coordinate system
Let K and k will be two oriented Euclidean spaces. A motion of K relative to k is
a mapping smoothly depending on t:
D t : K → k ,
which preserves the metric and orientation. Every motion can be uniquely written
as the composition of a rotation (D t which maps the origin of K into the origin
of k, i.e. D t is linear mapping) and a translation C t : k → k. Let call K and k
moving and stationary coordinate systems respectively. Let q(t) and Q(t) will be the
radius-vector of a point in a stationary and moving coordinate systems respectively.
Then
q(t) = D t Q(t) = B t Q(t) + r(t) .
} {{ } }{{}
rotation translation
Differentiating we get an addition formula for velocities
˙q =
ḂQ +B
}{{}
˙Q + ṙ .
transferred rotation
Suppose a point does not move w.r.t. to the moving frame, i.e.
r = ṙ = 0. Then
˙q = ḂQ = ḂB−1 q = Aq ,
˙Q = 0 and also that
– 20 –

where A : k → k is a linear operator on k. Since B is a rotation, it is an orthogonal
transformation: BB t = 1. Differentiating w.r.t to t we get
ḂB t + BḂt = 0 =⇒ ḂB −1 + (ḂB−1 ) t = 0 ,
i.e. A is skew-symmetric. On the other hand, every skew-symmetric operator from
R 3 to R 3 is the operator of vector multiplication by a fixed vector ω:
˙q = ω × q .
Generically ω depends on t. Thus, in the case of purely rotational motion with ˙Q ≠ 0
we will have
˙q = ω × q + B ˙Q = ω × q
} {{ }
transferred velocity
+ }{{} v ′
relative velocity
.
2.3.2 Rigid bodies
A rigid body is a system of point masses, constrained by holonomic relations expressed
by the fact that the distance between points is constant
|x i − x j | = r ij = const .
If a rigid body moves freely then its center of mass moves uniformly and linearly.
A rigid body rotates about its center of mass as if the center of mass were fixed at
a stationary point O. In this way the problem is reduced to a problem with three
degrees of freedom – motion of a rigid body around a fixed point O. The problem of
rotation of rigid body can be studied in more generality without assuming that the
fixed point coincides with the center of mass of a body. Since the Lagrangian function
is invariant under all rotations around O by Noether theorem the components of the
angular momentum M are conserved: Ṁ = 0 . The total energy which is equal to
the kinetic energy is also conserved. Thus, we see that
In the problem of motion of rigid body around a fixed point, in the absence of
outside forces, there are four integrals of motion: three components on M and the
energy. Thus, motion happens on a two-dimensional space inside the six-dimensional
phase-space (three rotation angles plus three velocities) :
M f = {M x = f 1 , M y = f 2 , M z = f 3 , E = f 4 > 0 }.
The phase space is a cotangent bundle to SO(3). The manifold M f is invariant: if
the initial conditions of motion give a point on M f then for all time of motion the
point in T SO(3) corresponding to the position and velocity of the body remains in
M f . The two-dimensional manifold M f admits a globally defined vector field (this
is the field of velocities of the motion on T SO(3)), it is orientable and compact (E
– 21 –

is the bounded kinetic energy). According to the known theorem in topology, a
two-dimensional compact orientable manifold admitting globally defined vector field
is isomorphic to a torus. This is our Liouville torus 3 . According to the Liouville
theorem motion on the torus will be characterized by two frequencies ω 1 and ω 2 . If
their ratio is not a rational number then the body never returns to its original state
of motion.
Consider a rigid body rotation around a fixed point O and denote by K a coordinate
system rotating with the body around O: in K the body is at rest. Every
vector in K is carried to k by an operator B. By definition of the angular momentum
we have
M = q × m ˙q = m q × (ω × q) .
Denote by J and by Ω the angular momentum and angular velocity in the moving
frame K. We have
J = m Q × (Ω × Q) .
This defines a linear map A: K → K such that AΩ = J. This operator is symmetric:
(AX, Y ) = (m Q × (X × Q), Y ) = m(Q × X, Q × Y )
because the r.h.s. is symmetric function of X, Y . The operator A is called the inertia
tensor. We see that taking X = Y = Ω we get
E = T = 1 2 (AΩ, Ω) = 1 2 (J, Ω) = m 2 (Q × Ω, Q × Ω) = m 2 ˙Q 2 = m 2 ˙q2 .
being a symmetric operator A is diagonalizable and it defines three mutually orthogonal
characteristic directions. In the basis where A is diagonal the inertia operator
and the kinetic energy take a very simple form
J i = I i Ω i ,
T = 1 3∑
I i Ω 2 i .
2
The axes of this particular coordinate system are called the principle inertia axes.
Problem. Rewrite expression for energy via the quantities of the stationary frame k.
We have
i=1
E = 1 2 (AΩ, Ω) = 1 2 (J, Ω) = 1 2 (M, ω) = m 2 (q × (q × ω), ω) = m (q × ω, q × ω)
2
= 1 (ω
2 m 2 q 2 − (ωq) 2) = 1 )
2 ω iω j m
(x 2 i δ ij − x i x j .
} {{ }
inertia tensor
3 We cannot use the Liouville theorem to derive this result, because the integrals M i do not
commute with each other and, therefore, the Frobenious theorem cannot be applied to deduce that
the level set is a smooth manifold. Nevertheless we can identify the Liouville torus by different
means.
– 22 –

2.3.3 Euler’s top
Consider the motion of a rigid body around a fixed point O. Let J and Ω will be
the vector of angular momentum and the angular momentum in the body, i.e. in the
moving coordinate system K. We have AΩ = J, where A is the inertia tensor. The
angular momentum M = B t J of the body in space is preserved. Thus, we have
0 = Ṁ = ḂJ + B ˙ J = ḂB−1 M + B ˙ J = ω × M + B ˙ J = B
(
Ω × J + J ˙
)
.
From here we find
dJ
dt = J × Ω = J × A−1 J .
These are the famous Euler equations which describe the motion of the angular momentum
insider the rigid body. If one takes the coordinate adjusted to the principle
axes then one gets the following system of equations
dJ 1
= a 1 J 2 J 3 ,
dt
dJ 2
= a 2 J 3 J 1 ,
dt
dJ 3
= a 3 J 1 J 2 .
dt
Here
a 1 = I 2 − I 3
, a 2 = I 3 − I 1
, a 3 = I 1 − I 2
.
I 2 I 3 I 1 I 3 I 1 I 2
In this way the Euler equations can be viewed as equations for the components of
the angular momentum insider the body.
Consider the energy
H = 1 2 (J, A−1 J) = 1 2
3∑
i=1
J 2 i
I i
.
It is easy to verify explicitly that it is conserved due to eoms:
3∑ J
(
i a1
Ḣ = J i = J 1 J 2 J 3 + a 2
+ a )
3
= 0 .
I i I 1 I 2 I 3
i=1
Verify the conservation of the length of the angular momentum
3∑
(
)
J˙
2 = J i J˙
i = J 1 J 2 J 3 a 1 + a 2 + a 3 = 0 .
i=1
This is of course agrees with the fact that M is conserved and that M 2 = J 2 . Thus,
we have proved that the Euler equations have two quadratic integrals: the energy
and M 2 = J 2 . Thus, J lies on the intersection of an ellipsoid and a sphere:
2E = J 2 1
I 1
+ J 2 2
I 2
+ J 2 3
I 3
, J 2 = J 2 1 + J 2 2 + J 2 3 .
– 23 –

One can study the structure of the curves of intersection by fixing the ellipsoid E > 0
and changing the radius J of the sphere.
Note that alternatively the Euler equations can be rewritten as the equations for
the angular velocity Ω:
dΩ 1
dt + I 3 − I 2
Ω 2 Ω 3 = 0 ,
I 1
dΩ 2
dt + I 1 − I 3
Ω 3 Ω 1 = 0 ,
I 2
dΩ 3
dt + I 2 − I 1
Ω 1 Ω 2 = 0.
I 3
We could express Ω 1 and Ω 3 from the conservation laws
Ω 2 1
(
)
1 =
(2EI 3 − J 2 ) − I 2 (I 3 − I 2 )Ω 2 2 ,
I 1 (I 3 − I 1 )
Ω 2 1
(
)
3 =
(J 2 − 2EI 1 ) − I 2 (I 2 − I 1 )Ω 2 2 .
I 3 (I 3 − I 1 )
Then plugging this into the Euler equation for Ω 2 we obtain
dΩ 2
dt
=
√
1 ( )(
)
√ (2EI 3 − J 2 ) − I 2 (I 3 − I 2 )Ω 2 2 (J 2 − 2EI 1 ) − I 2 (I 2 − I 1 )Ω 2 2 .
I 2 I1 I 3
We assume that I 3 > I 2 > I 1 and further that M 2 > 2EI 2 . Then making the
substitutions
√
√
(I 3 − I 2 )(J
τ = t
− 2EI 1 )
I 2 (I 3 − I 2 )
, s = Ω 2
I 1 I 2 I 3 2EI 3 − J 2
and introducing the positive parameter k 2 < 1 by
we obtain
k 2 = (I 2 − I 1 )(2EI 3 − J 2 )
(I 3 − I 2 )(J 2 − 2EI 1 )
τ =
∫ s
0
ds
√
(1 − s2 )(1 − k 2 s 2 ) .
The initial time τ = 0 is chosen such that for s = 0 one has Ω 2 = 0. Inverting the
last integral one gets the Jacobi elliptic function 4
Using two other elliptic functions
s = sn τ .
cn 2 τ + sn 2 τ = 1 , dn 2 τ + k 2 sn 2 τ = 1
4 Elliptic functions were first applied to this problem in Rueb, Specimen inaugural, Utrecht, 1834.
– 24 –

we obtain the solution
Ω 1 =
Ω 2 =
Ω 3 =
√
2EI 3 − J 2
I 1 (I 3 − I 1 ) cn τ ,
√
2EI 3 − J 2
I 2 (I 3 − I 1 ) sn τ ,
√
J 2 − 2EI 1
I 3 (I 3 − I 1 ) dn τ .
Period of all these three elliptic functions is given by 4K, where K is the complete
elliptic integral of the first kind:
K =
∫ 1
0
ds
√
(1 − s2 )(1 − k 2 s 2 ) .
Period in time t is therefore given by
√
I 1 I 2 I 3
T = 4K
(I 3 − I 2 )(J 2 − 2EI 1 ) .
After this time both Ω and J will return to their original values. Thus, Ω or J
perform a strictly periodic motion. What is remarkable, is that the top itself does
not return in its original position in the stationary coordinate system k.
We have obtained that the angular momentum J moves periodically with the
period T . On the other hand, we know that the Liouville torus has the dimension
two! This means that the actual motion of the body should be parameterized by two
frequencies ω 1,2 . Let us express the angular velocity Ω via the Euler angles and their
derivatives. Let x 1 , x 2 , x 3 be the axes of the moving frame k. Components of ˙θ on
x 1 are
˙θ 1 = ˙θ cos ψ , ˙θ2 = − ˙θ sin ψ , ˙θ3 = 0 .
The velocity ˙φ is directed along Z. Its projections are
˙φ 1 = ˙φ sin θ sin ψ , ˙φ2 = ˙φ sin θ cos ψ , ˙φ3 = ˙φ cos θ .
Finally, the velocity ˙ψ is directed along x 3 . Thus, we can write the components of
the angular velocity in the moving frame as
Ω 1 = ˙φ sin θ sin ψ + ˙θ cos ψ
Ω 2 = ˙φ sin θ cos ψ − ˙θ sin ψ
Ω 3 = ˙φ cos θ + ˙ψ .
– 25 –

Substituting these formula into the expression for the kinetic energy T = 1I 2 iΩ 2 i
obtain the kinetic energy in terms of the Euler angles.
we
Problem. By using Euler angles relate the angular momenta in the moving and the stationary
coordinate systems. The momentum M is directed along the Z axis of the stationary coordinate
system.
We have
M sin θ sin ψ = I 1 Ω 1 ,
M sin θ cos ψ = I 2 Ω 2 ,
M cos θ = I 3 Ω 3 .
From here
cos θ = I 3Ω 3
M , tan ψ = I 1Ω 1
I 2 Ω 2
.
Solution of the last problem allows one to find
√
I 3 (M
cos θ =
2 − 2EI 1 )
dn τ ,
M 2 (I 3 − I 1 )
√
I 1 (I 3 − I 2 ) cn τ
tan ψ =
I 2 (I 3 − I 1 ) sn τ .
Thus, both angles θ and ψ are periodic functions of time with the period T (the same
period as for Ω!). However, the angle φ does not appear in the formulas relating the
angular momenta in the moving and the stationary coordinate systems. We can find
it from
Ω 1 = ˙φ sin θ sin ψ + ˙θ cos ψ
Ω 2 = ˙φ sin θ cos ψ − ˙θ sin ψ .
Solving we get
˙φ = Ω 1 sin ψ + Ω 2 cos ψ
.
sin θ
This leads to the differential equation
dφ
dt = M I 1Ω 2 2 + I 2 Ω 2 2
.
I1Ω 2 2 1 + I2Ω 2 2 2
Thus, solution is given by quadrature but the integrand contains elliptic functions in
a complicated way. One can show that the period of φ, which is T ′ is not comparable
with T . This leads to the fact that the top never returns to its original state. The
periods T and T ′ are the periods of motion over the Liouville torus.
– 26 –

2.3.4 On the Jacobi elliptic functions
Consider a trigonometric integral
y = sin −1 x =
∫ x
0
∫
dy
arcsin φ
√ = d sin φ
√
1 − y
2 1 − sin 2 φ .
If −1 ≤ Rex ≤ 1 this integral coincides with the function y = arcsin x.
0
Upper−half plane
.>
>
−1
>
0
sin −1
>
+1
>
Image of the
upper−half plane
−
+
−
− Pi/2
> >
0
Pi/2
+
−
+
This integral maps the punctured (at ±1)upper-half plane one-to-one onto the shaded
strip. The integral is inverted by the function sin which is period with the period
∫ 1
dy
2π = 4 × complete integral √ .
0 1 − y
2
Thus, sin can be viewed as the function on the complex cylinder X = C/L with
L = 2πZ . It also gives a Riemann map of the strip |x| < π , y > 0 to the upper-half
2
plane, standardized by the values 0, 1, ∞ at 0, π, i∞. 2
It is remarkable discovery of Gauss and Abel that the same picture holds for the
incomplete integral of the first kind:
x →
∫ x
0
dy
√
(1 − y2 )(1 − k 2 y 2 ) .
The case k = 0 is trigonometric we have discussed above. The novel point is that
for k 2 ≠ 0, 1 the inversion of the integral now leads to an elliptic function, that is a
single-valued function having not just one but two independent complex periods.
– 27 –

Upper−half plane
01 01 0 0 1 1 01
−1/k
−1
1
1/k
sn x
K−i K’
K+i K’
+
−K
K
The rectangle region is mapped by the Jacobi function sn x one-to-one onto the
upper-half-plane with four punctures.
The mapping of the upper half-plane onto the rectangle is such that the points
0, 1, 1/k, ∞, −1/k, −1 have the images 0, K, K + iK ′ , iK ′ , K − iK ′ , −K respectively.
The function sn x repeats in congruent blocks of four rectangles and, therefore, is
invariant under translations by ω 1 = 4K(k) and ω 3 = 2iK ′ (k). Here K and K ′ are
complete elliptic integrals (K ′ is called complementary)
K =
K ′ =
∫ 1
0
∫ 1/k
1
dy
√
(1 − y2 )(1 − k 2 y 2 )
∫
dy
1
√
(1 − y2 )(1 − k 2 y 2 ) =
where k = √ 1 − k 2 is the complementary modulus.
Writing
x =
∫ sn x
and differentiating over x we will get
0
1 =
dy
√
(1 − y2 )(1 − k 2 y 2 )
sn ′ x
√
(1 − y2 )(1 − k 2 y 2 )
0
dy
√
(1 − y2 )(1 − k ′2 y 2 ) ,
or
(sn ′ x) 2 = (1 − y 2 )(1 − k 2 y 2 ) .
This is differential equation satisfied by the Jacobi elliptic function sn x.
– 28 –

2.3.5 Mathematical pendulum
The theory of elliptic functions finds beautiful applications in many classical problems.
One of them is the motion of the mathematical pendulum in the gravitational
field of the Earth.
Consider the mathematical pendulum (of mass M) in the gravitational field of
the Earth.
L
01
01
M
A pendulum in the gravitational field of the Earth. Here L is its length and G is
the gravitational constant.
G
First we derive the eoms. The radius-vector and the velocity are is
⃗r(t) = (L} sin {{ θ}
, L} cos {{ θ}
) , ⃗v(t) = (L cos θ ˙θ, −L sin θ ˙θ) .
x y
Projecting the Newton equations of the axes x and y we find
Differentiating we get
L d2 cos θ
dt 2 = mg , L d2 sin θ
dt 2 = 0 .
−L(cos θ ˙θ 2 + sin θ¨θ) = mg , − sin θ ˙θ 2 + cos θ¨θ = 0 .
Excluding from these equations ˙θ 2 we obtain the equations of motion
L¨θ = −mg sin θ .
This equation can be integrated once by noting that
i.e. that
d ˙θ 2
dt = 2 ˙θ¨θ = 2 ˙θ ( − mg
L sin θ) = − 2mg
L
sin θ ˙θ = 2mg d
L dt cos θ ,
(
d
˙θ 2 − 2mg )
dt L cos θ = 0 ,
– 29 –

Thus, the combination ˙θ 2 − 2mg
L
cos θ is an integral of motion. In fact, this is nothing
else as the total energy. Indeed, the total energy is (up to an additive constant which
can be always added)
E = m⃗v2
2 + U = mL2 ˙θ 2 + mgL(1 − cos θ) .
2
We rewrite the conservation law in the form
L 2 ˙θ2 = 2gh − 4gL sin 2 θ 2 ,
where h is an integration constant. Making the change of variables y = sin θ 2
arrive at
ẏ 2 = g ( h
)
L (1 − y2 )
2L − y2 .
We have now several cases to consider
we
• Under the oscillatory motion the point does not reach the top of a circle. This
h
means that ẏ terns to zero for some y < 1. Thus, < 1. Denoting h = 2L 2Lk2 ,
where k is a positive constant less then one we obtain
) ( )
ẏ 2 =
(1 gk2 − k 2 y2
1 − y2
.
L k 2 k 2
Solution to this equation is
( √ g
)
y = k sn
L (t − t 0), k .
The integration constants are√t 0 and k, they are determined from the initial
L
conditions. the period is T = K(k). g
• Rotatory motion. Here h > 2L. Thus, taking 2L = hk 2 we will have k 2 < 1.
Equation becomes
ẏ 2 =
g
Lk (1 − 2 y2 )(1 − k 2 y 2 )
whose solution is
( √ g t − t
)
0
y = sn
, k .
L k
• The point reaches the top. Here h = 2L and we get
ẏ 2 = g L (1 − y2 ) 2 → ẏ =
√ g
L (1 − y2 ) .
Solution is
(√ ) g
y = tanh
L (t − t 0) .
– 30 –

2.4 Systems with closed trajectories
The Liouville integrable systems of phase space dimension 2n are characterized by the
requirement to have n globally defined integrals of motion F j (p, q) Poisson commuting
with each other. Taking the level set
M f = {F j = f j , j = 1, . . . n}
we obtain (in the compact case) the n-dimensional torus. In general frequencies of
motion ω j on the Liouville torus are not rationally comparable and, as the result,
the corresponding trajectories are not closed.
A special situation arises if at least two frequencies become rationally comparable.
Such a motion is called degenerate. Here we will be interested in the situation
of the completely degenerate motion, i.e. when all n frequencies ω j are comparable.
In this case the classical trajectory is a closed curve and the number of global integrals
raises to 2n − 1. 5 They cannot Poisson-commute with each other because the
maximal possible number of commuting integrals can be n only. Below we will give
already accounted examples of degenerate motion.
Two-dimensional harmonic oscillator. The Hamiltonian is
H = 1 2 (p2 1 + p 2 2) + 1 2 (ω2 1q 2 1 + ω 2 2q 2 2) .
There are two independent and mutually commuting integrals
F 1 = 1 2 p2 1 + 1 2 ω2 1q 2 1 , F 2 = 1 2 p2 2 + 1 2 ω2 2q 2 2 ,
such that H = F 1 +F 2 . If the ratio ω 1 /ω 2 is irrational the trajectories are everywhere
dense on the Liouville torus. However, if
ω 1
ω 2
= r s ,
where r, s are relatively prime integers then there is a new additional integral of
motion
F 3 = ā s 1a r 2 ,
where
ā 1 = 1 √ 2ω1
(p 1 + iω 1 q 1 ) , a 2 = 1 √ 2ω2
(p 2 − iω 2 q 2 ) .
Indeed, we have
( )
F˙
3 = ā s−1
1 a r−1
2 sa2 ˙ā 1 + rā 1 ȧ 2 .
5 In quantum mechanics we have in this case the degenerate levels.
– 31 –

Then using the eoms ˙q = p and ṗ = −ω 2 q we find
Thus,
˙ā 1 = iω 1 ā 1 , ȧ 2 = −iω 2 a 2 ,
˙ F 3 = iā s 1a r 2(
sω1 − rω 2
)
= 0 .
This integral is homogenous function of degree r + s both over the coordinates and
momenta. The trajectories are closed. They are the so-called Lissajous figures. Find
the Poisson brackets between F and F i = 1 2 (p2 i + ω 2 i q 2 i ).
The Kepler problem. We know that the orbits in the Keplerian problem are closed
for E < 0. There exists an additional conserved Runge-Lenz vector:
⃗R = ⃗v × ⃗ J − k ⃗r r .
This vector is othogonal to the angular momentum:
( ⃗ J, ⃗ R) = ( ⃗ J, ⃗v × ⃗ J) − k r ( ⃗ J, ⃗r) = 0 − 0 = 0 .
Thus, there are five independent integrals of motion in the system with six phasespace
degrees of freedom. The Kepler Hamiltonian can be expressed via these five
quantities. Thus, the motion is completely degenerate.
The Euler top. The phase space has dimension six. We found four globally defined
conserved quantities: the Hamiltonian and three components of the angular momentum.
That is the reason why the Liouville torus has dimension two instead of three.
Since 6 − 4 = 2 ≠ 1 the motion is partially, but not completely degenerate.
3. Lax pairs and classical r-matrix
In this section we will study the cornerstone concepts of the modern theory of integrable
systems: the Lax pairs and classical r-matrix.
3.1 Lax representation
Let L, M be two matrices which are also functions on the phase space, i.e. L ≡ L(p, q)
and M = M(p, q), such that the Hamiltonian equations of motion can be written in
the form
˙L = [M, L] .
This is the Lax representation (the Lax pair) of the Hamiltonian equations. The
importance of this representation lies in the fact that it provides a straightforward
construction of the conserved quantities:
I k = trL k .
– 32 –

Indeed,
˙ I k = ktr(L k−1 ˙L) = ktr(L k−1 [M, L]) = tr[M, L k ] = 0 .
In fact solution of the Lax equation is
L(t) = g(t)L(0)g(t) −1 ,
where an invertible matrix g(t) is determined from the equation
M(t) = ġg −1 .
By the Newton theorem, the integrals I k are the functions of the eigenvalues of the
matrix L. The evolution of the system is called isospectral because the eigenvalues
of the matrix L are preserved in time. A Lax pair is not uniquely defined.
Problem. Show that if g is any invertible matrix then
L = gLg −1 ,
M = gMg −1 + ġg −1
also defines a Lax pair. We have
˙L = ġLg −1 + g[M, L]g −1 − gLg −1 ġg −1 = [gMg −1 + ġg −1 , gLg −1 ] = [M, L] .
A simple example of a dynamical system which possesses the Lax pair is provided
by the harmonic oscillator. One can take
( )
( p ωq
0 −
1
L =
, M =
ω )
2
1
ωq −p
ω 0 .
2
Indeed,
( ) ṗ ω ˙q
=
ω ˙q −ṗ
( 0 −
1
ω ) ( )
2 p ωq
1
ω 0 −
ωq −p
2
( p ωq
ωq −p
) ( 0 −
1
ω ) ( )
2 −ω 2 q ωp
1
ω 0 =
ωp ω 2 q
2
and we get the eoms of the harmonic oscillator ˙q = p and ṗ = −ω 2 q. The Hamiltonian
is H = 1 4 trL2 .
Obviously the Lax representation makes no reference to the Poisson structure.
We can find however the general form of the Poisson bracket between the matrix
elements of L which ensures that the conserved eigenvalues of L are in involution.
Suppose that L is diagonalizable
One has
L = UΛU −1 .
{L 1 , L 2 } = {U 1 Λ 1 U1 −1 , U 2Λ 2 U2 −1 } =
= {U 1 , U 2 }Λ 1 U1 −1 Λ 2U2
−1
} {{ } +U 1{Λ 1 , U 2 }U1 −1 Λ 2U2 −1 − U 1 Λ 1 U1 −1 {U 1, U 2 }U1 −1 Λ 2U2
−1
} {{ }
+U 2 {U 1 , Λ 2 }Λ 1 U1 −1 U 2 −1 − U 1 Λ 1 U 2 U1 −1 {U 1, Λ 2 }U1 −1 U 2
−1
− U 2 Λ 2 U2 −1 {U 1, U 2 }U 2 −1 Λ1U 1
−1
} {{ } −U 2Λ 2 U2 −1 U 1{Λ 1 , U 2 }U −1
2 U −1
1 + U 1 Λ 1 U1 −1 U 2Λ 2 U2 −1 {U 1, U 2 }U1 −1 U 2
−1
} {{ } ,
– 33 –

where we have assumed that the eigenvalues commute {Λ 1 , Λ 2 } = 0. Introducing
k 12 = {U 1 , U 2 }U1 −1 U2 −1 , q 12 = U 2 {U 1 , Λ 2 }U1 −1 U2 −1 , q 21 = U 1 {U 2 , Λ 1 }U1 −1 U2
−1
we could write
{L 1 , L 2 } = k 12 L 1 L 2 + L 1 L 2 k 12 − L 1 k 12 L 2 − L 2 k 12 L 1
This bracket can be further written as
− q 21 L 2 + q 12 L 1 − L 1 q 12 + L 2 q 21 .
{L 1 , L 2 } = [k 12 L 2 − L 2 k 12 , L 1 ] + [q 12 , L 1 ] − [q 21 , L 2 ]
= 1 2 [[k 12, L 2 ], L 1 ] − 1 2 [[k 21, L 1 ], L 2 ] + [q 12 , L 1 ] − [q 21 , L 2 ]
= [r 12 , L 1 ] − [r 21 , L 2 ] ,
where we have introduced the so-called r-matrix
r 12 = q 12 + 1 2 [k 12, L 2 ] .
Finally, the Jacobi identity for the bracket yields the following constraint on r:
[L 1 , [r 12 , r 13 ] + [r 12 , r 23 ] + [r 32 , r 13 ] + {L 2 , r 13 } − {L 3 , r 12 }] + cycl. perm = 0
Solving this equation for r is equivalent to classifying integrable systems. If r is
constant, i.e. independent of the dynamical variables, then only the first term is left.
In particular, the Jacobi identity is satisfied if
[r 12 , r 13 ] + [r 12 , r 23 ] + [r 32 , r 13 ] = 0 .
If r-matrix here is antisymmetric: r 12 = −r 21 then the corresponding equation is
called the classical Yang-Baxter equation.
3.2 Lax representation with a spectral parameter
Here we introduce the Lax matrices L(λ), M(λ) which depend analytically on a
parameter λ called a spectral parameter. We start by considering example of the
Euler top. Introduce two 3 × 3 anti-symmetric matrices
⎛
⎞
⎛
⎞
0 −J 3 −J 2
0 −Ω 3 −Ω 2
J = ⎝
−J 3 0 J 1
J 2 −J 1 0
⎠ , Ω = ⎝
Ω 3 0 Ω 1
Ω 2 −Ω 1 0
Then we can see that the Euler equations are equivalent to the following Lax representation
dJ
= [Ω, J] .
dt
⎠ .
– 34 –

i.e. L = J and M = Ω. However, trL n either vanish or are functions of J 2 and,
therefore, they do not contain the Hamiltonian. This can be cured by introducing
the diagonal matrix I:
⎛
1
(I ⎞
2 2 + I 3 − I 1 ) 0 0
I = ⎝
1
0 (I 2 1 + I 3 − I 2 ) 0 ⎠ .
1
0 0 (I 2 1 + I 2 − I 3 )
One can see that
J = IΩ + ΩI .
Assuming that all I i are different we introduce
Then we write the equation
which reduces to
We see that
L(λ) = I 2 + 1 J , M(λ) = λI + Ω .
λ
˙L(λ) = [M(λ), L(λ)]
1
J
λ ˙ = [λI + Ω, I 2 + 1 λ J] = [Ω, I2 ] + [I, J] + 1 [Ω, J]
λ
[Ω, I 2 ] + [I, J] = ΩI 2 − I 2 Ω + I(IΩ + ΩI) − (IΩ + ΩI)I = 0 .
Thus, vanishing of the 1/λ-term gives the Euler equations of motion. This Lax pair
produces the Hamiltonian among the conserved quantities. We have
trL(λ) 2 = trI 4 − 2 λ 2 J 2
trL(λ) 3 = trI 6 − 3 λ 2 ( 1
4 (trI)2 J 2 − I 1 I 2 I 3 H
)
.
The Euler-Arnold equations. The three-dimensional Euler top admits natural generalization
to the so(n) Lie algebra. Let Ω ∈ so(n) and I is a diagonal matrix.
Then
J = IΩ + ΩI .
is also skew-symmetric matrix: J t = −J. Assuming that all eigenvalues of I are
different we introduce
L(λ) = I 2 + 1 J , M(λ) = λI + Ω .
λ
Equations
˙ J = [J, Ω] ,
J = IΩ + ΩI
– 35 –

are called the Euler-Arnold equations. They are equivalent to the spectral-dependent
Lax equations
d
(I 2 + 1 )
dt λ J = [λI + Ω, I 2 + 1 λ J] .
The later are known as the Manakov equations.
The Kepler problem. Another interesting Lax pair can be found for the Kepler
problem (M.Antonowicz and S.Rauch-Wojciechowski). Introduce the following L
and M matrices which depend on three different parameters λ 1 , λ 2 , λ 3 :
⎛
L = 1 ⎜
⎝
2
− ∑ 3
i=k
− ∑ 3
i=k
x k ẋ k
λ−λ k
ẋ k ẋ k
λ−λ k
∑ 3
i=k
∑ 3
i=k
⎞
x k x k
λ−λ k
x k ẋ k
λ−λ k
⎟
⎠ , M =
( ) 0 1
,
k
0
r 3
where r = √ x 2 1 + x 2 2 + x 2 3 and x k are coordinates of the particle, while p k = x˙
k
are the corresponding conjugate momenta. Newton’s equation for x k arises as the
condition of vanishing of the residue of the pole λ = λ k .
3.3 The Zakharov-Shabat construction
There is no general algorithm how to construct a Lax pair for a given integrable
system. However, there is a general procedure of how to construct consistent Lax
pairs giving rise to integrable systems. This is a general method how to construct
the spectral dependent matrices L(λ) and M(λ) such that
˙L(λ) = [M(λ), L(λ)]
are equivalent to the eoms of an integrable system.
The basic idea of the Zakharov-Shabat construction is to specify the analytic properties
of the matrices L(λ) and M(λ) for λ ∈ C.
Let f(λ) be a matrix-valued function which has poles at λ = λ k ≠ ∞ of order
n k . We can write
f(λ) = f 0 }{{}
const
+ ∑ k
f k (λ),
f k (λ) =
} {{ }
polar part
∑−1
r=−n k
f k,r (λ − λ k ) r .
Around any λ k this function can be decomposed as
f(λ) = f + (λ) + f − (λ) ,
where f + (λ) is regular at λ = λ k and f − (λ) = f k (λ) is the polar part.
– 36 –

Assume that L(λ) and M(λ) are rational functions of λ. Let {λ k } be the set of
poles of L(λ) and M(λ). Assuming no poles at infinity we can write
L(λ) = L 0 + ∑ k
M(λ) = M 0 + ∑ k
L k (λ) , L k (λ) =
M k (λ) , M k (λ) =
∑−1
r=−n k
L k,r (λ − λ k ) r
∑−1
r=−m k
M k,r (λ − λ k ) r .
Here L k,r and M k,r are matrices and we assume that λ k do not depend on time.
Looking at the Lax equation we see that at λ = λ k the l.h.s. has a pole of order
n k , while the r.h.s. has a potential pole of the order n k + m k . Hence there are two
type of equations. The first type does not contain the time derivatives and comes
from setting to zero the coefficients of the poles of order greater than n k on the r.h.s.
of the equation. This gives m k constraints on the matrix M k . The equations of the
second type are obtained by matching the coefficients of the poles of order less or
equal to n k . These are equations for the dynamical variables because they involve
time derivatives.
Consider the matrix L(λ) around λ = λ k . Then the matrix Q(λ) = (λ − λ k ) n k L(λ)
is regular around λ k , i.e.
Q(λ) = (λ − λ k ) n k
L(λ) = Q 0 + (λ − λ k )Q 1 + (λ − λ k ) 2 Q 2 + · · ·
Such a matrix can be always diagonalized by means of a regular similarity transformation
g(λ)Q(λ)g(λ) −1 = D(λ) = D 0 + (λ − λ k )D 1 + · · · .
Indeed, regularity means that
and, therefore,
g(λ) = g 0 + (λ − λ k )g 1 + (λ − λ k ) 2 g 2 + · · ·
g(λ) −1 = h 0 + (λ − λ k )h 1 + (λ − λ k ) 2 h 2 + · · ·
I = g(λ)g(λ) −1 =
(
)(
)
= g 0 + (λ − λ k )g 1 + (λ − λ k ) 2 g 2 + · · · h 0 + (λ − λ k )h 1 + (λ − λ k ) 2 h 2 + · · ·
= g 0 h 0 + (λ − λ k )(g 0 h 1 + g 1 h 0 ) + · · ·
This allows to determine recurrently the inverse element
h 0 = g −1
0 , h 1 = −g −1
0 g 1 g −1
0 , etc.
– 37 –

Thus,
g(λ)Q(λ)g(λ) −1 =
(
= g 0 + (λ − λ k )g 1 + · · ·
(
= g 0 Q 0 g0 −1 + (λ − λ k )
Thus, we see that g 0 must diagonalize Q 0 :
and g 1 is found from the condition that
)(
)(
Q 0 + (λ − λ k )Q 1 + · · · g −1
g 0 Q 1 g0 −1 + g 1 Q 0 g0 −1 − g 0 Q 0 g0 −1 g 1 g0
−1
D 0 = g 0 Q 0 g −1
0
)
0 − (λ − λ k )g0 −1 g 1 g0 −1 + · · ·
)
+ · · ·
g 0 Q 1 g −1
0 + g 1 Q 0 g −1
0 − g 0 Q 0 g −1
0 g 1 g −1
0 = g 0 Q 1 g −1
0 + [g 1 g −1
0 , D 0 ]
is diagonal. The commutator of a diagonal matrix with any matrix is off-diagonal.
Thus, [g 1 g0 −1 , D 0 ] is off-diagonal and the matrix g 1 is found from the condition that
[g 1 g0 −1 , D 0 ] kills the off-diagonal elements of g 0 Q 1 g0 −1 . Thus,
D(λ) = D 0 + (λ − λ k ) ( )
g 0 Q 1 g0
−1 E ii ii + · · ·
Thus, we have shown that by means of a regular similarity transformation around
the pole λ = λ k the Lax matrix can be brought to the diagonal form
A(λ) =
∑−1
A
}{{} k,r
r=−n k diag
(λ − λ k ) r + regular
The diagonalizing matrix g(λ) is defined up to right multiplication by an arbitrary
analytic diagonal matrix.
Define the matrix B(λ) as
M(λ) = g(λ)B(λ)g(λ) −1 +
˙ g(λ)g(λ) −1 ,
where g(λ) is a regular matrix which diagonalizes L(λ) around λ = λ k . The Lax
representation implies that
Ȧ(λ) = [B(λ), A(λ)] .
Since A(λ) is diagonal then A(λ) ˙ = 0, i.e., A(λ) comprises integrals of motion!
Further, the consistency of the Lax equation implies that B(λ) is a diagonal matrix
as well.
We have
L k = (g (k) A (k) g (k)−1 ) − , M k = (g (k) B (k) g (k)−1 ) − .
– 38 –

We see that because g (k) is regular the matrices L k and M k depend only on the
singular part of A (k) and B (k) . Also expanding
g (k) =
n∑
k −1
r=0
g k,r (λ − λ k ) r + higher powers
we see that only terms with r = 0, · · · , n k − 1 contribute to the singular parts of L k
and M k .
The discussion above allows one to establish the independent degrees of freedom
of the Lax pair For every pole λ k these are two singular diagonal matrices
A (k)
− =
∑−1
A
}{{} k,r
r=−n k diag
(λ − λ k ) r , B (k)
∑−1
− = B
}{{} k,r
r=−m k diag
(λ − λ k ) r
and a regular matrix G (k) of the order n k − 1, defined up to right multiplication by
a regular diagonal matrix
G (k) =
n∑
k −1
r=0
g k,r (λ − λ k ) r ,
plus, in addition two constant matrices L 0 and M 0 .
reconstructed from these data as
The L and M matrices are
L(λ) = L 0 + ∑ k
M(λ) = M 0 + ∑ k
L k (λ) ,
M k (λ) ,
L k (λ) = (G (k) A (k)
− G (k)−1 ) −
M k (λ) = (g (k) B (k)
− g (k)−1 ) − .
Note that g (k) is determined by G (k) . In other words, with G (k) one constructs L(λ)
and then diagonalize it around pole λ k which produces the whole series g (k) . These
series is then used to build M k .
Since L(λ) and M(λ) are rational functions we can easily count the number
of independent variables and the number of equations. The independent variables
contained in L are L 0 and L k,r , r = 1, · · · , n k (i.e. for each k there are n k matrices).
The independent variables contained in M are M 0 and M k,r , r = 1, · · · , m k (i.e. for
each k there are m k matrices). Thus, a counting in units of N 2 , which is the size of
matrices, gives
number of variables = 2 }{{}
L 0 ,M 0
+ ∑ k
number of equations = 1 }{{}
constant part
+ ∑ k
n k + ∑ k
m k = 2 + l + m
(n k + m k ) = 1 + l + m .
} {{ }
number of poles
– 39 –

We see that the there is one more variable than the number of equations which
reflects the gauge invariance of the Lax equation. On the Riemann surfaces of the
higher genus the situation changes and the number of equations is always bigger than
the number of independent variables.
The general solution of the non-dynamic constraints on M(λ) has the form
M = M 0 + ∑ k
M k , M k = P (k) (L, λ) − ,
where P (k) (L, λ) is a polynomial in L(λ) with coefficients rational in λ and P (k) (L, λ) −
is its singular part at λ = λ k . Indeed, assuming that this is a solution we have
[M k , L] − = [P (k) (L, λ) − , L] − = [P (k) (L, λ) − P (k) (L, λ) + , L] − = −[P (k) (L, λ) + , L] −
but the r.h.s. here has poles of degree n k and less. Let us show that this is the
general solution. Recall that A (k) (λ) is a diagonal N × N matrix with all its matrix
elements distinct at λ = λ k . Its powers
( ) 0 ( ) N−1
A (k) (λ) , · · · , A (k) (λ)
span the space of all diagonal matrices. Thus,
B (k) (λ) = P (k) (A (k) (λ), λ) ,
where P (k) (A (k) ) is a polynomial of degree N − 1 in A (k) . Substituting this into the
formula for M k we get
M k = (g (k) B (k)
− g (k)−1 ) − = (g (k) P (k) (A (k) (λ), λ)g (k)−1 ) − = P (k) (L, λ) − .
The coefficients of P (k) are rational functions of the matrix elements of A (k) and B (k)
and therefore they admit the Laurent expansion in λ − λ k .
The following situation takes place:
• Dynamical variables are the elements of L. Choosing the number and the order
of poles of the Lax matrix amounts to specifying a particular model.
• Choosing the polynomials P (k) (L, λ) is equivalent to specifying the the dynamical
flows (one of the Hamiltonians).
The Euler top. For the Euler top we have
L(λ) = I 2 + 1 J , M(λ) = λI + Ω .
λ
– 40 –

We can add to M a polynomial of L to shift the pole in λ from infinity to the zero
point. In fact one has to take
P (L) = λ(αL 2 + βL + γ) ,
where
α = − 1
I 1 I 2 I 3
,
β = I2 1 + I 2 2 + I 3 3
2I 1 I 2 I 3
,
With this choice we get
γ = (I 1 + I 2 + I 3 )(I 2 + I 3 − I 1 )(I 1 + I 2 − I 2 )(I 1 + I 2 − I 3 )
16I 1 I 2 I 3
.
M(λ) → λI + Ω − P (L) = Ω − α(I 2 J + JI 2 ) − βJ − α } {{ } λ J 2 .
=0
Thus we have a new Lax pair
L(λ) = I 2 + 1 λ J , M(λ) = −α λ J 2 .
Check
˙L = 1 λ ˙ J = [M, L] = − α λ [I2 , J 2 ]
Thus, we should get
These are precisely the Euler equations
J ˙ = − 1 [I 2 , J 2 ] .
I 1 I 2 I 3
Here
dJ 1
dt = a 1J 2 J 3 ,
dJ 2
dt = a 2J 3 J 1 ,
dJ 3
dt = a 3J 1 J 2 .
a 1 = I 2 − I 3
I 2 I 3
, a 2 = I 3 − I 1
I 1 I 3
, a 3 = I 1 − I 2
I 1 I 2
.
The eigenvalues of J are (0, i√
⃗J 2
, −i√
⃗J 2
) and they are non-dynamical since ⃗ J 2
belongs to the center of the Poisson structure.
4. Two-dimensional integrable PDEs
Here we introduce some interesting examples of infinite-dimensional Hamiltonian
systems which appear to be integrable.
– 41 –

4.1 General remarks
Remarkably, there exist certain differential equations for functions depending on two
variables (x, t) which can be treated as integrable Hamiltonian systems with infinite
number of degrees of freedom. This is an (incomplete) list of such models
• The Korteweg-de-Vries equation
• The non-linear Schrodinger equation
∂u
∂t = 6uu x − u xxx .
i ∂ψ
∂t = −ψ xx + 2κ|ψ| 2 ψ ,
where ψ = ψ(x, t) is a complex-valued function.
• The Sine-Gordon equation
• The classical Heisenberg magnet
∂ 2 φ
∂t − ∂2 φ
2 ∂x + m2
sin βφ = 0
2 β
∂ ⃗ S
∂t = ⃗ S × ∂2 ⃗ S
∂x 2 ,
where ⃗ S(x, t) lies on the unit sphere in R 3 .
The complete specification of each model requires also boundary and initial conditions.
Among the important cases are
1. Rapidly decreasing case. We impose the condition that
ψ(x, t) → 0 when |x| → ∞
sufficiently fast, i.e., for instance, it belongs to the Schwarz space L (R 1 ), which
means that ψ is differentiable function which vanishes faster than any power
of |x| −1 when |x| → ∞.
2. Periodic boundary conditions. Here we require that ψ is differentiable and
satisfies the periodicity requirement
ψ(x + 2π, t) = ψ(x, t) .
– 42 –

The soliton was first discovered by accident by the naval architect, John Scott Russell,
in August 1834 on the Glasgow to Edinburg channel. 6 The modern theory originates
from the work of Kruskal and Zabusky in 1965. They were the first ones to call
Russel’s solitary wave a solition.
4.2 Soliton solutions
Here we discuss the simplest cnoidal wave type (periodic) and also one-soliton solutions
of the KdV and SG equations For the discussion of the cnoidal wave and
the one-soliton solution of the non-linear Schrodinger equation see the corresponding
problem in the problem set.
4.2.1 Korteweg-de-Vries cnoidal wave and soliton
By rescaling of t, x and u one can bring the KdV equation to the canonical form
u t + 6uu x + u xxx = 0 .
We will look for a solution of this equation in the form of a single-phase periodic
wave of a permanent shape
u(x, t) = u(x − vt) ,
where v = const is the phase velocity. Plugging this ansatz into the equation we
obtain
−vu x + 6uu x + u xxx = d ( )
− vu + 3u 2 + u xx = 0 .
dx
We thus get
−vu + 3u 2 + u xx + e = 0 ,
where e is an integration constant. Multiplying this equation with an integrating
factor u x we get
−vuu x + 3u 2 u x + u x u xx + eu x = d (
− v dx 2 u2 + u 3 + 1 )
2 u2 x + eu = 0 ,
6 Russel described his discovery as follows: “I believe I shall best introduce this phenomenon by
describing the circumstances of my own first acquaintance with it. I was observing the motion of a
boat which was rapidly drawn along a narrow channel by a pair of horses, when the boat suddenly
stopped-not so the mass of the water in the channel which it had put in motion; it accumulated
round the prow of the vessel in a state of violent agitation, then suddenly leaving it behind, rolled
forward with great velocity, assuming the form of a large solitary elevation, a rounded, smooth
and well-defined heap of water, which continued its course along the channel apparently without
change of form or diminution of speed. I followed it on horseback, and overtook it still rolling on
at a rate of some eight or nine miles an hour, preserving its original figure some thirty feet along
and a foot or foot and a half in height. Its height gradually diminished, and after a chase of one
or two miles I lost it in the windings of the channel. Such, in the month of August 1834, was my
first chance interview with that singular and beautiful phenomenon which I have called the Wave
of Translation, a name which it now very generally bears.
– 43 –

We thus obtain
u 2 x = k − 2eu + vu 2 − 2u 3 = −2(u − b 1 )(u − b 2 )(u − b 3 ) ,
where k is another integration constant. In the last equation we traded the integration
constants e, k for three parameters b 3 ≥ b 2 ≥ b 1 which satisfy the relation
Equation
v = 2(b 1 + b 2 + b 3 ) .
u 2 x = −2(u − b 1 )(u − b 2 )(u − b 3 ) ,
describes motion of a ”particle” with the coordinate u and the time x in the potential
V = 2(u − b 1 )(u − b 2 )(u − b 3 ). Since u 2 x ≥ 0 for b 2 ≤ u ≤ b 3 the particle oscillates
between the end points b 2 and b 3 with the period
l = 2
∫ b3
b 2
du
√
−2(u − b1 )(u − b 2 )(u − b 3 ) = 2 √ 2
K(m) ,
(b 3 − b 2 )
1/2
where m is an elliptic modulus 0 ≤ m = b 3−b 2
b 3 −b 1
≤ 1.
The equation
u 2 x = −2(u − b 1 )(u − b 2 )(u − b 3 ) ,
can be integrated in terms of Jacobi elliptic cosine function cn(x, m) to give
(√ )
u(x, t) = b 2 + (b 3 − b 2 ) cn 2 (b3 − b 1 )/2(x − vt − x 0 ), m ,
where x 0 is an initial phase. This solution is often called as cnoidal wave. When
m → 1, i.e. b 2 → b 1 the cnoidal wave turns into a solitary wave
u(x, t) = b 1 +
A
(√ ) .
cosh 2 A
(x − vt − x 2 0)
Here the velocity v = 2(b 1 + b 2 + b 3 ) = 2(2b 1 + b 3 ) = 2(3b 1 + b 3 − b 1 ) is connected to
the amplitude A = b 3 − b 1 by the relation
v = 6b 1 + 2A .
Here u(x, t) = b 1 is called a background flow because u(x, t) → b 1 as x → ±∞.
One can further note that the background flow can be eliminated by a passage to
a moving frame and using the invariance of the KdV equation w.r.t. the Galilean
transformation u → u + d, x → x − 6dt, where d is constant.
To sum up the cnoidal waves form a three-parameter family of the KdV solutions
while solitons are parametrized by two independent parameters (with an account of
the background flow).
– 44 –

4.2.2 Sine-Gordon cnoidal wave and soliton
Consider the Sine-Gordon equation
φ tt − φ xx + m2
β
sin βφ = 0 ,
where we assume that the functions φ(x, t) and φ(x, t) + 2π/β are assumed to be
equivalent. Make an ansatz
φ(x, t) = φ(x − vt)
which leads to
This can be integrated once
C = v2 − 1
φ 2 x − m2
2
(v 2 − 1)φ xx + m2
β
β cos βφ = v2 − 1
2 2
sin βφ = 0 .
φ 2 x + 2m2
β 2
βφ
sin2
2 − m2
β . 2
where C is an integration constant. This is nothing else as the conservation law of
energy for the mathematical pendulum in the gravitational field of the Earth! We
further bring equation to the form
φ 2 x = 2 (C + m2
v 2 − 1 β − 2m2
2 β 2
βφ sin2 2
As in the case of the pendulum we make a substitution y = sin βφ
2
(
C +
m 2
(y ′ ) 2 = m2
(v 2 − 1) (1 − y2 )
)
β 2
− y 2 .
2m 2
β 2
)
. (4.1)
which gives
This leads to solutions in terms of elliptic functions which are analogous to the cnoidal
waves of the KdV equation. However, as we know the pendulum has three phases
of motion: oscillatory (elliptic solution), rotatory (elliptic solution) and motion with
an infinite period. The later solution is precisely the one that would correspond to
the Sine-Gordon soliton we are interested in. Assuming v 2 < 1 we see 7 that such
a solution would arise from (4.1) if we take C = − m2 . In this case equation (4.1)
β 2
reduces to
2m
φ x =
β √ βφ
sin
1 − v2 2 .
This can be integrated to 8
4
( m(x − vt −
φ(x, t) = −ɛ 0
β arctan exp x0 )
)
√ .
1 − v
2
7 Restoring the speed of light c this condition for the velocity becomes v 2 < c 2 , i.e., the center
of mass of the soliton cannot propagate faster than light.
8 From the equation above we see that if φ(x, t) is a solution then −φ(x, t) is also a solution.
– 45 –

Here ɛ 0 = ±1. This solution can be interpreted in terms of relativistic particle moving
with the velocity v. The field φ(x, t) has an important characteristic – topological
charge
Q = β ∫
dx ∂φ
2π ∂x = β (φ(∞) − φ(−∞)) .
2π
On our solutions we have
Q = β 2π
(
4
)
− ɛ 0 ( π β 2 − 0) = −ɛ 0 ,
because arctan(±∞) = ± π and arctan 0 = 0. In addition to the continuous parameters
v and x 0 , the soliton of the SG model has another important discrete
2
characteristic – topological charge Q = −ɛ 0 . Solutions with Q = 1 are called solitons
(kinks), while solutions with Q = −1 are called ani-solitons (anti-kinks).
Here we provide another useful representation for the SG soliton, namely
m(x−vt−x
2i
φ(x, t) = ɛ 0
β log 1 + ie 0 )
√
1−v 2
1 − ie
m(x−vt−x 0
.
)
√
1−v 2
Indeed, looking at the solution we found we see that we can cast it in the form arctan α = z ≡
− β
m(x−vt−x
4ɛ 0
φ(x, t) or α = tan z = −i e2iz −1
e 2iz +1 , where α = e 0 )
√
1−v 2 . From here z = 1 1+iα
2i
log
1−iα
and the
announced formula follows.
Remark. The stability of solitons stems from the delicate balance of ”nonlinearity”
and ”dispersion” in the model equations. Nonlinearity drives a solitary wave to
concentrate further; dispersion is the effect to spread such a localized wave. If one
of these two competing effects is lost, solitons become unstable and, eventually,
cease to exist. In this respect, solitons are completely different from ”linear waves”
like sinusoidal waves. In fact, sinusoidal waves are rather unstable in some model
equations of soliton phenomena.
Sine-Gordon model has even more sophisticated solutions. Consider the following
( )
φ(x, t) = 4 β arctan ω sin mω1 (t−vx)
√
2
1−v 2
+ φ 0
) .
ω 1 cosh
(
mω2 (x−vt−x 0 )
√
1−v 2
This is solution of the SG model which is called a double-soliton or breaser. Except
motion with velocity v corresponding to a relativistic particle the breaser oscillates
both in space and in time with frequencies √ mvω 1
1−v 2
and √ mω 1
1−v 2
respectively. The parameter
φ 0 plays a role of the initial phase. In particular, if v = 0 the breaser is a
time-periodic solution of the SG equation. It has zero topological charge and can be
interpreted as the bound state of the soliton and anti-soliton.
– 46 –

4.3 Zero-curvature representation
The inverse scattering method (the method of finding certain class of solutions of
a non-linear integrable PDE) is based on the following remarkable observation. A
two-dimensional PDE appears as the consistency condition of the overdetermined
system of equations
∂Ψ
= U(x, t, λ)Ψ ,
∂x
∂Ψ
= V (x, t, λ)Ψ .
∂t
for a proper choice of the matrices U(x, t, λ) and V (x, t, λ). The consistency condition
arises upon differentiation the first equation w.r.t. t and the second w.r.t. x:
∂ 2 Ψ
(
)
∂t∂x = ∂ tU(x, t, λ)Ψ + U(x, t, λ)∂ t Ψ = ∂ t U(x, t, λ) + U(x, t, λ)V (x, t, λ) Ψ ,
∂ 2 Ψ
(
)
∂x∂t = ∂ xV (x, t, λ)Ψ + V (x, t, λ)∂ x Ψ = ∂ x V (x, t, λ) + V (x, t, λ)U(x, t, λ) Ψ ,
which implies the fulfilment of the following relation
∂ t U − ∂ x V + [U, V ] = 0 .
If we introduce a gauge field L α with components L x = U, L t = V , then the last
relation is the condition of vanishing of the curvature of L α :
F αβ (L ) ≡ ∂ α L β − ∂ β L α − [L α , L β ] = 0 .
Example: KdV equation. Introduce the following 2 × 2 matrices
( )
(
)
0 1
u
U =
, V =
x 4λ − 2u
λ + u 0
4λ 2 + 2λu + u xx − 2u 2 .
−u x
Show by direct computation that
(
)
0 0
∂ t U − ∂ x V + [U, V ] =
.
u t + 6uu x − u xxx 0
Example: Sine-Gordon equation. Introduce the following 2 × 2 matrices
U = β 4i φ tσ 3 + k 0
i
V = β 4i φ xσ 3 + k 1
i
βφ
sin
2 σ 1 + k 1 βφ
cos
i 2 σ 2
βφ
sin
2 σ 1 + k 0 βφ
cos
i 2 σ 2 ,
– 47 –

where σ i are the Pauli matrices 9 and
k 0 = m 4
(
λ + 1 )
, k 1 = m (
λ − 1 )
.
λ
4 λ
Show by direct computation that the condition of zero curvature is equivalent to the
Sine-Gordon equation.
The one-parameter family of the flat connections allows one to define the monodromy
matrix T(λ) which is the path-ordered exponential of the Lax component
U(λ):
∫ 2π
T(λ) = P exp dxU(λ) . (4.3)
0
Let us derive the time evolution equation for this matrix. We have
∂ t T(λ) =
=
∫ 2π
0
∫ 2π
0
dx Pe R 2π
x dyU (∂ t U) Pe R x
0 dyU
dx Pe R 2π
x dyU (∂ x V + [V, U]) Pe R x
0 dyU , (4.4)
where in the last formula we used the flatness of L α ≡ (U, V ). The integrand of the
expression we obtained is the total derivative
∂ t T(λ) =
∫ 2π
0
(
dx ∂ x Pe R 2π
x
Thus, we obtained the following evolution equation
dyU V Pe R x
0 dyU )
. (4.5)
∂ t T(λ) = [V (2π, t, λ), T(λ)] . (4.6)
This formula shows that the eigenvalues of T(λ) generate an infinite set of integrals of
motion upon expansion in λ. Thus, the spectral properties of the model are encoded
into the monodromy matrix.
The wording “monodromy” comes from the fact that T(t) represents the monodromy
of a solution of the fundamental linear problem:
9 The Pauli matrices are
σ 1 =
( ) 0 1
, σ 2 =
1 0
Ψ(2π, t) = T(t)Ψ(0, t) .
( )
( )
0 −i
1 0
, σ 3 = . (4.2)
i 0
0 −1
– 48 –

Indeed, if we differentiate this equation over t we get
∂ t Ψ(2π, t) = ∂ t TΨ(0, t) + T∂ t Ψ(0, t) ,
which, according to the fundamental linear system, gives
L t (2π, t)TΨ(0, t) = ∂ t TΨ(0, t) + TL t (0, t)Ψ(0, t) .
This leads to the same equation for the time evolution of the monodromy matrix as
found before:
∂ t T = [L t , T] .
4.4 Local integrals of motion
The Lax representation of the two-dimensional PDE allows one to exhibit an infinite
number of conservation laws. The procedure to derive the conservation laws from
the Lax representation is a direct analogue of the Zakharov-Shabat construction for
the finite-dimensional case. It is called the abelianization procedure.
Once again we start from the zero-curvature condition
∂ t U − ∂ x V − [V, U] = 0 .
We assume that the matrices U(x, t, λ) and V (x, t, λ) depend on the spectral parameter
λ in a rational way and they have poles at constant, i.e. x, t-independent, values
of λ k . Thus, we can write
U = U 0 + ∑ k
V = V 0 + ∑ k
U k , U k =
V k , V k =
∑−1
r=−n k
U k,r (x, t)(λ − λ r ) r ,
∑−1
r=−m k
V k,r (x, t)(λ − λ r ) r .
The same counting as in the finite-dimensional case shows that the zero-curvature
equations are always compatible: there is one more variable than the number of equations,
but there is a gauge transformation which leads the zero-curvature condition
invariant.
To understand solutions of the zero-curvature condition we will perform a local
analysis around a pole λ = λ k . Our aim is to show that around each singularity one
can perform a gauge transformation which brings the matrices U(λ) and V (λ) to a
diagonal form. Finally, to make the consideration as simple as possible we assume
that the pole is located at zero.
In the neighbourhood of λ = 0 the functions U and V can be expanded into
Laurent series
∞∑
∞∑
U(x, t, λ) = U r (x, t)λ r , V (x, t, λ) = V r (x, t)λ r .
r=−n
r=−m
– 49 –

Let g ≡ g(x, t, λ) be a regular gauge transformation around λ = 0 that is
∞∑
g = g r λ r , g −1 =
r=0
Consider the gauge transformation
∞∑
h r λ r .
r=0
Ũ = gUg −1 + ∂ x gg −1 ,
Ṽ = gV g −1 + ∂ t gg −1 .
Consider the transition matrix T(x, y, λ) which is a solution of the differential
equation (
)
∂ x − U(x, λ) T(x, y, λ) = 0
satisfying the initial condition T(x, x, λ) = I. Formally such a solution is given by
the path-ordered exponent
T(x, y, λ) = Pe R x
y dzU(z,λ) .
Under the gauge transformation we have
(
)
g(x, λ) ∂ x − U(x, λ) g −1 (x, λ)T g (x, y, λ) = 0 ,
where T g (x, y, λ) is the transition matrix for the gauged-transformed connection
which also obeys the condition T g (x, x, λ) = I. Thus, we obtain
T g (x, y, λ) = g(x, λ)T(x, y, λ)g −1 (y, λ) .
This formula shows how the transition matrix transforms under the gauge transformations
of the Lax connection. By means of a regular gauge transformation the
transition matrix can be diagonalized around every pole of the matrix U:
where
T(x, y, λ) = g(x, λ) exp(D(x, y, λ))g −1 (y, λ) ,
D(x, y, λ) =
∞∑
r=−n
D r (x, y)λ r
is the diagonal matrix. Below we consider a concrete example which illustrates the
abelianization procedure as well as the technique of constructing local integrals of
motion.
Example: The Heisenberg model. We start with the definition of the model classical
Heisenberg model. Consider a spin variable S(x):
S(x) = ∑ i
S i (x)σ i .
– 50 –

Clearly, S i (x) 2 = s 2 . Here σ i are the standard Pauli matrices obeying the relations
[σ i , σ j ] = 2iɛ ijk σ k , tr(σ j σ k ) = 2δ ij .
The spins S i (x) are the dynamical variables subject to the Poisson structure
{S i (x), S j (y)} = ɛ ijk S k (x)δ(x − y) .
The phase space is thus infinite-dimensional. Check in the class the Jacobi identity!
The Hamiltonian of the model is
H = − 1 4
∫ 2π
0
dx tr(∂ x S∂ x S)
Let us derive equations of motion (in the class!). We have
∂ t S(x) = {H, S(x)} = − 1 4
= −
∫ 2π
∫ 2π
0
dy {tr(∂ y S∂ y S), S(y)}
dy ∂ y S j (y){∂ y S j (y), S k (x)}σ k = −
∫ 2π
0
0
dy ∂ y S j (y)ɛ jki ∂ y (S i (y)δ(y − x))σ k =
= ɛ jki ∂xS 2 j (x)S i (x)σ k = ɛ ijk S i (x)∂xS 2 j (x)σ k = 1 2i [Si (x)σ i , ∂xS 2 j (x)σ j ]
Thus, equations of motion read
∂ t S = − i 2 [S, ∂2 xS] = − i 2 ∂ x[S, ∂ x S] .
If we introduce the non-abelian su(2)-current J with components
J x = S ,
J t = − i 2 [S, ∂ xS]
then the equations of motion take the form of the current conservation law:
∂ t J x − ∂ x J t = 0 ,
which is ɛ αβ ∂ α J β = 0. Equations of motion
∂ t S = − i 2 [S, ∂2 xS]
called the Landau-Lifshitz equations. In this form these equations can be generalized
to any Lie algebra. The integrability of the model relies on the fact that equations
of motion can be obtained from the condition of zero curvature:
Here
L x = − i λ S(x) ,
(∂ α − L α )Ψ(x, t) = 0 .
L t = − 2is2
λ 2 S(x) − 1
2λ [S(x), ∂ xS(x)] .
– 51 –

Indeed,
∂ t L x − ∂ x L t + [L x , L t ] = − i λ ∂ tS(x) + 2is2
λ ∂ xS(x)
2
+ 1
2λ ∂ x[S(x), ∂ x S(x)] +
i
2λ [S(x), [S(x), ∂ xS(x)]] = 0 .
2
Now one can compute the Poisson bracket between the components L x ≡ U(x, λ) of
the Lax connection. We have
{U(x, λ), U(y, µ)} = − 1
λµ {Si (x), S j (y)}σ i ⊗ σ j = − 1
λµ ɛijk S k (x)σ i ⊗ σ j δ(x − y) .
On the other hand, let us compute
[ σi ⊗ σ
]
[
i
λ − µ , U(x, λ) ⊗ I + I ⊗ U(y, µ) σi ⊗ σ i
δ(x − y) = −
λ − µ , i λ S(x) ⊗ I + I ⊗ i ]
µ S(y) δ(x − y)
= − i ( 1
λ − µ Sk (x)
λ [σ i, σ k ] ⊗ σ i + 1 µ σ i ⊗ [σ i , σ k ])
δ(x − y) =
= 2
λ − µ( 1
λ − 1 µ
)
ɛ ijk S k (x)σ i ⊗ σ j δ(x − y) = − 2
λµ ɛijk S k (x)σ i ⊗ σ j δ(x − y) .
We thus proved that the Poisson bracket between the components of the Lax connection
can be written in the form
{U(x, λ), U(y, µ)} =
[
]
r(λ, µ), U(x, λ) ⊗ I + I ⊗ U(y, µ) δ(x − y) ,
where the classical r-matrix appears to be
r(λ, µ) = 1 σ i ⊗ σ i
2 λ − µ .
This form of the brackets between the components of the Lax connection implies
that the Poisson bracket between the components of the monodromy matrix
[ ∫ 2π ]
T(λ) = P exp dx U(x, λ)
0
is
{T(λ) ⊗ T(µ)} =
[
]
r(λ, µ), T(λ) ⊗ T(µ) .
This is the famous Sklyanin bracket. It is quadratic in the matrix elements of the
monodromy matrix.
– 52 –

From the definition, T(λ) is analytic (entire) 10 in λ with an essential singularity at
λ = 0 11 It is easy to find the expansion around λ = ∞:
T(λ) = I + i λ
∫ 2π
dx S(x) − 1 ∫ 2π
dx S(x)
λ 2
∫ x
0
0
0
The development in 1/λ has an infinite radius of convergency.
dy S(y) + · · ·
To find the structure of T(λ) around λ = 0 is more delicate but very important as
it provides the local conserved charges in involution. Let us introduce the so-called
partial monodromy
[ ∫ x ]
T(x, λ) = P exp dy U(y, λ) .
The main point is to note that there exists a local gauge transformation, regular at
λ = 0, such that
T(x, λ) = g(x)D(x)g −1 (0) ,
where D(x) = exp(id(x)σ 3 ) is a diagonal matrix. We can choose g to be unitary,
and, since g is defined up to to a diagonal matrix, we can require that it has a real
diagonal part:
( )
1 1 v
g =
.
(1 + v¯v) 1 2 −¯v 1
Then the differenial equation for the monodromy
∂ x T = UT = − i λ ST
10 In complex analysis, an entire function is a function that is holomorphic everywhere on the
whole complex plane. Typical examples of entire functions are the polynomials, the exponential
function, and sums, products and compositions of these. Every entire function can be represented
as a power series which converges everywhere. Neither the natural logarithm nor the square root
function is entire. Note that an entire function may have a singularity or even an essential singularity
at the complex point at infinity. In the latter case, it is called a transcendental entire function.
As a consequence of Liouville’s theorem, a function which is entire on the entire Riemann sphere
(complex plane and the point at infinity) is constant.
11 Consider an open subset U of the complex plane C, an element a of U, and a holomorphic
function f defined on U − a. The point a is called an essential singularity for f if it is a singularity
which is neither a pole nor a removable singularity. For example, the function f(z) = exp(1/z) has
an essential singularity at z = 0. The point a is an essential singularity if and only if the limit
0
lim f(z)
z→a
does not exist as a complex number nor equals infinity. This is the case if and only if the Laurent
series of f at the point a has infinitely many negative degree terms (the principal part is an
infinite sum). The behavior of holomorphic functions near essential singularities is described by the
Weierstrass-Casorati theorem and by the considerably stronger Picard’s great theorem. The latter
says that in every neighborhood of an essential singularity a, the function f takes on every complex
value, except possibly one, infinitely often.
– 53 –

ecomes a differential equation for g and d:
g −1 ∂ x g + i∂ x dσ 3 + i λ g−1 Sg = 0 .
We project this equation on the Pauli matrices and get
∂ x v = − i λ (S − + 2vS 3 − S + v 2 )
∂ x d = 1
2λ (−2S 3 + vS + + ¯vS − ) .
The first of these equations is a Riccati equation for v(x).
functions v(x) and d(x) as
Expanding in λ the
∂ x d = − s ∞
λ + ∑
ρ n (x)λ n
v(x) =
n=0
∞∑
v n (x)λ n ,
n=0
v 0 = S 3 − s
S +
,
we rewrite the Riccati equation in the form
n∑
2isv n+1 = −v n ′ + iS + v n+1−m v m
m=1
and
ρ n = 1 2 (v n+1S + + ¯v n+1 S − ) .
Note that v(x) is regular at λ = 0. Equations above recursively determine the
functions v n (x) and ρ n (x) as local functions of the dynamical variables S i (x). This
describes the asymptotic behavior of T(λ) around λ = 0. The asymptotic series
become convergent if we regularize the model by discretizing the space interval!
Concerning the monodromy matrix T(λ), since g(x) is local and if we assume periodic
boundary conditions, we can write
where M(λ) = g(0)σ 3 g(0) −1 and
T(λ) = cos p(λ)I + i sin p(λ)M(λ) ,
p(λ) =
∫ 2π
0
dx ∂ x d .
The trace of the monodromy matrix, called the transfer matrix, is
trT(λ) = 2 cos p(λ) .
– 54 –

Thus, p(λ) is the generating function for the commuting local conserved quantities
The first three integrals are
I 0 = i
4s
I n =
∫ 2π
0
∫ 2π
I 1 = − 1
16s 3 ∫ 2π
I 2 =
0
dx ρ n (x) .
( S+
)
dx log ∂ x S 3 ,
S −
0
i ∫ 2π
64s 5
0
(
dx tr ∂ x S∂ x S
)
,
( )
dx tr S[∂ x S, ∂xS]
2 .
The integrals I 0 and I 1 correspond to momentum and energy respectively.
We conclude this section by outlining a general scheme known as Inverse Scattering
Method which allows one to construct explicitly the multi-soliton solutions of
integrable PDE’s.
PDE:
Initial data q(x,t)
Direct spectral problem
Lax representation
Monodromy
Local integrals of motion
Action−angle variables
Time evolution
in the original
configuration space
Time evolution
in the spectral space
(simple !)
Solution for t>0
q(x,t)
Inverse scattering problem
Riemann−Hlbert problem
dI
dt = 0
I −action variables
b −angle variables
b(t)=e−i
w t b(o)
INVERSE SCATTERING TRANSFORM −− NON−LINEAR ANALOG OF THE FOURIER TRANSFORM
5. Quantum Integrable Systems
In this section we consider certain quantum integrable systems. The basis tool to
solve them is known under the generic name “Bethe Ansatz”. There are several
different constructions of this type. They are
– 55 –

• Coordinate Bethe ansatz. This technique was originally introduced by H. Bethe
to solve the XXX Heisenberg model.
• Algebraic Bethe ansatz. It was realized afterwards that the Bethe ansatz can
be formulated in such a way that it can be understood as the quantum analogue
of the classical inverse scattering method. Thus, “Algebraic Bethe ansatz” is
another name for “Quantum inverse scattering method”.
• Functional Bethe ansatz. The algebraic Bethe ansatz is not the only approach
to solve the spectral problems of models connected with the Yang-Baxter algebra.
It is only applicable if there exists a pseudo-vacuum. For models like the
Toda chain, which has the same R-matrix as XXX Heisenberg magnet (spin-
1
chain), but has no pseudo-vacuum, the algebraic Bethe ansatz fails. For
2
these types of models another powerful technique, the method of “sepration of
variables” was devised by E. Sklyanin. It is also known as “Functional Bethe
Ansatz”
• Nested Bethe ansatz. The generalization of the Bethe ansatz to models with
internal degrees of freedom proved to be very hard, because scattering involves
changes of the internal states of scatters. This problem was eventually solved
by C.N. Yang and M. Gaudin by means of what is nowadays called “nested
Bethe ansatz”.
• Asymptotic Bethe anzatz Many integrable systems in the finite volume cannot
be solved by the Bethe ansatz methods. However, the Bethe ansatz provides the
leading finite-size correction to the wave function, energy levels, etc. for systems
in infinite volumes. Introduced and extensively studied by B. Sutherland.
• Thermodynamic Bethe ansatz. This method allows to investigate the thermodynamic
properties of integrable systems.
5.1 Coordinate Bethe Ansatz (CBA)
Here we will demonstrate how CBA works at an example of the so-called onedimensional
spin- 1 XXX Heisenberg model of ferromagnetism.
2
Consider a discrete circle which is a collection of ordered points labelled by the
index n with the identification n ≡ n + L reflecting periodic boundary conditions.
Here L is a positive integer which plays the role of the length (volume) of the space.
The numbers n = 1, . . . , L form a fundamental domain. To each integer n along the
chain we associate a two-dimensional vector space V = C 2 . In each vector space we
pick up the basis
( )
( )
1 0
| ↑〉 = , | ↓〉 =
0 1
– 56 –

We will call the first element “spin up” and the second one “spin down”. We introduce
the spin algebra which is generated by the spin variables S α n, where α = 1, 2, 3, with
commutation relations
[S α m, S β n] = iɛ αβγ S γ nδ mn .
The spin operators have the following realization in terms of the standard Pauli
matrices: S α n = 2 σα and the form the Lie algebra su(2). Spin variables are subject
to the periodic boundary condition S α n ≡ S α n+L .
Spin chain. A state of the spin chain can be represented as |ψ〉 = | ↑↑↓↑ · · · ↓↑〉
The Hilbert space of the model has a dimension 2 L and it is
H =
L∏
⊗V n = V 1 ⊗ · · · ⊗ V L
n=1
This space carries a representation of the global spin algebra whose generators are
S α =
L∑
I ⊗ · · · ⊗ Sn
α }{{}
⊗ · · · ⊗ I .
n−th place
n=1
The Hamiltonian of the model is
H = −J
L∑
SnS α n+1 α ,
where J is the coupling constant. More general Hamiltonian of the form
H = −J
n=1
L∑
J α SnS α n+1 α ,
n=1
where all three constants J α are different defines the so-called XYZ model. In what
follows we consider only XXX model. The basic problem we would like to solve is to
find the spectrum of the Hamiltonian H.
– 57 –

The first interesting observation is that the Hamiltonian H commutes with the spin
operators. Indeed,
[H, S α ] = −J
= −i
L∑
[SnS β n+1, β Sm] α = −J
n,m=1
L∑
n,m=1
L∑
[Sn, β Sm]S α β n+1 + Sn[S β n+1, β Sm]
α
n,m=1
( )
δnm ɛ αβγ SnS β γ n+1 − δ n+1,m ɛ αβγ SnS β γ n+1 = 0 .
In other words, the Hamiltonian is central w.r.t all su(2) generators. Thus, the
spectrum of the model will be degenerate – all states in each su(2) multiplet have
the same energy.
In what follows we choose = 1 and introduce the raising and lowering operators
S n ± = Sn 1 ± iSn. 2 They are realized as
( )
( )
0 1
0 0
S + = , S − = .
0 0
1 0
The action of these spin operators on the basis vectors are
S + | ↑〉 = 0 , S + | ↓〉 = | ↑〉 , S 3 | ↑〉 = 1 2 | ↑〉 ,
S − | ↓〉 = 0 , S − | ↑〉 = | ↓〉 , S 3 | ↓〉 = − 1 2 | ↓〉 .
This indices the action of the spin operators in the Hilbert space
S + k | ↑ k〉 = 0 , S + k | ↓ k〉 = | ↑ k 〉 , S 3 k| ↑ k 〉 = 1 2 | ↑ k〉 ,
S − k | ↓ k〉 = 0 , S − k | ↑ k〉 = | ↓ k 〉 , S 3 k| ↓ k 〉 = − 1 2 | ↓ k〉 .
The Hamiltonian can be then written as
H = −J
L∑
n=1
1
2 (S+ n Sn+1 − + Sn − S n+1) + + SnS 3 n+1 3 ,
For Lb = 2 we have
⎛ ⎞
1
)
2
0 0 0
H = −J
(S + ⊗ S − + S − ⊗ S + + 2S 3 ⊗ S 3 ⎜ 0 −
= −J⎝
1 2
1 0 ⎟
0 1 − 1 2 0 ⎠ .
0 0 0 1 2
This matrix has three eigenvalues which are equal to − 1J and one which is 3J.
2 2
Three states
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1
0
0
vs=1 hw = ⎜ 0
⎟
⎝ 0 ⎠ , ⎜ 1
⎟
⎝ 1 ⎠ ,
⎜ 0
⎟
⎝ 0 ⎠
0
} {{ }
h.w.
0
1
– 58 –

corresponding to equal eigenvalues form a representation of su(2) with spin s = 1
and the state
⎛ ⎞
0
vs=0 hw = ⎜ −1
⎟
⎝ 1 ⎠
}
0
{{ }
h.w.
which corresponds to 3 J is a singlet of su(2). Indeed, the generators of the global
2
su(2) are realized as
⎛ ⎞
⎛ ⎞
⎛ ⎞
0 1 1 0
0 0 0 0
1 0 0 0
S + = ⎜ 0 0 0 1
⎟
⎝ 0 0 0 1 ⎠ , S− = ⎜ 1 0 0 0
⎟
⎝ 1 0 0 0 ⎠ , S3 = ⎜ 0 0 0 0
⎟
⎝ 0 0 0 0 ⎠ .
0 0 0 0
0 1 1 0
0 0 0 −1
The vectors vs=1 hw and vs=0 hw are the highest-weight vectors of the s = 1 and s = 0
representations respectively, because they are annihilated by S + and are eigenstates
of S 3 . In fact, vs=0 hw is also annihilated by S − which shows that this state has zero
spin. Thus, we completely understood the structure of the Hilbert space for L = 2.
In general, the Hamiltonian can be realized as 2 L × 2 L symmetric matrix which
means that it has a complete orthogonal system of eigenvectors. The Hilbert space
split into sum of irreducible representations of su(2). Thus, for L being finite the
problem of finding the eigenvalues of H reduces to the problem of diagonalizing a
symmetric 2 L ×2 L matrix. This can be easily achieved by computer provided L is sufficiently
small. However, for the physically interesting regime L → ∞ corresponding
to the thermodynamic limit new analytic methods are required.
In what follows it is useful to introduce the following operator:
P = 1 2
(
I ⊗ I + ∑ α
) ( 1
σ α ⊗ σ α = 2
4 I ⊗ I + ∑ α
S α ⊗ S α )
which acts on C 2 ⊗ C 2 as the permutation: P (a ⊗ b) = b ⊗ a. Indeed, we have
It is appropriate to call S 3 the operator of the total spin. On a state |ψ〉 with
M spins down we have
S 3 |ψ〉 =
( 1
2 (L − M) − 1 ) ( 1
)
2 M |ψ〉 =
2 L − M |ψ〉 .
Since [H, S 3 ] = 0 the Hamiltonian can be diagonalized within each subspace of the
full Hilbert space with a given total spin (which is uniquely characterized by the
number of spins down).
Let M < L be a number of overturned spins. If M = 0 we have a unique state
|F 〉 = | ↑ · · · ↑〉.
– 59 –

This state is an eigenstate of the Hamiltonian with the eigenvalue E 0 = − JL 4 :
H|F 〉 = −J
L∑
SnS 3 n+1| 3 ↑ · · · ↑〉 = − JL 4 | ↑ · · · ↑〉 .
n=1
L!
Let M be arbitrary. Since the M-th space has the dimension one should
(L−M)!M!
find the same number of eigenvectors of H in this subspace. So let us write the
eigenvectors of H in the form
∑
|ψ〉 =
a(n 1 , . . . , n M )|n 1 , . . . , n M 〉
1≤n 1

i.e. the
L!
(L−1)!1!
= L allowed values of the pseudo-momenta are
p = 2πk with k = 0, · · · , L − 1 .
L
Further, we have the eigenvalue equation
H|ψ〉 = − JA L∑
]
e
[S ipm n + Sn+1 − + Sn − S n+1 + + 2S 3
2
nSn+1
3 |m〉 = E(p)|ψ〉 .
m,n=1
To work out the l.h.s. we have to use the formulae
as well as
S + n S − n+1|m〉 = δ nm |m + 1〉 , S − n S + n+1|m〉 = δ n+1,m |m − 1〉
2SnS 3 n+1|m〉 3 = 1 |m〉 , for m ≠ n, n + 1 ,
2
2SnS 3 n+1|m〉 3 = − 1 |m〉 , for m = n, or m = n + 1 .
2
Taking this into account we obtain
H|ψ〉 = − JA [ ∑ L (
)
e ipn |n + 1〉 + e ip(n+1) |n〉 + 1 2
2
− 1 2
n=1
L∑
e ipn |n〉 − 1 2
n=1
L∑
n=1
]
e ip(n+1) |n + 1〉 .
Using periodicity conditions we finally get
H|ψ〉 = − JA L∑ (
e ip(n−1) + e ip(n+1) + L − 4
2
2
n=1
From here we read off the eigenvalue
L∑ ( ∑ L
m=1
)
e ipn |n〉 = − J (
2
E − E 0 = J(1 − cos p) = 2J sin 2 p 2 ,
n=1
n≠m,m−1
)
e ipm |m〉
e −ip + e ip + L − 4
2
)
|ψ〉 .
where E 0 = − JL . Excitation of the spin chain around the pseudo-vacuum |F 〉
4
carrying the pseudo-momentum p is called a magnon 12 . Thus, magnon can be viewed
12 The concept of a magnon was introduced in 1930 by Felix Bloch in order to explain the reduction
of the spontaneous magnetization in a ferromagnet. At absolute zero temperature, a ferromagnet
reaches the state of lowest energy, in which all of the atomic spins (and hence magnetic moments)
point in the same direction. As the temperature increases, more and more spins deviate randomly
from the common direction, thus increasing the internal energy and reducing the net magnetization.
If one views the perfectly magnetized state at zero temperature as the vacuum state of the
ferromagnet, the low-temperature state with a few spins out of alignment can be viewed as a gas
of quasiparticles, in this case magnons. Each magnon reduces the total spin along the direction of
magnetization by one unit of and the magnetization itself by , where g is the gyromagnetic ratio.
The quantitative theory of quantized spin waves, or magnons, was developed further by Ted Holstein
and Henry Primakoff (1940) and Freeman Dyson (1956). By using the formalism of second
quantization they showed that the magnons behave as weakly interacting quasiparticles obeying
the Bose-Einstein statistics (the bosons).
– 61 –

as the pseudo-particle with the momentum p = 2πk , k = 0, . . . , L − 1 and the energy
L
E = 2J sin 2 p 2 .
The last expression is the dispersion relation for one-magnon states.
Let us comment on the sign of the coupling constant. If J < 0 then E k < 0
and |F 〉 is not the ground state, i.e. a state with the lowest energy. In other words,
in this case, |F 〉 is not a vacuum, but rather a pseudo-vacuum, or “false” vacuum.
The true ground state in non-trivial and needs some work to be identified. The
case J < 0 is called the anti-ferromagnetic one. Oppositely, if J > 0 then |F 〉 is a
state with the lowest energy and, therefore, is the true vacuum. Later on we will
see that the anti-ferromagnetic ground state corresponds M = 1 L and, therefore, it
2
is spinless. The ferromagnetic ground state corresponds to M = 0 and, therefore,
carries maximal spin S 3 = 1 2 L.13
where
Let us now turn to the more complicated case M = 2. Here we have
∑
|ψ〉 = a(n 1 , n 2 )|n 1 , n 2 〉 ,
1≤n 1

Here in the first bracket we consider the terms with n 2 > n 1 +1, while the last bracket
represents the result of action of H on terms with n 2 = n 1 + 1. Using periodicity
conditions we are allowed to make shifts of the summation variables n 1 , n 2 in the
first bracket to bring all the states to the uniform expression |n 1 , n 2 〉. We therefore
get
{
H|ψ〉 = − J ∑
a(n 1 − 1, n 2 )|n 1 , n 2 〉 +
∑
a(n 1 , n 2 − 1)|n 1 , n 2 〉
2
n 2 >n 1
+ ∑
n 2 >n 1 +2
⎧
⎨
− J 2 ⎩
∑
1≤n 1≤L
n 2>n 1+2
a(n 1 + 1, n 2 )|n 1 , n 2 〉 + ∑
a(n 1 , n 2 + 1)|n 1 , n 2 〉 + L − 8
2
n 2>n 1
[
a(n 1 , n 1 + 1)
|n 1 , n 1 + 2〉 + |n 1 − 1, n 1 + 1〉 + L − 4
2
∑
n 2 >n 1 +1
] ⎫ ⎬
|n 1 , n 1 + 1〉
⎭ .
a(n 1 , n 2 )|n 1 , n 2 〉
Now we complete the sums in the first bracket to run the range n 2 > n 1 . This is
achieved by adding and subtracting the missing terms. As the result we will get
H|ψ〉 =
{ ∑
− J 2
−
− J 2 ⎩
n 2>n 1
(
∑
1≤n 1 ≤L
⎧
⎨
a(n 1 − 1, n 2 ) + a(n 1 , n 2 − 1) + a(n 1 + 1, n 2 ) + a(n 1 , n 2 + 1) + L − 8 )
a(n 1 , n 2 ) |n 1 , n 2 〉
2
(
a(n 1 , n 1 )|n 1 , n 1 + 1〉 + a(n 1 + 1, n 1 + 1)|n 1 , n 1 + 1〉 +
+ a(n 1 , n 1 + 1)|n 1 , n 1 + 2〉
} {{ } + a(n 1, n 1 + 2)|n 1 , n 1 + 2〉
} {{ } +L − 8
) }
a(n 1 , n 1 + 1)|n 1 , n 1 + 1〉
2
∑
1≤n 1≤L
[
a(n 1 , n 1 + 1) |n 1 , n 1 + 2〉 + |n 1 − 1, n 1 + 1〉
} {{ } +L − 4
] ⎫ ⎬
|n 1 , n 1 + 1〉
2
⎭ .
The underbraced terms cancel out and we finally get
H|ψ〉 =
{ ∑
− J 2
⎧
⎨
+ J 2 ⎩
n 2 >n 1
(
∑
1≤n 1 ≤L
}
a(n 1 − 1, n 2 ) + a(n 1 , n 2 − 1) + a(n 1 + 1, n 2 ) + a(n 1 , n 2 + 1) + L − 8 )
a(n 1 , n 2 ) |n 1 , n 2 〉
2
⎫
(
) ⎬
a(n 1 , n 1 ) + a(n 1 + 1, n 1 + 1) − 2a(n 1 , n 1 + 1) |n 1 , n 1 + 1〉
⎭ .
If we impose the requirement that
a(n 1 , n 1 ) + a(n 1 + 1, n 1 + 1) − 2a(n 1 , n 1 + 1) = 0 (5.1)
then the second bracket in the eigenvalue equation vanishes and the eigenvalue problem
reduces to the following equation
2(E − E 0 )a(n 1 , n 2 ) = J [ 4a(n 1 , n 2 ) − ∑
a(n 1 + σ, n 2 ) + a(n 1 , n 2 + σ) ] . (5.2)
σ=±1
Substituting in eq.(5.1) the Bethe ansatz for a(n 1 , n 2 ) we get
Ae (p 1+p 2 )n + Be i(p 1+p 2 )n + Ae (p 1+p 2 )(n+1) + Be i(p 1+p 2 )(n+1)
(
)
− 2 Ae i(p 1n+p 2 (n+1)) + Be i(p 2n+p 1 (n+1))
= 0 .
}
– 63 –

This allows one to determine the ratio
B
A = −ei(p 1+p 2 ) + 1 − 2e ip 2
e i(p 1+p 2) + 1 − 2e ip 1 .
Problem. Show that for real values of momenta the ratio B A
is the pure phase:
B
A = eiθ(p 2,p 1 ) ≡ S(p 2 , p 1 ) .
This phase is called the S-matrix. We further note that it obeys the following relation
S(p 1 , p 2 )S(p 2 , p 1 ) = 1 .
Thus, the two-magnon Bethe ansatz takes the form
a(n 1 , n 2 ) = e i(p 1n 1 +p 2 n 2 ) + S(p 2 , p 1 )e i(p 2n 1 +p 1 n 2 ) ,
where we factored out the unessential normalization coefficient A.
Let us now substitute the Bethe ansatz in eq.(5.2). We get
(
) [ (
)
2(E − E 0 ) Ae i(p 1n 1 +p 2 n 2 ) + Be i(p 2n 1 +p 1 n 2 )
= J 4 Ae i(p 1n 1 +p 2 n 2 ) + Be i(p 2n 1 +p 1 n 2 )
−
)
)
−
(Ae i(p1n1+p2n2) e ip1 + Be i(p2n1+p1n2) e ip2 −
(Ae i(p1n1+p2n2) e −ip1 + Be i(p2n1+p1n2) e −ip2
)
)]
−
(Ae i(p1n1+p2n2) e ip2 + Be i(p2n1+p1n2) e ip1 −
(Ae i(p1n1+p2n2) e −ip2 + Be i(p2n1+p1n2) e −ip1 .
We see that the dependence on A and B cancel out completely and we get the
following equation for the energy
)
E − E 0 = J
(2 − cos p 1 − cos p 2 = 2J
2∑
k=1
sin 2 p k
2 .
Quite remarkably, the energy appears to be additive, i.e. the energy of a two-magnon
state appears to be equal to the sum of energies of one-magnon states! This shows
that magnons essentially behave themselves as free particles in the box.
Finally, we have to impose the periodicity condition a(n 2 , n 1 + L) = a(n 1 , n 2 ). This
results into
which implies
e i(p 1n 2 +p 2 n 1 ) e ip 2L + B A eip 1L e i(p 2n 2 +p 1 n 1 ) = e i(p 1n 1 +p 2 n 2 ) + B A ei(p 2n 1 +p 1 n 2 )
e ip 1L = A B = S(p 1, p 2 ) , e ip 2L = B A = S(p 2, p 1 ) .
– 64 –

The last equations are called “Bethe equations”.
quantization conditions for momenta p k .
They are nothing else but the
Let us note the following useful representation for the S-matrix.
We have
(
e
ip 2 − 1
)
+ 1 − e
ip 1
= − eip2 e i 2 p1 (
e
i
2 p1 − e − i 2 p1 )
+ e
i
S(p 2 , p 1 ) = − eip2 (
e
ip 1
− 1 ) + 1 − e ip2
e ip 1
= − e i 2 p2 sin p1
2 − e− i 2 p1 sin p2
2
e i 2 p 1 sin
p 2
2
− e − i 2 p 2 sin
p 1
2
=
e ip i
1 e 2 p 2
(
cos
p 2
= − cos p 2
2
sin p 1
2
− cos p 1
2
sin p 2
2
+ 2i sin p 1
2
sin p 2
2
cos p 1
2
sin p 2
2
− cos p 2
2
sin p 1
2
+ 2i sin p 1
2
sin p 2
Thus, we obtained
( i
e 2 p 2 − e
− i 2 p 2
)
sin
p 1
)
+ e
i
2 p 1
2
+ i sin p2
2 2
− ( cos p1
(
2
cos
p 1
)
2
+ i sin p1
2 sin
p 2
2
− ( cos p2
2
2
=
1
2
S(p 1 , p 2 ) =
cot p 1
− 1 cot p 2
+ i
2 2 2
1
cot p 1
− 1 cot p 2
− i .
2 2 2 2
1
( )
2 p2 e
− i 2 p2 − e i 2 p2
(
e
− i 2 p i )
1 − e 2 p 1
)
p1
− i sin
2 sin
p 2
) 2
sin
p 1
− i sin
p2
2
2 cot p 2
2
− 1 2 cot p 1
2
+ i
1
2 cot p 2
2
− 1 2 cot p 1
2
− i .
It is therefore convenient to introduce the variable λ = 1 cot p which is called
2 2
rapidity and get
S(λ 1 , λ 2 ) = λ 1 − λ 2 + i
λ 1 − λ 2 − i .
Hence, on the rapidity plane the S-matrix depends only on the difference of rapidities
of scattering particles.
2
Taking the logarithm of the Bethe equations we obtain
Lp 1 = 2πm 1 + θ(p 1 , p 2 ) , Lp 2 = 2πm 2 + θ(p 2 , p 1 ) ,
where the integers m i ∈ {0, 1, . . . , L − 1} are called Bethe quantum numbers. The
Bethe quantum numbers are useful to distinguish eigenstates with different physical
properties. Furthermore, these equations imply that the total momentum is
Writing the equations in the form
p 1 = 2πm 1
L } {{ }
P = p 1 + p 2 = 2π L (m 1 + m 2 ) .
+ 1 L θ(p 1, p 2 ) , p 2 = 2πm 2
+ 1
} {{ L } L θ(p 2, p 1 ) ,
we see that the magnon interaction is reflected in the phase shift θ and in the deviation
of the momenta p 1 , p 2 from the values of the underbraced one-magnon wave
numbers. What is very interesting, as we will see, that magnons either scatter off
each other or form the bound states.
– 65 –

The first problem is to find all possible Bethe quantum numbers (m 1 , m 2 ) for
which Bethe equations have solutions. The allowed pairs (m 1 , m 2 ) are restricted to
0 ≤ m 1 ≤ m 2 ≤ L − 1 .
This is because switching m 1 and m 2 simply interchanges p 1 and p 2 and produces
the same solution. There are 1 L(L + 1) pairs which meet this restriction but only
2
1L(L − 1) of them yield a solution of the Bethe equations. Some of these solutions
2
have real p 1 and p 2 , the others yield the complex conjugate momenta p 2 = p ∗ 1.
The simplest solutions are the pairs for which one of the Bethe numbers is zero,
e.g. m 1 = 0, m = m 2 = 0, 1, . . . , L − 1. For such a pair we have
Lp 1 = θ(p 1 , p 2 ) , Lp 2 = 2πm + θ(p 2 , p 1 ) ,
which is solved by p 1 = 0 and p 2 = 2πm.
Indeed, for p L
1 = 0 the phase shift vanishes:
θ(0, p 2 ) = 0. These solutions have the dispersion relation
E − E 0 = 2J sin 2 p 2 , p = p 2
which is the same as the dispersion for the one-magnon states. These solutions are
nothing else but su(2)-descendants of the solutions with M = 1.
One can show that for M = 2 all solutions are divided into three distinct classes
Descendents
} {{ } , Scattering States , Bound
} {{ }
} {{ States }
L
L(L−5)
L−3
+3 2
so that
L(L − 5)
L + + 3 + L − 3 = 1 L(L − 1)
2
2
gives a complete solution space of the two-magnon problem.
Pseudo−vacuum
F
L one−magnon states
01
01
01
01
01
01
01 01
01
01 L(L−1)
2
two−magnon states
0 0 1
01
01
0
0 1 0
01
1
101 0 1
1
01 0
01 01 01 01
0 1
1
01
1 01
0 1
– 66 –

The su(2)-multiplet structure of the M = 0, 1, 2 subspaces.
The most non-trivial fact about the Bethe ansatz is that many-body (multimagnon)
problem reduces to the two-body one. It means, in particular, that the
multi-magnon S-matrix appears to be expressed as the product of the two-body
ones. Also the energy is additive quantity. Such a particular situation is spoken
about as “Factorized Scattering”. In a sense, factorized scattering for the quantum
many-body system is the same as integrability because it appears to be a consequence
of existence of additional conservation laws. For the M-magnon problem the Bethe
equations read
M∏
e ipkL = S(p k , p j ) .
j=1
j≠k
The most simple description of the bound states is obtained in the limit when
L → ∞. If p k has a non-trivial positive imaginary part then e ip kL tends to ∞ and
this means that the bound states correspond in this limit to poles of the r.h.s. of the
Bethe equations. In particular, for the case M = 2 the bound states correspond to
poles in the two-body S-matrix. In particular, we find such a pole when
1
2 cot p 1
2 − 1 2 cot p 2
2 = i .
This state has the total momentum p = p 1 +p 2 which must be real. These conditions
can be solved by taking
The substitution gives
which is
cos 1 2
(
p
2 + iv )
sin 1 2
p 1 = p 2 + iv , p 2 = p 2 − iv .
(
p
2 − iv )
− cos 1 2
The energy of such a state is
(
E = 2J sin 2 p 1
2 + p )
sin2 2
2
We therefore get
E = 2J
(1 − cos p )
2 cosh v
(
p
cos p 2 = ev .
2 − iv )
sin 1 2
= 2i sin 1 2
(
p
2 + iv )
(
p
2 + iv )
sin 1 2
(
p
2 − iv )
,
( ( p
) ( p
))
= 2J sin 2 4 + iv + sin 2
2 4 − iv .
2
(
= 2J 1 − cos p 2
cos 2 p + 1 )
2
2 cos p = J sin 2 p 2 .
2
Thus, the position of the pole uniquely fixes the dispersion relation of the bound
state.
– 67 –

5.2 Algebraic Bethe Ansatz
Here we will solve the Heisenberg model by employing this time a new method called
the Algebraic Bethe ansatz. This method allows one to reveal the integrable structure
of the model as well as to study its properties in the thermodynamic limit.
Fundamental commutation relation. Suppose we have a periodic chain of length L.
The basic tool of the algebraic Bethe ansatz approach is the so-called Lax operator.
The definition of the Lax operator involves the local “quantum” space V i , which for
the present case is chosen to be a copy of C 2 . The Lax operator L i,a acts in V i ⊗ V a :
Explicitly, it is given by
L i,a (λ) : V i ⊗ V a → V i ⊗ V a .
L i,a (λ) = λI i ⊗ I a + i ∑ α
S α i ⊗ σ α ,
where I i , Si
α act in V i , while the unit I a and the Pauli matrices σ α act in an another
Hilbert space C 2 called “auxiliary”. The parameter λ is called the spectral parameter.
Another way to represent that the Lax operator is to write it as 2 × 2 matrix with
operator coefficients
( )
λ + iS
3
L i,a (λ) = i iS − i
iS + i λ − iSi
3 .
Introducing the permutation operator
P = 1 2
(
I ⊗ I +
3∑ )
σ α ⊗ σ α
α=1
we can write the Lax operator in the alternative form
(
L i,a (λ) = λ − i )
I i,a + iP i,a .
2
The most important property of the Lax operator is the commutation relations between
its entries. Consider two Lax operators, L i,a (λ 1 ) and L i,b (λ 2 ), acting in the
same quantum space but in two different auxiliary spaces. The products of these
two operators L i,a (λ 1 )L i,b (λ 2 ) and L i,b (λ 2 )L i,a (λ 1 ) are defined in the triple tensor
product V i ⊗ V a ⊗ V b . Remarkably, it turns out that these two product are related
by a similarity transformation which acts non-trivially in the tensor product V a ⊗ V b
only. Namely, there exists an intertwining operator R a,b (λ 1 , λ 2 ) = R ab (λ 1 − λ 2 ) such
that the following relation is true
R ab (λ 1 − λ 2 )L ia (λ 1 )L ib (λ 2 ) = L ib (λ 2 )L ia (λ 1 )R ab (λ 1 − λ 2 ) . (5.3)
This intertwining operator is called quantum R-matrix and it has the following explicit
form
R ab = λI ab + iP ab
– 68 –

The form of the L-operator and the R-matrix is essentially the same.
We check
”
”
“(λ 1 − λ 2 )I ab + iP ab L ia (λ 1 )L ib (λ 2 ) = L ib (λ 2 )L ia (λ 1 )
“(λ 1 − λ 2 )I ab + iP ab ,
which leads to
“
” “
”
(λ 1 − λ 2 ) L ia (λ 1 )L ib (λ 2 ) − L ib (λ 2 )L ia (λ 1 ) = iP ab L ia (λ 2 )L ib (λ 1 ) − L ia (λ 1 )L ib (λ 2 ) ,
It is easy to see that
L ia (λ 1 )L ib (λ 2 ) − L ib (λ 2 )L ia (λ 1 ) = P ib P ia − P ia P ib
and
”
iP ab
“L ia (λ 2 )L ib (λ 1 ) − L ia (λ 1 )L ib (λ 2 ) = (λ 1 − λ 2 )P ab (P ib − P ia ) = (λ 1 − λ 2 )(P ib P ai − P ia P ib )
This proves the statement.
The relation (5.3) is called the fundamental commutation relation.
Yang-Baxter equation. It is convenient to suppress the index of the quantum space
and write the fundamental commutation relation as
R ab (λ 1 − λ 2 )L a (λ 1 )L b (λ 2 ) = L b (λ 2 )L a (λ 1 )R ab (λ 1 − λ 2 ) .
We can think about L as being 2 × 2 matrix whose matrix elements are generators
of a certain associative algebra (operators). Relations (5.3) define the then the commutation
relations between the generators of this algebra. Substituting the indices
a and b for 1 and 2 we will write the general form of the fundamental commutation
relations
R 12 (λ 1 , λ 2 )L 1 (λ 1 )L 2 (λ 2 ) = L 2 (λ 2 )L 1 (λ 1 )R 12 (λ 1 , λ 2 ) .
What the R-matrix does is that it interchange the position of the matrices L 1 and
L 2 . Consider a triple product
L 1 L 2 L 3 = R12 −1 L 2 L 1 R 12 L 3 = R12 −1 L 2 L 1 L 3 R 12 =
= R12 −1 R13 −1 L 2 L 3 L 1 R 13 R 12 = R12 −1 R13 −1 R23 −1 L 3 L 2 L 1 R 23 R 13 R 12 .
Essentially, we brought the product L 1 L 2 L 3 to the form L 3 L 2 L 1 . However, we can
reach the same effect by changing the order of permutations
L 1 L 2 L 3 = R23 −1 L 1 L 3 L 2 R 23 = R23 −1 R13 −1 L 3 L 1 L 2 R 13 R 12 =
= R12 −1 R13 −1 L 2 L 3 L 1 R 13 R 12 = R23 −1 R13 −1 R12 −1 L 3 L 2 L 1 R 12 R 13 R 23 .
Thus, if we require that we do not generate new triple relations between the elements
of L we should impose the following condition on the R-matrix:
This is the quantum Yang-Baxter equation.
R 12 R 13 R 23 = R 23 R 13 R 12 . (5.4)
– 69 –

Semi-classical limit and quantization. Why the quantum Yang-Baxter equation is
called “quantum”? Assume that the R-matrix depends on the additional parameter
and when → 0 it expands into the power series starting with the unit:
R 12 = I 12 + r 12 + · · ·
Expanding quantum Yang-Baxter equation we see that the leading terms as well as
the terms proportional to cancel out. At order 2 we find
)
([r 2 12 , r 13 ] + [r 12 , r 23 ] + [r 13 , r 23 ] + O( 3 ) = 0 .
At order 2 we find the classical Yang-Baxter equation. Thus, the quantum Yang-
Baxter equation can be considered as the deformation (or quantization) of the classical
Yang-Baxter equation. Further we recall the relation between the Poisson bracket
of classical observables and the Poisson brachet of their quantum counterparts
{A, B} = lim [Â, ˆB]
→0
1
Now we notice that the fundamental computation relations can be written in the
equivalent form
[L 1 (λ 1 ), L 2 (λ 2 )] = i
λ 1 − λ 2
[P 12 , L 1 (λ 1 )L 2 (λ 2 )] .
This formula allows for the semi-classical limit
}{{} L → }{{} L
quantum classical
and it defines the Poisson bracket on the space of classical L -operators
1
[
{L 1 (λ 1 ), L 2 (λ 2 )} = lim
→0 [L i
]
1(λ 1 ), L 2 (λ 2 )] = P 12 , L 1 (λ 1 )L 2 (λ 2 ) .
λ 1 − λ 2
We see that r = i P appears to be the classical r-matrix. Thus, the semi-classical
λ
limit of the fundamental commutation relations is nothing else as the Sklyanin
bracket for classical Heisengerg magnetic. Inversely, we can think about the fundamental
commutation relations as quantization of the Poisson algebra of the classical
L-operators.
Monodromy and transfer matrix. For a chain of length L define the monodromy as
the ordered product of L-operators along the chain 14
T a (λ) = L L,a (λ) . . . L 1,a (λ).
14 Recall the definition of the monodromy as the path-ordered exponent in the classical case.
– 70 –

The monodromy is an operator on V L ⊗ V L−1 ⊗ . . . ⊗ V 1 ⊗ V a . It we take the trace
of the monodromy w.r.t. to its matrix part acting in the auxiliary space we obtain
an object which is called the transfer matrix and it is denoted as τ(λ) = tr a T a (λ).
Denote L = L i,a and L ′ = L i+1,a
R 12 L ′ 1L 1 L ′ 2L 2 = R 12 L ′ 1L ′ 2L 1 L 2 = L ′ 2L ′ 1R 12 L 1 L 2 = L ′ 2L ′ 1L 2 L 1 R 12 = L ′ 2L 2 L ′ 1L 1 R 12 .
This is because L 1 and L ′ 2 commute – they act both in different auxiliary spaces and
different quantum spaces. Thus, we deduce the commutation relation between the
components of the monodromy
R 12 (λ − µ)T 1 (λ)T 2 (µ) = T 2 (µ)T 1 (λ)R 12 (λ − µ) .
Now we can proof the fundamental fact about the commutation relations above.
Rewrite them in the form
T 1 (λ)T 2 (µ) = R 12 (λ − µ) −1 T 2 (µ)T 1 (λ)R 12 (λ − µ)
and takes the trace over the first and the second space. We will get
)
τ(λ)τ(µ) = tr 1,2
(R 12 (λ − µ) −1 T 2 (µ)T 1 (λ)R 12 (λ − µ) = τ(µ)τ(λ) .
Thus, the transfer matrices commute with each other for different values of the
spectral parameter
[τ(λ), τ(µ)] = 0 .
Hence, τ(λ) generates an abelian subalgebra. If we find the Hamiltonian of the model
among this commuting family then we can call our model quantum integrable. The
Hamiltonian must be
H = ∑ d k
c ka
dλ ln τ(λ)| k λ=λ a
.
a,k
for some coefficients c ka . This will ensute that the Hamiltonian belongs to the family
of commuting quantities. Since all the integrals from this family mutually commute
they can be simultaneously diagonalized.
Represent the monodromy as the 2 × 2 matrix in the auxiliary space
( )
A(λ) B(λ)
T (λ) =
,
C(λ) D(λ)
where the entries are operators acting in the space ⊗ L i=1V i . From the definition of
the monodromy and the L-operator it is clear that T is a polynomial in λ and
T (λ) = λ L + iλ L−1
L
∑
n=1
S α n ⊗ σ α + · · ·
– 71 –

Thus, the transfer matrix is also polynomial of degree L:
∑L−2
τ(λ) = tr a T a (λ) = A(λ) + D(λ) = 2λ L + Q j λ j .
Note that the subleading term of order λ L−1 is absent because Pauli matrices are
traceless. The coefficients Q j mutually commute
[Q i , Q j ] = 0 .
j=0
Hamiltonian and Momentum. It remains to find the Hamiltonian among the commuting
family generated by the transfer matrix. The L-operator has two special
points on the spectral parameter plane.
• λ = i 2 , where L i,a(i/2) = iP ia .
• λ = ∞. We see that
1 (λ)
ResT
i λ = ∑ L L
n=1
S α ⊗ σ α = S α
}{{}
su(2)
⊗ σ α .
This point will be related to the realization of the global su(2) symmetry of
the model.
Let us investigate the first point. We have
T a (i/2) = i L P L,a P L−1,a · · · P 1,a = i L P L−1,L P L−2,L · · · P 1,L P L,a =
= i L P L−2,L−1 P L−3,L−1 · · · P 1,L−1 P L−1,L P L,a = · · · = i L P 12 P 23 · · · P L−1,L P L,a .
Thus, we have managed to isolate a single permutation carrying the index of the
auxiliary subspace. Taking the trace and recalling that tr a P L,a = I L we obtain the
transfer matrix
τ(i/2) = i L P 12 P 23 · · · P L−1,L = U ← shift operator
Operator U is unitary U † U = UU † = I and it generates a shift along the chain:
U −1 X n U = X n−1 .
By definition an operator of the infinitezimal shift is the momentum and on the
lattice it is introduced as
U = e ip .
Now we differentiate the logarithm of the transfer matrix
dT a (λ)
dλ
| λ=i/2 = i ∑ L−1 n
P L,a · · · P
}{{} n,a · · · P 1,a = i ∑ L−1 n
absent
P 12 P 23 · · · P n−1,n+1 · · · P L−1,L .
– 72 –

This allows to establish that
dτ(λ)
dλ τ(λ)−1 | λ=i/2 =
= i −1 ( ∑
n
P 12 P 23 · · · P n−1,n+1 · · · P L−1,L
)(
P L,L−1 P L−1,L−2 · · · P 2,1
)
= 1 i
L∑
n,n+1
P n,n+1 .
On the other hand we see that
H = −J
L∑
SnS α n+1 α = − J 4
n=1
L∑
n=1
( 1
σnσ α n+1 α = −J
2
L∑
P n,n+1 − L 4
n=1
)
.
Hence,
( i dτ(λ)
H = −J
2
)
| λ=i/2 ,
dλ τ(λ)−1 − L 4
i.e. the Hamiltonian belongs to the family of L − 1 commuting integrals. To obtain
L commuting integrals we can add the operator S 3 to this family.
The spectrum of the Heisenberg model. Here we compute the eigenvalues of H by
using the algebraic Bethe ansatz. First we derive the commutation relations between
the operators A, B, C, D. The form of the R-matrix is
⎛
⎞
λ − µ + i 0 0 0
0 λ − µ i 0
R(λ − µ) = ⎜
⎟
⎝ 0 i λ − µ 0 ⎠ .
0 0 0 λ − µ + i
We compute
⎛
⎞
⎛
⎞
A(λ) 0 B(λ) 0
A(µ) B(µ) 0 0
0 A(λ) 0 B(λ)
T a (λ) = ⎜
⎟
⎝ C(λ) 0 D(λ) 0 ⎠ , T C(µ) D(µ) 0 0
b(λ) = ⎜
⎟
⎝ 0 0 A(µ) B(µ) ⎠ .
0 C(λ) 0 D(λ)
0 0 C(µ) D(µ)
Plugging this into the fundamental commutation relation we get
⎛
⎞
(α + i)A λ A µ (α + i)A λ B µ (α + i)B λ A µ (α + i)B λ B µ
αA λ C µ + iC λ A µ αA λ D µ + iC λ B µ αB λ C µ + iD λ A µ αB λ D µ + iD λ B µ
⎜
⎟
⎝ iA λ C µ + αC λ A µ iA λ D µ + αC λ B µ iB λ C µ + αD λ A µ iB λ D µ + αD λ B µ ⎠ =
(α + i)C λ C µ (α + i)C λ D µ (α + i)D λ C µ (α + i)D λ D µ
⎛
⎞
(α + i)A µ A λ αB µ A λ + iA µ B λ iB µ A λ + αA µ B λ (α + i)B µ B λ
(α + i)C
=
µ A λ αD µ A λ + iC µ B λ iD µ A λ + αC µ B λ (α + i)D µ B λ
⎜
⎟
⎝ (α + i)A µ C λ αB µ C λ + iA µ D λ iB µ C λ + αA µ D λ (α + i)B µ D λ ⎠ .
(α + i)C µ C λ αD µ C λ + iC µ D λ iD µ C λ + αC µ D λ (α + i)D µ D λ
– 73 –

To write down the fundamental commutation relations we have used the shorthand
notations A λ ≡ A(λ) and α = λ − µ. The relevant commutation relations are
[B(λ), B(µ)] = 0 ,
A(λ)B(µ) = λ − µ − i
λ − µ B(µ)A(λ) + i B(λ)A(µ) , (5.5)
λ − µ
D(λ)B(µ) = λ − µ + i
λ − µ B(µ)D(λ) − i
λ − µ B(λ)D(µ) .
The main idea of the algebraic Bethe ansatz is that there exists a pseudo-vacuum
|0〉 such that C(λ)|0〉 = 0 and the eigenvectors of τ(λ) with M spins down have the
form
|λ 1 , λ 2 , · · · , λ M 〉 = B(λ 1 )B(λ 2 ) · · · B(λ M )|0〉 ,
where {λ i } are “Bethe roots” which we will compare later on with the pseudomomenta
p i of the magnons in the coordinate Bethe ansatz approach. One can see
that the pseudo-vacuum can be identified with the state
Indeed, since we have
we find that
L n (λ)| ↑ n 〉 =
T (λ)|0〉 =
|0〉 = ⊗ L n=1| ↑ n 〉 .
( (λ +
i
)| ↑ )
2 n〉 i| ↓ n 〉
0 (λ − i )| ↑ 2 n〉
( )
(λ +
i
2 )L |0〉 ∗
0 (λ − i ,
2 )L |0〉
where ∗ stands for irrelevant terms. Thus, we indeed have
(
C(λ)|0〉 = 0 , A(λ)|0〉 = λ + i ) L|0〉 (
, D(λ)|0〉 = λ − i L|0〉
.
2
2)
Comparing with the coordinate Bethe ansatz we see that |0〉 ≡ |F 〉. We also see that
|0〉 is an eigenstate of the transfer matrix. The algebraic Bethe ansatz states that
the other eigenstates are of the form
|λ 1 , λ 2 , · · · , λ M 〉 = B(λ 1 )B(λ 2 ) · · · B(λ M )|0〉
provided the Bethe roots {λ i } satisfy certain restrictions.
restrictions.
We compute
A(λ)B(λ 1 )B(λ 2 ) · · · B(λ M )|0〉 =
Let us now find these
(
λ + i ) L ( ∏ M
λ − λ n − i
)
B(λ 1 )B(λ 2 ) · · · B(λ M )|0〉
2 λ − λ
n=1
n
M∑
M∏
+ Wn A (λ, {λ i })B(λ) B(λ j )|0〉 .
n=1
j=1
j≠n
– 74 –

Here the coefficients Wn A (λ, {λ i }) depend on λ and the set {λ i } M i=1. To determine
this coefficient we note that since the operators B(λ) commute with each other we
can write
M∏
|λ 1 , λ 2 , · · · , λ M 〉 = B(λ n ) B(λ j )|0〉 .
Thus,
A(λ)|λ 1 , λ 2 , · · · , λ M 〉 = λ − λ n − i
B(λ n )A(λ)
λ − λ n
j=1
j≠n
M∏
B(λ j )|0〉 +
j=1
j≠n
i
λ − λ n
B(λ)A(λ n )
M∏
B(λ j )|0〉 .
From this equation we see that only the second term on the r.h.s. will contribute to
Wn
A since this term does not contain B(λ n ). If we now mover A(λ) past B(λ j ) we
see that the only way to avoid the appearance of B(λ n ) is to use only the first term
on the r.h.s. of eq.(5.5). So the resulting term should have the form
i.e.
i
(λ n + i ) L ∏ M
λ − λ n 2
i=1
i≠n
W A n (λ, {λ i }) =
In the same way we obtain
D(λ)B(λ 1 )B(λ 2 ) · · · B(λ M )|0〉 =
and
λ n − λ i − i
B(λ)
λ n − λ i
M∏
B(λ j )|0〉 ,
j=1
j≠n
i
(λ n + i ) L ∏ M
λ n − λ j − i
.
λ − λ n 2 λ
j=1 n − λ j
j≠n
j=1
j≠n
(
λ − i ) L ( ∏ M
λ − λ n + i
)
B(λ 1 )B(λ 2 ) · · · B(λ M )|0〉
2 λ − λ
n=1
n
M∑
M∏
+ Wn D (λ, {λ i })B(λ) B(λ j )|0〉
n=1
Wn D (λ, {λ i }) = −
i (λ n − i ) L ∏ M
λ − λ n 2
Thus, we will solve the eigenvalue problem
j=1
j≠n
λ n − λ j + i
λ n − λ j
.
τ(λ)|λ 1 , · · · , λ M 〉 = Λ(λ, {λ n })|λ 1 , · · · , λ M 〉
j=1
j≠n
with
Λ(λ, {λ n }) =
(
λ + i ) L ∏ M
λ − λ n − i
(
+ λ − i ) L ∏ M
λ − λ n + i
2 λ − λ
n=1
n 2 λ − λ
n=1
n
– 75 –

provided W A n + W D n = 0 for all n, which means that
(
λ n + i ) L ∏ M
λ n − λ j − i
(
= λ n − i 2 λ
j=1 n − λ j 2
j≠n
We write the last equations in the form
( ) L λn + i/2
=
λ n − i/2
M∏
j=1
j≠n
) L M
∏
j=1
j≠n
λ n − λ j + i
λ n − λ j − i .
λ n − λ j + i
λ n − λ j
.
These are the Bethe equations. Introducing λ j = cot p j the Bethe equations take
precisely the same form as derived in the coordinate Bethe ansatz approach:
e ip iL =
M∏
S(p i , p j ) .
j=1
j≠i
Note that the parametrization λ j = cot p j has a singularity at k j = 0. From the
experience with the coordinate Bethe ansatz we know that all the eigenvectors for
which k j ≠ 0 are the highest weight states of the global spin algebra su(2). Thus, we
expect that the eigenvectors obtained in the algebraic Bethe ansatz approach have
the same property. Now we are going to investigate this issue in mode detail.
Realization of the symmetry algebra. Let us consider the fundamental commutation
relations in the limitimg case µ → ∞. We get
(
(λ − µ) + i 2 (I a ⊗ I b + ∑ α
) (
σa α ⊗ σb α ) T a (λ) µ L + iµ ∑ )
L−1 Sn α ⊗ σb α + · · · =
n,α
=
(
µ L + iµ ∑ ) (
L−1 Sn α ⊗ σb α + · · · T a (λ) (λ − µ) + i 2 (I a ⊗ I b + ∑
n,α
α
σ α a ⊗ σ α b ) .
The leading term of the order µ L+1 cancel out. The subleading term of order µ L
gives
−iT a (λ) ∑ Sn α ⊗ σb α + i 2 T a(λ) + i ( ∑ )
σa α ⊗ σb
α T a (λ) =
2
n,α
α
= i 2 T a(λ) + i (
2 T ∑ )
a(λ) σa α ⊗ σb
α − i ∑ Sn α ⊗ σb α T a (λ) .
α
n,α
Simplifying we get
∑
[T a (λ), S α + 1 2 σα a ] ⊗ σb α = 0 .
α
– 76 –

This results into the following equation which describes how the components of the
monodromy transform under the global symmetry generators
[S α , T a (λ)] = 1 2 [T a(λ), σ α a ] .
Thus, we end up with three separate equations
)
,
[S 3 , T a (λ)] = 1 2 [T a(λ), σa] 3 = 1 [ ( A(λ) B(λ)
2 C(λ) D(λ)
[S + , T a (λ)] = 1 [ (
2 [T a(λ), σ a + A(λ) B(λ)
] =
C(λ) D(λ)
and
[S − , T a (λ)] = 1 [ (
2 [T a(λ), σa − A(λ) B(λ)
] =
C(λ) D(λ)
)
,
)
,
( ) 1 0 ]
=
0 − 1
( ) 0 1 ]
=
0 0
( ) 0 0 ] (
=
1 0
Essentially, we need the following commutation relations
[S 3 , B] = −B , [S + , B] = A − D .
( )
0 −B(λ)
,
C(λ) 0
( )
−C(λ) A(λ) − D(λ)
,
0 C(λ)
B(λ) 0
D(λ) − A(λ) −B(λ)
The action of the symmetry generators on the pseudo-vacuum have been already
derived
S + |0〉 = 0 , S 3 |0〉 = L 2 |0〉 .
So the state |0〉 is the highest weight state of the symmetry algebra. Further, we find
and
S 3 |λ 1 , · · · , λ M 〉 =
( L
2 − M )
|λ 1 , · · · , λ M 〉
)
.
S + |λ 1 , · · · , λ M 〉 = ∑ j
B(λ 1 ) . . . B(λ j−1 )(A(λ j ) − D(λ j ))B(λ j+1 ) . . . B(λ M )|0〉
= ∑ j
O j B(λ 1 ) . . . B(λ j−1 ) ˆB(λ j )B(λ j+1 ) . . . B(λ M )|0〉 .
The coefficients O j are unknown for the moment. To calculate O j we will use the
arguments similar to those for computing Wj
A and Wj D . The only contributions to
O j will come from
B(λ 1 ) . . . B(λ k−1 )(A(λ k ) − D(λ k ))B(λ k+1 ) . . . B(λ M )|0〉 with k ≤ j.
If k = j this contribution will be
M∏
k j +1
λ j − λ k − i
(λ j + i ) L ∏ M
λ j − λ k + i
−
(λ j − i ) L
λ j − λ k 2
λ j − λ k 2
k j +1
– 77 –

and if k < j the contribution will be
Thus, adding up we obtain
O j =
−
=
−
k j +1
W A j (λ k , {λ} M k+1) + W D j (λ k , {λ} M k+1) .
M∏ λ j − λ k − i
(λ j + i ) j−1 L ∑
+ Wj A (λ k , {λ} M
λ j − λ k 2
k+1)
k j +1
k=j+1
k=1
M∏ λ j − λ k + i
(λ j − i ) j−1 L ∑
+ Wj D (λ k , {λ} M
λ j − λ k 2
k+1) =
k=1
(
M∏ λ j − λ k − i
(λ j + i ) j−1
j−1
L ∑ i ∏
1 +
λ j − λ k 2
λ k − λ j
k=1
p=k+1
(
M∏ λ j − λ k + i
(λ j − i ) j−1
j−1
L ∑ i ∏
1 −
λ j − λ k 2
λ k − λ j
k=1
k=j+1
Let us now note the useful identity
p=k+1
)
λ j − λ p − i
λ j − λ p
)
λ j − λ p + i
.
λ j − λ p
t n ≡ 1 +
j−1
∑
k=n
i
λ k − λ j
j−1
∏
p=k+1
j−1
λ j − λ p − i ∏ λ j − λ k − i
=
.
λ j − λ p λ j − λ k
k=n
We will prove this by induction over n. For n = j − 1 and n = j − 2 we have
i
t j−1 = 1 + = λ j − λ j−1 − i
,
λ j−1 − λ j λ j − λ j−1
i
i λ j − λ j−1 − i
t j−2 = 1 + +
λ j−1 − λ j λ j−2 − λ j λ j − λ j−1
Now we suppose that the formula holds for n = l, then we have
t l−1 = t l +
j−1
i ∏
λ l−1 − λ j
p=l
λ j − λ p − i
λ j − λ p
=
= λ j − λ j−1 − i λ j − λ j−2 − i
.
λ j − λ j−1 λ j − λ j−2
j−1
∏
p=l−1
λ j − λ p − i
λ j − λ p
,
which proves our assumption. With this formula at hand we therefore find
1 +
j−1
∑
i=1
i
λ i − λ j
j−1
∏
p=i+1
j−1
λ j − λ p − i ∏ λ j − λ k − i
=
.
λ j − λ p λ j − λ k
k=1
In the same way one can show that
1 −
j−1
∑
i=1
i
λ i − λ j
j−1
∏
p=i+1
j−1
λ j − λ p + i ∏ λ j − λ k + i
=
.
λ j − λ p λ j − λ k
k=1
– 78 –

It follows now from the Bethe equations that
(
O j = λ j + i ) L ∏ λ j − λ k − i
−
2 λ j − λ k
k=1
k≠j
(
λ j − i 2
) L ∏
k=1
k≠j
λ j − λ k + i
λ j − λ k
= 0 .
This proves that the eigenvectors obtained from the algebraic Bethe ansatz are the
highest weight vectors of the spin algebra.
Finally, we can compute the eigenvalues of the corresponding Bethe eigenvectors.
We obtain
( i dτ(λ)
E = −i
2 dλ τ(λ)−1 | λ=i/2 − L )
= E 0 + J L∑ 1
4 2 λ 2 j + 1 .
4
If we now use the parametrization λ j = 1 2 cot p j
2
E − E 0 = J
L∑
j=1
2
1 + cot 2 p j
2
we get
= 2
L∑
j=1
j=1
sin 2 p j
2 .
This expression agrees with the one obtained in the coordinate Bethe ansatz framework.
Let us summarize some important observations about the Bethe ansatz. First of
all, the Heisenberg model has su(2) symmetry which results into the fact that the
eigenvectors calculated by using the Bethe ansatz procedure splits into irreducible
representations of su(2). For finite values of λ j the eigenvectors of the algebraic Bethe
ansatz are the always the highest weight states of su(2). Descendents of the highest
weight vectors correspond to roots at infinity, correspondingly p j = 0. A second
observation is that the algebraic Bethe ansatz enables us to prove integrability of
the model and it gives an explicit construction of the Hilbert space of states in
terms of simultaneous eigenvectors of commuting integrals of motion. Comparing to
the classical inverse scattering method one can see that τ(λ) resembles the classical
action variables, while B(λ) corresponds to the angle variables.
5.3 Nested Bethe Ansatz (to be written)
Let g be an element from S M , the permutation group of of the integers 1 to M.
Obviously, there are M! permutations. Any such permutation is a collection of
integers
(
)
g = g1, g2, · · · , gM .
In other words, g puts g1 on the first place, etc. Every of M particles is characterized
by its position x i . We choose the fundamental region
x 1 ≤ x 2 ≤ · · · ≤ x M .
– 79 –

Now we have to specify which of M particles has a coordinate x 1 , which – coordinate
x 2 , etc. This is specified by fixing a permutation Q. The Bethe ansatz for the wave
function states that we look it in the form
Ψ(x|Q) = ∑
j=1 x jp π(j)
.
π∈S M
a(Q|π)e i P M
6. Introduction to Lie groups and Lie algebras
To introduce a concept of a Lie group we need two notions: the notion of a group
and the notion of a smooth manifold.
Definition of a group. A set of elements G is called a group if it is endowed with
two operations: for any pair g and h from G there is a third element from G which
is called the product gh, for any element g ∈ G there is the inverse element g −1 ∈ G.
The following properties must be satisfied
• (fg)h = f(gh)
• there exists an identity element I ∈ G such that Ig = gI = g
• gg −1 = I
Definition of a smooth manifold. Now we introduce the notion of a differentiable
manifold. Any set of points is called a differentiable manifold if it is supplied with
the following structure
• M is a union: M = ∪ q U q , where U q is homeomorphic (i.e. a continuous oneto-one
map) to the n-dimensional Euclidean space
• Any U q is supplied with coordinates x α q called the local coordinates. The regions
U q are called coordinate charts.
• any intersection U q ∩U p , if it is not empty, is also a region of the Euclidean space
where two coordinate systems x α q and x α p are defined. It is required that any
of these two coordinate systems is expressible via the other by a differentiable
map:
x α p = x α p (x 1 q, · · · x n q ) ,
α = 1, · · · n
x α q = x α q (x 1 p, · · · x n p) , α = 1, · · · n (6.1)
( )
∂x α
Then the Jacobian det
p
is different from zero. The functions (6.1) are
∂x β q
called transition functions from coordinates x α q to x α p and vice versa. If all the
transition functions are infinitely differentiable (i.e. have all partial derivatives)
the corresponding manifold is called smooth.
– 80 –

Definition of a Lie group: A smooth manifold G of dimension n is called a Lie
group if G is supplied with the structure of a group (multiplication and inversion)
which is compatible with the structure of a smooth manifold, i.e., the group operations
are smooth. In other words, a Lie group is a group which is simultaneously a
smooth manifold and the group operations are smooth.
The list of basic matrix Lie groups
• The group of n × n invertible matrices with complex or real matrix elements:
A = a j i , detA ≠ 0
It is called the general linear group GL(n, C) or GL(n, R). Consider for instance
GL(n, R). Product of two invertible matrices is an invertible matrix is
invertible; an invertible matrix has its inverse. Thus, GL(n, R) is a group. Condition
detA ≠ 0 defines a domain in the space of all matrices M(n, R) which is
a linear space of dimension n 2 . Thus, the general linear group is a domain in
the linear space R n2 . Coordinates in M(n, R) are the matrix elements a j i . If A
and B are two matrices then their product C = AB has the form
c j i = ak i b j k
It follows from this formula that the coordinates of the product of two matrices
is expressible via their individual coordinates with the help of smooth functions
(polynomials). In other words, the group operation which is the map
GL(n, R) × GL(n, R) → GL(n, R)
is smooth. Matrix elements of the inverse matrix are expressible via the matrix
elements of the original matrix as no-where singular rational functions (since
detA ≠ 0) which also defines a smooth mapping. Thus, the general Lie group
is a Lie group.
• Special linear group SL(n, R) or SL(n, C) is a group of real or complex matrices
satisfying the condition
detA = 1 .
• Special orthogonal group SO(n, R) or SO(n, C) is a group or real or complex
matrices satisfying the conditions
AA t = I , detA = 1 .
– 81 –

• Pseudo-orthogonal groups SO(p, q). Let g will be pseudo-Euclidean metric in
the space R n p,q with p + q = n. The group SO(p, q) is the group of real matrices
which preserve the form g:
AgA t = g , detA = 1 .
• Unitary group U(n) – the group of unitary n × n matrices:
UU † = I .
• Special unitary group SU(n) – the group of unitary n × n matrices with the
unit determinant
UU † = I , detU = 1 .
• Pseudo-unitary group U(p, q):
AgA † = g ,
where g is the pseudo-Euclidean metric. Special pseudo-unitary group requires
in addition the unit determinant detA = 1.
• Symplectic group Sp(2n, R) or Sp(2n, C) is a group or real or complex matrices
satisfying the condition
AJA t = J
where J is 2n × 2n matrix
and I is n × n unit matrix.
J =
( ) 0 I
−I 0
Question to the class: What are the eigenvalues of J? Answer:
J = diag(i, · · · i; −i, · · · , −i).
Thus, the group Sp(2n) is really different from SO(2n)!
The powerful tool in the theory of Lie groups are the Lie algebras. Let us see how
they arise by using as an example SO(3). Let A be “close” to the identity matrix
A = I + ɛa
is an orthogonal matrix A t = A −1 . Therefore,
I + ɛa t = (I + ɛa) −1 = I − ɛa + ɛ 2 a 2 + · · ·
– 82 –

From here a t = −a. The space of matrices a such that a t = −a is denoted as
so(3) and called the Lie algebra of the Lie group SO(3). The properties of this Lie
algebra: so(3) is a linear space, in so(3) the commutator is defined: if a, b ∈ so(3)
then [a, b] also belongs to so(3). A linear space of matrices is called a Lie algebra if
the commutator does not lead out of this space. Commutator of matrices naturally
arises from the commutator in the group:
ABA −1 B −1 = (I + ɛa)(I + ɛb)(I + ɛa) −1 (I + ɛb) −1
= (I + ɛa)(I + ɛb)(I − ɛa + ɛ 2 a 2 + · · · )(I − ɛb + ɛ 2 b 2 + · · · ) =
= I + ɛ(a + b − a − b) + ɛ 2 (ab − a 2 − ab − ba − b 2 + ab + a 2 + b 2 ) + · · · =
= I + ɛ 2 [a, b] + · · ·
The algebra and the Lie group in our example are related as
exp a =
∞∑
n=0
a n
n! = A ∈ SO(3)
Exponential of matrix. The exponent exp a of the matrix a is the sum of the
following series
∞∑ a m
exp a =
m! .
This series shares the properties of the usual exponential function, in particular it is
convergent for any matrix A. The following obvious properties are
• If matrices X and Y commute then
m=0
exp(X + Y ) = exp(X) exp(Y )
• The matrix A = exp X is invertible and A −1 = exp(−X).
• exp(X t ) = (exp X) t .
Definition of a Lie algebra: A linear vector space J (over a field R or C) supplied
with the multiplication operation (this operation is called the commutator) [ξ, η] for
ξ, η ∈ J is called a Lie algebra if the following properties are satisfied
1. The commutator [ξ, η] is a bilinear operation, i.e.
[α 1 ξ 1 + α 2 ξ 2 , β 1 η 1 + β 2 η 2 ] = α 1 β 1 [ξ 1 , η 1 ] + α 2 β 1 [ξ 2 , η 1 ] + α 1 β 2 [ξ 1 , η 2 ] + α 2 β 2 [ξ 2 , η 2 ]
2. The commutator is skew-symmetric: [ξ, η] = −[η, ξ]
– 83 –

3. The Jacobi identity
[[ξ, η], ζ] + [[η, ζ], ξ] + [[ζ, ξ], η] = 0
Let J be a Lie algebra of dimension n. Choose a basis e 1 , · · · , e n ∈ J . We have
[e i , e j ] = C k ije k
The numbers C k ij are called structure constants of the Lie algebra. Upon changing
the basis these structure constants change as the tensor quantity. Let e ′ i = A j i e i and
[e ′ i, e ′ j] = C ′k
ij e ′ k then C ′k
ij A m k e m = A r i A s j[e r , e s ] = A r i A s jC m rse m
Thus, the structure constants in the new basis are related to the constants in the
original basis as
C ′k
ij = A r i A s jC m rs(A −1 ) k m . (6.2)
Skew-symmetry and the Jacobi identity for the commutator imply that the tensor
C k ij defines the Lie algebra if and only if
C k ij = −C k ij , C m p[iC p jk] = 0 .
Classify all Lie algebras means in fact to find all solutions of these equations modulo
the equivalence relation (6.2).
Example. The Lie algebra so(3, R) of the Lie group SO(3, R). It consists of 3 × 3
skew-symmetric matrices. We can introduce a basis in the space of these matrices
⎛
0 0 0
⎞
⎛ ⎞
0 0 1
⎛ ⎞
0 −1 0
X 1 = ⎝ 0 0 −1 ⎠ , X 2 = ⎝ 0 0 0 ⎠ , X 3 = ⎝ 1 0 0 ⎠ .
0 1 0
−1 0 0
0 0 0
In this basis the Lie algebra relations take the form
[X 1 , X 2 ] = X 3 , [X 2 , X 3 ] = X 1 , [X 3 , X 1 ] = X 2 .
These three relation can be encoded into one
[X i , X j ] = ɛ ijk X k .
Example. The Lie algebra su(2) of the Lie group SU(2). It consists of 2 × 2 skewsymmetric
matrices. The basis can be constructed with the help of the so-called
Pauli matrices σ i
σ 1 =
( ) 0 1
, σ 2 =
1 0
( )
( )
0 −i
1 0
, σ 3 = .
i 0
0 −1
– 84 –

These matrices satisfy the relations
[σ i , σ j ] = 2iɛ ijk σ k , {σ i , σ j } = 2δ ij .
If we introduce X i = − i 2 σ i which are three linearly independent anti-hermitian matrices
then the su(2) Lie algebra relations read
[X i , X j ] = ɛ ijk X k
Note that the structure constants are real! Comparing with the previous example
we see that the Lie algebra su(2) is isomorphic to that of so(3, R):
su(2) ≈ so(3, R) .
With every matrix group we considered above one can associate the corresponding
matrix Lie algebra. The vector space of this Lie algebra is the tangent space at
the identity element of the group. For this case the operation “commutator” is the
usual matrix commutator. The tangent space to a Lie group at the identity element
naturally appears in this discussion. To understand why let us return to the case
of the Lie group GL(n, R). Consider a one-parameter curve A(t) ∈ GL(n, R), i.e, a
family of matrices A(t) from GL(n, R) which depend on the parameter t. Let this
curve to pass though the identity at t = 0, i.e., A(0) = I. Then the tangent vector
(the velocity vector!) at t = 0 is the matrix A(t)| ˙ t=0 . Other way around, let X be
an arbitrary matrix. Then the curve A(t) = I + tX for t sufficiently closed to zero
lies in GL(n, R). It is clear that
A(0) = I ,
˙ A(0) = X .
In this way we demonstrated that the space of vectors which are tangent to the group
GL(n, R) at the identity coincide with the space of all n × n matrices. This example
of GL(n, R) demonstrates a universal connection between Lie group G and its Lie
algebra: The tangent space to G at the identity element is the Lie algebra w.r.t. to
the commutator. This Lie algebra is called the Lie algebra of the group G.
Exercise to do in the class: making infinitesimal expansion of a group element
close to the identity compute the Lie algebras for the classical matrix groups discussed
above. The answer is the following list:
The list of basic matrix Lie algebras
• The general Lie group GL(n, R) or GL(n, C) has the matrix Lie algebra which
is M(n, R) or M(n, C), where M(n) is the space of all real or complex matrices.
– 85 –

• Special linear group SL(n, R) or SL(n, C) has the Lie algebra sl(n, R) or
sl(n, C) which coincides with the space of all real or complex matrices with
zero trace.
• Special orthogonal group SO(n, R) or SO(n, C) has the Lie algebra so(n, R) or
so(n, C) which are real or complex matrices satisfying the condition
X t = −X .
• Pseudo-orthogonal group SO(p, q) has the Lie algebra which is the algebra of
matrices X satisfying the condition
Xg + gX t = 0 .
We see that if we introduce the matrix u = Xg then the relation defining the
Lie algebra reads
u + u t = 0 .
Thus, the matrix u is skew-symmetric u t + u = 0. This map establishes the
isomorphism between so(p, q) and the space of all skew-symmetric matrices.
• Unitary group U(n) has the Lie algebra which is the space of all anti-hermitian
matrices
X † = −X .
• Special unitary group SU(n) has the Lie algebra which is the space of all antihermitian
matrices with zero trace
X † = −X , trX = 0 .
• Pseudo-unitary group U(p, q) has the Lie algebra which is the space of all
matrices obeying the relation
Xg + gX † = 0 .
The space u(p, q) is isomorphic to the space of anti-hermitian matrices. The
isomorphism is established by the formula u = Xg. Finally the Lie algebra of
the special pseudo-unitary group is defined by further requirement of vanishing
trace for X.
• The symplectic group Sp(2n, R) or Sp(2n, C) has the Lie algebra which comprises
all is the is a group or real or complex matrices satisfying the condition
XJ + JX t = 0
– 86 –

where J is 2n × 2n matrix
and I is n × n unit matrix.
J =
( ) 0 I
−I 0
Linear representations of Lie groups Consider an action of a Lie group a n-
dimensional vector space R n . This action is called a linear representation of Lie
group G on R n if for any g ∈ G the map
ρ :
g → ρ(g)
is a linear operator on R n . In other words, by a linear representation of G on
R n we call the homomorphism ρ which maps G into GL(n, R), the group of linear
transformations of R n . The homomorphism means that under this map the group
structure is preserved, i.e.
ρ(g 1 g 2 ) = ρ(g 1 )ρ(g 2 ) .
Any Lie group G has a distinguished element – g 0 = I and the tangent space T at
this point. Transformation
G → G :
g → hgh −1
is called internal automorphism corresponding to an element h ∈ G. This transformation
leaves unity invariant: hIh −1 = I and it transforms the tangent space T into
itself:
Ad(h) : T → T .
This map has the following properties:
Ad(h −1 ) = (Adh) −1 , Ad(h 1 h 2 ) = Adh 1 Adh 2 .
In other words, the map h → Adh is a linear representation of G:
where n is the dimension of the group.
Ad : G → GL(n, R) ,
Generally, one-parameter subgroups of a Lie group G are defined as parameterized
curves F (t) ⊂ G such that F (0) = I and F (t 1 +t 2 ) = F (t 1 )F (t 2 ) and F (−t) = F (t) −1 .
As we have already discussed for matrix groups they have the form
F (t) = exp(At)
– 87 –

where A is an element of the corresponding Lie algebra. In an abstract Lie group G
for a curve F (t) one defines the t-dependent vector
F −1 ˙ F ∈ T .
If this curve F (t) is one-parameter subgroup then this vector does not depend on t!
Indeed,
F ˙
dF (t + ɛ)
( dF (ɛ)
)
= | ɛ=0 = F (t)
,
dɛ
dɛ ɛ=0
i.e. F ˙ = F (t) F ˙ (0) and F −1 (t) F ˙ (t) = F ˙ (0) = const. Oppositely, for any non-zero
a ∈ T there exists a unique one-parameter subgroup with
F −1 ˙ F = a .
This follows from the theorem about the existence and uniqueness of solutions of
usual differential equations.
It is important to realize that even for the case of matrix Lie groups there are matrices
which are not images of any one-parameter subgroup. The exercise to do in the class:
Consider the following matrix:
( ) −2 0
g =
∈ GL + (2, R) ,
0 −3
where GL + (2, R) is a subgroup of GL(2, R) with positive determinant. Show that
there does not exist any real matrix ξ such that
e ξ = g .
The answer: it is impossible because since the matrix ξ is real the eigenvalues λ 1,2
of ξ must be either real of complex conjugate. The eigenvalues of e ξ are e λ 1
and e λ 2
.
If λ i are real then e λ i
> 0. If λ i are complex conjugate then e λ i
are also complex
conjugate.
It is also important to realize that different vectors ξ under the exponential map can
be mapped on the one and the same group element. As an example, consider the
matrices of the form
ξ = α
( ) 1 0
+ β
0 1
( ) 0 1
,
−1 0
where α, β ∈ R. Exponent e ξ can be computed by noting that
( ) 2 ( )
0 1 1 0
= − .
−1 0 0 1
Then we have
[ ( )
e ξ = e α 1 0
cos β +
0 1
( ) 0 1
]
sin β .
−1 0
– 88 –

It is clear that
( ) 1 0
α + β
0 1
( ) 0 1
, α
−1 0
( ) 1 0
+ (β + 2πk)
0 1
( ) 0 1
−1 0
has the the same image under the exponential map. In the sufficiently small neighbourhood
of 0 in M(n, R) the map exp A is a diffeomorphism. The inverse map is
constructed by means of series
for x sufficiently close to the identity.
ln x = (x − I) − 1 2 (x − I)2 + 1 3 (x − I)3 − · · ·
Linear representation of a Lie algebra. Adjoint representation. Let J be a
Lie algebra. We say that a map
ρ : J → M(n, R)
defines a representation of the Lie algebra J is the following equality is satisfied
for any two vectors ζ, η ∈ J .
ρ[ζ, η] = [ρ(η), ρ(ζ)]
Let F (t) be a one-parameter subgroup in G. Then g → F gF −1 generates a oneparameter
group of transformations in the Lie algebra
AdF (t) : T → T .
The vector d AdF (t)| dt t=0 lies in the Lie algebra. Let a ∈ T and let F (t) = exp(bt)
then
d
dt AdF (t)| t=0 a = d (
)
exp(bt)a exp(−bt) | t=0 = [b, a]
dt
Thus to any element b ∈ J we associate an operator ad b which acts on the Lie
algebra:
ad b : J → J , ad b a = [b, a] .
This action defines a representation of the Lie algebra on itself. This representation
is called adjoint. To see that this is indeed representation we have to show that it
preserves the commutation relations, i.e. that from [x, y] = z it follows that
We compute
[adx, ady] = adz .
[adx, ady]w = adx adyw − ady adxw = [x, [y, w]] − [y, [x, w]] = [x, [y, w]] + [y, [w, x]] =
− [w, [x, y]] = [[x, y], w] = [z, w] = adzw .
– 89 –

Here the Jacobi identity has been used.
Semi-simple and simple Lie algebras. General classification of Lie algebras is
a very complicated problem. To make a progress simplifying assumptions about the
structure of the algebra are needed. The class of the so-called simple and semi-simple
Lie algebras admits a complete classification.
A Lie subalgebra H of a Lie algebra J is a linear subspace H ⊂ J which is closed
w.r.t. to the commutation operation. An ideal H ⊂ J is a subspace in J such that
for any x ∈ J the following relation holds
[x, H] ⊂ H .
A Lie algebra J which does not have any ideals except the trivial one and the one
coincident with J is called simple. A Lie algebra which have no commutative (i.e.
abelian) ideals is called semi-simple. One can show that any semi-simple Lie algebra
is a sum of simple Lie algebras. Consider for instance the Lie algebra u(n) which is
the algebra of anti-hermitian matrices
u + u † = 0 .
The Lie algebra su(n) is further distinguished by imposing the condition of vanishing
trace: tru = 0. The difference between u(n) and su(n) constitute all the matrices
which are proportional to the identity matrix iI. Since
[λiI, u] = 0
the matrices proportional to iI form an ideal in u(n) which is abelian. Thus, u(n)
has the abelian ideal and, therefore, u(n) is not semi-simple. In opposite, su(n) has
no non-trivial ideals and therefore it is the simple Lie algebra.
A powerful tool in the Lie theory is the so-called Cartan-Killing from on a Lie algebra.
Consider the adjoint representation of J . The Cartan-Killing form on J is defined
as
(a, b) = −tr(ad a ad b )
for any two a, b ∈ J . The following central theorem in the Lie algebra theory can
be proven: A Lie algebra is semi-simple if and only if its Cartan-Killing form is
non-degenerate.
For a simple Lie algebra J of a group G the internal automorphisms Adg constitute
the linear irreducible representation (i.e. a representation which does not have invariant
subspaces) of G in J . Indeed, if Ad(g) has an invariant subspace H ⊂ J ,
i.e. gHg −1 ⊂ H for any g then sending g to the identity we will get
[J , H] ⊂ H
– 90 –

i.e. H is an ideal which contradicts to the assumption that J is the semi-simple Lie
algebra.
Cartan subalgebra. To demonstrate the construction of the adjoint representation
and introduce the notion of the Cartan subalgebra of the Lie algebra we use the
concrete example of su(3). The Lie algebra su(3) comprises the matrices of the form
iM, where M is traceless 3×3 hermitian matrix. The basis consists of eight matrices
which we chose to be the Gell-Mann matrices:
⎛ ⎞
⎛ ⎞
⎛ ⎞
0 1 0
0 −i 0
1 0 0
λ 1 = ⎝ 1 0 0 ⎠ , λ 2 = ⎝ i 0 0 ⎠ , λ 3 = ⎝ 0 −1 0 ⎠
0 0 0
0 0 0
0 0 0
⎛ ⎞
⎛ ⎞
⎛ ⎞
0 0 1
0 0 −i
0 0 0
λ 4 = ⎝ 0 0 0 ⎠ , λ 5 = ⎝ 0 0 0 ⎠ , λ 6 = ⎝
1 0 0
i 0 0
⎛ ⎞
⎛ ⎞
0 0 0
λ 7 = ⎝ 0 0 −i ⎠ , λ 8 = 1 1 0 0
√ ⎝ 0 1 0 ⎠ .
3
0 0 0
0 0 −2
0 0 1
0 1 0
There are two diagonal matrices among these: λ 3 and λ 8 which we replace by T z =
1
λ 2 3 and Y = √ 1
3
λ 8 . We introduce the following linear combinations of the generators
t ± = 1 2 (λ 1 ± iλ 2 ) , v ± = 1 2 (λ 4 ± iλ 5 ) , u ± = 1 2 (λ 6 ± iλ y ) .
One can easily compute, e.g.,
[t + , t + ] = 0 , [t + , t − ] = 2t z , [t + , t z ] = −t + , [t + , u + ] = v + , [t + , u − ] = 0 ,
[t + , v + ] = 0 , [t + , v − ] = −u − , [t + , y] = 0 .
Since the Lie algebra of su(3) is eight-dimensional the adjoint representation is eightdimensional
too. Picking up (t + , t − , t z , u + , u − , v + , v − , y) as the basis we can realize
the adjoint action by 8 × 8 matrices. For instance,
⎛ ⎞ ⎛
⎞ ⎛ ⎞
t + 0 0 0 0 0 0 0 0 t +
t −
0 0 2 0 0 0 0 0
t −
t z
−1 0 0 0 0 0 0 0
t z
ad t+ u +
u −
=
0 0 0 0 0 1 0 0
u +
0 0 0 0 0 0 0 0
u −
v +
0 0 0 0 0 0 0 0
v +
⎜ ⎟ ⎜
⎟ ⎜ ⎟
⎝ v − ⎠ ⎝ 0 0 0 0 −1 0 0 0 ⎠ ⎝ v − ⎠
y 0 0 0 0 0 0 0 0 y
} {{ }
matrix realization of t +
⎠
– 91 –

Note that both ad tz and ad y are diagonal. Thus, if x = at z + by then ad x is also
diagonal. Explicitly we find
⎛
⎞
a 0 0 0 0 0 0 0
0 −a 0 0 0 0 0 0
0 0 0 0 0 0 0 0
ad x =
0 0 0 − 1 2 a + b 0 0 0 0
1
0 0 0 0
2 a − b 0 0 0
.
1
0 0 0 0 0
2
⎜
a + b 0 0
⎝ 0 0 0 0 0 0 − 1a − b 0 ⎟
2
⎠
0 0 0 0 0 0 0 0
In other words, the basis elements (t + , t − , t z , u + , u − , v + , v − , y) are all eigenvectors
of ad x with eigenvalues a, −a, 0, − 1a + b, 1a − b, − 1 a − b and 0 respectively. The
2 2 2
procedure we followed in crucial for analysis of other (larger) Lie algebras. We found
a two-dimensional subalgebra generated by t z and y which is abelian. Further, we
have chosen a basis for the rest of the Lie algebra such that each element of the basis
is an eigenvector of ad x if x is from this abelian subalgebra. This abelian subalgebra
is called the Cartan subalgebra.
In general the Cartan subalgebra H is determined in the following way. An
element h ∈ H is called regular if ad h has as simple as possible number of zero eigenvalues
(i.e. multiplicity of zero eigenvalue is minimal). For instance, for su(3) the
element ad tz has two zero eigenvalues, while ad y has for zero eigenvalues. Thus, the
element ad tz is regular, while ad y is not. A Cartan subalgebra is a maximal commutative
subalgebra which contains a regular element. In our example the subalgebra
generated by t z and y is commutative and its maximal since there is no other element
we can add to it which would not destroy the commutativity.
Roots. It is very important fact proved in the theory of Lie algebras that any simple
Lie algebra has a Cartan subalgebra and it admits a basis where each basis vector
is an eigenstate of all Cartan generators; the corresponding eigenvalues depend of
course on a Cartan generator. In our example of su(3) for an element x = at z + by
– 92 –

we have
ad x t + = at +
ad x t − = at −
ad x t z = 0t z
ad x u + = (− 1 2 a + b)u +
ad x u − = ( 1 2 a − b)u −
ad x v + = ( 1 2 a + b)v +
ad x v − = (− 1 2 a − b)v −
ad x y = 0y .
We see that all eigenvalues are linear functions of the Cartan element x, in other
words, if we denote by e α the six elements t ± , v ± , u ± and by h i the two Cartan
elements t z , y we can write all the relations above as
[h i , h j ] = 0
[h i , e α ] = α(h i )e α ,
where α(h i ) is a linear function of h i . The generators e α , which are eigenstates of the
Cartan subalgebra, are called root vectors, while the corresponding linear functions
α(h) are called roots. To every root vector e α we associate the root α which is a
linear function on the Cartan sualgebra H. Linear functions on H, by definition,
form the dual space H ∗ to the Cartan subalgebra H.
The Cartan-Weyl basis. Now we can also investigate what is the commutator of
the root vectors. By using the Jacobi identity we find
[h, [e α , e β ]] = −[e α , [e β , h]] − [e β , [h, e α ]] = (α(h) + β(h))[e α , e β ] .
This clearly means that there are three distinct possibilities
• [e α , e β ] is zero
• [e α , e β ] is a root vector with the root α + β
• α + β = 0 in which case [e α , e β ] commutes with every h ∈ H and, therefore, is
an element of the Cartan subalgebra.
Thus,
[e α , e β ] = N αβ e α+β
– 93 –

if α + β is a root,
[e α , e −α ] ∼ h α
and [e α , e β ] = 0 if α + β is not a root. The numbers N αβ depend on the
normalization of the root vectors. The basis (h i , e α ) of a Lie algebra with the
properties described above is called the Cartan-Weyl basis.
– 94 –

7. Homework exercises
7.1 Seminar 1
Exercise 1. Consider a point particle moving in the potential U of the form depicted
in figure 1.
U
q
Fig. 1. Potential energy of a particle
Draw the phase curve of this particle. Hint: consult the case of the harmonic oscillator.
Exercise 2. Consider a point particle moving in the potential U of the forms depicted
in figure 2.
U
U
a)
q
b)
q
Fig. 2. Potential energies of a particle
Draw the corresponding phase curves.
Exercise 3. Consider a point particle of unit mass m which moves in one dimension
(the coordinate q and the momentum p) in the potential U(q), where
• case 1:
and g 2 is a (coupling) constant.
U(q) = g2
q 2 , E > 0
• case 2:
U(q) =
g2
sinh 2 q , E > 0
– 95 –

• case 3:
• case 4:
g2
U(q) = −
cosh 2 q , − g2 < E < 0
g2
U(q) = −
cosh 2 q , E > 0
Solve equations of motion for each of these potentials by quadratures. In which case
the motion if finite?
Exercise 3. Consider a linear space M with coordinates x k , k = 1, . . . , n. Show
that the expression
{F (x), G(x)} = C jk
l
x l ∂ j F ∂ k G
defines the Poisson bracket provided the constants C jk
l
constants of a Lie algebra.
7.2 Seminar 2
coincide with the structure
Exercise 1. Work out following the book three integrable tops: Euler, Lagrange and
Kowalewski tops. Work out equations of motion and check the conservation laws.
– 96 –

7.3 Seminar 3
Exercise 1. Consider a motion of a one-dimensional system. Let S(E) be the area
enclosed by the closed phase curve corresponding to the energy level E. Show that
the period of motion along this curve is equal to
T = dS(E)
dE .
Exercise 2. At the entry of the satellite into a circular orbit at a distance 300km
from the Earth the direction of its velocity deviates from the intended direction by
1 ◦ towards the earth. How is the perigee of the orbit changed?
Exercise 3. Find the principle axes and moments of inertia of the uniform planar
plate |x| ≤ q, |y| ≤ b, z = 0 with respect to 0.
Exercise 4. Find the inertia tensor of the uniform ellipsoid with the semi-axes a, b, c.
Exercise 5. Solve the Euler equations for the symmetric top: I 1 = I 2 .
Exercise 6. Consider the mathematical pendulum (of mass M) in the gravitational
field of the Earth. Integrate equations of motion in terms of Jacobi elliptic functions.
If the second (imaginary) period has any physical meaning? What is the elliptic
modulus k 2 ? Consider the limits k = 0 + and k = 1 − .
L
01
01
M
A pendulum in the gravitational field of the Earth. Here L is its length and G is
the gravitational constant.
G
– 97 –

7.4 Seminar 4
Exercise 1 (K. Bohlin). Consider the Kepler problem. Let x, y be the Cartesian
coordinates on the plane of motion. Introduce a complex variable z = x + iy and
show that the non-linear change of variables z → u 2 , t → τ given by
z = u 2 dt
,
dτ = 4|u2 | = 4|z|
maps the Kepler orbits with the constant energy E < 0 into the ones of the harmonic
oscillator with the complex amplitude u (a two-dimensional oscillator). Find the
period of oscillations.
Exercise 2 (Lissajous figures). Consider the two-dimensional harmonic oscillator.
Show that if
ω 1
ω 2
= r s ,
where r, s are relatively prime integers then there is a new additional integral of
motion
F = ā s 1a r 2 ,
where
ā 1 = 1 √ 2ω1
(p 1 + iω 1 q 1 ) , a 2 = 1 √ 2ω2
(p 2 − iω 2 q 2 ) .
The corresponding closed trajectories of the two-dimensional harmonic oscillator
are called the Lissajous figures. Find the Poisson brackets between F and F i =
1
2 (p2 i + ωi 2 qi 2 ), i = 1, 2.
Exercise 3. Consider the Kepler problem. Show that the components of the angular
momentum J i and the components of the Runge-Lenz vector R i form w.r.t. the
Poisson bracket the Lie algebra so(4). Recall that a 4 × 4 matrix X belongs to the
Lie algebra so(4) if it is skew-symmetric, i.e. X t + X = 0. Express the Kepler
Hamiltonian in terms of the conserved quantities J i and R i .
Exercise 4. Prove that the Poisson bracket
{L 1 , L 2 } = [r 12 , L 1 ] − [r 21 , L 2 ]
between the components of the matrix L implies that the quantities I k = trL k are
in involution, i.e. that {I k , I m } = 0.
Exercise 5 (Calogero model). Consider a dynamical system of n particles with
the coordinates q j and momenta p j , where j = 1, . . . n. The Hamiltonian of the
– 98 –

system is
H = 1 2
n∑
p 2 j + g ∑ 2 i

7.5 Seminar 5
Exercise 1 (Open Toda chain)
Consider a system of n interacting particles described by coordinates q j and the
corresponding conjugate momenta p j , where j = 1, . . . , n. The Hamiltonian of the
system has the form
H = 1 n∑ ∑n−1
p 2 j + exp[2(q j − q j+1 )] .
2
1
j=1
Show that equations of motion are equivalent to the Lax equation ˙L = [L, M], where
n∑ ∑n−1
L = p j E jj + exp[(q j − q j+1 )](E j,j+1 + E j+1,j ) ,
j=1
j=1
∑n−1
M = exp[(q j − q j+1 )](E j,j+1 − E j+1,j ) .
j=1
Here E jk is a matrix which has only one non-zero matrix element equal to 1 standing
in the intersection of j’s row with k’s column.
Exercise 2 (Differential equations for Jacobi elliptic functions).
Using the differential equation for the Jacobi elliptic function sn(x, k):
(sn ′ (x, k)) 2 = (1 − sn(x, k) 2 )(1 − k 2 sn(x, k) 2 ) .
and the identities relating sn(x, k) with other two functions cn(x, k) and dn(x, k)
derive the differential equations for cn(x, k) and dn(x, k).
Exercise 3 (The cnoidal wave and soliton of the NLS equation)
Consider the non-linear Schrodinger (NLS) equation
i ∂ψ
∂t = −ψ xx + 2κ|ψ| 2 ψ ,
where ψ = ψ(x, t) is a complex-valued function and we assume that κ < 0. By making
single-wave propagating ansatze for the modulus and the phase of ψ(x, t) determine
the cnoidal wave solution of the NLS equation. Show that upon degeneration of the
elliptic modulus the cnoidal wave turns into a soliton solution of the NLS equation.
Exercise 4 (Hamiltonian formulation of KdV equation)
Consider the following two Poisson brackets on the space of Schwarzian functions
u(x):
{u(x), u(y)} = −∂ x δ(x − y) .
– 100 –

Show that the KdV equation can be viewed as the Hamiltonian equation
where
H =
u t = {H, u} ,
∫ ∞
−∞
(
1
2 u2 x + u 3 )
.
Show that the Poisson structure is degenerate and
Q =
∫ ∞
−∞
is the central element of the Poisson bracket.
dx u(x)
Exercise 5 (Sine-Gordon Lagrangian)
Consider the Sine-Gordon model with the Lagrangian density
L = 1 2 ∂ µφ∂ µ φ + m2
(1 − cos βφ)
β2 over two-dimensional Minkowski space-time. Using the canonical formalism construct
the Hamiltonian (the generator of time translations) of the model. Using the
Noether theorem construct the momentum P (the generator of space translations)
and the generator K of Lorentz rotations.
Remark. The generators H, P, K form the Poincáre algebra of two-dimensional spacetime.
– 101 –

7.6 Seminar 6
Exercise 1
Prove that the commutators
[e 2 , e 3 ] = e 1 , [e 1 , e 5 ] = 2e 1 , [e 2 , e 5 ] = e 2 + e 3 [e 3 , e 5 ] = e 3 + e 4 [e 4 , e 5 ] = e 4
endows the space R 5 with the structure of a Lie algebra. Find the structure tensor
(structure constants) of this Lie algebra.
Exercise 2
In the space R 3 define a multiplication
[e a , e b ] = 0 , [e 3 , e a ] = B b ae b ,
where a, b = 1, 2 and B b a is a 2 × 2 matrix. Show that this commutator table endows
the space R 3 with the structure of a Lie algebra. Show that this construction allows
one to obtain any three-dimensional Lie algebra.
Exercise 3 (The exponential map)
Let X be an element of the Lie algebra sl(2, R). Show that
• if detX < 0 then
e X = cosh √ −detX I + sinh √ −detX
√
−detX
X .
• if detX > 0 then
e X = cos √ detX I + sin √ detX
√
detX
X .
Exercise 4 (The exponential map)
Prove an equality
⎛
⎞ ⎛
⎞
λ 1 0 · · · 0 e λ e λ eλ
e
· · · λ
2! (n−1)!
0 λ 1 · · · 0
0 e λ e λ e
· · · λ
(n−2)!
exp
⎜
. .
=
⎟
.
.
.
⎝ 0 1 ⎠ ⎜
⎟
⎝
e
0 λ
⎠
1!
0 · · · λ 0 · · · e λ
– 102 –

Exercise 5
Prove that for any matrix A the following identity is valid
det(exp A) = exp(trA) ,
or, equivalently,
exp(tr ln A) = detA .
Remark. This is very important identity which enters into the proofs of many
formulas from various branches of mathematics and theoretical physics. It must
always stay with you. Learn it by heart by repeating the magic words ”exponent
trace of log is determinant”.
Exercise 6
Let
⎛
0 −c 3 c 2
⎞
A = ⎝ c 3 0 −c 1
⎠ .
−c 2 c 1 0
Show that the matrices
O(c 1 , c 2 , c 3 ) = (I + A)(I − A) −1
belong to the Lie group SO(3). Show that the multiplication operation in SO(3)
written in coordinates (c 1 , c 2 , c 3 ) takes the form
where
O(c)O(c ′ ) = O(c ′′ )
c ′′ = (c + c ′ + c × c ′ )/(1 − (c, c ′ )) .
Here c = (c 1 , c 2 , c 3 ) is viewed as three-dimensional vector.
Exercise 7
Let G = SU(2) and H 2j is the space of all homogenious polynomials of degree 2j,
j = 0, 1/2, 1, · · ·
with a n ∈ C. Show that
f(z 1 , z 2 ) =
n=j
∑
n=−j
a n z j−n
1 z j+n
2
T j (g)f(z 1 , z 2 ) = f(αz 1 + γz 2 , βz 1 + δz 2 )
– 103 –

is a representation of the group SU(2). Here
g =
( ) α β
γ δ
with α = ¯δ and γ = − ¯β is a group element of SU(2).
Exercise 8
Prove that
⎛
1 + c
2
2 1t 2 c 1 c 2 t 2 − c 3 t c 1 c 3 t 2 ⎞
+ c 2 t
α(φ) = −I + ⎝ c
1 + c 2 t 2 2 c 1 t 2 + c 3 t 1 + c 2 2t 2 c 2 c 3 t 2 − c 1 t ⎠
c 3 c 1 t 2 − c 2 t c 3 c 2 t 2 + c 1 t 1 + c 2 3t 2
is a one-parameter subgroup in SO(3), where tan φ 2 = ct, c2 = c 2 1 + c 2 2 + c 2 3 and
⃗c = (c 1 , c 2 , c 3 ) is a constant vector.
Exercise 9
Let
⎛
cos ϕ − sin ϕ
⎞
0
⎛
1 0
⎞
0
B ϕ = ⎝ sin ϕ cos ϕ 0 ⎠ , C θ = ⎝ 0 cos θ − sin θ ⎠ .
0 0 1
0 sin θ cos θ
Show that any matrix A ∈ SO(3) can be represented in the form
A = B ϕ C θ B ψ .
Write the one-parameter subgroup from the exercise 8 in the coordinates (ϕ, θ, ψ).
– 104 –

7.7 Seminar 7
Exercise 1
Consider the classical Heisenberg model. Show that the formula for the Poisson
brackets between the components of the Lax matrix
[
]
{U(x, λ), U(y, µ)} = r(λ, µ), U(x, λ) ⊗ I + I ⊗ U(y, µ) δ(x − y) ,
with the classical r-matrix
r(λ, µ) = 1 σ i ⊗ σ i
2 λ − µ .
implies that the Poisson bracket between the components of the monodromy matrix
is of the form
[ ∫ 2π ]
T(λ) = P exp dx U(x, λ)
0
{T(λ) ⊗ T(µ)} =
[
]
r(λ, µ), T(λ) ⊗ T(µ) .
Exercise 2
Show that the Jacobi identity for the Poisson bracket
[
]
{T(λ) ⊗ T(µ)} = r(λ, µ), T(λ) ⊗ T(µ) .
implies the classical Yang-Baxter equation for the r-matrix sckew-symmetric r 12 (λ, µ) =
−r 21 (µ, λ):
[r 12 (λ, µ), r 13 (λ, ν)] + [r 12 (λ, µ), r 13 (µ, ν)] + [r 13 (λ, ν), r 23 (µ, ν)] = 0
Check (e.g. by using Mathematica) that the r-matrix
solves the classical Yang-Baxter equation.
r(λ, µ) = 1 σ i ⊗ σ i
2 λ − µ .
Exercise 3
Consider the zero-curvature representation for the KdV equation:
( )
(
)
0 1
u
U =
, V =
x
4λ − 2u
λ + u 0
4λ 2 + 2λu + u xx − 2u 2 − u x
Using abelianization procedure around the pole λ = ∞ find the first four integrals
of motion.
.
– 105 –

Exercise 4
Consider the non-linear Schrodinger equation:
i ∂ψ
∂t = −∂2 ψ
∂x 2 + 2κ|ψ|2 ψ ,
where ψ ≡ ψ(x, t) is a complex function. Show that this equation admits the following
zero-curvature representation
where
and
U = U 0 + λU 1 , V = V 0 + λV 1 + λ 2 V 2 ,
U 0 = √ κ( ¯ψσ + + ψσ − ) , U 1 = 1 2i σ 3
V 0 = iκ|ψ| 2 σ 3 − i √ κ(∂ x ¯ψσ+ − ∂ x ψσ − ) , V 1 = −U 0 , V 2 = −U 1 .
Using the abelianization procedure around λ = ∞ find the first four local integrals
of motion. What is the physical meaning of the first three integrals?
– 106 –

7.8 Seminar 8
Exercise 1
Consider XXX Heisenberg model. For the chain of length L = 3 find the matrix form
of the Hamiltonian as well as its eigenvalues. Construct the corresponding matrix
representation of the global su(2) generators. How many su(2) multiplets the Hilbert
space of the L = 3 model contains?
Exercise 2
Carry out an explicit construction of the Bethe wave-function a(n 1 , n 2 , n 3 ) for threemagnon
states of the Heisenberg model. Derive the corresponding Bethe equations.
Exercise 3
Show that on the rapidity plane λ = 1 cot p the S-matrix of the Heisenberg model
2 2
takes the form
S(λ 1 , λ 2 ) = λ 1 − λ 2 + i
λ 1 − λ 2 − i .
Hence, it depends only on the difference of rapidities of scattering particles.
Exercise 4
Show that L two-magnon states of the Heisenberg model with p 1 = 0 and p 2 = 2πm
L
with m = 0, 1, . . . , L − 1 are su(2)-descendants of the one-magnon states.
– 107 –