Spontaneous symmetry breaking in a
one-dimensional spin chain by reflecting
boundary conditions
Annelene Schulze
July 16, 2019
Bachelorscriptie Wiskunde en Natuur- & Sterrenkunde Begeleiding: prof. dr. Jasper Stokman, dr. Jasper van Wezel
Institute of Physics
Korteweg-de Vries Instituut voor Wiskunde
Abstract
In this thesis we solve the XXX1/2 spin chain and show that spontaneous symmetry breaking occurs. We do this by using the Algebraic Bethe Ansatz, from which you can derive the bethe equations. With these it is possible to find the possible eigenvalues of the Hamiltonian. We first do this for the periodic case, where H is given as
H = −J
n
X
i
Si· Si+1
To measure sponteaneous symmetry breaking we impose reflective boundary conditions by putting a magnetif field B in the Sz direction on one side of the chain and −Sz
direction on the other side. The Hamiltonian is then given by H = N −1 X n=1 (σn1σn+11 + σn2σn+12 + σn3σn+13 ) + 1 ξ− σ31+ 1 ξ+ σN3 The Bethe equations for the periodic case are given by
λj+2i λj−2i !N = l Y k6=j λj− λk+ i λj− λk− i
and for the system with reflective boundary conditions by −(µm+ ξ+− 1 2) (µm− ξ++12) (2µm− 1∆+) ∆−(µm) = M Y k=1,k6=m (µm− µk+ 1) (µm− µk− 1) (µm+ µk+ η) (µm+ µk− 1)
Titel: Spontaneous symmetry breaking in a one-dimensional spin chain by reflecting boundary conditions
Auteur: Annelene Schulze, Annelene Schulze 10793046 Begeleiding: prof. dr. Jasper Stokman, dr. Jasper van Wezel
Tweede beoordelaars: dr. Maris Ozols, prof. dr. Jean-Sebastien Caux Einddatum: July 16, 2019
Institute of Physics
University van Amsterdam
Science Park 904, 1098 XH Amsterdam http://www.iop.uva.nl
Korteweg-de Vries Instituut voor Wiskunde University van Amsterdam
Science Park 904, 1098 XH Amsterdam http://www.kdvi.uva.nl
Contents
1 Introduction 1
1.1 Acknowledgements . . . 2
2 Lie groups, Lie algebras and their representations 3 2.1 Definition of Lie Groups and Lie algebras . . . 3
2.2 The connection between SO(3) and SU(2) . . . 5
2.3 Representations of Lie algebras . . . 7
2.3.1 Representations of sl(2, C) . . . 7
3 Spin, and the XXX1/2 quantum spin chain 11 3.1 Spin . . . 11
3.2 The Heisenberg model . . . 14
3.3 Ferromagnetism and antiferromagnetism . . . 15
3.4 A notion of integrable systems . . . 16
3.5 A specific R-matrix . . . 17
4 Symmetry breaking in quantum mechanics 20 4.1 The concept of symmetry . . . 20
4.2 Spontaneous symmetry breaking in a quantum system . . . 21
5 Algebraic Bethe Ansatz 25 5.1 The Lax operator . . . 26
5.2 T-operator . . . 27
5.3 Proof that the Hamiltonian is a element of F (λ) in span{Q0,Q1,· · · .} . . . 31
5.4 The Bethe Ansatz Equations for the XXX1/2 model . . . 34
6 Imposing reflective boundary conditions on the XXX1/2 spin chain 40 6.1 The K-matrix . . . 41
6.2 Boundary conditions for the open chain . . . 42 6.3 Bethe equations and Bethe eigenvalues for reflective boundary conditions 47 6.4 Measure the spontaneous symmetry breaking in the quantum spin chain . 49
7 Conclusion 51
8 Popular summary 52
1 Introduction
In physics, we try to understand the universe and the world around us. For simple systems with only a small amount of variables it is often possible to solve the system, whilst for larger and more complex system, it becomes very difficult if not impossible to solve the system completely. One example of a system that is diffucult to solve is the one dimensional quantum spin chain. In 1931, Hans Bethe , whilst working at the laboratory of Enrico Fermi, wrote the famous article ”Zur Theorie der Metalle” in which he described a method to find to find all the eigenvalues and eigenvectors of an one di-mensional chain of electrons. This was a revolutionary article because it made it possible to find all the energy states of specific many body systems.
The notion of integrability in classical mechanics means that a system can be exactly solved and all the eigenstates are found. But by the differences in quantum and classical mechanics that same notion will have to be defined a bit differently for quantum sys-tems. In 1979, Ludvig Faddeev developed with his students in Sint Petersburg a method to solve integrable quantum systems which is better known under the name ”Algebraic Bethe Ansatz. This mathematical approach made it possible to solve in a mathematical way integrable systems in 1 + 1 dimensions, for which the integrability is governed by a solution of the Yang-Baxter equation, which is also called Bethe integrable. The one dimensional spin chain is an interesting model to solve as it these ”toy models” can tell us more about quantum field theoretic models of higher dimensions, as they are easier to solve. For instance is it possible to look at symmetry breaking which can occur in the spin chain.
Symmetry is an important concept in both physics and mathematics. But just as inter-esting, maybe even more interesting is what happens when symmetry is broken. Most objects in our physical world experience some sort of symmetry breaking. For infinitely large systems, an infinitesimal small perturbation can break the symmetry of the system. The spin chain can be influenced by an external magnetic field which could break it’s rotational symmetry.
In this thesis we will take a look at the XXX1/2 quantum spin chain which is Bethe integrable. The XXX1/2 spin chain is a many body quantum system. We will solve it
by using the Algebraic Bethe Ansatz. Before that, we will recall some definitions and theorems from Lie theory and quantum physics to understand the mechanics of the spin chain better. In chapter 5 we will discuss symmetry and symmetry breaking. After that, we will use the Algebraic Bethe Ansatz to solve the spin chain. First we will do that in the periodic case without boundary conditions and finally with reflective boundary
conditions. This way we will be able to find a method to measure spontaneous symmetry breaking in the XXX1/2 spin chain.
1.1 Acknowledgements
I would like to thank professor Jasper Stokman and dr. Jasper van Wezel for their supervision of my thesis and helping me out when I got stuck on the problems. I enjoyed the topic of my thesis very much and am grateful that they wanted to work with me on this.
2 Lie groups, Lie algebras and their
representations
Before we dive into quantum spin chain systems, let us first look at some important mathematical concepts which are underlying and give us a better understanding of the system. The theory of Lie groups and Lie algebras has some deep connections with properties of this system and we will recall some definitions and consequences of this theory and give the definitions of some specific Lie groups and algebras. Finally, we will also look at representations of these Lie algebras. In this chapter we will follow the book “An introduction to Lie groups and Lie algebras” by Kirillov and the paper “Representations of sl(2, C) and Clebsch-Gordan Rule” by Huan T. Vo.
2.1 Definition of Lie Groups and Lie algebras
Recall that a group G is a set together with an operation working on the elements of G. This operation is binary, meaning that it acts on two elements of the set, and returns a third element. A Lie group is a special kind of group as it is not only a group, but also a smooth differentiable manifold 1, which means that it is locally diffeomorphic to a Euclidean space. Because of this extra property, Lie groups are often referred to as continuous groups, although this does not capture the whole structure. This description however makes the use of Lie groups for understanding physical systems more intuitive since in physics most system operations are smooth.
Definition 1. A Lie group is a set G with two structures: G is a group and G is a smooth manifold. These structures agree in the sense that the multiplication map G × G → G and the inversion map G → G are smooth as maps between manifolds .
Group operations can be used to describe symmetries in physical systems. Continuous groups can be used for continuous symmetries. Examples of continuous symmetries are translational and rotational symmetry. The definition of a rotation is For the purpose of our thesis we will focus on two Lie groups with rotation operations. The first one is SU(2), which is a matrix group that is defined as follows.
Definition 2 (SU(2)). The Lie group SU(2) is a matrix group which is given by, SU(2) = {A ∈ GL(2, C) : A∗A = I, det(A) = 1} .
1
A topological manifold M is a Hausdorff space, second countable and locally Euclidean of dimension n. If it smooth, a function F on the local euclidean space is C∞.
where A∗ is the transposed complex conjugate matrix of A. That is, (A∗)ij = Aji. It
can also be written as the set of the matrices of the form SU(2) = α β −β α α, β ∈ C, |α| 2+ |β|2 = 1 .
Intuitively speaking, unitary transformation can be seen as rotations in a complex inner product space, and SU(2) gives the rotations in the complex plane. This is the complex analog of orthogonal transformations which are rotations in a real inner product space. We therefore also define the group of rotations in R3, which is called SO(3). Definition 3. SO(3), the 3D rotation group, is given by all invertible real 3 × 3 matrices A with det(A) = 1, and A−1= AT,
SO(3) = {A ∈ GL(3, R)| det(A) = 1, AT = A−1}.
Next to Lie groups we will also need the associated Lie algebras which are defined as Definition 4 (Lie Algebra). A Lie algebra is a vector space g over R or C together with a binary operation [·, ·] : g × g → g called the Lie bracket that satisfies the following axioms:
1. Bilinearity
[ax + by, z] = a[x, z] + b[y, z], (2.1) [z, ax + by] = a[z, x] + b[z, y] (2.2) for all scalars a, b in R or C and all elements x, y, z ∈ g.
2. Antisymmetry
[x, y] = −[y, x] (2.3) for all x and y in g.
3. The Jacobi identity,
[x, [y, z]] + [z, [x, y]] + [y, [z, x]] = 0 (2.4) for all x, y, z ∈ g.
A Lie algebra is often related to a Lie group as it’s tangent space at the identity element. Because Lie algebras are vector spaces rather than groups with also a smooth differentiable manifold structure on them, it is often easier to use the algebras since the properties of a vector space are easier to work with. The associated Lie algebras for SU(2) and SO(3) are given by
Definition 5. The Lie algebra su(2), associated to SU(2), is given by anti-hermitian 2 × 2 matrices with trace zero;
su(2) = {A ∈ Mat2(C) : A∗ = −A, Tr(A) = 0},
and a basis of su(2) is given by τ1 = 0 i i 0 , τ2= 0 1 −1 0 , τ3 = i 0 0 −i .
Definition 6. The Lie algebra so(3), associated to SO(3), is given by all skew symmetric 3 × 3 matrices, for which AT = A−1 and det(A) = 1.
A basis for so(3) is given by Lx = 0 0 0 0 0 −1 0 1 0 , Ly = 0 0 1 0 0 0 −1 0 0 , Lz = 0 −1 0 1 0 0 0 0 0 ,
2.2 The connection between SO(3) and SU(2)
It has been proven that there exists a remarkable relationship between the two Lie groups SU(2) and SO(3) which gives deep results in quantum physics. In this section we will give an elementary proof of the isomorphism between SO(3) ' SU (2)/Z2. Since the Lie
algebra su(2) is spanned by three matrices it is possible to write the elements of the Lie algebra as vectors in R3 by We will define a homomorphism between them to see their connection. R33 ~x = x1 x2 x3 ↔ 3 X i=1 xτi ∈ su(2)
Theorem 1. Consider the map
R : SU(2) → SO(3)
g 7→ Adg in the basis of {τi}i for su(2)
where Adg is the adjoint action given by Adg = gvg−1, gives a surjective Lie
homomor-phism, where v ∈ su(2) and g ∈ SU(2) and Proof. To prove that R maps g = α β
−β α
into SO(3), we represent it as a matrix given by the basis of su(2).
gτ1g−1 = i(βα + αβ) i(α2− β2) i(α2− β2) −i(βα + αβ) ! = Re(α2− β2)τ 1− Im(α2− β2)τ2+ 2Re(αβ)τ3 gτ2g−1 = αβ − βα α2+ β2 −(α2+ β2) −(αβ − βα) !
= Im(α2+ β2)τ1+ Re(α2+ β2)τ2+ 2Im(αβ)τ3
gτ3g−1 =
i(αα − ββ) −2iαβ −2iαβ i(ββ − αα)
Moreover, the matrix R(g) in the basis of τi for su(2) is given by
R(g) =
Re(α2− β2) Im(α2+ β2) −2Re(αβ)
−Im(α2− β2) Re(α2+ β2) 2Im(αβ)
2Re(αβ) 2Im(αβ) |α|2− |β|2
A painful but elementary matrix calculation shows that this has det(R(g)) = 1, and RT = R−1. As the matrix entries are quadratic in the real and imaginary parts of α, β,
it is also smooth.
To prove that it is an homomorphism, we see that
Ad(g1g2)(v) = g1· g2v(g1· g2)−1= (g1· g2)(v)(g2· g1)
= g1(g2vg−12 )g −1
1 = g1 Ad(g2)g−11
= Ad(g1) ◦ Ad(g2).
And we have proven that R is a Lie group homomorphism of SU(2) to SO(3)
We can use this fact to establish a relationship between the Lie groups SU(2) and SO(3
Theorem 2. The homomorphism R is surjective on SO(3), and SU(2)/Z2 where Z2=
{−1, 1} and Z2 = Ker(R), is isomorphic to SO(3):
SU(2)/Z2 ∼= SO(3)
Proof. The kernel of R is given by the matrices which send SU(2) to the identity: Ker(R) = {g ∈ SU(2) | gvg−1 = v ∀v ∈ su(2)} = {g ∈ SU(2) | gv = vg}. As ev-ery element of su(2) is a linear combination of the matrices τ1, τ2, τ3, we check which
matrix h ∈ SU(2) will leave the basis of su(2) unchanged under conjugation. Observe that hvh−1= v if and only if hv = vh. For h = a b
−b a : hτ1 = iβ iα iα −iβ =−iβ iα iα iβ = τ1h hτ2= −β α −α −β =−β α α β = τ2h hτ3 = iα −iβ −iβ −iα =iα iβ iβ −iα = τ3h
And we see that b = −b and a = a, so a ∈ R and b = 0. As all matrices in SU(2) have |a|2 + |b|2 = 1, it must be the case that a = ±1 and thus the kernel is given by Ker(R) = {1, −1} = Z2
To prove surjectivity we first notice that the homomorphism R is a 2 : 1 morphism as we can see by its kernel. So gvg−1 and (−g)v(−g)−1 are sent to the same element in SO(3). It is also clear that there does not exist a g ∈ SU(2) for which g = −g, as the 0 map is
not in SU(2). SU(2) has the structure of a smooth differentiable manifold and as such is locally Hausdorff. Therefore each g has an open neighborhood U and −g has an open neigbhorhood V s.t U ∩ V = ∅. This means that R restricts to a local homeomorphism on an open neighborhood U of any element in SU(2) and is an open map. Thus Im(R) is an open set, but also closed since SU(2) is compact [5] and SO(3) is Hausdorff, because it is a manifold. Since SO(3) is a connected group and Im(R) clopen, Im(R) = SO(3) and we conclude that R is surjective.
From group theory we remember the first isomorphism theorem, which says that when G and H are groups and φ : G → H is a surjective homomorphism, then H is isomorphic to G/ ker(φ) This also holds for Lie groups, so we find
SU(2)/Z2 ∼= SO(3)
and we have proven the theorem.
SU(2) is a double cover of SO(3) as we can see by the 2 : 1 homomorphism. The results of the last theorems give insight into some deep results in physics which we will come back to in the next chapter.
Corollary 1. The Lie algebras su(2) and so(3) are isomorphic
su(2) ∼= so(3) (2.5)
2.3 Representations of Lie algebras
The Lie groups and Lie algebras are still abstract mathematical concepts which we would like to link to a physical system. We do this by using a representation, which is defined as:
Definition 7. A representation of a Lie algebra g is a linear morphism of Lie algebras ρ : g → gl(V ), where ρ satisfies the condition:
ρ([X, Y ]) = ρ(X)ρ(Y ) − ρ(Y )ρ(X)
and gl(V ) is the general Lie algebra of degree n with the commutator as Lie bracket. This definition give us a tool to relate the properties of lie groups to physical systems which are described in (complex) vector spaces, and give us a notion on how to work with Lie algebras on our vector space. Representations can be build out of elementary building blocks, called irreducible representation. A Lie algebra representation ρ is called irreducible if the only g-invariant subspaces are V and {0}.
2.3.1 Representations of sl(2, C)
We are especially interested in representations of su(2) since these will give us under-standing about rotational operations on a complex vector space. The Lie algebra su(2)
is a real algebra and for our uses later in this thesis it is more convenient to work with the complexification sl(2), which is related to su(2) by:
sl(2, C) = su(2) + isu(2).
By expanding the Lie bracket of su(2) linearly it becomes a complex Lie algebra, where the complex representations of sl(2, C) are equivalent to su(2). Explicitly, sl(2, C) is given by the following definition.
Definition 8. The Lie algebra sl(2, C) is given by
sl(2, C) = {A ∈ Mat2(C) | Tr(A) = 0} (2.6)
A basis of this Lie algebra is given by e =0 1 0 0 , f =0 0 1 0 h =1 0 0 −1 , (2.7)
and the Lie brackets are given by:
[e, f ] = h, [h, e] = 2e, [h, f ] = −2f, (2.8) Another basis of the Lie algebra is given by the matrices
σ1= 0 1 1 0 , σ2= 0 −i i 0 , σ3 = 1 0 0 −1 (2.9) which are better known as the ”Pauli matrices”, and their importance will later become clearer. It is easily seen that these give an alternative basis, since σ1 = e+f , σ2 = i(f −e)
and σ3 = h.
Lemma 1. The Lie bracket for the Pauli matrix basis is given by [σi, σj] = 2ii,j,kσk
where i,j,k, the Levi-civita symbol, is an antisymmetric symbol which is 1 if only if i, j
and k are all distinct and otherwise 0.
Proof. We prove this by working out the Lie bracket given in the definition of sl(2, C) [σ1, σ2] = σ1σ2− σ2σ1= 2i[e, f ] = 2ih = 2iσ3
[σ2, σ3] = σ2σ3− σ3σ2= 2i([f, h] + [e, h]) = 2i(e + f ) = 2iσ1
[σ1, σ3] = σ1σ3− σ3σ1= [e, h] + [f, h] = −2e + 2f = 2iσ2
For physical systems defined in complex vector spaces it is preferred to work with sl(2, C) instead of su(2). The complex representations are equal but since it is the com-plexification the basis for sl(2, C) is given in one more real basis element. Observables, which are used for measuring properties in quantum physics, are Hermitian operators. These are real operators and by using a more real basis easier defined. If a group G is connected and simply connected the categories of G and it’s associative algebra g are the same. This means that the complex representations of SU(2) and sl(2, C) are the same.
The last part of this chapter will be about a specific representation of sl(2, C). Let (V, ·) be a representation of sl(2, C) by:
sl(2, C) × V → V (x, v) → (ρ(x)) · (v) We define the subspace V [λ] for λ ∈ C by:
V [λ] := {v ∈ V | hv = λv} (2.10) where h is the basis element of the basis (2.7 for sl(2, C). When V [λ] 6= 0, we call λ a weight of V , and accordingly V [λ] is called the weight space. A vector in V [λ] is called a weight vector and is an eigenvector of h so V [λ] is an eigenspace of h. If Re λ ≤ Re λ0 for all λ0 of V, we call λ the highest weight of V . It is interesting to see what the action of the other two basis elements are on V [λ]
Lemma 2. The actions of e and f on V [λ] are given by eV [λ] ⊂ V [λ + 2] f V [λ] ⊂ V [λ − 2].
Proof. We compute these actions by using the Lie brackets and that V is a representation of sl(2, C)
h(ev) = (he)(v) = (he − eh + eh)v = ([h, e] + eh)v = (2e + eλ)v = (2 + λ)ev
h(f v) = (hf )v = (hf − f h + f h)v = ([h, f ] + f h)v = (−2f + f λ)v = (λ − 2)f v
The space V [λ] is not invariant under the actions of e and f . In fact e works as a raising operator and f as a lowering operator as they respectively send the elements to the eigenspaces of h for λ − 2 and λ + 2.
Theorem 3. (1) For any n ≥ 0, let Vn be the finite dimensional vector space with basis
{v0, v1, ..., vn}. Define the action of sl(2, C) by
hvk = (n − 2k)vk, 0 ≤ k ≤ n
f vk = vk+1, 0 ≤ k < n; f vn= 0
evk = k(n + 1 − k)vk−1, 0 < k ≤ n; ev0 = 0
Corollary 2. Every finite dimensional representation V of sl(2, C) is the direct sum of it’s weight spaces. In other words,
V =M
n∈Z
V [n] where only finitely many V [n] are non-zero
Proof. We are only going to give a sketch of the proof, for the full proof the reader can consult the paper the paper by Huan. T. Vo [11]. Every finite dimensional representation of sl(2, C) is completely reducible. This follows from the fact that the representations of SU(2) and sl(2), and it suffices to prove that a finite dimensional representation of SU(2) with an inner product defined is completely reducible. By corollary 1.4 in [11], every irreducible representation of sl(2, C) with highest weight n can be written as a direct sum of the weight spaces. Now it follows directly since every completely reducible representation can be written as a sum of irreducible representations, and we conclude that we can write it as a direct sum of its weight spaces. [11]
3 Spin, and the XXX
1/2
quantum spin
chain
3.1 Spin
As a first step to understand the XXX1/2 spin chain let us first recall some facts about single spin systems. Conserved quantities, like mass, energy and charge of a quantum particle are called quantum numbers. One of those quantum numbers is spin, which has quantum number s and from the name related to the angular momentum. A spin state of a vector can be represented as a vector in a complex Hilbert space1, and the space is often called a spin space. Observables, which are Hermitian operators working on the spin space and measure it are called spin operators. For the purpose of this thesis we are interested in particles with spin s = 1/2, which are for example electrons. The highest weight representation of SL(2, C) with highest weights labeled by half integer s = 0, 1/2, 1, ... can by corollary 2 be written as a direct sum of it’s weight spaces. The quantum number s corresponds to the highest weigh representation of sl(2, C) and the Lie group SU(2) was the group of rotations in C2 and as we represent a spin state of s = 1/2 in C2 representations of this algebra give the spin operators which give the rotation of a spin state. This is the closest the analogy with the angular momentum will get. We recall the Pauli matrices which give a basis for sl(2, C)
σ1 = 0 1 1 0 , σ2= 0 i −i 0 , σ3 = 1 0 0 −1 (3.1) ANd we can write the representations of sl(2, C) as a trivial representation π1/2 of the
Pauli matrices.
π1/2: sl(2, C) → gl(C2), characterized by π1/2(σi) = 2Si.
which gives the spin operators S1, S2 and S3
S1 = 1 2 0 1 1 0 , S2 = 1 2 0 i −i 0 , S3 = 1 2 1 0 0 −1
. For a more intuitive picture of the spin we can represent the spin states as vectors in R3 since the su(2) ∼= so(3) Because of this, even though its only on an intuitive level, as the particles with spin are point particles, we can actually see the spin operators as a rotation around the axis.
0 |↑i |↓i |ψi z x y
Figure 3.1: The Bloch’s sphere, a often used graphical representation of spin in 3-dimensions with spin state ψ as vector shown in the basis |↑i and |↓i. [12] Lemma 3. The commutation relation of the spin operators is given by
[Smα, Snβ] = iαβγSnγδnm
Proof. Since S is a representation of sl(2, C) and is given by Sα = 12σi, we can write
[Smα, Snβ] = SmαSnβ− Sβ nSmα = 1 4σiσj− 1 4σjσi = 1 4[σi, σj] = i 2i,j,kσk = iα,βγSγδm,n
where the Dirac delta operator just emphasises that if the spin operators work on dif-ferent states in difdif-ferent Hilbert spaces, they will always commute.
It is convention to ues the eigenstates of Sz as basis for the spin space, where the notation of spin up is given by |↑i and the spin down is given by |↓i.
|↑i =1 0 , |↓i =0 1 (3.2) Note here that the Sz operator is the trivial representation of the h basis element of (2.7) which was the underlying basis element for the weight spaces while the other two basis elements were used as raising and lowering operators. This is why instead of using S1 and S2, we can define the raising and lowering operators S+ and S− which are defined as
S+ = S1+ iS2 S−= Sx− iSy
The following result confirms that the spin operators S+ and S− do indeed act as raising/lowering operators, respectively.
Lemma 4. Let S+ and S− be defined as in (3.3), where S+ works as raising operator and S− as lowering operator
Proof. We first write the operators in their matrix form. S+= 1 2 0 1 1 0 + i 2 0 −i i 0 =0 1 0 0 (3.3) S−= 1 2 0 1 1 0 − i 2 0 −i i 0 =0 0 1 0 (3.4) Observe that S+ and S− have the same form as e and f , which respectively work as lowering and raising operators on the weight spaces of V given in chapter 3.
At last, in the last chapter we found a remarkable isomorphism between the lie groups SO(3) and SU(2) but it was left in the middle why. As it turns out, all the represen-tations of SO(3) are also represenrepresen-tations of SU(2). The fact that particles with half integer spin were actually physically possible and made sense has been supported by this isomorphism. Next to that, because of the double cover, if you rotate a spin for 360◦ degrees it will have a sign flip, since 1 and - 1 are the same in SU(2). If we rotate the quantum state 760 degrees, it will have the same sign again!
3.2 The Heisenberg model
A spin chain is an example of a quantum many-body system. It is defined on a one dimensional lattice with N lattice sites, with on every site a spin space of dimension C2n+1. The spin space of the whole system is given by the tensor product of the lattice sites H = N O n=1 hn.
We are interested in the XXX1/2 spin chain which is an example of a spin chain of spin
1/2 (for example, a one dimensional chain of electrons), so for each n, hn = C2. The
one-dimensionality of the spin chain means that every site n on the lattice has only two neighbor sites, n + 1 and n − 1. Graphically we could represent the spin chain as
n
Figure 3.2: One-dimensional spin chain, with n spins on n lattice sites
In quantum mechanics the most important operator to describe a system is the Hamil-tonian. For a spin chain there are several models to describe its Hamiltonian and we will focus on two very often used ones. The first and mainly used one in this thesis is the Heisenberg model, which only assumes nearest neighbor coupling between spin states on the chain. Even though this is a simplification of reality it is still a powerful enough to find meaningful results about a spin system, since the effect of spins goes down exponentially for each lattice site m > n + 1 removed [1].
The Hamiltonian described by the Heisenberg model is given by H =X
α,n
JαSαnSn+1α (3.5) J , historically called the London-Heitlerische Austauschintegral, describes the strength of the nearest neighbor coupling of spins. When Jx 6= Jy 6= Jz the system is called
a XY Z model. The XXX1/2 is the isotropic case where Jx = Jy = Jz = J . The
Hamiltonian for the XXX system for a chosen J is given by H =X α,n SnαSn+1α −1 4 (3.6)
is a local operator working on one hnin H, its action on the nth tensor leg in the tensor
space is given by
Snα ≡ I1⊗ I2· · · ⊗ Snα.. ⊗ IN
The spin state on each lattice site n is defined as in the last section, and will be repre-sented by arrows which point up or down, as this is basis given by the eigenvectors of Sz. If a spin system is close to a magnetic field, the spin system is influeced by it
3.3 Ferromagnetism and antiferromagnetism
There are two special alignments of the spin chain that are distinguishable by their magnetization. These interesting cases are foremost indicated by the nearest neighbor coupling factor J , When J < 0 the groundstate of the energy is given when all the spins point in the same direction. This is called a ferromagnet. The ferromagnet is the only case in which the ground state of the system is given by an eigenstate of the Hamiltonian. A more common occurrence of the state of the spin chain is given when J > 0. The
n
Figure 3.3: One-dimensional ferromagnet, with n spins on n lattice sites
energy is then minimized when the magnetization is minimized, which is done by having anti parallel neighboring spins which gives a resulting magnetisation M = 0. To measure
n Figure 3.4: one dimensional ferromagnet
and find the eigenstates of the antiferromagnet We can split the antiferromagnet into two sublattices A and B with A having all the spins |↑i and B having all the spins |↓i. We note that the concept of spin is closely related with magnetism. In fact, for a quantum system, the magnetization in the z-direction is given by
Mz = N 2 − N X i Siz where Sz =Pn
3.4 A notion of integrable systems
Integrability as a mathematical concept comes in many definitions. The basic underly-ing idea of all integrable systems is that for a dynamical system, you can find the exact solutions for the system from its initial conditions. This notion works well for classical systems, but because quantum systems have different properties we will also need an-other more suitable definition for integrability for quantum systems. For 1-dimensional quantum systems integrability is characterized by the the Yang-Baxter equations, a com-mutation relation for a 1-parameter family of linear operators which are endomorphisms of a tensor product of vector space V ⊗ V .
Definition 9 (Yang-Baxter Equation). Let V be a complex vector space. Let R(u) be a function of u ∈ C taking values in EndC(V ⊗ V ). R(u) is said to be a solution of the
Yang-Baxter equation if:
R12(u − v)R13(u)R23(v) = R23(v)R13(u)R12(u − v) (3.7)
Here Rij is the matrix which works on the tensor product V⊗3, acting on the first leg
of R as linear endomorphism R(u) on the ith leg and the jth components and as an identity on the other component. A solution of the Yang-Baxter equation is called an R-matrix.
The XXX1/2 spin chain which we introduced in the last section and is the main focus point of this thesis is also an example of an integrable quantum system, as there are R-matrices which describe this system and also satisfy (3.7) [4]. Physically speaking, the Yang-Baxter equations can be interpreted as the scattering between the scattering of three elements in an one dimensional system, when the scattering is the result of change of initial position whilst keeping the order of the spin chain intact. We can also show this graphically:
t
Figure 3.5: Graphical representation of the Yang-Baxter equations In (3.5), we graphically see the Yang-Baxter equation for
which is equivalent to Equation 3.7 and how the equality of the left and right side of the equation occurs. The lines give the rapidity2 of x1, x2 and x3 and scattering between
two particles occurs when two lines cross.
If R is a solution satisfying the Yang-Baxter equations, it has the following properties, from [8]
• Regularity: R(0; η) = P
• Unitarity: R(λ; η)R(−λ; η) = I • Permutation symmetry:
• Crossing unitarity: R(λ)R(−λ − 2η) = ˜ρ(λ) Here, P is the permutation operator defined as
P (v ⊗ w) = w ⊗ v with v ∈ Vi and w ∈ Vj
Lemma 5. The commutation relations for the permutation operator are given by: P3,1P3,2 = P1,2P3,1= P3,2P2,1
P1,2 = P2,1
for 123 indication V1, V2, V3
Proof.
3.5 A specific R-matrix
We will now prove for a specific endomorphism R that it satisfies the Yang-Baxter equations. Let R : C → End(Vi⊗ Vj) act on Vi⊗ Vj with the explicit expression
R(λ) = λ(Ii⊗ Ij) + iPij (3.8)
where λ ∈ C and the permutation operator P (v1⊗ v2) = v2⊗ v1
The two operators I and P which are used to construct the chosen R-matrix could also be represented graphically. Let V1 and V2 be two vector spaces in a tensor product,
and let u ∈ V1 and v ∈ V2. Then, the identity operator lets the v and u whilst the
permutation operator will interchange their position in space.
Theorem 4. The linear operator R 3.8 satisfies the Yang-Baxter equations on the three-tensor product space (C2)⊗3
R12(λ − µ)R31(λ)R23(µ) = R23(µ)R31(λ)R12(λ − µ) (3.9)
u
v
(a) Identity operator
u
v
(b) Permutation operator
Figure 3.6: Graphical representation of the working of the identity and the permutation operator on a tensor product space V ⊗ W , with u ∈ V and v ∈ W .
Proof. To prove the theorem we will take the product of the three R operators. For easier notation we rewrite all Ii⊗ Ij = Ii,j.
Ri,j(λ) = λ(Ii,j) + iPi,j.
The LHS of the commutation relation is (λ − µ)I12+ iP12
λI13+ iP13
µI23+ iP23.
Which, when satisfying the Yang-Baxter equation, is equal to the RHS: (λ − µ)I23+ iP23 λI13+ iP13 µI12+ iP12 expanding the product on the LHS gives:
µλ(λ − µ)I + iµλP12+ iµ(λ − µ)P13− µP12P12
+iλ(λ − µ)P23− λP23P12− (λ − µ)P23P13− iP23P13P12
where we set I123 = I = I312, and we set PijI = Pij. The expansion of the RHS is equal
to
λµ(λ − µ)I + λ(λ − µ)iP3,2+ iµ(λ − µ)P3,1− (λ − µ)P3,1P3,2
+iµλP1,2− µP1,2P3,1− λP1,2P3,2− iP1,2P3,1P3,2
We immediately see that the only terms which are different are the terms with two or more permutation operators. It remains to prove:
−(λ − µ)P3,1P3,2= −(λ − µ)P3,2P3,1 (3.10)
−λP1,2P3,2= −λP3,2P1,2 (3.11)
−µP1,2P3,1= −µP3,2P1,2 (3.12)
and
But remembering the commutation relations for the permutation operators we can see easily that these are equal. For (3.10), it is clear that P1,2 = P2,1, and Pn,1,2= Pn,1,2 =
Pn,2,1 = P3,2P3,1. (3.11, 3.12) are directly solved by the permutation commutation
relations. For (3.13), we remember the nature of the permutation relations and picture them as cycles in S3:
P1,2P3,1P3,2(x ⊗ y ⊗ z) = P1,2P3,1(x ⊗ z ⊗ y) = P1,2(y ⊗ z ⊗ x) = (z ⊗ y ⊗ x)
P3,2P3,1P1,2(x ⊗ y ⊗ z) = P3,2P3,1(y ⊗ x ⊗ z) = P3,2(z ⊗ x ⊗ y) = (z ⊗ y ⊗ x),
so they are the same operators. So we conclude that all the terms are equal to each other and that
(λ − µ)(I1,2+ iP1,2)(λ)(I3,1+ iP3,1)(µ)(I3,2+ iP3,2) =
(λ − µ)(I3,2+ iP3,2)(λ)(I3,1+ iP3,1)(µ)(I1,2+ iP1,2).
We can conclude that
R1,2(λ − µ)R1,3(λ)R2,3(µ) = R2,3(µ)R1,3(λ)R1,2(λ − µ),
and we have proven that (3.8) satisfies the Yang-Baxter equation.
4 Symmetry breaking in quantum
mechanics
4.1 The concept of symmetry
Symmetries are a concept which you often come across in physics as it can tell you a lot about a system. For instance, symmetries are very closely related to the conservation laws. A system is symmetric if it is invariant under a transformation. We make the distinction between discrete and continuous symmetries and will focus fully on the latter one. In Chapter 2 we introduced Lie groups as groups with continuous operations, which can be used to study continuous symmetries. The breaking of symmetry would mean that the system is not invariant under a transformation anymore. Most things in our physical world have undergone some sort of symmetry breaking since symmetry breaking is very closely related to the localisation of an object.
Definition 10. A system is in it’s thermodynamic limit if it exists of N particles, and N → ∞.
For such a system an infinitesimal small perturbation can be enough to break the symmetry. This is called spontaneous symmetry breaking and will be the main focus of this chapter. We measure it by making use of non commuting limits, better known under the term singular limit.
Definition 11. Let A be a system and let i and j be properties of A, and let a and b in R.
lim
i→aj→blimA 6= limj→blimi→aA, (4.1)
that is, the limits are non-commuting, and we call the limit singular.
If a system A undergoes a singular limit, the symmetry is spontaneously broken. This can occur in both classical and quantum systems. We are interested in the quantum case, but since it is often easier to visualize classical systems we will try to understand it first by a classical example.
Imagine a sharp cylindrical pencil which is balanced on its tip. The tip has a width b, and the pencil makes an angle θ with the z-axis. If θ 6= 0, the pencil will fall and break its rotational symmetry around the z-axis, but if θ = 0, the width of the tip can
be any size without the pencil falling over. Let us formalize that in a more concrete way. To understand whether symmetry breaking occurs, we will take the limits of θ and b
lim
b→0θ→0limz > 0
lim
θ→0b→0limz = 0
where z is the vertical position of the end of the pencil. We see that the limits don’t commute so the symmetry is spontaneously broken.
One thing which comes to mind, is that as long as the pencil is in upright position it has a potential energy > 0, whilst when symmetry is broken and it is localized on the ground it is in its ground state. The symmetric state is a meta stable state which has an infinitely degenerate ground state. This gives us another hint that spontaneous symmetry happens very often as systems prefer to be in their ground state.
Before we turn to the quantum case we will discuss one more classical case which will later help us to understand what happens on quantum level. Assume a classical one di-mensional magnet with N microscopic bar magnets. The energy of the magnet is given by
E =X
x,δ
−|J |Sx· Sx+δ
where, in the thermodynamical limit with an infinitesimal perturbation lim
N →∞B→0lim hM i /N = 0
lim
B→0N →∞lim hM i N = sˆn
Which is proven in the lecture notes of Jasper van Wezel, [7], and we see that symmetry is broken spontaneously as the limits don’t commute. Classical systems are the thermo-dynamical limit of a quantum system. To fully understand such macroscopic systems it is important to focus on the spontaneous symmetry breaking which occurs in quantum systems.
4.2 Spontaneous symmetry breaking in a quantum system
Just like in the classical case a quantum system can be invariant under transformation, and more importantly, a symmetry is called spontaneously broken if a symmetric system is always found in a symmetry-broken state rather than it’s symmetric ground state. But in the quantum case, we make a difference between the symmetry of a state and symmetry of an operator
Definition 12. A state |φi is called symmetric under an operator U, if it is invariant under U , that is,
ˆ
Definition 13. An operator ˆA is symmetric under transformation ˆU if the expectation value of the operator is unaffected by transformations of the state:
hψ| ˆU†A ˆˆU |ψi = hψ| ˆA |ψi (4.2) Or, because it holds for any state of S,
ˆ
U†A ˆˆU = ˆA (4.3) In classical mechanics, all observables commute and the commutator would be zero. However, this is not true in quantum mechanics and because of different quantum prop-erties, not all observables of a quantum system commute. This leads to the following definition.
Definition 14. A quantum system is symmetric under a certain translation X if and only if the commutation relation between the Hamiltonian H and X vanishes, that is, if [H, X] = 0.
The result of this definition is that H and X share the same set of eigenstates. Let ˆH satisfy the condition ˆU†H ˆˆU = ˆH. Then this can also be written as:
ˆ
H ˆU = ˆU ˆH (4.4) ⇒ [ ˆU , ˆH] = 0 (4.5) Since classical systems are the limit of quantum systems we wonder what happens that we can localize object in the universe. After all, a system would be completely sym-metric, it would have to be spread out in the entire universe to satisfy the translational invariance. This however not the case since it is possible to localize classical objects. This is where spontaneous symmetry breaking was postulated to understand what hap-pens in the thermodynamic limit of quantum systems, since for a symmetric quantum system the eigenstates need to be shared with the Hamiltonian and all configurations of eigenstates need to be equivalent, while a classical system can have a definite position. [9].
Let H be the Hamiltonian of a system. We rewrite it in reciprocal space, and sepa-rate it in two parts:
H = Hcoll+
X
k6=0
Hint(k)
where k is the internal dynamics of the crystal. By defining the Hamiltonian as a collective H, we reduce the problem to a single particle problem. The system we are interested in, a quantum spin chain, is a system in the thermodynamical limit. The spin operator a representation of sl(2, C), the complexified associated Lie algebra of the Lie group SU(2). When our quantum spin chain is rotationally symmetric, [H, S2] = 0 and the system will be rotational invariant. First we assume that the lattice of the spin
chain exists of two sub lattices A and B which both have an independent spin SA and
SB. The total spin on A is given by SA and on B by SB where SA =
P
i∈ASi and
SB = Pj∈BSj. which both have their own collective Hamiltonian which leaves the
other sub lattice invariant. The collective Hamiltonian for the spin chain is then given by Hcoll = J NSA· SB= J N(S 2− S2 A+ SB2)
So in the thermodynamic limit, the excitation energy of the collective part of the Hamil-tonian goes to zero.
The energy, in eigenstates of the S operator is given by E(S, SA, SB) = J ~
2
2N S(S + 1) − SA(SA+ 1) − SB(SB+ 1)
The ground state of the spin chain is reached when S = 0. If you would just look at one of the two sublattices, the total spin of the lattice would change by flipping one spin. This costs a certain amount of energy which is why we don’t see this in most cases. But for the collective Hamiltonian, which is proportional to NJ, for N → ∞, the required energy to change the spin of the whole system will go to 0. Thus for a small perturbation it is possible to prefer a certain spin alignment, in fact, an infinitesimal perturbation will be enough to give the system a certain spin in a direction. Because of this the rotational symmetry is broken in the thermodynamic limit.
Ferromagnetism and antiferromagnetism
The ferromagnetic case of the Heisenberg model is very special as all the states point in the same direction and for this case we will just look at the whole lattice and not for sublattices A and B as they would point in the same direction. It is rotational invariant as [H, S2] commute. To show spontaneous symmetry breaking we introduce a perturbation by a magnetic field Bz which we will later take to 0. Then H gets a
potential by HB = H + BzStotz . In the thermodynamic limit, if limB→0 on the spin
chain, the result will be that all the spins will point in the z-direction and rotational symmetry is broken.
lim
N →∞B→0lim hM i /N = 0
lim
B→0N →∞lim hM i /N = 1/2
In case of the antiferromagnet, we divide the system in two subsystems A and B. The total spin of the system is given by the sum of these two spins, S = SA+ SB.
For the unperturbed system the Hamiltonian commutes with the spin; [H, S2] = 0. The
energy is minimized when all neighboring spins have spin pointing in opposite directions, and SA and SB have their maximum values. Just as in the ferromagnetic case we will
given by a magnetic field [HB, S2] 6= 0 where HB is the Hamiltonian with a magnetic
field in the z-direction, with direction up in the A sub lattice, and down in the B sub lattice. In the thermodynamic limit the system will become a classical magnet. This one is characterized by the alignment of spin up spin down spin up... If symmetry is spontaneously broken by an infinitesimal perturbation we will end up in this case, where the spin of both sub lattices is maximal. Using the singular limit gives us
lim N →∞B→0limh(SA− SB)i/N = 0 lim B→0N →∞lim h(SA− SB)i/N = 1 2
and we see that symmetry is broken spontaneously as they do not commute, and we get a perfectly odered spin chain as the classical antiferromagnet.
Another way to measure if the symmetry is broken broken symmetry is to look at the ground state wavefunction in the Stot basis. For the symmetric Hamiltonian this was
given by an eigenstate of this Hamiltonian, and there is only one which commutes with the total spin operator.
hφGS| Stot|φGSi = δ(Stot) (4.6)
Where φGS gives the ground state of the system for the unperturbed Hamiltonian.
As there is only one symmetric ground state for the Hamiltonian. As the perturbed system has an extra potential it doesn’t commute with the unperturbed Hamiltonian, and following will also not commute with the total spin anymore.
hφGSB| Stot|φGSBi = f (Stot).
For this chapter, we used the lecture notes of Jasper van Wezel and two papers about spontaneous symmetry breaking in quantum systems, [7, 9, 10].
5 Algebraic Bethe Ansatz
In 1931, Hans Bethe wrote the paper “Zur Theorie der metalle”, which gave a method of finding the eigenvalues an eigenstates of a one dimensional spin chain with spin 1/2. Years later around 1980 the Leningrad group with Ludvig Faddeev came up with an quantum counterpart of solving integrable systems, which gave an algebraic approach to solving the Bethe Ansatz for 1 + 1 dimensional space time models. This method, called the “Algebraic Bethe Ansatz” (short: ABA) can be used to solve the one dimensional XXX1/2 spin chain. The one dimensional spin chain, (C2)⊗N, is assumed to be on a
periodic lattice, more specifically, a discrete circle. We label each lattice site with n ∈ N, with n ≡ n + N . We assume here periodic boundary conditions meaning that moving through the spin chain won’t have an effect when passing N . Recalling the Hamiltonian for the spin chain (5.1)
H =X α,n SnαSn+1α −1 4 (5.1) The spin chain with periodic boundary conditions can be graphically represented as
Figure 5.1: Periodic spin chain with n lattice sites, each lattice site having one spin state By using the Algebraic Bethe ansatz we can find the eigenvalues of the Hamiltonian of the system. We consider the space of our XXX1/2 chain as a representation of sl(2, C), as explored in the last couple of chapters, given by
H =
N
O
n
C2
The core of the method is the Lax operator, which will work as a local operator and satisfies the Yang-Baxter equations. With it will possible to define a global generating operator which will give rise to the operators work on the space H
5.1 The Lax operator
The Lax operator is a local operator working on a local space in the tensor product of the spin chain. It is an endomorphism Ln,a working on a tensor product of hn⊗ V ,
where V is an auxiliary space equal to C2. The n denotes the leg of the tensor product working on hn, and the a notes the leg of the tensor product working on V . It is defined
as
Ln,a: C → End(C2⊗ C2)
Ln,a(λ) = λIn,a+ i
X
α
Snα⊗ σaα.
Inand Snα work on the space hn, Iaand σaαwork on the auxiliary space V . We can write
the Lax operator as a 2 × 2 matrix with entries in End(hn) with respect to the standard
basis of V = C2.
Ln,a= λIn,a+ i
X α Sα⊗ σα =λ 0 0 λ + i 0 S 1 n Sn1 0 + 0 −iS2 n iSn2 0 +S 3 n 0 0 −S3 n =λ 0 0 λ + i S3 Sn1− iS2 n Sn1+ iSn2 −S3 n =λ + iS 3 n iSn− iS+n λ − iS3n
We can rewrite the Lax operator in terms of the permutation operator P = 1 2 I ⊗ I + X α σα⊗ σα (5.2) We can prove that it is a permutation operator by writing it in its matrix representation. Let e+, e− be the standard basis for C2. Then the standard basis for C2⊗ C2 is given
by e1 = e+⊗ e+, e2 = e+⊗ e−, e3= e−⊗ e+, e4 = e−⊗ e− (5.3) where e+= 1 0 , e−=0 1 (5.4)
P = 1 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 + 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 + 0 0 0 −1 0 0 1 0 0 1 0 0 −1 0 0 0 + 1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 1 = 1 2 2 0 0 0 0 0 2 0 0 2 0 0 0 0 0 2 = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 ,
which is a permutation operator in C2⊗ C2, and we can rewrite the Lax operator to
Ln,a(λ) = (λ −
i
2)In,a+ iPn,a. (5.5) In section 3.5, we proved for a specific linear operator R on C2⊗ C2 that it satisfied the
Yang-Baxter equations and mentioned that it would be of importance in this chapter. It turns out that the R matrix which was given in 3.5 works as R operator for the XXX1/2
spin system, which has also been proven in literature [4]. The L-operator we just defined looks almost equal to the R-operator with just a translation of λ → (λ − 2i). But since that is just a linear transformation, it will not affect the commutation properties of the operator. To be able to satisfy the Yang-Baxter equations, consider two Lax operators working on the same quantum space hn and two different auxiliary spaces V1 and V2
which will be indicated by a1 and a2. Let hn⊗ V1⊗ V2 be the triple tensor product. L
works as an endomorphism on the space, and the specific Ln,a1 and Ln,a2 are defined as
Ln,a1(λ) : C → End(hn⊗ V1⊗ V2) = (λ −
i
2)In,a+ iPn,a ⊗ Ia2,
with Ln,a2 being defined similarly on the same tensor product space. These will satisfy
the YB-equation for the R- and L-operator given by
Ra1,a2(λ − µ)Ln,a1(λ)Ln,a2(µ) = Ln,a2(µ)Ln,a1(λ)Ra1,a2(λ − µ).
This form of the YB-equation is often called a RLL-relation, which is equivalent to the YB-equation for R(λ). The R-operator works as an intertwiner operator of the two auxiliary spaces. We can give the Lax operator a geometric interpretation as gives a connection along the chain by transporting quantum states along the chain.
φn+1= Lnφn
for a spin state φn=
φ1 n φ2n .
5.2 T-operator
We are interested in operators defined on the whole chain and we can define a global operator on the space H by taking the ordered product of all the local Lax operators for all n in the spin chain. This global operator is called the T -operator, which is an endomorphism from H ⊗ V → H ⊗ V by
Definition 15. Let T be an endomorphism on the space H ⊗ V , being the full product TN,a(λ) = LN,a(λ)...L1,a(λ)
with N indicating H and a indicating V , given by
TN,a: C → End((C2)⊗N ⊗ V ).
The endomorphism TN,a is a monodromy1 around the circle and we will refer to T
more often by “the monodromy”. We can represent T as a matrix in the auxiliary space by taking the basis of C2. The matrix is given by
TN,a(λ) =
A(λ) B(λ) C(λ) D(λ)
, (5.6)
where A(λ), B(λ), C(λ), D(λ) are operators defined on H, and are a result of the product of the Lax operator. Just as the Lax operator satisfies the Yang-Baxter equations locally on each n, we will prove that the T also satisfies this equation globally. We therefore consider two T operators TN,1 and TN,2 on H with 1 and 2 indicating two distinct
auxiliary spaces V1 and V2.
Theorem 1. Let TN,a1 and TN,a2 be two monodromy operators on the space H ⊗ V1⊗ V2
where V1 and V2 are auxiliary spaces with respectively indices 1 and 2. Then TN,a also
satisfies the RTT-relation
Ra1,a2(λ − µ)TN,a1(λ)TN,a2(µ) = TN,a2(µ)TN,a1(λ)Ra1,a2(λ − µ), (5.7)
Proof. From now on we will write the monodromy without the index N . Since T is a product of Lax operators, we prove the commutation relation (5.7) for two adjacent sites. It will then follow for all other adjacent sites as N is an ordered product. So we set N = 2 and Ta1 = L2,a1L1,a1 and Ta2 = L2,a2L1,a2. Simplyfing the notation even more
by defining Ra1,a2(λ − µ) := R12, L1,a1(λ) = L1 and L2,a1(λ) = L
0 1, L1,a2(µ) = L2 and L2,a2(µ) = L 0 2. R12T1T2 = R12L1L01L2L02 = R12L1L2L01L 0
2 as L2 and L01 work on different sites
= L2L1R12L01L 0
2 follows from RLL relation
= L2L02L1L01R12 follows from commutativity ,
and we have concluded the proof.
1
In topology, if p : ˜X → X is a covering map, then the monodromoy transformation induced by a path γ is the homeomorphism from the fibers over γ(0) to the fibers over γ(1), given by lifting the path γ to a path starting at a given point x in the fiber over γ(0) and taking as output the endpoint of that lifted path.
We can show that TN,a is a polynomial in λ of order N , since
TN,a(λ) = LN,a(λ) · · · L1,a(λ)
= λIN ⊗ Ia+ i X α SNα ⊗ σaα · · · λI1⊗ Ia+ i X α S1α⊗ σaα = λNIN + iλN −1X α (Saα⊗ σα a) + · · ·
where we see the total spin of the operator appear in the second term of the polynomial expansion. Since the monodromy is working on the tensor product of H and auxiliary space V , we will need the following definition to be able to generate operators only defined on H.
Definition 16. Let T (λ) be the product of N Lax operators. The family of operators F (λ) is given by the partial trace, being the trace taken over the auxiliary space V
F (λ) = TrV(T (λ)) = A(λ) + D(λ) (5.8)
which is a linear operator on H, and is called the transfer operator.
We want to be able to find the Hamiltonian as part from F (λ). Since H is a commuting operator it is necessary that F (λ) is also commuting.
Theorem 5. The family of operators F (λ) is commuting: [F (λ), F (µ)] = 0
Proof. To prove that F (λ) is commuting we use the commutation relation (5.7) and the properties of the trace. Since R has the property of unitarity we know that it is invertible by definition. We can rewrite the RTT relation (5.7) by taking the inverse of R(λ − µ) on the left side to
T1(λ)T2(µ) = R−1(λ − µ)T2(µ)T1(λ)R(λ − µ)
By taking then the partial trace over the two auxiliary spaces V1 and V2
TrV1⊗V2(T1(λ)T2(µ)) = TrV1⊗V2 R −1 (λ − µ)T2(µ)T1(λ)R(λ − µ) = TrV1⊗V2 (T2(µ)T1(λ)R(λ − µ)R −1(λ − µ) = TrV1⊗V2 T (µ)T (λ),
where we use the cyclic property of the trace. T (λ) is an operator on H ⊗ V1 and T (µ)
is an operator on H ⊗ V2. Because T1 and T2 work on these two different auxiliary
spaces, we can separate the trace into a product such that TrV1⊗V2 T2(µ)T1(λ)
= TrV1(T (µ)) TrV2(T (λ)). Now we can prove that the commutation relation and find
F (λ)F (µ) − F (µ)F (λ) = TrV1 T (λ) TrV2(T (µ) − TrV2 T (µ) TrV1(T (λ)
= TrV2 T (µ) TrV1(T (λ) − TrV2 T (µ) TrV1(T (λ)
= 0
Theorem 6. Let F (λ) be a family of operators given by definition 16. Then its expansion as polynomial in λ beginning with power λN −2 is given by
F (λ) = 2λN + N −2 X l=0 Qλλl with Qλ ∈ End(H)
Proof. Since F (λ) is the trace of T (λ) and we take the partial trace over its polynomial expansion, and taking the trace over the first term will giveTr(λNIN) = 2λNIN. Since the Pauli matrices are all trace less, the second term iλN −1TrV(Pα(S)αN −1⊗ σaα)) = 0.
The next term in the polynomial, −1λN −2P
α(SN −2α ⊗ σαa)(Sα⊗ σα), and we get a sum
of N − 2 commuting operators; F (λ) = 2λN+ N −2 X l=0 Qlλl (5.9)
Corollary 3. The N − 1 operators Ql are commuting
Proof. This follows directly from the fact F (λ) is a commuting operator, so all the operators in F (λ) need to be commuting.
By taking the trace over the auxiliary space we have found a set of commuting oper-ators which work on H. In the next section we will prove that the hamiltonian H of the XXX1/2 spin chain, is in the commutative algebra generated by Q1, . . . , QN −1. In the
5.3 Proof that the Hamiltonian is a element of F (λ) in
span{Q
0,Q
1,· · · .}
Before we continue to use the Algebraic Bethe Ansatz to derive the eigenvalues and eigenvectors we will prove that the chosen L-matrix works as a generating function for the spin chain, by proving that the hamiltonian H of the XXX1/2 spin chain, is in
the commutative algebra generated by Q1, . . . , QN −1 given by the family of operators
generated by the chosen Lax operator. The point λ = i/2 is rather special, as when plugged into the Lax operator that becomes
Ln,a(i/2) = iPn,a (5.10)
So for the T -operator and familiy of operators F (λ) at λ = 2i become TN,a(i/2) = iNPN,aPN −1,aPN −2,a· · · P1,a
= iNP1,2P2,3P3,4· · · PN −1,NPN,a
F (i/2) = TrV(TN,a) = iNTrV(PN,aPN −1,aPN −2,a· · · P1,a)
where we can write TN,a in the second way because of the commutation relations of the
permutatino operator. Let us investigate F (λ) further. Pn,a= 12(In⊗ Ia+PαSnα⊗ σαa),
where σαa is an operator working on V . So, for the the trace of F (λ) is given by TrV(T (i/2)) = TrV(P1,2P2,3P3,4· · · PN −1,NPN,a) = (P1,2P2,3P3,4· · · PN −1,N) TrV 1 2(IN ⊗ Ia) + X α σNα ⊗ σaα = (P1,2P2,3P3,4· · · PN −1,N) 1 2(IN ⊗ 2) = (P1,2P2,3P3,4· · · PN −1,N)IN
By taking the trace over the auxiliary space we eradicate the effect of it and are left with only an operator in the spin space H. Now we can define a shift operator U in H by
U = TrV(T (i/2)) = iNP1,2P2,3P3,4· · · PN −1,NIN
The operator U is a shift operator on a discrete one dimensional system with N sites. U ∈ End((C2)⊗N), where X ∈ End(C2). U satisfies the condition that U−1XnU = Xn−1
with n ∈ N a site in the system, where we read n mod N Lemma 6. For a one dimensional system, the operator
U = i−NtraTN(i/2) = P1,2P2,3· · · PN −1,N (5.11)
Proof. We need to prove that U Xn= XnU , with Xn is an operator which works on the
space n ∈ N of the lattice.
XnU = XnP1,2P2,3P3,4· · · Pn−1,nPn,n+1Pn+1,n+2· · · PN −1,N = P1,2P2,3P3,4XnPn−1,nPn,n+1Pn+1,n+2· · · PN −1,N = P1,2P2,3P3,4Pn−1,nXn−1Pn−1,nPn−1,nPn,n+1Pn+1,n+2· · · PN −1,N = P1,2P2,3P3,4Pn−1,nXn−1Pn,n+1· · · PN −1,N = P1,2P2,3P3,4Pn−1,nPn,n+1· · · PN −1,NXn−1 = U Xn−1
The operator U is an unitary operator since it is a product of permutation operators which are unitary. By definition, the momentum operator P produces an infinitesimal shift and we can rewrite the momentum operator as the shift operator U
eiP = U (5.12)
For any λ we can take the derivative of the Lax operator d
dλLn,a(λ) = In,a (5.13) And we can take the derivative of T
d
dλTa(λ) = d dλ(i
NN P
N,aPN −1,aPN −2,a· · · P1,a) (5.14)
for λ = i/2 = iN −1X
n
N PN,aPN −1,aPN −2,a· · · (
d
dλPn,a) · · · P1,a (5.15) Where from now on we write ˆPn,a = dλdPn,a = In,a. Just like we did before we can
rearrange the permutations to a sequence of P1,2P2,3· · · Pn−1,nPn,n+1· · · PN −1,N. Then
the derivative of F (λ) becomes: d dλF (λ) λ=i/2 = i N −1X n P1,2P2,3· · · Pn−1,n+1Pn+1,n+2· · · PN −1,N (5.16)
The logarithmic derivative is defined as (ln f )0= f0f−1. We use this calculation trick to simplify the expression of F (λ)
d dλln (F (λ)) λ=i/2= d dλFa(λ) Fa(λ)−1 λ=i/2 =X n P1,2P2,3· · · Pn−1,n+1· · · PN −1,N iNP1,2· · · Pn−1,nPn,n+1· · · PN −1,N −1 = −iX n P1,2P2,3· · · Pn−2,n−1Pn−1,n+1· Pn,n+1Pn−1,nPn−2,n−1· · · P2,3P1,2 = −iX n P1,2P2,3· · · Pn−2,n−1Pn,n+1Pn−2,n−1· · · P2,3P1,2 = −iX n Pn,n+1
Where the last two steps use the commutation relations of permutation operators. Reme-bering the Hamiltonian H =P
α,nSnαSn+1α −14 and the operator P = 1
2(I⊗I+
P
ασα⊗σα)
we can rewrite the Hamiltonian as a function of P. H =X α,n SnαSn+1α −1 4 =X α,n SnαSαn+1−X n 1 4 =X α,n Snα⊗ Sα n+1− N 4 = 1 4 X α,n σnα⊗ σα n+1− N 4 Rewriting the operator P gives
Pn,n+1= 1 2(In⊗ In+1+ X α σnα⊗ σαn+1) 1 2 X α σnα⊗ σn+1α = Pn,n+1− 1 2(In⊗ In+1) X α Snα⊗ Sn+1α = Pn,n+1− 1 2(In⊗ In+1) X α SnαSn+1α = Pn,n+1− 1 2(In⊗ In+1) Substituting this in H gives
H =X n (Pn,n+1− 1 2(In⊗ In+1)) =X n Pn,n+1− N 2(In⊗ In+1) = i 2 d dλln F (λ)|λ=i/2− N 2
So we conclude that H is part of the family of operators which is generated by the trace of the monodromy T , by being its logarithmic derivative. So our chosen R-matrix and equal Lax operator work as generating operators for the Hamiltonian of the spin chain [2].
5.4 The Bethe Ansatz Equations for the XXX
1/2model
The Algebraic Bethe Ansatz gives a method by using the (5.7) relation to diagonalize F (λ) to be able to find the eigenvalues and eigenstates. The relevant commutation relations for A,B,C,D from the RTT relation are
[B(λ), B(µ)] = 0 (5.17)
A(λ)(B(µ)) = f (λ − µ)B(µ)A(λ) + g(λ − µ)B(λ)A(µ) ; (5.18) D(λ)B(µ) = h(λ − µ)B(µ)D(λ) + k(λ − µ)B(λ)D(µ) ; (5.19) where f (λ) = λ − i λ ; g(λ) = i λ; (5.20) h(λ) = λ + i λ ; k(λ) = − i λ; (5.21)
To prove (5.20)-(5.21), we use the matrix representation of 5.7, again in the standard basis of C2⊗ C2. We can then write the R-matrix in its matrix form as
R(λ) = a(λ) b(λ) c(λ) c(λ) b(λ) a(λ) (5.22)
where we say that a = λ + i, b = λ and c = i for easier notation.
Ta1(λ) = A(λ) B(λ) A(λ) B(λ) C(λ) D(λ) C(λ) D(λ) (5.23)
Now, for showing the commutation relations we will use the RTT-relation again. We define Ta1 and Ta2, where Ta1 = TH⊗ Ia2 and Ta2 = Ia1⊗ TH
Ta2(µ) = A(µ) B(µ) C(µ) D(µ) A(µ) B(µ) C(µ) D(µ) (5.24) Ta1(λ)Ta2(µ) =
A(λ)A(µ) A(λ)B(µ) B(λ)A(µ) B(λ)B(µ) A(λ)C(µ) A(λ)D(µ) B(λ)C(µ) B(λ)D(µ) C(λ)A(µ) C(λ)B(µ) D(λ)A(µ) D(λ)B(µ) C(λ)C(µ) C(λ)D(µ) D(λ)C(µ) D(λ)D(µ) (5.25)
When writing the RTT-relation out in their matrices we get on the LHS a(λ − µ) b(λ − µ) c(λ − µ) c(λ − µ) b(λ − µ) a(λ − µ)
A(λ)A(µ) A(λ)B(µ) B(λ)A(µ) B(λ)B(µ) A(λ)C(µ) A(λ)D(µ) B(λ)C(µ) B(λ)D(µ) C(λ)A(µ) C(λ)B(µ) D(λ)A(µ) D(λ)B(µ) C(λ)C(µ) C(λ)D(µ) D(λ)C(µ) D(λ)D(µ) . (5.26)
and the product gives
αA(λ)A(µ) αA(λ)B(µ) αB(λ)A(µ) αB(λ)B(µ) βA(λ)C(µ) + γC(λ)A(µ) βA(λ)D(µ) + γC(λ)B(µ) βB(λ)C(µ) + γD(λ)A(µ) βB(λ)D(µ) + γD(λ)B(µ) γA(λ)C(µ) + βC(λ)A(µ) γA(λ)D(µ) + βC(λ)B(µ) γB(λ)C(µ) + βD(λ)A(µ) γB(λ)D(µ) + βD(λ)B(µ)
αC(λ)C(µ) αC(λ)D(µ) αD(λ)C(µ) αD(λ)D(µ)
where α = a(λ − µ), β = b(λ − µ), and γ = c(λ − µ).
The right hand side of the commutation relation can also be written in its matrix form to be able to see the commutation relation. The RHS in matrix form looks like
Ta2Ta1R(λ − µ) =
A(λ)A(µ) B(µ)A(λ) A(µ)B(λ) B(µ)B(λ) C(µ)A(λ) D(µ)A(λ) C(µ)B(λ) D(µ)B(λ) A(µ)C(λ) B(µ)C(λ) A(µ)D(λ) B(µ)D(λ) C(µ)C(λ) D(µ)C(λ) C(µ)D(λ) D(µ)D(λ) a(λ − µ) b(λ − µ) c(λ − µ) c(λ − µ) b(λ − µ) a(λ − µ) And its product is given by
αA(µ)A(λ) βB(µ)A(λ) + γA(µ)B(λ) γB(µ)A(λ) + βA(µ)B(λ) αB(µ)B(λ) αC(µ)A(λ) βD(µ)A(λ) + γC(µ)D(λ) γD(µ)A(λ) + βC(µ)B(λ) αD(µ)B(λ) αA(µ)C(λ) βB(µ)C(λ) + γA(µ)D(λ) γB(µ)C(λ) + βA(µ)D(λ) αB(µ)D(λ) αC(µ)C(λ) βD(µ)C(λ) + γC(µ)D(λ) γD(µ)C(λ) + βC(µ)D(λ) αD(µ)D(λ)
To show that (7.1) are part of the commutation relations forced by the RTT relation (5.17), we look at the relevant elements in the matrices where we can see the equal-ities. The upper right corner of both matrices shows the commutation relation for [B(λ), B(µ)] = 0. Because the matrices have to be equal by the commutation relation, a(λ − µ)B(λ)B(µ) = a(λ − µ)B(µ)B(λ), so B(λ)B(µ) = B(µ)B(λ). Then for the next relationship at third element of the first row of both matrices
a(λ − µ)B(λ)A(µ) = c(λ − µ)B(µ)A(λ) + b(λ − µ)A(µ)B(λ) (5.27) we get, when interchanging λ ↔ µ,
and by rearranging terms the expression becomes A(λ)B(µ) = a(µ − λ)
b(µ − λ)B(µ)A(λ) −
c(µ − λ)
b(µ − λ)B(λ)A(µ) (5.29) Now remembering a, and b, we substitute their formulas and find
A(λ)B(µ) = λ − µ + i
λ − µ B(µ)A(λ) − i
λ − µB(λ)A(µ). (5.30) We found the functions which were given by (5.17) and can do almost the same derivation for the other terms, where we look at the commutation relation of the fourth entry in the second row.
Remembering the highest weight representation of the harmonic oscillator we want a generalized analogous description here.The highest weight is when S+ω = 0 with ω ∈ C2. Theorem 7. There exists a vector Ω such that,
C(λ)Ω = 0 where Ω = e⊗N+
Proof. So, Sn+(λ)ωn = 0. This is the case when ωn = e+. The matrix form of the Lax
operator was given by:
Ln,a(λ) =
λ + iS3
n iSn−
iSn+ λ − iSn3
Using the matrix representation again, where Ln,a∈ End (hn⊗ V ).
Ln,aωn=(λ + iS 3 n)ωn (iSn−)ωn (iSn+)ωn (λ − iSn3)ωn = (λ + iSn3)1 0 (iSn−)1 0 (iSn+)1 0 (λ − iSn3)1 0 Ln,aωn= (λ + i 2)+ i− 0 (λ − 2i)+ =(λ + i 2) ∗ 0 (λ − 2i) ωn
The ∗ indicates an operator which we don’t need for our calculation. H is a tensorproduct of hn, so the vector H 3 Ωn=Nnωn. As T (λ) =Q1n=NLn,a we get
T (λ)Ω =Y n Ln,aΩ (5.31) =Y n (λ + i 2) ∗ 0 (λ − 2i) Ω (5.32) =Y n αN(λ) ? 0 δN(λ) Ω (5.33)
Where α(λ) = λ +2i and δ(λ) = λ −2i and, as we wanted, C(λ)Ω = 0
Corollary 4. Ω is an eigenvector of A(λ) and D(λ) simultaneously, and F = A + D. The eigenvalues of A(λ) and D(λ) are given by
A(λ)Ω = αN(λ)Ω; D(λ)Ω = δN(λ)Ω,
Theorem 8. Let T (λ) be a monodromy matrix, and Ω is an eigenvector of F (λ). Other eigenvectors of F (λ) are given by
Φ({λ}) = B(λ1)...B(λl)Ω (5.34)
Where {λ} is a set of complex parameters.
Proof. Ω is not an eigenvector of B(λ). We need to prove that A(λ)Φ({λ}) = αN(λ)Φ({λ}) and D(λ)Φ({λ}) = δN(λ)Φ({λ}) satisfy the Bethe ansatz equations, since F = A + D. By using the relevant set of the commutation relations (5.18) and the exchange relations (5.21) we will get a set of algebraic relations on the complex parameters λ1, ...λl
A(λ)Φ({λ}) = A(λ)B(λ1)...B(λl)Ω
= (f (λ − λ1)B(λ1)A(λ) + g(λ − λ1)B(λ)A(λ1))(B(λ2)..B(λl))Ω
= ((f (λ − λ1)B(λ1)A(λ)B(λ2)...B(λl) + g(λ − λ1)B(λ)A(λ1)(B(λ2)...B(λl))Ω
The first term of the RHS looks already like something we would like to use and can be obtained by using only the first term of the commutation relation of A(λ)B(µ) in (5.18). But with every commutation new terms will arise which are combinations of 2l−1 terms. The coefficients of these can be very complicated and difficult to calculate. But we can write it down as:
l Y k=1 f (λ − λk)αN(λ)B(λ1)...B(λl)Ω + l X k=1 Mk(λ, {λ})B(λ1)... ˆB(λk)B(λ)Ω (5.35)
But for the first commmutation of A(λ)B(λ1) in the second term of the RHS and after
that only using the first term of (5.18) we get M1(λ, {λ}) = g(λ − λ1)
l
Y
k=2
Because all the B(λi), B(λj) for i 6= j commute we can easily substitute λ1 → λk, as
they are only permutations which don’t alter the summation, so we can find a general expression for Mj(λ, {λ}) Mj(λ, {λ}) = g(λ − λj) l Y k6=j f (λj− λk)αN(λj)
We can do the same derivation for D(λ) with the commutation relation (5.19) and we get: D(λ)Φ({λ}) = D(λ)(B(λ1)...B(λl))Ω = λ Y k=1 h(λ − λk)δN(λ)B(λ1)...B(λl)Ω + l X k=1 Nk(λ, {λ})B(λ1)... ˆB(λk)...B(λl)B(λ)Ω, where Nj(λ, {λ}) = k(λ − λj) Y k6=j h(λj− λk)δN(λj)
From (5.21), we realize that g(λ − λj) = −k(λ − λj) and we set the complex parameters
in the set of {λ} in a way such that they satisfy the equations
l Y k6=j f (λj − λk)αN(λj) = l Y k6=j h(λj− λk)δN(λj) (5.37)
we get F (λ)Φ({λ}) = (A(λ) + D(λ))Φ({λ}) = Yl k=1 f (λ − λk)αN(λ)B(λ1)...B(λl) + g(λ − λj) l X k=1 l Y k6=j f (λj− λk)αN(λj)B(λ1)... ˆB(λk)B(λ) + λ Y k=1 h(λ − λk)δN(λ)B(λ1)...B(λl) − g(λ − λj) l X k=1 Y k6=j f (λj− λk)δN(λj)B(λ1)... ˆB(λk)...B(λl)B(λ) Ω = αN(λ) l Y k=1 f (λ − λk)B(λ1)...B(λl)Φ({λ}) + δN(λ) l Y k=1 h(λ − λk)B(λ1)...B(λl)Φ({λ}) = Λ(λ, {λ})Φ({λ}) (5.38)
Finally we rewrite the expressions (5.37) for α(λ) and δ(λ) and (5.21) by substitution and get l Y k6=j f (λj− λk)αN(λj) = l Y k6=j h(λj− λk)δN(λj) l Y k6=j (λj− λk) − i (λj− λk) + i (λj+ i 2) = l Y k6=j λj− λk λj− λk (λj− i 2) l Y k6=j (λj− λk) − i (λj− λk) + i l Y k6=j (λj+ i 2) = l Y k6=j λj− λk λj− λk l Y k6=j (λj− i 2) λj +2i λj −2i !N = l Y k6=j λj− λk+ i λj− λk− i . (5.39)
This is the main result of the Algebraic Bethe ansatz, where 5.39 is called the Bethe equations, and 5.38 is the eigenvalue of F (λ) for the eigenstate Φ({λ})
