by Nambu and Goldstone.1,2 One of consequences of SSB is the presence of ... Thus it avoids both the massless Goldstone bosons and the massless...

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Spontaneous Symmetry Breaking and Chiral Symmetry∗ Ling-Fong Li, Physics Department, Carnegie Mellon University, Pittsburgh, PA 15213, USA

Abstract In this introductory lecture, some basic features of the spontaneous symmetry breaking are discussed. More specifically, σ-model, non-linear realization, and some examples of spontaneous symmetry breaking in the non-relativistic system are discussed in details. The approach here is more pedagogical than rigorous and the purpose is to get some simple explanation of some useful topics in this rather wide area. .

∗ Lecture given at VII Mexico Workshop on Particles and Fields, Merida, Yucatan Mexico, Nov 10-17 1999

1

I.

Introduction

The symmetry principle is perhaps the most important ingredient in the development of high energy physics. Roughly speaking, the symmetries of the physical system lead to conservation laws, which give many important relations among physical processes. Many of the symmetries in Nature are however approximate symmetries rather than exact symmetries and are also very useful in the understanding of various phenomena in high energy physics. Among the broken approximate symmetries, the most interesting one is the Spontaneous Symmetry Breaking (SSB) which seems to have played a special role in high energy physics. Many important progress has come from the understanding of the SSB. The SSB is characterized by the fact that symmetry breaking shows up in the ground state rather than in the basic interaction. This makes it difficult to uncover this kind of approximate symmetries. Historically, SSB was first discovered around 1960 in the study of superconductivity in the solid state physics by Nambu and Goldstone.1,2 One of consequences of SSB is the presence of the massless excitation,3 called the Nambu-Goldstone boson, or just Goldstone boson for short. Later, Nambu1 applied the idea to the particle physics. In combination with SU (3) × SU (3) current algebra, SSB has been quite successful in the understanding of the chiral symmetry in the low energy phenomenology of strong interaction. More importantly, in 1964 it was discovered by Higgs5 and others6,7 that in the context of gauge theory, SSB has the remarkable property that it can convert the long range force in the gauge theory into a short range force. Thus it avoids both the massless Goldstone bosons and the massless gauge bosons. Weinberg,8 and Salam,9 then applied this ideas to construct a model of electromagnetic and weak interactions. The significance of this model was not realized until t’Hooft10 show in 1971 that it was renomalizable. Since then this model has enjoyed remarkable experimental success and now called the “Standard Model of Electroweak Interactions”.11 Undoubtedly, this will serve as benchmark for any new physics for years to come. In this article I will give a simple introduction to the spontaneous symmetry breaking and its application to chiral symmetries in the hadronic interaction. The emphasis is on the qualitative understanding rather than completeness and mathematical rigor. Eventhough SSB has been quite successful in explaining many interesting phenomena, its implementation in the theoretical framework is more or less put in by hand and it is not at all clear what is the origin of SSB. Here I will also discuss some non-relativistic example where the physics is more tractable in the hope that they might give some hints about the true nature of SSB. Maybe good understanding of SSB might extend its applicability to some new frontier.

II.

SU (2) × SU (2) σ-Model

The σ-model has a long and interesting history. It was originally constructed in 1960’s as a tool to study the chiral symmetry in the system with pions and 2

nucleons.4 Later the spontaneous symmetry breaking and PCAC (partially conserved axial current) were incorporated.. Eventhough this model is not quite phenomenologically correct it remains the simplest example which realizes many important aspects of broken symmetries. Even though the strong interaction is now described by QCD, the σ-model of pions and nucleons is still useful as an effective interaction in the low energies where it is difficult to calculate directly from QCD. In addition, the σ-model has also been used quite often as a framework to test many interesting ideas in field theory and string theory. Here we will discuss the most basic features of the σ-model. The Lagrangian for SU (2) × SU (2) σ-Model is given by λ →2 µ2 1 →2 →2 2 2 L = + (∂µ σ) + ∂µ π σ2+ π − σ2 + π (1) 2 2 4 → → +Niγ µ ∂µ N + gN σ + iγ 5 τ · π N →

where π = (π 1 , π 2 , π 3 ) is the isotriplet pion fields, σ is the isosinglet field, and N is the isodoublet nucleon field. To discuss the symmetry property, it is more useful to write this Lagrangian as :

L =

2 µ2 λ 1 ΣΣ† tr ∂µ Σ∂ µ Σ† + tr ΣΣ† − 2 4 8 +N L iγ µ ∂µ NL + N R iγ µ ∂µ NR + g(N L ΣNR + N R Σ† NL )

(2)

where →

→

Σ = σ + i τ · π,

NL =

1 (1 − γ 5 ) N, 2

NR =

1 (1 + γ 5 ) N 2

(3)

This Lagrangian is now clearly invariant under transformation, Σ → Σ′ = LΣR† , where

NL → NL′ = LNL ,

NR → NR′ = RNR

→ → R = exp −i τ · θ R

→ → L = exp −i τ · θ L ,

(4)

(5)

are two arbitrary 2 × 2unitary matrices. Thus the symmetry group is SU (2)L × SU (2)R and representation contents under this group are 1 1 1 1 Σ∼ , NL ∼ , ,0 , NR ∼ 0, 2 2 2 2 Remark:The nucleon mass term N L NR +h.c.transforms as 12 , 21 representation and is not invariant. One way to construct invariant nucleon mass term is to introduce another doublet of fermions with opposite parity, 1 1 ′ ′ NL ∼ 0, , NR ∼ ,0 2 2 3

′ ′ so that the term N L NR + N R NL + h.c. is invariant. This will give same mass to both doublets and is usually called parity doubling. As we shall see later, another way to give mass to nucleon is by spontaneous symmetry breaking which does not require another doublet. The general form of Noether current is of the form Jµ ∼

X i

∂L δφ ∂ (∂µ φi ) i

where δφi is the infinitesimal change of the fields under the symmetry transformations. We have for the left-handed transformation, δL σ

=

δ L NL

=

→ → θ L · π, → → θL · τ

−i

2

→

→

→

→

δ L π = − θ L σ+ θ L × π , NL ,

(6)

δ L NR = 0

and a JLµ = εabc π b ∂µ π c + [σ∂µ π a − π a ∂µ σ] + N L γ µ

τa NL 2

(7)

Similarly, δRσ δ R NL

= =

→

0,

→

→

→

→

→

δ R π = θ R σ+ θ R × π ,

− θ R · π,

δ R NR = −i

→ θR

(8)

→

·τ NR 2

and a JRµ = εabc π b ∂µ π c − [σ∂µ π a − π a ∂µ σ] + N R γ µ

τa NR 2

The corresponding charges are given by Z Z a a QaL = d3 xJL0 , QaR = d3 xJR0 . Using the canonical commutation relations, we can derive a b a b a b QL , QL = 0 QR , QR = iεijk QcR , QL , QL = iεijk QcL ,

(9)

(10)

(11)

which is the SUL (2) × SUR (2) algebra. The vector and axial charges are given by

Qa5 = QaR − QaL

Qa = QaR + QaL ,

In particular,the axial charges are Z Z h i σi Q5i = d3 xA0i (x) = d3 x i (σ∂0 π i − π i ∂0 σ) + N † γ 5 N 2 4

(12)

(13)

Remark: Another way to describe the symmetry of the σ-model is the O (4) symmetry, which is isomorphic to SU (2) × SU (2) locally and is characterized by 4 × 4 orthogonal matrix, RRT = RT R = 1 The infinitesimal transformation is Rij = δ ij + εij ,

with

εij = −εji

The scalar field φi = (π 1 , π 2 , π 3 , σ) transform as 4-dimensional vector, φi → φ′i = Rij φj ≃ φi + εij φj →2

The combination φi φi = σ 2 + π is just the length of the vector φi and is clearly invariant under the rotations in 4-dimension. If we take εij = εijk αk ,

ε4i = β i ,

we get →′

→

→

→

→

π = π + α × π + β σ,

i, j, k = 1, 2, 3 →

→

σ ′ = σ+ β · π →

Thus we see from Eqs(6,8) that the parameters α correspond to vector trans→

formations and β the axial transformation.

A.

Spontaneous Symmetry Breaking

The classical ground state is determined by minimum of the self interaction of scalars, → µ2 2 →2 λ 2 →2 2 V σ, π = − σ +π + σ +π (14) 2 4 The minimum of the potential is located at →2

σ2 + π =

µ2 ≡ v2 λ

(15)

which is a 3-sphere, S 3 in the 4-dimensional space formed by the scalar fields. Each point on S 3 is invariant under O (3) rotations. For example, the point (0, 0, 0, v) is invariant under the rotations of the first 3 components of the vector. Then, after a point on S 3 is chosen to be the classical ground state, the symmetry is broken spontaneously from O (4) to O (3) . Note that different points on S 3 are related to each other by the action of those rotations which are in O (4) but not in O (3) . These rotations are usually denoted by O (4) /O (3) .(This is called the coset space.). Thus we can identify 3-sphere with O (4) /O (3) . For the quantum theory, we need to expand the fields around the classical values →′ → σ = v + σ ′ , π = π , where < σ >= v (16) 5

Here v is usually called the vacuum expectation value (VEV). Then we see that

→2 2 λ ′2 ′ + V σ, π = µ σ + λvσ σ + π σ +π 4 → → → → gN σ + iγ 5 τ · π N = gvNN + gN σ ′ + iγ 5 τ · π ′ N

→

2 ′2

′

′2

→2 ′

(17)

Thus π ′ s are massless and N ’s are massive. Remark:If we had made another choice for VEV,e.g. < π 3 >= v,

< π 1 >=< π 2 >=< σ >= 0

The physics is still the same as we will now illustrate. In this case, we write ′ π 3 = π 3 + v to get → V σ, π = µ2 π ′2 + (cubic terms and higher) → → gN σ + iγ 5 τ · π N = gvN τ 3 γ 5 N + · · ·

Thus we still have 3 massless scalar fields. For the nucleon, if we define ′ τ3 ′ NL = exp −iπ NL , NR = NR 2

we get

′

gvNτ 3 γ 5 N = gvN N

′

which the mass term for the new field. It is easy to see that Q15 , Q25 , Q 3 form an unbroken SU (2) algebra. There are several interesting features worth noting: → (1) π ’s are massless. This is a consequence of the Goldstone theorem which states that spontaneous symmetry breaking (SSB)of a continuous symmetry will give massless particle or zero energy excitation. This theorem will be discussed in more detail in next subsection. → →′ (2)After SSB, the original multiplet σ, π splits into massless π s and massive σ. Also the nucleons become massive. Thus eventhough the interaction is SU (2)L × SU (2)R symmetric the spectrum is only SU (2) symmetric. This is the typical consequence of SSB. In some sense, the original symmetry is realized by combining the SU (2) multiplet, e.g. N ,with the massless Goldstone bosons to form the multiplets of SU (2)L × SU (2)R . This, as we will discuss later, is the basis of the low energy theorem. (3)The axial current in Eq(13) after the SSB will have a term linear in π field, (18) Aµi = iv∂µ π i + · · · which is responsible for the matrix element, < 0|Aµi |π j (p) >= ipµ v 6

(19)

Using this matrix element in π decay, we can identify the VEV v with the pion decay constant fπ . This coupling between axial current Aµi and π i will give rise to a massless pole. →2 ′ and the mass term (4)The appearance of the cubic term σ σ ′2 + π ′

gvNN is the result of the spontaneous symmetry breaking . Since these terms have dimension 3, they are usually called soft breaking, in contrast to the dimension 4 hard breaking terms. (5)In the scalar self interaction, quartic, cubic, and quadratic terms have only 2 parameters, λ, µ. This means that these 3 terms are not independent, and there is a relation among them. This is an example of low energy theorem for theory with spontaneous symmetry breaking.

B.

Low energy theorem

The SSB leads to many relations which are quite different from the usual symmetry breaking. The most distinct ones are relations among amplitudes involving Goldstone bosons in low energies. As we have mentioned before, these relations are consequence of the fact that Goldstone bosons are massless and can be tagged on to other particles to form a larger multiplet. Since Goldstone bosons do carry energies, this is possible only in limit that Goldstone bosons have zero energies. Consider the following processes involving the Goldstone bosons in the external states. (i)π 0 (p1 ) + σ (p2 ) → π 0 (p3 ) + σ (p4 ) The tree-level contributions are coming from diagrams in Fig1,

Fig 1 Tree graphs for ππ scattering 7

The amplitudes for these diagrams are given by, i , s

i 2 i . Mc = (−2iλv) , Md = −2iλ t − m2σ u (20) 1 3 1 2 2 1 M = Ma + Mb + Mc + Md = 4iλ v (21) + + + s t − m2σ u 2λv 2 2

Ma = (−2iλv)

Mb = 3 (−2iλv)

2

Here s, t, and u are the usual Mandelstam variables, s = (p1 + p2 )2 , t = 2 2 (p1 − p3 ) , u = (p1 − p4 ) . In the limit where pions have zero momenta, p1 = p3 = 0, we get s = u = m2σ , t = 0 and 1 3 1 1 =0 (22) + − + M = 4iλ2 v 2 m2σ m2σ m2σ m2σ

where we have used m2σ = 2λυ 2 . Thus the amplitude vanishes in the soft pion limit, i.e. pπ → 0. (ii)π 0 π 0 → π 0 π 0 Similar calculation gives M = Ma + Mb + Mc + M4 = −2iλ

s t u + + s − m2σ t − m2σ u − m2σ

In the soft pion limit, pi → 0, we get M≃

i 2iλ (s + t + u) = 2 (s + t + u) → 0. m2σ v

(23)

This is the same as the limit, m2σ → ∞, because soft pion means pion momentum much smaller than m2σ . These are simple examples of the low energy theorem which says that physical amplitudes vanish in the limit where myomata of Goldstone bosons go to zero. In examples above, the vanishing of these amplitudes results from some cancellation among different contributions. Since this is a general property of the Goldstone boson, there should be a better way of getting this. It turns that one can change the variables representing the scalar fields such that Goldstone bosons always enter with derivative coupling. Then the vanishing of the amplitudes involving Goldstone boson is manifest. This can be accomplished by the field redefinition which we will now describe briefly(? ). Suppose we start from a Lagrangian with field φ and make a transformation to a new field η, with the relation, φ = ηF (η) where F (η)is some power series in η.If we impose the condition that F (0) = 1, the free Lagrangian for η will be the same as that for φ. Then according to a general theorem valid with rather weak restrictions on the Lagrangian and F (η) , the on-shell matrix elements calculated with η fields and with φ fields 8

are the same. We will use this field redefinition to write the Goldstone boson interaction in terms of derivative coupling. Consider a simplified Lagrangian given by i µ2 λ 2 2 1h (∂µ σ)2 + (∂µ π)2 + (24) σ2 + π2 − σ + π2 2 2 4 which is just the O (2) version of the σ-model without the nucleon. As before , the SSB will require the shift of the σ-field, as in Eq(16) and π field is massless (Goldstone boson). Equivalently, we can use a complex field defined by φ = √1 (σ + iπ) so that the Lagrangian is of the form 2 L=

2 L = ∂µ φ† ∂ µ φ + µ2 φ† φ − λ φ† φ

(25)

The symmetry transformation is then φ → φ′ = eiα φ, α is some constant .Now we use the polar coordinates for the complex field iθ (x) 1 √ (ρ (x) + v) exp (26) φ (x) = v 2 to write the Lagrangian in the form, 2

L=

1 µ2 (ρ + v) λ 2 (∂ θ) + (∂µ ρ)2 + (ρ + v)2 − (ρ + v)4 µ 2 2v 2 2 4

(27)

This clearly shows that θ (x) is massless and has only derivative couplings. This follows from the fact that the U (1) (or SO (2)) symmetry φ → eiα φ, corresponds to θ → θ + vα, which is inhomogeneous. So θ (x) needs to have a derivative in order to be invariant under such inhomogeneous transformation. Note that this Lagrangian, due to the presence of terms like (∂µ θ)2 ρ2 is not renormalizable. But this Lagrangian will be used only as an effective theory to study the low energy phenomenology while the renomalizability deals with high energy behavior.

C.

Goldstone Theorem →

From Eq(17) we see that the pions π are massless. This is a consequence of the Goldstone theorem which states that spontaneous breaking of a continuous symmetry will give a massless particle or zero energy excitation. We will first → illustrate this by showing that quadratic terms in π are absent in the tree level as a consequence of spontaneous symmetry breaking of the original chiral symmetry.. We will use the O (4) notation, φi = (π 1 , π 2 , π 3, σ)The invariance under the chiral transformation implies that δV =

∂V ∂V δφ = εij φj = 0 ∂φi i ∂φi

9

(28)

Differentiating Eq(28) with respect to φk and then evaluating this at the minimum, we see that ∂2V |min εij hφj >= 0, (29) ∂φi ∂φk For the case < φj >= δ i4 v, we see that in the expansion of V around the minimum, σ = v, π i = 0, there are no terms of the form, π i π j , σ ′ π i , with σ ′ = σ − v. Therefore π ′i s are massless in the tree level. We can extend this to more general case where the effective potential is written as V (φi ) . This potential is invariant under some symmetry group G, which transforms φi as ′

φi → φi = φi + αa taij φj

or δφi = αa taij φj

(30)

where |αi | ≪ 1 are the parameters for the infinitesimal transformations and ta matrix for the representation where φi belongs. The invariance under these transformation implies that ∂V a a α tij φj = 0 ∂φi

(31)

The minimum of the potential is located at φi = vi, which satisfies the equation, ∂V =0 (32) ∂φi φ=v Differentiating Eq (31) with respect to φk and evaluating this at the minimum, φi = vi, we get 2 ∂ V (33) taij vj = 0 ∂φk ∂φi φ=v

This means that the vector uai = taij vj , if non-zero, is an eigenvector of the mass matrix 2 ∂ V m2ij = (34) ∂φk ∂φi φ=v with zero eigenvalue(massless). Thus the number of massless Goldstone bosons is just the number of independent vectors of the form, uai = taij vj . In other words, if uai 6= 0, the combination X φi taij vj (35) χa = ij

is the Goldstone boson, up to a normalization constant. → These arguments only show that π ’s are massless in the tree level. It turns out that this property is true independent of perturbation theory and can be illustrated in case of σ-model as follows. The axial charge is of the form, Z 5 Qi = i d3 x [π i ∂0 σ − σ∂0 π i + · · ·] (36) 10

Q Since ∂0 π i and ∂0 σ are just the momenta conjugate to π i and σ, we can derive, 5 Qi , π j (0) = δ ij σ (0) . (37) Between vacuum states, this yields < 0| Q5i , π j (0) |0 >= δ ij < 0|σ (0) |0 > .

(38)

For the case of SSB, we have

< 0|σ (0) |0 >6= 0

(39)

Note that this condition implies thatRthe axial charges Q5i ’s do not annihilate the vacuum, Q5i |0 >6= 0.Using Q5i = A0i (x) d3 x we can write the LHS of Eq (38) as Z X → h0|[Q5i , π j (0)]|0i = i d3 xh0| A0i (x) , π j |0i =i δ 3 pn [h0|A0i (0) |ni hn|π j |0ie

−iEn t

−

n hn|π j |0ih0|A0i (x) |nieiEn t ]

(40)

This has explicit dependence on t, while the right hand side, < 0|σ|0 >, is independent of time. The only way these two features can be consistent is to have a state with the property that En → 0 as

− p→ n → 0.

(41)

This is the content of the Goldstone theorem. q For the relativistic system, the →2 energy and momentum is related by En = pn +m2n .Then Eq (41) implies the existence of massless particle in the system. More specifically, there are physical states |π l i with the property that h0|A0i (0) |π l ihπ l |π j |0i = 6 0 and are massless from Goldstone theorem. It is convenient to choose the normalization such that < π l |π j |0 >= δ ij and write h0|Aµi (0) |π l (p)i = ifπ pµ δ il

with fπ a constnat

(42)

It is easy to see that fπ = h0|σ|0i

(43)

For the non-relativistic system Eq(41) simply says that the dispersion relation E (p)has zero energy excitation.

D.

Non-linear σ-model

In the σ-model without the nucleons, we have 3 massless π’s and a massive σ field. For the energies much smaller than mσ , the massless Goldstone bosons are 11

the important physical degrees of freedoms and it is desirable to write down an effective theory with π’s only. As we have seen, the theory with SSB has many physical consequences, e.g. low energy theorem for the Goldstone bosons. The removal of σ-field should preserve symmetry so that these results are maintained. Also, phenomenologically there are no good evidence for the existence of the σ meson which is the partner of π ′ s in the chiral symmetry. We now discuss the explicit steps for carrying out this process. Write the scalar fields as a vector in 4-dimensional space, → φi = φ1, φ2, φ3, φ4, = π , σ (44) We want to parametrize the φ fields in such a way that the non-Goldstone field to be eliminated later is O (4) invariant. One simple parametrization for this purpose is φi = Ri4 (x) s (x) , i = 1, · · · 4 (45)

where Rab a 4 × 4 orthogonal matrix, RRT = RT R = 1, which gives Ri4 Ri4 = 1 and φi φi = s2 . (46)

So s (x) is the magnitude of the vector φi and is clearly O (4) invariant. Thus it can be eliminated without effecting the symmetry. One simple choice for Ri4 is, (? ) − → 1 − η2 2η a (x) a = 1, 2, 3 R44 (x) = (47) Ra4 (x) = − →, − → 1 + η2 1 + η2 Note that we can invert these relations to get →

η=

The Lagrangian is of the form L=

′

1 2 (∂µ s) + 4s2 2

→

π σ+s → 2

∂µ η

→2 2

1+ η

(48)

1 2 2 λ 4 + µ s − s 2 4

(49)

So η i s are the massless Goldstone bosons. To study the physics of Goldstone bosons at low energies, E ≪ mσ , we can replace the s field by a constant, s (x) = v to get →2 ∂µ η (50) L = 2v 2 →2 2 1+ η →

→′

π In order to get the correct nomoralization we rescale, η = 2v so that →′ 2 1 ∂µ π 1 →′ 2 1 →′ 2 →′ 2 L= = − ∂ π + ······ ∂µ π π µ 2 →′2 2 2 4v 2 π 1 + 4v 2

12

(51)

Here the interaction terms will always contain derivatives and amplitudes in→′ volving π s will vanish in the limit of zero momenta (low energy theorem). According to the theorem of field redefinition, this describes the same physics as the usual σ-model Lagrangian in the Goldstone sector. For example, we can check that in the simple case of π 0 π 0 scattering in the tree level the amplitude from this Lagrangian in Eq (51) is M=

i 2i [p1 · p2 − p1 · p3 − p1 · p4 ] = 2 (s + t + u) . 2 v v

(52)

This is the same result as in Eq(23 ), obtained in the σ-model with mσ → ∞. The Lagrangian in Eq (51) which contains on the Goldstone boson fields, is one example of non-linear realization of chiral symmetry, which will be discussed in detail in the next section. Here we want to mention a useful geometric interpretation of Lagrangian in Eq (51). When we eliminate the O (4) invariant ′ field by setting s (x) = v , φi s satisfy the relation, φ21 + φ22 + φ23 + φ24 = v 2 which is just the sphere with radius v in 4-dimensional Euclidean space, S 3 . The variables η 1 , η 2 , and η 3 are just one particular choice of the coordinates of the space S 3 . The transformation of Lagrangian from φ fields to η fields can be understood in terms of metric tensor in S 3 . For simplicity, consider just the kinetic terms in L , 1 ∂µ φi ∂µ φj gij L= 2 where gij = δ ij is the trivial metric in the 4-dimensional Euclidean space. Then the transformation φi → φi (η) gives ∂φi ∂µ η a ∂η a

∂µ φi = and L= where

1 1 ∂φi ∂φj δ ij a b (∂µ η a ) ∂ µ η b = gab (η) (∂µ η a ) ∂ µ η b 2 ∂η ∂η 2 gab (η) = gij

∂φi ∂φj 4v 2 δ ab = 2 ∂ηa ∂η b (1 + η 2 )

(53)

(54)

(55)

is the induced metric on S 3 . Thus in the Lagrangian the coefficient of (∂µ η a ) ∂ µ η b is the metric of the space S 3 , which is just the coset space O (4) /O (3) . In the general case where the symmetry breaking is of the form, G → H, the non-linear Lagrangian can be written down with the metric on the manifold G/H as in Eq ( 54). It is interesting to see how the transformations of SU (2)×SU (2) are realized on this manifold S 3 . For the infinitesimal isospin rotation, we have →

→

→

δ π =α × π ,

δσ = 0 13

α : group parameters

(56)

which implies from Eq(48) that →

→

→

δ η =α × η

(57)

→

This is just a rotation on the vector η and the Lagrangian in Eq(??) with metric given in Eq(55) is clearly invariant under such transformation. The axial → transformation on π and σ is of the form →

→

→

δ π = β σ,

→

δσ = − β · π ,

which gives

→

→

δ η=

→ β 1 − η2 + η 2

→

β : group parameters

(58)

(59)

→

→

β · η

.

This transformation is non-linear and inhomogeneous.. But we can get simple transformation for the combination, ! ! → → → ∂µ η ∂µ η → δ = η ×β × (60) 1 + η2 1 + η2 This looks very much like an isospin rotation except that the parameters for the rotation now depend on the fields η and it is easy now to see that the Lagrangian in Eq(54) is invariant under the axial transformations. Remark: We can transform the metric in Eq(55) into the more familiar RobertsonWalker metric used in cosmology as follows. First we use the spherical coordi→ nates for η to write the metric in the line element as 4v 2

2

(dl) = gab dη a dη b =

(1 + η 2 )

2

i h 2 2 2 (dη) + η 2 (dθ) + η 2 sin2 θ (dφ) .

Define the new variable r by r=

2η . 1 + η2

(61)

(62)

In terms of new variable the line element is of form, 2

(dr) + r2 (dθ)2 + r2 sin2 θ (dφ)2 (63) 1 − r2 which is just the usual Robertson-Walker metric for the case of positive curvature. (dl)2 =

III.

Non-linear Realization

As we have seen in the last section, it is useful to write down a Lagrangian with only Goldstone bosons as an effective theory to describe physics at low energies. In this section we will discuss the general description of this procedure, which is usually called the non-linear realization. The usual discussion of 14

this subject is generally rather formal and abstract. The discussion here will emphasize the intuitive understanding rather than the mathematical rigor. In order to make the discussion here somewhat self contained, we will first discuss some simple results from group theory (17 ) which are useful for the understanding of non-linear realization. Then we discuss the general features of non-linear realization.

A.

Useful results from group theory

Here we will recall the rearrangement theorem which is central to most of the group theoretical result and then discuss the concept of coset space which forms the basis of the non-linear realization. (i)Rearrangement Theorem Let G = {g1 , · · · gn } be a finite group. If we multiply the whole group by an arbitrary group element gi ,i.e. {gi g1 , · · · gi gn }, the resulting set is just the group G itself. (ii)Coset space Coset space decomposes a group into non-overlapping sets with respect to a subgroup. Let H = {h1,···hl } be a non-trivial subgroup of G. For any element gi in G but not in H, the left coset gi H, or coset for short, is just {gi h1,··· gi hl }. The coset gi H will not have any element in common with the subgroup H, and any two such cosets are either identical or have no elements in common. This can be seen as follows. Consider cosets g1 H, and g2 H. Suppose that there is one element in common, g 1 hi = g 2 hj

for some i, j

Then we can write g1−1 g2 = hi h−1 j which means g1−1 g2 is one of the element of subgroup H. Then by the rearrangement theorem applied to the subgroup H, we get g1−1 g2 H = H,

⇒

g1 H = g2 H

i.e. these two cosets have the same group elements. The group G is now decomposed into these non-overlapping cosets and the collection of all the distinct cosets g1 H, · · · gk H, together with H, will contain all the group elements of G. This is denoted by G/H. This can be generalized to Lie group where rearrangement theorem is valid. Example of coset space: Consider points on 2-dimensional plane which form a group under the addition, (x1 , y1 ) · (x2 , y2 ) = (x1 + x2 , y1 + y2 ) Clearly, points on the y-axis, (0, y) , form a subgroup, denoted by H. Then the vertical line of the form, (a, y)with a fixed is a coset with respect to the 15

subgroup H. It is clear that the whole 2-dimensional plane can be decomposed into collection of such vertical lines. We can label these cosets, vertical lines, by choosing one element from each coset. Clearly there are many ways to choose such representatives. One convenient parametrization is to choose those points on the x-axis, so that the cosets are of the form xi H. Each group element (x, y) can be written as the product, (x, y) = (x, 0) · (0, y) where (0, y) ∈ H. Under the action of an arbitrary group element g = (a, b) this will give g (x, y) = =

(a, b) · (x, y) = (a, b) · (x, 0) · (0, y) = (a + x, 0) · (0, b) · (0, y) (a + x, 0) · (0, b + y)

(the computation here is organized in such a way that it parallel to the more complicate case in the non-linear realization.) Thus the group element g = (a, b) will move the points in the coset xH to points in the coset (a + x) H. In terms of coset parameters, we have g : x → x + a.

B.

Non-linear Realization of Symmetries

We will first discuss the general machinery of non-linear realization (1415 ) and then take up the special case of chiral symmetry where the parity symmetry will make the realization much simpler.. 1.

General case

Suppose the symmetry group G is spontaneously broken to a subgroup H. where both G and H are Lie groups. Choose the generators of G to be of the form {V1 , , . . . Vl , A1, . . . Ak } such that {V1 , V2 , . . . Vl , } are the generators of the subgroup H. As usual, the group elements in H can be written in the form, − → → exp i− α ·V

where α1 , α2 , . . . , αl , are the group parameters for H and are taken to be real. From the coset decomposition, we can write an arbitrary group element in G as − →− →

→ − →−

g = ei ξ · A ei α · V

(64) − →− →

Here ξ 1 , · · · ξ l ,are the parameters which label the coset ei ξ · A H. Note that since the vacuum is invariant under H, we have − →− →

→ − →−

− →− →

g|0 >= ei ξ · A ei α · V |0 >= ei ξ · A |0 > 16

(65)

− → i.e. ξ also labels the different vacua, which are degenerate. Recall that the different vacua form the manifold G/H. Thus the coset parameters are also the parameters for the manifold G/H. Under the action of an arbitrary group element g1 ∈ G, we have the combination − →− →

→ − →−

g1 g = g1 ei ξ · A ei α · V . Since g1 e

− →− → i ξ ·A

is also a group element in G, we can write a coset decomposition, − →− →

− →′ − → →′ − → · A i− α ·V

g1 ei ξ · A = ei ξ and then

(66)

e

− − →′ − → − → − → − →′ − → − → →− → − → →′ − →− →′′ − →− g1 ei ξ · A ei α · V = ei ξ · A ei α · V ei α · V = ei ξ · A ei α · V

(67)

(68)

where we have used the group property of H to write → − − → − →′ ·− V i→ α ·V

ei α

e

→ − →′′ ·− V

= ei α

(69)

− → →′ Note that the new coset parameters ξ ′ and parameters α for the subgroup H,all − → depend on the original coset parameters ξ , − →′ − → − − → − → − → → ξ = ξ ′ ξ , g1 , α′ = α′ ξ , g1 (70)

− → − → In this way, the group element g1 transform the coset parameters from ξ → ξ ′ in the coset space G/H. As we will see later, these coset parameters will be − → → identified with the Goldstone bosons. These transformations on ξ , and − α induced by the group elements will have the same group properties as the group elements and are called the non-linear realization of the group. This is in contrast to the usual representation of the group where group elements are − → represented by matrices. In the transformation in Eq(70) ξ ′ is generally not a − → linear function of ξ . But for the special case where g = h is a group element from the unbroken subgroup H, we get, from Eq(64), − − − − →′ − →− →− →− → − → → → → →− →− →− hg = hei ξ · A ei α · V = hei ξ · A h−1 hei α · V = hei ξ · A h−1 ei α ·V (71)

where

→ − →−

→′ − →−

hei α · V = ei α ·V

(72)

→

In general, the broken generators A transform as some representation D with respect to the subgroup H, hAi h−1 = Aj Dji (h)

(73)

For example, in the case of G = SU (2) × SU (2) model, the broken generators, A1 , A2 , A3 transform as triplet under the unbroken subgroup H = SU (2) . We can then write ′ − →− → (74) hei ξ · A h−1 = exp iξ i hAi h−1 = exp (iξ i Aj Dji (h)) = exp iξ j Aj 17

where

′

ξ j = Dji (h) ξ i

(75)

This means that ξ ′i s transform linearly under the subgroup H. Also it is easy → to see that the parameters α are independent of the coset parameters ξ i . But − → → ′ −

if the group element is of the form ei ξ · A the transformation law for the coset parameters ξ is non-linear and quite complicate. 2.

Chiral symmetry

For the case of chiral symmetry, there is significant simplification due to parity operation. We will illustrate this in the simple case of SU (2)L × SU (2)R symmetry. The parity operation is of the form, →

→

P : V →V ,

→

→

A→ − A

Consider the case where group element g consists of left-handed transformation, → → → g = exp i θ · V − A ≡L (76) and write the transformation of coset parameters as − →− →

− → →− → − → ′ − · A i α′ · V

− →− →

gei ξ · A = Lei ξ · A = ei ξ

e

(77)

Under the parity transformation, the left-handed transformation is changed into right-handed one, → → → → → → i θ · V −A i θ · V +A P (L) = P e ≡R (78) =e Then applying the parity transformation to Eq(77), we get − → →− → − → ′ − · A i α′ · V

(79)

h = ei α

− →′ − → ·V

(80)

Σ′ = LΣh† = hΣR†

(81)

− →− →

R e−i ξ · A = e−i ξ Using the notation

− →− →

Σ ≡ ei ξ · A ,

e

we can combine the Eqs(77,79) into

− → Note that as we have mentioned before, the parameters α′ depend on the coset − → parameters ξ , the transformation law for Σ here is non-linear because the →′ factor h depends on α . However, the combination U = Σ2 will have a simple transformation law, U ′ = LU R† . (82) 18

− → Since L, R are independent of the coset parameters ξ ,this transformation law is linear and will be useful for constructing Lagrangian. We are interested in the cases where spontaneous symmetry breaking is generated by the scalar fields. Some of these scalar fields become the massless Goldstone bosons, like the pions in the σ-model and others remain massive, like σ-field. Consider the scalar fields in the σ-model, where we will use the notation, → (83) φ = (π 1 , π 2 , π 3 , σ) = φ1, φ2, φ3, φ4 The vacuum expectation value which gives the classical ground state is of the form, 0 0 < φ >0 = (84) 0 v

Suppose we make a very general assumption that from a given point in φ-space we can reach any other point by some group transformation.(Space is transitive). Then the general field configuration can be written as φ1 (x) 0 φ2 (x) 0 φ (x) = for some g ∈ G (85) φ3 (x) = g 0 φ4 (x) σ

Remark:Strictly speaking we should use the representation matrices D (g) of the group element g rather than g itself. However for simplicity of notation, g here is a shorthand for D (g) . From the coset decomposition in Eq(64),we can write → − →−

→ − →−

g = ei π · A ei α · V

Here we have chosen coset parameters to be the pion fields. Then the scalar fields as 0 0 0 − → 0 − → − − → 0 0 − → − → → i π ·A i π ·A i α ·V e φ (x) = g 0 0 =e 0 =e σ σ σ

(86) we can write

(87)

where we have used the fact that the vector (0, 0, 0, σ) is proportional to the vac→ − →− uum configuration (0, 0, 0, v) and is invariant under the subgroup H = {ei α · V }. Since the Goldstone bosons are identified as the coset parameters, they trans→

form the same way as ξ in Eq(70), i.e. → − →−

− →′ − →− → − → · A i α′ · V

g1 ei π · A = ei π where

→

→

e

→

π ′ =π ′ (π, g1 ) ,

→

α′ =α′ (π, g1 ) 19

(88)

(89)

For the case g = h ∈ H,we have from Eq(75), ′

π j = Dji (h) π i

(90)

Remark: The scalar fields here have property that after separating out the Goldstone bosons, the remainders are proportional to the vacuum expectation value and is then invariant under the subgroup H. This is true only for scalar fields in the vector representation in O (n) or SU (n) groups and is not true for scalars in more general representation. In more general cases, we can separate out the Goldstone bosons by writing φ (x) as → − →−

φ (x) = ei π · A χ (x)

(91)

where χ (x) contains all the massive fields. Under the action of group element g1 , we have − →′ − →− → − → · A i α′ · V

→ − →−

g1 φ = g1 ei π · A χ (x) = ei π

e

where

− →′ − → ·V

χ′ (x) = ei α

− →′ − → ·A ′

χ (x) = ei π

χ (x)

χ (x)

(92) (93)

To see how the Goldstone bosons transform under the axial transformation, we →

− →

set L = R† = ei θ · A ,in Eq(81) to get − →′ − → ·A

ei2 π

− →− →

→ − →−

− →− →

= ei θ · A ei2 π · A ei θ · A

(94)

To understand this equation better, we first take the infinitesimal transforma− → tion, | θ | ≪ 1, − →− → → → → − → − → → − → → → → i2− π ·A i2− π ·A i2 π ′ · A i2− π ·A (95) i θ · A + ··· +e e =e + i θ ·A e →′

and then expand this in powers of π s, →′

→

→

→

→

→

1 + 2i π · A + · · · = 1 + 2i π · A +2i θ · A + · · · Comparing both sides we see that →′

→

→

π =π + θ + · · ·

(96)

Thus there is a inhomogeneous term in the transformation law for the Goldstone → bosons π and this is why Goldstone bosons have derivative coupling. → − →−

Since the transformation law for U = ei2 π · A is linear and simple, it is easier to construct the chirally invariant interaction in terms of U rather than Σ. It is easy to see that the only invariants without derivatives will involve trace of some powers of U U † ,which is just an identity matrix. (This also implies that the Goldstone boson coupling will involve derivatives). Thus the interaction with lowest numbers of derivative is of the form L = tr (∂µ U ∂ µ U ) 20

(97)

Covariant derivative: The parity symmetry in the σ-model is responsible for getting the simple combination U (x) which transforms linearly. For the more general case where there is no such simplification, to construct invariant terms involving derivatives is quite complicate because the non-linear transformation law will involve Golstone boson fields which are space-time dependent. This means that we need to construct the covariant derivatives. Futhermore, to exhibit the low energy explicitly we need to couple Golstone field with derivative to other matter fields. We will now discuss briefly in the simple case of σ-model. Write the scalar fields φ (x) in the form, 0 − → − → 0 φ (x) = ei π (x)· A (98) 0 = Σ (x) χ (x) σ (x) − →

− →

with Σ (x) = ei π (x)· A .As before under the action of group element g, we have gΣ (x) = Σ′ (x) h (x)

with

→ →

h (x) = ei α ·V

(99)

Since φ transforms linearly, we have φ′ = gφ which implies Σ′ χ′ = gΣχ = Σ′ hχ

or

χ′ = hχ

(100)

→

This means that χ transforms non-linearly because α in h depend on π fields. Making use of the simplification for the case of chiral symmetry we have, from transformation law,(81) ∂µ Σ′ = L(∂µ Σh−1 + Σ∂µ h−1 ) = (∂µ Σh + Σ∂µ h)R† and

If we define

Σ′−1 ∂µ Σ′ = h Σ−1 ∂µ Σ h−1 + h∂µ h−1 ∂µ Σ′ Σ′−1 = h ∂µ ΣΣ−1 h−1 − h∂µ h−1 1 −1 Σ ∂µ Σ − ∂µ ΣΣ−1 2 1 −1 Σ ∂µ Σ + ∂µ ΣΣ−1 aµ = 2 vµ =

we get

(101) (102) (103) (104) (105)

vµ′ = hvµ h−1 + h∂µ h−1 aµ = haµ h−1 This means that vµ transforms like ”gauge field”, while aµ transforms as global adjoint field. Therefore vµ can be used to construct the covariant derivative and aµ is like a global axial vector field. 21

Nucleon Field The nucleon field in the linear σ-model has the transformation properties, ′

NL → NL = LNL ,

′

and

NR → NR = RNR

Thus to couple nucleon fields to the Goldstone bosons, we could write down the following SU (2) × SU (2) invariant coupling, Lint = g N L ΣNR + N R Σ† NL .

However, this is not of the form of derivative coupling which exhibits the low energy theorem explicitly. For this purpose and general non-linear realization, we define a new nucleon field by, → →

e N = ei π · A N

Under the action of the group element g,we get → →

where

→′ →

e = ei π gN = gei π · A N

→ →

· A i α ·V

→ →

e

→′ →

e = ei π N

e ′ = ei α · V N e = hN e N

·A

e′ N

e transforms according to the representation of the subgroup H but with Thus N → → the group parameters depend on π fields, α π , g . Then from the transformation properties given in Eqs(??), we can write down the derivative coupling as e e µ (i∂µ − vµ ) N e + gN e γ µ aµ N LN = Nγ which will yield the low energy theorem explicitly.

IV.

Examples in the Non-relativistic System

In the frame work of relativistic field theory, e.g. in SU (2) × SU (2) σmodel, spontaneous symmetry breaking seems to be put in by hand, i.e. setting the quadratic terms to have negative sign in the scalar potential in order to develope vacuum expectation value. This is rather ad hoc and no physical reason is given for why this is the case. We will now discuss some simple nonrelativistic examples of spontaneous symmetry breaking in order to shed some light on this (18 ).

A.

Infinite range Ising model

Consider a system of N spins on an one dimensional lattice with Hamiltonian, N N X J X si (106) si sj − B H=− N i

where si = ±1, J is the coupling constant for spin-spin interaction and B is the external magnetic field. In this Hamiltonian, for calculational simplicity we allow every spin to interact with every other spin, while more realistic situation will be the short range nearest neighbor interaction. But the interest here is to see how the spontaneous symmetry breaking come about and we will ignore this. The partition function is given by !2 X X X βJ (107) si + βB si Z = T r e−βH = exp 2N s =±1 i i i

Using the identity for the Gaussian integral, r Z +∞ π − b2 −ax2 +bx dxe = e 4a a −∞

we can write the partition function as !# " r Z X X N βJ +∞ N βJ 2 Z= si x + β (Jx + B) dx exp − 2π −∞ 2 s =±1 i

(108)

i

Now we can sum over each si independently, !! X X = exp{N log [2 cosh β (Jx + B)]} si exp β (Jx + B) si =±1

i

and Z=

r

N βJ 2π

Z

+∞

−∞

N βJ 2 x + N log [2 cosh β (Jx + B)] dx exp − 2

(109)

From the partition function Z, we can compute the average spin, S=

1 ∂ 1 X si >= − < ln Z. N βN ∂B i

(110)

If S 6= 0 in the limit the external field vanishes, B → 0, then we have spontaneous symmetry breaking. Since we are interested in the case where N is very large, we can use saddle point method to compute Z. Write r Z N βJ +∞ Z= dx exp{−N βf (x)} (111) 2π −∞ where f (x) =

Jx2 1 − log [2 cosh β (Jx + B)] 2 β

(112)

The minimum of f (x) is given by f ′ (x) = 0,

⇒

x = tanh β (Jx + B) 23

(113)

Let xi , i = 0, 1, 2, · · · be the solutions of this transcendental equation, then s r N βJ X 2π Z= exp{−N βf (xi )} (114) ′′ (x ) 2π N βf i i Suppose x0 is the smallest of these solutions, it will dominate the partition function for large N. Then the average magnetization is then S=−

1 ∂ ln Z = tanh β (Jx0 + B) = x0 βN ∂B

(115)

where we have used the equation satisfied by x0 .Thus the minimum of f (x) will correspond to the average magnetization. To study spontaneous symmetry breaking, we set B = 0, in Eq(113), and get x = tanh βJx (116) It turns out that this equation has only the trivial solution, x = 0 if βJ < 1 and non-trivial solution exists only for βJ > 1. To understand this feature, we expand f (x) in powers of x for the case B = 0, Jx2 1 1 1 2 4 f (x) = − log 1 + (βJx) + (βJx) + · · · (117) 2 β 2 4! J 1 = (1 − βJ) x2 + β 3 J 4 x4 + · · · 2 12 Thus βJ > 1 corresponds to negative quadratic term, which is the familiar situation in the scalar potential in the σ-model and the like. In terms of temperature this condition, we have J > kT. (118) Here J is the coupling which wants to align the spins in the same direction while the effect of temperature is to randomize spins. Thus the condition in Eq(118) simply means that the interaction of spins has to overcome the thermalization in order to produce significant spin alignment. The temperature Tc = Jk is usually called the critical temperature and spontaneous symmetry breaking is possible only for T < Tc . In this simple example, the non-zero magnetization S breaks the symmetry, si → −si . and originates from the competition between spin-spin interaction which aligns the spin and the thermalization which tends to destroy the alignment. Remarks:(1)From the partition function in Eq(114) we see that the probability to find the system to have xi is given by the Boltzmann factor exp(−βN f (xi )) P (xi ) = P exp(−βN f (xi ))

(119)

i

If x0 is the absolute minimum for f (x) , then in the thermodynamic limit N → ∞, we have P (x0 ) → 1 and the probability for all the other xi will be 24

zero. (2)For the case T is near Tc ,the minimum of f (x) is located at small values of x. Thus we can expand f (x) in power series, β 1 β3 4 1 1− x2 + f (x) = x + ··· (120) 2β c βc 12 β 4c The minimum is then x0 =

s

3β 3c β3

s 3 β Tc 3T 1− = 1− βc Tc3 T

(121)

This means that near the critical temperature T → T0 ,the dependence of average magnetization on (T − T0 ) is non-analytic. This is a typical behavior of physical quantities near the critical point.

B.

Superfluid

The superfluid He4 provides a simple example of Goldstone excitation where the excitation energy, ε (k) goes to zero when the wave number k → 0. The helium atoms are tightly bounded and the long-distance attractive force between atoms are very weak while the short distance is strongly repulsive. Thus a system of helium atoms can be described as a gas of weakly interacting bosons with Hamiltonian, Z Z 1 1 † 2 3 H=− d xψ ∇ ψ + d3 xd3 yψ† (x) ψ † (y) v (x − y) ψ (x) ψ (y) (122) 2m 2 Here v (x) is the potential describes the effective interaction between helium atoms and ψ (x) is the field operator for the helium atom and satisfies the commutation relation, h i ψ (x) , ψ † (y) = δ 3 (x − y)

We will assume that the system is in a large box of volume Ω with periodical boundary condition. Clearly, this Hamiltonian is invariant under the transformation, ψ (x) → ψ ′ (x) = eiα ψ (x) (123) This is just a U (1) symmetry which says that the number of He atoms is conserved. The conserved charge is just the number operator, Z Q = d3 xψ † (x) ψ (x) (124) with the commutation relations, [Q, ψ (x)] = ψ (x) ,

h i Q, ψ† (x) = −ψ† (x) . 25

(125)

We can expand ψ (x) in plane waves, → → 1 X ψ (x) = √ a → ei k · x Ω → k

(126)

k

†

where a→ and a→ are the usual creation and annihilation operators satisfying k

k

the commutation relations, i h i h a→ , a→′ = 0, a→ , a†→ = δ → ,→ k

k

k

k′

(127)

k k′

The Hamilton is then of the form, H=

X ℏ2 k 2 † 1 X v (k1 − k3 ) δ k1 +k2 ,k3 +k4 a†k1 a†k2 ak3 ak4 ak ak + 2m 2Ω k

(128)

ki

where v (k) =

Z

→ →

d3 xei k · x v (x)

(129)

In these two equations and there after we have, for notational simplicity, ne′ glected the vector symbol for the wave vectors ki s. Since v (x) is real we have v (k) =v (−k) . For the trivial case where there is no interaction, v (x) = 0, the →

ground state is just the one in which all particles are in the k = 0 state, N a†0 |Ψ0 >v=0 = √ |0 > where ak |0 >= 0 ∀k (130) N! It is clear that if the interaction is small enough, in the ground state and low→ lying excited states, most of the particles will be in the k = 0 state , i.e. < n0 >≫< nk >

with

k 6= 0.

(131)

where nk = a†k ak . We are interested in the cases where N, the total number of particles, is very large. Thus n0 ˜N is very large. From the properties of the creation and annihilation operators, √ √ (132) a0 |n0> = n0 |n0 − 1 >, a†0 |n0> = n0 + 1|n0 + 1 > √ we will make the assumption that the matrix elements of a0 are of order n0 √ and a†0 of order n0 + 1. Thus in the limit n√ 0 ˜N → ∞, commutator of a0 and † † a0 , is of order unity while a0 , a0 are of order N , i h √ a0 , a†0 = 1 ≪ a0 or a†0 ˜ n0 (133) Thus we can neglect the commutator . Since a0 , and a†0 commute with all the other operators, i h for k 6= 0. [a0 , ak ] = a†0 , ak = 0 26

We can then take a0 and a†0 to be c-numbers (Schur’s lemma), √ a0 = a†0 = n0

(134)

√ Thus we will replace a0 and a†0 by n0 . Then the coefficients of terms quadratic in ak and a†k , k 6= 0, in the interaction will be of order n0 and those of quartic term is of order 1. Therefore we can make the approximation that neglect the quartic terms and get the Hamiltonian in the form, H

=

X ℏ2 k 2 † n0 X ak ak + [v (k) (a†k a†−k + ak a−k ) 2m 2Ω

(135)

+2 v (0) a†k ak + 2 v (k) a†k ak ] +

(136)

k6=0

or H=

X

k6=0

ωk a†k ak +

k6=0

where

n20 v (0) 2Ω

n0 X N2 v (k) (a†k a†−k + ak a−k ) + v (0) 2Ω 2Ω

(137)

k6=0

ωk =

ℏ2 k 2 n0 v (k) + 2m Ω

(138)

and we have used

N 2 = n0 +

X

k6=0

2

a†k ak ≈ n20 + 2n0

X

a†k ak

(139)

k6=0

Note that this Hamiltonian does not conserve the particle number but it conserves the momentum because the removal of k = 0 mode effects the particle number but not the momentum. Since this Hamiltonian contains only quadratic terms, we can solve this by Bogoliubove transformation as follows. Define the quasi-particle operators by

then we have

αk = cosh θk ak + sinh θk a†−k

(140)

α†k = cosh θk a†k + sinh θk a−k

(141)

where θk is an arbitrary parameter at our disposal. We now write the Hamiltonian in terms of the quasi particle operators by inverting the relations in Eq(??), ak = cosh θ k αk − sinh θk α†−k ,

a†−k = − sinh θ k αk + cosh θk α†−k

(142)

and θk so that the coefficient of the non-diagonal terms choose the parameter † † αk α−k + αk α−k is zero. The computation is straightforward and the result is tanh 2θk = 27

n0 v Ω

ωk

(143)

and the Hamiltonian is X N 2 v (0) 1 X H= εk α†k αk + + (εk − ω k ) 2Ω 2 k6=0

where εk =

s

ω 2k

−

n0 v Ω

(144)

k6=0

2

=

s

ℏ2 k 2 2m

2

+2

ℏ2 k 2 2m

n0 v (k) Ω

(145)

This is just the Hamiltonian for the uncoupled harmonic oscillators and the eigenvalues are X E= nk εk (146) k

The quai-particle energy excitation has the property that εk → 0,

as

k→0

(147)

which is just the Goldstone excitation. Clearly, the ground state |Ψ0 > is the one which is annihilated by all quasi particle operators αk, αk |Ψ0 >= 0

∀k

(148)

and the excited states are of the form, n1 n2 α†k1 α†k2 · · · |Ψ0 >

(149)

It is straightforward to show that the quasi particle ground state can be written in terms of the original creation operators a†k as X √ (150) tanh θ ki a†ki a†−ki }|0 > |Ψ0 >= Z exp{− ki

where Z=

Y ki

1 − tanh2 θki

(151)

This shows that the quasi particle ground state is a complicate combination of the vacuum, 2 particle states, 4 particle states,· · ·etc. To elucidate the Goldstone theorem, we note that the vacuum expectation value of ψ (0) in the ground state is non-zero, r 1 n0 < Ψ0 |ψ (0) |Ψ0 >= √ < Ψ0 |a0 |Ψ0 >= 6= 0. (152) Ω Ω where we have used the fact that < Ψ0 |ak |Ψ0 >= 0, for k 6= 0,from the momentum conservation. From the commutation relation in Eq( 125) we see that this is the symmetry breaking condition which implies that Q|Ψ0 >6= 0.

(153)

The quasi particle excitation which has the property that its energy εk goes to zero in the limit k → 0, is the Goldstone excitation implies by the Goldstone’s theorem. 28

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2

J. Goldstone, Nuovo Cimento 19, 154 (1961)

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J. Goldstone, A. Salam and S. Weinberg, Phys. Rev. 127, 965 (1962)

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M. Gell-Mann and M. Levy, Nouovo Cimento 16, 705 (1960)

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P. W. Higgs, Phys. Lett., 12 132 (1964), Phys. Rev. Lett., 13 508 (1964)

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11

See for example, Ta-Pei Cheng, and Ling-Fong Li, Gauge Theory of Elementary Particle Physics, Oxford University Press, (1984).

12

See for example, S. Alder, and R. Dashen, Current Algebra, W. A. Benjamin (1968), H. Georgi, Weak Interaction and Modern Particle Physics, Bejamin/Cummings, (1984), J. Donoghue, E. Golowich, and B. Holstein , Dynamics of Standard Model, Cambridge University Press, (1992).

13

:R. Haag, Phys. Rev. ,112 669, (1958)

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:S. Coleman, J. Wess, and B. Zumino, Phy Rev 177, 2239, (1969).

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:C. Callan, S. Coleman, J. Wess, and B. Zumino, 177, 2247, (1969)

16

S. Weinberg, The Quantum Field Theory II, Cambridge University Press, (1996)

17

see for example, M. Tinkham, Group Theory and Quantum Mechanics, McGraw-Hill, (1964).

18

J. Negele and H. Orland, Quantum Many-Particle Systems, Addison-Wesley (1998).

29

This figure "fig1.jpg" is available in "jpg" format from: http://arXiv.org/ps/hep-ph/0001116v1

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