Introduction to Spontaneous Symmetry Breaking

Ling-Fong Li

Carnegie Mellon University

2011 BCVSPIN, 25, July 2011

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 1 / 30 Outline

1 Introduction 2 Symmetry and Conservation Law 3 Explicit vs Spontaneous Symmetry Breaking 4 Goldstone Theorem 5 Spontaneous symmetry breaking in global symmetry 6 Spontaneous symmetry breaking in Local symmetry-Higgs Phenomena 7 Standard Model of Electroweak Interaction

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 2 / 30 Introduction

Fundamental Interactions in nature

1 Strong interactions–Quantum Chromodynamics(QCD) : gauge theory based on SU (3) symmetry

Electromagnetic interaction 2 Electroweak interaction–gauge theory based on Weak interaction  SU (2) U (1) symmetry with spontaneous symmetry breaking 

3 Gravitational interaction: gauge theory of local Lorentz symmetry

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 3 / 30 Symmetry and Conservation Law Noether’s Theorem: Any continuous transformation which leaves the action

S = d 4x Z L invariant, will give a conserved current µ ∂ Jµ = 0 which gives conserved charge dQ = 0, Q = d 3x J dt 0 Z Symmetry Transformation Conserved Charge time translation t t + a Energy ! space translation !x !x + !b Momentum rotation ! Angular momentum





Other conserved quantities: Electric charge, Baryon number,...

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 4 / 30 Remarks

Conservation laws = stability of particles e.g. Baryon number conservation) = proton is stable, Charge conservation = electron is stable, Dark Matter ? ) )


Conservation laws might change as we gain more knowledge e.g. muon number is violated only when ν oscillations were observed Parity violation was discovered only in late 50’s, CP violation ... In quantum system, symmetries = degenercies of energy levels e.g. rotational symmetry = (2l +)1) degeneracies )

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 5 / 30 2) Symmetry Breaking Most of the symmetries in nature are approximate symmetries. (a) Explicit breaking–

H = H0 + H1, H1 does not have the symmetry of H0

Example: Hydrogen atom in external magnetic …eld B!

2 ! 2 p Ze 1 ! H0 = , H1 = !µ B 2m 4πε0 r

! H0 is invariant under all rotations while H1 is invariant only for rotation along direction of B Add small non-symmetric terms to the Hamiltonian= degeneracies reduced or removed )

(b) Spontaneous breaking:

Hamiltonian has the symmetry, [Q, H ] = 0 but ground state does not, Q 0 = 0 (Nambu 1960, Goldstone 1961) j i 6 Example: Ferromagnetism T > TC (Curie Temp) all magnetic dipoles are randomly oriented–rotational symm T < TC all magnetic dipoles are in the same direction-not invariant under rotation

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 6 / 30 Ginsburg-Landau theory: Write the free energy u as function of magnetization M! in the form,

! ! 2 ! ! ! ! 2 u(M ) = (∂t M ) + α1 (T ) (M M ) + α2 (M M ) ,

where α2 > 0, α1 = α (T TC ) α > 0 Here u has O (3) rotational symmetry. The ground state is at

! ! ! M (α1 + 2α2M M ) = 0

! T > TC minimum at M = 0. ! α1 ! T < TC minimum at M = = 0 . If we choose M to be in some direction the 2α2 6 r rotational symmetry is broken.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 7 / 30 Goldstone Theorem Noether theorem: continuous symmetry = conserved current, ) dQ ∂ J µ = 0, Q = d 3xJ 0 (x ) , = 0 µ dt Z Suppose A (x ) and B (x ) are some local operators and transform into each other under the symmetry charge Q, [Q, A (0)] = B (0)

Suppose

0 [Q, A (0)] 0 = 0 B (0) 0 = v = 0, symmetry breaking condition h j j i h j j i 6 This implies Q 0 = 0, j i 6 and Q is a broken charge. Then

En = 0, as !p n = 0, for some state n

This means zero energy excitations.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 8 / 30 To show this, write 0 [Q, A (0)] 0 = d 3x 0 J 0 (x ) , A (0) 0 h j j i Z   Inserting a complete set of states and integating over x

0 [Q, A (0)] 0 h j j i 3 3 0 iEn t 0 iEn t = ∑ (2π) δ (!p n ) 0 J (0) n n A 0 e 0 A n n J (0) 0 e = υ = 0 n h j j i h j j i 6 n o

iEn t RHS is time independent while LHS depends explicitly on time from e  . This relation can be satis…ed only if there exists an intermediate state n for which j i

En = 0, for !p n = 0

For relativistic system, energy momentum relation yields

2 2 En = !p + m mn = 0, Goldstone boson n n ) q

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 9 / 30 Remarks

1 State n with property j i 0 A (0) n = 0, h j j i 6 is the massless Goldstone boson while local …eld B corresponds to some massive particle. This state will also have the property

n Q 0 = 0 h j j i 6 This means n can connect to vacuum through broken charge Q. j i 2 Spectrum Here Q is a broken symmetry charge. In general, each broken charge will have a Goldstone boson,

# of Goldstone bosons= # of broken generators

In many cases, there are symmetry charges which remain unbroken,

Qi 0 = 0, i = 1, 2, , l j i

These unbroken charges Q1, Q2, Ql form a group H, a subgroup of original symmetry group G . As a consequence, the
particles

will form multiplets of symm group H. For example, if symmetry SU (2) SU (2) is spontaneously broken to SU (2) , particles will form SU (2) multiplets 

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 10 / 30 3 In relativistic …eld theory, massless particle = long range force. In real world , no massless particls except photon. ) In strong interaction, we have approximate spontaneous symmetry breaking. = almost Goldtone bosons ) 4 Goldstone bosons can either be elementary …elds or composite …elds a) Elementary …elds Here local operators A, B are …elds in the Lagrangian–this is the case in Standard electroweak theory b) Composite …elds In this case A, B correspond to some bound states of elementary …elds. For example in QCD the ‡avor chiral symmetry is broken by quark condensates. Here we have 5 _ _ Q , qγ5q = qq

5   Q some_ chiral charge, q quark …eld _ If 0 qq 0 = 0 ( quark condensate), then qγ5q will correspond to a Goldstone boson. This is usually6 referred to as dynamical symmetry breaking.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 11 / 30 Global symmetry We now discuss the spontaneou symmetry in some simple relativistic …eld theory. As an example, consider 1 = [(∂ σ)2 + (∂ π)2 ] V (σ2 + π2 ) L 2 µ µ with µ2 λ V (σ2 + π2 ) = (σ2 + π2 ) + (σ2 + π2 )2 (1) 2 4 This potential has O (2) symmetry

σ σ cos α sin α σ 0 = π ! π0 sin α cos α π         where α is a constant, global symm

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 12 / 30 To get the minimum we solve the equations

∂V = µ2 + λ(σ2 + π2 ) σ = 0 ∂σ   ∂V = µ2 + λ(σ2 + π2 ) π = 0 ∂π Solution   µ2 σ2 + π2 = = ν2 circle in σ π plane λ

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 13 / 30 Choose σ = ν, π = 0 which are …elds con…gurations for classical vacuum. New quantum …elds

σ0 = σ ν , π0 = π correspond to oscillations around the minimum. Note that

σ = v = 0 h i 6 is the symmetry breaking condition in Goldstone’s theorem. The new Lagrangian is

1 2 2 2 2 2 2 λ 2 2 2 = [(∂ σ0 + (∂ π) ] µ σ0 λνσ0(σ0 + π0 ) (σ0 + π0 ) L 2 µ µ 4

There is no π 2 term, π massless Goldstone boson 0 ) 0

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 14 / 30 Local symmetry: gauge symmetry Maxwell Equations:

ρ ! !E = , ! B! = 0 r
ε0 r

! ! ! ! ∂B 1 ! ! ∂E ! E + = 0, B = ε0 + J r  ∂t µ0 r  ∂t

Source free equations can be solved by introducing scalar and vector potentials, φ, !A,

∂!A B! = ! !A, !E = !φ r  r ∂t

These solutions are not unique because of gauge invariance,

∂α φ φ , !A !A + !α ! ∂t ! r

Or A A ∂ α (x ) µ ! µ µ Classically, it is not so clear what the physical role is played by gauge invariance.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 15 / 30 In quantum mechanics, Schrodinger eq for charged particle is,

1 h 2 ∂ψ ! e !A eφ ψ = i h 2m i r ∂t "   #

This requires transformation of wave function,

e ψ exp i α (x ) ψ ! h   to get same physics. Thus gauge invariance is now connected to symmetry (local) transformation.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 16 / 30 Spontaneous symmetry breaking in local symmetry Local symmetry : symm parameter depends on space-time. Consider a …eld theory with U (1) local symmetry,

2 1 L = D φ † (D µφ) + µ2 φ†φ λ φ†φ F F µν µ 4 µν
  where D φ = ∂ igA φ, F = ∂ A ∂ A µ µ µ µν µ ν ν µ This local U (1) phase transformation is of
the form,

iα φ (x ) φ0 (x ) = e φ (x ) , α = α (x ) ! The derivative transforms as

iα ∂ φ (x ) ∂ φ0 (x ) = e ∂ φ (x ) i∂ α µ ! µ µ µ   which is not a phase U (1) transformation.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 17 / 30 Introduce Aµ to form covariant derivative,

D φ = ∂ igA φ µ µ µ
which will a simple U (1) transformation

iα Dµφ 0 = e Dµφ

if we require A (x ) A0 (x ) = A (x ) ∂ α (x ) µ ! µ µ µ which is the gauge transformation in Maxwell’s equations. Important features of local symmetry: gauge coupling is universal. Also gauge …eld (photon) is µ massless(long range force), because mass term AµA is not gauge invariant.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 18 / 30 Symmetry Breaking When µ2 > 0, minimum of potential

2 V (φ) = µ2 φ†φ + λ φ†φ   is v 2 µ2 φ†φ = , with v 2 = 2 λ Write 1 φ = (φ + iφ ) p2 1 2 Quantum …elds are the oscillations around the minimum,

0 φ 0 = v , 0 φ 0 = 0 h j 1 j i h j 2 j i

As before φ2 is a Goldstone boson. New feature : covariant derivative produces mass term for gauge boson,

2 2 2 2 g v D φ = ∂ igA φ ... AµA + ... µ µ µ ' 2 µ
with mass M = gv

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 19 / 30 Write the scalar …eld in polar coordinates

1 φ = [v + η (x )] exp(iξ/v ) p2

We can now use gauge transformation to transform away ξ. De…ne

1 φ" = exp ( iξ/v ) φ = [v + η (x )] p2

and 1 B = A ∂ ξ µ µ gv µ

ξ disappears. In fact ξ becomes the longitudinal component of Bµ.

massless gauge boson + Goldstone boson=massive vector meson

all long range forces disappear. This was discovered in the 600s by Higgs, Englert & Brout, Guralnik, Hagen & Kibble independently and is usually called Higgs phenomena The extension to non-abelian local symmetry(Yang-Mills …elds) is straightforward and no new features emerge.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 20 / 30 Connection with superconductivity Equation of motion for scalar …eld interacting with em …eld,

! B! = !J r  with !J = ie φ† ! ie !A φ ! + ie !A φ†φ r r       Spontaneous symmetry breaking φ = v and )

!J = e 2v 2!A

This is London equation. Then

! ! B! = ! !J , = 2B! = e 2v 2B! r  r  r  ) r   This gives Meissner e¤ect. ! ! For the static case, ∂0 A = 0, A0 = 0, we get E = 0 and from Ohm’slaw

!E = ρ!J , ρ resistivity

we get ρ = 0, i.e. superconductivity.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 21 / 30 Standard Model of Electroweak Interactions Important features of Weak Interactions : Universality of coupling strength 5 10 e.g. β decay and µ decay can be described by same coupling GF 2 in 4-fermion ' Mp theory, _ GF µ † wk = J J + h.c. where Jµ = νγ (1 γ ) e + L p2 µ µ 5


  h i

Short range interaction If weak interaction is mediated by a massive vector bosons W

2 µ g GF wk =g JµW + h.c. , 2 = L Mw p2

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 22 / 30 Then W boson is massive

These features suggest a theory gauge interaction with massive gauge boson.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 23 / 30 Standard Model (Weinberg 1967, ’t Hooft 1971) This is a gauge theory with spontaneous symmetry breaking. Gauge group: SU (2) U (1) gauge bosons: !A , B –includes both weak and em interaction  µ µ Spontaneous Symmetry Breaking :SU (2) U (1) U (1)em Simple choice for scalar …elds  ! φ+ φ = , φ0   The Lagrangian containing φ is

= D φ † (D µφ) V (φ) Lφ µ
where ig ig D φ = ∂ !τ !A 0 B φ µ µ 2
µ 2 µ   2 V (φ) = µ2 φ†φ + λ φ†φ     Spontaneous symmetry breaking: SU (2) U (1) U (1)em Minimum  ! ∂V 2 † = µ + 2(φ φ) φi = 0 ∂φi h i = ) µ2 + 2(φ†φ) = 0

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 24 / 30 Simple choice 1 0 µ2 φ 0 = ν = h i p2 ν λ   r Expand the quanutm …elds around the minimum

+ φ0 = + = 0 v φ φ0 φ 0 φ0 + h i p2 !

the quadratic terms in V are

2 2 2 2 † λv † 0 2 0 2 V2 (φ) = µ φ0 φ0 + φ0 φ0 + 2 Re φ0 v = 2 Re φ0 v 2     + 0 Thus φ0 and Im φ0 are Goldstone bosons and the symmetry breaking is

SU (2) U (1) U (1)  ! em From covariant derivative in Lφ

2 ~ 0 0 0 µ µ v ~τ Aµ g Bµ ~τ ~A0 g 0 B 0 0 = χ†(g
+ )(g
+ )χ + , χ = Lφ 2 2 2 2 2


1  

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 25 / 30 we get the mass terms for the gauge bosons,

v 2 = g 2 [(A1 )2 + (A2 )2 ] + (gA3 g 0 B )2 + Lφ 8 f µ µ µ µ g


2 +µ 1 2 µ = M W W + M Z Z + W µ 2 Z µ

where 1 g 2v 2 W + = (A1 iA2 ), M 2 = µ p2 µ µ W 4

1 g 2 + g 02 0 3 2 2 Zµ = (g A gBµ), MZ = v g 2 + g 02 4 The …eld p 1 0 3 Aµ = (g Aµ + gBµ) g 2 + g 02 is massless photon. p We can write the scalar …elds in the form

0 φ = exp i !τ !ξ (x ) /v
v + η (x )    

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 26 / 30 where !ξ (x ) are Goldstone bosons. We can make a " gauge transformation"

0 φ = U (!ξ )φ = 0 v + η (x )  

0 !τ !A ! ! µ ! τ Aµ 1 ! i ! 1 !
= U ( ξ )
U ( ξ ) ∂ U ( ξ ) U ( ξ ) 2 2 g µ  

Bµ0 = Bµ where U (!ξ ) = exp i !τ !ξ (x ) /v
  Again the Goldstone bosons !ξ (x ) will be eaten up by gauge bosons to become massive. The left over …eld η (x ) is usually called Higgs Particle. Massive gauge bosons:

2 2 W = 1 A1 iA2 M 2 = g v µ p2 µ  µ W 4 3 2 g 2 2 Zµ = cos θW A sin θW Bµ M = 2 v µ Z 4 cos θW 3 Aµ = sin θW Aµ + cos θW Bµ Mγ = 0

where g tan θ = 0 W g

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 27 / 30 Note that M 2 W 1 2 2 = MZ cos θW The mass for the Higgs particle is 2 2 mη = 2λv From comparing Standard Model with 4-fermion theory, we get

1 v = 246 Gev ' p2GF q Since magnitude of Higgs self coupling λ is not known, we do not know the mass of Higgs particle.

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 28 / 30 Fermion mass and spontaneous symmetry breaking

Another role played by spontaneous symmetry breaking is to give masses to the fermions. Fermions: a) Leptons

ν ν ν L = e , µ , τ , R = e , µ , τ , i e µ τ i R R R  L  L  L b) Quarks (Glashow, Iliopoupos, and Maiani, Kobayashi and Maskawa)

u c t q = 0 , 0 , 0 , U = u , c , t , D = d , s , b iL d s b iR R R R iR R R R  L  L  L All left-handed fermions are in SU (2) doublets and right-handed fermions are all singlets. Yukawa coupling: _ Y = fij Li Rj φ + h.c. + L


Fermions get their masses from spontaneous symmetry breaking through Yukawa couplings,

mij = fij v

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 29 / 30 This implies that the Yukawa couplings ∝ masses. This an example of spontaneous breaking of global symmetry(chiral symmetry),

!τ !α Li L0 = exp( i
)Li , Ri R 0 = Ri ! i 2 ! i

Remark: The scalar particles φ plays 2 important roles in Standard Model:

1 It breaks the gauge symmetry through the universal gauge coupling 2 It give mass to fermions through Yukawa couplings which are not universal

Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking2011 BCVSPIN, 25, July 2011 30 / 30