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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 Breaking 2011 BCVSPIN, 25, July 2011 1 / 30 Outline 1 2 3 4 5 6 7 Introduction Symmetry and Conservation Law Explicit vs Spontaneous Symmetry Breaking Goldstone Theorem Spontaneous symmetry breaking in global symmetry Spontaneous symmetry breaking in Local symmetry-Higgs Phenomena Standard Model of Electroweak Interaction Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 2 / 30 Introduction Fundamental Interactions in nature 1 2 3 Strong interactions–Quantum Chromodynamics(QCD) : gauge theory based on SU (3 ) symmetry Electromagnetic interaction Electroweak interaction–gauge theory based on Weak interaction SU (2 ) U (1 ) symmetry with spontaneous symmetry breaking Gravitational interaction: gauge theory of local Lorentz symmetry Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 3 / 30 Symmetry and Conservation Law Noether’s Theorem: Any continuous transformation which leaves the action S = Z d 4x L invariant, will give a conserved current ∂µ J µ = 0 which gives conserved charge dQ = 0, dt Q = Symmetry Transformation time translation t ! t +a ! ! ! space translation x ! x + b rotation Z d 3 x J0 Conserved Charge Energy Momentum Angular momentum Other conserved quantities: Electric charge, Baryon number,. . . Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 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 Breaking 2011 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 H 0 ! Example: Hydrogen atom in external magnetic …eld B H0 = !2 p 2m Ze 2 1 , 4πε0 r H1 = ! ! µ B ! H 0 is invariant under all rotations while H 1 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 j0 i 6= 0 (Nambu 1960, Goldstone 1961) Example: Ferromagnetism T > T C (Curie Temp) all magnetic dipoles are randomly oriented–rotational symm T < T C all magnetic dipoles are in the same direction-not invariant under rotation Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 6 / 30 Ginsburg-Landau theory: ! Write the free energy u as function of magnetization M in the form, ! ! ! α2 > 0, α1 = α (T ! ! ! u (M ) = ( ∂t M )2 + α1 (T ) (M M ) + α2 (M M )2 , where TC ) α>0 Here u has O (3 ) rotational symmetry. The ground state is at ! ! ! M (α1 + 2α2 M M ) = 0 ! T > T C minimum at M = 0. r ! T < T C minimum at M = ! α1 6= 0 . If we choose M to be in some direction the 2α2 rotational symmetry is broken. Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 7 / 30 Goldstone Theorem Noether theorem: continuous symmetry =) conserved current, ∂µ J µ = 0, Q = Z d 3 xJ 0 (x ) , dQ =0 dt 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 h0 j[Q , A (0 )]j 0 i = h0 jB (0 )j 0 i = v 6= 0, symmetry breaking condition This implies Q j0 i 6= 0, and Q is a broken charge. Then E n = 0, as ! p n = 0, for some state n This means zero energy excitations. Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 8 / 30 To show this, write h0 j[Q , A (0 )]j 0 i = Z d 3 x 0 J 0 (x ) , A (0 ) 0 Inserting a complete set of states and integating over x h0 j[Q , A (0 )]j 0 i = ! ∑ (2π )3 δ3 ( p n ) n n 0 J 0 (0 ) n hn jA j 0 i e iE n t h0 jA j n i n J 0 (0 ) 0 e iEn t RHS is time independent while LHS depends explicitly on time from e satis…ed only if there exists an intermediate state jn i for which E n = 0, for iE n t . o = υ 6= 0 This relation can be ! pn = 0 For relativistic system, energy momentum relation yields En = q !2 pn + m n2 ) m n = 0, Goldstone boson Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 9 / 30 Remarks 1 State jn i with property h0 jA (0 )j n i 6= 0, is the massless Goldstone boson while local …eld B corresponds to some massive particle. This state will also have the property hn jQ j 0 i 6 = 0 2 This means jn i can connect to vacuum through broken charge Q . 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, Q i j0 i = 0, i = 1, 2, ,l These unbroken charges Q 1 , Q 2 , Q l 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 Breaking 2011 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 _ _ Q 5 , qγ5 q = qq Q 5 some chiral charge, q quark …eld _ _ If 0 qq 0 6= 0 ( quark condensate), then qγ5 q will correspond to a Goldstone boson. This is usually referred to as dynamical symmetry breaking. Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 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 L = [(∂µ σ)2 + (∂µ π )2 ] V (σ2 + π 2 ) 2 with V ( σ2 + π 2 ) = λ µ2 2 ( σ + π 2 ) + ( σ 2 + π 2 )2 2 4 (1) This potential has O (2 ) symmetry σ π ! σ0 π0 = cos α sin α sin α cos α σ π where α is a constant, global symm Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 12 / 30 To get the minimum we solve the equations ∂V = ∂σ µ2 + λ ( σ 2 + π 2 ) σ = 0 ∂V = ∂π µ2 + λ ( σ 2 + π 2 ) π = 0 Solution σ2 + π 2 = µ2 = ν2 λ circle in σ π plane Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 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 hσi = v 6= 0 is the symmetry breaking condition in Goldstone’s theorem. The new Lagrangian is L= 1 [(∂µ σ02 + (∂µ π )2 ] 2 µ2 σ 02 λνσ0 (σ02 + π 02 ) λ 02 ( σ + π 02 )2 4 There is no π 02 term, ) π 0 massless Goldstone boson Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 14 / 30 Local symmetry: gauge symmetry Maxwell Equations: ! ! r E = ! r ! ! ρ , ε0 ! r B =0 ! ∂B E+ = 0, ∂t 1 ! r µ0 ! ! ! ∂E B = ε0 +J ∂t ! Source free equations can be solved by introducing scalar and vector potentials, φ, A, ! ! B =r ! A, ! E = ! ! rφ ∂A ∂t These solutions are not unique because of gauge invariance, φ !φ ∂α , ∂t ! A ! ! ! A + rα Or Aµ ! Aµ ∂ µ α (x ) 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 Breaking 2011 BCVSPIN, 25, July 2011 15 / 30 In quantum mechanics, Schrodinger eq for charged particle is, " 1 2m h! r i ! 2 eA # eφ ψ = i h ∂ψ ∂t This requires transformation of wave function, ψ ! exp i e 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 Breaking 2011 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, L = Dµ φ † (D µ φ ) + µ2 φ † φ λ φ† φ 2 1 F µν F µν 4 where D µ φ = ∂µ F µν = ∂µ A ν igA µ φ, ∂ν A µ This local U (1 ) phase transformation is of the form, φ (x ) ! φ 0 (x ) = e iα φ (x ) , α = α (x ) The derivative transforms as ∂ µ φ (x ) ! ∂ µ φ 0 (x ) = e iα ∂ µ φ (x ) i ∂µ α which is not a phase U (1 ) transformation. Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 17 / 30 Introduce A µ to form covariant derivative, D µ φ = ∂µ igA µ φ which will a simple U (1 ) transformation Dµ φ 0 =e iα Dµ φ if we require A µ (x ) ! A 0µ (x ) = A µ (x ) ∂ µ α (x ) which is the gauge transformation in Maxwell’s equations. Important features of local symmetry: gauge coupling is universal. Also gauge …eld (photon) is massless(long range force), because mass term A µ A µ is not gauge invariant. Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 18 / 30 Symmetry Breaking When µ2 > 0, minimum of potential V (φ) = is φ† φ = Write µ2 φ † φ + λ φ † φ v2 , 2 with v2 = 2 µ2 λ 1 φ = p ( φ1 + i φ2 ) 2 Quantum …elds are the oscillations around the minimum, h0 j φ1 j 0 i = v , h0 j φ2 j 0 i = 0 As before φ2 is a Goldstone boson. New feature : covariant derivative produces mass term for gauge boson, Dµ φ 2 = ∂µ igA µ φ 2 ' ... g 2v 2 µ A Aµ + . . . 2 with mass M = gv Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 19 / 30 Write the scalar …eld in polar coordinates 1 φ = p [v + η (x )] exp (i ξ/v ) 2 We can now use gauge transformation to transform away ξ. De…ne 1 φ" = exp ( i ξ/v ) φ = p [v + η (x )] 2 and Bµ = Aµ 1 ∂µ ξ gv ξ disappears. In fact ξ becomes the longitudinal component of B µ . massless gauge boson + Goldstone boson=massive vector meson all long range forces disappear. This was discovered in the 60 0 s 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 Breaking 2011 BCVSPIN, 25, July 2011 20 / 30 Connection with superconductivity Equation of motion for scalar …eld interacting with em …eld, ! ! r with ! J = ie φ† ! r ! B =J ! ie A ! ! r + ie A φ† φ φ Spontaneous symmetry breaking ) φ = v and ! ! J = e2v 2 A This is London equation. Then ! r ! r ! B ! =r This gives Meissner e¤ect. ! ! J, =) ! ! r2 B = e 2 v 2 B ! For the static case, ∂0 A = 0, A 0 = 0, we get E = 0 and from Ohm’s law ! ! E = ρJ , ρ resistivity we get ρ = 0, i.e. superconductivity. Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 21 / 30 Standard Model of Electroweak Interactions Important features of Weak Interactions : Universality of coupling strength e.g. β decay and µ decay can be described by same coupling G F ' theory, G Lwk = pF J µ J µ† + h.c . 2 h_ where J µ = νγµ (1 10 5 M p2 in 4-fermion i γ5 ) e + Short range interaction If weak interaction is mediated by a massive vector bosons W Lwk =g J µ W µ + h.c . , g2 G = pF M w2 2 Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 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 Breaking 2011 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 µ , B µ –includes both weak and em interaction Spontaneous Symmetry Breaking :SU (2 ) U (1 ) ! U (1 )em Simple choice for scalar …elds φ+ φ= , φ0 The Lagrangian containing φ is where Dµ φ = V (φ) = ∂µ † (D µ φ ) V (φ) ig ! ! τ Aµ 2 ig 0 Bµ 2 Lφ = D µ φ µ2 φ † φ + λ φ † φ φ 2 Spontaneous symmetry breaking: SU (2 ) U (1 ) ! U (1 )em Minimum h i ∂V = µ2 + 2 ( φ † φ ) φi = 0 ∂φi =) µ2 + 2 ( φ † φ ) = 0 Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 24 / 30 Simple choice 1 h φ i0 = p 2 0 ν ν= r µ2 λ Expand the quanutm …elds around the minimum φ = φ 0 + h φ i0 = φ0+ v φ +p ! 00 2 the quadratic terms in V are V 2 (φ) = µ2 φ 0 † φ 0 + 0 λv 2 0† 0 φ φ + 2 Re φ 0 2 2 0 v 2 = 2 Re φ 0 2 v2 0 Thus φ0+ and Im φ 0 are Goldstone bosons and the symmetry breaking is SU (2 ) U (1 ) ! U (1 )em From covariant derivative in Lφ Lφ = 0 0 0 0 0 0 g Bµ ~τ ~A µ v 2 † ~τ ~A µ g Bµ χ (g + )(g + )χ + 2 2 2 2 2 , χ= 0 1 Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 25 / 30 we get the mass terms for the gauge bosons, Lφ = = v2 2 fg [(A 1µ )2 + (A 2µ )2 ] + (gA 3µ 8 1 2 MW W +µ W µ + M Z2 Z µ Z µ + 2 0 g B µ )2 g + where 1 W µ+ = p (A 1µ 2 The …eld Zµ = p 1 g2 + g 02 iA 2µ ), 0 (g A 3 Aµ = p gB µ ), 1 g2 + g 2 MW = 02 M Z2 = g 2v 2 4 g2 + g 4 02 v2 0 (g A 3µ + gB µ ) is massless photon. We can write the scalar …elds in the form ! ! φ = exp i τ ξ (x ) /v 0 v + η (x ) Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 26 / 30 ! where ξ (x ) are Goldstone bosons. We can make a " gauge transformation" ! 0 v + η (x ) φ0 = U ( ξ )φ = ! !0 τ Aµ 2 ! = U(ξ ) ! ! τ Aµ U 2 1 ! i g (ξ ) ! ∂µ U ( ξ ) U 1 ! (ξ ) B µ0 = B µ where ! U ( ξ ) = exp ! ! iτ 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: Wµ = p1 2 A 1µ g2v2 4 g2 2 4 cos θ W iA 2µ 2 = MW Z µ = cos θ W A 3µ sin θ W B µ A µ = sin θ W A 3µ + cos θ W B µ where tan θ W = M Z2 = Mγ = 0 v2 g0 g Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 27 / 30 Note that M Z2 2 MW =1 cos2 θ W The mass for the Higgs particle is m η2 = 2λv 2 From comparing Standard Model with 4-fermion theory, we get 1 v = qp ' 246 Gev 2G F 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 Breaking 2011 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 Li = νe e , L νµ µ , L ντ τ , L R i = e R , µR , τ R , b) Quarks (Glashow, Iliopoupos, and Maiani, Kobayashi and Maskawa) q iL = u0 d , L c0 s , L t0 b , U iR = u R , cR , tR , D iR = d R , sR , b R L All left-handed fermions are in SU (2 ) doublets and right-handed fermions are all singlets. Yukawa coupling: _ LY = fij L i R j φ + h.c . + Fermions get their masses from spontaneous symmetry breaking through Yukawa couplings, m ij = fij v Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking 2011 BCVSPIN, 25, July 2011 29 / 30 This implies that the Yukawa couplings ∝ masses. This an example of spontaneous breaking of global symmetry(chiral symmetry), Li ! L i0 = exp ( i ! ! τ α )L i , 2 Ri ! R i0 = R 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 Breaking 2011 BCVSPIN, 25, July 2011 30 / 30