chapter 3 Spontaneous Symmetry Breaking and Nambu-Goldstone boson History 1961 Nambu: SSB of chiral symmetry and appearance of zero mass boson Goldstone s s theorem in general 1964 Higgs (+others): consider SSB with gauge field, obtain massive gauge boson which absorb would-be Goldstone boson 1968 Weinberg-Salam: developed the work of Glashow, and construct SU(2) U(1) electro-weak theory 1971 t Hooft: proved the renormalizability of the electroweak theory
Spontaneous Symmetry Breaking: example Consider the linear sigma model, in which the symmetry of the Lagrangian g is not shared by the ground state solution., is invariant under global U(1) transformation (φ=σ,π) Choosing the ground state by the energy minimization(μ real), and putting (fluctuation), one obtains π-mode becomes mass-less. Spontaneous Symmetry Breaking: example (cont.) example: magnon (spin (p wave) as collective zero mode (interact short range spin-spin) phonon: SSB due to translational ti l invariance i (etc.) in fluids& solids Note: If there is a long-range force like a Coulomb ~1/r, it still costs a finite energy to excite collective mode for even long wave length limit Goldstone boson becomes massive!
Goldstone theorem (proof) If the Lagrangian is invariant under a continuous symmetry transformation, there exists a conserved Noether current and hence conserved charge We also have a commutation relation, where is the structure constant of the Lie algebra. If the vacuum in invariant under group transformation with it is true for trivial vacuum, but not for vacuum with SSB SO(2)~U(1) model + -
Goldstone theorem (cont.) However, if we have degenerate vacua, it is not the case. Using (symmetry) y) and (SSB by condensation), one obtains eg e.g. SO(2) sigma model : σ&π Noether current & conserved charge one can show Hence 0 Q, π(0) 0 = i 0 σ(0) 0 0 Goldstone theorem (cont.) Using the last equation and inserting the intermediate states, one has Owing to the translational invariance, Hence, one obtains *
Goldstone theorem (cont.) Here, we would like to prove (*) is independent of y0. Consider the current conservation Thus, one obtains =0 Goldstone theorem (cont.) If M 0, it becomes sin(m ny0) which contradicts the independence of y0. Therefore, M must be zero. QED Q.E.D NB: N.B: intermediate states n does not include vacuum itself 0>!, because, if we took n=vacuum 0>, LHS of (*) would trivially give zero RHS.
number of NG bosons Consider the SO(3) ~ SU(2) theory (U: unitary matrix) potential is in fact invariant under this group trans. If one choose the vacuum with phi-3 condensation, is not invariant under the full G symmetry, but invarianti under subgroup H of G, In this case, using one obtains 2 Goldstone bosons, Goldstone theorem In this example, the vacuum with the condensation remains symmetric under a residual H: SO(2)~U(1) ( ) symmetry!. y conclusion Spontaneous breakdown of continuous symmetry zero-mass boson (Nambu-Goldstone boson) (zero mode=energy zero # of the non-zero mass boson = # of generator of H # of NG bosons = # of generators of G - # of generators of H (dimension of coset G/H) NB: generally not true for lower dimensional system not exactly true for gauge theory with long-range interaction
quiz 2 Consider SU(2) SU(2) linear sigma model where N is iso-doublet fermion. (1)Show that the fermion-boson interaction term invariant under the transformation. is (2)Derive the conserved vector and axial-vector currents (3) Prove that the Lie algebra between conserved charges, chapter 4 chapter 4 Higgs mechanism
spontaneous breakdown of the gauge symmetry 1 Consider the scalar field φ with the gauge field which is invariant under the U(1) gauge tras. as well as the internal SO(2) local symmetry (potential), φ1, φ2 (= σ and π) In the absence of the gauge field, one obtains the vacuum state for i.e. the scalar field condensation. Choosing one finds spontaneous breakdown of the gauge symmetry 2 φ1 and gauge field A: massive, φ2: massless But changes A (massive) φ φ2 (massless) unphysical!? We consider the new (unitary) gauge transformation then obtains NOT original SO(2) symmetry!! (translation?) where φ1 and A are massive, but φ2 disappears! In this case, the local gauge symmetry is spontaneously broken.
spontaneous breakdown of the gauge symmetry 3 This scalar fields are called Higgs field (particle) π σ It is important to note that degrees of freedom is conserved! Higgs model without SSB 1(σ)+1(π)+2(gauge, )+2( transverse polarizations) Higgs model with SSB 1(σ) +3 (gauge, transverse and longitudinal polarizations) non-abelian case 1 Consider the scalar field with a rotational SO(3)~SU(2) symmetry is invariant under local trans (in isospin space, for example) where the gauge field is given by We choose the vacuum (energy minimum) setting The original SO(3) symmetry is spontaneously broken to SO(2) (U(1)) symmetry about axis. (2 NG boson + 1 massive boson in the absence of gauge fields)
non-abelian case 2 Defining, one obtains We make use of the local gauge trans. (unitary gauge) which provides non-abelian case 3 We summarize the results below check # of degrees of freedom NB: 1 massless vector field is still invariant under the U(1) gauge symmetry
Physical Example: Superconductor Sperconductivity provides a nice illustration of the Abelican model. *unique feature of the SC: Meissner effect magnetic field cannot penetrate into the superconductor We discuss the origin of the Meissner effect on the Higgs model We rely on the Abelian Higgs model, and consider the static case,, The Hamiltonian density is given by which in fact corresponds to the Ginzburg-Landau potential (free energy). macroscopic order parameter φ=cooper pair Physical Example: Superconductor (cont.) Setting the potential parameter near the critical temperature If the potential has a minimum with a condition The Lagrangian is invariant under the U(1) gauge transformation but this symmetry is spontaneously broken due to the Cooper-pairpair condensation. Now we introduce the conserved (Noether) vector current Because spatial dependence of φ is weak, it becomes
Physical Example: Superconductor (cont.) which is so called London equation. With the Maxwell eqs. one can show which indicates the Meissner effect.