Week 6: Nambu-Goldstone bosons and the Higgs Mech- anism 1 Reading material from the books Burgess-Moore, Chapter 3,6 • Donoghue, Golowich, Holstein Chapter 4, 15 • Peskin and Schroeder, Chap 11 • Weinberg, Chap. 19 • 1 Spontaneously broken symmetries Let us consider first an example of a theory with a continuous symmetry such that the vacuum state is not invariant under the symmetry. The idea is to begin with a theory with a single complex scalar field φ, such that we have a symmetry under phase rotations of φ, namely φ exp(iθ)φ → is a symmetry of the lagrangian. A lagrangian density with that property is given by L = ∂µφ∗∂ φ V (φ∗φ) (1) − µ − and we will expand V to second order λ V (φ∗φ)= m2φ∗φ + (φ∗φ)2 (2) 4 A vacuum is usually characterized by being the minimum of the potential. If m2 > 0, then the vacuum is located at φ = 0. Such a state is called one where the symmetry is unbroken. This can be characterized by saying that δ S = 0 for all local operators θh i S. Usually, we think of δθS = [Q, S] as an operator equation, where Q is the charge of the corresponding symmetry (this is the integral of the j0 compo- nent of the associated Noether current). This is sometimes also written as Q 0 = 0. However, this is somewhat misleading because Q is not necessarily normalizable| i (it is an integral that goes all the way to infinity). 1 c David Berenstein 2009 1 Since the field φ transforms under the symmetry, we have that for an unbroken symmetry state δ φ(x) = i φ(x) =0 (3) θh i h i Now, let us consider the case where we find a classical solution of the potential, where m2 < 0. The field then satisfies 2m2φ + λφ(φφ∗)=0 (4) This can be solved by φ∗φ = 2m2/λ. However, the phase of φ is undeter- mined. − Let us expand around a real value of φ, namely φ = v + φ1/√2+ iφ2/√2, where φ1 and φ2 are real fields, and v = 2m2/λ. We find that, to quadratic order p− V (φ) m2(φ1)2 (5) ∼ 2 2 2 So that m1 =2m and m2 = 0, this is, one of the particles ends up being massless. This extra massless particle is called a Nambu-Goldstone boson. Notice that the state is characterized by [Q, φ] =0 (6) h i 6 Technically, this implies that Q 0 = 0, but this is exactly the case where Q is not really expected to be normalizable.| i 6 One would like to know if the massless property of φ1 is true always, or if this argument does not follow once we take into account loops. In order for the argument to be true, we need to make sure that the renormalization program does not spoil the symmetry. For cases of unbroken symmetry it is easy to find a regulator that respects the symmetry: we just need to write the propagators for φφ∗ in a way that does not spoil the symmetry. This requires that the following vanish when we introduce a regulator Tφ(x)φ(0) , Tφ∗(x)φ0 (7) h i h i Both dimensional regularization and Pauli-Villars regulators guarantee this, if we choose the Pauli-Villars regulator to just subtract terms from the standard propagator. 2 It is easy to show that only invariant terms will receive renormalizations (finite or infinite). This is a graphical argument. Any vertex has equal numbers of incident + charge legs and charge legs. (This is charge conservation). Any Feynman line will connect− an incident + to an incident . One can describe the flow of charge this way. − The total violation of charge of any diagram is N+(ext) N−(ext). The pairing of propagators shows that any internal + is paired wi− th an internal , so one finds that the total external charge violation is related to a sum over− the internal vertices V V V N (ext) N−(ext)= N N− =0 (8) + − + − XV So one has that all amplitudes with N N− = 0 vanish identically. This + − 6 means that all infinite and finite counterterms can be chosen to have the same property and the symmetry of the effective action or scattering amplitudes is preserved. One needs to make the same argument when the symmetry is sponta- neously broken. If the argument goes through, one finds again that in the spontaneously broken symmetry case the particle φ2 is massless after taking into account the renormalization of the vev. A simple way to understand this is to write φ in polar coordinates, φ = (v + φ/˜ √2) exp(iθ(x)). One finds that the kinetic term is (∂φ˜)2 +(v + φ/˜ √2)2(∂θ)2 (9) and that so long as v = 0, the kinetic term for θ is invertible. One can easily also show that the potential6 is only a function of φ˜ classically. Thus, in this model the symmetry is characterized by θ(x) θ(x)+ χ. The invariants are ∂θ(x). → Indeed, any external line with θ shows up with a derivative attached, so Lorentz invariance guarantees that the corrections will always have deriva- tives in θ, unless the amplitude has compensating 1/p2 behavior. 1.1 Some amplitudes ∗ ↔ Consider the current operator jµ(x) φ (x)∂µ φ(x) expanded around the spontaneously broken vacuum (there∼ are factors of i in the definition of the current if one wants it to be hermitian). It is easy to see that j (x) v∂ φ2(x)+ . (10) µ ∼ µ 3 where the other terms are quadratic on the fields or higher. We find that jµ acting on the vacuum has a non-zero probability of creating a one particle state where the Boson φ2 (the Nambu-Goldstone boson) is excited. We get therefore that p J (x) 0 f p exp(ipx) (11) h | µ | i ∼ π µ where p is a one-particle state. This will work similarly in the θ(x) repre- sentation,| i where the current is identified with j ∂ θ. µ ∼ µ Remember that we are using a representation of the states where the norm of p is Lorentz invariant and independent of p. | i The quantity fπ is called the pion decay constant. So we find that in a spontaneously broken phase (we use the terminology of condensed matter systems) the current acting on the vacuum can produce single particle scalar states. We will take this as the general definition of what it means to be in a spontaneously broken phase. 1.2 Goldstone’s theorem Assume that you have a quantum field theory that is invariant under some global symmetry group G. The Noether current associated to this symmetries is characterized by a Jµ (x). (12) where a =1,... dim(G). These currents are conserved at the quantum level, meaning that µa ∂µJ (x)=0 (13) µ for all x as operator equations. In particular the matrix elements of ∂µJ have to vanish between all physical states. Consider the amplitudes p J a(x) 0 f p exp(ipx) (14) h | µ | i ∼ πa µ where we have a non-zero amplitude to create a new state that corresponds to a scalar state with momentum p. We know that the amplitude can only depend on p as described above, and since the scalar state has no polarization in the little group, the vector indices can only be made of p. Using the current conservation we have that p ∂µJ a(x) 0 =0= f p pµ exp(ipx) (15) h | µ | i πa µ 4 2 So, if fπa = 0, we find that p = 0. This is, we predict a massless particle. This argument6 is correct for all currents that generate a new state. The currents that do not generate a new state form a subalgebra of the Lie algebra of G (let us call it H). Classically, this is the Lie algebra of invariants of a configuration ( we called this the little group in Lorentz representations), this is the subalgebra that leaves the vevs describing the vacuum invariant under rotations: those such that δav = 0. The coset G/H can be written as a set of irreducible representations of H. For each element of G/H there is one Nambu-Goldstone boson. The states that do not annihilate the vacuum expectation values δ v =0 a 6 give rise to some linearly independent set of vectors that can be associated to quantum fields by making the transformation dependent on x. These fields also create the Nambu-Goldstone bosons. The idea is that if φ = v, then h i δ φ = [ d3xa(x)j0(x),φ(x)] [ d3xa(x)j0, v] (16) a(x) Z a ∼ Z a 0 and this gives rise to an insertion of ja (x) at some x, thus creating a Nambu- Goldstone boson. On the other hand, if we evaluate the commutator explic- itly, we get some combination of scalar fields φ. 1.3 The Higgs mechanism We can try a similar toy model where we have a gauged U(1) field theory and try to follow the Nambu-Goldstone theorem in this case as well. The idea is to start with λ L = Dµφ∗D φ m2φ∗φ (φ∗φ)2 (17) − µ − − 4 The calculation of the vacuum, etc, proceed in the same way. Let us consider the radial representation of the system, where we focus on the phase of φ, this is φ(x) (v + h/√2) exp(iθ(x)) (18) ∼ 0 We find that the kinetic term for θ is given by (v + h/√2)2(∂ θ eA )2 (19) µ − µ while the kinetic term for h has no coupling to the photon.