Nambu-Goldstone Bosons and the Higgs Mech- Anism 1 Reading Material from the Books

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 component 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 spontaneously 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 derivatives 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) representation,| 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.

5 We should notice that the linearized current for the symmetry now takes the form jµ ∂ θ(x) eA (x) (20) ∼ µ − µ While under gauge transformation φ(x) exp(ieΛ(x)), Aµ Aµ ∂µ(x). We ﬁnd that the current is gauge invariant.→ Also, θ(x) θ(x→)+ eΛ(− x), so → we can in principle choose a gauge where θ(x) is constant. The current acting on the ground state can take us to a one particle state, but the one particle state can also have a photon in it, rather than just a Nambu-Goldstone boson. Thus, we ﬁnd that if we take into account current conservation for a spin one particle, we get the following relations

p, ǫ J µ(x) 0 fǫµ exp(ipx) (21) h ρ| | i ∼ The result has to be linear in the polarization of the spin one particle (this is what superposition in quantum mechanics tells us). µ Current conservation then implies that k ǫµ = 0. Notice that this is an instance where the gauge invariance lets us use Aµ instead of Fµν (as we would have for an ordiinary photon, where inserting Aµ would not be gauge invariant). Hence, we get that the vector field has to be massive. If the field Aµ was massless, the polarization would not be gauge invariant, because ǫµtoǫµ +αkµ gives the same physical state. Instead, we find that this is not allowed, because k2 = 0. 6 Also notice that in the Procca lagrangian we have that the current term is proportional to Aµ itself. Thus we find that the vector field is massive. Notice that under usual circumstances we would have had a massless Nambu-Goldstone boson, but that now it can be completely eliminated by gauge transformations. Thus the would-be Nambu-Goldstone boson is ab- sent: there is no such state. Instead we say that the Nambu-Goldstone boson has been eaten up by the massive vector particle. This is called the Higgs mechanism. If we count physical polarizations of states, the new polarization of the massive vector particle comes at the expense of a scalar particle. The asymp- totic number of degrees of freedom in the UV is the same. One can try to understand better how does the Nambu-Goldstone mode disappears from amplitudes. One can check that there is a cancellation between contributions from the Nambu-Goldstone mode and the time-like polarization of the vector field. This requires us to study the problem of gauge

6 invariance more carefully by going to a path integral formalism and making various choices of gauge conditions. These choices will make the cancellations more apparent. This Higgs mechanism is responsible for the phenomenon of supercon- ductivity. The U(1) symmetry is spontaneously broken by a condensate of Cooper-pairs and these give rise to a massive photon inside the supercon- ducting material. They also require the notion of BRST quantization, which we will describe in detail for the U(1) gauge ﬁeld.

2 The Rξ gauges Part of the process of renormalizing Gauge field theories requires us to use a gauge fixing procedure and introduce Faddev-Popoov ghosts in the process. We should also be able to do that in the spontaneously broken gauge theory setup. The standard BRST techniques give us a full lagrangian that computes the Faddeev-Popov determinant and gives us an effective action. This is usually done in the path integral formalism for convenience. However, given the lagrangian and using the procedure of canonical quantization gives the same results. Since the ghosts have the wrong statistics, but we preserve locality and Lorentz invariance, we have to give away man- ifest unitarity of the description. The set of physical states is then selected by the BRST procedure. Let us consider first the case of massless electrodynamics for a non- interacting photon. The BRST procedure usually has us pick a gauge. Lorentz invariance µ gives us an obvious choice f = ∂µA = 0. The BRST lagrangian will be given by

ξB2 L = L +¯cδ f ifB (22) total standard BRST − − 2 Where B is an auxiliary field and f is our gauge fixing function. The variation δBRST f is a gauge transformation of f with parameter c (the ghost field), whilec ¯ is the antighost field. If we integrate out B (it is a

7 Gaussian path integral), then we get that

f 2 L = L +¯cδ f (23) total standard BRST − 2ξ which is another standard form of the lagrangian. In this case we have that

δBRST Aµ = ∂µc (24) so that the ghost piece is given by

µ c∂¯ ∂µc (25) and the ghost fields decouple from matter. Notice that the presence of ξ modifies the Lagrangian for the gauge con- nection, making A0 dynamical (in the sense that the action now depends on time derivatives of A0. The process of canonical quantization now has a canonical conjugate to A0. Now we can ask what is the new propagator for A. This can be read from looking the quadratic term in the action and inverting the partial differential operator in momentum space. We find this way that

2 µν µ ν −1 µ ν ν (k η + k k + ξ k k )∆µρ(k)= δρ (26)

The end result is 1 2 1 ∆µρ = (k2η + k k + ξk k )= ηµν (ξ 1)kµkν/k2 (27) k2 µρ µ ρ µ ρ k2 − − Notice that now the propagator scales as 1/k2 when we take k large. The gauge ξ = 1 is called Feynman gauge. The gauge ξ = 0 is Landau Gauge, which is somewhat singular. The value of ξ should not aﬀect physical observables (those that are BRST invariant). The same procedure works for non-abelian gauge theories, except that the ghost ﬁelds do not decouple. This is because

δ (∂ A ) ∂ (g[c, A]+ ∂ c) (28) BRST µ µ ∼ µ µ 8 However, the propagators computed with diﬀerent values of ξ coincide with the propagators above. For spontaneously broken non-abelian gauge theories, we get quadratic terms in the action that involve Goldstone bosons and the gauge ﬁeld. These are of the form 1 f 2(∂ θ + gA )2 (29) −2 π µ µ We see here that if we try to invert the general propagator in one f the gauges above, the Propagator is in general some complicated matrix because θ and A mix. It is convenient to choose a gauge function so that the term f 2/ξ cancels this mixing when integrating by parts. That gauge is the ’t Hooft-Feynman gauge. µ The idea is to modify ∂µA =0 to

µ ∂ Aµ + Ωθ =0 (30) so that expanding we get that f 2/2ξ ξ−1(∂µA )2/2 ξ−1Ω∂µA θ ξ−1Ω2/2 (31) − ∼ − µ − µ − −1 µ If we integrate the second term by parts, we get ξ ΩAµ∂ θ and this can 2 be made to cancel with fπ gAµ∂µθ if we choose 2 Ω= fπ gξ (32) In this way we can calculate the propagator more easily, We find, by doing a BRST transformation, that δBRST θ gc, so that the 2 2 ∼ ghost field acquires a mass of order fπ g ξ. This is the same mass that the would-be Goldstone boson acquires in this gauge, and it is the same mass that the extra (scalar ) component of the gauge field acquires. Because these masses are equal, the one loop contribution to the self- energy of the ground state cancels between the scalar component of the gauge field, the Goldstone boson and the ghost fields. If we look into the general setup, we have that there is a mixing between the gauge field and the scalars in the kinetic term of the spontaneously broken theory. This mixing is given by ∂µφ( igAa T a) φ (33) − µ h i 9 Again, we want to consider canceling this kind of mixing in general, so we want to choose a gauge of the form

∂ Aµ + ΩφT a φ (34) µ a h i where we are considering real scalars for simplicity. In this gauge, if T a φ = h i 0, then we recover the standard gauge condition for an unbroken gauge symmetry, while if this term is not zero, we get a mass term for the would be Goldstone boson that goes like

ξg2 T a φ 2 (35) | h i| and similarly for the scalar component of the gauge field and the ghosts. Again, we see the cancellation in one loop vacuum energies. More to the point, in physical scattering amplitudes whenever we can produce a Goldsone boson, then we can also produce a scalar component of the gauge field. In tree diagrams these two effects should cancel each other. IN the BRST formalism we find that the scalar mode of the gauge field has negative norm and would in principle violate unitarity. If we combine it with the positive norm of the Goldstone boson, we get a null state being produced. Null states do not change the value of probabilities. HOwever, this cancellation only happens on summing over Feynman diagrams.