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The Higgs Mechanism I: Abelian Models

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The Higgs Mechanism I: Abelian Models PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 Dr. Anosh Joseph, IISER Mohali LECTURE 45 Wednesday, April 8, 2020 (Note: This is an online lecture due to COVID-19 interruption.) Topic: The Higgs Mechanism I: Abelian Models. We can ask the question: What happens if we include both local gauge invariance and spon- taneous symmetry breaking in the same theory? This will result in a rather remarkable type of spontaneous symmetry breaking in which the gauge bosons acquire mass; and this procedure is known as the Higgs mechanism. The interactions of these massive bosons are still constrained by the underlying gauge symmetry. Let us note that, from a historical point of view, the Higgs mechanism is not properly named. The same idea was discovered and understood by many people in different contexts, including Anderson (who proposed it first in a non-relativistic context in 1962), as well as Brout, Englert, Ginzburg, Guralnik, Hagan, Kibble, Landau and, of course, Higgs. In elementary particle physics, the principal application of spontaneously broken local symmetry is in the currently accepted model of weak interactions. The Higgs Mechanism Let us look at some simple examples of gauge theories with spontaneous symmetry breaking. Abelian Higgs Model Consider a complex scalar field coupled both to itself and to an electromagnetic field 1 L = − (F )2 + jD φj2 − V (φ); (1) 4 µν µ with Dµ = @µ + ieAµ. This Lagrangian is invariant under the local U(1) transformation 1 φ(x) ! eiα(x)φ(x);A (x) ! A (x) − @ α(x): (2) µ µ e µ PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 If we choose the potential in L to be of the form λ V (φ) = −µ2φ∗φ + (φ∗φ)2; (3) 2 with µ2 > 0, the field φ will acquire a vacuum expectation value, and the U(1) global symmetry will be spontaneously broken. The minimum of this potential occurs at r µ2 hφi = φ = ; (4) 0 λ or at any other value related by the U(1) symmetry given in Eq. (2). Let us expand the Lagrangian Eq. (1) about the vacuum state Eq. (4). Decompose the complex field φ(x) as 1 φ(x) = φ0 + p (φ1(x) + iφ2(x)): (5) 2 The potential Eq. (3) is rewritten 1 1 V (φ) = − µ4 + · 2µ2φ2 + O(φ3); (6) 2λ 2 1 i p so that the field φ1 acquires the mass m = 2µ and φ2 is a massless Goldstone boson. Let us now consider how the kinetic energy term of φ is transformed. Inserting the expansion Eq. (5), we rewrite 1 1 p jD φj2 = (@ φ )2 + (@ φ )2 + 2eφ · A @µφ + e2φ2A Aµ + ··· ; (7) µ 2 µ 1 2 µ 2 0 µ 2 0 µ where we have omitted terms cubic and quartic in the fields Aµ, φ1, and φ2. The last term, written explicitly in Eq. (7), is a photon mass term 1 ∆L = m2 A Aµ; (8) 2 A µ where the mass 2 2 2 mA = 2e φ0 (9) arises from the nonvanishing vacuum expectation value of φ. Notice that the sign of this mass term is correct; the physical spacelike components of Aµ appear in Eq. (8) as 1 ∆L = − m2 (Ai)2; (10) 2 A with the correct sign for the potential energy term. However, a model with a spontaneously broken continuous symmetry must have massless Gold- stone bosons. These scalar particles can appear as intermediate states in the vacuum polarization. Let us look at the third term in Eq. (7). It couples the gauge boson directly to the Goldstone boson 2 / 4 PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 p µ µ φ2. This gives the vertex of the form: i 2eφ0(−ik ) = mAk . (See Fig. 1.) Figure 1: The vertex diagramp that directly couples the gauge boson to the Goldstone boson φ2. It µ µ gives the contribution i 2eφ0(−ik ) = mAk . If we also treat the mass term, Eq. (8), as a vertex in perturbation theory, then the leading-order contributions to the vacuum polarization amplitude give the expression (see Fig. 2) i kµkν im2 gµν + (m kµ) (−m kν) = im2 gµν − : (11) A A k2 A A k2 Figure 2: Treating the mass term as a vertex in perturbation theory, we get the above leading-order contributions to the vacuum polarization amplitude for the gauge boson. We see that the Goldstone boson supplies exactly the right pole to make the vacuum polarization amplitude properly transverse. Although the Goldstone boson plays an important formal role in this theory, it does not appear as an independent physical particle. The easiest way to see this is to make a particular choice for the gauge, called the unitary gauge. Under the infinitesimal form of the gauge transformation we have φ0(x) = eiα(x)φ(x) iα(x) 1 = e φ0 + p (φ1(x) + iφ2(x)) 2 1 = (1 + iα(x)) φ0 + p (φ1(x) + iφ2(x)) 2 1 1 p = φ0 + p [φ1(x) − α(x)φ2(x)] + ip [φ2(x) + α(x)φ1(x) + 2α(x)φ0]: (12) 2 2 This gives, 0 φ1(x) = φ1(x) − α(x)φ2(x); (13) p 0 φ2(x) = φ2(x) + α(x)φ1(x) + 2α(x)φ0; (14) We may choose α(x) to make φ2(x) = 0, (unitary gauge) so that φ(x) becomes real-valued at every point x. (Note that the unitarity gauge makes the manifest number of scalar degrees of freedom minimal.) With this choice, the field φ2 is removed from the theory. The Lagrangian Eq. 3 / 4 PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 (1) becomes 1 L = − (F )2 + (@ φ)2 + e2φ2A Aµ − V (φ): (15) 4 µν µ µ If the potential V (φ) favors a nonzero vacuum expectation value of φ, the gauge field acquires a mass; it also retains a coupling to the remaining, physical field φ1. As mentioned earlier, this mechanism, by which spontaneous symmetry breaking generates a mass for a gauge boson, is known as the Higgs mechanism. We can see that spontaneous breaking of U(1) symmetry gives the following particle spectrum (i:) Goldstone mode (SSB of global U(1) symmetry): 2 massive scalar fields ! 1 massive scalar field + 1 massless scalar field. (ii:) Higgs mode (SSB of gauge U(1) symmetry): 2 massive scalar fields + 1 photon ! 1 massive scalar field + 1 massive photon. The number of degrees of freedom is preserved under these transformations. The involvement of a Goldstone boson is necessary in order for the gauge boson to acquire a mass. Though the Goldstone boson can be formally eliminated from the theory, its presence is still there. A massless vector boson has only two physical polarization states: a longitudinal polarization state cannot be produced. It appears in the formalism only to cancel the other unphysical contributions. However, a massive vector boson must have three physical polarization states. It is thus tempting to say that the gauge boson acquired its extra degree of freedom by eating the Goldstone boson. References [1] M. E. Peskin and D. Schroeder, Introduction to Quantum Field Theory, Westview Press (1995). [2] L. H. Ryder, Quantum Field Theory, Cambridge University Press (1996). 4 / 4.
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