(g 2) in QED and Beyond − 1 A brief introduction Anomalous magnetic moment of the electron is one of the crowning achievements of Quantum Electrodynamics (QED) in particular, and of Quantum Field Theory in general. Experimen- tal and theoretical calculations agree to about one part in a trillion [1]. The anomalous magnetic moment of the muon is sensitive to states beyond the Standard Model. These states could be heavier than what we can produce at colliders; they might couple dominantly to leptons, making harder to produce at hadron colliders. The muon (g 2) measurement then provides an interesting way to probe these new states, by measuring − their tiny effect with a highly sensitive experiment. To compare theory and experiment, highly technical calculations are involved. QED calculations have to be performed at high order, where the number of Feynman diagrams increase to hundreds. Contributions from hadronic physics becomes important at such high precision. These cannot be calculated using Feynman diagrams and require input from either experimental measurement, or novel theoretical techniques (a lot of progress is being made in this direction using lattice QCD, which will be covered in detail in forthcoming lectures in this series). In this note we will focus on the calculation of first radiative corrections to the lepton g 2, and briefly comment on estimating one-loop corrections from new physics. − 2 What to calculate? Let’s start with a classical system. The magnetic moment of a classical current carrying loop of wire is given by, ~µ = I~s (1) where I is the current through the loop, and ~s is the vector area. In presence of magnetic field, the loop experiences a torque, and there is a potential energy associated with the orientation of the loop relative to the magnetic field, U = ~µ.B~ (2) − 1 That is, the configuration where the loop is aligned with the magnetic field has lower potential energy. For simplicity, let us imagine a circular loop with a charge q traveling along the loop with angular velocity ω, as shown in figure 1. Then, q ~µ = πa2zˆ (3) 2π/ω since the current is given by a charge q traversing the loop per revolution. We can rewrite this as, q q ~µ = ωma2zˆ = L~ (4) 2m 2m where L is the angular momentum of the charged particle about the origin of the loop. Figure 1: A charge running around a loop of wire. We see that classically, the magnetic dipole moment is proportional to the angular mo- mentum of charged particles. This continues to be true in quantum mechanics (including the spin angular momentum for particles), with a small difference. For example, for an electron at rest, the magnetic dipole moment is given by, e ~µe = g S~ (5) 2me 2 Figure 2: Tree level diagram for a lepton scattering off a background classical field. where S~ is the spin operator. Analogously, in presence of a classical magnetic field, the spin of the electron prefers to be aligned with the magnetic field. The interaction Hamiltonian is, H = ~µ B~ (6) int − · In order to calculate the QED prediction for the muon (electron) magnetic moment, we need to compare the QED interaction of the electron with a classical background field. Let us focus on the tree level contribution first (figure 2). The interaction Lagrangian density at tree level is given by, = eℓ¯(x)γµℓ(x)A (x) (7) Lint − µ where we have denoted the classical field by A, and the lepton (electron/muon) by ℓ. The way to connect this relativistic Lagrangian density with the non-relativistic quantum mechanical potential is to compare the scattering amplitudes. The tree-level scattering amplitude in momentum space is given by, i =u ¯(p′)( ieγµ)u(k) A (q) (8) M − µ where u, u¯ are plane-wave solutions to the Dirac equation,e and Aµ(q) is the Fourier-transform of the classical field, and q = p′ p. Since we are interested in time-independent field − configurations, e 0 Aµ(q)= Aµ(~q)δ(q ) (9) e e 3 The vector potential corresponding to a static magnetic field is given by Aµ = (0,~A), with the magnetic field given by B~ (x)= ~A (10) ∇ × B (~q)= iǫijkqiA (~q) (11) ⇒ k − j We now set the scattering amplitude frome the non-relativistic theory equal to the one above, obtained from the relativistic theory. The electron couples in two different ways to the magnetic field (or the 3-vector potential): one via its spin, and the other via its motion (~p, which can also be thought of as orbital angular momentum). To isolate the spin coupling, we set ~p = 0. Since the magnetic field is linear in ~q, we retain a non-zero ~q, but work to linear order. The non-relativistic answer is easily obtained using the Born approximation. Comparing the two, i µk B (~q)=+ie u¯(p′)γku(k) A (q) (12) − h i k k with p′ =(m, ~q/2),k =(m, ~q/2). (Note p′2 = k2 = m2 is satisfied to linear order in q.) − e e Therefore, in order to pick out the magnetic moment, we need to pick out the term proportional to ~q from the spinor, i µk iǫijkqi A (~q) =+ie u¯(p′)γju(k) A (q) (13) − h i − j j Let us expand the spinor on the r.h.se in the non-relativistice limit √k σξ u(k)= · (14) √k σξ¯ · where as usual, the square root of a matrix is understood to pick out the square root of eigenvalue when acting on an eigenvector. In the non-relativistic limit, 1 1 √k σ = m ~q ~σ √m(1 + ~q ~σ/2m) (15) · − −2 · ≈ 2 · s 1 1 p′ σ = m ~q ~σ √m(1 ~q ~σ/2m) (16) · − 2 · ≈ − 2 · s p 4 Therefore, † √p′ σξ′ 0 1 0 σµ √k σξ u¯(p′)γku(k)= · · (17) √p′ σξ¯ ′ 1 0 σ¯µ 0 √k σξ¯ · · 1 1 1 1 = mξ′† (1 ~q ~σ)¯σk(1 + ~q ~σ)+(1+ ~q ~σ)σk(1 ~q ~σ) ξ − 4m · 4m · 4m · − 4m · (18) 1 1 mξ′† σ¯k + σk + ~q ~σ(σk σ¯k) (σk σ¯k) ~q ~σ ξ (19) ≈ 4m · − − − 4m · 1 = ξ′†q σiσk σkσi ξ (20) 2 i − 1 = ξ′† 2iǫikjσj q ξ (21) 2 i σj =2iǫikjq ξ′† ξ (22) i 2 i = ǫijkq Sj (23) −m ih i remembering that the relativistic states are normalized as, Sj =2mξ†Sjξ (24) h i due to the relative factor of 2m in the normalization of relativistic states. Comparing with the non-relativistic result, 1 i µk iǫijkqi A (~q) = ie Sk iǫijkq A (q) (25) − h i − j − mh i − i j e e e e µk = Sk = g Sk (26) h i mh i 2mh i which is to say, that the leading order prediction of QED is g = 2. 2.1 Higher order corrections The radiative corrections to the calculation above can be compactly written as, i = ie u¯(p′)Γµu(k)A (27) M − µ 5 Using equations of motion and gauge invariance, the most general form of Γµ can be deduced to be, iσµν q Γµ(p, q)= F (q2)γµ + F (q2) ν (28) 1 2 2m Remember we need to work to linear order in qi. i i i u¯(p′)σkνq u(k)= ξ′† − ǫijkqjσk ξ (29) 2m ν 2m 2m Therefore, we only need the form factors at 0 momentum transfer, ′ µ i ′† i ijk j k u¯(p )Γ (p, q)u(k)= ξ − ǫ q σ ξ[F1(0) + F2(0)] (30) 2m 2m From above, it is clear that, g = 2(F1(0) + F2(0)) (31) Due to gauge invariance, to all orders F1(0) = 1. Therefore, the contribution from F1 is always g = 2. The correction, g 2 then arises solely from F2. The anomalous magnetic moment is written as, − g 2 a = − = F (0) (32) 2 2 2.2 One loop diagrams The one loop diagrams relevant are shown in figure 3. To calculate the one loop contribution to the matrix element, we use the following master formula, 2 ′ S =( Z2) u¯(p )[Amputated one loop diagram]u(k) (33) h i p where Z2 is the wavefunction renormalization, calculated at one loop, for the electron. Note that there is no wavefunction renormalization included for the external classical field. The wavefunction renormalization cancels the renormalization of the F1(0) form factor of the vertex function to every order in perturbation theory. 6 Figure 3: One loop corrections to the magnetic moment. Wavefunction renormalization diagram for the lepton (left) and amputated vertex correction. 3 Feynman diagram calculation The tree-level contribution to F2 is absent. Therefore, the wavefunction renormalization does not contribute. The entire one-loop correction to g 2 then involves calculation of the − amputated vertex correction, and extracting the form factor F2(0). In this section we will focus on the Feynman diagram calculation: the calculation of of the lowest order radiative correction to the muon magnetic moment. This is a famous calculation by Schwinger [4, 5], first performed for the electron (although he did not use Feynman diagrams for his calculation). As we shall see, at this order, the magnetic moments of the electron, muon and the tau lepton are identical. The Feynman diagram is shown in figure 4. This particular calculation involves three aspects which we will try to isolate: 1. Symmetry: The result of the diagram is constrained by gauge invariance, and the charge conjugation and the parity symmetries of the QED Lagrangian. 2.