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External Fields and Measuring Radiative Corrections

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External Fields and Measuring Radiative Corrections Quantum Electrodynamics (QED), Winter 2015/16 Radiative corrections External fields and measuring radiative corrections We now briefly describe how the radiative corrections Πc(k), Σc(p) and Λc(p1; p2), the finite parts of the loop diagrams which enter in renormalization, lead to observable effects. We will consider the two classic cases: the correction to the electron magnetic moment and the Lamb shift. Both are measured using a background electric or magnetic field. Calculating amplitudes in this situation requires a modified perturbative expansion of QED from the one we have considered thus far. Feynman rules with external fields An important approximation in QED is where one has a large background electromagnetic field. This external field can effectively be treated as a classical background in which we quantise the fermions (in analogy to including an electromagnetic field in the Schrodinger¨ equation). More formally we can always expand the field operators around a non-zero value (e) Aµ = aµ + Aµ (e) where Aµ is a c-number giving the external field and aµ is the field operator describing the fluctua- tions. The S-matrix is then given by R 4 ¯ (e) S = T eie d x : (a=+A= ) : For a static external field (independent of time) we have the Fourier transform representation Z (e) 3 ik·x (e) Aµ (x) = d= k e Aµ (k) We can now consider scattering processes such as an electron interacting with the background field. We find a non-zero amplitude even at first order in the perturbation expansion. (Without a background field such terms are excluded by energy-momentum conservation.) We have Z (1) 0 0 4 (e) hfj S jii = ie p ; s d x : ¯A= : jp; si Z 4 (e) 0 0 ¯− µ + = ie d x Aµ (x) p ; s (x)γ (x) jp; si Z Z 4 3 (e) 0 µ ik·x+ip0·x−ip·x = ie d x d= k Aµ (k)u ¯s0 (p )γ us(p) e ≡ 2πδ(E0 − E)M where 0 (e) 0 M = ieu¯s0 (p )A= (p − p)us(p) We see that energy is conserved in the scattering (since the external field is static) but in general the initial and final 3-momenta of the electron can be different. The recoil is absorbed by the background field. 111 Quantum Electrodynamics (QED), Winter 2015/16 Radiative corrections We can represent the amplitude by the Feynman diagram ¡A(e) e− e− where the “×” denotes the external photon field. One can calculate the perturbative expansion of the S-matrix using Feynman diagrams just as before, except for two modifications in the Feyman rules (i) The matrix element has the form Sfi = hfj S jii = hfjii + 2πδ(Ef − Ei)M 4 (4) so that four-momentum conservation (2π) δ (pf − pi) is replaced by simply energy conser- vation 2πδ(Ef − Ei). (ii) Instead of r(k) for external photons we attach a factor (e) external field: Aµ (k) ¢ µ for external fields Let us concentrate on the scattering of electrons from an external field. In calculating the differ- ential cross-section we consider a single incoming electron with four-momentum (E; p) and several 0 0 outgoing states with four-momenta (Ei; pi). We have the scattering probability per unit time jS j2 fi = 2πδ(E − E)jMj2 T f where T = 2πδ(0). Allowing as usual for the normalization of in- and outgoing states we have the differential scattering probability per unit spacetime volume Z 3 0 3 0 2 d= p1 d= pn jMj dw = 0 ··· 0 2πδ(Ef − E) ∆ 2E1 2En V · 2EV The electromagnetic field will typically be the potential around a single heavy nucleus. Thus the four-vector flux for the external field is one particle per unit volume (1=V; 0), giving the differential cross-section Z 3 0 3 0 2 d= p1 d= pn jMj dσ = 0 ··· 0 2πδ(Ef − E) ∆ 2E1 2En F where F = Ev with v = jpj=E is the incoming velocity. If we simplify to the case of only a single p outgoing electron, we have d3p0 = jp0j2djp0jdΩ. Changing variables to E0 = p2 + m2 we get dσ 1 = jMj2 dΩ 16π2 222 Quantum Electrodynamics (QED), Winter 2015/16 Radiative corrections As an example consider scattering from a heavy nucleus at leading order. We have the Coulomb potential Ze A(e)(x) = ; 0 ; µ 4πjxj where Z is the atomic number of the nucleus and we have Fourier transform Ze A(e)(q) = ; 0 µ jqj2 Averaging over incoming spins and summing over outgoing spins we have at leading order 1 X X = jMj2 2 ss0 1 X = e2A(e)(p0 − p)A(e)(p0 − p) u¯(p0)γµu(p)¯u(p)γνu(p0) 2 µ ν ss0 Evaluating the sum X u¯(p0)γµu(p)¯u(p)γνu(p0) = Tr γµ(p= − m)γν(p=0 − m) ss0 = 4 p0µpν + pµp0ν − gµν(p0 · p − m2) (e) Since only the time-component A0 (q) is non-zero we get 2Z2e4 X = EE0 + p0 · p + m2 jp0 − pj4 and hence dσ 2α2Z2 = EE0 + p0 · p + m2 dΩ jp0 − pj4 where α = e2=4π. Turning to the kinematics we define the scattering angle θ by p0 · p = jp0jjpj cos θ Energy conservation implies E0 = E. Since the masses of the incoming and outgoing states are the same jp0j = jpj ≡ Ev. We then have 0 2 2 2 2 1 jp − pj = 4E v sin 2 θ 0 0 2 2 2 2 1 EE + p · p + m = 2E 1 − v sin 2 θ and the cross-section gives the Mott scattering formula dσ α2Z2 = 1 − v2 sin2 1 θ 2 4 4 1 2 dΩ 4E v sin 2 θ At low energies v 1 this approximates to the Rutherford scattering formula dσ α2Z2 = 2 4 4 1 dΩ 4m v sin 2 θ 333 Quantum Electrodynamics (QED), Winter 2015/16 Radiative corrections Anomalous electron magnetic moment and the Lamb shift We start by returning to the form of the lowest-order matrix element for electron scattering from an external field. Using the fact that p=u(p) = mu(p) we have, writing q = p0 − p, (e) 0 µ M = ieAµ (q)u ¯(p )γ u(p) ie = A(e)(q)u ¯(p0) p=0γµ + γµp= u(p) 2m µ ie = A(e)(q)u ¯(p0) p0µ + pµ + iq σµν u(p) 2m µ ν µν 1 µ ν µ ν µν µν (e) where σ = 2 i[γ ; γ ] so that γ γ = g −iσ . We have, for the Fourier transform, iqνAµ (q) = (e) (e) (e) µν @νAµ (q) where @νAµ is the position space derivative of Aµ . Since σ is antisymmetric and Fµν = @µAν − @νAµ we can then write ie h i M = u¯(p0) p0 · A(e)(q) + p · A(e)(q) − 1 F (e)(q)σµν u(p) 2m 2 µν Consider the case where the in- and outgoing electrons are at rest p = p0 = 0 but with a static background field. Using the standard basis where ! ! 1 0 0 σi γ0 = γi = for i = 1; 2; 3 0 −1 −σi 0 where σi are the Pauli matrices, in the p = 0 limit, the spinor u(0) has the form ! χ~ u(0) = 0 Thus the amplitude becomes ie M = ieφ(e)χ~yχ~ − B(e) · χ~yσχ~ 2m (e) (e) where φ is the background electric potential and Bi the background magnetic field. Note that M R 4 is none other than i times the expectation value of the interaction energy d xLI . For an electron at rest this will have two contributions: the electrostatic potential energy, and the interaction energy of the spin of the electron with the background magnetic field. Comparing with the expression above, we see these two terms are reproduced with χ~yχ~ the electron density and χ~yσχ~ the spin-density and we read off the electron magnetic moment e γ = 2m 444 Quantum Electrodynamics (QED), Winter 2015/16 Radiative corrections Now consider the one-loop corrections to the electron scattering from a background field. We have four£ Feynman diagrams¤¥¦ (a) (b) (c) (d) The last two diagrams simply give external leg renormalizations for the incoming and outgoing elec- trons. These lead only to wavefunction renormalizations and no finite radiative corrections. Diagram (b) gives a finite correction to the effective form of the external field and leads to the Lamb shift. For the magnetic moment we will be interested in diagram (a). 0 Diagram (a) has a finite contribution from Λc(p; p ), the radiative correction in the vertex modi- 0 fication. By explicitly doing the integrals defining Λc(p; p ) one can show, after some considerable work, that the amplitude is given by (e) 0 2 0 M(a) = ieAµ (q)u ¯(p ) e Λc(p; p ) u(p) e h α i = i A(e)(q)u ¯(p0) i q σµν u(p) 2m µ 2π µ Comparing with the expression for the low-order amplitude we see that the magnetic moment of the electron has been corrected. We have e α γ = 1 + 2m 2π This is the famous anomalous magnetic moment predicted by QED. It has been measured to extremely high accuracy. One also finds small corrections due to other particles in the theory and so this also gives a precision test of the particle content of the standard model. Finally we turn to the Lamb shift. Diagram (b) has a finite contribution coming from the photon self-energy, given by (e) 0 µ 2 2 M(b) = ieAµ (q)u ¯(p )γ u(p) −e Πc(q ) (e) The effect is to replace the background field Aµ by the combination corr 2 2 (e) Aµ (q) = 1 − e Πc(q ) Aµ (q) As before, suppose we take the background field of a heavy nucleus.
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