Introduction to the physics of highly charged ions Lecture 11: Feynman rules and vacuum polarization Zoltán Harman [email protected] Universität Heidelberg, 27.01.2014 Recapitulation of propagator theory Defining equation for the Feynman propagator for electrons and positrons (in relativistic units: ~ = c = 1, and with Feynman slash notation: A/ = γµ Aµ ): / 0 − eA/0 − m)GF (x0 , x; A) = δ (4) (x0 − x) . (i∇ The free (A = 0) relativistic propagator satisfies the equation 0 / − m)GF (x0 − x) = δ (4) (x0 − x) . (i∇ Final result with a positive imaginary part in the denominator: Z p+m d4 p −ip(x0 −x) / 0 GF (x −x) = e . 4 2 (2π) p − m2 + i The full propagator and S matrix Iterative solution for the full propagator: Z 0 0 / 1 )GF (x1 − x) GF (x , x; A) = GF (x − x) + e d4 x1 GF (x0 − x1 )A(x Z 2 / 1 )GF (x1 − x2 )A(x / 2 )GF (x2 − x) + . . . +e d4 x1 d4 x2 GF (x0 − x1 )A(x Integral equation for the full wave function is: Z / ψ(x) = ψ0 (x) + d4 yGF (x − y)A(y)ψ(y) . The S-matrix is given by: Sfi = lim hψ0f |ψi i = t→±∞ Z lim hψ0f (x)|ψ0i (x)i + hψ0f (x)| t→±∞ / d4 yGF (x − y)eA(y)|ψ i (y)i , yielding a perturbative expansion in powers of the elementary charge e Feynman rules for QED processes Goal: formulation of a perturbative approach: writing down rules for perturbation theory: Feynman rules – without rigorous derivation Exemplified on the following processes: Coulomb scattering of an electron Electron-proton scattering Electron-positron scattering Coulomb scattering of an electron For electrons (with positive energy): Sfi = δfi − i ∞ X en Z d 4 y1 . . . Z d 4 yn n=1 (+E) / n )GF (yn − yn−1 )A(y / n−1 ) . . . GF (y2 − y1 )A(y / 1 )ψu(+E) (y1 ) . ψ¯f (yn )A(y What are here the free wave functions? – Solutions of the free Dirac equation for an electron or positron (r = ±1) with a well-defined four-momentum p: ψ r = ω r (~p)e−ir px , with r=1,2,3,4 ; ~p ω r (0) , ω r (~p) = ˆ S − E with the Lorentz transformation matrix and the unit 4-spinors  r ~p E+m ˆ S − = E 2m 1  0   pz  E+m p+ E+m 0 1 p− E+m −pz E+m pz E+m p+ E+m 1 0 p−  E+m −pz  E+m      1 0 0    , ω 2 (0) = 1 , etc.  , ω 1 (0) =  0 0 0  0 0 1 Solutions for incoming and outgoing e− in Coulomb scattering (normalization in volume V): r r m m ¯ ψi (x) = u(pi , si )e−ipi x , ψ¯f (x) = u(pf , sf )e+ipf x ; Ei V Ef V with the notations u(p, s) = ω 1 (~p) ; u(p, −s) = ω 2 (~p). (Note: for positrons, the wave functions would be v(p, s) = ω 3 (~p) ; v(p, −s) = ω 4 (~p).) Coulomb potential: A0 (x) = A0 (~x) = −Ze/|~x|; ~A(~x) = ~0; thus the S-matrix element is Z m 1 1 (1) ¯ u(¯ pf , sf )γ0 u(pi , si ) d4 xei(pf −pi )x Sfi = +iZe2 p V Ei Ef |~x| The time component of the integral: Z dx0 e+i(Ef −Ei )x0 = 2πδ(Ef − Ei ) , expresses energy conservation. The 3-dimensional space-like integral with ~q = ~pf − ~pi : Z (1) Sfi = +iZe2 dx3 1 −i~q~x 4π e = 2 ; ~q |~x| so we obtain 1 m 4π ¯ p u(¯ pf , sf )γ0 u(pi , si ) 2 2πδ(Ef − Ei ) . V Ei Ef |~q| What is with 3-momentum conservation? – The infinitely heavy nucleus (source A0 ) does not recoil, it can absorb an arbitrary amount of momentum from ~pi and change it into ~pf . Electron-proton scattering Previously: the nucleus was infinitely heavy, M = ∞; now we describe the nucleus as a finite-mass particle (p+ ) with a wave function. It is a spin- 21 particle just like the e− . p+ current J µ (x): source of a 4-potential Aµ (x), which can be obtained solving Maxwell’s equations; S-matrix as before: Z Sfi = −ie / d4 xψ¯f (x)A(x)ψ i (x) . Wave equation (D’Alembert equation) for the 4-potential of the proton current, written with the quabla () operator: Aµ (x) ≡ ∂ν ∂ ν Aµ (x) ≡ (This holds in the Lorentz gauge: ∂µ Aµ = ∂2 − ∇2 ∂t2 ∂ φ ∂t Aµ (x) = 4πJ µ (x) . − ∇~A = 0.) Feynman Green’s function for this equation: DF (x − y) DF (x − y) = 4πδ (4) (x − y) , (1) in analogy to the e− e+ case before. We may also define the photon Green’s function in momentum space (Fourier transform): Z DF (x − y) = d4 q −iq(x−y) e DF (q) . (2π)4 Writing on the RHS of Eq. (1) the Dirac-delta as δ (4) (x − y) = Z DF (x − y) = d4 q ∂ν ∂ ν e−iq(x−y) DF (q) = (2π)4 Z R d4 q −iq(x−y) e , (2π)4 and with d4 q (−q2 )e−iq(x−y) DF (q) , (2π)4 we obtain the momentum representation of the photon propagator: DF (q2 ) = − 4π . q2 + i This is the Feynman (causal) Green’s function for a virtual photon. Since the photon is chargeless, it is its own antiparticle, and we don’t have to care about backward-in-time propagation. The 4-potential induced by the 4-current of the proton can be written as Aµ (x) = Z d4 yDF (x − y)J µ (y) . Let us substitute this into the formula of the S-matrix: (2) Sfi Z = −i d4 x eψ¯f (x)γµ ψi (x) DF (x − y)J µ (y) . Expression in the [. . . ]: 4-current of the e− ; in analogy we construct the 4-current for the p+ : s Jfiµ (y) = ep ψ¯fp (y)γ µ ψip (y) , with ψip (y) = M u(Pi , Si )e−iPi y . Eip V Thus the S-matrix can be further calculated as s s e2 m2 M2 ¯ u(pf , sf )γµ u(pi , si ) = +i 2 V Ef Ei Efp Eip Z Z Z d4 q −iq(x−y) −i(pf −pi )x i(Pf −Pi )y −4π ¯ e e e u(Pf , Sf )γ µ u(Pi , Si ) . d4 x d4 y (2π)4 q2 + i (2) Sfi Here, the integrals can be calculated easily (delta functions); we get (2) Sfi = −ie2 (2π)4 δ (4) Pf − Pi + pf − pi V2 × ¯ u(pf , sf )γµ u(pi , si ) s m2 Ef Ei s M2 Efp Eip 4π ¯ u(Pf , Sf )γ µ u(Pi , Si ) . (pf − pi )2 + i The delta function δ (4) (Pf − Pi + pf − pi ) expresses the conservation of 4-momentum in the process Symmetric expression in the electronic and protonic current (as expected): e− feels the field induced by the p+ current ⇔ p+ feels the field induced by the e− current Feynman rules Motivated by the calculations to the previous processes, we can write down general rules, using which any process can be calculated to arbitrary order (in principle). S-matrix element for a process 1 + 2 → 10 + 20 + 30 + . . . N: Sfi = i(2π)4 δ (4) p1 + p2 − N X k=1 The transition amplitude Mfi = P (n) n Mfi ! p0k Mfi 2 Y j=1 s N Nj Y 2Ej V k=1 s Nk0 2Ek0 V . can be constructed according to the following rules: 1 In order n, draw all Feynman diagrams with n vertices, containing the right number of initial and final particles 2 For external lines, write down the following expressions as factors: 3 For internal lines: 4 For all vertices, the following factor has to be included: 5 6 The contracted with the index of the internal or external photon line: P indexµ µ isP µν µ γµ or µ γµ DF P P At all vertices, 4-momentum conservation holds: a factor δ (4) ( j pj − k pk ); and R d4 p integration has to be performed over the undetermined momentum variable p: (2π) 4 The amplitudes of all contributing diagrams have to be added coherently, with the phase factors -1 for incoming positrons; -1 for fermion exchange graphs; -1 for all closed e− /e+ loops. The S matrix at higher orders 1 So far: calculations at the lowest orders in e or α. Since α ≈ 137 , this is often a useful approximation. However, with a good theory, one should be able to corrections of higher order (at least in principle). Also, many experiments are more accurate than the leading-order theory only. At first, the small higher-order corrections turned out to be infinitely large. Renormalization is necessary to arrive to finite results. Electron-positron scattering at higher orders Corrections of order α2 : Example: e− –e+ scattering. At leading order (∝ α): The vacuum polarization (VP) correction in diagram (i): MfiVP = 4πi +e4 ¯ u(p01 , s01 )(−iγ µ )u(p1 , s1 ) − 2 q + i " Z # 4 d k i i × Tr (−iγ ) (−iγ ) ν µ /k − / (2π)4 /k − m + i q − m + i 4πi v(p2 , s2 )(−iγ ν )v(p02 , s02 ) × − 2 q + i Here, "Tr" stands for calculating the trace of the matrix (i.e. sum over diagonal elements). It comes from the cyclicity of the fermionic loop: Tr(GγGγ) = X Gαβ γβδ Gδ γα αβδ Counting the powers of k: R d4 k k2 ⇒ quadratic divergence, ∼ k2 : infinite expression! Physical interpretation: the possibility of creating a virtual electron-positron pair influences the properties of the photon propagator iD0Fµν (q) = iDFµν (q) + iDFµλ (q) iΠλσ (q) iDFσν (q) , 4π with the vacuum polarization tensor iΠλσ (q) = −e2 4π Z " # 1 d4 k 1 Tr γ γ σ λ /k − m + i /k − / (2π)4 q − m + i Mathematically ill-defined due to the ultraviolet divergence. How to force convergence? introduce an upper limit K on the integral – what is the physical meaning of K? What should be the value of K? Why should be the results depend on K? introduce a smooth damping factor in the integrand, e.g. still problems with the physical interpretation K2 k2 +K 2 ⇒ finite expressions, but One possible way of properly regularizing the expressions: Pauli-Villars renormalization. Basic idea: subtract an integrand of the same asymptotic k → ∞ behavior to make the integral convergent Z Πµν (q) = ¯ µν (q) Π = Z d4 k fµν (q, k, m) d4 k ⇒ fµ,ν (q, k, m) + N X ! Ci fµν (q, k, Mi ) i=1 ¯ is convergent. At the end, the limit The constants {Ci , Mi } are to be chosen such that Π Mi → ∞ has to be taken. Physical observables cannot depend on the {Ci , Mi }. So, the regularized polarization tensor has the form: ¯ µν (q) Π = = ( Tr γµ (/k + m)γν (/k − / q + m) d4 k (2π)4 (k2 − m2 + i)((k − q)2 − m2 + i) ) N X Tr γµ (/k + Mi )γν (/k − / q + Mi ) = ... + Ci 2 k − Mi2 + i (k − q)2 − Mi2 + i i=1 Z d4 k kµ (k − q)ν + kν (k − q)µ − gµν (k2 − qk − m2 ) + reg. . 16πie2 4 2 2 2 2 (2π) (k − m + i) ((k − q) − m + i) 4πie2 Z After some long calculation, one obtains for the regularized VP tensor: 2 2 ¯ µν (q) = (q2 gµν − qµ qν ) − e ln Λ + ΠR (q2 ) , Π 3π m2 with the average cut-off momentum Λ and the regularization term ΠR (q2 ): N X Mi2 m2 ≡ − ln ΠR (q2 ) = 2e2 π Ci ln i=1 Λ2 m2 Z 1 and q2 dββ(1 − β) ln 1 − β(1 − β) 2 m 0 e2 q2 1 1 q2 + − + ... . π m2 15 140 m2 ≈ For small q ("weak" scattering), ΠR (q2 ) → 0, and the 2nd-order amplitude can be written in terms of the 1st-order amplitude as (2) Mfi (1) = Z3 Mfi , with Z3 = 1 − e2 Λ2 ln . 3π m2 Thus, the effective, physical charge of the electron is renormalized (modified) to eR = with e being the bare charge. Only eR is observable, and it only weakly depends on Λ. √ Z3 e, Effect of VP on an atomic electron Modification of the Coulomb potential A0 (~x) = −Ze/|~x| in momentum space: A00 (~q) = A0 (~q) + ΠR (−~q2 )A0 (~q) , or in real space A00 (~x) Z = ≈ d3 q 0 A (~q) = (2π)3 0 A0 (~x) − Z d3 q i~q~x 1 + ΠR (−~q2 ) A0 (~q) e (2π)3 e2 4 (3) Ze ∇2 A0 (~x) = − − e(Zα) δ (~x) , 15πm2 |~x| 15m2 using ∇2 (1/|~x|) = −4πδ (3) (~x). Non-relativistic energy shift of a bound hydrodenic state ψnl in first-order perturbation theory: VP ∆Enl = hψnl |e∆A0 |ψnl i = −α(Zα) 4m 4 |ψnl (0)|2 = − α(Zα)4 δl0 . 15m2 15πn3 Value for 2s electrons: VP ∆E2s = −1.122 × 10−7 eV = −27.1 MHz ↔ contradicts the experiment!