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(1)Quantum Field Theory-II Problem Set n. 5 - Solutions UZH and ETH, FS-2020 Prof. G. Isidori Assistants: C. Cornella, D. Faroughy, F. Kirk, J. Pagès, A. Rolandi www.physik.uzh.ch/en/teaching/PHY552. 1. Renormalization of QED. The QED Lagrangian is:. 1 (Fµ⌫ )2 + ¯(iD / m) , 4 where Dµ = @µ + ieAµ . The Feynman rule for the electron-photon interaction is given by: L=. Change everywhere µ. 1.1. Ta. s s. =. Q. ie. µ. .. (1). (2). µ. Electron self-energy. In order to determine the electron anomalous dimension (e) =. 1 @ µ Z2 , 2Z2 @µ. r. (e), defined by. = Z2. 1/2. bare. ,. (3). where µ is the scale at which we impose the renormalization conditions. We shall compute the electron self-energy at one loop, and then impose the renormalization condition h0|. ¯. r (p) r (. p)|0i. = p 2 = µ2. i /p. ,. (4). p 2 = µ2. in the limit µ2 m2 . The p2 = µ2 condition gives rise to the µ-dependence. The relevant diagram, with amputated external legs, is =. i⌃2 (p) .. (5). We can already see that the diagram has a superficial linear divergence: 4 2 1 = 1 ( 2 for the gauge propagator and 1 for the fermion propagator). ✓ ◆ Z i(p/ + k/ + m) d4 k i kµ k⌫ µ ⌫ i⌃2 = ( ie ) ( ie ) 2 gµ⌫ (1 ⇠) 2 , (6) (2⇡)4 (p + k)2 m2 + i✏ k + i✏ k where we wrote the photon propagator in a general R⇠ gauge, the Feynman gauge is the special case ⇠ = 1. From here on we will make this gauge choice. Since the diagram is both ultraviolet (UV) and infrared (IR) divergent we need to regulate both behaviours. The IR divergence can be regulated by adding a fictious mass mA to the photon, the subtleties regarding the interesting. 1.

(2) physics of IR divergencies are not relevant for the present discussion. In order to regulate the UV divergence in a way consistent with the gauge symmetry we generalise the loop integral to d = 4 ✏ dimensions. The -matrices algebra is slightly di↵erent in this case (see Peskin&Schroeder, sect.7.5), in particular: {. µ. ,. µ ⌫. ⌫ µ. } = 2g µ⌫ , =. (2. ✏). Tr[1] = 4 , ⌫. ,. gµµ = d = 4. µ ⌫ ⇢. µ. = 4g. ⌫⇢. ✏. ✏ ! ⌫ ⇢. ,. µ. µ = (4 µ ⌫ ⇢. ✏)1 , µ. =. ⇢ ⌫. 2. x. 4g ⌫⇢ + ✏. ⌫ ⇢. . (7). Moreover, in d dimensions the gauge coupling e becomes dimensionful, [e] = 2 d/2 = ✏/2, so we redefine e ! eµ̃✏/2 , where µ̃ is some energy scale. Our amplitude becomes Z (2 ✏)(p/ + k/) + (4 ✏)m dd k 2 ✏ = i⌃2 = e µ̃ d (2⇡) [(p + k)2 m2 + i✏][k 2 m2A + i✏] Z 1 Z (2 ✏)(p/ + k/) + (4 ✏)m dd k 2 ✏ = e µ̃ dx = (2⇡)d [x((p + k)2 m2 ) + (1 x)(k 2 m2A )]2 0 (8) Z 1 Z dd ` 1 2 ✏ = e µ̃ dx (2 ✏)(1 x)p/ + (4 ✏)m = (2⇡)4 (`2 (p2 ))2 0 Z 1 Z d d È 1 2 ✏ = ie µ̃ dx (2 ✏)(1 x)p/ + (4 ✏)m , 2 4 (2⇡) (È + (p2 ))2 0 where ` = k + xp, (p2 ) = x(1 x)p2 + xm2 (1 x)m2A , and in the last step we performed the Wick rotation by substituting `0 = i`0E . The momentum integral can be evaluated with standard methods (see Sec.7.5 and App.A.4 of Peskin&Schroeder): ✓ ◆n d2 Z d d È 1 1 (n d/2) 1 I1 (n) = = , (2⇡)d (`2E + )n (4⇡)d/2 (n) (9) ✓ ◆n d2 1 Z d d È `2E 1 d (n d/2 1) 1 d I2 (n) = = = I1 (n) . (2⇡)d (`2E + )n (4⇡)d/2 2 (n) 2 n d/2 1 This gives us finally Z 1 2 ✏ i⌃2 = e µ̃ dx (2 0 ⇢ Z ↵ 1 2 ✏!0 ' i dx 2⇡ 0 ✏. ✏)(1 E. i ✏)m (4⇡)d/2. x)p/ + (4. + log 4⇡. log. µ̃2. ((x. ✓. 2. d 2. 1)p/ + 2m). ◆. d 2. 2. = (10). ((x. 1)p/ + m). ,. where ↵ = e2 /4⇡ and in the last step we took the ✏ ! 0 limit and retained only the non-vanishing terms (ie. only up to O(1) since O(✏) terms vanish with ✏ ! 0). From the presence of the 1✏ term we see that the amplitude is UV divergent, and thus the need to renormalise. The relevant two-point function, computed up to O(e2 ), is given by ✓ ◆ i(p / + m) i(p / + m) i(p / + m) i ⌃ (p) iZ2 2 h0| (p) ¯( p)|0i = 2 + ( i⌃ (p)) +. . . = 1 + = , 2 p m2 p2 m2 p2 m2 /p m /p m /p mr (11). 2.

(3) where Z2 = 1 + Z2 is the wave-function renormalisation coefficient and mr = m + m is the renormalised mass. Both these terms are contained in ⌃2 , in particular Z2 is given by the coefficient of /p m inside ⌃2 and the rest renormalises the electron mass: ⌃2 (p) = (p/. m) Z2 + Bm .. (12). In fact, using this in eq. (11) one has, up to higher order terms: ✓ ◆ i Bm ¯ h0| (p) ( p)|0i ' (1 + Z2 ) 1 + ' /p m /p m /p. iZ2 , m(1 + B). (13). where mr = m(1 + B). In more detail, our renormalisation condition requires Z2 to be defined for p 2 = µ2 m2 , so (from eq. (12)): ✓ ◆ Z d⌃2 (p) ↵ 1 2 ( µ2 ) Z2 = 1 + =1+ dx(x 1) 1 log (14) E + log 4⇡ dp/ p2 = µ2 2⇡ 0 ✏ µ̃2 Let us now compute the electron anomalous dimension, then take the limit µ2 m2 , m2A , and finally perform the integration in x: Z Z 1 µ @ ↵ 1 x(1 x)2 µ2 ↵ µ2 m 2 ↵ (e) ' Z2 = dx ! dx(1 x) = . (15) 2 2 2 2 @µ 2⇡ 0 x(1 x)µ + xm (1 x)mA 2⇡ 0 4⇡ One can notice that this is exactly 1/2 of the coefficient of the 1/(d 4) = 1/✏ pole in eq. (14). Also, the same result can be obtained by deriving Z2 with respect to the scale µ̃: (e) = 1 @ µ̃ Z . 2Z2 @ µ̃ 2 1.2. Photon self-energy. Let us now compute the photon anomalous dimension A (e). =. 1 @ µ Z3 , 2Z3 @µ. Aµr = Z3. 1/2. Aµbare .. (16). As for the electron, we have to compute the photon self-energy at one loop and than impose the renormalization condition h0|Aµr (q)A⌫r ( q)|0i in the limit µ2. = q2 =. µ2. i q2. g µ⌫ q2 =. µ2. + O(q µ q ⌫ ) ,. m2 . The relevant diagram, with amputated external legs, is = i⇧µ⌫ 2 (q) =. i⇧µ⌫ 2. (17). (18). I.  i(p/ + k/ + m) i(k/ + m) d4 k = ( 1) Tr ( ie µ ) 2 ( ie ⌫ ) = 4 2 (2⇡) k m + i✏ (p + k)2 m2 + i✏ Z d4 k k µ (k + q)⌫ + k ⌫ (k + q)µ g µ⌫ (k · (k + q) m2 ) 2 = 4e , (2⇡)4 (k 2 m2 )((k + q)2 m2 ) Z. 3. (19).

(4) where we just evaluated the well-known Dirac traces in the numerator. Now, as done before, we cure the UV behaviour of the integral by generalising to d = 4 ✏ dimensions, introduce the Feynman parameter to combine the denominators, shift the integrating momentum to ` = k + xq, and finally neglect terms linear in ` in the numerator since they give a vanishing integral: Z 1 Z dd ` 2`µ `⌫ g µ⌫ `2 2x(1 x)q µ q ⌫ + g µ⌫ (m2 + x(1 x)q 2 ) µ⌫ 2 ✏ i⇧2 = 4e µ̃ dx , (20) (2⇡)d (`2 )2 0 where = m2 x(1 x)q 2 . By Lorentz symmetry, the first term in the numerator can be substituted by `2 g µ⌫ /d.1 As usual, to perform the integral we Wick rotate to Euclidean space by substituting `0 = i`0E Z 1 Z d d È (1 2/d)g µ⌫ `2E 2x(1 x)q µ q ⌫ + g µ⌫ (m2 + x(1 x)q 2 ) µ⌫ 2 ✏ i⇧2 = 4ie µ̃ dx . (21) (2⇡)d (`2E + )2 0 Using the integrals in eq. (9) we get Z 1 1 (2 d/2) µ⌫ 2 ✏ i⇧2 = 4ie µ̃ dx 2x(1 d/2 2 d/2 (4⇡) 0. x)(q 2 g µ⌫. q µ q ⌫ ) ⌘ (q 2 g µ⌫. q µ q ⌫ ) i⇧2 (q 2 ) . (22). Let us now take the ✏ ! 0 limit and keep only terms which do not go to zero: ✓ ◆ Z 2↵ 1 2 m2 x(1 x)q 2 2 ⇧2 (q ) ' dx x(1 x) log . E + log 4⇡ ⇡ 0 ✏ µ̃2. (23). Our renormalization condition requires h0|Aµ (q)A⌫ ( q)|0i. iZ3 µ⌫ g q2. = q2 =. µ2. q2 =. µ2. + O(q µ q ⌫ ) ,. (24). in the limit µ2 m2 (notice that here the condition is expressed in terms of the bare fields). µ⌫ Focussing on the g term, this implies that Z 3 = 1 + ⇧2 ( µ2 ) .. (25). Deriving first in µ and integrating then in x we get A (e). which, again, is exactly. 1 2. of the 1/(d. 4) =. =. ↵ , 3⇡. (26). 1/✏ pole in ⇧2 (q 2 ).. R Since the integral I µ⌫ = d4 ` `µ `⌫ f (`2 ) transforms as a tensor and it only depends on scalar quantities, the R result will necessarily be Rproportional to the metric: I µ⌫ = Ag µ⌫ . By contracting this with gµ⌫ we get gµ⌫ I µ⌫ = dA = d4 ` `2 f (`2 ), that is A = d1 d4 ` `2 f (`2 ). 1. 4.

(5) 1.3. Vertex function. Let us compute the one-loop diagram for the QED vertex A(q) (p1 ) ¯(p2 ) in Feynman gauge, Mµ (A ! ¯) = ie( µ + µ (p0 , p)), where =. and µ. 0. (p , p) =. Z. d4 k (2⇡)4 (p. ie. µ. 0. (p0 , p). ig⌫⇢ i(k/ + m) ( ie ⌫ ) 02 2 k) + i✏ k m2 + i✏. (27). µ. i(k/ + m) ( ie ⇢ ) , m2 + i✏. k2. (28). where k 0 = k + q (for the first part we follow Peskin 6.3 but using dim reg). Since the diagram is logarithmically divergent (4 2 1 1 = 0), let us introduce dimensional regularization. In working on the numerator we have to remember the relations between matrices valid for d dimensions. We get: Z ⌫ /0 dd k (k + m) µ (k/ + m) ⌫ µ 0 2 ✏ = (p , p) = ie µ̃ (2⇡)d (p k)2 (k 02 m2 )(k 2 m2 ) Z 0 0 dd k ⌫ k/ µ k/ ⌫ + m( ⌫ k/ µ ⌫ + ⌫ µ k/ ⌫ ) + m2 ⌫ µ ⌫ 2 ✏ = ie µ̃ = (2⇡)d (p k)2 (k 02 m2 )(k 2 m2 ) Z 0 0 0 dd k 2k/ µ k/ + ✏k/ µ k/ + 4m(k 0 + k)µ ✏m(k/ µ + µ k/) (2 ✏)m2 µ 2 ✏ = ie µ̃ = (2⇡)d (p k)2 (k 02 m2 )(k 2 m2 ) Z 0 0 dd k k/ µ k/ 2m(k 0 + k)µ + m2 µ 2✏ (k/ m) µ (k/ m) 2 ✏ = 2ie µ̃ (2⇡)d (p k)2 (k 02 m2 )(k 2 m2 ) (29). missinglater. Now we introduce the Feynman parameters: (p. k)2 (k 02. 1 m2 )(k 2. m2 ). =. Z. 1. dxdydz (x + y + z 0. 1). 2 , D3. (30). where D = k 2 + 2k · (yq zp) + yq 2 + zp2 (x + y)m2 . We complete the square in k by defining ` = k+yq zp, so that D = `2 and = xyq 2 +(1 z)2 m2 . Since we will be interested in fixing a renormalization condition for a scale q 2 = µ2 m2 , we can neglect all terms proportional to the fermion mass in the numerator. Also all terms odd in ` in the numerator will vanish. The remaining terms are h i ✏ ✏ ↵ µ (1 )`↵ ` + ( yq + zp)↵ ((1 y)q + zp) ((1 y)q + zp)↵ ( yq + zp) ' 2 2  1 2✏ 2 ↵ µ ` g↵ + ( yq + zp)↵ ((1 y)q + zp) = (31) d  1 2✏ 2 1 2✏ 2 µ µ ` (2 ✏) + ( y /q + z p/) ((1 y)/q + z p/) = ` (2 ✏) µ + C µ d d 5.

(6) Since only the term with `2 in the numerator is divergent, in the first step we neglected the last term proportional to ✏ as it will give terms which go to zero in the ✏ ! 0 limit. Putting all together, and performing the Wick rotation, we have µ. 42. Z. 1. ✏. dd È d 2 `2E (2 ✏) µ + C µ (p , p) = 2e µ̃IE dxdydz (x + y + z 1) = (2⇡)d (`2E + )3 0  Z ✏ 2e2 µ̃2IE 1 ✏ (✏/2) Cµ 2 + =41 d/2 dxdydz (x + y + z 1) µ (1 )(2 ✏) ' (4⇡) 2 4 2 0  ✓ ◆ Z ↵ 1 2 Cµ µ = dxdydz (x + y + z 1) 2 + log 4⇡ log 2 + + E 4⇡ 0 ✏ µ̃ 2 0. 2. 1. (32). 2. In the q 2 = µ2 m2 limit, the term proportional to C µ / goes to a constant, instead the term proportional to log increases logarithmically. In this limit we can therefore neglect the C µ term and spare ourselves a lot of Dirac algebra. Let us then impose the renormalisation condition in this limit, 1/2. µ. M (A !. ¯) =. ie(. µ. +. µ. 0. (p , p)) q 2 =µ2. ⌘. ier. µ. = q 2 =µ2. Z2 Z3 ie Z1. µ. ,. (33). q 2 =µ2. 1/2. in the limit µ2 m2 , where the numerator Z2 Z3 , which we already computed, exactly cancels the divergencies in the non-1PI diagrams with loop corrections on the external legs. So Z1 has to reabsorb the vertex corrections we just computed: ✓ ◆ Z ↵ 1 2 (µ2 ) Z1 (µ) ' 1 dxdydz (x + y + z 1) 2 + log 4⇡ log + , (34) 4⇡ 0 ✏ µ̃2 µ2 m 2 so that µ 1.4. @Z1 @µ. µ2. m2. =. ↵ @Z2 =µ . 2⇡ @µ. (35). QED -function. The -function, defined from the Callan-Symanzik equation, is given by the scale dependence of the coupling: ! ✓ ◆ 1/2 d d Z 2 Z3 d 1 µ d Z3 e3 (e) = µ er = µ e ' eµ Z2 Z1 + Z3 = e = e A (e) = . dµ dµ Z1 dµ 2 2 dµ 12⇡ 2 (36). 6.

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