Srednici Chapter 51 QFT Probems & Soutions A. George September 7, 13 Srednici 51.1. Derive the fermion-oop correction to the scaar proagator by woring through equation 5., and show that it has an extra minus sign reative to the case of a scaar oop. Note: I don t ie this probem very much because it is poory expained. Given what we have done up to this point, it woud be reasonabe to expect the diagram in question to be the foowing: 7 7 In fact, however, we are not going to consider the ines eading to the sources as propagators; we are rather going to consider the ines themseves as externa sources. This maes sense when phrased in this way, but this seems to contradict the instructions to foow equation 5.. In fact we must change equation 5. to account for our new vertex. Our diagram is: This corresponds to a vertex ining two propagators with one externa φ-fied. We must then modify equation 5. to account for this: [ ( )( ) [ Z(η,η,J) = exp ig d δ 1 δ x φ(x) i exp i d xd yη(x)s(x y)η(y) δη(x) i δη(x) [ i exp d xd yj(x) (x y)j(y) Now we need to write these exponents as a Tayor Series, and eep ony those terms corresponding to our diagram above. In particuar, we can eep ony the owest-order term (1) 1
from the fina exponentia, since we have no scaar propagators (the externa scaar fieds do not count as propagators, as discussed above). We have two vertices and two Dirac propagators, so we eep ony the second-order terms from the expansion of the first two exponentias. This gives: Z(η,η,J) = (ig) ( d x φ(x) i δ δη(x) )( 1 i δ δη(x) ) ( d y φ(y) δ i δη(y) (i) d a d b η(a)s(a b)η(b) (i) d c d d η(c)s(c d)η(d) )( 1 i δ δη(y) Now we have to be a itte bit carefu. Before we coud just start mady differentiating stuff, but this time we have to worry about commutation. Let s organize, and pu the scaars to the front. Z(η,η,J) = g d x d y d a d b d c d d φ(x)φ(y) η(a)s(a b)η(b)η(c)s(c d)η(d) ( δ )( δ )( δ )( δ ) δη(x) δη(x) δη(y) δη(y) To start the differentiation process, et s move the ast two functiona derivatives past the η(a)s(a b). We get a negative sign from the anticommutation with the functiona derivative. S(a b) is made up of scaars and therefore commutes, see equation 5.. Then: Z(η,η,J) = g d x d y d a d b d c d d φ(x)φ(y) ( δ )( δ ) δη(x) δη(x) ( )( ) δ δ η(a)s(a b) η(b)η(c)s(c d)η(d) δη(y) δη(y) Now we can tae these two functiona derivatives: reca that functiona derivatives give deta functions, which we integrate over. Further, we coud have chosen to use these functiona derivatives on the other propagators; the resut is the same up to dummy indices, so we mutipy by four (two for η and two for η): Z(η,η,J) = g d x d y d a d d φ(x)φ(y) η(a)s(a y)s(y d)η(d) ( δ )( δ ) δη(x) δη(x) Now we can tae the remaining functiona derivatives: Z(η,η,J) = g d x d y φ(x)φ(y)s(x y)s(y x) Sure enough, this has a negative sign. In the case of a scaar oop, there woud be no anticommutation, and so there woud be no negative sign. Now notice that since we started with equation 5., we ve ept ony one term, but we ve never formay associated that term with the diagram above. Now we use equation 5.6 to do so. The extra minus sign means that we have to buid our minus sign into the vaue of )
the diagram, ie into the Feynman Rues. But from this point, the ony way forward (that we ve covered) is to use our Feynman rues to assess the vaues of the diagrams, and then to use the Lehmann-Käen form to turn the 1PI diagrams into a propagator. This is what Srednici did in the chapter, so we are done. The resut is equations 51.5 and 51.11. Srednici 51.. Finish the computation of V Y (p,p), imposing the condition V Y (,) = igγ 5. Start with 51.7: iv Y (p,p) = g3 8π [ (1 ε 1 1 ( D og µ )) γ 5 + 1 Ñ Z g gγ 5 D Now we need to determine D and Ñ given p = p = : D = (x 1 +x )m +x 3 M = (1 x 3 )m +x 3 M = 1 Ñ = m γ 5 dx 1 dx dx 3 δ(x 1 +x +x 3 1) Since there is no dependence on x 1 or x, we can simpify: 1 1 x3 (...) = dx 3 dx (...) Doing the inner integra gives: Putting a this together gives: [{ iv Y (,) = g3 1 8π ε 1 1 Reordering: [{ iv Y (,) = g3 1 8π ε 1 1 Soving these integras, we have: [{ iv Y (,) = g3 1 8π ε 1 ( M og µ (...) = 1 1 dx 3 (1 x 3 )(...) ( )} (1 x3 )m +x 3 M dx 3 (1 x 3 )og γ 5 + µ m γ 5 dx 3 (1 x 3 ) Z (1 x 3 )m +x 3 M g gγ 5 1 ( )} m +x 3 (M m ) dx 3 (1 x 3 )og γ 5 + µ m γ 5 dx 3 (1 x 3 ) Z m +x 3 (M m g gγ 5 ) )+ 3 + m (M m ) m M m (M m ) og 3 ( )} M γ 5 + m
m γ 5 M og(m/m) m γ 5 Z (M m ) (M m g gγ 5 ) Two of these terms cance, another two combine: [{ iv Y (,) = g3 1 8π ε + 1 ( M ) m og M m µ (M m ) og m γ 5 M og(m/m) Z (M m ) g gγ 5 Canceing another two terms: which is: iv Y (,) = g3 8π γ 5 Now we use 51.5: which gives: [ 1 ε + 1 og [ V Y (,) = iz g gγ 5 i g3 1 8π γ 5 ε + 1 og igγ 5 = iz g gγ 5 i g3 8π γ 5 ( M ) m M m µ (M m ) og [ 1 ε + 1 og [ Z g = 1+ g 1 8π ε + 1 og ( ) M µ ( ) M µ ( ) M µ m M m og m M m og m M m og This concurs with 51.8, a good sign. Next we have: [ (1 iv Y (p,p) = g3 8π ε 1 1 ( )) D og γ µ 5 + 1 Simpifying: [ (1+ g 1 8π ε + 1 og iv Y (p,p) = gγ 5 + g3 8π ( ) M µ [ ( 3 1 m M m og ( D og µ ( )} M γ 5 + m ( ) M Z g gγ 5 m ( ) M m ( ) M m ( ) M m Ñ D ( )) M gγ 5 m )) γ 5 + 1 Ñ D [ ( ) ( ) + g3 M m M 8π γ 5 og + µ M m og m Now notice that in the first integra, we can write this as: 1 df3 ogd + 1 df3 ogµ. This second term is then ogµ 1 (1 x)dx = ogµ. This then cances with the denominator of the first term on the second row. Thus: [ ( iv Y (p,p) = gγ 5 + g3 3 8π 1 ( )) m M ogd+og(m)+ M m og γ 5 + 1 Ñ m D
which is: V Y (p,p) = igγ 5 + ig3 8π [ (3 + 1 ogd og(m) m M m og ( )) M γ 5 1 m Ñ D Note that const = const. Thus, we can combine two of these terms: [ (3 V Y (p,p) = igγ 5 + ig3 8π + 1 ( ) D og M m M m og ( )) M γ 5 1 m Ñ D which is the fina answer. Note: In Srednici s soutions, the first + sign on the right hand side is a negative sign. Someone is off by a sign, I suspect it is him. He aso has the γ 5 matrix distributed to the Ñ term, which is definitey incorrect. Srednici 51.3. Consider maing φ a scaar rather than a pseudoscaar, so that the Yuawa interaction is L Y u = gφψψ. In this case, renormaizabiity requires us to add a term L φ 3 = 1 6 Z κκφ 3, as we as a term inear in φ to cance tadpoes. Find the one-oop contributions to the renormaizing Z factors for this theory in the MS scheme. Let s rewrite equation 51.5 (the other equations sti hod, though we must add, as mentioned, a Y φ term to the counterterm Lagrangian). L 1 = Z g gφψψ+ 1 6 Z κκφ 3 + 1 Z λλφ These three terms represent the ony possibe interaction terms with mass dimensions. Note that we added a renormaization factor to the Yuawa interaction, as in the text. Now we have seven renormaization factors (three above, and four in equation 51.6. Y is a rea number, as discussed on page 66; we coud sove for it, but the resut woud not be very interesting, and has nothing to do with the renormaizing Z factors we are ased to find). The rest of the probem is to figure out these seven Z-factors. Reca that we are using the MS renormaization scheme, so we need ony to cance a the divergent terms. The finite terms do not affect the Z factors. We start by trying to correct our scaar propagator at the one-oop eve. The diagrams are: 5
+ + X Now we need to assess the vaue of each diagram. Diagram 1 We have: Then: (-1) because there is a Fermion oop ( 1 i) S(/ + /) S(/) from the oop, aong with an integra over. (ig) from the two vertices, since Z g = 1+O(g ). Π = g d (π) S(/ +/) S(/) We can tae a trace over these propagators, just be writing in index notation and reordering. Thus: Π = g d [ S(/ (π) Tr + /) S(/) Now we use equation 51.13. Let s oo at the numerator: numer = Tr [ ( / / +m)( / +m) Dropping those terms with an odd number of gamma matrices we have: which is: Simpifying: which we define to be which we define in anaogy to equation 51.1. numer = Tr [ // +// +m numer = [ ( ) ( )+m numer = [ m (+) numer = N As for the denominator, we use equation 51.15. Putting a this together, we have: 1 iπ = g d N dx π (q +D) 6
where q = +x and D = x(1 x) +m. Now et s change the integration variabe q. iπ = g π 1 Now et s put in our N, in terms of q rather than : iπ = g π 1 The terms inear in q integrate to zero, so: iπ = g π N dx d q (51.3.1) (q +D) dx d q m q +xq x q +x (q +D) 1 dx d q m q x +x (q +D) Next et s mae g g µ ε/, shifting the mass dimensionaity off of g. Π = g π iπ = g π µε 1 Now we use 51.18 and 51.19, and simpify: 1 { dx (m x +x ) dx d q m q x +x (q +D) ( ε og ( )) D µ ( +D ε + 1 og ( ))} D Now a we reay care about in the MS scheme is the divergent part, so we can write this as: 1 { } Π = g (m x +x )+D dx +(finite) π ε Now et s put the D in, and simpify: 1 { Π = g 6m +6x 6 x dx π ε Doing the integra: Diagram { } Π = g 6m + +(finite) π ε } +(finite) Nothing has changed from the text, so we can quote the answer (equation 51.1): Π = λ [ 1 (π) ε + 1 1 ( ) M og M µ µ Dropping the finite part: Π = λ [ 1 (π) ε +(finite) M 7
Diagram 3 This is the same diagram as in φ 3 theory. However, we cannot quote the answer from section 1, because that was done in six dimensions, but now we are woring in four dimensions. We can start from scratch, but everything we did up to equation (51.3.1) sti hods with the foowing modifications: There is no negative sign, since there is no fermion oop The vertex factor is iκ There is a symmetry factor of g g µ ε/ This gives: Using equation 51.18: This is: Diagram iπ = κ µ ε Π = κ 1 1 dx d q (π) 1 (q +D) dx 1 ( 16π ε og Π = κ 16π ε +(finite) ( )) D µ This is just a vertex, so we can tae the vaue straight from the chapter 5 Feynman Rues (or more recenty, equation 51.). Π = (Z φ 1) (Z M 1)M Now we sum a these sef-energies, and choose the Zs to cance the infinite parts. which impies: Simiary: (Z φ 1) g π ε = Z φ = 1 g π ε (Z M 1)M + κ 16π ε + λ 16π ε M g 6m π ε = Dividing through by M : (Z M 1)+ κ 16π εm + λ 16π ε g 6m π M ε = 8
Soving this: Z M = 1+ 1 [ κ 16π ε M +λ g m M Note: this is different from Srednici s soution, but it agrees with an independent soution from Andre Schneider at the University of Indiana. Moreover, Srenici s soution is necessariy incorrect because the mass φ 3 vertex forces κ to have a mass dimension of 1; Srednici s soution therefore adds a term with mass dimension = to terms of mass dimension =, which is obviousy wrong. Now for the fermion propagator. The diagrams are: + X Diagram Using the counterterm Lagrangian, we change the / to i/ and rub out the fieds, we get the vertex factor for the counterterm vertex (don t forget to mutipy by i, as with a vertex factors) as: iπ = i(z Ψ 1)/ i(z m 1)m Then: Diagram 1 Π = (Z Ψ 1)/ +(Z m 1)m Diagram one has two vertices, two propagators, and an integra over the oop; the resut is: ( ) 1 iπ = (ig) d i (π) S(/ +/) ( ) which is: iπ = g d / / +m (π) [( +) +m [ +m Using 51.15 to combine these denominators: iπ = g d 1 dx / / +m (π) (q +D) where q = +x and D = x(1 x) +m. Now we change the integration variabe q. Thus: iπ = g d q 1 dx / /q +x/ +m (π) (q +D) 9
We can drop the terms in the numerator that are odd in q, since those integrate to zero. Further, we tae g gµ ε/, shifting the mass dimensionaity onto µ. Then: iπ = g µ ε d q (π) We use equation 51.18 to do the q-integra: iπ = ig 16π 1 1 dx[m+/(x 1) dx m+/(x 1) (q +D) [ ε og ( ) D µ Doing the x integra: This is: [ iπ = ig m / [ 16π ε og ( ) D µ ( Π = g m / ) +(finite) 8π ε Combining these two diagrams, and requiring the m terms to cance, we have: Thus: and simiary with the / terms: 1 ε g 8π m (Z m 1)m = Z m = 1+ g 8π ε Thus: (Z Ψ 1)/ / 1 ε g Z Ψ = 1 g 16π ε 8π = Now we consider the correction to the φψψ vertex. The diagrams are: p p p + p+ p p p + p+ p p Assessing the vaues of these diagrams, we have: 1
Diagram 1 The ony contribution comes from the vertex, so: iπ = iz g g so: Diagram 3 Π = Z g g Note that there is ony one fermion propagator, so the numerator wi at most have terms of ony O(). There are three tota propagators, so the denominator wi be of O( 6 ). Thus, this integra goes as d 1 5. Even after four integras, this wi remain convergent assuming reasonabe boundary conditions. This therefore does not contribute to the Z-factors in this renormaiation scheme. Diagram Assessing the vaue of this diagram, we have: iπ = (ig) 3 ( 1 i ) 3 d (π) S(/p +/) S(/p+/) ( ) Writing these propagators: iπ = g 3 d (π) ( /p / +m)( /p / +m) ((p+) +m ))((p+) +m )( M ) Comparing this to equation 51.1, ony the definition of N has changed. Thus: iπ = g 3 d N df (π) 3 (q +D) 3 We coud cacuate N, but this is a ot of messy agebra. Reca that we ony care about the divergent part. By equation 1.7, the ony divergent part has a q in the numerator. We have: N / /q = q Aso, we shift the mass dimensionaity onto g: g gµ ε/. This gives: iπ = g 3 µ ε/ d q df (π) 3 (q +D) +(finite) 3 Now we use 1.7 to sove the integra (reca that the dimensionaity is ε).. We aso perform a Wic Rotation, which adds a factor of i. This gives: ( iπ = ig3 ε df (π) 3 Γ +(finite) ) 11
which is: Using equation 1.11, we have: Π = g3 8π ε +(finite) Π = g3 8π ε +(finite) Putting these together, we have: iv Y = iz g g g3 8π ε +(finite) Choosing Z g to absorb this divergence, we achieve, up to order g : Z g = 1+ g 8π ε Next we turn to the φ 3 vertex. The diagrams are: p p p p 1 p 3 +p +p p 1 p 1 p 3 p 3 p 1 p 1 p p 1 p 3 p 1 Diagram 1 The ony contribution comes from the vertex factor. Dropping the i, we have: Diagram Π = Z κ κ This is the same diagram as in φ 3 theory. Since we are now woring in four dimensions, we cannot simpy quote the resut from section 16. We can, however, use equation 16.6: Π = g d d q 1 (π) d (q +D) 3 1
Equation 16.7 gives: Now tae d = ε: Taing ε : Now we use 16.5: Π = g (π) Π = g Π = g Π = g Γ(3 d ) (π) d/d (3 d/) Γ ( 1+ ε D 1+ε/ (π) ) 1 (π) D 1 [ x3 x 1 1 +x 3 x +x 1 x 3 +m 1 Now we can set the externa momenta equa to zero, since they don t contribute to the divergent part (actuay we can set a of this equa to zero, since it is aready manifesty convergent, but et s stic with our usua set of trics): Dropping the convergent part, we have: Diagram 3 Π = g 3π m Π = (finite) Here we have to start from scratch. We have: ( ) 3 1 iπ = ( 1)(ig) 3 d i (π) S() S(p +) S( p 1 ) where the negative sign is necessary because we have a fermion oop. Expanding this: iπ = g 3 d ( / +m)( /p / +m)( / +/p 1 +m) (π) [( p 1 ) +m [(p +) +m [ +m Consider the numerator. We set the externa momenta to zero. The terms odd in integrate to zero, so we have: numer = 3m// +m 3 Using our usua tric (Feynman s formua), we can combine the denominator. Then: iπ = g 3 d (π) 1 1m +m 3 (q +D) 3 Switching our integration variabe to q: iπ = g 3 d q 1 1mq +(finite term of O(q )) df (π) 3 (q +D) 3 13
The constant term is of O(q 6 ), which wi ceary be convergent. Then: iπ = 1mg 3 d q 1 q df (π) 3 (q +D) 3 Now we tae a Wic Rotation and use equation 1.7: iπ = 1mg 3 i 1 Keeping the divergent terms ony, we have: which is: Doing the integra: Diagram iπ = 1mg 3 i 1 Π = mg3 16π ε Γ ( ε ) Γ(3) (π) Γ(3)Γ() D ε/ 1 (π) 1 [ ε +(finite) +(finite) Π = 3mg3 π ε +(finite) There are actuay three such diagrams: due to the two different vertices, the choice of which vertex goes on the eft is a substantia difference. Due to the oop, each diagram has a symmetry factor of two. Swapping the two externa propagators on the right does not yied a substantivey different diagram. Thus, we assess the vaues of these diagrams at: iπ = 3! (iκ)( iλ) ( ) 1 i Our usua manipuations give: iπ = 3κλ Using equation 51.18, we have: iπ = 3iκλ (π) d 1 (π) ( +M )((+p 1 ) +M ) d 1 1 (π) (q +D) [ ε og ( D µ ) which is: Π = 3κλ [ 1 16π ε +(finite) Now we combine these: Z κ κ+ 3mg3 π ε 3κλ 16π ε 1
We want Z κ to absorb these divergences ony. These give the O(ε 1 ) contributions to Z κ : Z κ = 1+ 1 ε Finay, we have the φ vertex. The diagrams are: ( ) 3λ 16π 3mg3 π κ p p 1 p p 3 p p p 1 +p +p 3 p p +p p 1 p 3 p +p +p 3 p +p p 3 In addition, there are the diagrams of figure 31.5. Diagram 1 Ony the vertex factor contributes. Diagram 3 Π = Z λ λ Fortunatey, this is exacty the same diagram that was considered in the chapter. The factors of γ 5 in the numerator are different, but since we ony care about determining the divergent part which Srednici tes us is, none of this matters. We coud introduce a negative sign because we do not need to anticommute γ 5 across S, but since this happens twice, any effect wi be ost. We can therefore quote Srednici s resut, equation 51.5: Diagrams of figure 31.5 Π = 3g π ( 1 ε +(finite) Fortunatey for us, φ theory naturay uses four dimensions, and so we can quote equation 51.51: ( ) Π = 3λ 1 16π ε +(finite) Diagram Here we wi have terms of O( ) in the numerator since these are a scaars, but of O( 8 ) in the denominator. Even after four integras, this wi be convergent. Then: Π = (finite) ) 15
We combine these in the usua way: Π = Z λ λ 3g π ε + 3λ 16π ε Choosing the contributions to Z λ at O(ε 1 ) to cance these divergences, we have: Z λ = 1+ 1 ε ( ) 3λ 16π 3g λπ 16