Disclaimer. [disclaimer]

Transcription

1 Disclimer This is prolem set (s turne in) for the moule physics750. This prolem set is not reviewe y tutor. This is just wht I hve turne in. All prolem sets for this moule cn e foun t If not stte otherwise in the ocument itself: This work y Mrtin Ueing is license uner Cretive Commons Attriution-ShreAlike 4.0 Interntionl License. [isclimer]

2 physics750 Avnce Quntum Fiel Theory Prolem Set 0 Mrtin Ueing mu@mrtin-ueing.e Tutor: Thorsten Schimnnek Prolem Achieve points Possile points Renormliztion of φ 3 -theory in six imensions 25 Totl 25 This ocument consists of 6 pges... Dimensionlity Four imensions The ction is imensionless s it is fe into the pth integrl. There in the exponentil it must not hve ny unit. As the Lgrngin ensity is integrte over the whole four imensionl spce-time n spce hs negtive energy imension, it must hve energy imension of four. Six imensions In our cse here we re in six imensions. Therefore the Lgrngin ensity must hve mss imension of six. We enote the mss/energy imension with the function M. We now nee M( ) = where = 6 s the integrtion over 6 x ecuse M( 6 x) = 6. Looking t the iliner mss term n tking M(m) = for grnte, we must hve M(φ) = 2. From tht we conclue M(c) = 4 n M(g) = 0. This is ll comine in Tle. Rel imension Now we will go into = 6 2ε imensions. The integrtion mesure is now of mss imensions n the Lgrngin ensity hs to e of mss imension. Strting gin with the mss term we see tht M( φ 2 ) = M( ) + M(φ 2 ) = 2 + 2M(φ)! =.

3 Term M for = 6 M for = 6 2ε x ε 6 6 2ε m φ 2 2 ε c 4 4 ε g 0 0 µ µ x 0 ε Tle : Mss imensions From there we must hve M(φ) = 2 ε. With tht we cn look t c. We hve M(cφ) = M(c) + M(φ) = M(c) + 2 ε! =. We conclue tht M(c) = 4 ε. Finlly we cn look t g. Here it is M(gφ 3 ) = M(g) + 3M(φ) = M(g) + 6 3ε! =. Solving for M(g) gives us M(g) = ε. We now hve to introuce the µ to sor the mss imension of g n mke g imensionless constnt gin. We ssume M(µ) = s µ hs een use s regulting photon mss efore. We nee M(µ x g) = ε such tht M(g) cn sty zero. This rings us to M(µ x g) = x M(µ) + M(g) = x + 0! = ε n we conclue tht x = ε. All the mss imensions for oth cses re collecte in Tle for esy review..2. Feynmn rules n superficil ivergence Feynmn rules There re two interctions in the Lgrngin. Those re µx g 3! φ3 = iµ x g, cφ = ic. We mrk the g-vertices with ot n the c-vertices with squre in orer to istinguish the tpole igrm from the simple propgtor. Mrtin Ueing 2 Tutor: Thorsten Schimnnek

4 Degree of ivergence Ech propgtor shoul give mss imension of 2 s there is q 2 in the enomintor. At ech loop we hve to integrte over moment n therefore mss imensions there. Ech g-vertex gives us M(g) = ε n ech c-vertex gives us M(c) = 4 ε. Comining ll this we cn give the mss imension of igrm. This irectly is the superficil egree of ivergence, D = 2N P + N L + εn g + [4 ε]n c. This expression is not tht helpful yet s we re suppose to express this egree of ivergence in terms of the numer of propgtors n vertices only. Hence we nee some expression to get ri of the numer of loops. Tle 2 shows couple of igrms with their respective numer of externl lines N e, loops N L, propgtors N P, g-vertices N g n c-vertices N c. We mke the nstz + Ñ e N e + Ñ L N L + Ñ P N P + Ñ g N g + Ñ c N c = 0 where the Ñ re the coefficients in this eqution. They re to e etermine. The numers in the tle provie system of equtions. Incluing ll igrms mkes it solvle with one-imensionl solution spce. We choose normliztion n therefore otin the eqution 2 N e 2N L + N g N c = 0. This reltion contins the numer of loops, so we cn express tht numer in terms of the other quntities. Interestingly the numer of propgtors oes not pper in the reltion. There is prolem tht the system of equtions only hs one-imensionl solution spce s we only extrct one eqution. Using the one reltion we cn eliminte N L from the superficil egree of ivergence. This gets us to D = 2N P Ne + N g N c + εng + [4 ε]n c. We cn simplify further y inserting = 6 2ε. = 2N P + [3 ε] 2 N e + N g N c + εng + [4 ε]n c. An then fctor out to get = 6 2ε 2N P [3 ε]n e + 3N g + N c..3. Superficilly ivergent igrms In this prt of the prolem we re ske to rw ll igrms with re one-prticle-irreucile n only consist of one loop n re superficilly ivergent. All igrms contining c-vertex re reucile. One cn lwys cut off the c-tpole ( ) s the c-tpole is vli igrm in itself. It is prticle isppering into the vcuum which is quite strnge thing. In conense mtter theory such igrm is the scttering with n impurity. Due to momentum conservtion no momentum cn e exchnge n it only contriutes in q = 0. Applying Fourier trnsform this gives the monopole moment n therefore is the verge impurity ensity. As fr s I cn Mrtin Ueing 3 Tutor: Thorsten Schimnnek

5 Nicknme Digrm N e N L N P N g N c triple tpole c one loop two loops oule loop fork n loop c c tree two ntenns three ntenns propgtor Tle 2: Digrms with their respective numer of externl lines N e, loops N L, propgtors N P, g-vertices N g n c-vertices N c. Externl legs re lelle to clerly istinguish them from the c-tpoles. Mrtin Ueing 4 Tutor: Thorsten Schimnnek

6 () One leg (c) Two legs () No legs c () Three legs (e) Four legs c Figure : -PI igrms with only one loop. Figure N e D S c e Tle 3: Superficil ivergences of the igrms shown in Figure. The symmetry fctor is lso inclue. recll we took the tpole s n irreucile igrm in orer to otin self-energy correction ue to exctly this impurity scttering. Figure contins couple of ifferent irreucile igrms with exctly one loop. There is not so much one cn o without the c-tpoles. We cn then look t the superficil egree of ivergence using either form of the formul erive ove. One hs to e it creful s oth igrms with no n one externl leg consist of one propgtor only. Our results re summrize in Tle 3. In the tle we cn see tht only the first four igrms re superficilly ivergent. The more externl lines we, the less ivergent (superficilly) the igrm ecomes. At some point the igrms re superficilly convergent. The igrm in Figure without ny externl legs is vcuum ule n will e cncele in ny oservle y the expnsion of the enomintor. Therefore this igrm exists ut it is cncele. We re not sure whether this is to e tken ccount. If it is not to e inclue, we re lcking one superficilly ivergent igrm. The next step is the symmetry fctor for ll the igrms. Their symmetry groups re the iherl groups Mrtin Ueing 5 Tutor: Thorsten Schimnnek

7 s they hve the sme symmetry s regulr polygon. Ech of these groups looks like this: n D n = { c i, c i } = {, c, c 2,..., c n,, c, c 2,..., c n }, i=0 where is the ientity, c is the rottion roun 2π/n n is the reflection long some ritrry ut fixe xis. In ech cse the group hs crinlity 2n. The symmetry group for the vcuum igrm seems to e SO(2) from the igrm (if it hn t een rwn like n ellipse). This will proly le to the wrong result. We hve no contrctions in this igrm so the symmetry fctor is proly just one..4. Tylor expnsion No legs Without ny legs, we hve no externl momentum. Therefore we just hve p i 0 = i [2π] p 2 m. 2 This is ivergent y D = 2, which seems to e qurtic for ε = 0. This term cnnot e oserve, it is just shift in the vcuum energy n therefore is not of interest to the renormliztion proceure. One leg One externl leg still oes not give us ny itionl momentum in the loop. Only the loop momentum cn go roun in the loop, no momentum cn e trnsporte into the g-tpole. Therefore we just hve i = i 2 µ x p g [2π] [p 2 ] 2 which hs ivergence of D = 4 which is qurtic. Two legs The igrm with two legs hs one loop momentum. Momentum cn e trnsfere through the loop. We choose the following convention for the moment: k + p k k p We cn write own the mplitue for this igrm: i 2 = [ iµ x g] 2 i 2 p [2π] p 2 [p + k] 2 Now we cn expn this roun k = 0 in the next step. Mrtin Ueing 6 Tutor: Thorsten Schimnnek

8 Just to svor this moment: We hve n integrl which is logrithmiclly ivergent (for ε = 0). A ivergence lrey mens tht we re on the very ege of vliity for this theory. An to cure this ivergence we now expn the integrn into n infinite power series. Then we exchnge the summtion n integrtion in orer to get series of integrls. This exchnge of limits shoul e shown crefully. Tht the integrl iverges in the first plce shoul e ig ft hint tht the exchnge will e highly uious. To mke it even worse, we clim tht the integrl oes not relly iverge for lrge enough ε R. At the next opportunity I will tell some mthemticins out this escltion of things n see how they cringe. However, I ssume tht this cn e trete rigorously (just like imensionl regulriztion in polr coorintes) n is not s s it relly souns. To mke the expnsion more visile, we efine the k-epenent term to e f (k). Then we cn mke the following tleu: f (k) = f (0) = [p + k] 2 p 2 f 2[p + k] (k) = [[p + k] 2 ] 2 f 2p (0) = [p 2 ] 2 f (k) = 8[p + k] 2 [[p + k] 2 ] 3 2 [[p + k] 2 ] 2 f 8p 2 (0) = [p 2 ] 3 2 [p 2 ] 2 From this we cn ssemle the power series expnsion. We hve inclue the egree of ivergence D for ech term. The first frction outsie of the rcket is inclue in this s well. We otin i 2 = [µ x g] 2 p p k 4p 2 [2π] p 2 p 2 m }{{ 2 } [p 2 ] 2 + [p }{{} 2 ] 3 [p 2 ] 2 k 2 +O(k 4 ). }{{} D= 4 D= 5 D= 6 It is pprent tht in ech term the power of p ecreses. The terms o in p rop out ue to the symmetric integrtion ouns. This mens tht we only hve epenence on k 2, the Lorentz-invrint mgnitue, s efore. The first term is ly ivergent for = 6. The lst two terms re just logrithmiclly ivergent. All higher terms (omitte in the ove expression) re convergent in the p-integrtion. There seem to e two ivergent terms here, one qurtic n one logrithmic term. Together with the two ivergent terms from the previous two igrms, we lrey hve four ivergent terms now. Three legs The igrm with three externl legs still hs only one loop momentum ut now two externl moment tht enter the loop: p k 3 c k p + k p + k + k 2 k 2 Mrtin Ueing 7 Tutor: Thorsten Schimnnek

9 The mplitue is just like in the cse with two externl legs, there is just n itionl propgtor n vertex in the mplitue. It is i 3 = [ iµ x g] 3 i 3 p [2π] p 2 [p + k ] 2 [p + k + k 2 ] 2. Luckily we lrey hve ivergence of D = 6 in the first term without ny expnsion. This mens tht the only iverging term is the one we lrey hve. We cn still expn in k n k 2 n otin the following result: i 3 = [µ x g] 3 p [2π] p 2 p 2 p 2 m + O k k2 2. The terms liner in k n k 2 rop out since those terms re lso o in p. Due to the symmetric integrtion ouns this will rop out. This is our fifth ivergent term. We re still missing one term. Perhps the term o in p of the igrm with two legs is to e inclue in the counting to six? In orer to get own to four we nee to rop one or two terms epening on the counting. Which one shll it e? The igrm with zero legs shoul not e inclue, relly. When we exclue the igrm for physicl resons n rop the term liner in p in the igrm with two legs, we re ctully own to four terms. Those terms re: i = µ x p g [2π] [p 2 ] 2, i 2 (k) = [µ x g] 2 p [2π] [p 2 ] 2 + i 3 (k, k 2 ) = [µ x g] 3 p [2π] [p 2 ] 3 + O k 2 + k2 2. 4p 2 [p 2 ] 4 [p 2 ] 3 k 2 + O(k 4 ),.5. Dimensionl regulriztion The formul for the imensionl regulriztion on the prolem set oes not inclue the p 2 epenency in the numertor t ll. This mens tht this term is not inclue or tht we hve to fin tht formul ourselves? Applying Eqution (2) from the prolem set llows us to regulrize the four terms. The term contining p 2 in the numertor hs to e one seprtely. Eqution (2) lso seems to e missing n imginry unit with respect to the imensionl regulriztion fctors given y Peskin n Schroeer (995, A.4). We chose to inclue the imginry unit from the ook to mke it consistent with the p 2 formul tken from the it. Also the ook hs minus sign epening on the power of the numertor. This is lso not inclue in Eqution (2) on the prolem set. For the cse given on the prolem set this shoul e ( ) r n for the cse with the extr p 2 tht shoul e ( ) r. We will nee tht lter on to cncel some terms in the igrm with two legs. Mrtin Ueing 8 Tutor: Thorsten Schimnnek

10 One leg We otin for the term from the g-tpole igrm i = iµx gπ /2 m 4 Γ [2π] We insert explicitly Γ (2) = iµx g Γ ( + ε) π 3 ε m2 2ε 26 2ε Γ (2) An we expn the Γ -function for smll ε. = iµε g 2ε ε π 3 ε 2 + γ + O(ε) 6 2ε In orer to etermine the resiue t ε = 0 we cn use the resiue formul given in Eqution (3) on the prolem set. This gives us resiue of igm2 π Two legs The igrm with two legs gives the following fter imensionl regulriztion: i 2 (k) = i[µx g] 2 π /2 m 4 Γ m 6 Γ 3 2 k 2 + [µ x g] 2 p 4p 2 [2π] Γ (2) Γ (3) [2π] [p 2 ] 4 k2 + O(k 4 ). We expn the rcket first s this will mke it it simpler. We hve = i[µx g] 2 π /2 m 4 Γ [2π] 2 2 Γ (2) i[µx g] 2 π /2 m 6 Γ [2π] + [µ x g] 2 p [2π] 4p 2 [p 2 ] 4 k2 + O(k 4 ). 3 2 k 2 Γ (3) We will expn the first n secon term for smll ε irectly. The secon summn lso ppers (with ifferent power of µ ε g) in the mtrix element for the igrm with three legs. We therefore just skip to the result here n sve the etils for the next mtrix element. Tht wy it will not e tht cluttere here n you cn still look through the etils. = ig2 2 6 π 3 4µ 2 π ε + ln γ + ig2 4µ π 3 k2 ε + ln π γ + [µ x g] 2 k 2 p [2π] 4p 2 [p 2 ] 4 + O(k4 ). Mrtin Ueing 9 Tutor: Thorsten Schimnnek

11 The lst two terms now epen on k 2 wheres the first one oes not. In the lst term we hve the itionl compliction of fctor p 2 in the integrl. This is not covere y the formul given on the prolem set. Either this term oes not contriute or just something we hve to look up. With typicl clcultion for imensionl regulriztion we en up with = ig2 2 6 π 3 + ig2 k π 3 2µπ ε + 2 ln m γ + ig2 k π 3 ε γ + [ + γ] ln 3 4µ 2 ε + ln π γ 4µ 2 π + O(k 4 ). Tht is not very pretty. After fixing up ll the phses using only the imensionl integrtion formuls of the ook we cn ctully comine the secon n thir term. During tht process couple terms cncel out. = ig2 2 6 π 3 4µ 2 π ε + ln γ + Now tht is result one ctully ecently work with. ig2 k 2 4µ π γ ln π + O(k 4 ). Three legs For the igrm with three legs we cn use imensionl regulriztion irectly. i 3 (k, k 2 ) = i[µx g] 3 π /2 m 6 Γ [2π] = ig3 4µ 3 π 2 7 π 3 = ig3 2 7 π 3 = ig3 2 7 π Γ (3) ε Γ (ε) + O k 2 + k2 2 4µ 3 π + O k 2 + k2 2 + ln ε + O(ε 2 ) ε γ + O(ε) + O k 2 + k2 2 4µ 3 ε + ln π γ + O ε + k 2 + k Counterterms So fr the Lgrngin hs een = 2 [ φ]2 c 0 φ m2 0 2 φ2 g 0 3! φ3. We hve not explicitly written the suscript zero yet s this ws not one in Eqution () on the prolem set either. Now we will hve to write them s we introuce the renormlize prmeters. The four prmeters of this theory re. Fiel strength renormliztion Z, 2. Mss m, 3. Coupling constnt g, 4. Coupling constnt c. Mrtin Ueing 0 Tutor: Thorsten Schimnnek

12 Term Digrm Vlue of Digrm 2 [ φ r] 2 m2 2 φ2 r i p 2 + iε cφ r ic g 3! φ3 r ig c 2 δ Z[ φ r ] 2 δ m 2 φ2 r ip 2 δ Z iδ m δ c φ r iδ c δ g 3! φ3 r iδ g Tle 4: New Feynmn rules with counterterms. c Renormlize Lgrngin First we introuce the fiel strength renormliztion fctor Z n hve φ = Zφ r where φ is the renormlize fiel. The Lgrngin then ssumes the form = 2 Z[ φ r] 2 c 0 Zφr Zm2 0 2 φ2 r Z3/2 g 0 φ 3 r 3!. We wnt the norml Lgrngin ck in orer to otin the Feynmn rules tht we re use to. We o this y splitting the constnts into the usul physicl term n counterterm. We o this using the following efinitions: δ Z := Z, δ m := Z 0 m2, δ c = Zc 0 c, δ g = Z 3/2 g 0 g. The Lgrngin with those counterterms then is = 2 [ φ r] 2 cφ r m2 2 φ2 r g 3! φ3 r + 2 δ Z[ φ r ] 2 δ c φ r δ m 2 φ2 r δ g 3! φ3 r. We see tht we hve recovere the Lgrngin we h efore with reefinition of the fiels. Also there re lot of new terms in the Lgrngin. Those cn e interprete s itionl couplings n therefore new Feynmn rules. The erivtion of those Feynmn rules from the Lgrngin is s usul. The terms tht look like mssive propgtor re to e tken like n interction. We hve liste ll terms n counterterms in Tle 4. Mrtin Ueing Tutor: Thorsten Schimnnek

13 Renormliztion conitions Now tht we hve the new Feynmn rules in terms of the counterterms, we ctully nee to compute those counterterms to the one-loop orer of our theory here. In orer to o so, we nee some renormliztion conitions. For the φ 4 theory Peskin n Schroeer (995, p. 325) rgue tht the norml propgtor shoul hve pole t the physicl mss m. This mens tht the propgtor with corrections to ll orers shoul hve the form of = i p 2 + (terms regulr t p2 = ). In our cse here we only look up to first loop-orer in perturtion theory n therefore cn compute tht lo explicitly. For the term δ g we nee some wy to relte the ll-orer scttering mplitue to the physicl cse. For the φ 4 theory this seems to e possile to look t the scttering mplitue t zero momentum. In our cse this shoul work s well. We only hve triple vertex here, so our conition woul e p p 3 = ig, t s = 4, t = u = 0. p 2 The δ c nees to e efine in similr wy s well. However, there cnnot e ny momentum trnsfer into tht c-tpole. The igrm for the g-vertex is to e tken s n mputte igrm. There is only one incoming leg for the c-vertex, therefore there is no lo -version of the c-vertex n we en up with = ic. This constrint is irectly fulfille if we set δ c = 0. This might e too simple, though. Computtion of δ g In the computtion of δ g we nee to inclue ll igrms up to single loop n the counterterm. In the φ 4 theory there is the tree level igrm n the s-, t- n u-chnnel igrms. Here we only hve triple interction vertex n therefore hve three ifferent vertex corrections. A glnce t the QED section y Peskin n Schroeer (ii., 0.3) shows tht conition with the vertex correction is inee the right wy to renormlize such vertex. Up to one loop we therefore hve the following expnsion: = Mrtin Ueing 2 Tutor: Thorsten Schimnnek

14 The mile three terms with the loops re topologiclly equivlent to the cse with three legs just with the roles of the moment shifte sutly. In the cse of vnishing externl momentum ll those igrms shoul e exctly the sme. This result hs een compute lrey n is the leing term in the Tylor expnsion of the three-legge loop igrm. The tree level igrm just gives ig ( iµ ε g in rel imension) s cn e seen from the new Feynmn rules in Tle 4. This is exctly the term tht we wnt to give t zero momentum. Therefore the counterterm is exctly the negtive of those three loop igrms tken together. Luckily we hve lrey compute those n even regulrize them. So we simply hve δ g = 3i 3. Previously we hve erive this mtrix element n therefore lrey hve the counterterm, δ g = 3ig3 4µ π 3 ε + ln π γ. Computtion of δ m n δ Z The clcultion of the mss n of the fiel strength renormliztion goes y the self-energy. We cn o oth t the sme time. The sic ie of the self-energy ws tht the pole of the propgtor is shifte y geometric series of -PI los in the propgtor. We therefore nee to look t the -PI lo tht we put into tht series n see wht p 2 epenence tht hs. This -PI lo hs vlue of im 2 (p 2 ) in the nottion of Peskin n Schroeer (995, p. 328). In the geometric series tht will go into the enomintor to give the moifie propgtor s i p 2 M 2 (p 2 ). Our renormliztion conition from the propgtor emns tht the pole of the propgtor is t p 2 =, the physicl mss. Therefore t p 2 = we must hve M 2 (0) = 0. Also for the resiue of unity we must hve p 2 M 2 (p 2 ) p 2 = = 0. Next we nee to compute M 2 to the esire orere which is the one loop orer. The re propgtor is not to e inclue s it is not irreucile in the pproprite sense. Therefore we strt with the single loop. Our expnsion therefore is: = + The loop with two externl legs hs een clculte n expne in externl momentum lrey. This Tht coul e monster chsing someoy in horror movie. Mrtin Ueing 3 Tutor: Thorsten Schimnnek

15 mens we hve im 2 (k 2 ) = i 2 + ik 2 δ Z iδ m. Here we cn see lrey tht only the term which contins the k 2 cn contriute to the fiel strength renormliztion counterterm δ Z. First we write out the term fully. It is im 2 (k 2 ) = ig2 4µ π 3 ε + ln π γ + ig2 k 2 4µ π γ ln π + ik 2 δ Z iδ m. We ifferentite with respect to k 2 n set this to zero. This then gives us δ Z = g2 4µ π γ ln π. It is it surprising tht this oes not epen on ε t ll. In the φ 4 theory one ctully otins δ Z = 0 in the first loop orer. We go ck to the eqution, set k 2 =. This gives us δ m = g2 4µ π 3 ε + ln π γ + g2 4µ π γ ln π + δ Z. Then we insert the freshly erive δ Z. We set the expression to zero n otin δ m from tht. The result is = g2 4µ π 3 ε + ln π γ + g2 4µ π γ ln π g2 4µ π γ ln π. We cn extrct the common fctor in the front n otin the shorter expression = g2 2 4µ π 3 ε + 2 ln π Grouping terms gives = g π 3 ε + 8 4µ 2 π + 2[ γ] ln 2γ. 3 As lst step we cncel out the fctor of two n otin = g2 2 6 π 3 ε + 4 4µ 2 π + [ γ] ln γ. 3 2γ µ 2 3 γ ln π + 4µ 2 3 γ ln π..7. Purpose of liner term The liner term i not contriute nything to the nlysis presente here. Therefore one coul hve omitte it. On the other hn this term is like constnt ckgroun current tht shoul t lest give n Mrtin Ueing 4 Tutor: Thorsten Schimnnek

16 energy shift. In Conense Mtter we were le to sctter from the sme impurity (tht ckgroun current) twice n ctully trnsfer some momentum long the interction lines to form loop. This looke like this: p p k k p k This proly leverges on the ckgroun lttice eing le to sor momentum n give it ck t some other point. In our cse we woul nee something like this to get n interesting loop contriution from the c-vertices: p p p k k p k We o not hve this interction etween the c-vertices, so there it not much we cn o. A posting on Physics Stck Exchnge 2 suggests tht this term oes hve n effect when there is some egenercy in the groun stte. 2 Mrtin Ueing 5 Tutor: Thorsten Schimnnek

17 A. Questions A. Questions Sie question In generl reltivity one writes for the norml prtil erivtive n for the covrint erivtive (or connection s it is clle in mthemtics). In guge theory one lso strts with ut clls the covrint erivtive D. Why is tht? To me the D looks like more generl form of, the exterior erivtive tht cts on ifferentil forms. In some wy we hve f f. Is there reson t ll, is it just convention or is it to istinguish the guge covrint connection from fiel theory from the Levi-Civit connection of generl reltivity? References Peskin, Michel E. n Dniel V. Schroeer (995). An Introuction to Quntum Fiel Theory. Westview Press. ISBN: Mrtin Ueing 6 Tutor: Thorsten Schimnnek