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Lollipop diagrams in defect N=4 super Yang-Mills theory Guo, Bin 2017

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Lollipop diagrams in defect N = 4super Yang-Mills theorybyBin GuoB.Sc., Lanzhou University, 2012M.Sc., Lanzhou University, 2015A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Physics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)August 2017c© Bin Guo, 2017AbstractIn this thesis, we have studied the lollipop diagrams in defect N = 4 superYang-Mills field theory with nontrivial background, which is dual to the D3-D5 brane system with the probe D5 brane carrying k units of flux. Usingthe framework for performing loop computations for this system built byBuhl-Mortensen, Leeuw, Ipsen, Kristjansen and Wilhelm [2, 3], we provethat for arbitrary N and k, the contribution of the lollipop diagrams to theone-point function is zero. This improves their result, where they take theplanar limit N  1 and the probe brane limit k/N  1.iiLay SummaryAll particles are either bosons, such as photons, or fermions, such as elec-trons. Supersymmetry is a symmetry relating bosons and fermions. In asupersymmetric theory, there are some cancellations of quantum effects be-tween bosons and fermions. In this thesis, we generalize such a proof that fora particular system, even the supersymmetry is partially broken, a specialkind of cancellation still exists.iiiPrefaceChapter 1 contains chosen material done by others. Chapters 2, 3 andthe appendix are based on my research under the supervision of ProfessorGordon Semenoff.ivTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . vAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 AdS/dCFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The action and the background solution . . . . . . . . . . . . 21.3 Fuzz sphere decomposition . . . . . . . . . . . . . . . . . . . 41.4 Diagonalization of the mass terms . . . . . . . . . . . . . . . 71.5 The propagators . . . . . . . . . . . . . . . . . . . . . . . . . 81.6 Dimensional reduction regularization . . . . . . . . . . . . . 102 Lollipop diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1 A useful formula of the fuzz sphere Harmonics . . . . . . . . 122.2 One-loop effective action for one-point function . . . . . . . . 142.2.1 Contribution from the Y ml part . . . . . . . . . . . . 142.2.2 Contribution from the Ena (Ean) part . . . . . . . . . 172.2.3 one-loop one-point function . . . . . . . . . . . . . . . 183 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20AppendicesA Useful expectation values contributed from the Y ml part . 21vTable of ContentsB Useful expectation values contributed from the Ena (Ean)part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25viAcknowledgementsI would like to thank my supervisor Gordon Semenoff. He has helped methroughout my research and I have benefited tremendously from his courses.I am grateful to have worked with someone with such a strong physicalinsight. I would like to thank Mark Van Raamsdonk for being the secondreader of this thesis. I would like to thank Shuailiang Ge and KrishanSaraswat for useful discussion. I would also like to thank all my friends atUBC. Finally, I would like to thank my family for their support.viiChapter 1Introduction1.1 AdS/dCFTThe AdS/CFT conjecture [1, 7] relates 4-dimensional N = 4 SU(N) super-symmetric Yang-Mills (SYM) theory, which is a conformal field theory, withYang-Mills coupling gYM to type IIB string theory with string length ls andcoupling gs on AdS5×S5 with radius L and N units of F(5) flux on S5. Thestrongest version states that these two theories are dynamically equivalentif the parameters satisfyg2YM = 2pigs, 2g2YMN = L4/l4s . (1.1)Usually we define the ’t Hooft coupling λ = g2YMN . However it is difficultto do calculation on both sides to check the strongest version of the corre-spondence. So people usually consider the weaker form of the conjecture bytaking some limits.In the strong version, we take the limit gs → 0 and hold L/ls fixed. Thelimit gs → 0 allows us to do perturbative calculation in string theory and theleading order is the classical string theory. This limit also implies N →∞,which is the planar limit of the SU(N) gauge field theory. To go back to thestrongest version, notice that for fixed λ we have 1/N ∼ gs. Then the 1/Nexpansion on the field theory side should correspond to the gs expansion instring theory side, which is the genus expansion of the string worldsheet.In the weak form we further take the limit λ→∞. In the string theoryside L/ls  1, then we have an effective type IIB supergravity on weaklycurved background. The field theory side becomes a strong coupled fieldtheory. So it is usually called strong/weak duality.The field theory living on the boundary can be derived from the lowenergy excitation of open string attached to a stack of N D3-branes. For thebosonic part, the massless excitations parallel to the brane correspond togauge field Aµ and the excitations transversal to the brane correspond to thesix scalar fields φi. Further, there are four Majorana fermions, which makethe bosons and fermions have the same degrees of freedom. The SO(4, 2)symmetry of the AdS5 corresponds to the 4-d conformal symmetry, which is11.2. The action and the background solutionSO(4, 2), in field theory side. The SO(6) symmetry of the S5 correspondsto the rotational symmetry SO(6) among the six real scalars and the SU(4)symmetry among the four fermions since SU(4) ∼ SO(6).The generalization of AdS/CFT with probe branes are widely studied[4–6]. The classical one is to add a probe D5 brane. If the coordinates ofthe 10-dimensional space are labeled by x0, x1, . . . , x9, the D3-branes oc-cupy x0, x1, x2, x3 and the D5-brane with geometry AdS4 × S2 occupiesx0, x1, x2, x4, x5, x6. Then the field theory side is a defect conformal fieldtheory with a co-dimension one defect. Because of the defect, the symmetrySO(4, 2) breaks to SO(3, 2), which is the conformal symmetry of the defect,and symmetry SO(6) breaks to SO(3)× SO(3).An interesting case is to let the D5 brane carry k units of Dirac monopoleflux on the S2. It can be realized as k of N D3 branes get dissolved in theD5 brane. Then on the field theory side, the defect connecting SU(N) gaugetheory in the region x3 > 0 and SU(N-k) gauge theory in the region x3 < 0.Further, three of the six scalar fields carry nontrivial background.The extra parameter k gives the system interesting new features [8]. Alarge enough k allows us to have both λ 1, which ensure a valid supergrav-ity description on string theory side, and λ/k2  1, which gives us a smallexpansion parameter on field thoery side and allows us to do perturbativecalculations. Then we can do calculations on both side and compare themdirectly. Further we need k  N to ensure the probe brane limit, where theadded D5 brane would not change the supergravity background.In the recent papers [2, 3], they provide a systematic way to do pertur-bative calculations on the field theory side. They prove that the one-loopone point effective action is zero in the probe brane limit k  N . Noticingthat half of the original supersymmetries remain when a defect is added tothe field theory, it is natural to guess the remaining supersymmetries wouldensure a zero one-loop one point effective action for arbitrary k and N . Inthis thesis, we prove that the one-loop one point effective action is zero forarbitrary k and N .1.2 The action and the background solutionThe field theory side is a defect CFT consists of a 4-dimensional bulk fieldand fields living on the codimension-one defect. The defect fields have noeffect to the one-loop lollipop diagram that we will consider. Hence we start21.2. The action and the background solutionwith the 4-dimensional N = 4 SYM with the actionSN=4 =2g2YM∫d4x tr[− 14FµνFµν − 12DµφiDµφi +i2Ψ¯ΓµDµΨ+12Ψ¯Γ˜i[φi,Ψ] +14[φi, φj ][φi, φj ]].(1.2)Here the indices are µ, ν = 0, 1, 2, 3 and i, j = 1, 2, . . . , 6. The field Ψ is a10-d Majorana-Weyl fermion and Γµ, Γ˜i are the 10-d gamma matrices.The background is the classical solutionφcli = −1x3t(k)i ⊕ 0(N−k), for i = 1, 2, 3, (1.3)where t(k)i is the k-d irreducible representation of su(2) and 0(N−k) is a(N − k)-d zero matrix. All the other fields have vanishing background.Following papers [2, 3], after expanding around the background φi → φcli +φiand gauging it, the perturbative action is given bySN=4 + Sgh = Skin + Sm,b + Sm,f + Scubic + Squartic, (1.4)where Sgh is the ghost action. The quartic terms are not relevant if we areonly interested in the one-loop one-point effective action. The kinetic termsareSkin =2g2YM∫d4x tr[12Aµ∂ν∂νAµ+12φi∂ν∂νφi+i2ψ¯γµ∂µψ+c¯∂µ∂µc]. (1.5)The mass terms of bosonic fields areSm,b =2g2YM∫d4x tr[12[φcli , φclj ][φi, φj ] +12[φcli , φj ][φcli , φj ] +12[φcli , φj ][φi, φclj ]+12[φcli , φi][φclj , φj ] +12[Aµ, φcli ][Aµ, φcli ] + 2i[Aµ, φi]∂µφcli].(1.6)The mass terms of fermionic fields areSm,f =2g2YM∫d4x tr[12ψ¯Gi[φcli , ψ]− c¯[φcli [φcli , c]]]. (1.7)31.3. Fuzz sphere decompositionThe cubic vertices areScubic =2g2YM∫d4x tr[i[Aµ, Aν ]∂µAν + [φcli , φj ][φi, φj ] + i[Aµ, φi]∂µφi+ [Aµ, φcli ][Aµ, φi] +12ψ¯γµ[Aµ, ψ] +3∑i=112ψ¯Gi[φi, ψ]+6∑i=412ψ¯Gi[φi, γ5ψ] + i(∂µc¯)[Aµ, c]− c¯[φcli , [φi, c]]].(1.8)Here the 10-d Majorana-Weyl fermion Ψ is reduced to four 4-d Majoranafermions ψ1, . . . , ψ4 and Gi are the resulting matrices whose properties canbe found in [3].1.3 Fuzz sphere decompositionIn order to diagonalize the mass terms (1.6) and (1.7) that involve traceof matrix fields, it is convenient to introduce a method called fuzz spheredecomposition. Noticing that these traces contain commutator [t(k)i , φ], itis natural to look at eigenstates of the operator Li, which acts on a k × kmatrix φ asLiφ = [ti, φ]. (1.9)Li forms a representation of su(2) because[Li, Lj ]φ = [ti, [tj , φ]]− [tj , [ti, φ]] = [[ti, tj ], φ] = iεijkLkφ. (1.10)This representation is highly reducible and can be reduced by explicitlyconstructing the basis Y ml of a k × k matrix, which is the eigenstate of theoperators L2 = LiLi and L3L2Y ml = l(l + 1)Yml ,L3Yml = mYml .(1.11)Here the indices are l = 0, 1, . . . , k − 1 and m = −l, . . . , l. Hence the totalnumber of Y ml is k2. Further they have the propertiesLiYml = Ym′l [t(2l+1)i ]m′,m, (1.12)41.3. Fuzz sphere decompositionwhere t(2l+1)i is the (2l + 1)-d representation of su(2). By scaling the basisproperly, the orthogonality relation can be normalized and hencetr[Y m†l Ym′l′ ] = δll′δmm′ , (1.13)whereY m†l = (−1)mY −ml , (1.14)which is consistent withL3Ym†l = [t3, Ym†l ] = − ([t3, Y ml ])† = −mY m†l . (1.15)For a general complex k-d matrix φ, it has k2 complex degrees of freedom.It can be decomposed asφ = φl,mYml , (1.16)with k2 independent complex coefficients φl,m. Becauseφ† = φ∗l,mYm†l = φ∗l,m(−1)mY −ml , (1.17)for a Hermitian matrix φ, φ† = φ impliesφ∗l,m(−1)m = φl,−m. (1.18)From now on we consider φ as a Hermitian matrix. Notice that this de-composition is nondegenerate and hence by specifing the eigenvalues of L2and L3 we can determine a unique degree of freedom in the matrix φ. Forexample, fromL2I = 0, L3I = 0,tr(1√kI1√kI) = 1,(1.19)we find Y m=0l=0 =1√kI.Now we consider the case with the definition of LiLi = [t(k)i ⊕ 0(N−k), ], (1.20)where t(k)i is the k-d representation of su(2) and 0(N−k) is (N − k)-d zeromatrix. ThenLiφ =[(ti 00 0),(A BB† C)]=([ti, A] tiB−B†ti 0). (1.21)51.3. Fuzz sphere decompositionWe decompose A, which is Hermitian, by basis Y ml ,A = φl,mYml . (1.22)We decompose(0 B0 0)by basis Ena,B = φn,aEna, (1.23)where the indices are n = 1, 2, . . . , k and a = k + 1, k + 2, . . . , N . Ena is aN -d matrix defined as(EMN )P,Q = δM,P δN,Q, (1.24)and henceEMNEPQ = δPNEMQ. (1.25)The orthogonality relation istr(EMNEPQ)= δPNδMQ. (1.26)Equivalently, we havetr(EN†MEPQ)= δNP δMQ, (1.27)that istr(En†aEn′a′)= δnn′δaa′ . (1.28)These matrices have the property(LiEna)pq = (ti)pk(Ena)kq = (ti)pnδaq = (ti)n′nδn′pδaq = (En′a)pq(ti)n′n.(1.29)Therefore, matrices Ena with n = 1, 2, . . . , k form a k-d representation ofsu(2). Similarly for Ean, which is used to decompose(0 0B† 0),(LiEan)pq = −(Ean)pk(ti)kq = −δap(ti)nq = −(ti)nn′δapδn′q = −(ti)nn′(Ean′)pq.(1.30)For a particular a, these matrices also form a k-d representation of su(2).The minus sign on the right hand side is consistent with ti on the left,[Li, Lj ]Ena = En′a[ti, tj ]n′n = En′aiεijk(tk)n′n = iεijkLkEna,[Li, Lj ]Ean = [−tj ,−ti]nn′Ean′ = −iεijk(tk)nn′Ean′ = iεijkLkEan.(1.31)61.4. Diagonalization of the mass termsIn summary,LiEna = En′a(ti)n′n,LiEan = −(ti)nn′Ean′ .(1.32)We can decompose(0 00 C)by basis Eaa′ . However since Li annihilatesit, it is not necessary. Further, these bases are orthogonal to each other,tr(Y ml Ean) = 0 = tr(Yml Ena). (1.33)Here and in the following, the Y ml should be undertstood as (Yml )(k)⊕0(N−k).A general Hermitian matrix φ can be decomposed asφ = φl,mYml + φn,aEna + φa,nEan + φa,a′Eaa′ . (1.34)The coefficients satisfyφ∗l,m(−1)m = φl,−m,φ∗a,n = φn,a,φ∗a,a′ = φa′,a.(1.35)Consider a mass termtr[φL2φ]= tr[φ†L2φ]= l(l + 1)φ∗l,mφl,m +k2 − 14φ∗n,aφn,a +k2 − 14φ∗a,nφa,n= l(l + 1)φ†l,mφl,m + 2k2 − 14φ†n,aφn,a.(1.36)In the third line, to make it field theory we change ∗ to †. We write the 2 inthe second term explicitly because usually there is a 2 in the kinetic term.1.4 Diagonalization of the mass termsIn the papers [2, 3], they diagonalized the mass terms (1.6) and (1.7) ex-plicitly. By using the operator Li, the mass term of the bosonic part can bewritten asSm,b =2g2YM∫d4x1x23tr[− 12φiL2φi − 12A3L2A3 + iεijkφiLjφk+ iφiLiA3 − iA3Liφi − 12φαL2φα − 12AiL2Ai].(1.37)71.5. The propagatorsHere, repeated indices are summed over i, j, k = 1, 2, 3 and α = 4, 5, 6 im-plicitly and Ai should be understood as the components A0, A1, A2. Thelast two terms are decoupled from the rest of the mass terms, then they arediagonalized after using the fuzz sphere decomposition. Hence we call fieldsφ4, φ5, φ6, A0, A1 and A2 the easy fields and call fields φ1, φ2, φ3 and A3,which are coupled together, the complicated fields.To diagonalize the mass terms of the complicated fields, define a fieldCT = (φ1, φ2, φ3, A3) and Hermitian matrices Si(Si)jk = −12iεijk,(Si)j,4 =12iδij , (Si)4,j = −12iδij .(1.38)Then the mass terms can be written asSm,b =2g2YM∫d4x1x23tr[CT (−12L2 + 2SiLi)C]. (1.39)Notice the operator Li, which acts on the color indices of the field C, hasthe su(2) algebra [Li, Lj ] = iijkLk. The operator Si acting on the flavorindices also satisfies the su(2) Lie algebra[Si, Sj ] = iεijkSk. (1.40)A general degree of freedom of C can be labeled by using the eigenvalues ofL2, L3, S2 and S3. Because both Li and Si satisfy the su(2) algebra, we canintroduce operator Ji = Li+Si, which satisfies the su(2) algebra. Thereforewe can specify a degree of freedom by the eigenvalues of J2, J3, L2 and S2.Then the operator −12L2+2SiLi = −12L2+J2−L2−S2 are diagonalized andthe mass coupled problem can be resolved. Similar method can be appliedto the fermionic part.1.5 The propagatorsAfter resolving the mass coupled problem, the propagators can be foundexplicitly [2, 3] and we list the result here. The fuzz sphere decompositionof a field isφ = φlmYml + φnaEna + φanEan + φaa′Eaa′ . (1.41)81.5. The propagatorsConsider the degrees of freedom coming from Y ml . For the bosonic easyfields φ4, φ5, φ6, A0, A1 and A2, propagators are the same as〈(φ4)lm(φ4)†l′m′〉 = δll′δmm′K l(l+1). (1.42)Here we make the spacetime coordinates implicit and the above formulashould be understood as 〈(φ4)lm(x)(φ4)†l′m′(y)〉 = δll′δmm′K l(l+1)(x, y). K l(l+1)is a function whose property will be introduced later. For bosonic compli-cated fields, the propagators are〈(φi)lm(φj)†l′m′〉 = δijδll′δmm′(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))−iijk(tk)mm′δll′ 12l + 1(K l(l−1) −K(l+1)(l+2)),(1.43)〈(A3)lm(A3)†l′m′〉 = δll′δmm′(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2)), (1.44)〈(φi)lm(A3)†l′m′〉 = iδll′(ti)mm′12l + 1(K l(l−1) −K(l+1)(l+2)). (1.45)The propagators for fermions are〈(ψi)lm(ψj)l′m′〉 = δijδmm′δll′(l + 12l + 1K−lF +l2l + 1K l+1F)−δll′ [Gk]ij(tk)mm′ 12l + 1(K−lF −K l+1F).(1.46)Now consider the degrees of freedom come from Ena and Ean. For thebosonic easy fields, the propagators are〈(φ4)na(φ4)†n′a′〉 = δnn′δaa′Kk2−14 . (1.47)For bosonic complicated fields, the propagators are〈(φi)na(φj)†n′a′〉 = δijδnn′δaa′(k + 12kK(k−2)2−14 +k − 12kK(k+2)2−14)−iijk(tk)nn′δaa′ 1k(K(k−2)2−14 −K (k+2)2−14),(1.48)〈(A3)na(A3)†n′a′〉 = δnn′δaa′(k + 12kK(k−2)2−14 +k − 12kK(k+2)2−14), (1.49)〈(φi)na(A3)†n′a′〉 = iδaa′(ti)nn′1k(K(k−2)2−14 −K (k+2)2−14). (1.50)91.6. Dimensional reduction regularizationThe propagators for fermions are〈(ψi)na(ψj)n′a′〉 = δaa′δijδnn′1k(k + 12K− k−12F +k − 12Kk+12F)−δaa′ [Gk]ij(tk)nn′ 1k(K− k−12F −Kk+12F).(1.51)Notice that there is no correlation between the degrees of freedom fromY ml and Ena(or Ean). Hence they contribute to the one-loop one-point func-tion separately. We don’t need the propagators from Eaa′ since they don’tcontribute to the one-loop one-point function.1.6 Dimensional reduction regularizationIn SYM, the dimensional regularization breaks the supersymmetry since itonly change the number of degrees of freedom of gauge bosons, then makesthe number of degrees of freedom of fermions and bosons unequal. In ourcase the 4-dimensionalN = 4 SYM, consider the on-shell degrees of freedom.There are one vector (matrix) gauge field Aµ with degrees of freedom of 2,six scalar fields with degrees of freedom 6 and four Majorana fermions withdegrees of freedom of 4×2 = 8. Therefore we have equal numbers of fermionsand bosons. However, after dimensional regularization to D = 4 − 2, thedegrees of freedom of the vector field is changed to 2 − 2 and other fieldsare the same. So the numbers of bosons and fermions are not equal since8− 2 6= 8.Dimensional reduction regularization is a usual way to resolve this prob-lem. The D = 4 N = 4 SYM can be obtained from a dimensional reductionof D = 10 N = 1 SYM. By dimensional reducing it to D = 4 − 2, thenumber of scalar fields is 6 + 2. Hence the fermions and bosons have thesame degrees of freedom 8 = 6 + 2+ 2− 2.In our case, the dimensional reduction should be adapted to the presentof the defect. We keep the defect as codimension one with the dimension3 − 2. The gauge field four components have been separated into easyfields, which are A0, A1, A2 and complicated field, which is A3, with thetotal number nA = nA,easy + nA,com = 3 + 1. In the dimensional reduction,since the defect is codimension one, we set nA,easy = 3 − 2 and nA,com =1. Then nA = nA,easy + nA,com = 4 − 2. Perturbations of the scalarsthat have nonvanishing backgrounds are complicated fields and those havevanishing backgrounds are easy fields. If we keep the number of scalarsthat have nonvanishing backgrounds as 3, the separations are nφ,com = 3101.6. Dimensional reduction regularizationand nφ,easy = 3 + 2. It is a choice consistent with the classical equation ofmotion and the Nahm condition [3].In dimension D = 4 − 2, the regularized bosonic propagators have theproperty [3]Kν(x, x) =g2YM2116pi2x23(m2[−1−log(4pi)+γE−2 log(x3)+2Ψ(ν+12)−1]−1),(1.52)where γE is the Euler-Mascheroni constant and Ψ is the digamma function.For propagators of fermions, we havetrKmF (x, x) =sign(m)g2YM214pi2x33[|m|3 + |m|2 − 3|m| − 1+ |m|(|m|2 − 1)(− 1− log(4pi) + γE − 2 log(x3) + 2Ψ(|m|)− 2)].(1.53)11Chapter 2Lollipop diagram2.1 A useful formula of the fuzz sphereHarmonicsAny complex k-d matrix can be decomposed by the basis Y ml , where l =0, 1, . . . , k − 1, as a complex vector space. Now consider the matrix∑mY m†l Ym′l′ Yml , (2.1)where the indices l are not summed over. By using (1.12), which is[ti, Yml ] = Ym′l (ti)m′m, (2.2)where (ti)m′m should be understood as (t(2l+1)i )m′m, we have[ti, Ym†l ] = −[ti, Y ml ]† = −Y m′†l (ti)∗m′m = −(ti)mm′Y m′†l . (2.3)Therefore[ti,∑mY m†l Ym′l′ Yml ]=∑mm′′−(ti)mm′′Y m′′†l Ym′l′ Yml +∑mm′′(ti)m′′mYm†l Ym′l′ Ym′′l+∑mY m†l [ti, Ym′l′ ]Yml=∑mY m†l [ti, Ym′l′ ]Yml .(2.4)It implies that∑m Ym†l Ym′l′ Yml has the same transformation property asY m′l′ . From the nondegenerate of the eigenvalues l,m of Yml , we have∑mY m†l Ym′l′ Yml = C(l, l′,m′)Y m′l′ , (2.5)122.1. A useful formula of the fuzz sphere Harmonicswhere C(l, l′,m′) is a constant, which may depend on l, l′ and m′. Byapplying a unitary transformation U = e−iλiti ,C(l, l′,m′)Y m˜′l′ = U(∑mY m†l Ym′l′ Yml)U †=∑mUY m†l U†UY m′l′ U†UY ml U†=∑m˜Y m˜†l Ym˜′l′ Ym˜l = C(l, l′, m˜′)Y m˜′l′ ,(2.6)which implies that C(l, l′,m′) can not depend on m′. Then equation (2.5)becomes ∑mY m†l Ym′l′ Yml = C(l, l′)Y m′l′ . (2.7)In our calculation, we are interested in∑m[Y m†l , [ti, Yml ]] = Cliti. (2.8)To calculate Cli, we first define a constantl2 + (l − 1)2 + ...+ (−l + 1)2 + (−l)2 = cl. (2.9)We havetr(t(2l+1)i t(2l+1)i ) = tr(t(2l+1)3 t(2l+1)3 ) = cl. (2.10)Here indices i are not summed over. It implies that the trace doesn’t dependon indices i. Fromtr(tN−12i tN−12i ) = cN−12, (2.11)and∑mtr(ti[Ym†l , [ti, Yml ]]) =∑mtr([ti, Ym†l ][ti, Yml ])=∑m(−1)(ti)mm′(ti)m′′mtr(Y m′†l Ym′′l ) = (−1)tr(t2l+1i t2l+1i ) = −cl,(2.12)we have ∑m[Y m†l , [ti, Yml ]] = −clcN−12ti = Clti. (2.13)Notice that the constant Cl doesn’t depend on i, which is consistent with(2.7).132.2. One-loop effective action for one-point function2.2 One-loop effective action for one-pointfunctionBy contracting two fields in the cubic interaction terms, we will get theeffective action that will contribute to the one-point function. Because thereis no propagator that correlates the degrees of freedom from the Y ml and theEna (Ean) parts, the contributions from these two parts can be calculatedseparately.2.2.1 Contribution from the Y ml partBased on the expectation values calculated in Appendix (A), we summa-rize the contribution from the Y ml part in this subsection. Consider thecontribution from tr(i[Aµ, Aν ]∂µAν). We havetr(i[Aˆµ, Aˆν ]∂µAν) = 0. (2.14)Here and in the following we use hats to indicate the two fields contractedtogether by a propagator. Because 〈AµAν〉 6= 0 only when µ = ν, antisym-metry of µν makes the above contribution zero. Further, we havetr(i[Aµ, Aˆν ]∂µAˆν) = tr(iAµ[Aˆν , ∂µAˆν ]) = tr(iAµ 12∂µ[Aˆν , Aˆν ]) = 0, (2.15)and similarlytr(i[Aˆµ, Aν ]∂µAˆν) = 0. (2.16)Then the total contribution from tr(i[Aµ, Aν ]∂µAν) is zero.Consider the contribution from term tr([φcli , φj ][φi, φj ]). We havetr([φcli , φˆj ][φi, φˆj ])easy = −1y3K l(l+1)Cltr(φiti)nφ,easy, (2.17)andtr([φcli , φˆj ][φi, φˆj ])com = −1y3tr(φi[φˆj [ti, φˆj ]])com= − 1y3(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))Cltr(φiti)nφ,com.(2.18)Further, we havetr([φcli , φj ][φˆi, φˆj ]) =1y3tr([φˆi, φˆj ][ti, φj ])com= − 1y3(−1)nφ,com − 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(tiφi).(2.19)142.2. One-loop effective action for one-point functionHere, the contribution only comes from the complicated part. Furthermore,tr([φcli , φˆj ][φˆi, φj ]) = −1y3tr(φˆi[φj [ti, φˆj ]])= − 1y3{−(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))Cltr(φiti)− (nφ,com − 1) 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(φiti)}.(2.20)Consider the contribution from tr([Aµ, φcli ][Aµ, φi]). We havetr([φcli , Aˆµ][φi, Aˆµ])easy = −1y3K l(l+1)Cltr(φiti)nA,easy, (2.21)andtr([φcli , Aˆµ][φi, Aˆµ])com= − 1y3(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))Cltr(φiti)nA,com.(2.22)Further, we havetr([φcli , Aµ][φˆi, Aˆµ]) = 0, (2.23)tr([φcli , Aˆµ][φˆi, Aµ]) = tr(φcli [Aˆµ[φˆi, Aµ]]) = 0. (2.24)The last equality is due to the Jacobi identity.Consider the contribution from tr(i[Aµ, φi]∂µφi). We havetr(i[Aˆµ, φˆi]∂µφi) = tr(i[Aˆ3, φˆi]∂3φi)nA,com= − 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(ti∂3φi)nA,com.(2.25)Further, we havetr(i[Aˆµ, φi]∂µφˆi) = tr(i[Aˆ3, φi]∂3φˆi)nA,com=1212l + 1∂3(Kl(l−1) −K(l+1)(l+2))Cltr(tiφi)nA,com.(2.26)Furthermore,tr(i[Aµ, φˆi]∂µφˆi) = tr(iAµ[φˆi, ∂µφˆi]) = tr(iAµ 12∂µ[φˆi, φˆi]) = 0. (2.27)Consider the contribution from tr(−c¯[φcli , [φi, c]]). We havetr(−ˆ¯c[φcli , [φi, cˆ]]) = tr([φcli , ˆ¯c][φi, cˆ]) = −1y3K l(l+1)Cltr(φiti)(−nc). (2.28)152.2. One-loop effective action for one-point functionIt is similar to tr([φcli , φˆj ][φi, φˆj ])easy except the minus sign for the fermionloop.Consider the contributiontr(i∂µˆ¯c[Aµ, cˆ]) = 0, (2.29)which is similar to tr(i[Aµ, φˆi]∂µφˆi) = 0.Consider the contribution from vertices123∑i=1tr(ψ¯Gi[φi, ψ]) +123+nφ,easy∑i=4tr(ψ¯Gi[φi, γ5ψ]) +12tr(ψ¯γµ[Aµ, ψ]). (2.30)Only the first term has contribution12tr( ˆ¯ψGi[φi, ψˆ]) = −12nψ12l + 1(K−lF −K l+1F )Cltr(φiti). (2.31)Finally the one-loop effective action from the Y ml part that will con-tribute to the one-point function isV lmeff (y) =−1y3K l(l+1)Cltr(φiti)neasy− 1y3(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))Cltr(φiti)(nφ,com + nA,com − 1)− 1y3(−32)12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(tiφi)(nφ,com − 1)− 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(ti∂3φi)nA,com+1212l + 1∂3(Kl(l−1) −K(l+1)(l+2))Cltr(tiφi)nA,com− 12nψ12l + 1(trK−lF − trK l+1F )Cltr(φiti),(2.32)where neasy = nφ,easy + nA,easy − nc. The function K here should be un-derstood as the regularized K(y, y) (1.52)(1.53). Further, the fourth termcan be integrated by parts, since we consider the spacetime integral of theeffective action.162.2. One-loop effective action for one-point functionThen the effective action becomesV lmeff = −1y3Cltr(φiti)1l{[− 1− log(4pi) + γE − 2 log(x3) + 2Ψ(l)− 1]×[(l3 + l2)(neasy + nφ,com + nA,com − 2nψ − 1)+ 3l(nφ,com − 2nA,com − 1)]+2l2(neasy + nφ,com + nA,com − 2nψ − 1)+l(neasy + 6nφ,com − 3nA,com − 2nψ − 12)+6(nφ,com − 2nA,com − 1)}.(2.33)In the dimensional reduction regularization, we have neasy = nφ,easy +nA,easy − nc = 3 + 2 + 3− 2− 1 = 5, nφ,com = 3, nA,com = 1 and nψ = 4.Notice that all these don’t depend on . After substituting these, we havethe one-loop effective action, which will contribute to one-point function,V lmeff = 0. (2.34)2.2.2 Contribution from the Ena (Ean) partFrom Appendix (B), we find the contribution from the Ena (Ean) part is thesame as the contribution from the Y ml part if we make the correspondencek = 2l + 1, − 2(N − k) = Cl. (2.35)Therefore, from the result of the Y ml part we haveV naeff (y) =2(N − k)y3Kk2−14 tr(φiti)neasy+N − ky3(k + 1kK(k−2)2−14 +k − 1kK(k+2)2−14 )tr(φiti)(nφ,com + nA,com − 1)− 3N − ky31k(K(k−2)2−14 −K (k+2)2−14 )tr(φiti)(nφ,com − 1)+2(N − k)k(K(k−2)2−14 −K (k+2)2−14 )tr(ti∂3φi)nA,com− (N − k)k∂3(K(k−2)2−14 −K (k+2)2−14 )tr(tiφi)nA,com+N − kk(trK− k−12F − trKk+12F )tr(tiφi)nψ,(2.36)172.2. One-loop effective action for one-point functionand finally the contribution isV naeff = 0. (2.37)2.2.3 one-loop one-point functionThe total contribution to the one-loop effective action isV lmeff = Vlmeff + Vnaeff = 0. (2.38)By connecting the field in the one-loop effective action to an external field,we get the one-loop one-point function. This diagram is called lollipopdiagram. From equation (2.38), we find for any field the one-loop one-pointfunction (lollipop diagram) is zero〈φ〉1−loop = 0. (2.39)18Chapter 3ConclusionIn this thesis, we have studied the lollipop diagram in a defect field the-ory with nontrivial background. This field theory is originated from theAdS/dCFT correspondence, where the gravity side is a D3-D5 brane systemwith the probe D5 branes carrying k units of flux. Due to the existence ofthe defect and the nontrivial background, half of the supersymmetries arebroken. However, there may still be enough supersymmetries to make thecontribution of lollipop diagrams to the one-point function zero. In [3], theyprove that the contribuiton is indeed zero in the planar limit N → ∞ andthe probe limit kN  1. Further, based on their explicit result for N, k < 9,they argued that the contribution should be zero for arbitrary N and k.In this thesis, we proved the contribution of lollipop diagrams is zero forarbitrary N and k.In the usual supersymmetric QFT, the zero contribution is due to thecancellation between the boson loops and fermion loops with the same mass.In our calculation, we calculated the contributions from the degrees of free-dom from the Y ml part and the Ena (Ean) part separately. We found thatfor the Y ml part, the cancellation is between the boson loops and fermionloops with the same l and for the Ena (Ean) part, the cancellation is betweenthe boson loops and fermion loops with same a. This is partly because thecommutators with ti mix Yml of different m and mix Ena (Ean) of differentn.In our calculation, we used a modified version of dimensional reductionregularization from [3]. An explicit regularization is needed in this method.It would be interesting to find a proof based on the remaining supersymme-tries and prove these results without any explicit regularization.19Bibliography[1] Martin Ammon and Johanna Erdmenger. Gauge/gravity duality. Cam-bridge Univ. Pr., Cambridge, UK, 2015.[2] Isak Buhl-Mortensen, Marius de Leeuw, Asger C. Ipsen, Charlotte Krist-jansen, and Matthias Wilhelm. One-loop one-point functions in gauge-gravity dualities with defects. Phys. Rev. Lett., 117(23):231603, 2016.[3] Isak Buhl-Mortensen, Marius de Leeuw, Asger C. Ipsen, Charlotte Krist-jansen, and Matthias Wilhelm. A Quantum Check of AdS/dCFT. JHEP,01:098, 2017.[4] Oliver DeWolfe, Daniel Z. Freedman, and Hirosi Ooguri. Holographyand defect conformal field theories. Phys. Rev., D66:025009, 2002.[5] Andreas Karch and Lisa Randall. Localized gravity in string theory.Phys. Rev. Lett., 87:061601, 2001.[6] Andreas Karch and Lisa Randall. Open and closed string interpretationof SUSY CFT’s on branes with boundaries. JHEP, 06:063, 2001.[7] Juan Martin Maldacena. The Large N limit of superconformal fieldtheories and supergravity. Int. J. Theor. Phys., 38:1113–1133, 1999.[Adv. Theor. Math. Phys.2,231(1998)].[8] Koichi Nagasaki and Satoshi Yamaguchi. Expectation values of chi-ral primary operators in holographic interface CFT. Phys. Rev.,D86:086004, 2012.20Appendix AUseful expectation valuescontributed from the Y mlpartIn this appendix, we calculated some useful expectation values from thecontribution of the degrees of freedom of the Y ml part. They will be usefulto calculate one-loop effective action for one-point function.• Contribution to tr([ti, φˆ4][φi, φˆ4])The hats means contraction of the two φ4tr([ti, φˆ4][φi, φˆ4])= 〈(φ4)†l′m′(φ4)lm〉tr(φi[Y m′†l′ [ti, Yml ]])=∑lK l(l+1)∑mtr(φi[Ym†l [ti, Yml ]])= K l(l+1)Cltr(φiti).(A.1)In the last line, there is a sum over l implicitly.• Contribution to tr(Aµ[φˆ4, ∂µφˆ4])Becauselimx→y〈φ4(x)∂µφ4(y)〉 =12∂µ limx→y〈φ4(x)φ4(y)〉 =12∂µ〈φ24〉, (A.2)and the antisymmetry of the commutator, we havetr(Aµ[φˆ4, ∂µφˆ4]) = 0. (A.3)• Contribution to tr(φi[φˆ1[ti, φˆ1]])We havetr(φi[φˆ1[ti, φˆ1]]) =(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))Cltr(φiti). (A.4)• Contribution to tr(φˆi[φˆj [ti, φj ]])21Appendix A. Useful expectation values contributed from the Y ml partBecause of the ti and the sum over i, the i, j here label the complicatedfieldstr(φˆi[φˆj [ti, φj ]]) = tr([φˆi, φˆj ][ti, φj ])= 〈(φi)†l′m′(φj)lm〉Atr([Y m′†l′ , Yml ][ti, φj ])= iijk(tk)mm′12l + 1(K l(l−1) −K(l+1)(l+2))tr([Y m′†l , Y ml ][ti, φj ])= iijk12l + 1(K l(l−1) −K(l+1)(l+2))tr(Cltk[ti, φj ])= iijk12l + 1(K l(l−1) −K(l+1)(l+2))Clikiptr(tpφj)= −nφ,com − 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(tiφi).(A.5)In the fourth line we use(tk)mm′ [Ym′†l , Yml ] = [Ym′†l [tk, Ym′l ]] = Cltk. (A.6)In the sixth line we usekijkip = 2δjp = (nφ,com − 1)δjp, (A.7)since in our regularization method we have nφ,com = 3.• Contribution to tr(φˆi[φj [ti, φˆj ]])tr(φˆi[φj [ti, φˆj ]]) = −tr(φj [φˆi[ti, φˆj ]])= −〈(φi)†l′m′(φj)lm〉tr(φj [Y m′†l′ [ti, Yml ]]).(A.8)For the complicated fields〈(φi)†l′m′(φj)lm〉 = δijδll′δmm′Scom + iijk(tk)mm′δll′Acom, (A.9)where the symmetric and antisymmetric part isScom =l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2),Acom =12l + 1(K l(l−1) −K(l+1)(l+2)).(A.10)For the symmetric part,〈(φi)†l′m′(φj)lm〉Str(φj [Y m′†l′ [ti, Yml ]]) = Scom Cltr(φiti). (A.11)22Appendix A. Useful expectation values contributed from the Y ml partFor the antisymmetric part,〈(φi)†l′m′(φj)lm〉Atr(φj [Y m′†l′ [ti, Yml ]])= iijk(tk)mm′tr(φj [Ym′†l [ti, Yml ]])Acom= iijktr(φj [Ym′†l [ti[tk, Ym′l ]]])Acom= iijktr(φj [Ym′†l [tq, Ym′l ]])iikqAcom= (nφ,com − 1)Cltr(φiti)Acom,(A.12)where in the fourth line we use the antisymmetry of i, k. The total contri-bution istr(φˆi[φj [ti, φˆj ]]) =− Scom Cltr(φiti)− (nφ,com − 1)Cltr(φiti)Acom=−(l + 12l + 1K l(l−1) +l2l + 1K(l+1)(l+2))Cltr(φiti)− (nφ,com − 1) 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(φiti).(A.13)• Contribution to tr(φˆi[Aˆ3[ti, A3]])tr(φˆi[Aˆ3[ti, A3]]) = tr([φˆi, Aˆ3][ti, A3])= 〈(φi)lm(A3)†l′m′〉tr([Y ml , Y m′†l′ ][ti, A3]).(A.14)Because〈(φi)lm(A3)†l′m′〉 ∼ δll′(ti)mm′ , (A.15)we havetr(φˆi[Aˆ3[ti, A3]]) ∼ tr(ti[ti, A3]) = 0. (A.16)Due to the Jacobi identity, we havetr(φˆi[A3[ti, Aˆ3]]) = tr(φˆi[Aˆ3[ti, A3]]) = 0. (A.17)• Contribution to tr(i[Aˆ3, φˆi]∂3φi)tr(i[Aˆ3, φˆi]∂3φi)= i〈(φi)lm(A3)†l′m′〉tr([Y m′†l′ , Yml ]∂3φi)= − 12l + 1(K l(l−1) −K(l+1)(l+2))tr([Y m†l [ti, Yml ]∂3φi)= − 12l + 1(K l(l−1) −K(l+1)(l+2))Cltr(ti∂3φi).(A.18)23Appendix A. Useful expectation values contributed from the Y ml part• Contribution to tr(i[Aˆ3, φi]∂3φˆi)tr(i[Aˆ3, φi]∂3φˆi) = itr([∂3φˆi, Aˆ3]φi)= i〈∂3(φi)lm(A3)†l′m′〉tr([Y ml , Y m′†l′ ]φi)=i2∂3(〈(φi)lm(A3)†l′m′〉)tr([Y ml , Y m′†l′ ]φi)=i212l + 1∂3(Kl(l−1) −K(l+1)(l+2))i(ti)mm′tr([Y ml , Y m′†l′ ]φi)= −1212l + 1∂3(Kl(l−1) −K(l+1)(l+2))(−1)Cltr(φiti)=1212l + 1∂3(Kl(l−1) −K(l+1)(l+2))Cltr(φiti).(A.19)• Contribution to 12tr( ˆ¯ψGi[φi, ψˆ])12tr( ˆ¯ψj(Gi)jk[φi, ψˆk]) =12(Gi)jk〈(ψ¯j)l′m′(ψk)lm〉tr(Y m′l′ [φi, Y ml ])=12(Gi)jk〈(ψ¯j)l′−m′(ψk)lm〉tr(φi[Y ml , Y −m′l′ ]).(A.20)From the propagators of fermions (1.46), we have〈(ψi)lm(ψ¯j)l′−m′〉(−1)m′ = δijδmm′δll′Sf − δll′(Gk)ij(tk)mm′Af. (A.21)The symmetric part is12(Gi)jk(−1)δjktr(φi[Y ml , Y m†l ])Sf = 0, (A.22)since tr(Gi) = 0. The antisymmetric part is12(Gi)jk(Gh)kj(th)mm′Atr(φi[Yml , Ym′†l ])=14tr({Gi, Gh})A(−1)tr(φi[Y m′†l [th, Ym′l ]])=14tr(2Inψ×nψ)A(−1)Cltr(φiti)= −12nψ12l + 1(K−lF −K l+1F )Cltr(φiti),(A.23)where nψ = 4. The total contribution is12tr( ˆ¯ψGi[φi, ψˆ]) = −12nψ12l + 1(K−lF −K l+1F )Cltr(φiti). (A.24)24Appendix BUseful expectation valuescontributed from the Ena(Ean) partIn this appendix, we calculated some useful expectation values from thecontribution of the degrees of freedom of the Ena and Ean part. They willbe useful to calculate one-loop effective action for one-point function.• Contribution to tr([ti, φˆ4][φi, φˆ4])tr([ti, φˆ4][φi, φˆ4])= 〈(φ4)†a′n′(φ4)an〉tr(φi[Ea′†n′ [ti, Ean]]) + 〈(φ4)†n′a′(φ4)na〉tr(φi[En′†a′ [ti, Ena]])= 〈(φ4)†a′n′(φ4)an〉tr(φi[En′a′ ,−Eanti]) + 〈(φ4)†n′a′(φ4)na〉tr(φi[Ea′n′ , tiEna])= 〈(φ4)†a′n′(φ4)an〉tr(φi(−δaa′En′nti + EantiEn′a′))+ 〈(φ4)†n′a′(φ4)na〉tr(φi(Ea′n′tiEna − tiδa′a Enn′))= Kk2−14 tr(φi(−2(N − k)ti))= −2(N − k)K k2−14 tr(φiti).(B.1)where we use∑n,aEantiEna = (Ean)pq(ti)qh(Ena)hl= δapδnq(ti)qhδnhδal =∑n(N − k)(ti)nn = 0.(B.2)• Contribution to tr(Aµ[φˆ4, ∂µφˆ4])Similar to our discussion for the Y ml part, the contribution is zero.• Contribution to tr(φi[φˆ1[ti, φˆ1]])tr(φi[φˆ1[ti, φˆ1]]) = −2(N − k)(k + 12kK(k−2)2−14 +k − 12kK(k+2)2−14)tr(φiti).(B.3)25Appendix B. Useful expectation values contributed from the Ena (Ean) part• Contribution to tr(φˆi[φˆj [ti, φj ]])Here i, j label the complicated fieldstr(φˆi[φˆj [ti, φj ]]) = tr([φˆi, φˆj ][ti, φj ])= 〈(φi)n′a′(φj)an〉Atr([En′a′ , Ean][ti, φj ]) + 〈(φi)a′n′(φj)na〉Atr([Ea′n′ , Ena][ti, φj ])= −iijk(tk)n′nAtr([En′a, Ean][ti, φj ]) + iijk(tk)nn′Atr([Ean′ , Ena][ti, φj ])= −2iijkAtr(tk[ti, φj ])(N − k)= −2iijkAikihtr(thφj)(N − k)= 2(N − k)(nφ,com − 1)1k(K(k−2)2−14 −K (k+2)2−14 )tr(tiφi),(B.4)where we use(ti)n′n[En′a, Ean] = (ti)n′nEn′nδaa − (ti)n′nδn′n Eaa = (N − k)ti. (B.5)• Contribution to tr(φˆi[φj [ti, φˆj ])tr(φˆi[φj [ti, φˆj ]) = tr([φˆi, φj ][ti, φˆj ]) = −tr(φj [φˆi[ti, φˆj ]]). (B.6)The symmetric part istr(φiti)(N − k)(k + 12kK(k−2)2−14 +k − 12kK(k+2)2−14). (B.7)To calculate the antisymmetric part, first we find(tk)n′n[En′a[ti, Ean]] = (tk)n′n[En′a,−Eanti]= (tk)n′n(−En′aEanti + EantiEn′a)= (tk)n′n(−δaaEn′nti + (N − k)(ti)nn′PN−k)= −(N − k)tkti + (N − k)tr(tkti)PN−k,(B.8)where(EantiEn′a)pl = (Ean)pq(ti)qh(En′a)hl= δapδnq(ti)qhδn′hδal= (N − k)(ti)nn′(PN−k)pl.(B.9)Here PN−k is the projection operator to the (N − k) × (N − k) subspace.Similarly(tk)nn′ [Ean′ [ti, Ena]] = −(N − k)titk + (N − k)tr(tkti)PN−k. (B.10)26Appendix B. Useful expectation values contributed from the Ena (Ean) partThen the antisymmetric part is− 〈(φi)n′a′(φj)an〉Atr(φj [En′a′ [ti, Ean]])− 〈(φi)a′n′(φj)na〉Atr(φj [Ea′n′ [ti, Ena]])= iijk(tk)n′ntr(φj [En′a[ti, Ean]])A− iijk(tk)nn′tr(φj [Ean′ [ti, Ena]])A= iijktr(φj [−(N − k)tkti + (N − k)tr(tkti)PN−k]− φj [−(N − k)titk + (N − k)tr(tkti)PN−k])A= −iijk(N − k)tr(φj [tk, ti])A= (N − k)(nφ,com − 1)tr(φiti)A=N − kk(nφ,com − 1)(K(k−2)2−14 −K (k+2)2−14 )tr(φiti)(B.11)• Contribution to tr(φˆi[Aˆ3[ti, A3]])tr(φˆi[Aˆ3[ti, A3]]) = tr([φˆi, Aˆ3][ti, A3])= 〈(φi)n′a′(A3)an〉tr([En′a′ , Ean][ti, A3]) + 〈(φi)a′n′(A3)na〉tr([Ea′n′ , Ena][ti, A3])=1k(K(k−2)2−14 −K (k+2)2−14)×(i(ti)n′ntr([En′a, Ean][ti, A3])− i(ti)nn′tr([Ean′ , Ena][ti, A3]))= 2i1k(K(k−2)2−14 −K (k+2)2−14)tr(ti[ti, A3])= 0.(B.12)• Contribution to tr(i[Aˆ3, φˆi]∂3φi)tr(i[Aˆ3, φˆi]∂3φi)= i〈(A3)n′a′(φi)an〉tr([En′a′ , Ean]∂3φi) + i〈(A3)a′n′(φi)na〉tr([Ea′n′ , Ena]∂3φi)= i(−i)(ti)n′ntr([En′a, Ean]∂3φi)Aii(ti)nn′tr([Ean′ , Ena]∂3φi)A= (N − k)tr(ti∂3φi)A+ (N − k)tr(ti∂3φi)A= 2N − kk(K(k−2)2−14 −K (k+2)2−14 )tr(ti∂3φi).(B.13)27Appendix B. Useful expectation values contributed from the Ena (Ean) part• Contribution to tr(i[Aˆ3, φi]∂3φˆi)tr(i[Aˆ3, φi]∂3φˆi) = itr([∂3φˆi, Aˆ3]φi]) = − i2tr(∂3[Aˆ3, φˆi]φi)= −N − kk∂3(K(k−2)2−14 −K (k+2)2−14 )tr(tiφi).(B.14)• Contribution to 12tr( ˆ¯ψGi[φi, ψˆ])12tr( ˆ¯ψj(Gi)jk[φi, ψˆk])=12(Gi)jk〈(ψ¯j)n′a′(ψk)an〉tr(En′a′ [φi, Ean])+12(Gi)jk〈(ψ¯j)a′n′(ψk)na〉tr(Ea′n′ [φi, Ena])=12(Gi)jk〈(ψ¯j)n′a′(ψk)an〉tr([En′a′ , Ean]φi)(−1)+12(Gi)jk〈(ψ¯j)a′n′(ψk)na〉tr([Ena, Ea′n′ ]φi).(B.15)The symmetric part is zero because tr(Gi) = 0. The antisymmetric part is12(Gi)jk(Gl)kj(tl)n′nAtr([En′a, Ean]φi)+12(Gi)jk(Gl)kj(tl)nn′Atr([Ena, Ean′ ]φi)= (Gi)jk(Gl)kjAtr(tlφi)(N − k)=12tr({Gi, Gl})Atr(tlφi)(N − k)= tr(Inψ×ψ)Atr(tiφi)(N − k)= nψN − kk(K− k−12F −Kk+12F )tr(tiφi).(B.16)28

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