Srednicki Chapter 34

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1 Srednck Chapter 3 QFT Problems & Solutons A. George January 0, 203 Srednck 3.. Verfy that equaton 3.6 follows from equaton 3.. We take Λ = + δω: U + δω ψu + δω = + δωψ[ + δω] x Next we use equaton 3.3, and U + δω = I + δωm: 2 I 2 δωmψi + 2 δωm = I + 2 δωs ψx xδω We do the multplcaton on the left sde, droppng the term wth two dfferentals. On the rght, we dstrbute: ψ + 2 ψδωm 2 δωmψ = Iψx xδω + 2 δωs ψx xδω On the left, we rewrte usng ndex notaton. On the rght, we expand n a Taylor Seres: ψx xδω = ψx xδω µ ψxδω: ψ + 2 δω µν[ψ, M µν ] = ψx xδω µ ψx + 2 δω µνs µν a ψ b x xδω The ψ terms cancel. Further, the last term already has a dfferental, so we keep only the leadng term n the expanson of ψx xδω: 2 δω µν[ψ, M µν ] = xδω µ ψx + 2 δω µνs µν Here comes the subtelty. Each of these terms s multpled by δω µν. Ths s an arbtrary antsymmetrc tensor. Any symmetrc coeffcents are rrelevent, snce those terms must be equal to zero. Ant-symmetrc coeffcents result n nonzero terms, and so can be equated. In ths equaton, M and S are already antsymmetrc. All that remans s to antsymmetrze the frst term on the rght sde. We do ths by wrtng the term twce: 2 δω µν[ψ, M µν ] = 2 xν δω µν µ ψx 2 xν δω µν µ ψx + 2 δω µνs µν

2 Now swtch the dummy ndces n the second term: 2 δω µν[ψ, M µν ] = 2 xν δω µν µ ψx 2 xµ δω νµ ν ψx + 2 δω µνs µν Now we swtch them back n the δω. Ths costs a mnus sgn, due to ant-symmetry: Ths s: 2 δω µν[ψ, M µν ] = 2 xν δω µν µ ψx + 2 xµ δω µν ν ψx + 2 δω µνs µν [ 2 δω µν[ψ, M µν ] = 2 xν µ ψx + ] 2 xµ ν ψx δω µν + 2 δω µνs µν a b ψ b x Now all the coeffcents of δω are antsymmetrc, so we can equate them: Fnally, we multply by 2 : 2 [ψ, M µν ] = 2 xν µ ψx + 2 xµ ν ψx + 2 Sµν a b ψ b x [ψ, M µν ] = xν µ x µ ν ψx + S µν Usng the defnton of µν, and nsertng the remanng atn ndces for consstency: whch s equaton 3.6. [ψ a, M µν ] = µν ψ a x + S µν Srednck 3.2. Verfy that equatons 3.9 and 3.0 obey equaton 3.. Usng equaton 3.9: [S j, Smn ] = εjk ε mno [σ k, σ o ] et k be ether m or n we can neglect the case where k s o, snce the commutator wll vansh. Then: [S j, Smn ] = ε jm ε mno [σ m, σ o ] + ε jn ε mno [σ n, σ o ] Smlarly, we let o be ether or j: [S j, Smn ] = ε jm ε mn [σ m, σ ] + ε jn ε mn [σ n, σ ] + ε jm ε mnj [σ m, σ j ] + ε jn ε mnj [σ n, σ j ] Playng wth the ndces: [S j, Smn ] = ε jm ε nm [σ m, σ ] + ε jn ε mn [σ n, σ ] + ε jm ε njm [σ m, σ j ] ε jn ε mjn [σ n, σ j ] Now ths gets a lttle bt subtle. Note that we cannot use the summed propertes of the ev-cevta symbol such as ε abc ε dbc = 2δ ad, because these ndces are not beng summed 2

3 over. The ndces of the ev-cevta symbol are set once and for all by the choces on the left hand sde. Instead, consder the frst term. There are only three spatal dmensons, so, j, and m must be, 2, or 3. Duplcates are not allowed by the defnton of the ev-cevta symbol. It follows then that j must equal n. Snce the two symbols are dentcal, they can only yeld a postve one or zero. Notce, therefore, that all the nformaton of the ev-cevta symbols s contaned n ths statement: the term s nonzero only f j = n, 2 j, and 3 m. We can repeat ths analyss for all terms: the ndces n condtons and 3 wll change, condton 2 wll reman the same. Note that 3 s redundant: ths nformaton s already encoded n the commutator. for, we can more easly state that n a δ. et s try to do so: [S j, Smn ] = δ jn [σ m, σ ] + δ jm [σ n, σ ] + δ n [σ m, σ j ] δ m [σ n, σ j ] As How do we mpose the remanng condton, that j? It s already taken care of: f = j, the frst and thrd terms cancel, as do the thrd and fourth. Hence, we get a nonzero result f and only f j, just as requred. Next, recall that [σ a, σ b ] = 2ε abc σ c = S ab. So: [S j, Smn ] = δ jn S m + δ jm S n + δ n S mj δm S nj Fnally, recall that for spatal atn ndces, g = δ. Also, the S matrces are antsymmetrc. Thus: [S j, Smn ] = g m S nj gjm S n g n S jm + gjn S m whch s equaton 3. for spatal ndces. Next we use equaton 3.0: [S k0, S l0 ] = [σ k, σ l ] The answer from 3. s: [S k0, S l0 ] = g kl S 00 [S k0, S l0 ] = 2 εklm σ m [S k0, S l0 ] = S kl g 0l S k0 g k0 S 0l + g 00 S kl The frst term vanshes snce S s antsymmetrc. The mddle terms vansh because g s dagonal recall that atn ndces represent spatal ndces. g 00 =. Hence, So those match up. [S k0, S l0 ] = S kl 3

4 Fnally, we must consder the case: [S j, S k0 ] = εjl [σ l, σ k ] = 2 εjl ε lkm σ m = 2 εjl ε lk σ 2 εjl ε lkj σ j = 2 εjl ε kl σ 2 εjl ε kjl σ j As before, we replace the two ε functons wth logcal requrements, and fnd that these can be better mplemented wth delta functons. Specfcally: [S j, S k0 ] = 2 δ jk σ δ k σ j = g k S j0 gjk S 0 Is ths what we would expect? Recall that, j, k represent spatal ndces only; g s dagonal, so any terms nvolvng g 0, g j0, or g k0 must vansh. Equaton 3. therefore gves exactly ths. Srednck 3.3. Show that the ev-cevta symbol obeys: ε µνρσ ε αβγσ = δ µ αδ ν βδ ρ γ δ µ β δν γδ ρ α δ µ γδ ν αδ ρ β + δµ β δν αδ ρ γ + δ µ αδ ν γδ ρ β + δµ γδ ν βδ ρ α ε µνρσ ε αβρσ = 2δ µ αδ ν β δ µ β δν α ε µνρσ ε ανρσ = 6δ µ α For the frst: f µνρ s an even permutaton of αβγ, then we wll get - superscrpts are nverse of subscrpts. An odd permutaton wll gve +. Ths s exactly what the frst lne shows. For the second: the same logc holds, except ths tme ρ and σ can be exchanged, gvng a factor of 2. For the thrd: the same logc holds, except ths tme ν, ρ, and σ can be exchanged, gvng a factor of 6. Srednck 3.. Consder a feld C a...c ȧ...ċ x, wth N undotted ndces and M dotted ndces, that s furthermore symmetrc on exchange of any par of undotted ndces, and also symmetrc on exchange of any par of dotted ndces. Show that ths feld corresponds to a sngle rreducble representaton 2n+, 2n + of the orentz Group, and dentfy n and n. et s start by recognzng what ths s. We re beng asked to deal wth the followng: 2, 2,..., 2, 2... Ths s, of course, equal to: ,

5 where there are N terms n the frst part and M n the second. Next we need to determne the number of rreducble representatons of ths, and the dmensonalty thereof. To determne the rreducble represntatons, we use Young Tableaux for a more complete ntroducton, see Sakura secton 6.5. In a young tableaux, each box represents one ndex, whch n SU2 can be 0 or. Boxes are then combned: a horzontal strng of boxes represents a symmetrc state; a vertcal strng of boxes represents an ant-symmetrc state; boxes n any other arrangement represent a mxed symmetry. To determne 2 2, then, we have: = The frst of these s symmetrc and the second s antsymmetrc. What about the dmensonalty? The rule s that rows must be nondecreasng; the columns must ncrease. So the symmetrc dagram s a trplet, t can be 0 0, 0, or. The antsymmetrc dagram can only be 0. Thus, 2 2 = 3 S A. Smlarly, let s determne Ths s: = = = Notce that s not a legal dagram s SU2, per the rules above. What about the dmensonalty? The frst can be 0 0 0, 0 0, 0, or. The others can be 0 0 or 0. Thus, = S + 2 M + 2 M. In the case at hand, we are absolutely requred to consder only the symmetrc case. The symmetrc case wll have N boxes n a horzontal row. As n the examples above, ths horzontal row wll have dmensonalty N+, snce there are exactly N+ ways to buld the box wth no s, wth one,..., wth all s. Hence, the dmensonalty s N +, M +. It s obvous that N = 2n and M = 2n. It remans to show that ths representaton s rreducble. But that s the whole pont of Young Tableaux: when all the symbols are replaced by, the remanng Tableaux are a pror rreducble. 5