Contributions à l'Etude du Vertex Topologique en Théorie ... - Toubkal

342 L.B. Drissi et al. / Nuclear Physics B 801 [FS] (2008) 316–345By using the fact that q L 0 acts as translation operator on Γ − (z), we get([ s∏Γ −(2)( ((z) = lim Γ− q k z ))] )q sL 0.s→∞k=0(A.17)For simplicity, we consider the action of Γ − (2) (z), on the vacuum, that reads as( (z)|0〉= ∏ ∞(Γ − q k z )) |0〉.(A.18)Γ (2)−k=0Substituting Γ − (q k z) by its expression (A.13), we find(Γ − (2) (z)|0〉=exp i ∑ )∞∑ q knn zn J −n |0〉n>0 k=0(= exp i ∑ 1 z n )n (1 − q n ) J −n |0〉.n>0(A.19)Notice that Γ − (2) (z)|0〉 can be decomposed as follows, (Γ − (2) (z)|0〉=exp i ∑ ) (1(A.20)n zn J −n exp i ∑ ∞ )∑q n q knn zn J −n |0〉n>0n>0 k=0or equivalently like(Γ − (2) (z)|0〉=exp i ∑ ) (1n zn J −n exp i ∑ 1 (qz) n )n (1 − q n ) J −n |0〉n>0n>0(A.21)showing that we have:Γ (2)−(1) (2)(z)|0〉=Γ (z)Γ (qz)|0〉, z∈ C.−t 2 =1−(b) Case Γ −(3) (z)Here, we have∞Γ −(3) (z)|0〉= ∏ [ (2) ] ∞∏ (Γ − (z)qL 0 (2)( |0〉= Γ − q k z )) |0〉.k=0Substituting Γ −(2) (qk z) by its expression (A.2), we find(Γ − (3) (z)|0〉=exp i ∑ )1∞∑q kn z nn (1 − q n ) J −n |0〉n>0 k=0(= exp i ∑ 1 z n )n (1 − q n ) 2 J −n |0〉.n>0Finally, if we rewrite the above relation as follows( ∑Γ − (3) (z)|0〉=exp iz n ) (J −n ∑ i(qz) n )J −nn(1 − q n exp) n(1 − q n ) 2 |0〉n>0n>0(A.22)(A.23)(A.24)(A.25)