S[Ψ]/V 26. Ψ λ=1. Schnabl s solution. 1 2π 2 g 2

Transcription

1

2

3 Schnabl s solution Ψ λ=1 S[Ψ]/V 26 Ψ λ=1 Ψ λ=1 Ψ 1 2π 2 g 2

4 S[Ψ] = 1 g 2 ( 1 2 Ψ,QΨ Ψ, Ψ Ψ ) Ψ = φ(x)c A μ (x)α μ 1 c ib(x)c ( dz Q = ct m + bc c + 3 ) 2πi 2 2 c QΨ + Ψ Ψ =0 δ Λ Ψ=QΛ + Ψ Λ Λ Ψ

5 O V (Ψ) = I V (i) Ψ = Φ V, Ψ V (i) = c(i)c( i)v m (i, i) matter primary, dim (1,1) I V (i) On-shell closed string state: Φ V = V (i) I

6 O V (Ψ) QΦ V =0, Φ V, Ψ Λ = Φ V, Λ Ψ on-shell V: dim (0,0), midpoint O V (δ Λ Ψ) = 0 O V (e Λ Qe Λ )=0

7 Ψ λ = = λ r f n (λ) ψ λe r r=0 = n r 1 r n! ψ r r=0 n=0 ( ) N lim ψ N+1 r ψ r r=n (λ =1) N n=0 λ n+1 r ψ r r=n (λ 1) n=0 ψ r 2 π U r+2 U r+2 [ 1 ] π (B 0 + B 0 ) c(πr 4 ) c( πr 4 )+1 2 ( c( πr 4 )+ c(πr 4 )) 0 U r+2 = ( 2 r +2 ) L0

8 i z arctan z = z i z φ( z) = (cos z) 2h φ(tan z) for primary with dim h B 0 = b 0 + k=1 2( 1) k+1 4k 2 1 b 2k L 0 = {Q, B 0 } = L 0 + k=1 2( 1) k+1 4k 2 1 L 2k

9 S[Ψ λ ]/V 26 = 1 2π 2 g 2 (λ =1) 0 ( λ < 1)

10 O V (Ψ) O V (Ψ λ ) = Φ V, Ψ λ = I Φ V Ψ λ

11 Φ V = ζ mn c(i)v m (i)c( i)v n ( i) I m,n = m,n ζ mn U 1 U 1 c(i )Ṽ m (i ) c( i )Ṽ n ( i ) 0 ±i ±im Φ V,M m,n ζ mn U 1 U 1 c(im)ṽ m (im) c( im)ṽ n ( im) 0

12 U r U r φ 1 ( x 1 ) φ n ( x n ) 0 U s U s ψ 1 (ỹ 1 ) ψ m (ỹ m ) 0 = U r+s 1 U r+s 1 φ 1 ( x 1 ) φ n ( x n ) ψ 1 (ỹ 1 ) ψ m (ỹ m ) 0 ((B 0 + B 0 )Ψ 1) Ψ 2 =(B 0 + B 0 )(Ψ 1 Ψ 2 )+( 1) Ψ 1 π 2 Ψ 1 B 1 Ψ 2,

13 V m (y)v n (z) ψ r v mn + finite (y z) (y z) 2 Φ V,M,ψ r = C ( V sinh 4M 2πi r +1 4M π sin π )( cosh 4M r +1 r +1 cos = mat 0 0 mat ζ mn v mn. C V Φ V,ψ r = m,n lim Φ V,M,ψ r = C V M + 2πi π r +1 )( sinh 4M ) 2, r +1 independent of r O V (Ψ λ )= k=0 f k (λ) k r k! Φ V,ψ r r=0 = f 0 (λ) Φ V,ψ 0 = { CV 2πi (λ =1) 0 (λ = 1)

14 λ = 1 O V (Ψ λ=1 )= lim N ( Φ V,ψ N+1 N n=0 r Φ V,ψ r r=n ) lim N ( ) lim Φ V,M,ψ N+1 M + ( ) lim lim Φ V,M,ψ N+1 M + N = 0

15 λ = 1 Ψ λ=1 = lim N = lim N = ψ 0 + ( ( ψ N+1 ψ 0 + N n=0 N n=0 r ψ r r=n ) (ψ n+1 ψ n r ψ r r=n ) (ψ n+1 ψ n r ψ r r=n ) n=0 )

16 Φ k = 1 4i lim θ π 2 c(e iθ )c(e iθ ):e ik X(eiθ,e iθ) : I = 1 4 ee m+e gh c 0 c 1 0 E m = ( 1) n 2n α n α n n=1 E gh = ( 1) n c n b n n=1 n=1 2i 2α ( 1) n k i α i 2n+1 2n 1, k i k i k i =4/α Q Φ k = 0

17 Φ η = = E = 1 52α i η μν lim c(e iθ ) X μ (e iθ )c(e iθ ) X ν (e iθ ) I θ π 2 ( ) (m n)π mn cos α m α n e E c 0 c 1 0, 13 2 n,m=1 ( ( 1) n 1 ) 2n α n α n + c n b n n=1 Q Φ η = 0

18 V c = c cv m 0 ˆγ(1 c, 2) V c 1c = 2 Φ V V m = e ik ix i Φ k V m = 1 X X 26 Φ η

19 ˆγ(1 c, 2) φ c 1c ψ 2 = h 1 [φ c (0, 0)]h 2 [ψ(0)] h i M h i O h 1 (w) = i w 1 w +1 h 2 (w) = 1 2 (w 1 w )

20 ˆγ(1 c, 2) (Q (1) c + Q (1) c + Q (2) )=0 ˆγ(1 c, 2) (L (2) ( = ˆγ(1 c, 2) 2m 1 + L(2) 2m+1 ) 4(2m 1)i( 1) m (L (1) 0 L (1) 0 ) ) (f 2m 1,k L (1) k 1 + f 2m 1,k L (1) k 1 k=2 ˆγ(1 c, 2) (L (2) ( = ˆγ(1 c, 2) 2m L(2) 2m ) ( 1) m m c 2 8m( 1)m (L (1) 0 + L (1) 0 ) ) (f 2m,k L (1) k 1 + f 2m,k L (1) k 1 k=2

21 ψ r ψ r 2 = k=1, + r 2π 2 e u 2k(r)L 2k ( sin 2π r [ 1 π sin 2π r ) 2 s 2;s:even ( 1 r 2π sin 2π r ) ( 1) s 2 +1 ( ) 2 s s 2 1 r p 1;p:odd p,q 1;p+q:odd ( 2 r cot π r ) p c p 0 ( 2 ( 1) q r cot π ) p+q ] b s c p c q 0 r u 2 (r) = r2 4 3r 2, u 4(r) = r r 4, u 6(r) = 16(r2 4)(r 2 1)(r 2 +5) 945r 6,... ψ N+1 = O(N 3 ) (N )

22 LMathematica O η (Ψ λ,l )= λ n+1 r Φ η,ψ r,l r=n n=0 1 λ 1 λ = 1

23 O L=8 L=10 L=12 L=14 L=0 L=2 L=4 L=

24 0.040 L= L= L=4 L= L=12 L=14 L=8 L=

25 0.16 L=12 L=14 L=6 L=8 L=10 L=4 L=2 L=

26 O η (Ψ λ ) = L 1 2π (λ =1) 0 (λ 1) O η (Ψ λ=1,l )

27 S[Ψ λ=1 ]/(V 26 T 25 ) S[Ψ λ ]/(V 26 T 25 ) λ

28 b 0 Ψ N = 0 O η (Ψ N ) O η (Ψ N ) 1 2π = O η (Ψ N ) O η (Ψ λ=1 )

29 Table 1 2π 2 g 2 S[Ψ N ]/V 26 2π 2 g 2 S[Ψ N ]/V 26 O η (Ψ N )

30 L 0 L O k (Ψ λ L ) = O η (Ψ λ L ) O k (Ψ N L ) = O η (Ψ N L ) ψ L = O k (ψ L ) = O η (ψ L ) p,q 0,n i 2,j i 1,k i 0 n 1 + +np+j 1 + +j l +k 1 + +kq =L C (L) n i,j i,k i L (m) n 1 L (m) n p b j1 b jq c k1 c kq c 1 0 (L (m) 2n L(m) 2n ) Φ c cv m =( 1) n 3n Φ c cvm, (L (m) 2n 1 + L(m) 2n+1 ) Φ c cv m = 0

31 Ψ λ λ = 1 { { 1 1 (λ =1) (λ =1) 2π 2 g 2 O 0 ( λ < 1) k/η (Ψ λ )= 2π 0 ( 1 λ<1) S[Ψ λ ]/V 26 = Ψ λ=1 Ψ λ <1 Ψ λ=1 Ψ N S[Ψ λ=1 ] S[Ψ N ] O k/η (Ψ λ=1 ) O k/η (Ψ N ) Ψ λ=1 Ψ N

32 O V (Ψ λ=1 )= ˆγ(1 c, 2) ψ 0 2 c (1) 1 c (1) 1 V m 1c B N c 0 c 1 c 1 V m ˆγ(1 c, 2) ψ 0 2 P 1c = 1 2π B N c 0 O V (Ψ) = A disk Ψ (V ) Adisk 0 (V )

33 ˆγ(1 c, 2) Ψ λ=1 2 Pb 0 = 1 2π B N + ˆγ(1 c, 2) χ 2 Pb 0 Ψ λ=1 = ψ 0 + (ψ n+1 ψ n r ψ r r=n ) ψ 0 + χ n=0 Q( Ψ λ=1 ) + ( Ψ λ=1 ) ( Ψ λ=1 ) = 0 Q Q +ad Ψλ=1 Ψ λ=1