� � Lecture 18 8.251 Spring 2007 Lecture 18 -Topics • Open Strings Still for open string:Heisenberg operators: XI (τ,σ), x−0 , PτI (σ), p+[XI (σ), PτJ (τ,σ�)] = iηIJ δ(σ − σ�) [x−0 ,p +]= −i ∂∂=2α�p + +2α�p + p− = H ∂τ ∂X+ ⇔ � � � Hamiltonian, H 1 ∂X− p− = dσ(P−τ =2πα� ∂τ ) H =2α�p+p− = L+0 from analysis of classical string Are we sure H =2α�p+p−? After all, p− is the product of lots of operators, which can be ill-defined. Must be careful in our quantum case. X¨ I − XI�� =0 IXI (τ,σ)= x0 + √2α�α0I τ + i√2α� � 1 αIn cos(nσ)e−inτ n n=0 1 ∂xJ τJ =P2πα� ∂τ (X˙I + XI �)(τ,σ)= √2α� � αnI e(−in(τ +σ)) σ ∈ [0,π] (1) n∈Z 1 Lecture 18 8.251 Spring 2007 XI αI (−in(τ −σ))(˙− XI �)(τ,σ)= √2α� � eσ ∈ [0,π] (2)nn∈Z This is an important computation. Later, we will do this for closed strings too, and we’ll see very similar (though not same). Best way to select Fourier modes is in [0, 2π] but σ ∈ [0,π]. σ →−σ (X˙I − XI �)(τ, −σ)= √2α� � αI e(−in(τ+σ)) (2’)nn∈Z This makes sense when σ ∈ [−π, 0]. αI (−in(τ +σ))AI (τ,σ)= √2α� � eσ ∈ [−π,π]n� n∈Z (X˙I + XI �)(τ,σ) σ ∈ [0,π]= (X˙I − XI �) σ ∈ [−π, 0] Now have σ defined over [−π,π]. [XI (τ,σ),X˙I (τ,σ�)] = 2πα�iηIJ δ(σ − σ�) [X˙I (τ,σ),X˙J (τ,σ�)] = 0 [XI �(τ,σ),XJ �(τ,σ�)] = 0 X’s commute at different σ’s so can then differentiate. [(X˙I ± XI �)(τ,σ), (X˙J ± XJ �)(τ,σ)] = [( X˙I ± XI �)(τ,σ), (X˙J ± XJ �)(τ,σ)] = ±4πα�iηIJ d (σ − σ�)dσ 2 � � Lecture 18 8.251 Spring 2007 [AI (τ,σ),AJ (τ,σ�)] = 2α� e(−im�(τ+σ))e(−in�(τ +σ� ))[αmI � ,αnJ � ] m�,n� ⎧ ⎨ 4πα�iηIJ d δ(σ − σ�) σ,σ� ∈ [0,π]dσ 4πα�iηIJ d= ⎩ dσ δ(σ − σ�)=0 σ ∈ [0,π],σ� ∈ [−π, 0] −4πα�iηIJ d δ(σ� − σ)=4πα�iηIJ d δ(σ − σ�) σ,σ� ∈ [−π, 0]d(−σ) d(σ) e(−im�(τ +σ))e(−in�(τ+σ�))[αIm� ,αJ ]=2πiηIJ dδ(σ − σ�) σ,σ� ∈ [−π,π]n� dσ m�,n� Apply the following integral operations: 1 � π dσe(imσ) 1 � σ dσ�e(inσ) 2π · 2π−π −σ Divide by e(−i(m+n)τ) on both sides: [αI ,αI ]= −nηIJ δm+n,0e(i(m+n)τ ) mn[αI ,αI ]= mδm+n,0ηIJ mnCommutation relation proved in book: [x0I ,p J ]= iηIJ Note: αI = √2α�p I 0 [αI ,αI ]= mηIJ δm,nmnαµ = aµ√n n> 0nn 3 � � � � Lecture 18 8.251 Spring 2007 αµ = aµ+√n =(αµ )+ n< 0−nn +nOpposite signs for m and n [amI ,a nJ ]=0 [amI+ ,a nJ+]=0 m> 0,n> 0: IJ[a √m,a √n]= mηIJ δm,nmn [amI ,a mJ+]= ηIJ δm,n +√2α�αn−=1 Ln⊥n=0 2pp− =1 L0⊥p+ → α� 1 � L⊥= αI αI n 2 n−pp p∈Z Don’t have to worry if n = 0. Might have to worry if �n = 0. But what we want is: H = L⊥=2α�p+p−.= 1 αI αI but α’s don’t 0 L0⊥2 p∈Z −pp commute so don’t know if this is right. 1 M2 = −p 2 =2p + p− − p I p I = L⊥0 − p I p I α� 11 ∞L⊥= αI 0αI (αI αI + αpαI )0 2 0 +2 −pp −pp=1∞1 = α�p I p I + α−pαI + 2(D − 2) pp p=1 p=1 Note αp>0 is destruction operation convention. αp<0 is creation operation conventtion 4 � � � � � � � � � � Lecture 18 8.251 Spring 2007 1 � ∞1 ∞� M2 I+ I= α� pa a + 2(D − 2) ppp p=1 p=1 In classical theory, had 1 � ∞1 ∞� M2 I+ I= na a + 2(D − 2) pα� nn n=1 p=1 Showed all states of string had mass ¿ 0. Couldn’t get anything intersting withoou mass. ∞Would be great here if 1 (D − 2) p=1 p = −1. Then: 2 M2 = α1 � ( � na In + anI − 1) Now want oscillation states without mass ∞1 p =1+2+3+4+ ... = −12 p=1 Crazy, huh? Not true in general, of course, but almost true in one sense. Since we want: 1 ∞2(D − 2) p = −1 p=1 11 2(D − 2) − 12 = −1 ⇒ D = 26(dimension of string) Now how is ∞ 1 ?! p=1 p = − 12 Recall Riemann Zeta Function: 5 � � � � Lecture 18 8.251 Spring 2007 ∞1 ζ(s)= ns n=1 1 ∞1 ∞ζ(s = −1) = − == n12 n−1 n=1 n=1 ζ(s) well-defined and convergent for s ≥ 2. Doesn’t converge for s = 1(pole). ζ defined on complex plane. The beauty of analytic functions: If you know it is defined in a very small finite regin, you know it everywhere by the Cauchy-Riemann. 12p + p− = α� (L⊥0 + a) a = constant Define for once and for all: 1 ∞L⊥0 =2 α0I α0 I + α−IpαpI p=1 [M−I (a,D),M−I (a,D)] = 0 Set standards of messy computation. All books omit at least some details. M−J αJ ≈ [L+,L+ ]=(m − n)L+ + dim. of spacetime ≈ α−nmnmm+n So need to find algebra of Viroso operators. 6