String Theory- Closed Strings I

Lecture 20 8.251 Spring 2007 Lecture 20 -Topics • Closed Strings Recall: For closed strings in light cone gauge: σ ≈ σ +2π x + = α�p +τ Pτ +constant X˙± X�− = 12p1 + (X˙I ± X�I )2 α� H = α�p + p− τµ 1 X˙µP=2πα� Open Strings: [XI (τ,σ), Pτ,J (τ,σ�)] = iηIJ δ(σ − σ�) X˙I ± X�I = √2α� � αne(−in(τ±σ)) [αI ,αI ]= mδm+n,0δIJ mnGraviton States: IJ+ξIJ a p+,pτ |Ω�ξIJ = ξJI = ξJI =0Solve and Find Mode Expansion of Closed Strings Xµ(τ,σ)= Xµ(τ + σ)+ Xµ LR(τ − σ) Left and Right (XLµ Rand Xµ ) both solve wave equation, as goes their sum. Let: u = τ + σ, v = τ − σ 1 � � � � � Lecture 20 8.251 Spring 2007 xµ(τ,σ +2π)= xµ(τ,σ)True σ ≈ σ +2π except when the wold has a compact dimension and x goesaroudn a circle -even though back at same σ after 2π, at a different x coordinate.XL(u)+ XR(v)= XL(u +2π)+ XR(v − 2π)XR(v) − XR(v − 2π)= XL(u +2π) − XL(u) This is the periodicity condition. XL and XR are independent variables. XL�and X�are periodic. R XL�µ(u)= α� � αnµe(−inu) 2 n∈Z X�µ α� � αµ (−inv)(v)= eRn2 n∈Z Have 2 independent sets of oscillators. Not related to the open string oscillators. XLµ(u)= 21 XLµ 0 + α2 � α0µu + i√... 1 α�Xµ (v)= Xµ + αµ 0 v + ... RR022 All terms in Xµ and Xµ have e−inu component except first two terms of each. LR Periodicity Condition: αµ 0 = α0 µ∀µ 1(Xµ + Xµ α�Xµ(τ,σ)= L0 R0)+ √2α�αµ 0 τ + i � ... 22 2 � Lecture 20 8.251 Spring 2007 Let: xµ = 12 (Xµ + Xµ )0 L0 R0Momentum of String: � 2πpµ = τµdσP0 � 2π1 ∂xµ = dσ2πα� 0 ∂τ =1 √2α�αµ(2π)2πα� 0 2 = αµ α� 0 X˙µ = Xµ�(τ + σ)+ Xµ �(τ − σ)LRXµ� = Xµ�(τ + σ) − Xµ LR�(τ − σ) X˙µ + Xµ� =2Xµ� = √2α� � n ∈ Zαµe(−in(τ +σ)) LnX˙µ − Xµ� =2Xµ � = √2α� � n ∈ Zαµe(−in(τ−σ)) Rn� + 1 � (X˙I + XI �)2 =4α� Ln e(−in(τ +σ)) L⊥ n = αpI αnI −p⇒ 2 p � + 1 � (−in(τ −σ))(X˙I − XI �)2 =4α� Ln e⇒ L⊥ n =2 αIpαnI −pp˙11 � L+(−in(τ +σ))X− + X−� =4α� n eα� 2p+ 2 � + = Le(−in(τ +σ))p+ n= √2α� � α−n e(−in(τ +σ))X˙− − X−� =2 � L⊥n e(−in(τ −σ)) p+ 3 � � � Lecture 20 8.251 Spring 2007 √2α�α−=2 L⊥ nn p+ √2α�α−n =2 L⊥n p+ |L⊥0 = L⊥o | constraint on state space of theory. Differences between closed and open string. Hilber space: can’t just double everything for closed strings. X0 doesn’t double and momentum doesn’t double. L⊥0 = L⊥0 constant. L⊥0 = 12 αI 0αI 0 + N⊥ ← number operator = ∞ p=1 αI αpI −pL⊥0 = α4 � pI pI + N⊥ (For open strings, did not have factor 14 L⊥= α� pI pI + N⊥, N⊥ = � αI αI 04 p=1 −pp L⊥0 = L⊥ 0 ⇒ N⊥ = N⊥ Recall: √2α�α−= 2(L⊥ − 1)nn p+ √2α�α−= 2(L⊥− 1)nn p+ 1 +√2α�α−0 = p+ (L⊥0 + L0 − 2) = α�p− +H = α�pp− = L⊥0 + L⊥0 − 2 2 M2 = N⊥ + N⊥ − 2 α� Recall open string: M2 = 1 (−1+ N⊥)α� ∂xI [L⊥+ L⊥0 ,XI (τ,σ)] = −i ∂τ 0 4 � � � ��� ��� ��� ��� Lecture 20 8.251 Spring 2007 ∂xI [L⊥0 ,XI (τ,σ)] = i0 − L⊥∂τ ∂XI XI (τ,σ+) = XI (τ,σ)+ ∂σ = XI (τ,σ)+[−i�(L⊥0 − L⊥0 ),XI (τ,σ)] ∞25n=1 I=2 m=1 J=2 N⊥ = N⊥ +4Ground state: N⊥ = N⊥ = 0, |p,pτ �, M2 = − . Close string tachyon. Not α�well understood. Next state: M = RIJ a1 I+ a1 J+ |p+,pτ �, N⊥ = N⊥ = 1, M2 = 2 (1 + 1 − 2) = 0. � �αWe have a (D − 2)x(D − 2) matrix. Any matrix can be split into symmetric and antisymmetric: RIJ = SIJ + AIJ R + RT S = 2 R − RT A = 2 11 RIJ =(SIJ − D − 2 δIJ SK )+( D − 2 δIJ SK KK + AIJ )= SˆIJ + S�δIJ + AIJ∞25(anI+)λn,I λm,J + p ,pτ + p ,pτ I+ J+SˆIJ a I+ J+ I+ I++ + p ,pτ + S�ap ,pτ + AIJ a1 aM = a a1 1 1 1 1 ˆIJ SIJ ↔ ξIJ ap+ pτ |Ω� graviton states AIJ : “Kalbra-mon” statesS� just one single state. No Lorentz index. Massless scalar. A real troublemaker.Called dilation scalar. Tells us how strong strings interact.5