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� � � � � � Lecture 22 8.251 Spring 2007 Last Time: dynamical variables needed for describing superstrings ψI α=1,2 ψ1 I = ΨI 1(τ − σ) ψ2 I = ΨI 2(τ + σ) BCs: ψ1 I (τ, σ )δψ1 I (τ, σ ) − ψ2 I (τ, σ )δψ2 I (τ, σ ) = 0 ∗ ∗ ∗ ∗Suppose: ψI (τ, 0) = 0 then ΨI (τ) = 0 by ψI = ΨI (τ − σ). Bad! Instead relate 1 1 1 ψ1 and ψ2, assemble full spin or ΨI (τ, σ) ΨI (τ, σ) = ψψ1 II (τ, σ) σ ∈ [0, π] (τ, −σ) σ ∈ [−π, 0]2 ΨI continuous at σ = 0 because of BC ψ1(τ, π) = ±ψI (τ, π).1 2 Either periodic or antiperiodic. Take periodic and get Rammond. (Actually more complicated). Take antiperiodic and get Neven Schwarz BCS (2 years after Rammond) NS BC: ΨI (τ, π) = −ΨI (τ, −π). ΨI (τ, σ) = bI e(−ir(τ −σ)).rr∈Z+1 2 Creation Operations: bI , bI , bI −5/2 −3/2 −1/2 Destruction Operations: bI 5/2, bI 3/2, bI 1/2 ΨI is anticommutative, all bIr operations anticommutative. bI , bJ = δr+s,0δIJ r s NS State: � 9�� 9�∞(αI )λn,I (bI )ρr,I +, �pτ−n −r·|NS� ⊗ p I=2 n=1 J=2 r=1232, 52, 1� � � �� � � � Lecture 22 8.251 Spring 2007 ρr,I ∈ {0, 1} Recall for open bosonic string, normal ordered: 1 � 1 �� M2 αI αI= α� 2 −p p p∈Z 12 Now: 1 � 1 � 1 �� M2 = αI αI + rbI bI −p p 2 −r r2α� p∈Zr∈Z+1 � 1 � rbI bI rbI bI 2 −r r r −r ,−31 2= − 231r=− r= ,2 2 2 1 � = rbI bI 2 −r r r 1 1 3 5 = − 2(D − 2) 2+2+2+ . . . 1 12 1 1 1 = − 2(D − 2)2 12 1 = − 48(D − 2) For boson, aB 1 .= − 24 In open bosonic string, M2 = 1 (. . . + 1) where a = −1 but 24 contributions so α�1 aB = − 24 . 1Here, aNS = − 48 for antiperiodic fermion. 1 ∞M2 = ( αI αI + rbI bI + (D − 2)(aB + aNS ))α −p p −r r p=1 r= 31 2,2 2� ���� � Lecture 22 8.251 Spring 2007 1 1 M2 = (N⊥α� tot − 2) Add text here. In early 1970s, confusion over whether these are bosons or fermions. It’ll turn out that these are photons. Count states of a given N⊥ Given: ∞+ n a1 : f(x) = a(n) x n=0 number of states with N⊥=n |0N⊥ = 0, a +|0N⊥ = 1, (a +)2|0N⊥ = 2 1 1 1 f1(x) = 1 + x + x 2 + . . . = 1 − x Given: 1+ a2 : f2(x) = 1 + x 2 + x 4 + . . . = 1 − x2 + +|0N⊥ = 0, a2 |0N⊥ = 2, (a2 )2|0N⊥ = 4 1 f1(x) = 1 + x + x 2 + . . . = 1 − x Given: 1 1+ + a1 , a 2 : f1(x)f2(x) = 1 − x · 1 − x2 + + +Now can do full open bosonic string with a1 , a 2 , a 3 , . . . Generating Function: ∞1 fos = (1 − xn)n=1 3� � � � � � ���� � Lecture 22 8.251 Spring 2007 ∞1 ∞fos = = p(n) x n (1 − xn) ���� n=1 n=0 partitions of n Partitions of 4: ({4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}) = number of ways to get N⊥ = 4 Nln p(N) ≈ 2π 6 1 N p(N) ≈ 4N√3 exp(2π 6) M For full open string: ∞1 fos = (1 − xn)24 n=1 This gives you degeneracy of any level of open string. 2 1 Given b+1 : f1(x) = 1 + x Given b+: f2(x) = 1 + x2 Given b+1 , b+2 : f12(x) = f1(x)f2(x) = (1 + x)(1 + x2) Given b+ (x) = 1 + √x: f2 12fNS (x) = a(r) x r = 1−12+ 8 x 0 + 36 x 12· · + (#)x1 r # of states with α� M2=r � n−11 ∞� 1 + x�8 = √x 1 − xn n=1 Ramond: ΨI (τ, σ) = dI exp(−in(τ − σ)). dI , dI = δm+n,0δIJ .n m n dI 0 → 8: 4 creation and 4 destruction = ξ1, ξ2, ξ3, ξ4 4� � � � Lecture 22 8.251 Spring 2007 Vacuum State: |0� |0�: 1 ξI ξJ 0�: 6 ξ1ξ2ξ|3ξ4 |0�: 1 This yields 8, |R1 a�. ξI |0�: 4 ξI ξJ ξK |0�: 4 This yields 8, |R2 a�. 8 + 8 = 16 gound states. Total set of vacua states: �RA , A = 1 . . . 16, split into 2 types. Ramond mass formula: 1 ∞ ∞M2 = ( αI αpI + ndI dIn)α�−p −np=1 n=1 1Substraction constant is equal to zero since aR = 24 α�M2 = 0 R1 a� |R2 a�αI Ra Raα�M2 = 1 αI Ra|−1 |1 �,d−I 1 |Ra 2 � −1 |2 �,d−I 1 |1 � Why is this supersymmetry? Left and right columns have opposite fermionic states. Don’t know if Ra 1 is a boston or a fermion, but know R2 a is the opposite. No bosons that look like |R1 a� in real world Partition function in Raman Sector � n∞� 1 + x�8 fR(x) = 16 1 − xn n=1 5� � � � Lecture 22 8.251 Spring 2007 To get supersymmtry, throw out half of the states from each sector and put them together. 1 + xn−12 �8 �∞�8�∞1 1 − x 1−n2 1 − xn ftruncated = NS 2√x 1 − xn − n=1 n=1 Anything with an odd number of fermions will change the sign. � n ?∞� 1 + x�8 fNS = 8 1 − xn n=1 Do we or don’t we have supersymmetry? 1829: German Mathematician Jacoby wrote treatise on elliptic function with this identity, labelled “a very strange identity”. Critical dimension D = 10 for supersymmetry.


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MIT 8 251 - Lecture Notes

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