Lecture 22 8 251 Spring 2007 Last Time dynamical variables needed for describing superstrings I 1 2 1I I1 2I I2 BCs 1I 1I 2I 2I 0 Suppose 1I 0 0 then 1I 0 by 1I I Bad Instead relate 1 and 2 assemble full spin or I I 1I 2I 0 0 I continuous at 0 because of BC 11 2I 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 I I I r Z 12 bIr e ir Creation Operations bI 5 2 bI 3 2 bI 1 2 Destruction Operations bI5 2 bI3 2 bI1 2 I is anticommutative all bIr operations anticommutative bIr bJs r s 0 IJ NS State 9 I n n I n 1 I 2 9 J 2 r 12 32 52 1 bI r r I N S p p Lecture 22 8 251 Spring 2007 r I 0 1 Recall for open bosonic string normal ordered M2 1 1 I p pI 2 p Z Now M2 1 2 1 1 1 I p pI rbI r bIr 2 2 1 rbI r bIr r 12 32 p Z r Z 2 1 rbI bI 2 1 3 r r r 2 2 1 2 rbI r bIr r 1 D 2 2 1 3 5 2 2 2 1 12 1 1 1 D 2 2 12 2 1 D 2 48 1 For boson aB 24 In open bosonic string M 2 1 aB 24 1 1 where a 1 but 24 contributions so 1 Here aN S 48 for antiperiodic fermion M2 1 I I p p rbI r bIr D 2 aB aN S p 1 1 3 r 2 2 2 Lecture 22 8 251 Spring 2007 M2 1 1 N 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 a 1 f x n 0 xn a n number of states with N n 0N 0 a 1 a1 2 0N 2 1 0N f1 x 1 x x2 1 1 x Given 2 4 a 2 f2 x 1 x x 1 1 x2 2 0N 0 a 2 a 4 2 0N 2 0N f1 x 1 x x2 Given a 1 a2 f1 x f2 x 1 1 x 1 1 1 x 1 x2 Now can do full open bosonic string with a 1 a2 a3 Generating Function fos 1 1 xn n 1 3 Lecture 22 8 251 Spring 2007 fos 1 1 xn n 0 n 1 p n xn 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 ln p N 2 N 6 1 exp 2 p N 4N 3 N M 6 For full open string fos 1 1 xn 24 n 1 This gives you degeneracy of any level of open string Given Given Given Given b 1 f1 x 1 x 2 b 2 f2 x 1 x b1 b2 f12 x f1 x f2 x 1 x 1 x2 b 1 f 12 x 1 x 2 fN S x r 1 1 xr 1 2 8 x0 36 x 2 x1 a r of states with M 2 r 8 1 1 xn 1 x n 1 1 xn Ramond I I dIn exp in dm dIn m n 0 IJ dI0 8 4 creation and 4 destruction 1 2 3 4 4 Lecture 22 8 251 Spring 2007 Vacuum State 0 0 1 I J 0 6 1 2 3 4 0 1 This yields 8 R1a I 0 4 I J K 0 4 This yields 8 R2a 8 8 16 gound states Total set of vacua states RA A 1 16 split into 2 types Ramond mass formula M2 1 I I I I nd d p 1 p p n 1 n n Substraction constant is equal to zero since aR M 2 0 M 2 1 R1a I 1 R1a dI 1 R2a 1 24 R2a I 1 R2a dI 1 R1a Why is this supersymmetry Left and right columns have opposite fermionic states Don t know if R1a is a boston or a fermion but know R2a is the opposite No bosons that look like R1a in real world Partition function in Raman Sector fR x 16 8 1 xn n 1 5 1 xn Lecture 22 8 251 Spring 2007 To get supersymmtry throw out half of the states from each sector and put them together 1 8 1 8 1 1 xn 2 1 xn 2 truncated fN S 1 xn 2 x n 1 1 xn n 1 Anything with an odd number of fermions will change the sign fN S 8 8 1 xn n 1 1 xn 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 6