1 Superstrings. 1.1 Classical theory

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1 Contents 1 Superstrings 1.1 Classical theory ANTI-COMMUTING ψ S FINAL ACTION Eq.m. and b.c NS/R in general Symmetries CONFORMAL SYMMETRY SUPERCONFORMAL SYMMETRY Quantization of ψ µ Constrains VIRASORO CONSTRAINTS GSO projection NS R D-branes at angles c (R, R) CONSTRAINTS SUMMERY OF THE SPECTRA CONSISTENT CLOSED SUPERSTRINGS CLOSED IB = IIB/Ω Open I OPEN UNORIENTED CHAN-PATON FACTORS Appendices M. Canonical quantization of fermions N. RR sector O. Some identities a) CONVENTIONS b) DIRAC ALGEBRA IDENTITIES c) FORMS

2 1 Superstrings As we shall see the SUPERSTRING is quite complex object consisting of several sectors (Hilbert spaces) of the following supersymmetric action. S = 1 ( 1 4π α ax µ a X µ + iψ µ γ a a ψ µ ) (1.1) 1.1 Classical theory ANTI-COMMUTING ψ S d Minkowski ( 1, 1) Spinors {γ a, γ b } = η ab = diag( 1, 1) ab (1.) ψ M = C(γ 0 ) T ψ, CC = 1 (1.3) ψ M γ a a ψ M = ψ γ a a ψ C γ a C = ±γ at (1.4) d: γ 0 = σ, γ 1 = iσ 1, C = γ 0, C γ a C = γ at ψ M = γ 0 γ 0 ψ = ψ, ψ M = ψ ψ = S = 1 4π d σ ψ γ a a ψ = 1 π ( ψ The above has (also for arbitry backgroud) Z Z symmetry 1 call ( 1) F (for the left Z ). ψ + ) (1.5) d σ i(ψ + ψ + ψ + ψ + ) (1.6) with generators which we shall ψ + ψ +, ψ ψ (1.7) 1.1. FINAL ACTION S = 1 ( + X π X + iψ + ψ + iψ + ψ + ) (1.8) 1 Because ψ s do not have any direct physical meaning (contrary to X which has clear meaning for some limiting cases) we could have ψ µ (σ +π) = t µν ψ ν (σ). But t must commute with target Lorentz generators and be symmery of S thus in this case it must be ±1 for all µ.

3 1. Eq.m. and b.c. + X = 0, + φ = 0, ψ + = 0 (1.9) c : We can consider the folowing R: (Ramond), NS (Neveu-Schwarz) boundary conditions (we suppress here ±) (R) ψ(σ + π) = ψ(σ) r Z (1.10) (NS) ψ(σ + π) = ψ(σ) r Z + 1 (1.11) NS sector can be called twisted sector: ψ(σ + π) = g ψ(σ) for any g in global symmetry group of the action. Thus eq.m. ψ + = 0, + ψ = 0 has solutions ψ µ + = r ψ µ r e irσ+, ψ µ = r ψ µ r e irσ r Z (R), r Z + 1 (NS) (1.1) Derivation of the eq.m. from (??) gives the following boundary contributions dτ π 0 dσ σ( ψ + δψ + + ψ δψ ) yielding Possible b.c. ψ + δψ + ψ δψ = 0 for x = 0, π (1.13) 1.a. σ = 0, we set δψ + = δψ what is always possible by the Z symmetry (??). This gives ψ + (τ, 0) = ψ (τ, 0). Mode expansion (??) r ψµ r e irτ = r ψµ r e ir τ sets {r} = {r } and ψ r = ψ r. Thus ther is only one fermion ψ so that ψ ± (σ ± ) = ψ(σ ± ) and 1.b. σ = π, δψ + = ±δψ i.e. ψ(τ + π) = ψ(τ π) (R) r Z ψ(τ + π) = ψ(τ π) (NS) r Z + 1 (1.14) Reality ψ + r = ψ r, r 0 and CCR dictate creation-anihilation splitting. There is problem with ψ 0 which is real see. 3

4 1.3 NS/R in general One chooses NS/R b.c. the same for all of the space components. R: ψ s follow the mode expansion of X fields. It always gives anti-commuting fields. NS: ψ s have 1/ shift. a R = 0 but a NS varies for different situations. GSO changes (how???) Spectra of intersecting branes/instantons. [] 1.4 Symmetries CONFORMAL SYMMETRY X X(f(σ + ), σ ), ψ + ( + f) 1/ ψ + (f(σ + ), σ ), ψ ψ (f(σ + ), σ ) (1.15) Current from terms linear in f(σ +, σ ) from = f + + T ++ = 1( +X + X + iψ + + ψ + ) (1.16) 1.4. SUPERCONFORMAL SYMMETRY Global (1,1) SUSY is part of it δx = iɛ + (σ + )ψ +, δψ + = ɛ + (σ + ) + X (1.17) Current from terms linear in ɛ + (σ +, σ ) from X and ψ + gives G ++ = iψ + + X (1.18) For ɛ + = ±ɛ i.e. on the boundary ɛ σ = 0.???? 1.5 Quantization of ψ µ π + = i 4π ψ + are constraints, thus we introduce we calculate {F, F } = 1/π Dirac bracket to define CCR 3 F = π + i 4π ψ + (nd class constraints (see e.g. Lust-Theisen)) thus we need {ψ, π} D = {ψ, π} {ψ, F }{F, F } 1 {F, π} = i (1.19) This anti-commutator here would be classically Poisson bracket. {a, b} a P B df = q a π b + q b π a thus {F, F } a P B = i/(π). 3 It gives {ψ, F } D = {F, π} D = 0 4

5 what gives {ψ µ +(τ, σ), ψ ν +(τ, σ )} = πδ(σ σ )η µν (1.0) 1.a. {ψ r µ, ψs ν } = δ r+s,0 η µν, ψ r + ψ r 0 = 0, r > 0. 1.b. For r = 0, = ψ r. For r 0 this defines creation-anihilation operators: {ψ µ 0, ψ ν 0} = η µν (1.1) we get Dirac algebra for the construction of the relevant Fock space: out of n real (for simplicity, Euclidean) ψ 0 s we form n complex b i = 1 (ψ i 0+iψ i+n 0 ), b i = 1 (ψ i 0 iψ i+n 0 ) They respect {b i, b j } = δ ij. We define a state Ω >, b i Ω >= 0. Fock space S = m n=0 b i 1...b i n Ω >, its dimension dim C (S) = m = dim. of Dirac spinor rep. We then split into eigenspace of ( 1) F of dim R = m, S = S + S where are. S + = m n=0 even b i 1...b i n Ω >, S = m 1 n=1 odd b i 1...b i n Ω > (1.) In other words (1.1) has the following reps 4 ψ µ 0 = Γ µ. (1.3) Notice that (as always) Γ µ : (S +, S ) (S, S + ). DO carefully the Dirac algebra: change notation to {Γ, Γ} = η and then write down the Γ s expicitely. D=10. In (1,9) Minkowski space these Weyl spinors can be real (Majorana) so dim R (S ± ) = 16. More notation: S + S + = ξ α α, S S = ξ α α. (1.4) We shall see that all these states has the same level i.e. the same mass shall condition. 1.6 Constrains (R) r Z, (NS) r Z + 1. L m = 1 : α m n α m : + 1 (r m) : ψ m r ψ r : 4 m r G r = α n ψ r n. (1.5) n 4 The other reps = Γ µ T, Γ µ, Γ µ + are equivalent (which (?)[] ) or related by outer automorphism of the Lie algebra (which is charge conjugation). 5

6 Correspondingly, the Virasoro algebra is enlarged, with the non zero (anti) commutators being [L m, L n ] = (m n)l m+n + c 1 A(m)δ m+n {G r, G s } = L r+s + c 1 B(r)δ r+s [L m, G r ] = 1 (m r)g m+r, (1.6) where A NS = (m 3 m), B NS = (4r 1), A R = m 3, B R = 4r. For the free strings in d-dim Minkowski c = d + d/. Shifting in R sector L 0 L 0 + c we get the same algebra as for NS: 4 N=1 SCA VIRASORO CONSTRAINTS (NS) (L m a NS δ m,0 ) phys = 0, G m phys = 0 m > 0 (1.7) (R) (L m a R δ m,0 ) phys = 0, G m phys = 0 m 0 (1.8) and tilde for closed strings a NS = 1/, a R = 0 (1.9) Calculate a for arbitrary number of D and N b.c. G 0 = iψ 0 α i p µ Γ µ +... (Generalized Dirac operator) (1.30) (NS) L 0 phys = 0 (p + m ) phys = 0 ( x m )φ(x) = 0 (1.31) (R) G 0 phys = 0 (γ µ p µ +...) phys = 0, (Generalized Dirac eq.) (1.3) In (RR) sector we deal with forms GSO projection Z Z for c and Z for commute with SCA. (Left) Fermion number F is defined mod. 5 F = m>0 ψ µ mψ µ m + (R only) b i b i + c (1.33) [F, X] = 0, ( 1) F ψ(σ)( 1) F = ψ(σ) (1.34) thus ψ µ 0 p µ = ψ i p i + ψ i p i. ψ i : Ω n Ω n+1, ψ i : Ω n Ω n 1 thus they represents d, d (Hodge). All together it leads to (d + d )F n = 0. We have (d + d ) = according to S-Virasoro. Do it: use gamma identities?????. For closed strings we have Hodge decomposition: ψ µ 0 p µ = +, ψµ 0 p µ = +. 6

7 We take 7 ( 1) F α R = α R S +, ( 1) F α R = α R S (1.35) ( 1) F = e iπf = e iπf is well defined 6. We can take physical states to be its eigenstates because it leaves invariant the constraits: ( 1) F phys = ± phys, ( 1) F phys = ± phys. For the NS vacuum we have ( 1) F 0 NS = 0 NS (1.36) ( 1) in (NS) comes from ghosts. This is reasonable because colliding two massless spinors of the same chirality we get odd forms of ( 1) F = 1 so e.g. NS vector and there is such a state for m = 0 with the above choice. With the opposite chice one would get ( 1) F = 1 vector at massive level!!! NO SUSY!!! One can project the spectrum to states with ( 1) F interaction. There is SUSY. Show it by counting states. = 1 and this is preserved by the NS masses (NS ), 0 p = 1, (NS+), ξ µψ µ 1/ 0 m = 0,... (1.37) Virasoro constraints: for ξ µ only from G 1/. It gives p µ ξ µ = 0. Null states exists for ξ µ and provide gauge symmetry. corresponds to δa µ = µ χ(p). χ(p)g 1/ 0 > χ(p)p µ φ µ 1/ 0 > (1.38) 1.8. R (R+), ξ α α, (R ), ξ α α, p = 0 (1.39) Virasoro constraints: for ξ α, ξ α only from G 0 = iα 0 ψ By (1.1) we have 8 ψ µ 0 α = 1 Γ µ βα β (1.40) 6 Notice that in (R) ( 1) F generalizes the γ 5. ( 1) F s k=1 bi k 0 = ( 1) s s k=1 bi k 0. Also [F, ψ m ] = sign(m)ψ m. 7 8 This fixes the irrelevant constant c in (R) in (1.33). It is irrelevant because it switches the chirality of spinors. One can check that Γψ µ 0 α = ψµ 0 α i.e. it is proportional to β. 7

8 Thus G 0 ξ α α = 0 = p µ ξ α Γ µ βα β, p µ Γ µ βα ξ α (p) = 0, Dirac eq. (1.41) NO null states. NS+ ξ ψ 1/ 0 m = 0 (odd # of fermions) 0 NS 0 m = 1/4 (even # of fermions) 0 R+ S + = α m = 0 (even # of fermions) α (odd # of fermions ψ r ) α (1.4) R S = α m = 0 (α α in above) The rightmost states are massive. Take ( 1) F = 1 massless states ξ ψ 1/ 0, S + = α. In vector state there are 10 = 8 degrees of freedom, in spinor field there are 16 8 = 8 d.o.f. ( 8 due to Dirac eq.). 1.9 D-branes at angles for 4 N-D BC the GSO changes sign!!! Polchinski II, p c Formally we have 4 possible sectors (NS±, NS±), (NS±, R±), (R±, NS±), (R±, R±) (1.43) but some must be eliminated by L 0 = L 0 e.g. NS can apear only with NS. Sectors (N S, N S ) contains tachyon (N S+, N S+) contains string gravity (NS, R) contains superpartners of gravity Ψ µ α = α, p, R ψ µ 1 0, p decomposes to χ µ α and λ α = (Γ µ ) β α Ψ µ β. For critical string gravitino has null state: corresponds to δχ µ µ = µ χ. Local SUSY (R, R) contains form (gauge fields) G 1/ 0 > α p µ ψ µ 1/ 0 > α (1.44) 8

9 (R, R) We decompose tensor product of spinors S + S + ξ αβ α β or S + S into irrps of the Lorentz group. We need Clebsch-Gordon coeffs (CGS). Denote µ 1...µ k [k] Notice that (Γ 0 = C = C T = C + ) S Γ [k] S = S T CΓ [k] S and S T Γ [k] C + S are irreps and both equivalent (?)[] (see...) thus Γ [k] C + are CGC of S S = anti symmetric tensors. Similarly we can use CGC: ΓΓ [k] C + = ( ) k(k+1)/ ɛ [k] [10 k] Γ [10 k] /((10 k)!). We denote F [10 k] Γ [10 k] C + = ( ) k(k+1)/ F [10 k] (ΓΓ [k] C + ) Λ 10 k M (forms) ξ α α = k (F k ) α α = k 10/ By ( 1) F α = Γ βα β = α thus F [k] = ( ) k(k+1)/ F [10 k]. ξ α α = k 10/ (R+, R+). k is odd. F 5 = F 5. S + S + = F 1 F 3 F + 5. (F [k] Γ [k] C + ( ) k(k+1)/ F [10 k] ΓΓ [k] C + ) ( is conventional) F [k] (1 + Γ)Γ [k] C + (1.45) Counting: S + S + : lhs = 4 4, rhs = ! ! = = 56. (R+, R ). k is even. S S + = F 0 F F 4. S S + : lhs = 4 4 = 56 = rhs = ! = CONSTRAINTS By (1.30) we get (picking only terms without Γ) 0 = G 0 Fk = p µ Γ µ F k = p F k (p, F k ) k 1 0 = G 0 Fk = p µ F k Γ µ = ( 1) k (p F k + (p, F k ) k 1 ) [] 9 (we used (1.67)). Thus all coefficients of the k-form must vanish i.e. (p F k ) (p, F k+ ) = 0 and (p F k ) + (p, F k+ ) = 0 p [µ F µ1...µ k ] = 0 = p µ 1 F µ1...µ k (1.46) 9

10 SUMMERY OF THE SPECTRA We suppress momentum p from indexing the states. sector # (NS+, NS+) ψ µ ψ ν G µν, B µν, φ (R+, R+) α β F 1 = dc 0, F 3 = dc, F 5 + = dc 4 (R+, R ) α β F = dc 1, F 4 = dc 3 (R+, NS+) α ψ 1 0 µ χ α µ chirality λ α chirality + (NS+, R ) ψ 1 0 µ β χµ α chirality + λ α chirality (NS, NS ) 0, m = 1 T Moreover we need more then one sector to produce modular invariant theory CONSISTENT CLOSED SUPERSTRINGS type GSO sectors IIA ( ) F L (NS+, NS+), (NS+, R ), (R+, NS+), (R+, R ) IIB ( ) F L (NS+, NS+), (NS+, R+), (R+, NS+), (R+, R+) 0A? (NS+, NS+), (NS, NS ), (R+, R ), (R, R+) 0B ( ) F L+F R (NS+, NS+), (NS, NS ), (R+, R+), (R, R ) (1.47) 10

11 CLOSED IB = IIB/Ω Ω : σ π σ α n α n ψ r ( ) r ψr α α (1.48) Invariant states: G µν, φ (1.49) C (1.50) χ µ α, λ α (1.51) (1.49): one must take into account that NS vacuum carries the fermion number (1.36). (1.50): only Γ µν (M. -l.c. = 8d E) is a-symmetric matrix due to (1) C=1 for M-W E-spinors () (Γ µν ) T = CΓ νµ C = Γ µν. Thus from a-commuting of spinor states α, β we get Ω-invatiant state (Γ µν ) αβ α β. (1.51): symmetric combination of IIB (NS+,R+), (R+,NS+) Open I Consists of closed IB = IIB/Ω and open unoriented superstring OPEN UNORIENTED Ω : ψ(σ + ) ψ Ω (σ + ) = ψ(σ + π) ψ(σ ) ψ Ω (σ ) = ψ(σ + π) (1.5) It follows that Ω : (R) (R) (N S) (N S) (1.53) so ψ Ω (σ + ) = ±ψ Ω (σ ) σ=0, and ψ Ω (σ + ) = ψ Ω (σ ) σ=π thus in order to get to the original situation we have to make Z transform in the (NS) sector. Thus Ω : ψ r = ( 1) r ψ r = { ψr (R) ψ r (NS) (1.54) The spectrum is modified only in the (NS) sector: only (even # of ψ r ) 0 survaive so e.g. the U(1) gauge boson is lost CHAN-PATON FACTORS 11

12 1.1 Appendices Appendix M. Canonical quantization of fermions. Out of n real (Euclidean) ψ s we form n complex b i = 1 (ψ i + iψ i+n ), b i = 1 (ψ i iψ i+n ) (1.55) i These are independent variables and ψ τψ = ib i τ b i πb i = ibi CCR {b i, b i } = δ ii, {b i, b j } = 0, {b i, b j } = 0 {ψ µ, ψ ν } = η µν (1.56) We define a state Ω >, b i Ω >= 0 Then m n=0 even b i 1...b i n Ω > rep.of one chirality m 1 n=1 odd b i 1...b i n Ω > rep. of another chirality (1.57) ψ µ = i (b i + b i ), for µ = i = 1,...m ψ µ = i i(b i b i ), for µ = m + i = m + 1,...m {ψ µ, ψ ν } = δ µν (1.58) In this algebra the Dirac matrices are Γ µ = ψ µ. N. RR sector α, p, R β, p, R or α, p, R β, p, R plus (undoted doted) which are equivalent. {Γ µ, Γ ν } M = η µν E = η µν so (Γ µ ) M = η µν (Γ ν ) E = η µν (Γ ν ) Γ µ 1,...µ k df = 1 k! permut sgn(µ 1,...µ k )Γ µ 1...Γ µ k Lorentz generators: M µν = i Γµν. For even dimensions there are Weyl reps S ± ; ψ α S, ψ α S + thus we can do projections of Γ µν on appropriate spaces using ΓS ± = ±S ±, {Γ, Γ µ } = 0, Γ = Γ (1.59) Γ µt also respect (N.). (?) This is an equivalent rep thus there is a unitary operation C CΓ µ C = ±Γ µt (1.60) C is a nontrivial metric on S ± i.e. ψ T Cψ is an Lorentz invartiant due to (M µν ) T = CM µν C. 1

13 The charge (Marojana) conjugate spinor ψ c df = Cψ T M = C(Γ 0 ) T ψ E = Cψ (1.61) transforms the same way as ψ. Moreover ψ is an eigenstate of Γ then also ψ c 10 ψ ψ is an Lorentz invatiant due to (M µν ) [Γ 0 ] = [Γ 0 ]M µν. Γ(S + ) c = sign( ) (S + ) c (1.6) ψ Γ µ ψ and ψcγ µ ψ are Lorentz vectors Dirac operator V µ Γ µ : S ± S (V is a vector) due to (1.59) V µ CΓ µ : S ± S Majorana-Weyl spinors in M space (GSW vol.i,p.00,88) we take (Γ µ ) = η µν Γ ν, moreover for M-W we can take (Γ µ ) T = η µν Γ ν. C = Γ 0 CΓ µ C = Γ µt, Ψ M = Ψ M (1.63) RR fields S + CΓ µ 1...µ ks+ for odd k, S CΓ µ 1...µ ks+ for even k (dut to (1.59)) is anitsymmetric tensor of the rank k (we can change (+ ) above) Due to (1.59) and the Hodge duality ΓΓ µ 1...µ k ɛ µ 1...µ k µk+1...µ d Γ µ k+1...µ d the above tensors are linearly independent only for k [d/]. For k = d/ the tensor is (a-) self-dual for for S + (S ). O. Some identities A) CONVENTIONS we take for M-W according to (1.63) (Γ µ ) = η µν, Γ ν (Γ µ ) T = η µν Γ ν, (1.64) CΓ µ C = Γ µt, C = Γ 0, Ψ M = Ψ M (1.65) B) DIRAC ALGEBRA IDENTITIES Γ µ 1,...µ k df = 1 k! σ S k sgn(σ)γ σ(µ1)...γ σ(µk) (1.66) The following holds [[?] what convention??(n.)] 10 Γ µ 1...µ k Γ ν = Γ µ 1...µ k,ν k Γ [µ 1...µ k 1 g µ k]ν = ( 1) k (Γ νµ 1...µ k + k g ν[µ 1 Γ µ...µ k ] ) (1.67) Γ ν Γ µ 1...µ k = Γ νµ 1...µ k k g ν[µ 1 Γ µ...µ k ] (1.68) d-dim of the space, lhs = ΓC(Γ 0 ) T (Γ T ) (S + ) = ΓC(Γ 0 ) T Γ T (S + ) = ΓCΓ T (Γ 0 ) T (S + ) = ( 1) d(d 1)/ (sign(1.60)) d (S + ) c. For E space the overal sign is changed. 13

14 C) FORMS 1.a. A k = 1 k! A µ 1...µ k dx µ 1...dx µ k 1.b. F n df = d(a (n 1) ) = 1 (n 1)! µ n A µ1...µ (n 1) dx µn dx µ 1...dx µ (n 1) (as forms) and F n = 1 n! F µ 1...µ n dx µ 1...dx µn. Thus: F µ1...µ n = n [µ1 A µ...µ n], 11 F 0...(n 1) = 0 A 1...(n 1) + cycles sign(cycle) µ 1 A µ...µ n (altogether d-terms) e.g. F 1 = 1 A A 1, F 13 = 1 A 3 + A A 1. 1.c. Hodge star 1 1.d. [a 1...a n ] df = 1 n! (all sign permutations of a 1...a n ) dx µ 1...dx µ k df = g (d k)! ɛµ 1...µ k ν1...ν d k dx ν 1...dx ν (d k) (1.69) a k = sign(det(g)) ( ) k(d k) a k (1.70) a b = ( ) k(d k) a b = k! 1 g aµ1,...µ k b µ 1,...µ k dv = 1 (a b)(1.71) k! 1.e. kinetic terms L = 1 F n F n = 1 n! F n F n = 1 [(F 0... ) (space indices)] (1.7) diff [a 1...a n ] = df 1 n! perm sign(permutation) a 1...a n. df = sign(µ 1...µ k ), ɛ µ1...µ k = sign(det(g)) det(g) for even perm. ɛ µ1...µ k 14