1 Canonical quantization conformal gauge

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1 Contents 1 Canonical quantization conformal gauge 1.1 Free field space of states Constraints VIRASORO ALGEBRA Hph = H ph null SPURIOUS AND NULL STATES Physical states P H TECHNICAL TOOL N= N= N= Effective fields c Spectra Appendices A. Quantum symmetries a) L-C QUANTIZATION b) NORMAL ORDERING CONSTANTS a c) UNORIENTED STRINGS

2 1 Canonical quantization conformal gauge String theory in Mink. space in the conf. gauge is 1 FREE D CFT + CONSTRAINTS S = T d σ[( τ X) ( σ X) ] (1.1) Σ (L n δ n,0 ) ph = 0 = ( L n δ n,0 ) ph (1.) Set of constraints on physical states: naively T ++ ph = T ph = 0, but this is not good. Canonical momenta L (p X ) µ = ( τ X µ ) = T τx µ (1.3) CCR : [X µ (τ, σ), (p X ) ν (τ, σ )] = iη µν δ(σ σ ) (1.4) Open strings: [α µ m, α ν n] = mδ m, n η µν, [x µ, p ν ] = iη µν, α 0 p (1.5) Closed string: as above + the same for α & [α, α] = 0,, α 0 p. (1.6) 1.1 Free field space of states. Cartan subalgebra i.e. maximal set of the hermitian commuting operators is e.g. α 0. q = e iqx 0, p µ q = q µ q α m p µ = 0, m > 0, α 0 p = p p (1.7) (unphysical) space of states { i α µ i p, m i > 0 Some of the states has negative norms: e.g. (m > 0) } (1.8) α 0 m p = p α 0 mα 0 m p CCR = m p p Similarly for c. { i αµ i i αν i n i p, n i, m i > 0 } 1 In the Old Covariant Quantization (OCQ) one does not need ghosts. Virasoro is anomalous.

3 1. Constraints 1..1 VIRASORO ALGEBRA. According to sec.?? for all type of strings: Notice that L 0 = α0/ + α n α n, L m = 1 α m n α n, L m = (L m ) (1.9) n>0 n [L m, L n ] = (m n)l m+n + c 1 m(m 1)δ m, n (1.10) 1.a. all operators must be normal ordered: non-trivial ordering occures only for: L 0 leads to the constant a below in constraints (1.11). 1.b. normal ordering is essential to get the central term in [L m, L n ]. 1.c. L m generates symmetries (see App.A.. 1.d. L 0 = 1 dσt00 is sth-like the energy. π (Below: m i > 0). 1.3 Hph = H ph null For. All the physical states respect (in agreement with (1.10)) : L m ph = 0, m > 0, (L 0 a) ph = 0, (mass-shell) (1.11) { ph } H ph (1.1) All states in H fph have non-negative norm. 1.a. Among { ph } there are still zero-norm states called null states which should be removed by some farther constraints i.e. by n ξ = 0 for some n below.!!!! 1.b. For some a (a 1) and c (c 6) Virasoro constraints cut out all unphysical, negative norm states from the unconstraint Hilbert space. 1.c. For the critical case a = 1, c = 6 the spectrum consists of the same states as for the l-c gauge. This is the CRITICAL STRING. derive a=1 from l-c quantization. It is important fo the fermionic string too []. 1.d. No consistent interaction has been found for the non critical case. 3

4 1.e. There also are physical (?) operators corresponding to closed strings -HOW TO SEE IT???? c (L 0 L 0 ) ph = 0, (L 0 + L 0 a) ph = (L 0 a) ph = 0 (1.13) SPURIOUS AND NULL STATES : spur = i L any s 1.a. For all { ph }: spur ph = 0 There are no other states with the above property. 1.b. null states { null } = { spur } { ph }: All { null } are orthogonal to each other, because they are spurious and physical. any ph null ph ph = ph + null is a gauge transformation; [Pol.I p.13]. It does not change any sclar product with other physical states it is also property of the interaction. 1.4 Physical states P H Here it is defined to be free from negative norm states. One gets physical Hilbert H space (of positive norm states) after eliminating all the null states from P H. L 0 are mass shell relations L n gives perpendicularity of polarization tensors. null gives gauge-like freedom (L 0 a) i αµ i 0, p = (α / + i m i a) i αµ i 0, p respect constraints iff (α / + i m i a) = 0 mass-shell (1.14) TECHNICAL TOOL Level eigenstates ˆN state = N state (L 0 = α 0/ + N), N = n=1 α nα n ˆN i αµ i 0, p = ( i m i) i αµ i 0, p ˆN i αµ i 0, p = [ ˆN, i αµ i ] 0, p = ( i m i) i αµ i 0, p ˆN any s = s any s ˆN i L any = ( i m i + s) i L any ˆN i L any = [ ˆN, i L ] any s + i L ˆN any s = (s+ i m i) i L any 4

5 At the given level there are only finite number of the non-trivial constraints. Some constraints are trivial: we use ([L n, α m ] = mα n m, n > 0) then L n i αµ i 0, p = [L n, i αµ i ] 0, p 0, iff n > sup (i,j) (m i, m j ) The state at level N is a linear combinaton of the above that in general L n N = 0 trivially for n > N (1.15) 1.4. N=0 p α 0/ = a, tachyon for a= N=1 α 0/ = a 1 (1.16) L 1 ξ µ α µ 1 0, p = p µ ξ µ 0, p, p µ ξ µ (p) = 0 (1.17) Null state: L 1 0, p = p µ α µ 1 0, p. This state has ξ µ = p µ thus is physiacal for p = 0 gauge symmetry only for a = N= ξ µ1,.. i αµ i 0, p + have the same momentum as i αµ i 0, p symmetry gauge 5

6 1.5 Effective fields mass shell condition (p + m ) p = 0. Is equivalent to free field ( m )φ(x) = 0 which has decomposition thus d 3 p φ(x)= [a( p)e ipx + a ( p)e ipx ] p0 =ω ω p (1.18) p p = a(p) 0 (1.19) ( m )Φ a (x) = 0? L φ(x)( m )φ(x) (1.0) We see that there is a correspondence between string quanta and states of a QFT. N=1 ξµα i 1 0, µ p A µ (x) massless vector particle 3 A µ (x) = dp[ ξµ(p)a i i (p)e ipx + h.c], p 0 = p (1.1) i=1 in the gauge µ A µ = 0 (p µ ξµ i = 0), ξµα i 1 0, µ p = a i (p) 0 Null states residual gauge symmetry: at massless level only δa µ (x) = µ ζ(x), ζ(x) = 0 (1.) The state is physical if p = 0 i.e. ζ(x) = 0. Effective action: U(1) gauge theory with the gague choice µ A µ = 0 we get exactly free Lagrangian. L = 1 4 (F µν) = 1 ( µa ν ) + 1 ( µa ν )( ν A µ ) = 1 ( µa ν ) + 1 ( µa µ ) (1.3) Residual gauge symm. is δa µ = µ ζ, δl = µ A ν µ ν ζ integrating by parts ν we get invariance. This gauge choice has residual gauge symmetry as in (1.). The residual gauge symmetry gauges away the p µ â part. In field theory one farther fixes the gauge choice by choosing a vector n µ, n p = 1, n = 0 and imposing n ξ i (p) = 0. Counting of the degrees of freedom (D): D- polarizations of gauge bosons. 6

7 1.6 c (L 0 L 0 ) i αµ i k αν k n k 0, p = 0, i m i = k n k Level 1. ξ µν α µ 1 α 1 0, ν p. L 1, L 1 p µ ξ µν = p ν ξ µν = 0 one of these d equations is not independent p µ ξ µν p ν = 0; null= ξ µν = p µ ζ ν + ζ µ p ν iff p µ ζ µ = p µ ζµ = 0; ζ µ = ζ µ = p µ gives the same null state. Counting: d (d 1) ((d 1) 1) = d 4d + 4. GOOD. Easier: take p = (p, p, 0,...) i.e. p + = p all other =0; then p + ξ +ν = p + ξ µ+ = 0 (eq. for ξ ++ is the same) (d-1) constraints; null: p + ζ + = p + ζ+ = 0 i.e. ζ + = ζ + = 0 (d-) null but one p p is common. Effective theory N=1:masssless states: graviton h µν, B µν, dilaton φ. The action: Exapnsion around flat background: g µν interaction is given by the following Lagrangian For h µ µ = 0, µ h µν = 0 we get 1 G (R 1 κ ( φ) 11 ) H (1.4) = η µν + κ 0 h µν. The linearized part of the graviton L = 1 λh λ µ µ h ν ν + 1 λh λ µ ν h νµ 1 4 λh µν λ h µν λh µ µ λ h ν ν (1.5) L = 1 4 λh µν λ h µν (1.6) ( 1 4 is becuase the metric tensor is symmetric) with the gauge gauge invariance h µν h µν + µ ξ ν + ν ξ µ. (1.7) Antisymmetric tensors We have 1 1 H where H is a field strenght for B fields. In general: A k = 1 k! A µ 1...µ k dx µ 1...dx µ k (1.8) 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) = 1 n! F µ 1...µ n dx µ 1...dx µn Thus: F µ1...µ n = n [µ1 A µ...µ n], 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, H 13 = 1 B 3 + B B 1. So 1 1 H µνρ = 1 1 ( µb νρ +cycl.) = 1 4 ( µb νρ ) 1 µb νρ ν B ρµ 1 µb νρ ρ B µν +one more [a 1...a n ] = df 1 n! perm sign(permutation) a 1...a n. 7

8 Analize null states for the massive modes [] The spectrum is ghost free for a=1 and d=6 or a 1 and d 5 GSWI, Spectra Closed oriented - CO, Closed unoriented - CUO, Open oriented - OO, Open unoriented - OUO. closed CO CU O open OO OU O m = d m = d , p T T 0, p T T m = d m = 1 1 d 48 α (µ 1 α 1 0, ν) p G µν G µν α 1 0, µ p A µ α 1( α µ 1 ) µ 0, p φ φ α [µ 1 α ν] 1 0, p B µν (1.9) 8

9 1.8 Appendices A. Quantum symmetries Classically a J a = 0 t Q = 0 so in Hamilton formalism Q = Q(p, q) and t Q = 0, {Q, H} P B = 0. Thus Q is generator of symmetry (see below). In quantum theory [Q, H] = 0. Transformation laws under the symmetry for fields are generated by the charge because the commutator is a differentiation i.e. for H = AφB we have [Q, H] = [Q, AφB] = [Q, A]φB + A[Q, φ]b + Aφ[Q, B]. We set: φ = e iɛq φe iɛq δ ɛ φ = iɛ[q, φ] (1.30) In our case from (??) we get ɛq = 1 π ɛ(τ)t (τ) = ɛm L m. For the scalar field δ ɛ + X = m ɛ + me imσ+ + X thus [L m, α n ] = (m + n)α m+n gives normalization of L n. A) L-C QUANTIZATION We solve constraints, thus we should work only with physical states, thus there should not be ghosts. 1.a. solution for X + = x p + τ is conformally equivalent to the most general solution (p + > 0). Thus we can farther fix the gauge choosing X +. 1.b. We analize T ++ = 0 p + + X = ( + X ). Its zero mode p + p = (p ) + α n α m (1.31) must be normaled ordered and it gives mass condition. The nonzero modes give relation between X and X oscilations not interesting. Normal ordering influences the CR for generators of the Lorentz group (??) (which looks like broken here) so there might be anomaly. The vanishing of the anomaly leads to the critical string (above). 1.c. the Hilbert space is H l c = {i,p} ( i αj i m i ) p +, p, m i < 0 1.d. the masses of states we read from the zeroth component of (??). Normal ordering is important. We denote c. As above N = α nα n, n=1 n = 1 1, α 0 = p (1.3) n=1 p + p = N + (a 0 ) (d ) 1 M = 1 (d ) (N ) (1.33) 4 1.a. X + = x p + τ is conformally equivalent to the most general solution (p + > 0). 1.b. The above choice determine X by solving T ++ = T = 0 to give p + + X = ( + X ), p + X = ( X ) 9

10 1.c. there is two constraints left over because the above constraint have the same l.h.s for the zeroth Fourier component: N = Ñ and mass shell M = N (d ). The Hilbert space 1 is H l c = α k i n i α k j m j ) p +, p, m i < 0 (1.34) {i,j,p}( i 1.d. for a = 1, c = 6 we have H OCQ = H l c. j 10

11 B) NORMAL ORDERING CONSTANTS a Contribution from a single field to a is a = ± 1 { (m + α) = ± 1 + bosons ζ( 1, α) = (±)(α α )/4, m=0 where α > 0 depends on b.c.. fermions (1.35) For SST ψ s follow the pattern of X (mod 1/) due to superconformal symmetry (s-conf. parameter ɛ can be periodic (R sector) or anti-per. (NS sector). CLARIFY. (?) which directions do not enter a? ψ in R sector has the same α as bosons, thus total a = 0, number of massless states depends on number of ψ 0 s. In extreme case there can be no ψ 0 s and only 0 R massless, which moreover can be projectd out by GSO. ψ in NS sector has α shifterd by 1 For D-D, N-N b.c. bosons have α = 0 a = 1 4,fermions have α = 1 a = 1 48 sum= a = 1 16 For N-D b.c. bosons have α = 1 a = 1 the ψ 0 s appears!!!!! sum= a = fermions have α = 0 a =. In this case 4 In the case # of N-D=4 (preserving Susy) one has (we removed D-D directions from a, why (?) ) a R = a NS = 0 (1.36) there appears massless spinor reps in NS sector!!!!! - contributions from 4 NN or DD cancels that from 4 ND (in NS sector). For two D-branes rotated by θ in X i, X i+1 (Z i = X i +ix i+1 ) we got a = (±) 1 m=0 (m+ α) with α = θ/π. 3 For one complex boson and fermion in NS (as above) a = (m+α) (m+α+ 1 ) = ζ( 1, α) ζ( 1, α+ 1 ) = α (1.37) m=0 m=0 Generalized Riemann function or Hutwitz function ζ( 1) = 1 1, ζ(0) = 1 1 n=1 n = 1 ζ( 1) = 1 4, 1 (r + 1 ) s = 1 (s 1)ζ(s) s= 1 = 1 48 r=0 (1.38) 3 see arxiv:hep-th/ v and for more detailed presentation 11

12 ζ(s, 1 ) r=0 (r + 1 ) s = s r=0 (r + 1) s = s ( r=0 (r + 1) s r=1 (r) s ) = ( s 1)ζ(s). r=0 (r + α) s = ζ(s, α) is so-called Generalized Riemann function or Hutwitz function. ζ( 1, α) = (α α )/. C) UNORIENTED STRINGS We can get unoriented strings requiring that states are invariant under Ω (below). This reduces number of states. c : σ π σ, Ω : α n α n : σ π σ, Ω : α n ( 1) n α n Also the only consistent (with the scattering amplitudes) choice is invariance of tachyon under the reflection Ω 0, p = 0. For unorineted strings one needs: Ω 0, p = 0, p 1