e θ 1 4 [σ 1,σ 2 ] = e i θ 2 σ 3

Transcription

1 Fermions Consider the string world sheet. We have bosons X µ (σ,τ) on this world sheet. We will now also put ψ µ (σ,τ) on the world sheet. These fermions are spin objects on the worldsheet. In higher dimensions, we can take a local orthonormal frame and spin around in a complete circle. In 3 space dimensions for example, this changes the fermion as e θ 4 [σ,σ ] = e i θ σ 3 () with θ = π each component of the spinor changes sign. ψ and ψ define the same fermion wavefunction. We will assume the same behavior for the + case we have now. In the Euclidean case we can rotate in the -d worldsheet and obtain this property immediately. Now consider what boundary conditions are appropriate for the fermion. Suppose we are on the cylinder, and we go around σ σ + π. We want to come back to the same configuration, but we do not know if we should represent this by ψ or ψ. there are two possibilities In the first case we get the modes In the second case we get ψ(σ + π) = ψ(σ) (Ramond = R) () ψ(σ + π) = ψ(σ) (Neveu Schwarz = NS) (3) ψ = ψ = n= n= ψ n e inσ (R) (4) ψ n+ e i(n+ )σ (NS) (5) Zero point energy For bosons the Hamiltonian is H = ω(a a + aa ) = ωa a + ω (6) For fermions we have H = ω(b b bb ) = ωb b ω (7) where we have noted that the zero point energy of the bosons gave us [a,a ] =, [b,b ] + = (8) [ ] = ( ) = 4 (9)

2 where we have used that S = = (0) which we had proved earlier. If we have the R sector, we see that the fermions give us ( ) = 4 If our theory is supersymmetric on the world sheet, then we will have as many bosons as fermions, so the total ground state energy is zero in the R sector. If we have the NS sector then we get () ( ) () we wish to compute T = (3) Suppose we have regularized this sum just like we did for S. Then we can write Adding in the even integers can be done by writing T = (4) S = (5) S = = T + S (6) T = S = (7) 4 If we have supersymmetry on the worldsheet then we have as many bosons and fermions. One boson and one fermion would give 4 48 = (8) 6 the lowest state in the NS sector will be tachyonic. Let there be d spacetime dimensions transverse to the string world sheet. Then the vacuum energy of the NS sector will be d 6 (9) The lowest excitation that we can make is unit above this level. Suppose we require that this be massless. Then we have d 6 + = 0 (0) which gives d = 8 () Adding the two dimensions along the string worldsheet, we get the total spacetime dimension as D = d + = 0 ()

3 3 The -loop partition function Note: The discussion below follows Polchinski chapter 0. We have 0 bosons and 0 fermions. Because two directions are fixed along the world sheet, we have 8 transverse bosons and 8 transverse fermions. We do have the momentum from the remaining two directions however. 3. Bosons Let us first write the contribution of a boson. Overall our partition function is d 0 k Z = V 0 q 4 q 4 q α k 4 +N q α k 4 +Ñ (π) 0 H (3) where For one boson, we get q = e πiτ (4) dk πτ k π π e α = πτ α π = [4π τ α ] (5) For the oscillators we have overall for one boson we get q 4 n= q n = η (6) L[4π τ α ] η (7) where L is the length of the circle on which the boson is compactified, and we are assuming that L at the end. We also have two zero mode integrals from the directons along the string worldsheet, so we get from the bosons Z X = V 0 [4π τ α ] [ η [4π τ α ] ] 8 (8) 3. The theta functions We write z = e πiν (9) The theta functions of interest to us are θ 00 (ν,τ) = ( q m )( + zq m )( + z q m ) (30) m= 3

4 θ 0 (ν,τ) = ( q m )( zq m )( z q m ) (3) m= θ 0 = e πiτ/4 cos(πν) ( q m )( + zq m )( + z q m ) (3) m= θ (ν,τ) = e πiτ/4 sin(πν) ( q m )( zq m )( z q m ) (33) m= Focus on the holomorphic fermions. Take fermions at a time. On the σ circle, we can have NS or R sectors in the path integral. In the time direction, we also have two choices. If we have periodic fermions in τ, then we must insert ( ) F. If we have antiperiodic fermions, then we insert nothing. We thus get 4 path integrals, and we can add them in some way to get a modular invariant; this will define a theory. 3.3 Periodic and antiperiodic fermions Consider a path integral for fermions on a line τ, which is latticized to 4 points. The path integral is dc dc dc 3 dc 4 e i(c c +c c 3 +c 3 c 4 +c 4 c ) (34) where we have taken periodicity across the τ circle. There are two contributions: dc 4 dc 3 dc dc ( ic c )( ic 3 c 4 ) = ( ) (35) and dc dc dc 3 dc 4 ( ic c 3 )( ic 4 c ) = (36) so that the total Z vanishes. In the Hamiltonian description, we have two states, 0,b 0. We count these with ( ) F, with F = 0 for the first, and F = for the second, to get Z = tr( ) F e τh = = 0 (37) If the fermions were antiperiodic, then the path integral does not vanish, and we have Z = tre τh (38) 3.4 The fermionic oscillators In the NS sector we have ψ = π r=z+ d r e ir(τ+σ) (39) [ψ(σ),ψ(σ )] + = π e i(r+s)τ [d r,d s ] + e irσ+isσ (40) r,s 4

5 Setting we have Similarly, in the R sector with [ψ(σ),ψ(σ )] + = π = e i (σ σ ) π r,s [d r,d s ] + = δ r+s,0 (4) e i(r+s)τ δ r+s,0 e irσ+isσ = π r e ir(σ σ ) (4) e in(σ σ ) = e i (σ σ ) δ(σ σ ) = δ(σ σ ) (43) n ψ = d n e in(τ+σ) (44) π n [d m,d n ] + = δ m+n,0 (45) Note that the oscillators carry another index µ = 0,,... D. we actually have The zero modes have the following behavior [d µ m,d ν n] + = η µν δ m+n,0 (46) [d µ 0,dν 0] + = η µν (47) This is like the clifford algebra for dirac matrices, and we must find its representation in a similar way. Group the fermions in pairs. Consider any two fermions, d,d, and make the combinations d + = (d + id ), d = (d id ) (48) [d +,d + ] + = (d + ) = 0, [d,d ] + = (d ) = 0, [d +,d ] = (49) we can use D +,D as raising and lowering fermion operators. We can define a vacuum by d i 0 = 0, i =,,... 5 (50) and use d + i as raising operators. Since we can apply each raising operator at most once, we will get 5 = 3 states. Half of these will have an even number of d + applications, and half will have an odd number. These will give the two weyl components of the 3 dimensional spinor in 0-D. 3.5 Partition functions of fermions (a) First consider the NS sector. Let the fermions be antiperiodic across the τ circle as well, so we are in NS-NS. Then we should just take a trace with no insertion of ( ) F. For one fermion mode, we get ( + q n ) (5) 5

6 For two fermions, and counting all the modes we get ( + q n )( + q n ) (5) n= We wish to make this into a theta function. We multiply and divide by η = q 4 ( q n ) (53) n= We also had a vacuum energy 48 for each fermion. the total vacuum energy contribution was q 48q 48 = q 4 (54) This cancels against the contribution of the η, and we get η ( q n )( + q n )( + q n θ 00 (0,τ) ) = η(τ) n= Z 00 (τ) (55) Overall we will have 8 transverse fermions, so 4 such pairs of fermions, so we will get the contribution Z 4 00 (56) (b) Now still take the NS sector on the σ cycle, but take R in the τ direction. we must now insert ( ) F, which says that for each fermion mode we get The rest is the same as before, and we get ( q n ) (57) η ( q n )( q n )( q n ) = θ 0(0,τ) η(τ) n= Z 0 (τ) (58) and with 8 fermions we will get Z 4 0 (59) (c) Now take the R sector in the σ direction, and NS in the τ direction. the vacuum energy is now zero if we consider one boson and one fermion, but this means that for each fermion we have q 4. The two fermions give q, and we get another q 4 from η, so overall we will have q 8 = e πiτ/8 = e iπτ/4 (60) Also note that for the R sector we will have two ground states, since we have 0 and d + 0. we will get another factor of from these zero modes. The rest of the fermionic oscillators will give ( + q n )( + q n ) (6) n= 6

7 So overall we get η eiπτ/4 and with 8 fermions we will get ( q n )( + q n )( + q n ) = θ 0(0,τ) Z 0 (6) η(τ) n= Z 4 0 (63) (d) Now we take R in both directions. We have ground states as before, but with ( ) F these give = 0. the result vanishes. we can write this as since θ (0,τ) = 0. ( θ (0,τ) ) 4 Z 4 = 0 (64) η(τ) 4 Combining the sectors Consider the NS sector along the σ cycle. In this sector the lowest state has energy. This is a tachyon. The next level is obtained by applying ψ i, so it is at level zero. these two states differ by a half integer. Recall that for string states we must have level matching between the left and right movers L 0 = L 0 (65) If we kept all states on the left, and all states on the right, then we will get cases where from the oscillator levels we would get L oscillator 0 L oscillator 0 is a half integer. The difference needs to be made up from the unequal p L,p R. But recall that for noncompact directions, p L = p R, and for compact directions p L = πn p L Tn wl, p R = πn p L + Tn wl (66) We have p L p R = 4(π) πα n pn w = 4 α n p n w (67) p L + 8πTN L = p R + 8πTN R (68) N L N R = 8πT (p L p R) = πα 4 8π α n pn w = n p n w (69) which is an integer. we cannot have arbitrary matches of left and right oscillator levels since half integer differences are not allowed for N L N R. we must separate out the odd and even levels. Note that only the fermion has half integer modings. Suppose we say that each time we apply a fermion we change the fermion number by unity. Let the vacuum of the NS sector have fermion number. The fermion number of defined only mod. The next level above the vacuum will be ψ i 0 NS (70) 7

8 with m = 0. For the open string we have just one sector, so we see that we get 8 massless quanta. In 0-D, these give the physical degrees of freedom of a photon. We might be confined to a smaller dimension by having the ends of the open string lie on a D-brane. Suppose we have a p-brane. Then the directions i normal to the brane will give transverse vibrations of the brane, while those along the brane will give a gauge field. we learn that D-branes can vibrate, and that they carry a gauge field on their worldvolume. Returning to the string, we can separate the off and even levels by using the projection operators P ± = ( ± ( )F) (7) What states should we keep? The bosonic states of the string will come from the NS-NS and RR sectors. The fermions will come from the RR sectors. The RR sector has integer level states. To match onto the other sector, we will have to keep the integer level states from the NS sector. we discard the odd level states by using the projection operator ( ( )F ) (7) In particular this removes the tachyon from the spectrum. It also tells us that we should combine the first two terms in the path integral (a) and (b) as (a)-(b). In the R sector, we keep one of the two chiralities of the fermion. Each time we apply a d + i, we go from odd to even fermion number, but we also change chiralities, So if we are to keep even or odd fermion number, then we have to keep one chirality of the spinor from the ground states. For the left and right sectors, we can either keep the same chirality or opposite chiralities, In the first case we get IIB string theory, and in the second case we get IIA string theory. we should use in the R sector as well. ( ± ( )F ) (73) There is one last sign that we must understand. In field theory, if we have a fermion loop, then we have to include an extra minus sign. We do this by assigning one sector the R sector - a minus sign. Then for spacetime bosons which are NSNS or RR there is no sign, while for NSR and RNS we will have a minus sign. overall we see that we must write [Z4 00 Z 4 0 Z 4 0 Z 4 ] (74) The last term is zero, and the first three vanish because of the Jacobi abstruse identity θ 4 00 θ4 0 θ4 0 (75) This vanishing tells us that we have no vacuum energy for the superstring in flat space. 8

9 5 Excitations of a D-brane We have seen that an open string can end on a D-brane. An open string has only one sector, not L R sectors. The NS sector will have a GSO projection, and so the tachyon state will be removed. The Next level is made from ψ i p (76) We have m = p = α (N ) = 0 (77) so we have massless excitations. we have 8 transverse massless modes. The directions along the brane give a gauge field A a, while those normal to the D-brane give transverse vibrations. The R sector gives a spacetime fermion, which gives 8 superpartners of the bosonic degrees of freedom. We have a 6 comonent spinor to start with, but the equation of motion cuts these down by half to 8. 6 Gauge fields Now consider the closed string. From the bosonic excitations, NSNS sector, we have ψ i ψ j 0 (78) The transverse traceless part gives a graviton, the antisymmetric part gives the B ij, and the trace gives the dilaton. In the RR sector we have for IIB, and for IIA. 0 α 0 β (79) 0 α 0 β (80) We can classify these states by inserting gamma matrices and making linear combinations. For IIB we have C 0 : 0 α 0 α (8) etc. C µν : 0 α [γ µ γ ν ] α β 0 β (8) For IIA we have C µ : 0 α β αγ β 0 (83) etc. This gives the RR fields for the theory. C µνλ 3 : 0 α [γ µ γ ν γ λ α β ] β 0 (84) 9

10 The NSNS B µν field is produced by the elementary string, but the RR fields are produced by D-branes. Closed string theory is complete in the sense that all the fundamental particle states are such that they give all the background fields that the string can propagate in. 7 T-duality The oscillator expansion of the bosonic field X µ was We can write this as Suppose we consider X µ (τ,σ) = [ xµ 0 + α p µ α L (τ + σ) + i = [ xµ 0 + α p µ α R (τ σ) + i n n α n n ein(τ+σ) ] ᾱ n n ein(τ σ) ] (85) X µ = X µ (ξ + ) + X µ (ξ ) (86) X µ = X µ (ξ + ) X µ (ξ ) (87) This will also be a solution to the equation of motion. What is the significance of this new solution? Suppose we perform the following change of variables on the world sheet Then where we have used that τ has negative signature, and ǫ τσ =. and we achieve the change mentioned above. a X = ǫ ab b X (88) τ X = σ X (89) σ X = τ X (90) ( τ + σ )X = ( τ + σ )X (9) ( τ σ )X = ( τ σ )X (9) 7. Open strings First let us see the effect of the change X X on open strings. Suppose the endpoint of the open string has N boundary condition σ X(σ = 0) = 0 (93) This gives τ X (σ = 0) = 0 (94) 0

11 which gives X (σ = 0) = x 0 = constant (95) with a similar behavior at σ = π. we get fixed endpoints, which is a D boundary condition. under this change X X we go from N to D boundary conditions and vice versa. 7. Closed strings We had with Under the change X µ X µ we get The old winding implied a distance between endpoints The new momentum must be correctly quantized or X µ = x µ 0 + α p µ τ + w µ σ + oscillators (96) p µ = p µ L + pµ R (97) w µ = α (p µ L pµ R ) (98) p µ = p µ L pµ R = α wµ (99) w µ = α (p µ L + pµ R ) = α p µ (00) L πr = πw µ = πα (p µ L pµ R ) (0) p µ = α wµ = π L = R (0) R α = R (03) R = α (04) R This is the basic T-duality relation which relates large circles to small circles, while interchanging winding and momentum. 8 Action of T-duality on fermions Let X 9 be the direction that is T-dualized. Then for the right movers we have taken X R 9 = XR 9 (05) To preserve supersymmetry, we must also take ψ R 9 = ψr 9 (06)

12 This means that in the R sector, when we make the combinations ψr 8 ± iψ9 R, we will have ψ 8 R ± iψ 9 R = ψ 8 R iψ 9 R (07) the creation and annihilation gamma matrices have been interchanged. the vacuum has been changed from 0 to Γ + (5) 0. This implies a change of chirality. the right vacuum has switched chirality, while the left vacuum has remained unchanged. we go from IIB to IIA and vice versa. 9 The Weyl-Petersen measure We will show that the measure is modular invariant. We have d τ τ τ = τ (08) (09) We find that τ + iτ = τ + iτ = τ iτ τ + τ τ = τ Q τ = τ Q τ iτ Q (0) () () τ τ = Q + τ Q, τ τ = τ τ Q (3) τ τ = τ τ Q, τ τ = Q τ Q (4) (τ,τ ) (τ,τ ) = Q (5) dτ dτ τ = dτ dτ τ (6) 0 The overall partition function The overall partition function will have d τ Z = V 0 [4π τ α ] [ η [4π τ α ] ] 8 Z ψ Zψ (7) τ

13 Symmetries of the IIb string We look at the perturbative symmetries, which come from symmetries of the world sheet action. We have Ω, ( ) F L, ( ) F R (8) Each of these squares to unity. But These make an 8 parameter group Ω( ) F L Ω = ( ) F R (9),Ω,( ) F L,( ) F R,Ω( ) F L,Ω( ) F R,( ) F L ( ) F R,Ω( ) F L ( ) F R (0) where we can either apply or not apply each of the three elements. these map to D 4, the dihedral group with 4 elements, which are the symmetries of the square. These symmetries are,r x,r y,r x R y,s,s 3,R x S,R y S () where R x is reflection in the x axis, etc, and S is rotation by a right angle. Note that R x R y = S, and we can check that SR x S = R x. We can make the map so that Note that Ω = R x, Ω( ) F L = S () ( ) F L = R x S (3) (R x S) = R x SR x S = R x R x = (4) Nonperturbative symmetries The IIB theory also has a SL(, Z) nonperturbative symmetry ( ) a b M =, ad bc = (5) c d Define The theory has two -forms and a 4-form Under this symmetry we have λ = C 0 + ie φ (6) B ij, C ij (7) C ijkl (8) λ = aλ + b cλ + d ( ) ( ) B ij = M T Bij C ij C ij (9) (30) C ijkl = C ijkl (3) 3

14 3 Relation between the perturbative and nonperturbative symmetries 3. The element R Consider the element of SL(, Z) R = I (3) Then B ij,c ij change sign, while λ,c ijkl remain fixed. This is in fact the behavior of ( ) F LΩ. Under Ω, g ij,φ,c ij (33) are fixed, while C 0,B ij,c ijkl (34) change sign. To see this, note that for the NSNS fields, we interchange left and right movers, while for the RR fields, we interchange the two spinors coming from left and right. In C 0, we have the contraction η αβ ψ α Lψ β R (35) where η is antisymmetric. C 0 is odd. For C ij we get an extra negative sign from the permutation of order of the two gamma matrices, while in C ijkl the reversal of order is actually an even permutation so because of η it is odd. Under ( ) F L, the NS sector is unchanged, while the R sector changes sign. RR states are odd, and all C fields change sign. with ( ) F LΩ we find that change sign, while the other fields are fixed. this is R. B ij,c ij (36) 3. The element S Consider the SL(, Z) transformation Then we have S = iσ (37) S( ) F L S = Ω (38) To check this, note that S changes B ij to C ij, which changes sign under ( ) F L, and then we change back to B ij. B ij changes sign, which is a property of Ω. Similarly we can check the other elements. 4