Week 11 Reading material from the books

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1 Week 11 Reading material from the books Polchinski, Chapter 6, chapter 10 Becker, Becker, Schwartz, Chapter 3, 4 Green, Schwartz, Witten, chapter 7 Normalization conventions. In general, the most convenient normalization for masses is one that makes it easier to understand in terms of the dimensions of operators. For a closed string state with momentum k, the dimension associated to : exp(ikx) : is given by k2 (1) 2 Including the level, one gets that the condition for having a physical vertex operator is k L = 1 (2) which states that we have an operator of dimension one. This means that the masses of closed strings are integers multiplied by two. For the open string, including the images, one finds that the dimension is = 2k 2 + L = 1 (3) There are two factors of two that are important. The first one is from the fact that for the open string we have a Green s function that includes the images. This doubles the number. The second one (which I got confused in class about and didn t realize I was missing) is that we need to add the holomorphic plus anti-holomorphic dimensions and get it all to add up to one. This means that for the open string tachyon vertex operator, the mass squared is equal to 1/2, but it corresponds to dimension equal to (1). If we use dimension as a normalization for the mass squared, we need to rescale the momentum. No matter what normalization we use, the closed string ground state has twice the energy of an open string ground state, and when written in terms of the mass squared, there is a factor of four difference. 1

2 Also, because mass squared and dimension differ by a factor of 2 in standard conventions, it is easy to get confused. The important thing to consider is that in the expression for the Veneziano scattering amplitude A Γ( 2s 1)Γ( 2t 1) Γ( 2s 2t 2) + (s t u) (4) The poles in s correspond to the mass squared of the intermediate state, and this is a half integer. This corresponds to integer values of the level. Also, s = (k 1 + k 2 ) 2 = m m 2 2 2k 1 k 2 for two initial states and it is the center of mass energy squared. 1 Chan-Paton indices and gauge invariance. We have seen that at level one, we can have a massless particle, whose vertex operator is described by V ɛ µ X µ exp(ikx) (5) The conditions for this vertex operator to be of dimension one are that k 2 = 0 and that ɛ k = 0. Moreover, if we have ɛ µ αk µ, we get a null state, as the vertex operator is then a total derivative. If we have two tachyons and a massless particle, we can calculate the three point vertex from the ordering 1, 2, 3, with V 1 c(0) exp(ik 1 X(0)), V 2 c(1) exp(ik 2 X(1)), and V 3 c(y)ɛ µ ( X µ ) exp(ik 3 X) We also have that 2(k 1 ) 2 = 2(k 2 ) 2 = 1, and that (k 3 ) 2 = 0. It is easy to calculate the contractions for the c and for the X separately. We get 1y(y 1) (6) from the ghosts. We get for the X contractions 2(y 1 ɛ k 1 + (y 1) 1 ɛ k 2 )y 4k 1k 3 (1 y) 4k 2k 3 (7) Using momentum conservation we find that (k 1 + k 3 ) 2 = (k 2 ) 2, so that k 1 k 3 = 0 = k 2 k 3. Also notice that since k 3 ɛ = 0, we can use momentum conservation to calculate this vertex, and replace k 1 and k 2 by (k 1 k 2 )/2 and (k 2 k 1 )/2 respectively. 2

3 The end result is (y 1 (y 1) 1 )ɛ (k 1 k 2 ) 1 y(y 1) ɛ (k 1 k 2 ) (8) We notice that in the end the dependence on y vanishes, and the Feynman rule is proportional to ɛ (k 1 k 2 ). We notice that this is antisymmetric in the exchange of 1 2, so when we add the contribution from the ordering 2, 1, 3 we get a minus sign and thee amplitude would seem to vanish. However, we have already seen that ends of strings live on objects we have called D-branes. When we have various such objects, we need to keep track on whether the strings are joining consistently: a string from brane A to brane B, joining a string from brane B to brane C would give us a string going from brane A to C. These strings from A to B are characterized by having the label A on their left edge and the brane B on the right. From the point of view of the disc, near their vertex operator we should label the boundaries according to which brane they correspond to and the label B If we reorder the vertices on the disc diagram one would see that the boundaries would not be consistent. Thus, we need to dress up the amplitudes with the brane labels A a b, which indicate amplitudes for the a label being equal to A and the b label being equal to B. If we do this, we get that these labels are N N matrices, where N is the number of branes. We also find that the vertex is proportional to (A 1 ) a b(a 2 ) b c(a 3 ) c a (A 1 ) a b(a 3 ) b c(a 2 ) c a tr(a 1 A 2 A 3 A 1 A 3 A 2 ) (9) This ends up being equivalent to having a gauge group U(N) associated to N N matrices. The tachyon fields are in the adjoint representation, and the trace form is just a fancy way to write the Lie algebra structure functions f abc. The matrix labels are called Chan-Paton factors. Nowadays it is understood that they just label different branes where the strings can end and are not that mysterious. We can also do a calculation with three vertex operators for massless vector particles, to find a similar result. The advantage is that since k 2 i = 0 for the three massless particles, we can do an expansion around k = 0, and the vertex we get has two pieces. One is proportional to tr(a 1 A 2 A 3 3

4 A 1 A 3 A 2 )ɛ 3 (k 1 k 2 )ɛ 1 ɛ 2 plus it s permutations. This is the same as the physical vertex that one associates to a three vector particle vertex in ordinary Yang Mills theory. Thus we see that string interactions predict ordinary Yang Mills interactions: the associated scattering amplitudes would be the same. There is an extra term proportional to ɛ 1 (k 2 k 3 )ɛ 2 (k 1 k 3 )ɛ 3 (k 1 k 2 ). This would come from a term in the effective spacetime lagrangian of the form tr(f µ ν F ν ρ F ρ µ) (10) This is usually non-renormalizable (and an irrelevant operator in the renormalization group). In effective field theory one usually assigns to this vertex some value at some UV scale, that is of order 1/Λ 2 UV. Here we see that the value is fixed string theory, so one finds that the effective field theory is in general much more constraining than ordinary effective field theory, and the theory is predictive. 2 OPE s and factorization One of the important things that we need to understand is how poles in intermediate channels develop. Let us take an example of two vector particle vertex operators at positions 0 and y, and let us consider the process of integrating over y near zero, with however many other vertex operators the the right of y. The two vertex operators that we need are V 1,2 ɛ 1,2 X exp(ik 1,2 X) (11) If y is close to zero, we can replace the two vertex operators by their OPE, so we need to do an integral of the form 1 0 dyv 1 (y)v 2 (0) (12) where for convenience we have put the finite value of y 0 that is the end of the integration limit at one. Using the OPE, we find that V 1 (y)v 2 (0) c 123 y V 3 (0) (13) 4

5 where for simplicity we have taken only the most singular term in the OPE. We can now do the integral over y. The dimension of V 3 i not necessarily at a physical value, but instead it depends on the total momentum (k 1 + k 2 ) and it is linear in (k 1 + k 2 ) 2 s, the Mandlestam variable. Doing the integral over y is straightforward, and it gives 1 0 y A 2s dy 1 A + 2s 1 (14) in the region of the complex plane where s is very negative, so that the integral converges. The pole happens exactly when the integral becomes logarithmically divergent. One can continue around that point to define the scattering amplitude on the physical region, but the pole can not be removed. It is physical. We notice that this develops poles in the s channel exactly when A+s/2 1 = 0. If we use 1 = 2(k 1 ) 2 + L 1 = 1, (where L 1 1 ends up being twice the mass of the particle), etc, we find that it is exactly in the region where the dimension 3 1 that the pole develops. If we cutoff the integral at epsilon, we get that the integral is proportional to log(ɛ). On can easily show from the conformal mapping to the vertical strip that this is proportional to a long time on the Lorentzian worldsheet. A particle propagating for a very long time is one that is on-shell. Off-shell particles (not real particles) usually display exponentially damping behavior to indicate the uncertainty principle allowed violation of energy conservation when doing things at finite time. Having dimension equal to one is exactly what makes a primary operator into a physical state. Notice moreover that the residue of the pole is proportional to the OPE coefficient, so that we have that the amplitude would factorize into two disc diagrams, with the Feynmann rule at the 123 vertex being given exactly by the OPE coefficient on-shell. This is the standard statement that field theory ampltiudes factorize into products when one of the intermediate particles is on-shell, which is a property of any theory that can be defined using Feynman diagrams. What we see is that this prescription is consistent. ONe last thing to notice, is that if we take two of the vector particles and their OPE, on the right hand side we can get a tachyon vertex operator, and the coefficient of the OPE does not vanish when the tachyon is on-shell. To get rid of that tachyon and get a more meaningful theory, we would need to contrive a way to make that residue vanish. Another thing we would 5

6 want to have is spacetime fermions. As we have already seen, we can get spin representations from fermion zero modes on the worldsheet. however, if the fermions are independent of the bosons, we would have spin representations in some internal symmetry space. Instead, we want the fermions to be connected to the bosons, so that the fermion zero modes give rise to the gamma matrices of spacetime. Thus, we would want ψ i to be tied to X µ so that the two indices coincide. To do this, we want a symmetry that lets us turn a X µ into a ψ µ and viceversa. Such a symmetry is called supersymmetry. 6

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