Canonical quantization of a string describing $N$ branes at angles

We study the canonical quantization of a bosonic string in presence of N twist fields. This generalizes the quantization of the twisted string in two ways: the in and out states are not necessarily twisted and the number of twist fields N can be bigger than 2. In order to quantize the theory we need to find the normal modes. Then we need to define a product between two modes which is conserved. Because of this we need to use the Klein-Gordon product and to separate the string coordinate into the classical and the quantum part. The quantum part has different boundary conditions than the original string coordinates but these boundary conditions are precisely those which make the operator describing the equation of motion self adjoint. The splitting of the string coordinates into a classical and quantum part allows the formulation of an improved overlap principle. Using this approach we then proceed in computing the generating function for the generic correlator with L untwisted operators and N (excited) twist fields for branes at angles. We recover as expected the results previously obtained using the path integral. This construction explains why these correlators


Introduction and conclusions
Since their introduction, D-branes have been very important in the formal development of string theory as well as in attempts to apply string theory to particle phenomenology and cosmology. However, the requirement of chirality in any physically realistic model leads to a somewhat restricted number of possible D-brane set-ups. An important class is intersecting brane models where chiral fermions can arise at the intersection of two branes at angles. An important issue for these models is the computation of Yukawa couplings and flavour changing neutral currents.
Besides the previous computations many other computations often involve correlators of twist fields and excited twist fields. It is therefore important and interesting in its own to be able to compute these correlators. As known in the literature [1] and explicitly shown in [2] for the branes at angles case and in less precise way in [3] for the case of magnetized branes these computations boil down to the knowledge of the Green function in presence of twist fields and of the correlators of the plain twist fields. In many previous papers correlators with excited twisted fields have been computed on a case by case basis without a clear global picture, see for example ([5], [6], [7]).
In this technical paper we have analyzed the N excited twist fields amplitudes with L boundary vertices at tree level for open strings localized at D-branes intersections on R 2 (or T 2 ) and we rederive the results of [2].
We will nevertheless follow a different approach from most of the literature in a twofold way. Firstly, we use the so called Reggeon vertex [25], which allows to compute the generating function of all correlators, in particular we will use the formulation put forward in [26]. Secondly we use the canonical quantization approach while all the previous literature has used the classical path integral approach ( [1], [8]). In the case at hand the path integral approach is more efficient than the also classical sewing approach ( [9], [10]). This approach has been explored in many papers in the branes at angles setup as well as the T dual magnetic branes setup see for example ([11], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24]).
At the heart of the path integral approach is the idea that the interaction of a string with twisted strings in the fundamental state can be replaced by a discontinuity on the string boundary conditions. This is depicted in figure 1. We use this idea as the starting point of our computation based on canonical formalism. The fact that we have boundary conditions which change with the worldsheet time implies that we have a very mild worldsheet time dependent worldsheet metric. Hence the usual quantization cannot be applied in a straightforward way but we have to find a proper way to defined the product between modes. This is done using the Klein-Gordon product used in General Relativity. To have a well defined, worldsheet time independent product between two modes implies that we have to split the string into a classical and quantum part and quantize the quantum part only, exactly as in the path integral approach. Since this procedure is at variance with the usual one we check that we recover the standard results in the cases of the untwisted X(σ, τ ) Figure 1: The interacting strings are mapped into boundary condition discontinuities.
string and of the usualm twisted ones. All this is done in section 2 using the Klein-Gordon product used in General Relativity. In this section we discuss also the expression of the Hamiltonian in term of oscillators. We find that it is quadratic in oscillator but not diagonal since we have an almost free theory with worldsheet time dependent background. We derive also the modes for the three twists case. We are able to find many orthogonal basis but all of them are missing of one mode with respect to the basis of the in string. This fact can be partially understood as the consequence of the boundary "interactions" which break some symmetries of the original in string. Since we have not understood this issue completely, we resort to using an improved version of the standard overlap approach which is also used in quantum mechanics in presence of discontinuities of the Hamiltonian. In section 3 we tackle the problem of computing the in and out vacua which necessarily differ since we have a worldsheet time dependence in the boundary conditions. In principle it should be possible to compute them from the basic principles. Because of the not complete understanding of modes we compute the out vacuum as a kind of surface state (an exponential of an expression quadratic in the operators) of the in vacuum assuming the knowledge of the Green function. This is however not a big issue since it can be derived using the analytic properties and boundary conditions.
In section 4 we perform the actual computations of the generating functions for amplitudes involving plain and excited twisted states. This is done in steps. First considering the amplitudes with plain unexcited twisted fields and arbitrary untwisted states. Then considering amplitudes with excited twisted states without untwisted ones and finally, assembling all.
Our main result is to be able to rederive in a different way the generating function of correlators with N excited twists and L untwisted states found in [2]. It is given in eq. (129) which shows that all correlators can be computed once the N plain twist operators correlator together with the Green function in presence of these N twists are known. This expression requires the precise knowledge of the Green function 1 and its regularized versions. Luckily these are well known. From these expressions it is clear that the computation of amplitudes, i.e. moduli integrated correlators, with (untwisted) states carrying momenta are very unwieldy because Green functions can at best be expressed as sum of product of type D Lauricella functions. This should however not be a complete surprise since in [27] it was shown that twist fields correlators in orbifold setup are connected to loop amplitudes which, up to now, have not been expressed in term of simpler functions.

The setup of branes at angles
The Euclidean action for a string configuration is given by The complex string coordinate is a map from the upper half plane to a closed polygon Σ in C, i.e. X : H → Σ ⊂ C. For example in fig. 2 we have pictured the interaction of N = 4 branes at angles D t with t = 1, . . . N . The interaction between brane D t and D t+1 is in f t ∈ C. We use the rule that index t is defined modulo N . As shown in [3] given the number of twist fields N there are N − 2 different sectors. They are labeled by an integer M , 1 ≤ M ≤ N − 2 which is in correspondence with the number of reflex angles (the interior angles bigger 1 Note that the Green functions used in this paper are dimensionful and normalized as than π), more precisely M is N − 2 minus the number of reflex angles as shown in figure 3 in the case N = 6. The intuitive reason why they are different is that we need go through the straight line, i.e. no twist, if we want to go from a reflex angles to a convex one.

Splitting into classical and quantum part and Klein-Gordon metric
In order to proceed with the canonical quantization we want finding the normal modes associated with the equations of motion and the boundary conditions The previous constraints are simply stating that when x t < x < x t−1 a boundary of the string is on the brane D t . The brane D t is described in a well adapted coordinate system as √ 2iX 2t = e −iπαt X z − e iπαt Xz = 2ig t ∈ iR, i.e. it extends along X 1t with √ 2X 1t = e −iπαt X z + e iπαt Xz. Therefore the string has Dirichlet boundary condition in the X 2t direction and has Neumann boundary condition in the perpendicular direction X 1t . In particular √ 2|g t | is the distance of the brane from the origin and is the intersection point between D t and D t+1 . The configuration can be pictured as in figure 4 in the Euclidean case. This configuration corresponds to the Minkowskian configuration of figure 2 where both the incoming and the outgoing strings are untwisted. A similar configuration is drawn in figure  5. In this case the incoming string is twisted because one twist is sitting in the origin and the outgoing is untwisted. Obviously there is also a third possibility pictured in figure 8 where both the incoming and outgoing strings are twisted, this corresponds to the case where there is a twist at x = 0 and one at x = ∞. When there are N = 2 twists this is the usual twisted string. The issue is now to find a (non positive definite) product for the modes which is conserved in (Euclidean) time. This issue is less trivial than usual because our spacetime, i.e. the worldsheet, is changing with (Euclidean) time even if in a mild way through the change of the boundary conditions. The solution to this problem is well known in General Relativity. Since we are dealing with free fields satisfying the Klein-Gordon equation we know that there is a conserved current. Explicitly given any two solutions is conserved because of the equation of motion. This is nevertheless not sufficient to get a conserved (non positive definite) product, in fact we must deal with boundary contributions. In the Euclidean case we consider the surface S(r 0 , r 1 ) in the upper half plane delimited by two semicircles of "time" r 0 and r 1 (r 0 < r 1 ) and by the two segments on the x axis [r 0 , r 1 ] and [−r 1 , −r 0 ] then we find 2 In order to get a metric independent on time r we need rewriting the two integrals along the x axis as a difference of a function which depends only on the background fields at the initial and final times. While we know that we can write a definite integral as a difference of a function evaluated at final and initial times what it is not certain is that this difference does depend only the final and initial background fields since we have boundary conditions with discontinuities.
The two integrals along the x axis can be expressed as a difference of a function which depends only on the background fields at the evaluation time and actually vanish only if we consider solutions F q which satisfy the boundary conditions in eq.s (3), with g t = 0, i.e.
In the following we call these boundary conditions quantum boundary conditions. The quantum boundary conditions can be also written as Consider then one of the x boundary contributions 2 Our conventions are * du = −i du, * dū = i dū.
Let us consider the first term ∂ y F † 1 F 2 . We can split the integration interval into pieces where the boundary conditions are constant. Then using the quantum boundary conditions we have the identity F 2 = I+Rt 2 F 2 and therefore we can write If we would not use the quantum boundary conditions but the original boundary conditions the second identity in (8) would read (I − R t )F | y=0 = 2G t = −2ig t (e 2παt , e −iπαt ) T then the contribution from a piece of the integration interval where the boundary conditions are constant would be −∂ y F † This term can be integrated explicitly using the fact that F s split into a sum of left and right moving pieces and it is non vanishing. Hence the resulting boundary contribution (10) does depend not only on the fields at r 0 and r 1 but also on the fields at the discontinuities between r 0 and r 1 . This means that the would be product does depend on the history and not only on the background fields at the time where the metric is computed and hence there is no time independent product of modes. We conclude therefore that only for solutions satisfying the quantum boundary conditions (8) we have the non positive definite metric (actually Hermitian form) This means that we must proceed as in the path integral approach and split the string coordinate into a classical and quantum part and then quantize only the quantum part. Explicitly we write with X cl satisfying the original boundary conditions (3) and and X q satisfying the quantum conditions (7). Notice that only with the quantum boundary conditions (7) the two dimensional laplacian ∂ u∂ū is self-adjoint and it is certain to have a Green function. This is shown in appendix A. Finally, notice that the previous discussion can be and must be applied also to branes with magnetic field. In this case the Minkowskian boundary conditions read X I − F IJsẊ J | σ=s = 0 with s = 0, π and F IJs the magnetic fields. Then the boundary contribution corresponding to (10) is not anymore zero but evaluates to and therefore induces a product which is the "weird" metric used in [4].

Doubling trick and the metric
We have established that we must quantize the fluctuation around the classical solution which satisfy the quantum boundary conditions (7) we now look to solutions to equations of motion with these boundary conditions. As usual the general solution of eq. (2) is given by X I (u,ū) = X I L (u) + X I R (ū). Then because of the boundary conditions we are led to consider the two possible independent sets of quantum modes where (X z q,L (u), Xz q,R (ū)) T can be any element of the basis of solutions and is labeled by a further basis index which is suppressed in this section. Similarly for (X z q,R (ū), Xz q,L (u)) T . After this splitting the couple of quantum boundary conditions in eq. (8) become simply one condition, one for each set, explicitly 3 since the derivative boundary condition follows from the previous because In the case of the classical part the previous equations would have been stated using the derivatives because there is no obvious way of splitting the constants g t into a left and right part. Using derivatives we miss the information on the constants g t which has to be kept adding further conditions. Explicitly the boundary conditions for the classical part can be written as (when all x t = 0, ∞) These X (c) (u,ū) andX (a) (u,ū) can be combined into two sets of function defined on the whole complex plane minus the cut [x N , x 1 ] using the doubling trick as 4 3 The untwisted case requires a slightly more general solution because of the log |u| is possible and it is needed for , i.e. we could ask X z q,L (x + i0 + ) = e i2παt Xz q,R (x − i0 + ) + δ Lt , X z q,R (x − i0 + ) = e i2παt Xz q,L (x+i0 + )+δ Rt but then the non derivative boundary conditions imply that δ Lt +δ Rt = 0 and that there is a unique non chiral solution X I L (u) + X I R (ū). 4 Notice that in order to perform the gluing this way we need the D 1 to be the last brane on the real positive axis. withX (z) = (X (z)) * and very simple boundary conditions Obviously the gluing can be performed in different ways, i.e. we can glue along Dt brane instead of D 1 as again withX (t) (z) = (X (t) (z)) * and boundary conditions In the following we use always the gluing along D 1 if not otherwise stated. The metric (12) can then be calculated for any pairs of these functions using the doubled formalism as (for more details see appendix B) where the last equation is meaningful since the product (X (r 0 e −iθ )) * dX (r 0 e iθ ) dθ is continuous for θ = 0 despite the fact its factors are not. A direct computation similar to the one done to get the previous equation (24) gives which is obviously compatible with the Hermitian property of the form and the product (X (c) ,X (a) ) in eq. (24).

Radial canonical quantization
We want now quantize the Euclidean string action (1). In order to do so we split the string field into its classical and quantum part (13) so that the action becomes We take τ E = ln r to be the time but we write all expressions as functions of r, this means that the Hamiltonian rescales r = |u| = e τ E and not that it shifts r. The canonical momentum is then given by and the Euclidean Hamiltonian is by definition From the canonical commutation relation with δ(θ − θ ) the delta function with the appropriate boundary conditions, we get In order to write the Hamiltonian using the creation and annihilation operators we need a way to extract them from the quantum field X q by mean of a product of the quantum fluctuation X q with an appropriate solution F . This product can be written as In particular the commutation relation of two such products is given by where σ 1 is the Pauli matrix. When choosing the two solutions F 1,2 to be any of the basis elements the previous commutation relation become X (c)n andX (c)n are a basis for the quantum modes. In deriving the first equation in the first line we used the fact that any σ 1 X * (c)n has the same boundary conditions as any of X (c)m and hence they can be expanded on the X (c)m basis. Then we can use eq. (23) to set to zero the commutation relations involving two X (c) modes. Similarly forX (a)n .
We can now simplify the previous equations (33) if we notice that the functionχ •n (z) associated with σ 1X * (c)n by the doubling trick can be rewritten asχ and similarly for χ •n (z) = e i2πα 1 [χ n (z)] * which is associated with σ 1 X * (c)n . Using these relations, the hermiticity property of the product and the explicit expression in eq. (24) for the product in terms of the doubled mode functions we can then write the non vanishing commutation relations for the basis elements (33) as We now expand the quantum fluctuation as 5 We suppose that the (doubling of) quantum modes satisfy a reality condition like and a normalization condition with s an integer and N * n = −N n e −iβ−iβ . It follows that the creation and annihilation operators can be obtained as and that they satisfy the commutation relations On general ground the Euclidean Hamiltonian defined in eq. (28) is generically not diagonal in the mode operators since the Lagrangian is time dependent or that is the same in states are not equal to out states and hence they cannot be eigenstates of the Hamiltonian. Another way to understand this is to notice that the Hamiltonian rescales u but the generic mode function is not an homogeneous function of u. Explicitly the Hamiltonian can be expanded in modes as where h nm is constant despite the Lagrangian is time dependent because of the boundary conditions. This happens because r∂ r X (c)m are also solutions of the e.o.m. and therefore satisfy the same "selection rules" as X (c)m .
H is not diagonal in modes even if it is self-adjoint w.r.t. the usual L 2 metric because it is self-adjoint for any time τ but it does depend on time.
Finally, notice that we have not written any normal ordering since its definition depends on the vacuum and we have not specified any. Neither we will do it since we will use an overlap approach which uses different Hamiltonian for different worldsheet times. See sections 2.7 and 3.

The N = 0 case: the usual untwisted string
Since the previous product for the modes is different from the normal one it is of interest to see how it works in the usual and simplest case with ND boundary conditions. It is also worth to check that we get the same commutation relations as in the quantization with the normal product.
We consider a single D1 brane D t in R 2 . In this case the boundary conditions are simply for all The classical solution is simply given by The two sets of modes in eq.s (16) which obey the quantum boundary conditions are Notice however thatX 0 is different from all the others elements since its components are neither holomorphic nor antiholomorphic since neither the holomorphic nor the antiholomorphic parts satisfy the boundary conditions separately.
Furthermore X * is not proportional to the classical solution X cl , as it could at first glance look because of the sign of the second component.
Using the doubling trick the previous modes can be combined into the following functions defined in the whole complex plane Using these functions and eq.s (23) and (24) we can compute the non vanishing products of these elements where all products involvingX 0 cannot be computed using eq. (24). In fact this equation is derived under the hypothesis that the two functions can be assembled into (anti)holomorphic doubled functions defined on the complex plane. Therefore these products must be computed from the original definition given in eq. (12).
We can now expand the quantum fluctuations as The operators can be extracted from the previous expansion as and then we can compute the non vanishing commutation relations as These are exactly the usual commutation relations and expansion once we identify (n > 0) As usual the vacuum is defined as We can then compute the untwisted Green functions Notice that G zz U does not feel whether the brane is rotated while both G zz U and Gzz U do because of the phases. Finally, we consider two parallel branes non overlapping then the non derivative boundary conditions become while the derivative one is left unchanged. The classical solution is then while the quantum fluctuations remain unchanged. Notice however that the classical solution has infinite action because of the finite and constant energy density of the stretched string.

A N = 2 case: the usual twisted string
We now consider the quantization of two D1 branes at angles, D t and D t+1 . This is the usual setup where the string describes a twisted in and out state.
This means that the twist fields are located at x = 0 and x = ∞ so that the boundary conditions read The classical solution is then simply given by the constant which is a special case of eq. (4). The quantum fluctuations can be expanded on These can be combined into the function defined on the whole complex plane minus the real negative axis as follows from eq.
The non vanishing products of these elements are As in the previous case the quantum fluctuations can be expanded as The coefficients can then be extracted as and their non vanishing commutation relations are With the identification (n ≥ 0) we recover the usual commutation relations Finally, we can write the usual expansion for the quantum fluctuations as The vacuum is defined in the usual way by We can then compute the N = 2 twisted Green functions where we have used as follows from the general expression for the hypergeometric function 2 F 1 (a, b; c; x) = ∞ n=0 (a)n(b)n n!(c)n x n with (a) n = Γ(a + n)/Γ(a) the Pochhammer symbol. They have the following symmetry properties which follow from the hypergeometric transformation properties, in particular 2 F 1 (1, ; 1 + ; x) = ¯ x 2 F 1 (1,¯ ; 1 +¯ ; 1/x) implies the last equality.

A N = 3 case: in and out twisted strings
In this section we want to exam the next simplest example which corresponds to the case of N = 3 twists, one located in the origin, one at infinity and the third in an arbitrary point of the positive real axis. This example shows the main issues to be understood and solved. When we look after the classical solution with finite action it is naturale to consider a basis of the derivatives which can be used to compute the classical solution like ∂ z χ cl (z). Nevertheless when we try to use their integrated expressions dz∂ z χ cl (z) as a basis for the quantum fluctuations we realize immediately that they do not satisfy all the boundary conditions. We are therefore forced to consider a combination of them but in doing so it seems naively that we cannot find all possible asymptotics behaviors for u → 0. The apparent solution of this problem seems to be very simple. It amounts to consider from start a basis χ q (z) and not a basis for derivatives. This basis is however not apt to easily find the classical solution since the configuration with finite action is the combination of infinite basis elements. Moreover the simplest basis is not orthogonal and therefore we must orthogonalize it. In doing so combinations of infinite basis elements must be considered. It happens then that some new basis elements have a finite convergence radius and must be analytically continued.
In performing the analytic continuation we find that these new basis elements do not satisfy anymore the original and required boundary conditions. Hence one of the new basis element must be used to make further combinations which respect the boundary conditions and in order to preserve orthogonality it must be dropped as independent basis element. In this way we find again the original problem, i.e. we cannot find all possible asymptotics behaviors for u → 0. Since we are not yet in stand of understanding completely this issue we use the classical overlap approach to derive the correlators.
Let us now explain in more details the issues discussed above. In order to apply the general formula (19) we need setting D 1 as the last brane on the real positive axis therefore our setup is with D 3 for x 3 = −∞ < x < x 2 = 0, D 2 for x 2 = 0 < x < x 1 and D 1 for x 1 < x, i.e. we have the boundary conditions The setup is shown in figure 6.
x 1 x 1 x 2 = 0 Figure 6: Three different pictures of the setup with incoming and outgoing twisted strings with a discontinuity on the boundary at x 1 . In the pictures an incoming string between D 2 and D 3 is pictured.
A basis for the derivative of the classical solution on C − [x N , where the power of the cut at z = x 1 is dictated by the boundary conditions and the finiteness of the contribute from the area around u = x 1 to the Euclidean action S E . The same requirement from u = 0 implies n = m = 1. Since 3 t=1 t = M = 1 only ∂χ cl,0 has then a finite Euclidean action. The cuts of these basis elements are pictured in figure 7. Then the classical solution is given by 7 where the constant c 0 is fixed by the condition χ cl (x 1 ) = f 1 .
We can now try to use the integrated eq.s (72) to find a basis for the in quantum fluctuations, i.e. for the fluctuations which behave as the solutions (59) as z → 0. So we can for example write where the lower integration extreme is chosen to fulfill the quantum boundary condition χ cl (x 1 ) = 0 which can be understood as the fact that when g t = 0 for ts all branes intersect at the origin, i.e. f t = 0 for the quantum fluctuations.
The integrals with n ≥ 1 behave as const n,0 + const n,1 z −n−¯ 2 for z → 0 . This is not the desired behavior for z → 0 because of const n,0 . The required behavior needs const n,0 = 0 as follows from eq. (59) for z → 0. Therefore we must subtract the constants const n,0 . This can be achieved by making linear combinations in many different ways, for example between ∂χ cl,n (w) with n > 1 and ∂χ cl,n=1 (w). This approach produces a basis but gives the impression that we are missing the n = 1 mode. Even if we choose a different way of substracting the constants const n,0 we get one constraint and hence one mode "less" than expected.
To investigate better this point we start directly with a basis 8 for the quantum fluctuations which satisfies all the boundary conditions given by where cuts are running from −∞ to either x 2 = 0 or to x 1 as in figure 7. They are located in this way in order to satisfy the boundary conditions. To these functions defined on C − [x N , x 1 ] = C − (−∞, x 1 ] correspond the modes defined on the upper half plane H Notice that it is not immediate to use this basis to find the classical solution since all these modes have infinite Euclidean action and only a combination of them reproduce the classical solution. In particular implies that the derivative of the classical solution ∂χ cl,1 (z) is given by an infinite sum of quantum basis elements. We now compute the non vanishing products of two quantum basis elements and get This means that these modes are not orthogonal and we need to find combinations which are orthogonal. For any B ∈ Z we can then search orthogonal solutions in the form The reason of such an ansatz is that given the integer B, χ q,n = ξ q,n with n ≤ −B − 1 andχ q,m =ξ q,m with m ≥ B − 1 are automatically orthogonal and are the biggest set of orthogonal elements among the original ξ andξ. The explicit solution is then In particular we recognize that for n ≥ B when we setc n n = 1. This is one of the solution associated with the Papperitz-Riemann symbol of the hypergeometric The last form of the P symbol clearlu shows that the indeces around the singular points are the desired ones. The P symbols also shows that the other solution is simply the constant 1. This is important when performing the analytic continuation of the hypergeometric to the region |z/x 1 | > 1.
In fact in the region |z/x 1 | > 1 we find that the two independent solutions mix and the analytically continued solution does not satisfy anymore the required boundary conditions. This means that we need again to consider combinations of the solutions found above in the |z/x 1 | < 1 so that their continuation satisfies the proper boundary conditions. Therefore we find again one mode less than expeceted. Another way of seeing the problem is to notice that the hypergeometric equation associated with the symbol (82) has a = 0, b = n + 1 −¯ 3 and c = n + 1 + 2 so that the hypergeometric equation reads with w = z/x 1 . This can immediately integrated and gives exactly as for the classical solutions in eq. (74). It has therefore the same problems with the boundary conditions as before. This problem can be generalized to the generic N and it is that one asymptotic mode (either for u → 0 or equivalently for u → ∞) is missing for any twist field we insert at x t = 0, ∞. This is caused by the constraints X q (x t , x t ) = 0 we have to impose on the quantum fields. Physically we can partially understand this as a breaking of some symmetries, for example if the in string is untwisted the momentum conservation of the whole is broken because the other branes at angles. Another naive way of understanding is that the classical solution freezes these modes. It would however be nice to have a better understanding of this problem and of how to choose the basis. From the previous discussion we learn however a rule how to write a system of orthogonal modes for a generic N . The rule is simple up to the further linear combinations/subtractions in order to satisfy the boundary conditions in the whole upper plane when performing the analytic continuation. We start from a maximally orthogonal subset of the would be quantum modes χ q andχ q . This subset corresponds to the modes χ q,n = ξ q,n with n ≤ −B −1 andχ q,m =ξ q,m with m ≥ B − 1 in the case of this section. Since the product is roughly dzχ q (z) dχq dz , we can define the remaining orthogonal modes by choosing their derivatives dχq dz so that the products χ q (z) dχq dz have only simple poles. This practically means that dχq dz ∼ dχ cl dz . In the case of this section in the integrated form this corresponds to the modes given in eq.(84).

Normal modes and improved overlap approach
The issue is how to cope with the problem in finding a basis of orthogonal modes with the proper boundary conditions and havig all the possible asymptotic modes. From the discussion in the previous section this is probably impossible. The simplest way to proceed further is to consider the standard approach in quantum mechanics. The standard procedure in quantum mechanics in presence of discontinuities is using the overlap of the wave function and that the basis of the incoming wave functions is complete and orthogonal. This is essentially the same approach used by Cremmer, Gervais, Kato, Ogawa and Mandelstam in the early seventies in computing the three-string vertex ( [41], [42]) and the extended to the case of the D1 string in [43].
There is however a slight subtlety to be stressed and understood before we can apply the overlap method to our case. In previous papers there were three strings and the conditions imposed were the overlap of the string coordinates X and their momenta P . In our case we have only one string with discontinuous boundary conditions but we still want to impose the continuity of the coordinate. In the case N = 3 of section 2.6 we could think of simply using the in and out expansions and require at the transition point with X cl the constant given in eq. (57) and X q (u,ū) the quantum field given in eq. (66) for in and out strings. Nevertheless in the case where multiple transitions are encountered as we discuss in the next section it is not clear which is the classical part of the intermediate strings. We are therefore led to proceed as follows. First we split X(u,ū) into the classical solution X cl (u,ū) and the quantum part X q (u,ū). The classical solution has a global nature and must be computed for the whole evolution before proceeding to the second step. Second we require that only the quantum fluctuations overlap. This approach is a refined version of the overlap approach and is consistent with the naive idea of overlap of the whole string coordinate.
We formulate therefore the overlap approach as 9 where X (t) q (u,ū) is the quantum fluctuation of the string comprised in the half annulus with x t < |u| < x t−1 and having the appropriate boundary conditions.

Green function, in and out vacuum
From now on we adopt the strategy outlined in the previous section and focus on the configuration pictured in figure 8 where both in and out states are twisted. This means that x N −1 = 0 and x N = ∞.
In principle we should be able to derive the Green function from the canonical quantization. The derivative of the Green function is given in operatorial formalism by Figure 8: Map from the stripe to the upper half plane to the target polygon Σ with twisted in and out strings. but since we have to yet completely understood the global modes we assume we compute the Green function in the usual way i.e. using its singularities and boundary conditions. Notice that in the configuration we consider there is implicit the limit x N → ∞, however this limit does not require any special treatment or factor since the CFT definition of the derivative of the Green function gives Comparison between the two previous expression (86) and (87) suggest that we can identify for the configuration we consider. In the following we will assume such a natural identification. Given the Green function we can consider the region |u| < x N −2 . Using the overlap approach we can identify ∂ u X I q (u,ū) with the twisted string ∂ u X I T (D N −1 D N ),q (u,ū). This twisted string has the boundary conditions which follow from being attached to the D N −1 brane on x > 0 and to the D N brane on x < 0. The derivative of the Green function can then be written as In particular we can identify the in vacuum with the vacuum of the twisted Hilbert space After performing the normal ordering with respect to this vacuum the previous equation can then be written as where we have defined the Green function regularized at the twisted interaction point x t with t = N − 1 (in the case at hand From equation (91) we can then determine the coefficients which enter the operatorial expression for the out vacuum with B nz,mz = B mz,nz . A simple computation gives N = 0 out |0 in . If we take IJ = zz in eq. (91) we get the ∂ u ∂ v derivative of ∞ n,m=0 If we consider the analogous expressions of eq. (91) with the derivatives ∂ u ∂v, ∂ū∂ v and ∂ū∂v we find the corresponding derivatives of the previous equation (94). The derivative expressions can be integrated to obtain eq. (94) where the integration constant is fixed to be zero because of the cut structure. Similarly if we take IJ = zz we get the derivatives of ∞ n,m=0 where the integration constant is again fixed to be zero because of the cut structure. There is also the third possibility given by IJ =zz. Nevertheless all three expressions for the B coefficients are equivalent because of the ∆ IJ boundary conditions. The previous equation can also be rewritten in a more compact form as where ∆ IJ (N,M )(t=N −1)LL (z; w) is the holomorphic part in both u and v analytically continued in the complex plane minus cuts.
Finally let us comment on why we have not explicitly indicated the x N → ∞ limit in all the previous expressions. The reason is that we could repeat all the previous arguments for a configuration with a finite x N . In particular also for an outgoing untwisted vacuum. There would be two minor differences. The one difference with the previous results is not taking the x N → ∞ limit. The second one is the relation between the correlator of the plain twist fields and the product of the in and out vacua which would read

The generating function of amplitudes
In this section we would like to perform the actual computations of the generating functions for amplitudes involving plain and excited twisted states. This is done in steps. First considering the amplitudes with plain unexcited twisted fields and arbitrary untwisted states. Then considering amplitudes with excited twisted states without untwisted ones and finally, assembling all.

Amplitudes with untwisted vertices and plain twist fields
Given the previous knowledge of the out vacuum as a sliver constructed on the in vacuum we can now easily compute boundary amplitudes with vertices in the |u| < x N −1 region. In particular we want to compute the generating function of the amplitudes with |x i | < x N −1 . ξ i is a generic untwisted state living on D N −1 or D N and V ξ i (x i ) its emission vertex. To this end we need mapping the abstract vertex operator V ξ i (x i ) into its operatorial realization in the twisted Hilbert space . This mapping is realized using the SDS vertex S(x; {c I }). The SDS vertex for the twisted Hilbert space H T (D N −1 D N ) is explicitly given by where : · · · : is the normal ordering in the twisted Hilbert space. We defined the boundary regularized Green function for the twisted Hilbert space at the point x where t x is the index of the brane on which the vertex with coordinate x is.
In the case at hand we have actually only two possibilities, either 0 < In this expression G IJ N =2,T (D N −1 D N ) (u,ū; v,v; {0, α N −1 ; ∞, α N }) the same Green function used in eq. (92) and given in eq. (68). G IJ U (tx) (u,ū; v,v) is the Green function of the untwisted string with both ends on D tx and given in eq. (53).
Then we can realize the mapping as In order to compute the generating function of all the correlators like (98) it is then enough to insert a SDS for any untwisted operator into the radial ordered expression To compute explicitly the previous expression we have to split the full fields X I T (D N −1 D N ) into classical and quantum parts and then normal order the quantum parts. After these operations we get where bou(i) is a short hand for bou(x i ). The term in the last line can be evaluated using the analogous expression of 0|e βa 2 e αa † |0 = e βα 2 for an infinite number of oscillators and gives where ∆ IJ (N,M )(t=N −1) is the expression in eq. (92). When we assemble all contributions we find finally where we have introduced the boundary Green function in sector M regularized at the pointx i The expressions of the two terms are given in eq.s (92) and (100). We have now found the generating function for the emission of untwisted states from either D N −1 of D N and with |x i | < x N −1 . Eq. (105) is the same expression found in [2] (eq. 80 in section 4.1) when we drop the x N → ∞ limit, we let thex i be generic and not constrained by |x i | < x N −2 and we substitute . The expression in [2] is however valid without the constraint onx. The reason for this substantial equality is quite simple. Suppose we want compute the generating function for the emission of untwisted states from any of the brane of the configuration we consider. We need therefore computing the analogous expression of eq. (102) where the SDS vertices are the proper ones for the untwisted states. For example, for x t <x i < x t−1 the emission of an untwisted state is from the brane D t . The corresponding operatorial vertex is realized in the twisted Hilbert space H (DtD N ) using the full fields X I T (DtD N ) . The map abstract from operator to its operatorial realization is then performed by a SDS vertex analogous to (99) but defined in the twisted Hilbert space H (DtD N ) using the fields X I T (DtD N ) . To compute the analogous expression of eq. (102) but with no constraints on the location of the vertices we proceed as follows. We split all the full fields X I T (DtD N ) into classical and quantum parts. Then we can use the continuity equation (85) for the quantum part to perform an analytic continuation of the quantum part of any SDS vertex from an arbitrary region to the region |u| < x N −2 . This seems at first sight quite difficult because of the normal ordering which differs among the different twisted Hilbert spaces. Fortunately it is not so. The key observation is that the normal ordered SDS vertex in eq. (99) is obtained from a non normal ordered vertex as with t x = t when x > 0. We can then easily perform the analytic continuation and get where it is necessary to take the η → 0 + limit in the last line only after we have continued back the result to the original position. Only then the difference G IJ U (t) − G IJ U (N −1) which would otherwise be divergent in the limit η → 0 + combines with ∆ IJ (N,M )bou(i) given in eq. (106) and valid for |x| < x N −2 to give the finite boundary Green function regularized at the original point Notice that this analytic continuation is performed only on the operators and not on the possible polarizations and momenta which are still the ones allowed in the original region.

Amplitudes with chiral vertices and plain twist fields
In this section we would like to compute the correlators of L c chiral vertices as a warm up for the next section where we compute the correlators of N excited twists. In particular we want to compute the generating function of the amplitudes As in the previous section the SDS vertex for the emission of chiral untwisted states from a twisted string with ends on D t and D N is given by where we used the fact that ∂ n u 1 ∂ m u 2 G IJ U (t) is different from zero only when IJ = zz or IJ =zz. Moreover G zz U (t) is actually independent on t since it does not depend on the phase α t . The previous equation can be written in a more compact way by normal ordering the operatorial part and performing the limit. It then reads as where we have introduced the derivative of the regularized chiral Green function as Performing the same steps as in the previous section we get where we have written the dependence on the complex conjugate variables u c even if the derivatives of X I cl (u,ū) and G IJ (N,M ) (u,ū; v,v; {x t , α t }) are independent of it. We have done this in order to be consistent with the notation used in the boundary case. The regularized chiral Green function is defined as expected as Notice once more that the subtraction term is different from zero only when IJ = zz or IJ =zz because of the derivatives. There is actually an extra bonus in this approach: we can also prove the explicit form of amplitudes involving closed string states that were conjectured in [2]. This happens using the vertices for the emission of closed string states written in open string formalism. These vertices are written as a product of a chiral times an antichiral part (up to cocycles). Then we can apply the same procedure decribed before to these products.

Amplitudes with excited twist fields
In this section we would like to compute the correlators of N excited twists. In particular we want to compute the generating function of the amplitudes The excited twists in the previous correlator can be described using the operator to state correspondence. According to the notation introduced in [2] the operator to state correspondence is given by with k t = −i 1 2 √ 2α e iπαt and k¯ t = −i 1 2 √ 2α e −iπαt 10 . These states are built on the twisted vacuum |T T (D t−1 Dt) = σ t,ft (0)|0 SL(2) therefore the excited twists are naturally described in the twisted Hilbert space H T (D t−1 Dt) . However the H T (D t−1 Dt) space is generically not any of the twisted Hilbert spaces there are during the string propagation. They are H T (Dt D N ) for t = 1, . . . N − 1 since the x < 0 (σ = π in Minkowskian version) boundary of the string is always attached on D N . This is luckily not a problem since we can describe any excited twist as a limit of the product of a chiral operator and a plain twist field, for example In the previous expression the term with ∂ u ∂ v ∆ zz T (D t−1 Dt),chir is necessary to cancel the term which arises from the equation analogous to eq. (112) for the H T (D t−1 Dt) twisted Hilbert space.
In our case the plain twist field σ t,ft (x t ) is hidden into the boundary conditions. We can get any excited twist field at x t by choosing the appropriate chiral operator at u and then take the limit u → x t . It follows then that it is enough to represent the abstract chiral operator needed to create the wanted excited twisted field in the desired twisted Hilbert space which can be either As discussed in [2] the generating function for the abstract operators which is needed for generating any excited twists is given by Explicitly this means that The abstract generator (119) can be realized in any Hilbert space. In particular in the twisted Hilbert space The first two factors on the right hand side are nothing else but the chiral SDS vertex with u dependent c nI , roughly c nI → d nI (u − x t ) t I . Therefore the ∆ zz in the quadratic term depends on u and v only and not on x t . The last factor is the "normalization" of the abstract vertex, i.e. the second factor on the right hand side in eq. (119) and therefore depends on u−x t and v −x t . It is now enough to insert a SDS for any untwisted operator into the radial ordered expression (122) in order to compute the generating function of all the correlators like (116).
The computation of the expectation value is the same as before for the chiral vertices. What is interesting is to trace the contributions to self interactions. For any t we get two contributions: one from the expectation value and the other from the "normalization" of the abstract vertex, i.e. the third factor in eq. (121). We get therefore where c t is the index associated with the chiral vertex at u ct = u t and ∂ u ∂ v ∆ IJ (N,M )(ct) is the derivative of regularized Green function given in eq.
(115). This expression can be written in a more compact way as where we have defined the (derivative of the) regularized Green function at the position x t of the twist fields t to be and we used G IJ to write the last line. Notice that when x N −1 = 0 the previous equation becomes eq. (92). Actually because of the chiral derivatives the previous expression simplifies fot two combintations of indeces IJ to The third combination corresponds to ∂ u ∂ v ∆ zz (N,M )(t) and does not simplify. On general basis the regularized Green function is obtained by subtracting the divergent part with the proper monodromy at the point of regularization. At the point where a twist field is located the divergent part with the proper monodromy means G N =2,T (D t−1 Dt) while in all other points means G U . In particular both (u − x t ) tI (v − x t ) tJ ∂ u ∂ v ∆ IJ (N,M )(t) and (u − x t ) tI (v − x t ) tJ ∂ u ∂ v G IJ (N,M ) are analytic functions at u = x t . Assembling all pieces we get therefore the generating function for the excited twists correlator (127)

The generating function for for N excited twist fields and L plain vertices
We are now in the position of computing the desired generating function for N excited twist fields and L plain vertices. It simply amounts to the computation of This computation can be done as explained in the previous sections. The result is made of the product of three blocks: interactions between two excited twists, interactions between two plain vertices and interactions between one excited twists and one plain vertex. This structure is evident in the final result where the last line is exactly due to the interactions between one excited twist and one plain vertex.

A Self-adjointness of the laplacian
We want to show that ∂∂ is a self-adjoint operator only if we use the quantum boundary conditions. In particular we define ∂∂ = ∂ 2 x + ∂ 2 y as operator which acts on a couple of complex functions f I (u,ū) defined on the upper half plane. Then we take not only f I ∈ L 2 (H) but we require that ∂ x f I , ∂ y f I , ∂ 2 x f I and ∂ 2 y f I be defined almost everywhere and that the action H dx dy f I * (∂ 2 x + ∂ 2 y )f I be finite. Since we need to integrate by part we need b a dx∂ 2 x f I (u,ū) = ∂ x f I (b + iy, b − iy) − ∂ x f I (a + iy, a − iy) (and similarly for y) hence ∂ x f I and ∂ y f I must be absolutely continuous. The similar condition with a single derivative is a consequence of the existence almost everywhere of ∂ 2 x f I , ∂ 2 y f I which imply that ∂ x f I , ∂ y f I be almost everywhere continuous.
Now we can determine the domain of the dual operator, i.e. we determine the conditions we must impose on an arbitrary vector g I so that we can write (g, ∂∂f ) = (∂∂g, f ). In order to do so we compute using the previous boundary conditions from which we see that g I must satisfy the same boundary conditions as f I and hence the operator is not only Hermitian but self-adjoint.

B Details on the metric for modes
Consider for example the computation (X (a)1 ,X (a)2 ). In the following we writeX 1 =Ḡ andX 2 =F for notational simplicity. It is immediate to get which vanishes because of the boundary conditions which ensure that Ḡ (r 0 e iθ ) * F (r 0 e iθ ) θ=0 + = Ḡ (r 0 e −iθ ) * F (r 0 e −iθ ) θ=0 + . Similarly for (X (a) , X (c) ) we arrive to −i where it is meaningful write the integration interval as [−π, π] since both the terms are continuous at θ = 0, i.e. for example ∂ θḠ (r 0 e −iθ ) * F(r 0 e +iθ )| θ=0 + = ∂ θḠ (r 0 e iθ ) * F(r 0 e −iθ )| θ=0 + . Now integrating by part the first and using the boundary conditions to evaluate to zero the constant obtained from the integration by part we find −i