#  The generating function of amplitudes with $N$ twisted and $M$ untwisted
states

Igor Pesando1

1Dipartimento di Fisica Teorica, Università di Torino
and I.N.F.N. - sezione di Torino
Via P. Giuria 1, I-10125 Torino, Italy
<EMAIL_ADDRESS>
###### Abstract:

We show that the generating function of all amplitudes with $N$ twisted and
$M$ untwisted states, i.e. the Reggeon vertex for magnetized branes on
$\mathbb{R}^{2}$ can be computed once the correlator of $N$ non excited
twisted states and the corresponding Green function are known and we give an
explicit expression as a functional of the these objects.

D-branes, Conformal Field Theory

††preprint: DFTT-20-2011

##  1 Introduction and conclusions

In the late 80s a lot of work was done in computing the generating functions
of all amplitudes for the bosonic string and superstring. Many methods were
(further) developed such as the sewing method ([1]), the group theoretic
method ([2]) and conserved charges method ([3]). Following the main idea of
([4],[5]) in this paper we would like to compute the generating function for
$N$ generic excited twisted states and $M$ generic untwisted states on
$\mathbb{R}^{2}$ for the open string in presence of magnetic fields in the
upper half plane using the path integral approach. Much work has been already
done in computing non excited twisted states correlation functions, especially
on $T^{2}$ (see for example [6], [7], [8] and [9]) but not so much on the
computation of correlators involving excited twisted fields ([15], [16] see
for earlier work) which remain quite mysterious.

In this paper we want to show that there is a quite simple way of labeling
excited twisted states which is deeply connected with the operator-state map
and that few ingredients are actually needed for computing all correlators
involving excited twisted state and arbitrary untwisted ones on
$\mathbb{R}^{2}$. To obtain any correlator is only necessary the knowledge of
the full (i.e. classical and quantum) $N$ non excited twist correlator on the
disk111 The twist fields in this and the following correlators are actually
$\sigma_{\epsilon,\kappa=0}(x,\bar{x})$, see the in the main text.

$C(x_{1},\dots
x_{N})=\langle\sigma_{\epsilon_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N}}(x_{N},\bar{x}_{N})\rangle_{disk,~{}full}~{}~{}~{}~{}x_{t}\in\mathbb{R}$
(1)

and the boundary Green function in presence of such operators

$G^{ij}_{bou}(x;y;\\{x_{t}\\}_{t=1\dots
N})=G^{ji}_{bou}(y;x;\\{x_{t}\\}_{t=1\dots
N})=G^{ij}(x,\bar{x};y,\bar{y};\\{x_{t}\\}_{t=1\dots N})$ (2)

which can be derived from

$G^{ij}(z,\bar{z};w,\bar{w};\\{x_{t}\\}_{t=1\dots N})=\frac{\langle
X^{i}(z,\bar{z})X^{j}(w,\bar{w})\sigma_{\epsilon_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N}}(x_{N},\bar{x}_{N})\rangle_{disk}}{\langle\sigma_{\epsilon_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N}}(x_{N},\bar{x}_{N})\rangle_{disk}}.$
(3)

by setting $z=x,w=y\in\mathbb{R}$.

The main result of the paper is the generating function for the above
mentioned amplitudes given in eq.s (32) and (33). These two expressions have
exactly the same contain but the latter is written in a more usual way, i.e.
using auxiliary expansion variables while the former has an expression like
those used in the previous literature ([1]). Let us now explain the building
blocks of this last version of the main formula (32).

  * •

To any (excited) twisted operator inserted at $x_{t}$ ($t=1\dots N$) in the
amplitude we associate an auxiliary Hilbert space ${\cal H}_{t}$. On ${\cal
H}_{t}$ act the quantum fields $X^{i}_{(t)}(z,\bar{z})$222 In the following
quantum fields have attached the label of the Hilbert space they act on, e.g.
$X^{i}_{(t)}(z,\bar{z})$ while classical fields in path integral have no
label, i.e. $X^{i}(z,\bar{z})$.  ($i=1,2$ or $i=z,\bar{z}$)

$\displaystyle Z_{(t)}(z,\bar{z})$
$\displaystyle=X^{z}_{(t)}(z,\bar{z})=\frac{X^{1}_{(t)}+iX^{2}_{(t)}}{\sqrt{2}}=\frac{1}{2}\left(Z_{L(t)}(z)+Z_{R(t)}(\bar{z})\right),$
$\displaystyle\bar{Z}_{(t)}(z,\bar{z})$
$\displaystyle=X^{\bar{z}}_{(t)}(z,\bar{z})=\frac{X^{1}_{(t)}-iX^{2}_{(t)}}{\sqrt{2}}=\frac{1}{2}\left(\bar{Z}_{(t)L}(z)+\bar{Z}_{(t)R}(\bar{z})\right)$
(4)

which have expansions

$\displaystyle Z_{(t)L}(z)$ $\displaystyle=$ $\displaystyle
z_{(t)0}+i\sqrt{2\alpha^{\prime}}e^{-i\gamma_{t}}\sum_{n=0}^{\infty}\left[\frac{\bar{\alpha}_{(t)n+1-\epsilon_{t}}}{\,{n+1-\epsilon_{t}}}z^{-(n+1-\epsilon_{t})}-\frac{\alpha_{(t)n+\epsilon_{t}}^{\dagger}}{\,{n+\epsilon_{t}}}z^{+(n+\epsilon_{t})}\right]$
$\displaystyle Z_{(t)R}(\bar{z})$ $\displaystyle=$ $\displaystyle
z_{(t)0}+i\sqrt{2\alpha^{\prime}}e^{+i\gamma_{t}}\sum_{n=0}^{\infty}\left[\frac{\bar{\alpha}_{(t)n+1-\epsilon_{t}}}{\,{n+1-\epsilon_{t}}}{\bar{z}}^{-(n+1-\epsilon_{t})}-\frac{\alpha_{(t)n+\epsilon_{t}}^{\dagger}}{\,{n+\epsilon_{t}}}{\bar{z}}^{+(n+\epsilon_{t})}\right]$
(5)

and

$\displaystyle\bar{Z}_{(t)L}(z)$ $\displaystyle=$
$\displaystyle\bar{z}_{(t)0}+i\sqrt{2\alpha^{\prime}}e^{+i\gamma_{t}}\sum_{n=0}^{\infty}\left[-\frac{\bar{\alpha}_{(t)n+1-\epsilon_{t}}^{\dagger}}{\,{n+1-\epsilon_{t}}}z^{+(n+1-\epsilon_{t})}+\frac{\alpha_{(t)n+\epsilon_{t}}}{\,{n+\epsilon_{t}}}z^{-(n+\epsilon_{t})}\right]$
$\displaystyle\bar{Z}_{(t)R}(\bar{z})$ $\displaystyle=$
$\displaystyle\bar{z}_{(t)0}+i\sqrt{2\alpha^{\prime}}e^{-i\gamma_{t}}\sum_{n=0}^{\infty}\left[-\frac{\bar{\alpha}_{(t)n+1-\epsilon_{t}}^{\dagger}}{\,{n+1-\epsilon_{t}}}{\bar{z}}^{+(n+1-\epsilon_{t})}+\frac{\alpha_{(t)n+\epsilon_{t}}}{\,{n+\epsilon_{t}}}{\bar{z}}^{-(n+\epsilon_{t})}\right]$
(6)

The previous fields satisfy the boundary conditions333 These can also be
written as $e^{i\gamma_{t}}\partial
Z_{(t)L}(x)=\frac{1}{\cos\gamma_{t}}\partial Z_{(t)}(x,x)$ when $x>0$ and
$e^{i\gamma_{t-1}}\partial Z_{(t)L}(y)=\frac{1}{\cos\gamma_{t-1}}\partial
Z_{(t)}(y,{\bar{y}})$ when $y<0$. These expressions are those used to connect
the open string operators when naturally expressed as function of
$X(x,{\bar{x}})$ to their expressions as functional of $X_{L}(x)$.

$\displaystyle e^{i\gamma_{t}}\partial Z_{(t)L}|_{x}$ $\displaystyle=$
$\displaystyle
e^{-i\gamma_{t}}\bar{\partial}Z_{(t)R}|_{x}~{}~{}x\in\mathbb{R}^{+}$ (7)
$\displaystyle e^{+i\gamma_{t-1}}\partial Z_{(t)L}|_{y}$ $\displaystyle=$
$\displaystyle
e^{-i\gamma_{t-1}}\bar{\partial}Z_{(t)R}|_{y}~{}~{}y=|y|e^{i\pi}\in\mathbb{R}^{-}$
(8)

where we have defined the phases ($-\frac{\pi}{2}<\gamma_{t}<\frac{\pi}{2}$)

$\displaystyle e^{i\gamma_{t}}=\frac{1+iB_{t}}{\sqrt{1+B_{t}^{2}}}$
$\displaystyle\rightarrow$ $\displaystyle
B_{t}=\tan\gamma_{t}=2\pi\alpha^{\prime}~{}q_{(0)}F_{12(0)}$ $\displaystyle
e^{i\gamma_{t-1}}=\frac{1+iB_{t-1}}{\sqrt{1+B_{t-1}^{2}}}$
$\displaystyle\rightarrow$ $\displaystyle
B_{t-1}=\tan\gamma_{t-1}=2\pi\alpha^{\prime}~{}q_{(\pi)}F_{12(\pi)}$ (9)

where $B_{t-1}=2\pi\alpha^{\prime}~{}q_{(\pi)}F_{12(\pi)}$ and
$B_{t}=2\pi\alpha^{\prime}~{}q_{(0)}F_{12(0)}$ are the adimensional magnetic
fields which are on the $x<0$ ($\sigma=\pi$) and $x>0$ ($\sigma=0$)
boundaries. In the field expansion the shift $\epsilon_{t}$ is given by

$\epsilon_{t}=\left\\{\begin{array}[]{l
c}\frac{1}{\pi}\left(\gamma_{t}-\gamma_{t-1}\right)&\gamma_{t}>\gamma_{t-1}\\\
1+\frac{1}{\pi}\left(\gamma_{t}-\gamma_{t-1}\right)&\gamma_{t}<\gamma_{t-1}\end{array}\right.~{}~{}~{}~{}0\leq\epsilon_{t}<1$
(10)

The previous operators act on the ${\cal H}_{t}$ twisted ground state defined
by

$\bar{\alpha}_{(t)n+1-\epsilon_{t}}|T_{t}\rangle=\alpha_{(t)m+\epsilon_{t}}|T_{t}\rangle=x^{2}_{(t)0}|T_{t}\rangle=0$
(11)

and have the non vanishing commutation relations 444Since the annihilator and
creator operators have flat indexes this holds independently of our choice of
the taking the metric diagonal; in particular from definition of the complex
fields we have $(dX^{1})^{2}+(dX^{2})^{2}=2dZd\bar{Z}$, i.e. $G_{z\bar{z}}=1$.

$\displaystyle[\bar{\alpha}_{(t)n+1-\epsilon_{t}},\bar{\alpha}_{(t)m+1-\epsilon_{t}}^{\dagger}]$
$\displaystyle=$ $\displaystyle(n+1-\epsilon_{t})\delta_{n,m}~{}~{}n,m\geq 0$
$\displaystyle{}[\alpha_{(t)n+\epsilon_{t}},\alpha_{(t)m+\epsilon_{t}}^{\dagger}]$
$\displaystyle=$ $\displaystyle(n+\epsilon_{t})\delta_{n,m}~{}~{}n,m\geq 0$
$\displaystyle{}[z_{(t)0},\bar{z}_{(t)0}]$ $\displaystyle=$
$\displaystyle\frac{2\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$ (12)

Notice that the choice of the definition of the zero modes vacuum is somewhat
arbitrary since they do not change the energy, our choice is dictated by our
gauge choice for the background magnetic field $A=B~{}x^{1}~{}dx^{2}$ which
implies the translational invariance $X^{2}\rightarrow X^{2}+\epsilon$ and by
the observation that is almost the proper choice in toroidal
compactifications. The existence of the zero modes imply that the vacuum is
degenerate since all the states
$|T_{t},\kappa_{t}\rangle=e^{i\kappa_{t}x^{1}_{(t)0}}|T_{t}\rangle$ have
exactly the same energy of the vacuum and therefore there exists a one
parameter family of twist fields ([12])
$\sigma_{\epsilon_{t},\kappa_{t}}(x,{\bar{x}})$.

Given the previous vacuum definition we have the following twisted Green
functions

$\displaystyle G_{T(t)}^{zz}(z,\bar{z};w,\bar{w})$
$\displaystyle=[Z^{(+)}(z,\bar{z}),Z^{(-)}(w,\bar{w})]|_{an.cont}=\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$
$\displaystyle G_{T(t)}^{\bar{z}\bar{z}}(z,\bar{z};w,\bar{w})$
$\displaystyle=[\bar{Z}^{(+)}(z,\bar{z}),\bar{Z}^{(-)}(w,\bar{w})]|_{an.cont}=-\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$
$\displaystyle G_{T(t)}^{z\bar{z}}(z,\bar{z};w,\bar{w})$
$\displaystyle=[Z^{(+)}(z,\bar{z}),\bar{Z}^{(-)}(w,\bar{w})]|_{an.cont}$
$\displaystyle=\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$
$\displaystyle~{}~{}-\frac{\alpha^{\prime}}{2}\left[\,g_{\epsilon_{t}}\left(\frac{w}{z}\right)+\,g_{\epsilon_{t}}\left(\frac{\bar{w}}{\bar{z}}\right)+e^{-2i\gamma_{t}}\,g_{\epsilon_{t}}\left(\frac{\bar{w}}{z}\right)+e^{2i\gamma_{t}}\,g_{\epsilon_{t}}\left(\frac{w}{\bar{z}}\right)\right]$
$\displaystyle G_{T(t)}^{\bar{z}z}(z,\bar{z};w,\bar{w})$
$\displaystyle=[\bar{Z}^{(+)}(z,\bar{z}),Z^{(-)}(w,\bar{w})]|_{an.cont}$
$\displaystyle=-\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$
$\displaystyle~{}~{}-\frac{\alpha^{\prime}}{2}\left[\,g_{1-\epsilon_{t}}\left(\frac{w}{z}\right)+\,g_{1-\epsilon_{t}}\left(\frac{\bar{w}}{\bar{z}}\right)+e^{2i\gamma_{t}}\,g_{1-\epsilon_{t}}\left(\frac{\bar{w}}{z}\right)+e^{-2i\gamma_{t}}\,g_{1-\epsilon_{t}}\left(\frac{w}{\bar{z}}\right)\right]$
(13)

which can be obtained by analytically continuing their operatorial expression
from $|z|>|w|$ to the whole upper plane in such a way to preserve the symmetry
$G^{ij}(z,\bar{z};w,\bar{w})=G^{ij}(w,\bar{w};z,\bar{z})$. In the previous
expressions we have defined $\,g_{\nu}(z)$ as the analytic continuation of

$\,g_{\nu,s}(z)=-\sum_{n-\nu>0}\frac{1}{n-\nu}z^{n-\nu}~{}~{}~{}|z|<1,~{}~{}~{}-\pi+2\pi
s<\phi=arg(z)\leq\pi+2\pi s.$ (14)

in the properly chosen sheet $s$. Notice that the symmetry of the Green
function $G^{ij}(z,\bar{z};w,\bar{w})=G^{ji}(w,\bar{w};z,\bar{z})$ is not
obvious in the zero modes sector, i.e. for the constant terms but it holds due
to the $\,g$ transformation property ([12])

$\displaystyle\,g_{\nu,s}(z)=\,C_{\nu,s}(\phi)+\,g_{1-\nu,-s}\left(\frac{1}{z}\right),~{}~{}~{}~{}\,C_{\nu,s}(\phi)=\left\\{\begin{array}[]{cc}\frac{\pi
e^{-i\pi\nu}}{sin\pi\nu}e^{-i2\pi\nu s}&2\pi s<\phi<\pi+2\pi s\\\ \frac{\pi
e^{+i\pi\nu}}{sin\pi\nu}e^{-i2\pi\nu s}&-\pi+2\pi s<\phi<2\pi
s\end{array}\right.$ (17) (18)

This fact implies that we cannot really completely separate the zero modes and
non zero modes also for the twisted sector as it already happens for the
untwisted one.

For $x>0>y$ and $|y/x|<1$ the previous Green functions become on the
boundary555 When $|y/x|>1$ we must be more careful since we want to evaluate
the $\,g$ on a cut; for example when $0<x,y$ the expression which is valid for
all ranges is
$G_{T(t)~{}bou}^{z\bar{z}}(x;y)=\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}-\alpha^{\prime}~{}\cos\gamma_{t}[e^{-i\gamma_{t}}\,g_{\epsilon_{t}}\left(\frac{y}{x}e^{-i0}\right)+e^{i\gamma_{t}}\,g_{\epsilon_{t}}\left(\frac{y}{x}e^{+i0}\right)].$
In any case we can always use the symmetry property for the Green functions to
reduce the computation in the range where we can apply the given expressions.

$\displaystyle G_{T(t)~{}bou}^{zz}(x;y)$
$\displaystyle=\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$ $\displaystyle
G_{T(t)~{}bou}^{\bar{z}\bar{z}}(x;y)$
$\displaystyle=-\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}$ $\displaystyle
G_{T(t)~{}bou}^{z\bar{z}}(x;y)$
$\displaystyle=\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}-2\alpha^{\prime}~{}\cos\gamma_{t}~{}\cos\gamma_{t-1}~{}e^{i\gamma_{t}-i\gamma_{t-1}}~{}\,g_{\epsilon_{t}}\left(\frac{y}{x}\right)$
$\displaystyle G_{T(t)~{}bou}^{\bar{z}z}(x;y)$
$\displaystyle=-\frac{\pi\alpha^{\prime}}{B_{t}-B_{t-1}}-2\alpha^{\prime}~{}\cos\gamma_{t}~{}\cos\gamma_{t-1}~{}e^{-i\gamma_{t}+i\gamma_{t-1}}~{}\,g_{1-\epsilon_{t}}\left(\frac{y}{x}\right)$
(19)

The other cases can be obtained with the substitution rule
$x>0~{}~{}\cos\gamma_{t}~{}e^{i\gamma_{t}}\leftrightarrow
x<0~{}~{}\cos\gamma_{t-1}~{}e^{-i\gamma_{t-1}}$ and the same for $y$ in the
$G^{z\bar{z}}$ propagator. For the $G^{\bar{z}z}$ propagator one takes the
complex conjugate of the previous substitution rule.

  * •

In a similar way to any untwisted operator we insert in the amplitude we
associate an auxiliary Hilbert space ${\cal H}_{a,t_{a}}$. This Hilbert space
as well as the position where the untwisted vertex is inserted $x_{a,t_{a}}$
are better labeled by both a counting label $a=1\dots M$ and a further label
$t_{a}\in\\{1,\dots N\\}$ which specify which is the magnetic field felt by
the untwisted state (dipole string). This could seem irrelevant but it is
important in defining the regularized Green functions (30) and in computing
the non commutative phases. In the following we will use a lighter notation as
$x_{a,t_{a}}\rightarrow x_{a}$ when there is not possibility of confusion.

On the auxiliary Hilbert space ${\cal H}_{a,t_{a}}$ act the quantum fields

$Z_{(a,t_{a})}(z,\bar{z})=\frac{1}{\sqrt{2}}\left(X^{1}_{(a,t_{a})}(z,\bar{z})+iX^{2}_{(a,t_{a})}(z,\bar{z})\right)=\frac{1}{2}(Z_{(a,t_{a})L}(z)+Z_{(a,t_{a})R}(\bar{z}))$
(20)

which have expansions

$\displaystyle Z_{(a,t_{a})L}$ $\displaystyle=$ $\displaystyle
e^{-i\gamma_{t_{a}}}\Bigg{(}z_{(a,t_{a})0}-2\alpha^{\prime}{\bar{p}}_{(a,t_{a})}~{}i\ln(z)+i\sqrt{2\alpha^{\prime}}\sum_{n=1}^{\infty}+\frac{\bar{\alpha}_{(a,t_{a})n}}{\,{n}}z^{-n}-\frac{\alpha_{(a,t_{a})n}^{\dagger}}{\,{n}}z^{n}\Bigg{)}$
$\displaystyle Z_{(a,t_{a})R}$ $\displaystyle=$ $\displaystyle
e^{+i\gamma_{t_{a}}}\left(z_{(a,t_{a})0}-2\alpha^{\prime}{\bar{p}}_{(a,t_{a})}~{}i\ln(\bar{z})+i\sqrt{2\alpha^{\prime}}\sum_{n=1}^{\infty}+\frac{\bar{\alpha}_{(a,t_{a})n}}{\,{n}}\bar{z}^{-n}-\frac{\alpha_{(a,t_{a})n}^{\dagger}}{\,{n}}\bar{z}^{n}\right)$

and

$\displaystyle\bar{Z}_{(a,t_{a})L}$ $\displaystyle=$ $\displaystyle
e^{+i\gamma_{t_{a}}}\left(\bar{z}_{(a,t_{a})0}-2\alpha^{\prime}{p}_{(a,t_{a})}~{}i\ln(z)+i\sqrt{2\alpha^{\prime}}\sum_{n\neq
0}-\frac{\bar{\alpha}_{(a,t_{a})n}^{\dagger}}{\,{n}}z^{n}+\frac{\alpha_{(a,t_{a})n}}{\,{n}}z^{-n}\right)$
$\displaystyle\bar{Z}_{(a,t_{a})R}$ $\displaystyle=$ $\displaystyle
e^{-i\gamma_{t_{a}}}\left(\bar{z}_{(a,t_{a})0}-2\alpha^{\prime}{p}_{(a,t_{a})}~{}i\ln(\bar{z})+i\sqrt{2\alpha^{\prime}}\sum_{n\neq
0}-\frac{\bar{\alpha}_{(a,t_{a})n}^{\dagger}}{\,{n}}\bar{z}^{n}+\frac{\alpha_{(a,t_{a})n}}{\,{n}}\bar{z}^{-n}\right)$

The previous quantum fields satisfy the boundary conditions

$\displaystyle e^{+i\gamma_{t_{a}}}\partial Z_{(a,t_{a})}|_{x}$
$\displaystyle=$ $\displaystyle
e^{-i\gamma_{t_{a}}}\bar{\partial}Z_{(a,t_{a})}|_{x}~{}~{}x\in\mathbb{R}^{+}$
$\displaystyle e^{+i\gamma_{t_{a}}}\partial Z_{(a,t_{a})}|_{y}$
$\displaystyle=$ $\displaystyle
e^{-i\gamma_{t_{a}}}\bar{\partial}Z_{(a,t_{a})}|_{y}~{}~{}y=|y|e^{i\pi}\in\mathbb{R}^{-}$

where we have defined the angle $\gamma_{t_{a}}$, in a similar way for the
twisted scalar (dicharged string), as

$e^{i\gamma_{t_{a}}}=\frac{1+iB_{t_{a}}}{\sqrt{1+B_{t_{a}}}}\Rightarrow
B_{t_{a}}=\tan\gamma_{t_{a}},~{}~{}~{}~{}-\frac{\pi}{2}<\gamma_{t_{a}}<\frac{\pi}{2}$
(23)

The creation and destruction operators act on the dipole ground state defined
by

$\bar{\alpha}_{(a,t_{a})n}|0_{(a,t_{a})}\rangle=\alpha_{(a,t_{a})n}|0_{(a,t_{a})}\rangle={\bar{p}}_{(a,t_{a})}|0_{(a,t_{a})}\rangle={p}_{(a,t_{a})}|0_{(a,t_{a})}\rangle=0$
(24)

and have non trivial commutation relations

$\displaystyle[z_{(a,t_{a})0},\bar{z}_{(a,t_{a})0}]$ $\displaystyle=$
$\displaystyle 2\pi\alpha^{\prime}B_{t_{a}}$
$\displaystyle{}[z_{(a,t_{a})0},{p}_{(a,t_{a})}]$ $\displaystyle=$
$\displaystyle i$
$\displaystyle{}[\alpha_{(a,t_{a})n},{\alpha}^{\dagger}_{(a,t_{a})m}]$
$\displaystyle=$ $\displaystyle n\delta_{m,n}$
$\displaystyle{}[\bar{\alpha}_{(a,t_{a})n},\bar{\alpha}^{\dagger}_{(a,t_{a})m}]$
$\displaystyle=$ $\displaystyle n\delta_{m,n}$ (25)

The normal ordering is the usual one but it worth noticing that in the zero
modes sector is defined as

$:e^{i(\bar{k}Z_{(a)zm}+k\bar{Z}_{(a)zm})(x,{\bar{x}})}:=\left\\{\begin{array}[]{c
r}e^{i\cos\gamma_{t}(\bar{k}z_{(a)0}+k\bar{z}_{(a)0})}~{}e^{2\alpha^{\prime}\ln(|x|)\cos\gamma_{t}(\bar{k}{\bar{p}}_{(a)0}+k{p}_{(a)0})}&x>0\\\
e^{i\cos\gamma_{t}(\bar{k}\hat{z}_{(a)0}+k\hat{\bar{z}}_{(a)0})}~{}e^{2\alpha^{\prime}\ln(|x|)\cos\gamma_{t}(\bar{k}{\bar{p}}_{(a)0}+k{p}_{(a)0})}&x<0\end{array}\right..$
(26)

with $\hat{z}_{(a)0}=z_{(a)0}-i2\pi\alpha^{\prime}\tan\gamma_{t}~{}{\bar{p}}$
which have the property that their commutation relations are the opposite of
the $z_{(a)0}$ ones. Finally the untwisted Green functions in a magnetic
background $B_{t_{a}}$ are given by ($0<arg(z-\bar{w})<\pi$)

$\displaystyle G_{U(t_{a})}^{zz}(z,\bar{z},w,\bar{w})$
$\displaystyle=G_{U(t_{a})}^{\bar{z}\bar{z}}(z,\bar{z},w,\bar{w})=0$
$\displaystyle G_{U(t_{a})}^{z\bar{z}}(z,\bar{z},w,\bar{w})$
$\displaystyle=[Z^{(+)}(z,\bar{z}),\bar{Z}^{(-)}(w,\bar{w})]|_{an.cont}$
$\displaystyle=+\frac{1}{2}\pi\alpha^{\prime}\sin(2\gamma_{t_{a}})-\alpha^{\prime}\left[\ln|z-w|+\cos(2\gamma_{t_{a}})\ln|z-\bar{w}|+\sin(2\gamma_{t_{a}})arg(z-\bar{w})\right]$
$\displaystyle G_{U(t_{a})}^{\bar{z}z}(z,\bar{z},w,\bar{w})$
$\displaystyle=[\bar{Z}^{(+)}(z,\bar{z}),Z^{(-)}(w,\bar{w})]|_{an.cont}$
$\displaystyle=-\frac{1}{2}\pi\alpha^{\prime}\sin(2\gamma_{t_{a}})-\alpha^{\prime}\left[\ln|z-w|+\cos(2\gamma_{t_{a}})\ln|z-\bar{w}|-\sin(2\gamma_{t_{a}})arg(z-\bar{w})\right]$
(27)

The constant terms can be obtained by rewriting

$z_{(a)0}=z_{(a)00}+i\pi\alpha^{\prime}\tan\gamma_{t}~{}{\bar{p}}$ (28)

so that $[z_{(a)00},\bar{z}_{(a)00}]=0$ an considering the additional term
proportional to ${\bar{p}}$ coming from this rewriting as belonging to
$Z^{(+)}(z,\bar{z})$. Notice however once again that the constant terms are
needed to ensure the symmetry
$G^{ij}(z,\bar{z};w,\bar{w})=G^{ji}(w,\bar{w};z,\bar{z})$.

The previous Green functions become on the boundary $z=x,w=y\in\mathbb{R}$ 666
When using these Green functions in eq. (3) in absence of twist fields we
recover the results from the operatorial formalism.

$\displaystyle G_{U(t_{a}),~{}bou}^{z\bar{z}}(x;y)$
$\displaystyle=\frac{1}{2}\pi\alpha^{\prime}\sin(2\gamma_{t_{a}})-2\alpha^{\prime}\left[\cos^{2}(\gamma_{t_{a}})\ln|x-y|+\frac{1}{2}\sin(2\gamma_{t_{a}})arg(x-\bar{y})\right]$
$\displaystyle G_{U(t_{a}),~{}bou}^{\bar{z}z}(x;y)$
$\displaystyle=-\frac{1}{2}\pi\alpha^{\prime}\sin(2\gamma_{t_{a}})-2\alpha^{\prime}\left[\cos^{2}(\gamma_{t_{a}})\ln|x-\bar{y}|-\frac{1}{2}\sin(2\gamma_{t_{a}})arg(x-\bar{y})\right]$
(29)

From these expressions we can read the open string metric ${\cal
G}^{z\bar{z}}_{(t)}=\cos^{2}(\gamma_{t})$ and the non commutativity parameter
$\Theta^{z\bar{z}}_{(t)}=\frac{1}{2}\sin(2\gamma_{t})$, we can also read the
$\mathbb{R}^{2}$ vielbein ${\cal V}^{\underline{z}\bar{z}}_{(t)}={\cal
V}^{\underline{\bar{z}}z}_{(t)}=\frac{1}{\cos(\gamma_{t})}$ where
$\underline{z},\underline{\bar{z}}$ are the flat indexes which are also
implicit in the creation and destruction operators.

  * •

We define the boundary Green function regularized by the untwisted Green
function for a background $B_{t_{a}}$ as

$\displaystyle G_{bou,~{}reg~{}U(t_{a})}^{ij}(x;y;\\{x_{v}\\})=$
$\displaystyle
G^{ij}_{bou}(x;y;\\{x_{v}\\})-G_{U(t_{a}),~{}bou}^{ij}(x;y)~{}~{}~{}~{}x,y\in\mathbb{R}$
(30)

where $G_{U(t_{a}),~{}bou}^{ij}(x,y)$ are defined in eq.s (• ‣ 1). The choice
of the background $B_{t}$ in the regularization would seem arbitrary but it is
not since these regularized Green functions (and their derivatives) enter only
where an untwisted dipole state is emitted and this is on a well defined
interval of the boundary.

We also define the analogous twisted boundary Green function regularized by
the twisted Green function at the twist insertion point $t$ as777The
symmetrization is because we a symmetric function in $x\leftrightarrow y$,
i.e. independent on the way we take the limit $x>y$ or $y>x$.

$\displaystyle G^{ij}_{bou,~{}reg~{}T(t)}(x,y;\\{x_{v}\\})$
$\displaystyle=G^{ij}_{bou}(x,y;\\{x_{v}\\})-G^{ij}_{T(t)~{}bou}(x,y;\\{x_{0}=x_{t},x_{\infty}=\infty\\})$
(31)

where $G^{ij}_{t~{}bou}$ are given in eq.s (• ‣ 1) with the substitution
$\frac{y}{x}\rightarrow\frac{y-x_{t}}{x-x_{t}}$.

Given the previous building blocks the main formula is given by ($x_{t}\neq
x_{a}~{}~{}~{}\forall t,a$)

$\displaystyle\langle V_{N+M}(\\{x_{t}\\}_{t=1,\dots
N};\\{x_{a,t_{a}}\\}_{a=1,\dots M})|=C(x_{1},\dots x_{N})$
$\displaystyle\prod_{a=1}^{M}\langle
0_{(a)a},z_{(a)00}=\bar{z}_{(a)00}=0|\prod_{t=1}^{N}\langle
T_{(t)},x^{1}_{(t)}=0|\delta(i\sum_{a}(\alpha_{(a)0}-\bar{\alpha}_{(a)0})+i\sum_{t}(z_{(t)0}-\bar{z}_{(t)0}))$
$\displaystyle\prod_{a}\exp\Big{\\{}-\frac{1}{4\alpha^{\prime}}\alpha_{(a)0}^{2}~{}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}^{2}~{}G^{\bar{z}\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})-\frac{1}{4\alpha^{\prime}}\bar{\alpha}_{(a)0}^{2}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}^{2}~{}G^{zz}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a}\exp\Big{\\{}}-\frac{1}{2\alpha^{\prime}}\sum_{n,m=0}^{\infty}\alpha_{(a)n}~{}\bar{\alpha}_{(a)m}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n}_{x}}{n!}~{}\frac{\partial^{m}_{y}}{m!}~{}G^{z\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})\Big{\\}}\Big{|}_{x=y=x_{a}}$
$\displaystyle\prod_{t}\exp\Big{\\{}\frac{1}{2}\left(\frac{\tan\gamma_{t}-\tan\gamma_{t-1}}{2\pi\alpha^{\prime}}{x_{(t)0~{}i=2}}{}\right)^{2}~{}G^{22}_{bou,~{}reg~{}T(t)}(x;y;\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt-\frac{1}{2\alpha^{\prime}}\sum_{n,m=1}^{\infty}\frac{\bar{\alpha}_{(t)n}}{n-\epsilon_{t}}\frac{\alpha_{(t)m}}{m-1+\epsilon_{t}}~{}{\cal
V}_{(t)~{}\underline{z}\bar{z}}{\cal V}_{(u)~{}\underline{\bar{z}}z}$
$\displaystyle\hskip
70.0001pt~{}\frac{\partial^{n-1}_{x}}{(n-1)!}~{}\frac{\partial^{m-1}_{y}}{(m-1)!}\Big{[}(x-x_{t})^{1-\epsilon_{t}}(y-x_{t})^{\epsilon_{t}}~{}\partial_{x}\partial_{y}G^{z\bar{z}}_{bou,~{}reg~{}T(t)}(x,y;\\{x_{v}\\})\Big{]}\Big{\\}}\Big{|}_{x=y=x_{t}}$
$\displaystyle\prod_{a<b}\exp\Big{\\{}-\frac{1}{2\alpha^{\prime}}\alpha_{(a)0}~{}\alpha_{(b)0}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{b})~{}\underline{\bar{z}}z}~{}G^{zz}_{bou}(x;y;\\{x_{v}\\})-\frac{1}{2\alpha^{\prime}}\bar{\alpha}_{(a)0}~{}\bar{\alpha}_{(b)0}~{}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}{\cal
V}_{(t_{b})~{}\underline{z}\bar{z}}~{}G^{\bar{z}\bar{z}}_{bou}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}-\frac{1}{2\alpha^{\prime}}\sum_{n,m=0}^{\infty}\alpha_{(a)n}~{}\bar{\alpha}_{(b)m}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{b})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n}_{x}}{n!}~{}\frac{\partial^{m}_{y}}{m!}~{}G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}-\frac{1}{2\alpha^{\prime}}\sum_{n,m=0}^{\infty}\bar{\alpha}_{(a)n}~{}\alpha_{(b)m}~{}{\cal
V}_{(t_{b})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n}_{x}}{n!}~{}\frac{\partial^{m}_{y}}{m!}~{}G^{\bar{z}z}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}\Big{|}_{x=x_{a},y=x_{b}}$
$\displaystyle\prod_{t<u}\exp\Big{\\{}\left(\frac{\tan\gamma_{t}-\tan\gamma_{t-1}}{2\pi\alpha^{\prime}}{x_{(t)0~{}i=2}}{}\right)\left(\frac{\tan\gamma_{u}-\tan\gamma_{u-1}}{2\pi\alpha^{\prime}}{x_{(u)0~{}i=2}}{}\right)~{}G^{22}_{bou}(x;y;\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt-\frac{1}{2\alpha^{\prime}}\sum_{n,m=1}^{\infty}\frac{\bar{\alpha}_{(t)n}}{n-\epsilon_{t}}\frac{\alpha_{(u)m}}{m-1+\epsilon_{u}}~{}{\cal
V}_{(t)~{}\underline{z}\bar{z}}{\cal V}_{(u)~{}\underline{\bar{z}}z}$
$\displaystyle\hskip
70.0001pt~{}\frac{\partial^{n-1}_{x}}{(n-1)!}~{}\frac{\partial^{m-1}_{y}}{(m-1)!}\left[(x-x_{t})^{\epsilon_{t}}(y-x_{u})^{1-\epsilon_{u}}\partial_{x}\partial_{y}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})\right]$
$\displaystyle\hskip
30.00005pt-\frac{1}{2\alpha^{\prime}}\sum_{n,m=1}^{\infty}\frac{\alpha_{(t)n}}{n-1+\epsilon_{t}}\frac{\bar{\alpha}_{(u)m}}{m-\epsilon_{u}}~{}{\cal
V}_{(t)~{}\underline{\bar{z}}z}{\cal V}_{(u)~{}\underline{z}\bar{z}}$
$\displaystyle\hskip
70.0001pt~{}\frac{\partial^{n-1}_{x}}{(n-1)!}~{}\frac{\partial^{m-1}_{y}}{(m-1)!}\left[(x-x_{t})^{1-\epsilon_{t}}(y-x_{u})^{\epsilon_{u}}\partial_{x}\partial_{y}G^{z\bar{z}}_{bou}(x,y;\\{x_{v}\\})\right]\Big{\\}}\Big{|}_{x=x_{t},y=x_{v}}$
$\displaystyle\prod_{t,a}\exp\Big{\\{}-\frac{1}{2\alpha^{\prime}}\left(\frac{\tan\gamma_{t}-\tan\gamma_{t-1}}{\pi}\frac{x_{(t)0~{}i=2}}{\sqrt{2\alpha^{\prime}}}\right)\sum_{m=0}^{\infty}~{}\alpha_{(a)m}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}~{}\frac{\partial^{m}_{y}}{m!}G^{2z}_{bou}(x,y;\\{x_{v}\\})$
$\displaystyle-\frac{1}{2\alpha^{\prime}}\left(\frac{\tan\gamma_{t}-\tan\gamma_{t-1}}{\pi}\frac{x_{(t)0~{}i=2}}{\sqrt{2\alpha^{\prime}}}\right)\sum_{m=0}^{\infty}~{}\bar{\alpha}_{(a)m}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}~{}\frac{\partial^{m}_{y}}{m!}G^{2\bar{z}}_{bou}(x,y;\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt-\frac{1}{2\alpha^{\prime}}\sum_{n=1,m=0}^{\infty}\frac{\bar{\alpha}_{(t)n}}{n-\epsilon_{t}}~{}\alpha_{(a)m}~{}{\cal
V}_{(t)~{}\underline{z}\bar{z}}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}~{}\frac{\partial^{n-1}_{x}}{(n-1)!}~{}\frac{\partial^{m}_{y}}{m!}\left[(x-x_{t})^{\epsilon_{t}}\partial_{x}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})\right]$
$\displaystyle\hskip
30.00005pt-\frac{1}{2\alpha^{\prime}}\sum_{n=1,m=0}^{\infty}\frac{\alpha_{(t)n}}{n-1+\epsilon_{t}}~{}\bar{\alpha}_{(a)m}~{}{\cal
V}_{(t)~{}\underline{\bar{z}}z}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n-1}_{x}}{(n-1)!}~{}\frac{\partial^{m}_{y}}{m!}\left[(x-x_{t})^{1-\epsilon_{t}}\partial_{x}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})\right]\Big{\\}}\Big{|}_{x=x_{t},y=x_{a}}$
(32)

where the operator indexes are raised an lowered using the flat metric while
Green function indexes are raised an lowered using $\mathbb{R}^{2}$ metric.
The previous expression can also be written without using the auxiliary
operators as a more conventional generating function. In order to do so we
introduce the auxiliary parameters $d_{(t)n}$, $\bar{d}_{(t)n}$ and $c_{(a)n}$
and $\bar{c}_{(a)n}$ which roughly correspond to $\alpha_{(t)n+1-\epsilon}$,
$\bar{\alpha}_{(t)n-\epsilon}$ and $\alpha_{(a)n}$, $\bar{\alpha}_{(a)n}$ (see
eq.s (4) and (3) for a precise mapping) of the previous expression. Then we
can write the generating function as

$\displaystyle{\cal V}_{N+M}(\\{d_{(t)}\\}_{t=1,\dots
N};\\{c_{(a)}\\}_{a=1,\dots M};\\{x_{t}\\}_{t=1,\dots
N};\\{x_{a,t_{a}}\\}_{a=1,\dots M})=$ $\displaystyle=$
$\displaystyle\delta\left(Re\left(\sum_{t}d_{(t)0}+\sum
c_{(a)0})\right)\right)~{}C(x_{1},\dots x_{N})$
$\displaystyle\prod_{a}\exp\Big{\\{}\frac{1}{2}c_{(a)0}^{2}~{}G^{\bar{z}\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})+\frac{1}{2}\bar{c}_{(a)0}^{2}~{}G^{zz}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a}\exp\Big{\\{}}\sum_{n,m=0}^{\infty}c_{(a)n}~{}\bar{c}_{(a)m}~{}\partial^{n}_{x}\partial^{m}_{y}G^{z\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})\Big{\\}}\Big{|}_{x=y=x_{a}}$
$\displaystyle\prod_{t}\exp\Big{\\{}\frac{1}{2}d_{(t)0}^{2}~{}G^{\bar{z}\bar{z}}_{bou,~{}reg~{}T(t)}(x;y;\\{x_{v}\\})+\frac{1}{2}\bar{d}_{(t)0}^{2}~{}G^{zz}_{bou,~{}reg~{}T(t)}(x;y;\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt+d_{(t)0}~{}\bar{d}_{(t)0}~{}G^{z\bar{z}}_{bou,~{}reg~{}T(t)}(x;y;\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt+\sum_{n,m=1}^{\infty}\bar{d}_{(t)n}~{}d_{(t)m}~{}\partial^{n-1}_{x}\partial^{m-1}_{y}\Big{[}(x-x_{t})^{1-\epsilon_{t}}(y-x_{t})^{\epsilon_{t}}~{}\partial_{x}\partial_{y}G^{z\bar{z}}_{bou,~{}reg~{}T(t)}(x,y;\\{x_{v}\\})\Big{]}\Big{\\}}\Big{|}_{x=y=x_{t}}$
$\displaystyle\prod_{a<b}\exp\Big{\\{}\bar{c}_{(a)n}~{}\bar{c}_{(b)m}~{}G^{zz}_{bou}(x;y;\\{x_{v}\\})+c_{(a)n}~{}c_{(b)m}~{}G^{\bar{z}\bar{z}}_{bou}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}+\sum_{n,m=0}^{\infty}\bar{c}_{(a)n}~{}c_{(b)m}~{}\partial^{n}_{x}~{}\partial^{m}_{y}~{}G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}+\sum_{n,m=0}^{\infty}c_{(a)n}~{}\bar{c}_{(b)m}~{}\partial^{n}_{x}~{}\partial^{m}_{y}~{}G^{\bar{z}z}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}\Big{|}_{x=x_{a},y=x_{b}}$
$\displaystyle\prod_{t<u}\exp\Big{\\{}d_{(t)0}~{}d_{(u)0}~{}G^{\bar{z}\bar{z}}_{bou}(x_{t},x_{u};\\{x_{v}\\})+\bar{d}_{(t)0}~{}\bar{d}_{(u)0}~{}G^{zz}_{bou}(x_{t},x_{u};\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt+d_{(t)0}~{}\bar{d}_{(u)0}~{}G^{\bar{z}z}_{bou}(x_{t},x_{u};\\{x_{v}\\})+\bar{d}_{(t)0}~{}d_{(u)0}~{}G^{z\bar{z}}_{bou}(x_{t},x_{u};\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt+\sum_{n,m=1}^{\infty}d_{(t)n}~{}\bar{d}_{(u)m}~{}\partial^{n-1}_{x}\partial^{m-1}_{y}\left[(x-x_{t})^{\epsilon_{t}}(y-x_{u})^{1-\epsilon_{u}}\partial_{x}\partial_{y}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})\right]$
$\displaystyle\hskip
30.00005pt+\sum_{n,m=1}^{\infty}\bar{d}_{(t)n}~{}d_{(u)m}~{}\partial^{n-1}_{x}\partial^{m-1}_{y}~{}\left[(x-x_{t})^{1-\epsilon_{t}}(y-x_{u})^{\epsilon_{u}}\partial_{x}\partial_{y}G^{z\bar{z}}_{bou}(x,y;\\{x_{v}\\})\right]\Big{\\}}\Big{|}_{x=x_{t},y=x_{v}}$
$\displaystyle\prod_{t,a}\exp\Big{\\{}\sum_{m=0}^{\infty}d_{(t)0}~{}\bar{c}_{(a)m}~{}\partial^{m}_{y}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})+\bar{d}_{(t)0}~{}c_{(a)m}~{}\partial^{m}_{y}G^{z\bar{z}}_{bou}(x,y;\\{x_{v}\\})$
$\displaystyle\hskip
30.00005pt+\sum_{n=1,m=0}^{\infty}d_{(t)n}~{}\bar{c}_{(a)m}~{}\partial^{n-1}_{x}\partial^{m}_{y}\left[(x-x_{t})^{\epsilon_{t}}\partial_{x}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})\right]$
$\displaystyle\hskip
30.00005pt+\sum_{n=1,m=0}^{\infty}\bar{d}_{(t)n}~{}c_{(a)m}~{}\partial^{n-1}_{x}\partial^{m}_{y}\left[(x-x_{t})^{1-\epsilon_{t}}\partial_{x}G^{\bar{z}z}_{bou}(x,y;\\{x_{v}\\})\right]\Big{\\}}\Big{|}_{x=x_{t},y=x_{a}}$
(33)

Notice that all the previous expressions are meaningful because of the
behavior of the Green functions

$\displaystyle G^{zz}_{bou}(x;y;\\{x_{v}\\})$ $\displaystyle=const$
$\displaystyle G^{\bar{z}\bar{z}}_{bou}(x;y;\\{x_{v}\\})$
$\displaystyle=const$ $\displaystyle
G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})=G^{\bar{z}z}_{bou}(y;x;\\{x_{v}\\})$
$\displaystyle{\sim}_{x\rightarrow
x_{t}}const+(x-x_{t})^{\epsilon_{t}}[g^{z\bar{z}}_{0}(y;\\{x_{v}\\})+O(x-x_{t})]$
$\displaystyle
G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})=G^{\bar{z}z}_{bou}(y;x;\\{x_{v}\\})$
$\displaystyle\sim_{y\rightarrow
x_{u}}const+(y-x_{u})^{1-\epsilon_{u}}[g^{z\bar{z}}_{0}(x;\\{x_{v}\\})+O(y-x_{u})]$
$\displaystyle
G^{\bar{z}z}_{bou}(x;y;\\{x_{v}\\})=G^{z\bar{z}}_{bou}(y;x;\\{x_{v}\\})$
$\displaystyle\sim_{x\rightarrow
y;x,y\in(x_{t},x_{t+1})}const-2\alpha^{\prime}\cos^{2}\gamma_{t}log|x-y|+O(x-y)$
(34)

where $g^{z\bar{z}}_{0}$ and $g^{z\bar{z}}_{0}$ are some functions of the
given variables and the last line is strictly speaking true when $x\rightarrow
y$ but not at the same time when $x\rightarrow x_{t}$ and $y\rightarrow
x_{t}$. It is anyhow true that
$(x-x_{t})^{1-\epsilon_{t}}(y-x_{t})^{\epsilon_{t}}~{}\partial_{x}\partial_{y}G^{z\bar{z}}_{bou,~{}reg~{}T(t)}(x,y;\\{x_{v}\\})$
is well defined for $x=y=x_{t}$ as discussed in the appendix B.

The rest of the paper is organized in the following way: in the next section
we make some examples of the use of the previous formulae and we clarify the
operator to state mapping we use in the twisted sector. In section 3 we derive
the previous formulae for the case with non excited twisted matter and finally
in section 4 we consider excited twisted matter.

##  2 Examples

We want now apply the main formulae stated in the previous section to some
examples while elucidating the nature of excited twisted states.

We start from the simplest example and then move to some more complex ones
while in appendix A we check the $N=2$ not excited states and $M$ tachyons
amplitude against the result found in ([12]).

###  2.1 Example 1: $N$ not excited twisted states

From the operator to auxiliary state map

$\sigma_{\epsilon_{t},\kappa_{t}}(x_{t},\bar{x}_{t})\leftrightarrow|T_{t},\kappa_{t}\rangle=\lim_{x\rightarrow
0}\sigma_{\epsilon_{t},\kappa_{t}}(x,\bar{x})|0_{SL}\rangle$ (35)

we deduce that

$\displaystyle\langle\sigma_{\epsilon_{1},\kappa_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\rangle$
$\displaystyle=\langle V_{N+0}(\\{x_{t}\\}_{t=1,\dots
N}|\prod_{t}|T_{t},\kappa_{t}\rangle$
$\displaystyle=\delta(\sum_{t}\kappa_{t})e^{-\frac{1}{2}\sum_{t}\kappa_{t}^{2}~{}G^{22}_{bou,~{}reg~{}T(t)}(x_{t};x_{t})}e^{-\frac{1}{2}\sum_{u,t}\kappa_{t}\kappa_{u}~{}G^{22}_{bou}(x_{t};x_{u})}~{}C(x_{1},\dots
x_{N})$ (36)

where the phases proportional to $\kappa$ probably vanish as they do in the
$N=2$ case but this must be checked with an explicit computation of the Green
function which can, in principle, be extracted from ([6]) after T-dualizing.
The same computation can be performed using the more conventional generating
function as

$\displaystyle\langle\sigma_{\epsilon_{1},\kappa_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\rangle={\cal
V}_{N+M}\prod_{t}e^{i\kappa_{t}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
d_{(t)0}^{2}}}\Big{|}_{c=0;d=0}.$ (37)

###  2.2 Example 2: $N$ not excited twisted and $2$ untwisted states

Similarly to what done in the previous example from the maps

$\displaystyle\sigma_{\epsilon_{t}}(x_{t},\bar{x}_{t})\leftrightarrow|T_{t}\rangle=\lim_{x\rightarrow
0}\sigma_{\epsilon_{t}}(x,\bar{x})|0_{SL}\rangle$ $\displaystyle\partial
X^{z}(y_{a},\bar{y}_{a})\leftrightarrow-i\sqrt{2\alpha^{\prime}}\cos\gamma_{t_{a}}~{}\alpha^{\dagger}_{(a)1}|0_{(a)}\rangle=\lim_{y\rightarrow
0^{+}}\partial Z_{(a)}(y,\bar{y})|0_{SL}\rangle$ (38)

where we have made the choice $x_{t_{a}}<y_{a}<x_{t_{a}}+1$ (which fixes the
magnetic field felt by the untwisted state) , we deduce

$\displaystyle\langle\sigma_{\epsilon_{1},\kappa_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\partial_{y_{1}}Z(y_{1},\bar{y}_{1})\partial_{y_{2}}{\bar{Z}}(y_{2},\bar{y}_{2})\rangle$
$\displaystyle=\langle V_{N+2}(\\{x_{t}\\}_{t=1,\dots
N},\\{x_{a}\\}_{a=1.2}|\otimes_{t}|T_{t}\rangle~{}\otimes(-i\sqrt{2\alpha^{\prime}}\cos\gamma_{t_{1}})\alpha_{(1)1}^{\dagger}|0_{(1)}\rangle~{}\otimes(-i\sqrt{2\alpha^{\prime}}\cos\gamma_{t_{2}}~{})\bar{\alpha}_{(2)1}^{\dagger}|0_{(2)}\rangle$
$\displaystyle=\delta(\sum_{t}\kappa_{t})e^{-\frac{1}{2}\sum_{t}\kappa_{t}^{2}~{}G^{22}_{bou,~{}reg~{}T(t)}(x_{t};x_{t})}e^{-\frac{1}{2}\sum_{u,t}\kappa_{t}\kappa_{u}~{}G^{22}_{bou}(x_{t};x_{u})}~{}C(x_{1},\dots
x_{N})~{}\partial_{y_{1}}\partial_{y_{2}}G^{z\bar{z}}_{bou}(y_{1};y_{2};\\{x_{t}\\})$
(39)

where ${\bar{y}}$ is a function of $y$ as in eq. (46). The previous result is
an immediate consequence of the the definition of $G$ given in (3) but it can
also be interpreted as a “proof” of eq. (3) since the Green function entering
in the previous formula is the Green function obtained from the path integral.

###  2.3 Example 3: $N-1$ not excited twisted, $1$ excited twisted and $2$
untwisted states

We can now discuss the excited twisted states. The easiest way to denote an
excited twisted state is by writing from which untwisted state it can be
obtained by OPE, for example
$\left(\partial_{x}Z~{}\partial_{x}^{2}Z~{}\sigma_{\epsilon,\kappa}\right)(x,{\bar{x}})$
can be obtained by taking the finite part of the OPE
$(\partial_{y}Z~{}\partial_{y}^{2}Z)(y,{\bar{y}})\sigma_{\epsilon,\kappa}(x,{\bar{x}})$
as $y\rightarrow x$. This limit can be taken in a clearer and easier way when
we realize $(\partial_{y}Z~{}\partial_{y}^{2}Z)(y,{\bar{y}})$ as a normal
ordered operator in a twisted auxiliary Hilbert space where
$\sigma_{\epsilon,\kappa}(x,{\bar{x}})\leftrightarrow|T,\kappa\rangle$. In
particular we can define the finite part of the limit as

$\lim_{y\rightarrow
0^{+}}:\left[y^{1-\epsilon}\partial_{y}Z(y,y)~{}\partial_{y}[y^{1-\epsilon}\partial_{y}Z(y,y)]\right]:|T,\kappa\rangle=\left(-i\sqrt{2\alpha^{\prime}}\,\cos\gamma\right)^{2}\alpha^{\dagger}_{\epsilon}\alpha^{\dagger}_{1+\epsilon}|T,\kappa\rangle$
(40)

Similarly we can consider the map

$\displaystyle\left(\partial\bar{Z}\sigma_{\epsilon_{t},\kappa_{t}}\right)(x_{t},\bar{x}_{t})\leftrightarrow-i\sqrt{2\alpha^{\prime}}\,\cos\gamma_{t}\bar{\alpha}^{\dagger}_{1-\epsilon}|T_{t},\kappa_{t}\rangle$
(41)

and compute the correlator

$\displaystyle\langle\left(\partial_{x}{\bar{Z}}\sigma_{\epsilon_{1},\kappa_{1}}\right)(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\partial_{y_{1}}Z(y_{1},\bar{y}_{1})\rangle$
$\displaystyle=\langle V_{N+1}(\\{x_{t}\\}_{t=1,\dots
N},\\{x_{a}\\}_{a=1.2}|(-i\sqrt{2\alpha^{\prime}}\cos\gamma_{1}~{})\bar{\alpha}_{(1)\epsilon_{1}}^{\dagger}|T_{1},\kappa_{1}\rangle\otimes_{t>1}|T_{t},\kappa_{t}\rangle~{}\otimes(-i\sqrt{2\alpha^{\prime}}\cos\gamma_{t_{1}})\alpha_{(1)1}^{\dagger}|0_{(1)}\rangle$
(42)

to be

$\displaystyle\delta(\sum_{t}\kappa_{t})e^{-\frac{1}{2}\sum_{t}\kappa_{t}^{2}~{}G^{22}_{bou,~{}reg~{}T(t)}(x_{t};x_{t})}e^{-\frac{1}{2}\sum_{u,t}\kappa_{t}\kappa_{u}~{}G^{22}_{bou}(x_{t};x_{u})}~{}C(x_{1},\dots
x_{N})~{}$
$\displaystyle~{}~{}\partial_{y_{1}}\left[(x-x_{1})^{\epsilon_{1}}\partial_{x}G^{\bar{z}z}_{bou}(x,y_{1};\\{x_{v}\\})\right]|_{x=x_{1}}.$
(43)

The same computation can be performed using the generating function as

$\displaystyle\langle\left(\partial_{x}{\bar{Z}}\sigma_{\epsilon_{1},\kappa_{1}}\right)(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\partial_{y_{1}}Z(y_{1},\bar{y}_{1})\rangle={\cal
V}_{N+M}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
d_{(1)1}}e^{i\kappa_{1}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
d_{(1)0}^{2}}}\prod_{t>1}e^{i\kappa_{t}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
d_{(t)0}^{2}}}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
c_{(1)1}}\Big{|}_{c=0;d=0}$ (44)

##  3 Derivation for untwisted matter

The starting point is very similar to ([4],[5]) where it was recognized that
the generator for all closed (super)string amplitudes is a quadratic path
integrals. The idea in the previous papers is that the appropriate boundary
condition for R and/or NS sector can be obtained simply by inserting linear
sources with the desired boundary conditions. Because of this assumption the
quantum fluctuations are the same for all the amplitudes: from the purely NS
to the mixed ones. It was later realized that this prescription misses a
proper treatment of the quantum fluctuations ([10]) and that when this part is
considered the amplitudes factorize correctly ([11]).

Here we consider open strings and we realize the proper twisted boundary
conditions by quadratic boundary terms which are nothing else but the coupling
of the string to the magnetic field background. Therefore a non excited twist
field is realized by a discontinuity in the magnetic field888 And in
particular the transition from an eigenstate $|\alpha\rangle$ in magnetic
field $B_{t}$ to an eigenstate $|\beta\rangle$ in magnetic $B_{t+1}$ at
worldsheet time $\tau_{t}$ can be computed as in usual quantum mechanics as
$\langle\beta|\alpha\rangle=\int{\cal D}X~{}\langle
X(\sigma),\tau_{t}^{-}|\alpha\rangle\left(\langle
X(\sigma),\tau_{t}^{+}|\beta\rangle\right)^{*}$. .

The $N$ non excited twist field amplitudes with Euclidean worldsheet metric is
then computed by the quadratic path integral999 We define
$d^{2}z=2dz~{}d\bar{z}$ and $z=e^{\tau_{E}+i\sigma}\in H$.  101010This is the
path integral corresponding to all twist fields with zero “momentum” and
therefore it is proportional to a $\delta(0)$ which arises from $X^{2}$ zero
mode. The general case is treated in the next section.

$\displaystyle C(x_{1},\dots x_{N})\delta(0)$ $\displaystyle={\cal N}\int{\cal
D}X~{}e^{-\frac{1}{2\pi\alpha^{\prime}}\left[\int_{H}d^{2}z~{}\frac{1}{2}G_{ij}\partial_{z}X^{i}\partial_{\bar{z}}X^{j}-i\int_{\partial
H}dx~{}B(x)X^{1}(x,{\bar{x}})\partial_{x}X^{2}(x,{\bar{x}})\right]}$ (45)

where $dx\partial_{x}X^{2}(x,{\bar{x}})$ must be interpreted as the pullback
on the boundary of $dX^{2}$ in such a way that $\bar{x}$ depends on $x$ as

${\bar{x}}=\left\\{\begin{array}[]{ll}x&x=|x|>0\\\
xe^{-i2\pi}&x=|x|e^{i\pi}<0\end{array}\right.,$ (46)

$H$ is the superior half plane and the adimensional magnetic field $B(x)$ is
given by

$B(x)=2\pi\alpha^{\prime}~{}qF_{12}(x)=\sum_{t=1}^{N}B_{t}~{}\theta(x-x_{t})~{}\theta(x_{t+1}-x)$
(47)

so that the dicharged string at $x=x_{t}$ feels a magnetic field $B_{t-1}$ on
the left, i.e. $\sigma=\pi$ and $B_{t}$ on the right, i.e. $\sigma=0$. Here we
have set $B_{-1}=B_{N+1}$ and $x_{-1}=-\infty$, $x_{N+1}=+\infty$.

In the previous equation (45) we have chosen the gauge

$A_{1}=0,~{}~{}~{}~{}A_{2}=Bx^{1}$ (48)

in order to make clear that we have only one zero mode which is associated
with the shift $X^{2}\rightarrow X^{2}+\epsilon$ and therefore there is only
one conserved momentum as it is the case in the Landau levels problem on
$\mathbb{R}^{2}$.

Now we can add the untwisted states vertexes. This can be done by considering
the generating function of all untwisted vertexes at $x=x_{a}\in\mathbb{R}$
as111111 This formulation involving the full field on the boundary is only
right for NN boundary condition. For DD boundary condition one should use a
slightly different one ([13]).

$\displaystyle{\cal S}(c_{(a)})$
$\displaystyle=\exp\\{\sum_{n=0}^{\infty}c_{(a)\,n\,i}\partial^{n}_{x}|_{x=x_{a}}X^{i}(x,{\bar{x}})\\}=\exp\\{\int_{\partial
H}dx~{}J_{i}(x;x_{a})~{}X^{i}(x,{\bar{x}})\\}$ $\displaystyle J_{i}(x;x_{a})$
$\displaystyle=\sum_{n=0}^{\infty}c_{(a)ni}\partial^{n}_{x_{a}}\delta(x-x_{a})$
(49)

where $c$ are arbitrary complex numbers (or functions as we use in the next
section). To understand how the previous generating vertex work let us take
the example from ([12]). Consider the vertex which describes the fluctuations
of the gauge vector around the dipole string background we can derive it from
a generating functional for the dipole string as 121212 Notice that here we
are talking about the abstract path integral representation of the vertex and
not of the operatorial representation. The operatorial representation of the
vertex can be realized in an auxiliary space with both twisted and untwisted
boundary conditions. The untwisted auxiliary Hilbert space representation is
the usual operatorial representation while the twisted one is the one derived
in ([12]). Just because of these different realizations this auxiliary Hilbert
space must not be confused with ${\cal H}_{(a,t_{a})}$ introduced before which
is a way of representing the $c_{(a)ni}$, see (3).  131313 It is worth
noticing how vertexes for dipole strings have the same functional form
independently on the magnetic backgrounds $B_{t_{a}}$ nevertheless they differ
because different conditions for physical states, in the previous example we
have $k_{\mu}\eta^{\mu\nu}k_{\nu}+k_{i}{\cal
G}^{ij}(B_{t_{a}})k_{j}=\epsilon_{\mu}\eta^{\mu\nu}k_{\nu}+\epsilon_{i}{\cal
G}^{ij}(B_{t_{a}})k_{j}=0$ where $\mu,\nu\neq 1,2$.

$\displaystyle
V(x_{a};\epsilon,k)=\epsilon_{i}\partial_{x}X^{i}(x,{\bar{x}})e^{ik_{j}X^{j}(x,{\bar{x}})}\Big{|}_{x=x_{a}}=S(c_{(a)},x_{a})~{}\epsilon_{i}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
c_{(a)1i}}e^{ik_{j}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
c_{(a)0j}}}\Big{|}_{c_{(a)}=0}$ (50)

As a matter of facts the previous vertex gives an indefinite result when
inserted in the path integral even when $B=0$. We must therefore regularize it
and consider

$\displaystyle{\cal S}_{reg}(c_{(a)})$ $\displaystyle={\cal
N}_{(a)}(x_{a})\exp\Big{\\{}\sum_{n=0}^{\infty}c_{(a)\,n\,i}\partial^{n}_{x}|_{x=x_{a}}\langle
X^{i}(x,{\bar{x}})\rangle\Big{\\}}$ $\displaystyle={\cal
N}_{(a)}(x_{a})\exp\Big{\\{}\int_{\partial
H}dx~{}J_{i,reg}(x;x_{a})~{}X^{i}(x,{\bar{x}})\Big{\\}}$ (51)

where the regularized curred is given by

$\displaystyle J_{i,reg}(x;x_{a})$
$\displaystyle=\sum_{n=0}^{\infty}c_{(a)ni}\partial^{n}_{x_{a}}\delta_{reg}(x-x_{a}),$
(52)

the averaged field by

$\langle X^{i}(x,x)\rangle=\int_{\partial H}dy~{}\delta_{reg}(x-y)X(y,y)$ (53)

and the normalization factor is

$\displaystyle{\cal N}_{(a)}(x_{a})=\exp\Big{\\{}$
$\displaystyle-\frac{1}{2}\sum_{n,m=0}^{\infty}c_{(a)ni}~{}c_{(a)mj}$
$\displaystyle\int_{\partial H}dx\int_{\partial
H}dy~{}\partial^{n}_{x}\delta_{reg}(x-x_{a})~{}\partial^{m}_{y}\delta_{reg}(y-x_{a})~{}G_{U(t_{a}),~{}bou}^{ij}(x;y)\Big{\\}}\Big{|}_{x=x_{a},y=x_{a}}$
(54)

with $G_{U(t_{a}),~{}bou}^{ij}(x;y)$ the boundary Green functions in the
dipole case with magnetic field $B_{t_{a}}$ given in eq.s (• ‣ 1). There are
two reasons why we have introduced the previous definitions. The first is that
it works in reproducing the amplitudes for $N=2$ as discussed in appendix A.
The second is connected to the way the regularization terms is suggested from
the operatorial formalism. In operatorial formalism the simplest approach is
to consider a point splitting, i.e.
$[\exp\left(c_{(a)i}X^{i}_{(a)}(x_{a},x_{a})\right)]_{p.s.}=\exp\left(c_{(a)i}[X^{i(-)}_{(a)}(x_{a}e^{-\eta},x_{a}e^{-\eta})+X^{i(+)}_{(a)}(x_{a},x_{a})]\right)$
which implies a regularization factor ${\cal
N}_{(a)}(x_{a})=\exp\left(-\frac{1}{2}c_{(a)i}c_{(a)j}G^{ij}_{bou}(x;y)\right)\Big{|}_{x=x_{a},y=x_{a}e^{-\eta}}$.
When we smooth the fields the previous regularization factor becomes

$\displaystyle{\cal N}_{(a)}(x_{a})=$
$\displaystyle\exp\left(-\frac{1}{2}c_{(a)i}c_{(a)j}\int_{x>y}dx~{}dy~{}2~{}\delta_{reg}(x-x_{a})\delta_{reg}(y-x_{a})G^{ij}_{bou}(x;y)\right)$
$\displaystyle=$ $\displaystyle\exp\left(-c_{(a)i}c_{(a)j}\int
dx~{}dy~{}\delta_{reg}(x-x_{a})\delta_{reg}(y-x_{a})\frac{G^{ij}_{bou}(x;y)~{}\theta(x-y)+G^{ji}_{bou}(y;x)~{}\theta(y-x)}{2}\right)$
$\displaystyle=$ $\displaystyle\exp\left(-\frac{1}{2}c_{(a)i}c_{(a)j}\int
dx~{}dy~{}\delta_{reg}(x-x_{a})\delta_{reg}(y-x_{a})G^{ij}_{bou}(x;y)\right)$
(55)

where the factor $2$ in the first line is due to the fact we are using one of
the two $\delta$ just one half because of the constraint $x>y$ as it can be
directly verified by using a step function regularization of the delta. In the
last step we have used the property $G^{ji}_{bou}(y;x)=G^{ij}_{bou}(x;y)$.

In conclusion the path integral we want to compute in order to get the
generating function for all the $M$ untwisted correlators in presence of $N$
twists is

$\displaystyle Z(\\{x_{t}\\};\\{x_{a}\\})$ $\displaystyle={\cal N}\int{\cal
D}X~{}e^{-\frac{1}{2\pi\alpha^{\prime}}\left[\int_{H}d^{2}z~{}\frac{1}{2}G_{ij}\partial_{z}X^{i}\partial_{\bar{z}}X^{j}+i\int_{\partial
H}dx~{}B(x)X^{1}(x,{\bar{x}})\partial_{x}X^{2}(x,{\bar{x}})\right]}$
$\displaystyle~{}~{}~{}~{}\times\prod_{a=1}^{M}{\cal
N}_{(a)}(x_{a})e^{\int_{\partial H}dx~{}J_{i,reg}(x;x_{a})X^{i}(x,{\bar{x}})}$
(56)

This path integral can then be performed to get

$\displaystyle Z(\\{x_{t}\\};\\{x_{a}\\})$ $\displaystyle=C(x_{1},\dots
x_{N})~{}\delta(i\sum_{a}c_{(a)0\,i=2})$
$\displaystyle\prod_{a}\exp\Big{\\{}\frac{1}{2}\int_{\partial
H}dx\int_{\partial H}dyJ_{i,reg}(x;x_{a})~{}J_{j,reg}(y;x_{a})$
$\displaystyle\hskip
50.00008pt~{}[G^{ij}_{bou}(x;y;\\{x_{v}\\})-G_{U(t_{a})~{},bou}^{ij}(x;y)]\Big{\\}}$
$\displaystyle\prod_{a,b;a\neq b}\exp\Big{\\{}\frac{1}{2}\int_{\partial
H}dx\int_{\partial H}dyJ_{i,reg}(x;x_{a})~{}J_{j,reg}(y;x_{b})$
$\displaystyle\hskip 50.00008pt~{}G^{ij}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}$ (57)

where $G^{ij}_{bou}(x;y;\\{x_{v}\\})$ is the Green function of the quadratic
operator which turns out to be the one defined in (2). Notice that the
quadratic operator is not even hermitian exactly as it happens for Landau
levels in the plain quantum mechanics. The previous result can be also
rewritten after the regularization has been removed as

$\displaystyle Z(\\{x_{t}\\};\\{x_{a}\\})=$ $\displaystyle C(x_{1},\dots
x_{N})~{}\delta(i\sum_{a}c_{(a)0\,i=2})$
$\displaystyle\prod_{a}\exp\Big{\\{}\frac{1}{2}\sum_{n,m=0}^{\infty}c_{(a)ni}~{}c_{(a)mj}~{}\partial^{n}_{x}|_{x=x_{a}}~{}\partial^{m}_{y}|_{y=x_{a}}~{}G^{ij}_{bou~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})\Big{\\}}$
$\displaystyle\prod_{a,b;a\neq
b}\exp\Big{\\{}\frac{1}{2}\sum_{n,m=0}^{\infty}c_{(a)ni}~{}c_{(b)mj}~{}\partial^{n}_{x}|_{x=x_{a}}~{}\partial^{m}_{y}|_{y=x_{b}}~{}G^{ij}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}$
(58)

where we have defined the boundary Green function regularized by the untwisted
Green function for a background $B_{t_{a}}$

$\displaystyle G^{z\bar{z}}_{bou~{}reg~{}U(t_{a})}(x,y;\\{x_{v}\\})$
$\displaystyle=\left[G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})-G^{z\bar{z}}_{U(t_{a}),~{}bou}(x;y)\right]$
(59)

The previous expression can be simplified a little using the symmetry
$G_{bou}^{ij}(x;y)=G_{bou}^{ji}(y;x)$ and by rewriting it in the complex basis
as

$\displaystyle Z(\\{x_{t}\\};\\{x_{a}\\})=$ $\displaystyle C(x_{1},\dots
x_{N})~{}\delta\left(\sum_{a}(c_{(a)0}-\bar{c}_{(a)0})\right)$
$\displaystyle\prod_{a}\exp\Big{\\{}\frac{1}{2}c_{(a)0}^{2}~{}G^{\bar{z}\bar{z}}_{bou~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})+\frac{1}{2}\bar{c}_{(a)0}^{2}~{}G^{zz}_{bou~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a}\exp\Big{\\{}}\sum_{n,m=0}^{\infty}\bar{c}_{(a)n}~{}c_{(a)m}~{}\partial^{n}_{x}|_{x=x_{a}}~{}\partial^{m}_{y}|_{y=x_{a}}~{}G^{z\bar{z}}_{bou~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})\Big{\\}}$
$\displaystyle\prod_{a<b}\exp\Big{\\{}\bar{c}_{(a)n}~{}\bar{c}_{(b)m}~{}G^{zz}_{bou}(x_{a};x_{b};\\{x_{v}\\})+c_{(a)n}~{}c_{(b)m}~{}G^{\bar{z}\bar{z}}_{bou}(x_{a};x_{b};\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}+\sum_{n,m=0}^{\infty}\bar{c}_{(a)n}~{}c_{(b)m}~{}\partial^{n}_{x}|_{x=x_{a}}~{}\partial^{m}_{y}|_{y=x_{b}}~{}G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}+\sum_{n,m=0}^{\infty}c_{(a)n}~{}\bar{c}_{(b)m}~{}\partial^{n}_{x}|_{x=x_{a}}~{}\partial^{m}_{y}|_{y=x_{b}}~{}G^{\bar{z}z}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}$
(60)

with $\bar{c}_{(a)n}=c_{(a)nz}$ and $c_{(a)n}=c_{(a)n\bar{z}}$.

We can now give a different formulation of the previous result if we realize
the algebra

$[c_{(a)ni},\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
c_{(b)mj}}]=\delta_{i}^{j}~{}\delta_{n,m}~{}\delta_{a,b}$ (61)

on the untwisted scalar (dipole string) auxiliary Hilbert spaces ${\cal
H}_{a,t_{a}}$ with backgrounds $B_{t_{a}}$ introduced before as

$\displaystyle 1\rightarrow$ $\displaystyle\langle
z_{(a)00}=\bar{z}_{(a)00}=0|\langle 0_{(a)}|$
$\displaystyle\bar{c}_{(a)n\,}\rightarrow\frac{i}{\sqrt{2\alpha^{\prime}}\cos\gamma_{t_{a}}}\frac{\alpha_{(a)n}}{n!},$
$\displaystyle~{}~{}c_{(a)n\,}\rightarrow\frac{i}{\sqrt{2\alpha^{\prime}}\cos\gamma_{t_{a}}}\frac{\bar{\alpha}_{(a)n}}{n!}~{}~{}~{}n\geq
0$
$\displaystyle\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial\bar{c}_{(a)m}}\rightarrow-i\sqrt{2\alpha^{\prime}}\,(m-1)!\,\cos\gamma_{t_{a}}\,\alpha_{(a)m}^{\dagger},$
$\displaystyle~{}~{}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
c_{(a)m}}\rightarrow-i\sqrt{2\alpha^{\prime}}\,(m-1)!\,\cos\gamma_{t_{a}}\,\bar{\alpha}_{(a)m}^{\dagger}~{}~{}m>0$
(62)

where the “strange” choice of the normalization is due to the last expressions
which arise from the desire of identifying

$-i\sqrt{2\alpha^{\prime}}\,(m-1)!\,\cos\gamma_{t_{a}}\,\alpha_{(a)m}^{\dagger}\sim\partial^{m}X^{(-)z}_{(a)}(x,{\bar{x}})|_{x=0}.$
(63)

Using these auxiliary Hilbert spaces we can now rewrite the previous
expression for the $M$ untwisted correlators (3) as

$\displaystyle Z(\\{x_{t}\\};\\{x_{a}\\})$ $\displaystyle=C(x_{1},\dots
x_{N})~{}\delta\left(i\sum_{a}(\alpha_{(a)0}-\bar{\alpha}_{(a)0})\right)~{}\prod_{a=1}^{M}\langle
z_{(a)00}=\bar{z}_{(a)00}=0|\langle 0_{(a)}|$
$\displaystyle\prod_{a}\exp\Big{\\{}-\frac{1}{4\alpha^{\prime}}\alpha_{(a)0}^{2}~{}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}^{2}~{}G^{\bar{z}\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a}\exp\Big{\\{}}-\frac{1}{4\alpha^{\prime}}\bar{\alpha}_{(a)0}^{2}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}^{2}~{}G^{zz}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a}\exp\Big{\\{}}-\frac{1}{2\alpha^{\prime}}\sum_{n,m=0}^{\infty}\alpha_{(a)n}~{}\bar{\alpha}_{(a)m}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n}_{x}}{n!}~{}\frac{\partial^{m}_{y}}{m!}~{}G^{z\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x_{v}\\})\Big{\\}}\Big{|}_{x=y=x_{a}}$
$\displaystyle\prod_{a<b}\exp\Big{\\{}-\frac{1}{2\alpha^{\prime}}\sum_{n,m=0}^{\infty}\alpha_{(a)n}~{}\bar{\alpha}_{(b)m}~{}{\cal
V}_{(t_{a})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{b})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n}_{x}}{n!}~{}\frac{\partial^{m}_{y}}{m!}~{}G^{z\bar{z}}_{bou}(x;y;\\{x_{v}\\})$
$\displaystyle\phantom{\prod_{a<b}\exp\Big{\\{}}-\frac{1}{2\alpha^{\prime}}\sum_{n,m=0}^{\infty}\bar{\alpha}_{(a)n}~{}\alpha_{(b)m}~{}{\cal
V}_{(t_{b})~{}\underline{\bar{z}}z}{\cal
V}_{(t_{a})~{}\underline{z}\bar{z}}~{}\frac{\partial^{n}_{x}}{n!}~{}\frac{\partial^{m}_{y}}{m!}~{}G^{\bar{z}z}_{bou}(x;y;\\{x_{v}\\})\Big{\\}}\Big{|}_{x=x_{a},y=x_{b}}$
(64)

where ${\cal V}$ are the $\mathbb{R}^{2}$ vielbein which connect the $\alpha$
flat index with the Green function $G$ curved index.

##  4 Derivation for twisted matter

The strategy we are going to follow is to consider the amplitude derived in
previous section with $N+M$ untwisted states at the positions
$\\{x_{a}\\}_{a=1\dots M},\\{x_{f}\\}_{f=1\dots N}$ and unexcited twists at
positions $\\{x_{t}\\}_{t=1\dots N}$. Then we choose $N$ of untwisted states
at the positions $\\{x_{f}\\}_{f=1\dots N}$ for which we take the limit
$x_{f}\rightarrow x_{t}$. In order to get the desired amplitude with $M$
untwisted and $N$ excited twisted states we must choose in a proper way the
$c_{(f)ni}$. This amounts not only to choose $c_{(f)ni}$ in (3) as a function
of $x_{f}$ as in eq. (4) but to introduce a further normalization ${\cal
R}(x_{f})$ as in eq. (74) in such a way that we can “undo” the OPE and get a
result which is a generating function for the twisted states

$\exp\Big{\\{}d^{2}_{(t)0}x_{0}^{i=2(aux\,t)}-i\sqrt{2\alpha^{\prime}}\,\cos\gamma_{t}\,\sum_{n=1}^{\infty}\left[d_{(t)n}(n-1)!\bar{\alpha}^{\dagger(aux\,t)}_{n-\epsilon_{t}}+\bar{d}_{(t_{f})n}(n-1)!\alpha^{\dagger(aux\,t)}_{n-1+\epsilon_{t}}\right]\Big{\\}}|T_{(aux\,t)}\rangle$
(65)

when expressed in a chart where $x_{t}=0$. The “strange” normalization is
chosen because it is the easiest map from operators to states, f.x. the
twisted excited state which can be obtained by subtracting the divergences of
the limit $y\rightarrow x_{t}^{+}$ of
$[\partial_{y}^{3}Z(y,y)]^{2}\sigma_{\epsilon_{t}}(x_{t},x_{t})$ gives the
state

$\lim_{y\rightarrow
0^{+}}\left[\partial_{y}^{2}[y^{1-\epsilon_{t}}\partial_{y}Z^{(aux\,t)}(y,y)]\right]^{2}|T_{(aux\,t)}\rangle=\left(-i\sqrt{2\alpha^{\prime}}~{}\cos\gamma_{t}~{}2!~{}\alpha^{\dagger(aux\,t)}_{2+\epsilon_{t}}\right)^{2}|T_{(aux\,t)}\rangle$
(66)

thus making contact between eq. (65) and eq. (4).

Let us start studying the OPE ${\cal S}(c,x_{f})\sigma_{t}(0,0)$. This can be
studied in an auxiliary Hilbert space ${\cal H}_{aux\,t}$ (not to be confused
the the Hilbert space ${\cal H}_{t}$ which we introduced in the first section
and which is associated with coefficients $d_{(t)}$) where
$\sigma_{\epsilon_{t}}(0,0)$ is represented by the twisted vacuum
$|T_{(aux\,t)}\rangle$ and the generating function ${\cal S}(c,x_{f})$ as

$\displaystyle{\cal S}_{(aux\,t)}(c_{(f)},x_{f})=$ $\displaystyle
e^{-\frac{1}{2}c_{(f)0}^{2}~{}G^{zz}_{bou,~{}reg~{}U(t_{a})}(x_{f};x_{f};\\{x=0,x=\infty\\})-\frac{1}{2}\bar{c}_{(f)0}^{2}~{}G^{\bar{z}\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x_{f};x_{f};\\{x=0,x=\infty\\})}$
$\displaystyle
e^{-\frac{1}{2}\sum_{m,n=0}^{\infty}c_{(f)n}~{}\bar{c}_{(f)m}~{}\partial^{n}_{x}|_{x=x_{f}}~{}\partial^{m}_{y}|_{y=x_{f}}~{}G^{z\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x=0,x=\infty\\})}$
$\displaystyle:e^{\sum_{n=0}^{\infty}\bar{c}_{(f)n}~{}\partial^{n}_{x}|_{x=x_{f}}Z_{(aux\,t)}(x,\bar{x})+c_{(f)n}~{}\partial^{n}_{x}|_{x=x_{f}}\bar{Z}_{(aux\,t)}(x,\bar{x})}:$
(67)

In the previous equation the normal ordering is performed with respect to the
operators entering the expansion of the quantum fields
$Z_{(aux\,t)}(z,\bar{z})$ and $\bar{Z}_{(aux\,t)}(z,\bar{z})$ which act on
${\cal H}_{aux\,t}$. Then the OPE can be computed as

$\displaystyle{\cal S}(c,x_{f})\sigma_{\epsilon_{t}}(0,0)\leftrightarrow$
$\displaystyle
e^{-\frac{1}{2}c_{(f)0}^{2}~{}G^{zz}_{bou,~{}reg~{}U(t_{a})}(x_{f};x_{f};\\{x=0,x=\infty\\})-\frac{1}{2}\bar{c}_{(f)0}^{2}~{}G^{\bar{z}\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x_{f};x_{f};\\{x=0,x=\infty\\})}$
$\displaystyle
e^{-\frac{1}{2}\sum_{m,n=0}^{\infty}\partial^{n}_{x}|_{x=x_{f}}~{}\partial^{m}_{y}|_{y=x_{f}}\left[~{}c_{(f)n}~{}\bar{c}_{(f)m}~{}G^{z\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x=0,x=\infty\\})\right]}$
$\displaystyle
e^{\sum_{n=0}^{\infty}\partial^{n}_{x}|_{x=x_{f}}\left[\bar{c}_{(f)n}~{}Z_{(aux\,t)}^{(-)}(x,x)\right]+\partial^{n}_{x}|_{x=x_{f}}\left[c_{(f)n}~{}\bar{Z}_{(aux\,t)}^{(-)}(x,x)\right]}|T_{(aux\,t)}\rangle$
(68)

which is similar to a rewriting of eq. (65) as

$\displaystyle\lim_{x_{f}\rightarrow 0^{+}}$ $\displaystyle
e^{\bar{d}_{(t)0}~{}\left[Z_{(aux~{}t)}^{(-)}(x_{f},x_{f})\right]+d_{(t)0}~{}\left[\bar{Z}_{(aux~{}t)}^{(-)}(x_{f},x_{f})\right]}$
$\displaystyle
e^{\sum_{n=1}^{\infty}\bar{d}_{(t)n}~{}\partial^{n-1}_{x}|_{x=x_{f}}\left[x^{1-\epsilon_{t}}\partial_{x}Z_{(aux~{})}^{(-)}(x,x)\right]+d_{(t)n}~{}\partial^{n-1}_{x}|_{x=x_{f}}\left[x^{\epsilon_{t}}\partial_{x}\bar{Z}_{(aux~{})}^{(-)}(x,x)\right]}|T_{(aux\,t)}\rangle$
(69)

where in the second line we have written $\partial Z$ since we want to get rid
of zero modes and in the limit it is necessary to write $x_{f}\rightarrow
0^{+}$ since the the behavior of $\partial Z$ changes by an overall
normalization when $x<0$.

Comparison between the two previous expressions suggests to consider then the
operator acting on the Hilbert space ${\cal H}_{(aux~{}t)}$

$\displaystyle{\cal T}_{(aux~{}t)}(d_{(t)},x_{f})$ $\displaystyle={\cal
N}(d_{(t)},x_{f},x_{t})~{}e^{\bar{d}_{(t)0}~{}\left[Z_{(aux~{}t,reg)}(x_{f},x_{f})\right]+d_{(t)0}~{}\left[\bar{Z}_{(aux~{}t,reg)}(x_{f},x_{f})\right]}$
$\displaystyle
e^{\sum_{n=1}^{\infty}\bar{d}_{(t)n}~{}\partial^{n-1}_{x}\left[x^{1-\epsilon_{t}}\partial_{x}Z_{(aux~{}t,reg)}(x,x)\right]+d_{(t)n}~{}\partial^{n-1}_{x}\left[x^{\epsilon_{t}}\partial_{x}\bar{Z}_{(aux~{}t,reg)}(x,x)\right]}\Big{|}_{x=x_{f}^{+}}$
(70)

where $Z_{(aux~{}t,reg)}(x_{f},x_{f})$ is point split regularized of
$Z_{(aux~{}t)}(x_{f},x_{f})$ defined as

$Z_{(aux~{}t,reg)}(x_{f},x_{f})=Z_{(aux~{}t)}^{(-)}(x_{f}e^{-\eta},x_{f}e^{-\eta})+Z_{(aux~{}t)}^{(+)}(x_{f},x_{f}),$

no normal ordering is performed and the normalization factor is given by

$\displaystyle{\cal N}^{-1}(d_{(t)},x_{f},x_{t})$
$\displaystyle=\Big{\\{}e^{\frac{1}{2}\bar{d}_{(t)0}^{2}G^{zz}_{T(t)~{}bou}(x;y)+\frac{1}{2}d_{(t)0}^{2}G^{\bar{z}\bar{z}}_{T(t)~{}bou}(x;y)+\bar{d}_{(t)0}d_{(t)0}\frac{G^{z\bar{z}}_{T(t)~{}bou}(x;y)+G^{\bar{z}z}_{T(t)~{}bou}(x;y)}{2}}$
$\displaystyle
e^{\frac{1}{2}\sum_{n=1}^{\infty}\bar{d}_{(t)0}d_{(t)n}\big{[}\partial^{n-1}_{x}(x^{\epsilon_{t}}\partial_{x}G^{\bar{z}z}_{T(t)~{}bou}(x;y))+\partial^{n-1}_{y}(y^{\epsilon_{t}}\partial_{y}G^{z\bar{z}}_{T(t)~{}bou}(x;y))\big{]}}$
$\displaystyle
e^{\frac{1}{2}\sum_{n=1}^{\infty}d_{(t)0}\bar{d}_{(t)n}\big{[}\partial^{n-1}_{x}(x^{1-\epsilon_{t}}\partial_{x}G^{z\bar{z}}_{T(t)~{}bou}(x;y))+\partial^{n-1}_{y}(y^{1-\epsilon_{t}}\partial_{y}G^{\bar{z}z}_{T(t)~{}bou}(y;x))\big{]}}$
$\displaystyle
e^{\frac{1}{2}\sum_{n,l=1}^{\infty}d_{(t)l}\bar{d}_{(t)n}\big{[}\partial^{l-1}_{y}\partial^{n-1}_{x}(y^{\epsilon_{t}}x^{1-\epsilon_{t}}\partial_{x}\partial_{y}G^{z\bar{z}}_{T(t)~{}bou}(x;y))+\partial^{l-1}_{x}\partial^{n-1}_{y}(x^{\epsilon_{t}}y^{1-\epsilon_{t}}\partial_{x}\partial_{y}G^{\bar{z}z}_{T(t)~{}bou}(x;y))\big{]}}$
$\displaystyle\Big{\\}}\Big{|}_{x=x_{f};y=x_{f}e^{-\eta}}$ (71)

where
$G^{ij}_{T(t)~{}bou}(y;x)=G^{ji}_{T(t)~{}bou}(x;y)=G^{ij}_{bou}(y;x;\\{x_{1}=0,x_{2}=\infty\\})$
are the (analytic continuation of the) boundary Green functions defined in
eq.s (• ‣ 1).

The reason why we have written the previous expression in a non normal ordered
way is to understand the expression of the regularization factor of the
corresponding “classical” vertex (4) which we want to insert in the path
integral.

The previous operator can also be written in a way to make its connection with
the idea of undoing the OPE clearer as

$\displaystyle{\cal T}_{(aux~{}t)}(d_{(t)},x_{f})=$ $\displaystyle{\cal
R}(d_{(t)},x_{f})~{}{\cal S}_{(aux~{}t)}(c_{(f)}(d_{(t)},x_{f}),x_{f})$
$\displaystyle=$ $\displaystyle{\cal
R}(d_{(t)},x_{f})e^{\sum_{n=0}^{\infty}\bar{d}_{(t)n}~{}\partial^{n}_{x}|_{x=x_{f}}\left[x^{-\epsilon_{t}}Z_{(aux\,t)}(x,x)\right]+d_{(f)n}~{}\partial^{n}_{x}|_{x=x_{f}}\left[x^{-(1-\epsilon_{t})}\bar{Z}_{(aux\,t)}(x,x)\right]}$
(72)

where

$\displaystyle c_{(f)0}(d_{(t)},x_{f})=d_{(t)0}~{}~{}$
$\displaystyle~{}~{}\bar{c}_{(f)0}(d_{(t)},x_{f})=\bar{d}_{(t)0}$
$\displaystyle c_{(f)n}(d_{(t)},x_{f})=$
$\displaystyle\sum_{k=n}^{\infty}\binom{k-1}{n-1}d_{(t)k}\partial^{k-n}x_{f}^{-(1-\epsilon_{t})}$
$\displaystyle\bar{c}_{(f)n}(d_{(t)},x_{f})=$
$\displaystyle\sum_{k=n}^{\infty}\binom{k-1}{n-1}\bar{d}_{(t)k}\partial^{k-n}x_{f}^{\epsilon_{t}}$
(73)

and the normalization factor is

$\displaystyle{\cal R}(d_{(t)},x_{f})=$ $\displaystyle
e^{-\sum_{m,n=0}^{\infty}d_{(t)n}~{}\bar{d}_{(t)m}~{}\partial^{n-1}_{x}~{}\partial^{m-1}_{y}\left[x^{1-\epsilon_{t}}y^{\epsilon_{t}}~{}\partial_{x}\partial_{y}G^{z\bar{z}}_{bou,~{}reg~{}U(t_{a})}(x;y;\\{x=0,x=\infty\\})\right]}\Big{|}_{x=y=x_{f}}$
(74)

in order to undo the OPE and get the desired result as in eq. (4).

We shall now translate the previous operator (4) into an abstract operator we
can insert in the path integral at an arbitrary point $x_{t}$, therefore we
move it from $x_{t}=0$ to a generic $x_{t}$ and we consider a generating
vertex as

$\displaystyle{\cal T}(d_{(t)},x_{t})=$ $\displaystyle\lim_{x\rightarrow
x_{t}^{+}}{\cal N}_{\cal T}(d_{(t)},x,x_{t})~{}e^{\bar{d}_{(t)0}~{}\langle
Z(x,{\bar{x}})\rangle+d_{(t)0}~{}\langle\bar{Z}(x,{\bar{x}})\rangle}$
$\displaystyle
e^{\sum_{n=1}^{\infty}\bar{d}_{(t)n}~{}\partial^{n-1}_{x}\langle(x-x_{t})^{1-\epsilon_{t}}\partial_{x}Z(x,{\bar{x}})\rangle+d_{(t)n}~{}\partial^{n-1}_{x}\langle(x-x_{t})^{\epsilon_{t}}\partial_{x}\bar{Z}(x,{\bar{x}})\rangle}$
(75)

where $x\rightarrow x_{t}$ has to be understood as taking the limit after the
path integral has been computed. We have defined the averaged fields such as

$\langle(x-x_{t})^{1-\epsilon_{t}}\partial_{x}Z(x,{\bar{x}})\rangle=\int_{\partial
H}dy~{}\delta_{reg}(x-y)(y-x_{t})^{1-\epsilon_{t}}\partial_{y}Z(y,{\bar{y}})$
(76)

because we want a well defined regulated expression after performing the path
integral and introduced the normalization factor

$\displaystyle{\cal N}_{\cal T}(d_{(t)},x,x_{t})=$ $\displaystyle
e^{-\frac{1}{2}\bar{d}_{(t)0}^{2}\langle\langle
G^{zz}_{N=2~{}bou}(x;x)\rangle\rangle-\frac{1}{2}d_{(t)0}^{2}\langle\langle
G^{\bar{z}\bar{z}}_{N=2~{}bou}(x;x)\rangle\rangle-\bar{d}_{(t)0}d_{(t)0}\langle\langle
G^{z\bar{z}}_{N=2~{}bou}(x;x)\rangle\rangle}$ $\displaystyle
e^{-\frac{1}{2}\sum_{n=1}^{\infty}\bar{d}_{(t)0}d_{(t)n}\partial^{n-1}_{x}\langle\langle(x-x_{t})^{\epsilon_{t}}\partial_{x}G^{\bar{z}z}_{N=2~{}bou}(x;y)\rangle\rangle}$
$\displaystyle
e^{-\frac{1}{2}\sum_{n=1}^{\infty}d_{(t)0}\bar{d}_{(t)n}\partial^{n-1}_{x}\langle\langle(x-x_{t})^{1-\epsilon_{t}}\partial_{x}G^{z\bar{z}}_{N=2~{}bou}(x;y)\rangle\rangle}$
$\displaystyle
e^{-\frac{1}{2}\sum_{n,l=1}^{\infty}d_{(t)l}\bar{d}_{(t)n}\partial^{l-1}_{y}\partial^{n-1}_{x}\langle\langle(y-x_{t})^{\epsilon_{t}}(x-x_{t})^{1-\epsilon_{t}}\partial_{x}\partial_{y}G^{z\bar{z}}_{N=2~{}bou}(x;y)\rangle\rangle}\Big{|}_{y=x}$
(77)

where the doubly regularized Green functions are defined such as

$\displaystyle\langle\langle G^{zz}_{N=2~{}bou}(x;x)\rangle\rangle$
$\displaystyle=\int dy_{1}\int
dy_{2}~{}\delta_{reg}(x-y_{1})~{}\delta_{reg}(x-y_{2})$ $\displaystyle\hskip
70.0001ptG^{zz}_{bou}(y_{1};y_{2};\\{x_{1}=x_{t},x_{2}=\infty\\})$
$\displaystyle\langle\langle(y-x_{t})^{\epsilon_{t}}(x-x_{t})^{1-\epsilon_{t}}\partial_{x}\partial_{y}G^{z\bar{z}}_{N=2~{}bou}(x;y)\rangle\rangle$
$\displaystyle=\int dy_{1}\int
dy_{2}~{}\delta_{reg}(x-y_{1})~{}\delta_{reg}(y-y_{2})$
$\displaystyle(y_{2}-x_{t})^{\epsilon_{t}}~{}(y_{1}-x_{t})^{1-\epsilon_{t}}\partial_{1}\partial_{2}G^{z\bar{z}}_{bou}(y_{1};y_{2};\\{x_{1}=x_{t},x_{2}=\infty\\})$
(78)

Since the previous expression is in nuce the same as for the untwisted matter,
we can immediately deduce from (3) the result of inserting and integrating
over the $X$ to be eq. (33).

In analogy with what done for the untwisted states we can realize the algebra

$\left[d_{(t)n},\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
d_{(u)m}}\right]=\left[\bar{d}_{(t)n},\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial\bar{d}_{(u)m}}\right]=\delta_{m,n}`\delta_{u,t}$
(79)

with operators acting on the twisted scalar (dicharged string) auxiliary
Hilbert spaces ${\cal H}_{t}$ as

$\displaystyle 1\rightarrow$ $\displaystyle\langle
T_{\epsilon_{t}},x^{1}_{(t)0}=0|$
$\displaystyle\bar{d}_{(t)n\,}\rightarrow\frac{i}{\sqrt{2\alpha^{\prime}}\cos\gamma_{t}}\frac{\alpha_{(t)n-1+\epsilon_{t}}}{(n-1)!~{}(n-1+\epsilon_{t})},$
$\displaystyle~{}~{}d_{(t)n\,}\rightarrow\frac{i}{\sqrt{2\alpha^{\prime}}\cos\gamma_{t}}\frac{\bar{\alpha}_{(t)n-\epsilon_{t}}}{(n-1)!~{}(n-\epsilon_{t})}~{}~{}~{}n>0$
$\displaystyle\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial\bar{d}_{(t)m}}\rightarrow-i\sqrt{2\alpha^{\prime}}\,(m-1)!\,\cos\gamma_{t}\,\alpha_{(t)m-1+\epsilon_{t}}^{\dagger},$
$\displaystyle~{}~{}\frac{\stackrel{{\scriptstyle\leftarrow}}{{\partial}}}{\partial
d_{(t)m}}\rightarrow-i\sqrt{2\alpha^{\prime}}\,(m-1)!\,\cos\gamma_{t}\,\bar{\alpha}_{(t)m-\epsilon_{t}}^{\dagger}~{}~{}m>0$
(80)

which gives eq. (32) when substituted into eq. (33).

Acknowledgments We would like to thank P. Di Vecchia and F. Pezzella for
discussions. The author thanks the Nordita for hospitality during different
stages of this work.

##  Appendix A Check of the $N=2$ amplitudes

We would now check that the operatorial amplitudes with $N=2$ and the path
integral approach give the same result, phases included. Let us consider the
tachyonic amplitude

$\displaystyle\langle\sigma_{-\epsilon,\lambda}(x_{\infty},x_{\infty})~{}\sigma_{\epsilon,\kappa}(x_{0},x_{0})~{}V_{T}(x_{1};k_{(1)})\dots~{}V_{T}(x_{M};k_{(M)})\rangle$
(81)

with $x_{t=1}=x_{0}$, $x_{t=2}=x_{\infty}$ and $\gamma_{0}$ and $\gamma_{1}$
arbitrary but $\gamma_{2}=\gamma_{0}$ so that
$\pi\epsilon_{1}=-\pi\epsilon_{2}$. Using the results from ([12]) we can
compute it in the limit $x_{0}\rightarrow 0$ and
$x_{\infty}\rightarrow\infty$, for $x_{1}>\dots x_{M}>0$ and when multiplied
by the appropriate power of $x_{\infty}$ as

$\displaystyle\langle
T_{\epsilon},-\lambda|e^{-\frac{1}{2}R^{2}(\epsilon)~{}\Delta(k_{(1)})}x_{1}^{-\Delta(k_{(1)})}e^{i(\bar{k}_{(1)}z_{0}+k_{(1)}\bar{z}_{0})}e^{i\cos\gamma_{1}[\bar{k}_{(1)}Z_{nzm}(x_{1},x_{1})+k_{(1)}\bar{Z}_{nzm}(x_{1},x_{1})]}\dots|T_{\epsilon},\kappa\rangle$
$\displaystyle=$ $\displaystyle\delta(\kappa+\kappa+\sum_{a}k_{(a)2})$
$\displaystyle\prod_{a}e^{\frac{1}{2}\pi\alpha^{\prime}\frac{1}{\tan\gamma_{1}-\tan\gamma_{0}}(k_{(a)}^{2}-\bar{k}_{(a)}^{2})}$
$\displaystyle\prod_{a}\Big{[}e^{-\frac{1}{2}R^{2}(\epsilon)~{}\Delta(k_{(a)})}x_{a}^{-\Delta(k_{(a)})}\Big{]}$
$\displaystyle\prod_{a}e^{-\pi\alpha^{\prime}\frac{1}{\tan\gamma_{1}-\tan\gamma_{0}}\frac{\kappa}{\sqrt{2}}(k_{(a)}-\bar{k}_{(a)})}$
$\displaystyle\prod_{a<b}\Big{[}e^{\pi\alpha^{\prime}\frac{1}{\tan\gamma_{1}-\tan\gamma_{0}}(k_{(a)}-\bar{k}_{(a)})(k_{(b)}+\bar{k}_{(b)})}e^{\alpha^{\prime}\cos^{2}\gamma_{1}(k_{(a)}\bar{k}_{(b)}\,g_{1-\epsilon}\left(\frac{x_{b}}{x_{a}}\right)+\bar{k}_{(a)}k_{(b)}\,g_{\epsilon}\left(\frac{x_{b}}{x_{a}}\right))}\Big{]}$
(82)

where we have used the commutation relations (• ‣ 1),
$R^{2}(\epsilon)=\lim_{u\rightarrow
1^{-}}[\,g_{\epsilon}(u)+\,g_{1-\epsilon}(u)-2\log(1-u)]=-\left(\psi(\epsilon)+\psi(1-\epsilon)-2\psi(1)\right)$
and $~{}\Delta(k_{(a)})=2\alpha^{\prime}\cos^{2}\gamma~{}k_{(a)}\bar{k}_{(a)}$
is the conformal dimension of the tachyonic vertex.

We can now compare with the general expression (33) with the identifications
$c_{(a)0}\rightarrow i~{}k_{(a)}$, $\bar{c}_{(a)0}\rightarrow
i~{}\bar{k}_{(a)}$, $d_{(1)0}\rightarrow i~{}\frac{\kappa}{\sqrt{2}}$,
$\bar{d}_{(1)0}\rightarrow i~{}\frac{\kappa}{\sqrt{2}}$, $d_{(2)0}\rightarrow
i~{}\frac{\lambda}{\sqrt{2}}$ and $\bar{d}_{(2)0}\rightarrow
i~{}\frac{\lambda}{\sqrt{2}}$. We can also compare with the expression (32)
upon the product with the state…

in order to understand where the different terms come from the path integral
point of view. In matching these terms is important to be careful in rewriting
all the Green functions $G(x;y)$ in such a way that $x>y$ by using the
symmetry (2) since this is the natural way they appear from the operatorial
formalism. We recognize that

  * •

the factor
$\exp\\{\frac{1}{2}\pi\alpha^{\prime}\frac{1}{\tan\gamma_{1}-\tan\gamma_{0}}(k_{(a)}^{2}-\bar{k}_{(a)}^{2})\\}$
come from the $G^{zz}_{bou~{}reg~{}U(t_{a})}$ and
$G^{\bar{z}\bar{z}}_{bou~{}reg~{}U(t_{a})}$ terms.

  * •

The factors
$\prod_{a}\Big{[}e^{-\frac{1}{2}R^{2}(\epsilon)~{}\Delta(k_{(a)})}x_{a}^{-\Delta(k_{(a)})}\Big{]}$
come from the $G^{z\bar{z}}_{bou~{}reg~{}U(t_{a})}$ terms. In particular the
result follows from the following steps

$\displaystyle G^{z\bar{z}}_{bou~{}reg~{}U(t_{a})}(x_{a}^{+},x_{a})=$
$\displaystyle\frac{\pi\alpha^{\prime}}{\tan\gamma_{1}-\tan\gamma_{0}}-\pi\alpha^{\prime}\sin\gamma_{1}\cos\gamma_{1}+2\alpha^{\prime}\cos^{2}\gamma_{1}\ln|x_{a}|$
$\displaystyle-2\alpha^{\prime}\cos^{2}\gamma_{1}\left(\,g_{\epsilon}\left(\frac{x_{a}}{x_{a}^{+}}\right)-\log\left(1-\frac{x_{a}}{x_{a}^{+}}\right)\right)$
$\displaystyle=\frac{\pi\alpha^{\prime}}{\tan(\gamma_{1}-\gamma_{0})}\cos^{2}\gamma_{1}+2\alpha^{\prime}\cos^{2}\gamma_{1}\ln|x_{a}|$
$\displaystyle~{}~{}-2\alpha^{\prime}\cos^{2}\gamma_{1}\left(\psi(1-\epsilon)-\psi(1)\right)$
$\displaystyle=2\alpha^{\prime}\cos^{2}\gamma_{1}\ln|x_{a}|-\alpha^{\prime}\cos^{2}\gamma_{1}\left(\psi(\epsilon)+\psi(1-\epsilon)-2\psi(1)\right)$
(83)

where we have used the expression for the Green function given in eq. (• ‣ 1)
since we have chosen $x=x_{a}^{+}$ and in the last line we have used the
digamma property $\psi(1-\epsilon)=\psi(\epsilon)+\pi\cot(\pi\epsilon)$. It is
also worth stressing that the result is independent on setting $x_{a}^{+}$ in
the first argument since the function $G^{z\bar{z}}_{bou~{}reg~{}U(t_{a})}$ is
continued analytically at $x=y=x_{a}$ in such a way that
$G^{ij}(x;y)=G^{ji}(y;x)$ so that would we have chosen the first argument to
be $x_{a}^{-}$ we would have got the same result computing
$G^{\bar{z}z}_{bou~{}reg~{}U(t_{a})}(x_{a},x_{a}^{-})$.

  * •

The terms
$e^{-\pi\alpha^{\prime}\frac{1}{\tan\gamma_{1}-\tan\gamma_{0}}\frac{\kappa}{\sqrt{2}}(k_{(a)}-\bar{k}_{(a)})}$
arise from the $\prod_{t,a}$ terms. While rewriting the Green functions
$G(x;y)$ in such a way that $x>y$ we see that the terms with $t=2$ (at
$x_{\infty}$) cancel while those from $t=1$ (at $x_{0}$) do not and reproduce
the operatorial result.

  * •

The terms $\prod_{a<b}$ come trivially from the corresponding ones in eq.
(33).

  * •

The terms $\prod_{t}$ in (33) give a trivial result in a non trivial way. It
is immediate to find that
$G^{zz}_{bou,~{}reg~{}(t)}=G^{\bar{z}\bar{z}}_{bou,~{}reg~{}(t)}=0$. On the
other side we get for $0<\frac{y-x_{0}}{x-x_{0}}<1$ and
$0<\omega=\frac{y-x_{0}}{x-x_{0}}\frac{x-x_{\infty}}{y-x_{\infty}}<1$

$\displaystyle G^{z\bar{z}}_{bou,~{}reg~{}(t)}(x;y;\\{x_{0},x_{\infty}\\})$
$\displaystyle=-2\alpha^{\prime}\cos^{2}\gamma\left[\,g_{\epsilon}(\omega)-\,g_{\epsilon}(\frac{y-x_{0}}{x-x_{0}})\right]$
(84)

Now we can write $x=x_{0}+\alpha$, $y=x_{0}+\alpha(1-\delta)$ with $\alpha>0$
and $0<\delta<1$ and expand in $\delta$ to get

$\displaystyle G^{z\bar{z}}_{bou,~{}reg~{}(t)}(x;y;\\{x_{0},x_{\infty}\\})$
$\displaystyle=-2\alpha^{\prime}\cos^{2}\gamma\left[-\log\left(1+\frac{\alpha}{x_{\infty}-x_{0}}\right)+\frac{(\epsilon-1)\alpha}{x_{\infty}-x_{0}+\alpha}\delta+O(\delta^{2})\right]$
(85)

which vanishes when $\alpha=\delta=0$. In a similar way can be treated the
case when $x,y\rightarrow x_{\infty}$.

  * •

The terms $\prod_{t<u}$ in (33) give also a trivial result in a not completely
trivial fashion. We have to evaluate the Green functions for
$\omega=\frac{y-x_{0}}{x-x_{0}}\frac{x-x_{\infty}}{y-x_{\infty}}$ when
$x=x_{0}$ and $y=x_{\infty}$ so $\omega=0$ and we are left with only the
constant terms,as we expect from the general asymptotic (1) hence
$\prod_{t<u}=\exp\\{\frac{\pi\alpha^{\prime}}{\tan\gamma_{1}-\tan\gamma_{0}}\
[-d_{(1)0}d_{(2)0}+\bar{d}_{(1)0}\bar{d}_{(2)0}-d_{(1)0}\bar{d}_{(2)0}+\bar{d}_{(1)0}d_{(2)0}]\\}$
which vanishes when evaluated with the previously stated substitutions for
which $d_{(t)0}=\bar{d}_{(t)0}$.

##  Appendix B Behavior of the Green function when $x,y\rightarrow x_{t}$

In this section we follow and adapt the computation done in ([6]). We start
considering the derivative of Green function of the left moving part defined
as

$\partial_{z}\partial_{w}G^{z\bar{z}}_{LL}(z;w;\\{x_{t}\\}_{t=1\dots
N})=\frac{\langle\partial_{z}Z_{L}(z)\partial_{w}\bar{Z}_{L}(w)\sigma_{\epsilon_{1},\kappa_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\rangle_{disk}}{\langle\sigma_{\epsilon_{1},\kappa_{1}}(x_{1},\bar{x}_{1})\dots\sigma_{\epsilon_{N},\kappa_{N}}(x_{N},\bar{x}_{N})\rangle_{disk}}$
(86)

which has asymptotics

$\displaystyle-\frac{1}{2\alpha^{\prime}}\partial_{z}\partial_{w}G^{z\bar{z}}_{LL}(z;w;\\{x_{t}\\}_{t=1\dots
N})$ $\displaystyle\sim_{z\rightarrow w}\frac{1}{(z-w)^{2}}+O(1)$
$\displaystyle\sim_{z\rightarrow x_{t}}(z-x_{t})^{\epsilon_{t}-1}$
$\displaystyle\sim_{w\rightarrow x_{t}}(w-x_{t})^{-\epsilon_{t}}$ (87)

then we can write

$\displaystyle-\frac{1}{2\alpha^{\prime}}\partial_{z}\partial_{w}G^{z\bar{z}}_{LL}(z;w;\\{x_{t}\\}_{t=1\dots
N})=$
$\displaystyle\prod_{u}\left(\frac{w-x_{u}}{z-x_{u}}\right)^{1-\epsilon_{u}}\Big{[}\frac{1}{(z-w)^{2}}\sum_{u<v}a_{uv}\frac{(z-x_{u})(z-x_{v})}{(w-x_{u})(w-x_{v})}$
$\displaystyle+\sum_{u_{1}<u_{2}<u_{3}<u_{4}}\frac{b_{u_{1}u_{2}u_{3}u_{4}}}{(w-x_{u_{1}})(w-x_{u_{2}})(w-x_{u_{3}})(w-x_{u_{4}})}\Big{]}$
(88)

Now we can study the behavior $x,y\rightarrow x_{1}$ by setting

$z=x_{1}+\alpha,~{}~{}~{}~{}w=x_{1}+\alpha(1-\delta)$ (89)

and letting $\alpha,\delta\rightarrow 0^{+}$. A simple computation gives

$\displaystyle-\frac{1}{2\alpha^{\prime}}\partial_{z}\partial_{w}G^{z\bar{z}}_{LL}(z;w;\\{x_{t}\\}_{t=1\dots
N})\sim$
$\displaystyle\frac{\sum_{u<v}a_{uv}}{\delta^{2}}+\frac{1}{\alpha}\frac{1}{1-\delta}\sum_{1<u_{2}<u_{3}<u_{4}}\frac{b_{1u_{2}u_{3}u_{4}}}{(x_{1}-x_{u_{2}})(x_{1}-x_{u_{3}})(x_{1}-x_{u_{4}})}+O(1)$
(90)

from which we deduce the constraint

$\sum_{u<v}a_{uv}=1.$ (91)

Then the regularized Green function which is obtained by subtracting the
corresponding Green function with only two twist, one of which in $x_{1}$ and
the other at $\infty$ is of the form

$\displaystyle\partial_{z}\partial_{w}G^{z\bar{z}}_{LL,~{}reg}(z;w;\\{x_{t}\\}_{t=1\dots
N})\sim$ $\displaystyle\frac{B_{1}(x_{u})}{\alpha}\frac{1}{1-\delta}+O(1)$
(92)

hence the terms in $\prod_{t}$ are well defined since

$\displaystyle(z-x_{1})^{1-\epsilon_{1}}~{}(w-x_{1})^{\epsilon_{1}}~{}\partial_{z}\partial_{w}G^{z\bar{z}}_{LL,~{}reg}(z;w;\\{x_{t}\\}_{t=1\dots
N})\sim$
$\displaystyle\alpha~{}(1-\delta)^{\epsilon_{1}}\left[\frac{B_{1}(x_{u})}{\alpha}\frac{1}{1-\delta}+O(1)\right].$
(93)

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