BCFT and OSFT moduli: an exact perturbative comparison

Abstract

Starting from the pseudo-${\mathcal {B}}_0$ gauge solution for marginal deformations in OSFT, we analytically compute the relation between the perturbative deformation parameter $\tilde{\lambda }$ in the solution and the BCFT marginal parameter $\lambda$, up to fifth order, by evaluating the Ellwood invariants. We observe that the microscopic reason why $\tilde{\lambda }$ and $\lambda$ are different is that the OSFT propagator renormalizes contact-term divergences differently from the contour deformation used in BCFT.

1 Introduction and conclusion

In the recent years there has been published overwhelming evidence that the various consistent open string backgrounds (i.e. D-branes) can be described analytically as solitons of open string field theory (OSFT) [1,2,3,4,5,6,7,8,9,10,11].¹

A classical [15] yet not fully understood problem in this correspondence is how the D-branes moduli space is described in OSFT. Given an exactly marginal boundary field j, there is a corresponding family of OSFT solutions, which can be generically found in powers of a deformation parameter $\tilde{\lambda }$,

$$\begin{aligned} \Psi _{\tilde{\lambda }} = \tilde{\lambda }cj(0)|0\rangle \ +\sum _{k=2}^{\infty }\tilde{\lambda }^k \,\Psi _k, \end{aligned}$$

(1.1)

where $\Psi _k$ are perturbative contributions obeying the recursive relation

$$\begin{aligned} Q\Psi _k+\sum _{n=1}^{k-1}\Psi _n\Psi _{k-n}=0. \end{aligned}$$

(1.2)

Physically we expect that the deformation parameter ${\tilde{\lambda }}$, which we used to construct the solution, should be related to the natural parameter $\lambda$ in boundary conformal field theory (BCFT), given by the coefficient in front of the boundary interaction which deforms the original world-sheet action,

$$\begin{aligned} S_\lambda =\ S_0\ +\ \lambda \int _{-\infty }^{\infty }\mathrm{d}x\ j(x). \end{aligned}$$

(1.3)

On general grounds, $\tilde{\lambda }$ does not have a gauge invariant meaning, but nonetheless it is useful to understand how $\lambda$ and ${\tilde{\lambda }}$ are related for a given solution, because this can shed light on the different mechanisms by which a classical solution changes the world-sheet boundary conditions.

Analytic solutions for marginal deformations with nonsingular OPE ($j j \sim \hbox { reg}$) have been computed to all orders in [2, 3]. A different perturbative analytic solution for marginal currents with singular OPE has been constructed in [4] and generalized in [5].² An analytic solution for any self-local (hence exact [18]) marginal deformation has been constructed nonperturbatively in [9]. Conveniently, this solution is directly expressed in terms of the deformation parameter of the underlying BCFT, $\lambda$. In [19] this has been used to explicitly find the relation between the BCFT modulus $\lambda$ and the coefficient of the marginal field in the solution $\left\langle c\partial cj|\Psi (\lambda )\right\rangle$. It has been observed that this function of $\lambda$ starts linearly; then it has a local maximum, and finally it approaches zero for large values of $\lambda$. Nontrivial evidence that this behavior may also be present in Siegel gauge³ has been given in [20] in level truncation, but it has not been possible there to establish the validity of the full equations of motion for large BCFT moduli.

In this note we would like to study this problem in another analytic wedge-based example, which is quite close to Siegel gauge. We will analyze the observables of the solution proposed by Schnabl in [2], in the so-called pseudo-${\mathcal {B}}_0$ gauge

$$\begin{aligned} \Psi _{\tilde{\lambda }}= & {} \sum _{k=1}^{\infty } \tilde{\lambda }^k \widehat{U}_{k+1}\,\hat{\Psi }_k|0\rangle , \end{aligned}$$

(1.4)

$$\begin{aligned} {\mathcal {B}}_0\hat{\Psi }_k|0\rangle= & {} 0, \end{aligned}$$

(1.5)

for a chiral marginal current j(z) with OPE

$$\begin{aligned} j(z) j(0) = \frac{1}{z^2} + \hbox {regular}. \end{aligned}$$

(1.6)

Computing the Ellwood invariants and matching them with the BCFT expected answers, gives the following relation:

$$\begin{aligned} \tilde{\lambda }=\tilde{\lambda }(\lambda ) = \lambda -\ 3\log 2 \lambda ^3 +2.38996(7)\ \lambda ^5 +O(\lambda ^7). \end{aligned}$$

(1.7)

Let us comment on the relation found.

Perhaps the most interesting fact about (1.7) is the origin itself-of the found coefficients of $\lambda ^{2n+1}$. These coefficients are obtained by comparing the Ellwood invariants computed from the solution in powers of ${\tilde{\lambda }}$, with the coefficients of the Ishibashi states obtained from the marginally deformed boundary state expressed in powers of $\lambda$, see Eqs. (4.11)–(4.16). Naively these two quantities reduce to the same world-sheet calculation and therefore one would expect to find perfect match between $\lambda$ and ${\tilde{\lambda }}$, which is evidently not true. This is explained as follows. At order $\tilde{\lambda }^k$, the encountered Ellwood invariants have the structure of OSFT tree-level amplitudes between an on-shell closed string and k on-shell open strings given by the marginal field cj, with $\tilde{\lambda }$ playing the role of the open string coupling constant. These amplitudes are naively affected by infrared divergences due to the collisions of the marginal fields at zero momentum, which correspond to the propagation of the zero momentum tachyon. The propagator $\frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}$ gives a uniquely defined prescription to renormalize these singularities; see Sect. 3. On the other hand, in BCFT, the same contact-term divergences are renormalized by contour deformation [18], so that the renormalized boundary interaction $e^{-\lambda \oint \mathrm{d}s j(s)}$ acquires a topological nature. This difference in the renormalization procedure of contact-term divergences is the ultimate reason why $\lambda$ and $\tilde{\lambda }$ are different. Had the self-OPE between the currents been regular, we would have found no difference between the two quantities.

We also observe that the growing of the coefficients in (1.7) is in agreement with the findings from other non-perturbative approaches (although in different gauges) such as [19, 20], and it suggests that the power series in ${\tilde{\lambda }}$ may have a finite radius of convergence. It would be desirable to improve our calculation to be able to estimate the growing of the higher order coefficients and the nature of the singularity in the complex ${\tilde{\lambda }}$ space. This would be a complementary (perturbative) way of understanding the reason why (in Siegel gauge) the marginal solution breaks down at a critical value of ${\tilde{\lambda }}$. Indeed, it turns out that our computations in pseudo $\mathcal{B}_0$-gauge can be related to the analogous computations in Siegel gauge, whose direct evaluation is notoriously complicated. Work in this direction is in progress [21].

The paper is organized as follows. In Sect. 2 we review the needed material for constructing the boundary state in BCFT [18] and in OSFT [23]. Then we review the construction of the marginal solution in the pseudo-${\mathcal {B}}_0$ gauge [2], and we explicitly write it down up to the fifth order. Section 3 describes the regularization procedure implemented by the propagator $\frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}$. In Sect. 4 we write down the coefficients of the Ishibashi states in the boundary state in terms of the deformation parameter $\lambda$ using the standard BCFT prescription by Recknagel and Schomerus [18]. Then we compute the same quantities for the OSFT solution in the pseudo-${\mathcal {B}}_0$ gauge. Finally we compare the coefficients of the Ishibashi states in OSFT and BCFT and we obtain the function $\tilde{\lambda }=\tilde{\lambda }(\lambda )$ up to fifth order. An appendix contains useful formulas for the correlators encountered.

2 The boundary state and the marginal solution

Let us consider a deformation of a BCFT by a boundary primary operator j(x) of conformal weight one,

$$\begin{aligned} \delta S_{\text {BCFT}}=\ \lambda \int j(x)\ \mathrm{d}x. \end{aligned}$$

(2.1)

From the OSFT point of view the new theory can be described by a classical solution, a state in the original BCFT

$$\begin{aligned} \Psi _{\tilde{\lambda }} = \tilde{\lambda }\Psi _1 \ +\ O(\tilde{\lambda }^2), \end{aligned}$$

(2.2)

where

$$\begin{aligned} \Psi _1 = cj(0)\ |0\rangle \equiv |cj\rangle . \end{aligned}$$

(2.3)

The leading term in $\tilde{\lambda }$ satisfies the linearized equations of motion

$$\begin{aligned} Q_B\Psi _1 = 0. \end{aligned}$$

(2.4)

If j is exactly marginal, higher orders in $\tilde{\lambda }$ should exist,

$$\begin{aligned} \Psi _{\tilde{\lambda }} = \sum _{k=1}^\infty \tilde{\lambda }^k\ \Psi _k, \end{aligned}$$

(2.5)

and they can be found by solving the recursive equations of motion

$$\begin{aligned} Q_B\Psi _{\ell } = \sum _{k=1}^{\ell -1} \Psi _k \Psi _{\ell -k}, \end{aligned}$$

(2.6)

with the initial condition (2.3).

Notice that while in BCFT the perturbation is unique, the OSFT solution is not unique because it can be changed by gauge transformations. We can get rid of this gauge redundancy by computing observables. In particular the information on the marginal deformation can be effectively cast in the boundary state.

Boundary states in bosonic string theory can be written as a superposition of Ishibashi states $|{\mathcal {V}^m}\rangle \!\rangle$ [24]

$$\begin{aligned} |B\rangle = \sum _m n_m\ |{\mathcal {V}^m}\rangle \!\rangle \ \otimes \ |B_{\text {gh}}\rangle , \end{aligned}$$

(2.7)

where $|B_{\text {gh}}\rangle$ is the universal ghost part. When we deform a given world-sheet theory with an exactly marginal boundary deformation, the boundary state will be deformed to

$$\begin{aligned} \delta S_{\text {BCFT}} = \lambda \int j(x) \ \mathrm{d}x\ \longrightarrow \ |B(\lambda )\rangle , \end{aligned}$$

(2.8)

with

$$\begin{aligned} |B(\lambda )\rangle = \left[ e^{-\lambda \oint \mathrm{d}s\ j(s)} \right] _{\text {R}} |B_0\rangle , \end{aligned}$$

(2.9)

where $\left[ \ldots \right] _{\text {R}}$ means that a regularization is needed (and it will be reviewed later on), and $|B_0\rangle$ is the boundary state of the starting BCFT.

On the other hand, given an OSFT solution $\Psi _{\tilde{\lambda }}$, the boundary state will depend on $\tilde{\lambda }$

$$\begin{aligned} \Psi _{\tilde{\lambda }} \longrightarrow \ |B_{\Psi }(\tilde{\lambda })\rangle . \end{aligned}$$

(2.10)

The two boundary states should be the same by the Ellwood conjecture [25] and this induces a functional relation

$$\begin{aligned} \tilde{\lambda }= \tilde{\lambda }(\lambda ). \end{aligned}$$

(2.11)

To obtain this relation we can compare the coefficients of the Ishibashi states. From (2.7) it follows that⁴

$$\begin{aligned} n_m^{\text {BCFT}}(\lambda )= & {} \left\langle \mathcal {V}_m|B(\lambda )\right\rangle = \langle \mathcal {V}_m| \left[ e^{-\lambda \oint \mathrm{d}s\ j(s)} \right] _{\text {R}} |B_0\rangle \nonumber \\= & {} \left\langle \left[ e^{-\lambda \oint \mathrm{d}s\ j(s)} \right] _{\text {R}} \mathcal {V}_m(0,0)\right\rangle _{\text {Disk}}, \end{aligned}$$

(2.12)

where $\langle \mathcal {V}_m|$ is the BPZ conjugate of the Virasoro primary $|\mathcal{V}^n\rangle$ of the matter so that $\left\langle \mathcal {V}_m|\mathcal{V}^n\right\rangle =\left\langle \mathcal {V}_m|\mathcal{V}^n\right\rangle \!\rangle =\delta _m^n$ where we used the fact that Ishibashi states have the generic form

$$\begin{aligned} |\mathcal{V}^n\rangle \!\rangle =|\mathcal{V}^n\rangle +\text {Virasoro descendants}. \end{aligned}$$

(2.13)

The series expansion of the exponential in (2.12) gives rise to contact divergences and one needs to renormalize them properly. In the next section we will review the standard procedure of [18].

The way to compute the $n_m$ from OSFT was given in [23] by appropriately generalising the Ellwood invariant

$$\begin{aligned} n_m^{\text {SFT}}(\tilde{\lambda })= & {} \left\langle \mathcal {V}_m|B_{\Psi }(\tilde{\lambda })\right\rangle \ \nonumber \\= & {} \ 2 \pi i\ \langle \mathcal {I}| \ V_m^{(0,0)}(i,-i)\ |\Psi _{\tilde{\lambda }}-\Psi _{\text {TV}}\rangle , \end{aligned}$$

(2.14)

where $\Psi _{\text {TV}}$ is a tachyon vacuum solution and $V_m^{(0,0)}(i,-i)$ is a weight (0, 0) bulk field of the form

$$\begin{aligned} V_m^{(0,0)} \equiv c\bar{c}\ V_m^{(1,1)} = c\bar{c}\ {\mathcal {V}}_m^{(h_m,h_m)} \ \otimes \ {\mathcal {V}}_{\text {aux}}^{(1-h_m,1-h_m)}. \end{aligned}$$

(2.15)

As explained in detail in [23], the auxiliary bulk field ${\mathcal {V}}_{\text {aux}}^{(1-h_m,1-h_m)}$ lives in an auxiliary $\hbox { BCFT}_{\text {aux}}$of $c=0$ and has unit one-point function on the disk,

$$\begin{aligned} \left\langle \,{\mathcal {V}}^{(1-h_m,1-h_m)}_{\text {aux}}(0,0)\,\right\rangle _{\text {Disk}} = 1. \end{aligned}$$

(2.16)

In a similar way the open string fields entering in (2.14) are lifted to the extended BCFT

$$\begin{aligned} \hbox {BCFT}_0^{\text {new}} = \hbox {BCFT}_0\ \otimes \ \hbox {BCFT}_{\text {aux}}. \end{aligned}$$

(2.17)

For the solution we will be dealing with this lifting procedure is trivial and amounts to the substitution $L_n\rightarrow L_n+L_n^{(\mathrm aux)}$ in the equations that will follow. For this reason we will not distinguish between normal and lifted string fields in the sequel.

As far as the solution itself-is concerned, we search for it in the convenient pseudo-${\mathcal {B}}_0$ gauge [2], making the following ansatz:

$$\begin{aligned} \Psi _{\tilde{\lambda }} = \sum _{r=1}^\infty (\tilde{\lambda })^r\ U_{r+1}^* U_{r+1} \hat{\Psi }_{r} |0\rangle , \end{aligned}$$

(2.18)

with the gauge condition

$$\begin{aligned} {\mathcal {B}}_0 \hat{\Psi }_r \ |0\rangle = 0, \end{aligned}$$

(2.19)

where ${\mathcal {B}}_0$ is the zero mode of the b ghost in the Sliver frame, obtained from the UHP by the conformal transformation $z=\frac{2}{\pi }\arctan w$

$$\begin{aligned} {\mathcal {B}}_0 = \oint \frac{\mathrm{d}z}{2\pi i}\ z\ b(z), \end{aligned}$$

(2.20)

and the operators $U_r$ are the common exponentials of total Virasoro operators creating the wedge states [27] in the well known way

$$\begin{aligned} |r\rangle = U_r^* U_r |0\rangle = U_r^* |0\rangle = \underbrace{|0\rangle * \ldots * |0\rangle }_{r-1}. \end{aligned}$$

(2.21)

Solving order by order in $\tilde{\lambda }$ (2.6) we find

$O(\lambda ^2)$::

$Q\Psi _2 = -(cj)^2$.

The r.h.s. is explicitly given by

$$\begin{aligned} (cj)^2\equiv & {} cj(0)|0\rangle * cj(0)|0\rangle \ \nonumber \\= & {} U_3^* U_3\ cj\left( \tfrac{1}{2} \right) cj\left( -\tfrac{1}{2} \right) |0\rangle , \end{aligned}$$

(2.22)

where the cj insertions are written in the Sliver frame. The solution is therefore

$$\begin{aligned} \Psi _2= & {} U_3^* U_3\ \hat{\Psi }_2 |0\rangle \nonumber \\= & {} -\ U_3^* U_3\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ cj\left( \tfrac{1}{2} \right) cj\left( -\tfrac{1}{2} \right) |0\rangle , \end{aligned}$$

(2.23)

where ${\mathcal {L}}_0$ is the zero mode of the energy-momentum tensor in the Sliver frame,

$$\begin{aligned} {\mathcal {L}}_0 = \oint \frac{\mathrm{d}z}{2\pi i}\ z\ T(z). \end{aligned}$$

(2.24)

Note that inverting $Q_B$ using ${\mathcal {B}}_0/{\mathcal {L}}_0$ is only meaningful if the OPE of cj with itself-does not produce weight zero terms, otherwise we would find a vanishing eigenvalue of ${\mathcal {L}}_0$. As is well known this is the first nontrivial condition for j to generate an exactly marginal deformation.

$O(\lambda ^3)$::

$Q_B\Psi _3 + \left[ cj,\Psi _2\right] =0$.

At the third order the solution $\Psi _3$ is written in terms of $\Psi _2$. We write the state $\left[ cj,\Psi _2\right]$ as

$$\begin{aligned} \left[ cj,\Psi _2\right]= & {} \Big [ U_2^* U_2\ cj(0)|0\rangle ,\ U_3^* U_3\ \hat{\Psi }_2|0\rangle \Big ] \nonumber \\= & {} U_4^* U_4\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} |0\rangle , \end{aligned}$$

(2.25)

where in the second step we explicitly write the width of the wedge states using the $U_r$ operators. The inside insertions have to be placed according to (A.2): to lighten a bit the notation we have defined the graded commutator-like symbol $[\![\ldots ]\!]$ as

$$\begin{aligned}&\Big [\!\Big [ \psi (x),\phi (y)\Big ]\!\Big ]_{\text {(r,s)}} \nonumber \\&\quad \equiv \psi \left( x+\tfrac{s-1}{2} \right) \phi \left( y-\tfrac{r-1}{2} \right) \nonumber \\&\qquad -\ (-1)^{|\psi ||\phi |}\ \phi \left( y+\tfrac{r-1}{2} \right) \psi \left( x-\tfrac{s-1}{2} \right) ,\nonumber \\ \end{aligned}$$

(2.26)

coming from

$$\begin{aligned}&\left[ U_r^* U_r\ \psi (x)|0\rangle ,\ U_s^* U_s\ \phi (y)|0\rangle \right] \nonumber \\&\quad =\ U_{r+s-1}^* U_{r+s-1}\ \Big [\!\Big [ \psi (x),\phi (y)\Big ]\!\Big ]_{\text {(r,s)}}\ |0\rangle . \end{aligned}$$

(2.27)

Finally we can write the solution at the third order as

$$\begin{aligned} \Psi _3= & {} U_4^* U_4\ \hat{\Psi }_3 |0\rangle \ \nonumber \\= & {} - U_4^* U_4\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} |0\rangle . \end{aligned}$$

(2.28)

$O(\lambda ^4)$::

$Q_B\Psi _4 + \left[ cj,\Psi _3\right] + \Psi _2^2 =0$.

Again,

$$\begin{aligned} \Psi _4= & {} U_5^* U_5\ \hat{\Psi }_4 |0\rangle \ = -\ U_5^* U_5\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\nonumber \\&\times \;\left( \frac{1}{2}\ \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}}\ + \Big [\!\Big [ cj,\hat{\Psi }_3\Big ]\!\Big ]_{\text {(2,4)}} \right) |0\rangle .\nonumber \\ \end{aligned}$$

(2.29)

$O(\lambda ^5)$::

$Q_B\Psi _5 + \left[ cj,\Psi _4\right] + \left[ \Psi _2,\Psi _3\right] =0$.

We find the fifth order as

$$\begin{aligned} \Psi _5= & {} U_6^* U_6\ \hat{\Psi }_5 |0\rangle \ = - U_6^* U_6\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \nonumber \\&\times \; \left( \Big [\!\Big [ cj,\hat{\Psi }_4\Big ]\!\Big ]_{\text {(2,5)}} + \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_3\Big ]\!\Big ]_{\text {(3,4)}}\right) |0\rangle .\nonumber \\ \end{aligned}$$

(2.30)

This procedure can be continued to higher order.⁵ Although higher orders can easily be written down, their Ellwood invariants become more and more complicated because they involve a large number of multiple integrals which by themselves need to be properly renormalized, as we will see in the next section.

3 Contact-term divergences and the propagator

The computation of the Ellwood invariants for the solution we have just presented involves in general contact divergences due to the definition of the propagator ${\mathcal {B}}_0/{\mathcal {L}}_0$. As usual, we start by defining the inverse of ${\mathcal {L}}_0$ via the Schwinger representation

$$\begin{aligned} \frac{1}{{\mathcal {L}}}_0 = \int _0^1 \frac{\mathrm{d}s}{s}\ \ s^{{\mathcal {L}}_0}, \end{aligned}$$

(3.1)

which is well defined for eigenvalues of ${\mathcal {L}}_0$ with a strictly positive real part. The operator $s^{{\mathcal {L}}_0}$ is the generator of dilatations $z\mapsto s z$ in the Sliver frame. Its action on a primary field with conformal weight h in the Sliver frame is

$$\begin{aligned} s^{{\mathcal {L}}_0} \phi (z)= s^h\ \phi \left( s z \right) . \end{aligned}$$

(3.2)

The integral representation (3.1) is only valid for fields with a positive scaling dimension $h>0$; if we apply the above integral representation to a state $|\varphi _{-|h|}\rangle$ with negative weight $-|h|$,

$$\begin{aligned} \frac{1}{{\mathcal {L}}_0}\ |\varphi _{-|h|}\rangle= & {} \int _0^1 \frac{\mathrm{d}s}{s}\ s^{{\mathcal {L}}_0}\ |\varphi _{-|h|}\rangle \ \\= & {} |\varphi _{-|h|}\rangle \int _0^1 \frac{\mathrm{d}s}{s^{|h|+1}},\qquad \text {(incorrect)}, \end{aligned}$$

we find a divergence, as s approaches zero. But this just reflects that the integral representation (3.1) has been used outside its domain of validity. This can easily be remedied in the following way:

$$\begin{aligned} \frac{1}{{\mathcal {L}}_0} = \left. \frac{1}{{\mathcal {L}}_0+\epsilon } \ \right| _{\epsilon = 0} =\ \left. \int _0^1 \frac{\mathrm{d}s}{s^{1-\epsilon }}\ s^{{\mathcal {L}}_0} \ \right| _{\epsilon = 0}. \end{aligned}$$

(3.3)

This prescription amounts to computing the Schwinger integral in its region of convergence by assuming $\mathrm{Re}(\epsilon )> |h|$, and then analytically continuing to $\epsilon =0$ ⁶

$$\begin{aligned} \frac{1}{{\mathcal {L}}_0}\ |\varphi _{-|h|}\rangle= & {} \left. \int _0^1 \frac{\mathrm{d}s}{s^{1-\epsilon }}\ s^{{\mathcal {L}}_0}\ |\varphi _{-|h|}\rangle \right| _{\epsilon = 0} \nonumber \\= & {} \left. |\varphi _{-|h|}\rangle \int _0^1 \mathrm{d}s\ s^{\epsilon - |h| - 1}\ \right| _{\epsilon = 0} \nonumber \\= & {} \left[ |\varphi _{-|h|}\rangle \left. \frac{s^{\epsilon - |h|}}{\epsilon - |h|} \right| _{0}^{1}\ \right] _{\epsilon = 0} \nonumber \\= & {} \left. \frac{1}{\epsilon -|h|}\ |\varphi _{-|h|}\rangle \right| _{\epsilon =0}=-\frac{1}{|h|}\ |\varphi _{-|h|}\rangle . \end{aligned}$$

(3.4)

This analytic continuation allows one to define ${\mathcal {L}}_0^{-1}$ on every state we encounter during our computations except on weight zero states which remain as an obstruction, as it should be.⁷ Pragmatically, this procedure is equivalent to adding and removing the tachyon contribution from the OPE, for example,

$$\begin{aligned} \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [ cj(x)\ cj(-x)\Big ]\rightarrow & {} \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [ cj(x)\ cj(-x)\ +\ \frac{1}{2x}\ c\partial c (0)\Big ]\ \nonumber \\&- \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \ \frac{1}{2x}\ c\partial c (0) , \end{aligned}$$

(3.6)

and to defining $1/{\mathcal {L}}_0$ on the tachyon as

$$\begin{aligned} \frac{1}{{\mathcal {L}}_0}\ |c \partial c \rangle = -\ |c\partial c\rangle . \end{aligned}$$

(3.7)

4 Comparing $\lambda$ and $\tilde{\lambda }$

In this section we perturbatively compute the coefficients of the series expansion of $\tilde{\lambda }= \tilde{\lambda }(\lambda )$

$$\begin{aligned} {\tilde{\lambda }} = \tilde{\lambda }(\lambda ) = \sum _{k=0}^\infty b_k\ \lambda ^k, \end{aligned}$$

(4.1)

up to fifth order. On general grounds we expect that $b_0=0$ and $b_1=1$, and this will be verified in the next subsections. The $b_k$s are computed by equating the coefficients of the Ishibashi states in the boundary state in BCFT and OSFT [22, 23],

$$\begin{aligned} n_m^{\text {BCFT}}(\lambda ) = n_m^{\text {SFT}}(\tilde{\lambda }). \end{aligned}$$

(4.2)

In both cases one can expand the above coefficients in power series of the corresponding deformation parameter

$$\begin{aligned} n_m^{\text {BCFT}}(\lambda )\equiv & {} \sum _{k=0}^\infty B_{k,m}^{\text {BCFT}} \lambda ^k \ ,\quad n_m^{\text {SFT}}(\tilde{\lambda }) ,\nonumber \\\equiv & {} \sum _{k=0}^\infty B_{k,m}^{\text {SFT}} {\tilde{\lambda }}^k. \end{aligned}$$

(4.3)

The $B_{k,m}^{\text {BCFT}}$ coefficients can be found expanding the exponential in (2.12),

$$\begin{aligned} B_{k,m}^{\text {BCFT}}\equiv & {} \frac{(-1)^k}{k!}\ 2^{2h_m} \int _{-\infty }^\infty \mathrm{d}s_1 \dots \nonumber \\&\int _{-\infty }^\infty \mathrm{d}s_k\ \left\langle \,\mathcal {V}_m(i,-i)\ j(s_1) \dots j(s_k)\,\right\rangle _{\text {UHP}}, \end{aligned}$$

(4.4)

where the conformal factor $2^{2h_m}$ comes from the transformation of ${\mathcal {V}}_m$ under the map from the disk to the UHP.

These integrals need a renormalization, discussed by Recknagel and Schomerus [18]: thanks to the self-locality property of the current j, one can modify the path of each integral to be parallel to the real axis but with a positive imaginary part $\epsilon$, with $0< \epsilon<\!< 1$,

$$\begin{aligned} \int _{-\infty }^{\infty }\ \mathrm{d}x_{k-1} \ \longrightarrow \ \int _{-\infty +ik\epsilon }^{\infty +ik\epsilon }\ \mathrm{d}x_{k-1}. \end{aligned}$$

(4.5)

In such a way all the contact divergences between the currents are avoided and only the contraction of the currents with the closed string will give contribution. Thanks to this renormalization the loop operator $\left[ e^{-\lambda \oint j(s)\mathrm{d}s}\right] _{\text {R}}$ becomes a topological defect.

For the sake of simplicity we consider an exactly marginal deformation produced by the operator

$$\begin{aligned} j(z) = i \sqrt{2}\ \partial X(z), \end{aligned}$$

(4.6)

on an initial Neumann boundary condition of a free boson compactified at the self-dual radius $R=1$($\alpha '=1$).⁸ This deformation switches on a Wilson line in the compactified direction, which can be detected by a closed string vertex operator carrying winding charge,

$$\begin{aligned} \mathcal {V}_m(z,\bar{z}) = e^{i m {\tilde{X}}(z,\bar{z})}, \end{aligned}$$

(4.7)

where m is the winding number (which specifies the closed string index) and ${\tilde{X}}(z,\bar{z})=X(z) - \bar{X}(\bar{z})$ the T-dual field of $X(z,\bar{z})$.⁹ This closed string state has conformal weight $(\frac{m^2}{4},\frac{m^2}{4})$.

Performing the renormalized integral (4.4), one obtain with this choice of the current and closed string state (see Appendix 1 for conventions and basic correlators),

$$\begin{aligned} B_{k,m}^{\text {BCFT}} = \frac{\left( -\ \mathrm{im}\sqrt{2}\ \pi \right) ^k}{k!}, \end{aligned}$$

(4.8)

which can easily be resummed to

$$\begin{aligned} n_m^{\text {BCFT}}(\lambda ) = e^{-\mathrm{im}\sqrt{2}\ \pi \lambda }. \end{aligned}$$

(4.9)

In the OSFT framework, the analytic computation of coefficients of the Ishibashi state involves the Ellwood invariants and we compute them order by order in $\tilde{\lambda }$, starting from (2.14), which gives

$$\begin{aligned} B_{0,m}^{\text {SFT}}= & {} -\ 2 \pi i\ \langle \mathcal {I}| \ V_m^{(0,0)}(i,-i)\ |\Psi _{\text {TV}}\rangle , \nonumber \\ B_{1,m}^{\text {SFT}}= & {} 2 \pi i\ \langle \mathcal {I}| \ V_m^{(0,0)}(i,-i)\ |cj\rangle , \nonumber \\ B_{k,m}^{\text {SFT}}= & {} \ 2 \pi i\ \langle \mathcal {I}| \ V_m^{(0,0)}(i,-i)\ |\Psi _{k}\rangle \,\qquad k\ge 2. \end{aligned}$$

(4.10)

The relation between $\lambda$ and $\tilde{\lambda }$ must be universal, in the sense that it cannot depend on the particular choice of the closed string. In our specific computation we will see that this is the case by verifying that the relation is independent of the winding charge m.

Now rewriting (4.2) using (4.3) and (4.1) gives explicitly

$O(\lambda ^0)$ :

$$\begin{aligned} b_0 \ \propto \ B_{0,m}^{\text {SFT}} -B_{0,m}^{\text {BCFT}} \ (=\ 0), \end{aligned}$$

(4.11)

$O(\lambda ^1)$ :

$$\begin{aligned} b_1 = \dfrac{B_{1,m}^{\text {BCFT}}}{B_{1,m}^{\text {SFT}}}, \end{aligned}$$

(4.12)

$O(\lambda ^2)$ :

$$\begin{aligned} b_2 = \dfrac{ \left( B_{1,m}^{\text {SFT}} \right) ^2 B_{2,m}^{\text {BCFT}}\ -\ \left( B_{1,m}^{\text {BCFT}} \right) ^2 B_{2,m}^{\text {SFT}}}{\left( B_{1,m}^{\text {SFT}} \right) ^3}, \end{aligned}$$

(4.13)

$O(\lambda ^3)$ :

$$\begin{aligned} b_3= & {} \dfrac{ \left( B_{1,m}^{\text {SFT}} \right) ^3 B_{3,m}^{\text {BCFT}}\ -\ \left( B_{1,m}^{\text {BCFT}} \right) ^3 B_{3,m}^{\text {SFT}}}{ \left( B_{1,m}^{\text {SFT}} \right) ^4} \nonumber \\&+\; 2\ \dfrac{ B_{1,m}^{\text {BCFT}} B_{2,m}^{\text {SFT}} }{ \left( B_{1,m}^{\text {SFT}} \right) ^4}\ \left( \left( B_{1,m}^{\text {BCFT}} \right) ^2 B_{2,m}^{\text {BCFT}}\ \right. \nonumber \\&\left. -\; \left( B_{1,m}^{\text {BCFT}} \right) ^2 B_{2,m}^{\text {SFT}} \right) , \end{aligned}$$

(4.14)

$O(\lambda ^4)$ :

$$\begin{aligned} b_4= & {} \frac{\left( B_{1,m}^{\text {BCFT}} \right) ^4 \left( 5 B_{2,m}^{\text {SFT}} B_{3,m}^{\text {SFT}} -\left( B_{1,m}^{\text {SFT}} \right) B_{1,m}^{\text {SFT}} \right) }{\left( B_{1,m}^{\text {SFT}} \right) ^6} \nonumber \\&-\;3\frac{\left( B_{1,m}^{\text {BCFT}} \right) ^2 B_{1,m}^{\text {BCFT}} B_{3,m}^{\text {SFT}} }{ \left( B_{1,m}^{\text {SFT}} \right) ^4 } \nonumber \\&+\;\frac{ 6 \left( B_{1,m}^{\text {BCFT}} \right) ^2 B_{2,m}^{\text {BCFT}} \left( B_{2,m}^{\text {SFT}} \right) ^2}{\left( B_{1,m}^{\text {SFT}} \right) ^5} \nonumber \\&+\;\frac{ \left( B_{1,m}^{\text {SFT}} \right) ^6 B_{4,m}^{\text {BCFT}} -5 \left( B_{1,m}^{\text {BCFT}} \right) ^4 \left( B_{2,m}^{\text {SFT}} \right) ^3 }{\left( B_{1,m}^{\text {SFT}} \right) ^7} \nonumber \\&-\;\frac{ B_{2,m}^{\text {SFT}} \left( 2 B_{1,m}^{\text {BCFT}} B_{3,m}^{\text {BCFT}}+\left( B_{2,m}^{\text {BCFT}} \right) ^2\right) }{\left( B_{1,m}^{\text {SFT}} \right) ^3}, \end{aligned}$$

(4.15)

$O(\lambda ^5)$ :

$$\begin{aligned}&b_5 = \frac{14 \left( B_{1,m}^{\text {BCFT}} \right) ^5 \left( B_{2,m}^{\text {SFT}} \right) ^4}{\left( B_{1,m}^{\text {SFT}} \right) ^9} -\;\frac{21 \left( \left( B_{1,m}^{\text {BCFT}} \right) ^5 \left( B_{2,m}^{\text {SFT}} \right) ^2 B_{3,m}^{\text {SFT}}\right) }{\left( B_{1,m}^{\text {SFT}} \right) ^8} \nonumber \\&+\;\frac{6 \left( B_{1,m}^{\text {BCFT}} \right) ^5 B_{2,m}^{\text {SFT}} B_{4,m}^{\text {SFT}} + 3 \left( B_{1,m}^{\text {BCFT}} \right) ^5 \left( B_{3,m}^{\text {SFT}} \right) ^2 - 20 \left( B_{1,m}^{\text {BCFT}} \right) ^3 B_{2,m}^{\text {BCFT}} \left( B_{2,m}^{\text {SFT}} \right) ^3}{\left( B_{1,m}^{\text {SFT}} \right) ^7} +\;\frac{20 \left( B_{1,m}^{\text {BCFT}} \right) ^3 B_{2,m}^{\text {BCFT}} B_{2,m}^{\text {SFT}} B_{3,m}^{\text {SFT}}-\left( B_{1,m}^{\text {BCFT}} \right) ^5 B_{5,m}^{\text {SFT}}}{\left( B_{1,m}^{\text {SFT}} \right) ^6} \nonumber \\&+\;\frac{-4 \left( B_{1,m}^{\text {BCFT}} \right) ^3 B_{2,m}^{\text {BCFT}} B_{4,m}^{\text {SFT}}+6 \left( B_{1,m}^{\text {BCFT}} \right) ^2 \left( B_{2,m}^{\text {SFT}} \right) ^2 B_{3,m}^{\text {BCFT}}+6 B_{1,m}^{\text {BCFT}} \left( B_{2,m}^{\text {SFT}} \right) ^2 \left( B_{2,m}^{\text {SFT}} \right) ^2}{\left( B_{1,m}^{\text {SFT}} \right) ^5} \nonumber \\&+\; \frac{-3 \left( B_{1,m}^{\text {BCFT}} \right) ^2 B_{3,m}^{\text {BCFT}} B_{3,m}^{\text {SFT}}-3 B_{1,m}^{\text {BCFT}} \left( B_{2,m}^{\text {SFT}} \right) ^2 B_{3,m}^{\text {SFT}}}{\left( B_{1,m}^{\text {SFT}} \right) ^4} +\; \frac{-2 B_{1,m}^{\text {BCFT}} B_{2,m}^{\text {SFT}} B_{4,m}^{\text {BCFT}}-2 B_{2,m}^{\text {BCFT}} B_{2,m}^{\text {SFT}} B_{3,m}^{\text {BCFT}}}{\left( B_{1,m}^{\text {SFT}} \right) ^3}+\frac{B_5^{\text {BCFT}}}{B_{1,m}^{\text {SFT}}}. \end{aligned}$$

(4.16)

4.1 Zeroth order

As a starting consistency check, the zeroth order of the expansion of the coefficients of the Ishibashi states in OSFT is

$$\begin{aligned} B_{0,m}^{\text {SFT}} = -\ 2\pi i\ \langle \mathcal {I}| V_m^{(0,0)}(i,-i) | \Psi _{\text {TV}}\rangle , \end{aligned}$$

(4.17)

where, as explained in [23], the tachyon vacuum contribution can be replaced with $\Psi _{\text {TV}} \rightarrow \frac{2}{\pi } c(0) |0\rangle$. The amplitude then becomes¹⁰

$$\begin{aligned} B_{0,m}^{\text {SFT}}= & {} - 4 i\ \langle \mathcal {I}| V_m(i,-i) c(0) |0\rangle \nonumber \\= & {} -\ 4 i\ \left\langle \, V_m(i,-i)\ f_\mathcal {I} \circ c(0) \,\right\rangle _{\text {UHP}}, \end{aligned}$$

(4.18)

where we used the conformal map defining the identity string field $f_\mathcal {I} (z) = \frac{2z}{1-z^2}$. Then

$$\begin{aligned} B_{0,m}^{\text {SFT}}= & {} - 2 i\ \left\langle \,c(i)c(-i)c(0) \,\right\rangle _{\text {UHP}} \left\langle \,\mathcal {V}_m(i,-i)\,\right\rangle _{\text {UHP}} \nonumber \\&\times \;\left\langle \, \mathcal {V}_{\text {aux}}^{(1-h_m,1-h_m)}(i,-i)\,\right\rangle _{\text {UHP}} \nonumber \\= & {} - 2 i\ (2i)\ 2^{-2h_m} \ 2^{2(h_m-1)} = 1. \end{aligned}$$

(4.19)

Consistently we find

$$\begin{aligned} B_{0,m}^{\text {SFT}} = B_{0,m}^{\text {BCFT}} =1, \end{aligned}$$

(4.20)

which confirms that

$$\begin{aligned} b_0 = 0. \end{aligned}$$

(4.21)

4.2 First order

As an extra starting check, let us look at the first order, where we have to compute

$$\begin{aligned} B_{1,m}^{\text {SFT}}= & {} 2\pi i\ \langle \mathcal {I}| V_m(i,-i) |cj\rangle \nonumber \\= & {} 2\pi i\ \left\langle \, V_m(i\infty ,-i\infty )\ cj(0) \,\right\rangle _{C_1}, \end{aligned}$$

(4.22)

where in the last step we wrote the correlator on a cylinder of width one $C_1$, without any conformal factor because the conformal weight of all the insertions is zero.

Acting with the map

$$\begin{aligned} z \rightarrow \ e^{2\pi i z}, \end{aligned}$$

(4.23)

this two-point function on the cylinder becomes the two-point function on the disk D,

$$\begin{aligned} B_{1,m}^{\text {SFT}}= & {} 2\pi i\ \left\langle \, V_m(0,0)\ cj(1) \,\right\rangle _{D} \nonumber \\= & {} 2\pi i\ \left\langle \,c \bar{c}(0)c(1)\,\right\rangle _{D} \left\langle \, \mathcal {V}_m(0,0)\ j(1) \,\right\rangle _{D} \nonumber \\&\times \;\left\langle \,\mathcal {V}_{\text {aux}}^{(1-h_m,1-h_m)}(0,0)\,\right\rangle _{D} \nonumber \\= & {} 2\pi i\ \left\langle \, \mathcal {V}_m(0,0)\ j(1) \,\right\rangle _{D} = -\ i \pi m \sqrt{2}, \end{aligned}$$

(4.24)

which equals the amplitude computed from BCFT (4.8),

$$\begin{aligned} B_{1,m}^{\text {BCFT}} = B_{1,m}^{\text {SFT}}, \end{aligned}$$

(4.25)

and so the corresponding coefficient in the $\tilde{\lambda }/\lambda$ relation is

$$\begin{aligned} b_1 = 1. \end{aligned}$$

(4.26)

4.3 Second order

At second order the Ellwood invariant contains one ${\mathcal {B}}_0/{\mathcal {L}}_0$ propagator inside $\Psi _2$,

$$\begin{aligned}&B_{2,m}^{\text {SFT}} \nonumber \\&\quad =\ 2\pi i\ \langle \mathcal {I}| V_m(i,-i) |\Psi _2\rangle \nonumber \\&\quad =\ 2\pi i\ \left\langle \,V_m(i\infty ,-i\infty )\ \hat{\Psi }_2\,\right\rangle _{C_2} \nonumber \\&\quad =\ - \ 2\pi i \left\langle \,V_m(i\infty ,-i\infty )\ \left( \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ cj\left( \tfrac{1}{2} \right) \ cj\left( -\tfrac{1}{2} \right) \right) \,\right\rangle _{C_2}.\nonumber \\ \end{aligned}$$

(4.27)

This amplitude is depicted in Fig. 1.

The action of the propagator on the double insertion of cj follows the regularization (3.3), so this state can be written as

$$\begin{aligned} \Psi _2= & {} - U_3^* U_3\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}cj\left( \tfrac{1}{2} \right) cj\left( -\tfrac{1}{2} \right) \ |0\rangle \nonumber \\= & {} - U_3^* U_3 \int _0^1 \frac{\mathrm{d}s}{s^{1-\epsilon }}\ {\mathcal {B}}_0\ s^{{\mathcal {L}}_0}\ cj\left( \tfrac{1}{2} \right) cj\left( -\tfrac{1}{2} \right) \ |0\rangle \ \Bigg |_{\epsilon =0} \nonumber \\= & {} - U_3^* U_3 \int _0^1 \frac{\mathrm{d}s}{s^{1-\epsilon }}\ {\mathcal {B}}_0\ cj\left( \tfrac{s}{2} \right) cj\left( -\tfrac{s}{2} \right) \ |0\rangle \ \Bigg |_{\epsilon =0}, \end{aligned}$$

(4.28)

The ${\mathcal {B}}_0$ ghost acts on c(w) as the contour integral

$$\begin{aligned}{}[{\mathcal {B}}_0 ,c(w)] = \oint _{w} \frac{d z}{2\pi i}\ z\ b(z) c(w)=w, \end{aligned}$$

(4.29)

so that we have

$$\begin{aligned} \Big [ {\mathcal {B}}_0 , c\left( \tfrac{s}{2} \right) c\left( -\tfrac{s}{2} \right) \Big ] = \frac{s}{2} \ \Big [ c\left( -\tfrac{s}{2} \right) \ +\ c\left( \tfrac{s}{2} \right) \Big ]. \end{aligned}$$

(4.30)

Therefore (renaming $s/2\rightarrow s$) $\Psi _2$ simplifies to

$$\begin{aligned} \Psi _2= & {} -\ U_3^* U_3 \int _0^{\tfrac{1}{2}} \mathrm{d}s\ s^\epsilon \nonumber \\&\times \;\Big [ j\left( s \right) cj\left( -s \right) + cj\left( s \right) j\left( -s \right) \Big ]|0\rangle \Bigg |_{\epsilon =0}, \end{aligned}$$

(4.31)

and the amplitude becomes

$$\begin{aligned} B_{2,m}^{\text {SFT}} = -\ 4\pi i\ \int _0^{\frac{1}{2}} \mathrm{d}s\ s^\epsilon \ \left\langle \, V_m(i\infty )\ cj(s)\ j(-s) \,\right\rangle _{C_2} \ \ \Bigg |_{\epsilon =0},\nonumber \\ \end{aligned}$$

(4.32)

where we have used the obvious rotational invariance of the bc CFT on the cylinder.

Fig. 1

Diagram related to $\left\langle \,V_m\hat{\Psi }_2 \,\right\rangle$. The s variable is the Schwinger parameter (taking value in the interval [0, 1]) of the propagator and $\epsilon$ is the corresponding regulator

Using the Wick theorem (which is reviewed in Appendix A) and in particular (A.17) we obtain

$$\begin{aligned}&B_{2,m}^{\text {SFT}} = -\ 4\pi i\ \int _0^{\frac{1}{2}} \mathrm{d}s\ s^\epsilon \nonumber \\&\quad \times \;\left\langle \,c(i\infty )c(-i\infty )c(s)\ \mathcal {V}_m(i\infty )\ \mathcal {V}_{\text {aux}}^{(1-h_m,1-h_m)}(i\infty )\,\right\rangle _{C_2} \nonumber \\&\quad \times \; \Bigg \{ \left\langle \,j(x_2)j(x_2)\,\right\rangle _{C_2}-m^2\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle _{C_2} \Big )^2 \Bigg \},\nonumber \\ \end{aligned}$$

(4.33)

which using (A.13), (A.15) and (A.18) gives

$$\begin{aligned} B_{2,m}^{\text {SFT}} = 4\ \int _0^{\frac{1}{2}} \mathrm{d}s\ \Bigg \{ s^\epsilon \ \frac{\pi ^2}{8}\ \csc ^2 (\pi s) -\ m^2\ \frac{\pi ^2}{2} \Bigg \}, \end{aligned}$$

(4.34)

where the $\epsilon$ prescription acts only on the first term because it is the only one which is divergent. This gives

$$\begin{aligned} \int _0^{\frac{1}{2}} \mathrm{d}s\ s^\epsilon \ \csc ^2 (\pi s)\ \Bigg |_{\epsilon =0}= & {} \int _0^{\frac{1}{2}} \mathrm{d}s\ \left( \csc ^2 (\pi s)\ -\ \frac{1}{\pi ^2 s^2} \right) \nonumber \\&+\;\int _0^{\frac{1}{2}} \mathrm{d}s\ s^\epsilon \ \frac{1}{\pi ^2 s^2} \ \Bigg |_{\epsilon =0} \nonumber \\= & {} \frac{2}{\pi ^2} \ +\ \frac{2^{1-\epsilon }}{\pi ^2 (\epsilon - 1)} \ \Bigg |_{\epsilon =0} \nonumber \\= & {} 0; \end{aligned}$$

(4.35)

here we have used our analytic continuation which, as explained in Sect. 3, amounts to computing the integral in the region of the $\epsilon$-complex plane where it converges ($\text {Re}\,\epsilon > 1$) and then analytically continue to $\epsilon \rightarrow 0$. In doing this we have also took the freedom of ignoring convergent terms proportional to $\epsilon$ since we are only interested in the $\epsilon \rightarrow 0$ limit.

Computing also the other convergent integral, we obtain again a perfect match with the BCFT results,

$$\begin{aligned} B_{2,m}^{\text {SFT}} = -\ m^2\ \pi ^2 = B_{2,m}^{\text {BCFT}}, \end{aligned}$$

(4.36)

which leads to

$$\begin{aligned} b_2 = 0. \end{aligned}$$

(4.37)

4.4 Third order

At this level the amplitude we have to compute is

$$\begin{aligned} B_{3,m}^{\text {SFT}}= & {} 2\pi i\ \langle \mathcal {I}| V_m(i,-i) |\Psi _3\rangle \nonumber \\= & {} \ 2\pi i\ \left\langle \,V_m(i\infty ,-i\infty ) \hat{\Psi }_3 \,\right\rangle _{C_3}, \end{aligned}$$

(4.38)

where $\Psi _3$ is defined in (2.28). This amplitude is depicted in Fig. 2.

Fig. 2

Diagram related to $\left\langle \,V_m\hat{\Psi }_3 \,\right\rangle$. The first leg of cj from the left is to be understood as a commutator. The t and y variables are the Schwinger parameters (taking value in the interval [0, 1]) and $\epsilon _{1,2}$ the corresponding regulators

Explicitly we find

$$\begin{aligned} \Psi _3= & {} U_4^* U_4\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\left[ cj(0)|0\rangle , \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ cj\left( \tfrac{1}{2} \right) cj\left( -\tfrac{1}{2} \right) |0\rangle \right] \nonumber \\= & {} U_4^* U_4\ \int _0^1 \mathrm{d}y\ y^{\epsilon _2+1}\int _0^{1} \frac{ \mathrm{d}t}{2}\ t^{\epsilon _1} \nonumber \\&\times \; \Bigg \{ cj(y)\ j\left( y\left( \tfrac{t}{2}-\tfrac{1}{2} \right) \right) \ j\left( y\left( -\tfrac{t}{2}-\tfrac{1}{2} \right) \right) \nonumber \\&+\; j\left( y\left( \tfrac{t}{2}+\tfrac{1}{2} \right) \right) \ j\left( y\left( -\tfrac{t}{2}+\tfrac{1}{2} \right) \right) \ cj(-y) \nonumber \\&+\; j(y)\ \Big [ j\left( y\left( \tfrac{t}{2}-\tfrac{1}{2} \right) \right) \ cj\left( y\left( -\tfrac{t}{2}-\tfrac{1}{2} \right) \right) \nonumber \\&+\; cj\left( y\left( \tfrac{t}{2}-\tfrac{1}{2} \right) \right) \ j\left( y\left( -\tfrac{t}{2}-\tfrac{1}{2} \right) \right) \ \Big ] \nonumber \\&+\; \Big [ j\left( y\left( \tfrac{t}{2}+\tfrac{1}{2} \right) \right) \ cj\left( y\left( -\tfrac{t}{2}+\tfrac{1}{2} \right) \right) \nonumber \\&+\; cj\left( y\left( \tfrac{t}{2}+\tfrac{1}{2} \right) \right) \ j\left( y\left( -\tfrac{t}{2}+\tfrac{1}{2} \right) \right) \Big ] \nonumber \\&\times \;j(-y) \Bigg \} |0\rangle \Bigg |_{\epsilon _1=0} \ \Bigg |_{\epsilon _2=0}, \end{aligned}$$

(4.39)

where $\epsilon _1$ is the regulator for the most internal propagator (the one inside the lower order contribution $\hat{\Psi }_2$ (2.23)) and $\epsilon _2$ is the regulator for the external propagator. From the perturbative construction of the solution, it is clear that $\epsilon _1$ should be analytically continued to zero before

$\epsilon _2$. Using the symmetries of the correlator in the matter and ghost sector and renaming $t/2\rightarrow t$, the whole Ellwood invariant reduces to

$$\begin{aligned} B_{3,m}^{\text {SFT}}= & {} 12\pi i\ \int _0^1 \mathrm{d}y\ y^{\epsilon _2+1} \nonumber \\&\int _0^{\frac{1}{2}} \mathrm{d}t\ t^{\epsilon _1}\left\langle \,V_m(i\infty )\ cj\left( \tfrac{3}{2}y \right) \ j\left( yt \right) \ j\left( -yt \right) \,\right\rangle _{C_3}.\nonumber \\ \end{aligned}$$

(4.40)

It is useful to change variable with $x=\frac{3}{2}y$ and $s=ty=\frac{2}{3}xt$ so as to rewrite the integral as

$$\begin{aligned} B_{3,m}^{\text {SFT}}= & {} 8\pi i\ \int _0^{\frac{3}{2}} \mathrm{d}x\, x^{\epsilon _2-\epsilon _1} \nonumber \\&\times \;\int _0^{\frac{x}{3}} \mathrm{d}s\ s^{\epsilon _1}\ \left\langle \,V_m(i\infty )\ cj(x)\ j(s)\ j(-s) \,\right\rangle _{C_3}.\nonumber \\ \end{aligned}$$

(4.41)

Now we apply the Wick theorem (see (A.17) of Appendix A),

$$\begin{aligned} \begin{aligned} B_{3,m}^{\text {SFT}}&= 8\pi i\ \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle _{C_3} \int _0^{\frac{3}{2}} \mathrm{d}x\,x^{\epsilon _2-\epsilon _1} \int _0^{\frac{x}{3}} \mathrm{d}s\ s^{\epsilon _1}\ \\&\quad \times \; \Bigg \{\mathrm{im}\ \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle _{C_3} \Bigg [ \left\langle \,j(s)j(-s)\,\right\rangle _{C_3} \\&\quad +\; \left\langle \,j(x)j(s)\,\right\rangle _{C_3}\ +\ \left\langle \,j(x)j(-s)\,\right\rangle _{C_3} \Bigg ] \\&\quad -\mathrm{im}^3\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle _{C_3} \Big )^3 \Bigg \} \ , \end{aligned} \end{aligned}$$

(4.42)

and using the correlators (A.15) and (A.18), we end up with the following integral:

$$\begin{aligned} \begin{aligned}&B_{3,m}^{\text {SFT}} \\&\quad = -\ 12 \int _0^{\frac{3}{2}} \mathrm{d}x \ x^{\epsilon _2-\epsilon _1}\int _0^{\frac{x}{3}} \mathrm{d}s\ s^{\epsilon _1} \\&\qquad \times \; \Bigg \{\mathrm{im}\ \frac{\pi \sqrt{2}}{3}\ \Bigg [ \left\langle \,j(s)j(-s)\,\right\rangle _{C_3} \\&\qquad +\; \left\langle \,j(x)j(s)\,\right\rangle _{C_3}\ +\ \left\langle \,j(x)j(-s)\,\right\rangle _{C_3} \Bigg ] \\&\qquad -\;\mathrm{im}^3\ \left( \frac{\sqrt{2}\pi }{3} \right) ^3 \Bigg \} \ . \end{aligned} \end{aligned}$$

(4.43)

The first integral contains a divergence in $\left\langle \,j(s)j(-s)\,\right\rangle$ when s approaches zero. Explicitly using (A.13) this part of the amplitude is given by

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{3}{2}} \mathrm{d}x x^{\epsilon _2-\epsilon _1}\int _0^{\frac{x}{3}} \mathrm{d}s s^{\epsilon _1}\Bigg [ \left\langle \,j(s)j(-s)\,\right\rangle _{C_3} \\&\qquad +\;\left\langle \,j(x)j(s)\,\right\rangle _{C_3}\ +\ \left\langle \,j(x)j(-s)\,\right\rangle _{C_3} \Bigg ] \\&\quad = \int _0^{\frac{3}{2}} \mathrm{d}x\ x^{\epsilon _2-\epsilon _1}\int _0^{\frac{x}{3}} \mathrm{d}s\ s^{\epsilon _1}\Bigg [ \csc ^2\left[ \frac{2\pi }{3}s\right] \ \\&\qquad +\; \csc ^2 \left[ \frac{\pi }{3}(x+s)\right] \ +\ \csc ^2 \left[ \frac{\pi }{3}(x-s)\right] \Bigg ] \Bigg |_{\epsilon _1=0} \ \Bigg |_{\epsilon _2=0} \\&\quad =\ \frac{3}{2 \pi } \int _0^{\frac{3}{2}} \mathrm{d}x \ x^{\epsilon _2}\tan \left[ \frac{2\pi }{9} x \right] \Bigg |_{\epsilon _2=0} = \frac{27}{4 \pi ^2}\ \log 2. \end{aligned} \end{aligned}$$

(4.44)

Notice that the integral in x is convergent which tells us that we could have avoided the $\epsilon _2$ regulator. This is because the external propagator acts on a state which is in the fusion of three marginal operators and therefore it cannot contain the tachyon in its level expansion. We have

$$\begin{aligned} \begin{aligned}&B_{3,m}^{\text {SFT}} \\&\quad = i\frac{2 \sqrt{2}}{3!} m^3 \pi ^3 \ -\ 3i\sqrt{2}\ m\pi \ \log 2\ \\&\quad =\ B_{3,m}^{\text {BCFT}} \ +\ (3\log 2) \ B_{1,m}^{\text {BCFT}}. \end{aligned} \end{aligned}$$

(4.45)

Summarizing: from the BCFT side we found that the third order is proportional to $m^3$ (4.8) and there are no other terms. Instead, in the OSFT computation, at the third order we still get the same BCFT number proportional to $m^3$ but in addition to it there is another contribution coming from the peculiar renormalization implicitly defined by the propagator ${\mathcal {B}}_0/{\mathcal {L}}_0$. This is the first time that there appears a discrepancy between the two approaches. As a consequence the third order coefficient in the $\tilde{\lambda }(\lambda )$ equation (4.1) is

$$\begin{aligned} b_3 = -\ 3\log 2. \end{aligned}$$

(4.46)

4.5 Fourth order

The fourth order Ellwood invariant is given by

$$\begin{aligned} B_{4,m}^{\text {SFT}}= & {} 2\pi i\ \langle \mathcal {I}| V_m(i,-i)\ |\Psi _4\rangle \nonumber \\= & {} \ 2\pi i\ \left\langle \,V_m(i\infty ,-i\infty )\ \hat{\Psi }_4 \,\right\rangle _{C_4}; \end{aligned}$$

(4.47)

there are two contributions coming from the $\Psi _{\tilde{\lambda }}$ solution (2.29)

$$\begin{aligned} \hat{\Psi }_4|0\rangle= & {} \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \left( \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(2,4)}} \right. \nonumber \\&\left. -\;\frac{1}{2}\ \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}} \right) |0\rangle . \end{aligned}$$

(4.48)

The Ellwood invariant at this order is given by

$$\begin{aligned} B_{4,m}^{\text {SFT}}= & {} 2\pi i\ \Bigg ( \left\langle \,V_m(i\infty ,-i\infty )\frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(2,4)}} \,\right\rangle _{C_4} \nonumber \\&-\; \frac{1}{2}\ \left\langle \,V_m(i\infty ,-i\infty )\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}} \,\right\rangle _{C_4} \Bigg ) \nonumber \\\equiv & {} 2\pi i\ \left( \left\langle \,V_m\hat{\mathcal {A}}_{2,4}\,\right\rangle _{C_4}\ -\ \left\langle \,V_m\hat{\mathcal {A}}_{3,3} \,\right\rangle _{C_4} \right) . \end{aligned}$$

(4.49)

First term $\left\langle \,V_m\hat{\mathcal {A}}_{2,4}\,\right\rangle$

In the first term, as before, we need to compute the commutator of the insertions and apply the propagators,

$$\begin{aligned} \hat{\mathcal {A}}_{2,4} |0\rangle = U_5^* U_5\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(2,4)}} |0\rangle .\nonumber \\ \end{aligned}$$

(4.50)

The corresponding amplitude is depicted in Fig. 3.

Fig. 3

First diagram $\left\langle \,V_m\hat{\mathcal {A}}_{2,4}\,\right\rangle$. The first two legs of cj from the left are to be understood as commutators. The z, y and 2t variables are the Schwinger parameters (taking value in the interval [0, 1]). The integration variable s in (4.51) is related to the Schwinger parameters as $s=t y$

Applying the two propagators, the amplitude takes the form

$$\begin{aligned} \begin{aligned}&\left\langle \,V_m\hat{\mathcal {A}}_{2,4}\,\right\rangle _{C_4} \\&\quad = -\ 3\int _0^1 \mathrm{d}z\ z^{2+\epsilon _3}\int _0^1 \mathrm{d}y\ y^{\epsilon _2-\epsilon _1} \int _0^{\frac{y}{2}} \mathrm{d}s\ s^{\epsilon _1} \ \Bigg \langle \ V_m(i\infty ) \\&\qquad \times \; \Bigg \{ j\left( \tfrac{3}{2}z \right) \ j\left( z\left( y-\tfrac{1}{2} \right) \right) \ j\left( z\left( s-\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \\&\qquad \times \; j\left( z\left( -s-\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \times \Big [ c\left( \tfrac{3}{2}z \right) \ +\ c\left( z\left( y-\tfrac{1}{2} \right) \right) \\&\qquad +\;c\left( z\left( s-\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \ +\ c\left( z\left( -s-\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \Big ] \\&\qquad +\;j\left( \tfrac{3}{2}z \right) j\left( z\left( -y-\tfrac{1}{2} \right) \right) j\left( z\left( s+\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \\&\qquad \times \; j\left( z\left( -s+\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \times \Big [ c\left( \tfrac{3}{2}z \right) \ +\ c\left( z\left( -y-\tfrac{1}{2} \right) \right) \\&\qquad +c\left( z\left( s+\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \\&\qquad +\ c\left( z\left( -s+\tfrac{1}{2}y-\tfrac{1}{2} \right) \right) \Big ] \Bigg \} \ \Bigg \rangle _{C_4} \Bigg |_{\epsilon _1=0} \ \Bigg |_{\epsilon _2=0} \ \Bigg |_{\epsilon _3=0} . \end{aligned}\nonumber \\ \end{aligned}$$

(4.51)

Using the symmetries of the problem translating the correlators $\xi \rightarrow \xi +\frac{z}{2}$ and changing variables $w=sz$ and $x=yz$, the amplitude simplifies to

$$\begin{aligned} \begin{aligned}&\left\langle \,V_m\hat{\mathcal {A}}_{2,4}\,\right\rangle _{C_4}\\&\quad = - 6\int _0^2 \mathrm{d}z\ z^{\epsilon _3+\epsilon _2-2\epsilon _1} \int _0^{\frac{z}{2}} \mathrm{d}x \int _0^{\frac{x}{2}} \mathrm{d}w\ w^{\epsilon _1} \\&\qquad \times \;\Bigg \{ \left\langle \, V_m(i\infty )\ j(z)\ j(x)\ j\left( w-\tfrac{x}{2} \right) \ j\left( -w-\tfrac{x}{2} \right) \,\right\rangle _{C_4} \\&\qquad +\;\left\langle \, V_m(i\infty )\ j(z)\ j\left( w+\tfrac{x}{2} \right) \ j\left( -w+\tfrac{x}{2} \right) \ j(-x) \,\right\rangle _{C_4} \ \Bigg \}. \end{aligned} \end{aligned}$$

(4.52)

Second term $\left\langle \,V_m\hat{\mathcal {A}}_{3,3} \,\right\rangle$

The second term is the Ellwood invariant of $\hat{\mathcal {A}}_{3,3}$,

$$\begin{aligned} \hat{\mathcal {A}}_{3,3}|0\rangle = \frac{1}{2}\ U_5^* U_5\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}} |0\rangle , \end{aligned}$$

(4.53)

and it is depicted in Fig. 4.

Fig. 4

Second diagram $\left\langle \,V_m\hat{\mathcal {A}}_{3,3} \,\right\rangle$. z, 2t and 2s are the Schwinger parameters (taking values in the interval [0, 1])

Explicitly we have to compute the star product of two $\Psi _2$ and then act with a propagator ${\mathcal {B}}_0/{\mathcal {L}}_0$:

$$\begin{aligned} \begin{aligned} \left\langle \,V_m\hat{\mathcal {A}}_{3,3} \,\right\rangle _{C_4}&= \int _0^1 \mathrm{d}z\ z^{2+\epsilon _3}\ \int _0^\frac{1}{2} \mathrm{d}t\ t^{\epsilon _2}\int _0^{\frac{1}{2}} \mathrm{d}s\ s^{\epsilon _1} \\&\quad \times \; 2\Bigg \langle \ V_m(i\infty )\ j(z(t+1))\ j(z(-t+1)) \\&\quad \times \; j(z(s-1)) j(z(-s-1)) \\&\quad \times \; \Big [c(z(t+1)) + c(z(-t+1)) + c(z(s-1)) \\&\quad +\; c(z(-s-1)) \Big ] \ \Bigg \rangle _{C_4} \Bigg |_{\epsilon _{1,2}=0} \ \Bigg |_{\epsilon _3=0}. \end{aligned} \end{aligned}$$

(4.54)

Again the four different insertions of the ghosts contribute in the same way, and therefore

$$\begin{aligned} \begin{aligned} \left\langle \,V_m\hat{\mathcal {A}}_{3,3} \,\right\rangle _{C_4}&= 8\int _0^1 \mathrm{d}z\ z^{2+\epsilon _3}\ \int _0^\frac{1}{2} \mathrm{d}t\ t^{\epsilon _2}\int _0^{\frac{1}{2}} \mathrm{d}s\ s^{\epsilon _1} \\&\quad \times \; \Bigg \langle \ V_m(i\infty ) cj(z(t+1))\ j(z(-t+1)) \\&\quad \times \; j(z(s-1))\ j(z(-s-1))\ \Bigg \rangle _{C_4}. \end{aligned} \end{aligned}$$

(4.55)

With the change of variable $x=zt$ and $y=zs$,

$$\begin{aligned} \begin{aligned} \left\langle \,V_m\hat{\mathcal {A}}_{3,3} \,\right\rangle _{C_4}&= 8\int _0^1 \mathrm{d}z\ z^{\epsilon _3-\epsilon _2-\epsilon _1}\int _0^{\frac{z}{2}} \mathrm{d}x\ x^{\epsilon _2} \int _0^{\frac{z}{2}} \mathrm{d}y\ y^{\epsilon _1} \\&\quad \times \; \Bigg \langle \ V_m(i\infty )\ cj(z+x) \\&\quad \times \; j(z-x)\ j(-z+y))\ j(-z-y)\ \Bigg \rangle _{C_4}. \end{aligned} \end{aligned}$$

(4.56)

Complete $B_{4,m}^{\text {SFT}}$

The complete term at this order is given by summing the two integrals (4.52) and (4.56),

$$\begin{aligned}&B_{4,m}^{\text {SFT}} \nonumber \\&\quad = - 12\pi i \int _0^2 \mathrm{d}z\ z^{\epsilon _3+\epsilon _2-2\epsilon _1} \int _0^{\frac{z}{2}} \mathrm{d}x \int _0^{\frac{x}{2}} \mathrm{d}w\ w^{\epsilon _1} \nonumber \\&\qquad \Bigg \{ \left\langle \, V_m(i\infty )\ j(z) j(x) j\left( w-\tfrac{x}{2} \right) j\left( -w-\tfrac{x}{2} \right) \,\right\rangle _{C_4} \nonumber \\&\qquad +\ \left\langle \, V_m(i\infty )\ j(z)\ j\left( w+\tfrac{x}{2} \right) \ j\left( -w+\tfrac{x}{2} \right) \ j(-x) \,\right\rangle _{C_4} \Bigg \} \nonumber \\&\qquad -16\pi i \int _0^1 \mathrm{d}z\ z^{\epsilon _3-\epsilon _2-\epsilon _1}\int _0^{\frac{z}{2}} \mathrm{d}x\ x^{\epsilon _2} \nonumber \\&\qquad \int _0^{\frac{z}{2}} \mathrm{d}y\ y^{\epsilon _1}\ \langle \ V_m(i\infty )\ cj(z+x) \nonumber \\&\qquad j(z-x)\ j(-z+y))\ j(-z-y)\ \rangle _{C_4}. \end{aligned}$$

(4.57)

As in the previous order $\tilde{\lambda }^3$, here also we have contribution from three different powers of the winding number m coming from the different contractions in Wick theorem (A.17). This means that we can write $B_{4,m}^{\text {SFT}}$ in terms of the $B_{k,m}^{\text {BCFT}} \sim m^k$,

$$\begin{aligned} \begin{aligned} B_{4,m}^{\text {SFT}} = a_0\ B_{0,m}^{\text {BCFT}}\ +\ a_2\ B_{2,m}^{\text {BCFT}} \ +\ a_4\ B_{4,m}^{\text {BCFT}}. \end{aligned} \end{aligned}$$

(4.58)

From this consideration (setting to zero all the regulators as they are not important here) we find that

$$\begin{aligned} a_4\ B_{4,m}^{\text {BCFT}}= & {} \frac{6}{8}\ m^4 \pi ^4 \int _0^2 \mathrm{d}z \int _0^{\frac{z}{2}} \mathrm{d}x \int _0^{\frac{x}{2}} \mathrm{d}w \nonumber \\&+\; \frac{1}{2}\ m^4 \pi ^4 \int _0^1 \mathrm{d}z \int _0^{\frac{z}{2}} \mathrm{d}x \int _0^{\frac{z}{2}} \mathrm{d}w \nonumber \\= & {} \frac{m^4 \pi ^4}{6}, \end{aligned}$$

(4.59)

$$\begin{aligned} a_2\ B_{2,m}^{\text {BCFT}}= & {} - \frac{3}{16}\ m^2 \pi ^4 \int _0^2 \mathrm{d}z\ z^{\epsilon _3+\epsilon _2-2\epsilon _1} \int _0^{\frac{z}{2}} \mathrm{d}x \int _0^{\frac{x}{2}} \mathrm{d}w\ w^{\epsilon _1} \nonumber \\&\times \; \Bigg \{ 2 \csc ^2 \left[ \frac{\pi }{2} w \right] \ +\ 2\ \csc ^2 \left[ \frac{\pi }{4} \left( w-\frac{3}{2}x \right) \right] \nonumber \\&+\;2\ \csc ^2 \left[ \frac{\pi }{4} (w+\frac{3}{2}x) \right] \ +\ \csc ^2 \left[ \frac{\pi }{4} (z+x) \right] \nonumber \\&+\;\csc ^2 \left[ \frac{\pi }{4} (z-x) \right] + \csc ^2 \left[ \frac{\pi }{4} \left( z+w+\frac{x}{2} \right) \right] \nonumber \\&+\;\csc ^2 \left[ \frac{\pi }{4} \left( z+w-\frac{x}{2} \right) \right] \nonumber \\&+\;\csc ^2 \left[ \frac{\pi }{4} \left( z-w+\frac{x}{2} \right) \right] \nonumber \\&+\;\csc ^2 \left[ \frac{\pi }{4} \left( z-w-\frac{x}{2} \right) \right] \Bigg \} \Bigg |_{\epsilon _1=0}\Bigg |_{\epsilon _2=0} \Bigg |_{\epsilon _3=0} \nonumber \\&-\;\frac{1}{4} m^2 \pi ^4 \int _0^1 \mathrm{d}z\ z^{\epsilon _3-\epsilon _2-\epsilon _1}\int _0^{\frac{z}{2}} \mathrm{d}x\ x^{\epsilon _2} \nonumber \\&\times \;\int _0^{\frac{z}{2}} \mathrm{d}w\ w^{\epsilon _1} \Bigg \{ \csc ^2 \left[ \frac{\pi }{2} w \right] + \csc ^2 \left[ \frac{\pi }{2} x \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w+2z+x \right) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w+2z-x \right) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w-2z+x \right) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( -w+2z+x \right) \right] \Bigg \} \ \Bigg |_{\epsilon _{1,2}=0} \ \Bigg |_{\epsilon _3=0} \nonumber \\= & {} \big ( 2\ m^2 \pi ^2\big ) \ 3\ \log 2. \end{aligned}$$

(4.60)

$$\begin{aligned} a_0\ B_{0,m}^{\text {BCFT}}= & {} - \frac{3}{32}\ \pi ^4 \int _0^2 \mathrm{d}z\ z^{\epsilon _3+\epsilon _2-2\epsilon _1} \int _0^{\frac{z}{2}} \mathrm{d}x \nonumber \\&\times \; \int _0^{\frac{x}{2}} \mathrm{d}w\ w^{\epsilon _1}\ \Bigg \{ \csc ^2 \left[ \frac{\pi }{2} w \right] \ \csc ^2 \left[ \frac{\pi }{4} (z+x) \right] \nonumber \\&+\;\csc ^2 \left[ \frac{\pi }{2} w \right] \ \csc ^2 \left[ \frac{\pi }{4} (z-x) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w+\frac{3}{2}x \right) \right] \ \csc ^2 \left[ \frac{\pi }{4} \left( z+w-\frac{x}{2} \right) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w+\frac{3}{2}x \right) \right] \ \csc ^2 \left[ \frac{\pi }{4} \left( z-w+\frac{x}{2} \right) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w-\frac{3}{2}x \right) \right] \ \csc ^2 \left[ \frac{\pi }{4} \left( z-w-\frac{x}{2} \right) \right] \nonumber \\&+\; \csc ^2 \left[ \frac{\pi }{4} \left( w-\frac{3}{2}x \right) \right] \nonumber \\&\times \;\csc ^2 \left[ \frac{\pi }{4} \left( z+w+\frac{x}{2} \right) \right] \Bigg \} \ \Bigg |_{\epsilon _1=0} \ \Bigg |_{\epsilon _2=0} \Bigg |_{\epsilon _3=0} \nonumber \\&-\ \frac{1}{4} m^2 \pi ^4 \int _0^1 \mathrm{d}z\ z^{\epsilon _3-\epsilon _2-\epsilon _1}\int _0^{\frac{z}{2}} \mathrm{d}x\ x^{\epsilon _2} \nonumber \\&\times \; \int _0^{\frac{z}{2}} \mathrm{d}w\ w^{\epsilon _1} \Bigg \{ \csc ^2 \left[ \frac{\pi }{2} w \right] \ \csc ^2 \left[ \frac{\pi }{2} x \right] \nonumber \\&+\ \csc ^2 \left[ \frac{\pi }{4} \left( w+2z+x \right) \right] \ \csc ^2 \left[ \frac{\pi }{4} \left( w-2z+x \right) \right] \nonumber \\&+\ \csc ^2 \left[ \frac{\pi }{4} \left( w+2z-x \right) \right] \nonumber \\&\times \; \csc ^2 \left[ \frac{\pi }{4} \left( -w+2z+x \right) \right] \Bigg \} \Bigg |_{\epsilon _{1,2}=0} \Bigg |_{\epsilon _3=0} \nonumber \\= & {} 0 \ . \end{aligned}$$

(4.61)

Using these results we find that

$$\begin{aligned} B_{4,m}^{\text {SFT}} = B_{4,m}^{\text {BCFT}}\ +\ \left( 6 \log 2 \right) \ B_{2,m}^{\text {BCFT}}, \end{aligned}$$

(4.62)

which corresponds to

$$\begin{aligned} b_4 = 0. \end{aligned}$$

(4.63)

4.6 Fifth order

At the fifth order the solution is composed of three terms,

$$\begin{aligned} \begin{aligned}&\hat{\Psi }_5|0\rangle \\&\quad = \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Bigg ( -\ \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0) , \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(2,4)}} \Big ]\!\Big ]_{\text {(2,5)}} \ \\&\qquad -\ \Big [\!\Big [ \Psi _2, \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(3,4)}} \\&\qquad +\ \frac{1}{2}\ \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}} \Big ]\!\Big ]_{\text {(2,5)}} \Bigg )|0\rangle . \end{aligned} \end{aligned}$$

(4.64)

Then the Ellwood invariant we have to compute is

$$\begin{aligned} \begin{aligned} B_{5,m}^{\text {SFT}}&= 2\pi i\ \langle \mathcal {I}| V_m(i,-i)\ |\Psi _5\rangle \\&= 2\pi i\ \bigg \langle \,V_m(i\infty ,-i\infty )\ \hat{\Psi }_5 \,\bigg \rangle _{C_5}\ \\&= 2\pi i\ \Bigg (- \bigg \langle \,V_m(i\infty ,-i\infty )\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\\\Big ]\!\Big ]&{ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0) , \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(2,4)}} }_{\text {(2,5)}} \,\bigg \rangle _{C_5}\ \\&\quad -\; \bigg \langle \,V_m(i\infty ,-i\infty )\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ \Psi _2, \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(3,4)}} \,\bigg \rangle _{C_5} \ \\&\quad +\; \frac{1}{2}\ \bigg \langle \,V_m(i\infty ,-i\infty )\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}} \Big ]\!\Big ]_{\text {(2,5)}} \,\bigg \rangle _{C_5} \Bigg )\\&\equiv 2\pi i\ \left( -\ \bigg \langle \,V_m\hat{\mathcal {A}}_{2,5} \,\bigg \rangle _{C_5} \ -\ \bigg \langle \,V_m\hat{\mathcal {A}}_{3,4} \,\bigg \rangle _{C_5} \ + \ \frac{1}{2}\ \bigg \langle \,V_m\hat{\mathcal {A}}^{3,3}_{2,5} \,\bigg \rangle _{C_5} \right) . \end{aligned} \end{aligned}$$

(4.65)

First term $\left\langle \,V_m\hat{\mathcal {A}}_{2,5} \,\right\rangle$

The first term involves the state

$$\begin{aligned}&\hat{\mathcal {A}}_{2,5} |0\rangle \nonumber \\&\quad =\ U_6^* U_6\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0), \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0) , \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(2,4)}} \Big ]\!\Big ]_{\text {(2,5)}} |0\rangle .\nonumber \\ \end{aligned}$$

(4.66)

Fig. 5

First diagram $\left\langle \,V_m\hat{\mathcal {A}}_{2,5}\,\right\rangle$

The amplitude to compute is depicted in Fig. 5 and after some manipulations, involving changes of variables and conformal transformations, we get

$$\begin{aligned}&2\pi i\ \left\langle \,V_m\hat{\mathcal {A}}_{2,5} \,\right\rangle _{C_5} =\ -\ 48\pi i \int _0^{\frac{5}{2}} dT\ T^{\epsilon _4-\epsilon _3-\epsilon _2+\epsilon _1} \nonumber \\&\qquad \times \; \int _0^{\frac{T}{5}} dX\ X^{\epsilon _3+\epsilon _2-2\epsilon _1} \int _0^{2X} dY \int _0^{\frac{Y}{2}} dZ\ Z^{\epsilon _1} \nonumber \\&\qquad \times \; \Bigg \{ \left\langle V_m(i\infty )\ cj(T)\ j(3X)\ j(Y-X)\right. \nonumber \\&\qquad \times \; \left. j\left( Z-\tfrac{Y}{2}-X \right) j\left( -Z-\tfrac{Y}{2}-X \right) \right\rangle _{C_5} \nonumber \\&\qquad +\; \left\langle V_m(i\infty )\ cj(T)\ j(3X)\ j\left( Z+\tfrac{Y}{2}-X \right) \right. \nonumber \\&\qquad \times \;\left. j\left( -Z+\tfrac{Y}{2}-X \right) \ j(-Y-X)\right\rangle _{C_5} \nonumber \\&\qquad +\; \left\langle V_m(i\infty )\ cj(T)\ j\left( Z+\tfrac{Y}{2}+X \right) \right. \nonumber \\&\qquad \times \; \left. j\left( -Z+\tfrac{Y}{2}+X \right) \ j(-Y+X)\ j\left( -3X \right) \right\rangle _{C_5} \nonumber \\&\qquad +\; \left\langle V_m(i\infty )\ cj(T)\ j(Y+X)\ j\left( Z-\tfrac{Y}{2}+X \right) \right. \nonumber \\&\qquad \times \;\left. j\left( -Z-\tfrac{Y}{2}+X \right) \ j\left( -3X \right) \right\rangle _{C_5} \Bigg \}\Bigg |_{\epsilon _1=0} \Bigg |_{\epsilon _2=0} \Bigg |_{\epsilon _3=0} \Bigg |_{\epsilon _4=0}, \nonumber \\ \end{aligned}$$

(4.67)

where the Schwinger parameters $t_1,t_2,t_3,t_4$ are related to the integration variables as

$$\begin{aligned} \begin{aligned} T&=\ t_4, \\ X&=\ t_4 t_3, \\ Y&=\ t_4 t_3 t_2, \\ Z&=\ t_4 t_3 t_2 t_1. \end{aligned} \end{aligned}$$

(4.68)

Second term $\left\langle \,V_m\hat{\mathcal {A}}_{3,4} \,\right\rangle$

The state here is

$$\begin{aligned} \hat{\mathcal {A}}_{3,4} |0\rangle = U_6^* U_6\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ \Psi _2, \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\hat{\Psi }_2\Big ]\!\Big ]_{\text {(2,3)}} \Big ]\!\Big ]_{\text {(3,4)}} |0\rangle ,\nonumber \\ \end{aligned}$$

(4.69)

and its Ellwood invariant is (Fig. 6)

$$\begin{aligned} \begin{aligned}&2\pi i\ \left\langle \,V_m\hat{\mathcal {A}}_{3,4} \,\right\rangle _{C_5} \\&\quad =\ -\ 24\pi i \int _0^{\frac{5}{2}} dT\ T^{\epsilon _4-\epsilon _3-\epsilon _2} \\&\qquad \times \;\int _0^{\frac{2}{5}T} dX\ X^{\epsilon _3-\epsilon _2} \int _0^{\frac{X}{2}} dY\ Y^{\epsilon _2} \int _0^{\frac{T}{2}} dZ\ Z^{\epsilon _1} \\&\qquad \times \Bigg \{ \left\langle V_m(i\infty )\ cj(T+Z)\ j(T-Z)\ j(X)\right. \\&\qquad \times \;\left. j\left( Y-\tfrac{X}{2} \right) \ j\left( -Y-\tfrac{X}{2} \right) \right\rangle _{C_5} \\&\qquad +\;\left\langle V_m(i\infty )\ cj(T+Z)\ j(T-Z)\ j\left( Y+\tfrac{X}{2} \right) \right. \\&\qquad \times \;\left. j\left( -Y+\tfrac{X}{2} \right) \ j(-X)\right\rangle _{C_5} \Bigg \} \Bigg |_{\epsilon _1=0} \Bigg |_{\epsilon _2=0} \Bigg |_{\epsilon _3=0} \Bigg |_{\epsilon _4=0}, \end{aligned} \end{aligned}$$

(4.70)

where the integration variables are related to the Schwinger parameters as

$$\begin{aligned} \begin{aligned} T&=\ t_4, \\ X&=\ t_4 t_3, \\ Y&=\ t_4 t_3 t_2, \\ Z&=\ t_4 t_1. \end{aligned} \end{aligned}$$

(4.71)

Fig. 6

Second diagram $\left\langle \,V_m\hat{\mathcal {A}}_{3,4}\,\right\rangle$

Third term $\left\langle \,V_m\hat{\mathcal {A}}^{3,3}_{2,5} \,\right\rangle$

The last term involves the state

$$\begin{aligned} \hat{\mathcal {A}}^{3,3}_{2,5}|0\rangle = \frac{1}{2}\ U_5^* U_5\ \frac{{\mathcal {B}}_0}{{\mathcal {L}}_0}\ \Big [\!\Big [ cj(0),\Big [\!\Big [ \hat{\Psi }_2,\hat{\Psi }_2\Big ]\!\Big ]_{\text {(3,3)}}\Big ]\!\Big ]_{\text {(2,5)}} |0\rangle ,\nonumber \\ \end{aligned}$$

(4.72)

and it becomes (Fig. 7)

$$\begin{aligned} \begin{aligned}&2\pi i\ \left\langle \,V_m\hat{\mathcal {A}}^{3,3}_{2,5} \,\right\rangle _{C_5} \\&\quad =\ 32\pi i \int _0^{\frac{5}{2}} dT\ T^{\epsilon _4-\epsilon _3} \\&\qquad \times \; \int _0^{\frac{2}{5}T} dX\ X^{\epsilon _3-\epsilon _2-\epsilon _1} \int _0^{\frac{X}{2}} dY\ Y^{\epsilon _2} \int _0^{\frac{X}{2}} dZ\ Z^{\epsilon _1} \\&\qquad \times \; \Bigg \{ \left\langle V_m(i\infty )\ cj(T)\ j(X+Y)\ j(X-Y)\right. \\&\qquad \times \; \left. j\left( Z-X \right) \ j\left( -Z-X \right) \right\rangle _{C_5} \\&\qquad +\;\left\langle V_m(i\infty )\ cj(X+Y)\ j(X-Y)\ j\left( Z-X \right) \right. \\&\qquad \times \; \left. j\left( -Z-X \right) j(-T) \right. \rangle _{C_5} \Bigg \} \Bigg |_{\epsilon _1=0} \Bigg |_{\epsilon _2=0} \Bigg |_{\epsilon _3=0} \Bigg |_{\epsilon _4=0}, \end{aligned} \end{aligned}$$

(4.73)

where the Schwinger parameters are related to the integration variables as

$$\begin{aligned} \begin{aligned} T&=\ t_4, \\ X&=\ t_4 t_3, \\ Y&=\ t_4 t_3 t_2, \\ Z&=\ t_4 t_3 t_1. \end{aligned} \end{aligned}$$

(4.74)

Fig. 7

Third diagram $\left\langle \,V_m\hat{\mathcal {A}}^{3,3}_{2,5}\,\right\rangle$

Fig. 8

The three diagrams are divergent with a simple pole in $T=0$

Fig. 9

In a we show the three residues of the simple poles corresponding to the three diagrams. In b the sum of the three functions $F_i(t)$ is shown to be finite in $T=0$

Complete $B_{5,m}^{\text {SFT}}$

Now we have to compute the three Ellwood invariants as we have done in previous examples. Again the total Ellwood invariant can be expanded in terms of the first, third and fifth power of the winding number,

$$\begin{aligned} B_{5,m}^{\text {SFT}} = a_1\ B_{1,m}^{\text {BCFT}}\ +\ a_3\ B_{3,m}^{\text {BCFT}} \ +\ a_5\ B_5^{\text {BCFT}}. \end{aligned}$$

(4.75)

The $a_5$ coefficient is easily computed:

$$\begin{aligned} \begin{aligned}&a_5\ B_5^{\text {BCFT}} \\&= - 480\ \frac{(i m)^5}{5!}\ \Big ( \left\langle \,X(i\infty ) j(0)\,\right\rangle _{C_5} \Big )^5 \\&\quad \times \int _0^{\frac{5}{2}} dT \int _0^{\frac{T}{5}} dX \int _0^{2X} dY \int _0^{\frac{Y}{2}} dZ\ \\&-\ 300\ \frac{(i m)^5}{5!}\ \Big ( \left\langle \,X(i\infty ) j(0)\,\right\rangle _{C_5} \Big )^5 \\&\quad \times \int _0^{\frac{5}{2}} dT \int _0^{\frac{2}{5}T} dX \int _0^{\frac{X}{2}} dY \int _0^{\frac{T}{2}} dZ\ \\&-\ 80\ \frac{(i m)^5}{5!}\ \Big ( \left\langle \,X(i\infty ) j(0)\,\right\rangle _{C_5} \Big )^5 \\&\quad \times \int _0^{\frac{5}{2}} dT \int _0^{\frac{2}{5}T} dX \int _0^{\frac{X}{2}} dY \int _0^{\frac{X}{2}} dZ\ \\&=i \frac{4\sqrt{2}}{5!}\ \pi ^5 m^5 \ = \ B_5^{\text {BCFT}}, \end{aligned} \end{aligned}$$

(4.76)

which gives the usual exact match with the BCFT results. As far as $a_3$ is concerned the computation follows closely the fourth order (with one more integral) and everything can be analytically done giving the result

$$\begin{aligned} a_3 = 9 \log 2. \end{aligned}$$

(4.77)

This is precisely the number needed to ensure that $b_5$ is m-independent (4.16) so this is a consistency check.

Let us now address the $a_1$ coefficient, which is determined by the O(m) winding number contribution in (4.75). This is generated by the term from the Wick theorem with the maximal number of contractions between the j and computing the four dimensional integrals (coming from the three diagrams) analytically is not possible. Therefore we proceed analytically as far as we can and then we resort to numerics. The Y and Z integrals can be analytically computed in all of the three diagrams, including the subtraction of the tachyon divergence.

Rescaling the X variable in the first diagram $X\rightarrow 2X$, the O(m) contribution $E_i$ from each diagram is reduced to an expression of the form

$$\begin{aligned} a_1 \ B_1^{\text {BCFT}}= & {} E_1+E_2+E_3,\qquad O(m) \nonumber \\ E_i= & {} \int _0^{\frac{5}{2}} dT\ T^{\epsilon _4-\epsilon _3} \nonumber \\&\times \;\int _0^{\frac{2}{5}T} dX\ X^{\epsilon _3}\ f_i(T,X)\ \Bigg |_{\epsilon _3=0} \Bigg |_{\epsilon _4=0}, \quad i=1,2,3,\nonumber \\ \end{aligned}$$

(4.78)

where the function $f_i$ are known analytically. To renormalize the tachyon divergence in the X integration we explicitly subtract the second order pole in X from the function $f_i$ in the following way:

$$\begin{aligned} F_i(T)= & {} \int _0^{\frac{2}{5}T} dX\ X^{\epsilon _3}\ f_i(T,X)\ \Bigg |_{\epsilon _3=0} \nonumber \\= & {} \int _0^{\frac{2}{5}T} dX\ \Big ( f_i(T,X) \ -\ \frac{(f_i)_{-2}}{X^2} \Big ) \nonumber \\&+\;\int _0^{\frac{2}{5}T} dX\ X^{\epsilon _3}\ \frac{(f_i)_{-2}}{X^2}\ \Bigg |_{\epsilon _3=0}. \end{aligned}$$

(4.79)

It turns out that the coefficient of the $1/X^2$ pole, which in the above formula is indicated as $(f_i)_{-2}$, is T independent.

This treatment leaves us with three numerical functions of T, which have to be integrated in the interval $\left[ 0,\frac{5}{2}\right]$. Surprisingly each of these functions shows a nonvanishing 1 / T pole, as shown in Fig. 8,

$$\begin{aligned} F_i(T) = \frac{p_i}{T}\ +\ {\tilde{F}}_i(T), \end{aligned}$$

(4.80)

with ${\tilde{F}}_i$ finite in $T=0$. We explicitly find

$$\begin{aligned} {\left\{ \begin{array}{ll} p_1 = -\ 62.10989(8) ,\\ p_2 = -\ 2.45519(8) ,\\ p_3 = 64.56508(9). \end{array}\right. }. \end{aligned}$$

(4.81)

These poles are potentially problematic, and it is reassuring that the sum of them vanishes,

$$\begin{aligned} p_1\ +\ p_2\ +\ p_3 = 0; \end{aligned}$$

(4.82)

see also Fig. 9. This is an important consistency check, because a 1 / T pole would be an obstruction to the existence of the solution at the fifth order. Numerically computing the integral over T finally gives

$$\begin{aligned} a_1 = 10.58226(7). \end{aligned}$$

(4.83)

To conclude the total contribution to fifth order is given by

$$\begin{aligned} B_{5,m}^{\text {SFT}} = B_5^{\text {BCFT}}\ +\ \left( 9 \log 2 \right) \ B_{3,m}^{\text {BCFT}}\ +\ 10.58226(7)\ B_{3,m}^{\text {BCFT}},\nonumber \\ \end{aligned}$$

(4.84)

which corresponds to (4.1)

$$\begin{aligned} b_5 = 2.38996(7). \end{aligned}$$

(4.85)

Appendix A: Conventions and correlators

Witten’s star product

The Witten star product is best understood in the Sliver frame (with coordinate z), which is related to the UHP (with coordinate w) by the map

$$\begin{aligned} z = \frac{2}{\pi }\ \arctan w. \end{aligned}$$

(A.1)

In this frame the star product of wedge states with insertions is given by

$$\begin{aligned} \begin{aligned}&\Bigg ( U_{r}^* U_{r} \ \Phi _1( x_1)\ \ldots \ \Phi _n( x_n) |0\rangle \Bigg ) * \\&\quad \Bigg ( U_{s}^* U_{s}\ \Psi _1( y_1)\ \ldots \ \Psi _m( y_m) |0\rangle \Bigg ) \\&=\ U_{r+s-1}^* U_{r+s-1}\ \Phi _1\left( x_1+\tfrac{s-1}{2} \right) \ \ldots \\&\quad \Phi _n\left( x_n+\tfrac{s-1}{2} \right) \ \Psi _1\left( y_1-\tfrac{r-1}{2} \right) \ \ldots \Psi _m\left( y_m-\tfrac{r-1}{2} \right) |0\rangle , \end{aligned} \end{aligned}$$

(A.2)

where the coordinates are in the“ $2/\pi$” Sliver frame (A.1), which implies some rescaling w.r.t. to (2.24) of [1], where the $2/\pi$ factor in (A.1) was omitted.

Upper Half Plane

In the BCFT of a free boson X at the self-dual radius $R=1$, with Neumann boundary conditions, consider the bulk winding mode

$$\begin{aligned} \mathcal {V}_m(z,\bar{z}) = e^{i m {\tilde{X}}(z,\bar{z})}, \end{aligned}$$

(A.3)

where $\tilde{X}(z,\bar{z}) = X_L(z) - X_R(\bar{z})$ is the T-dual field of $X(z,\bar{z})$, and the boundary marginal field

$$\begin{aligned} j(x) = i\sqrt{2}\partial X(x)=j(z){\Big |}_{z=x}, \end{aligned}$$

(A.4)

which is defined as a bulk chiral field placed at the boundary. The chiral closed string field X(z) has the following two-point functions ($\alpha '=1$):

$$\begin{aligned} \left\{ \begin{aligned} \left\langle \,X(z) \ X(w)\,\right\rangle _{\text { UHP }}&= \ - \dfrac{1}{2}\ \log (z-w) ,\\ \left\langle \,X(z) \ \partial X(w)\,\right\rangle _{\text { UHP }}&= \ \dfrac{1}{2}\ \dfrac{1}{z-w} ,\\ \left\langle \, \partial X(z) \ \partial X(w)\,\right\rangle _{\text { UHP }}&= \ -\ \dfrac{1}{2}\ \dfrac{1}{(z-w)^2} \end{aligned}\right. . \end{aligned}$$

(A.5)

The current j(x) has the two-point function

$$\begin{aligned} \left\langle \,j(x) \ j(y) \,\right\rangle _{\text { UHP }} \ = \ \frac{1}{(x-y)^2}, \end{aligned}$$

(A.6)

and it has the following OPE with the bulk winding mode $\mathcal {V}_m (z,\bar{z}) = e^{i m {\tilde{X}}(z,\bar{z})}$:

$$\begin{aligned} \mathcal {V}_m (z,\bar{z}) \ j(x)\sim & {} -\ \frac{m}{\sqrt{2}}\ \frac{1}{z-x}\ \mathcal {V}_m (z,\bar{z}) \nonumber \\&+\; \frac{m}{\sqrt{2}}\ \frac{1}{\bar{z}-x}\ \mathcal {V}_m (z,\bar{z}). \end{aligned}$$

(A.7)

We also have

$$\begin{aligned} \left\langle \,{\tilde{X}}(i,-i) \ \partial X(x)\,\right\rangle _{\text { UHP }} \ = \ -\ \frac{i}{1+x^2}, \end{aligned}$$

(A.8)

and, using the Wick theorem,

$$\begin{aligned}&\left\langle \,\mathcal {V}_m (i,-i) \ j(x) \,\right\rangle _{\text { UHP }} \nonumber \\&\quad = i m \sqrt{2} \ \frac{1}{1+x^2} \ \left\langle \,\mathcal {V}_m (i,-i)\,\right\rangle _{\text { UHP}}. \end{aligned}$$

(A.9)

The one-point function is¹¹

$$\begin{aligned} \left\langle \,\mathcal {V}_m (i,-i)\,\right\rangle _{\text { UHP }} \ = \ 2^{-m^2/2}. \end{aligned}$$

(A.10)

In addition we have the following correlator for the auxiliary closed string field [23]:

$$\begin{aligned} \left\langle \,\mathcal {V}_{\text {aux}}^{(1-h_m,1-h_m)}(i,-i)\,\right\rangle _{\text { UHP }} = \frac{1}{4} \ 2^{m^2/2}. \end{aligned}$$

(A.11)

Cylinder

On the cylinder the chiral closed string field X(z) has the following correlators [1]:

$$\begin{aligned} \left\{ \begin{aligned}&\left\langle \,X(z) \ X(w)\,\right\rangle _{C_N} = \ -\ \dfrac{1}{2}\ \log \left[ \sin \left[ \dfrac{\pi }{N} (z-w) \right] \right] ,\\&\left\langle \,X(z) \ \partial X(w)\,\right\rangle _{C_N} = \ \dfrac{\pi }{2N}\ \cot \left[ \dfrac{\pi }{N} (z-w) \right] ,\\&\left\langle \, \partial X(z) \ \partial X(w)\,\right\rangle _{C_N} = \ -\ \dfrac{\pi ^2}{N^2}\ \csc ^2 \left[ \dfrac{\pi }{N} (z-w) \right] \ , \end{aligned} \right. \end{aligned}$$

(A.12)

and then

$$\begin{aligned} \left\langle \,j(x) \ j(y) \,\right\rangle _{C_N}\ = \ \dfrac{2\pi ^2}{N^2}\ \csc ^2 \left[ \dfrac{\pi }{N} (z-w) \right] . \end{aligned}$$

(A.13)

Wick theorem

In the main text we deal with correlators of the following form:

$$\begin{aligned}&\left\langle \,\mathcal {V}_m(z,\bar{z}) \prod _{i=1}^N j_i(x_i)\,\right\rangle \nonumber \\&\quad = \left\langle \,\mathcal {V}_m(z,\bar{z})\,\right\rangle \ \sum _{k=0}^N \frac{(\mathrm{im})^k}{k!}\ \left\langle \,:(\tilde{X}(z,\bar{z}))^k: \prod _{i=1}^N j_i(x_i)\,\right\rangle .\nonumber \\ \end{aligned}$$

(A.14)

This correlator significantly simplifies on a cylinder $C_n$, when the bulk operator (properly dressed with the ghosts and the auxiliary sector to acquire total weight zero; see (2.15)) is placed at the midpoint $i\infty$. In particular, thanks to the rotational invariance of $C_n$ we have

$$\begin{aligned} \left\langle \,V_m(i\infty )\ c(x)\,\right\rangle _{C_N}= & {} \left\langle \,V_m(i\infty )\ c(0)\,\right\rangle _{C_N} \nonumber \\= & {} \frac{N}{\pi }\ \left\langle \,V_m(i,-i)\ c(0)\,\right\rangle _{UHP} \nonumber \\= & {} \frac{iN}{2\pi }. \end{aligned}$$

(A.15)

The term containing the maximal number of contractions with the closed string is given by

$$\begin{aligned} \begin{aligned}&\frac{(\mathrm{im})^N}{N!}\ \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \\&\qquad \times \; \left\langle \,:(\tilde{X}(i\infty ))^N: \prod _{i=1}^N j(x_i) \,\right\rangle \ \\&\quad =\ (\mathrm{im})^N\ \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \prod _{i=1}^N \left\langle \,\tilde{X}(i\infty ) j(x_i)\,\right\rangle \ \\&\quad =\ (\mathrm{im})^N\ \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \prod _{i=1}^N \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \ \\&\quad =\ (\mathrm{im})^N\ \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \left( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \right) ^N. \end{aligned} \end{aligned}$$

(A.16)

Let us list for convenience the explicit correlators which are used in the main text:

$$\begin{aligned} \begin{aligned}&\underline{N=0} \qquad \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle , \\&\underline{N=1} \qquad in\ \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle , \\&\underline{N=2} \qquad \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \Bigg \{ \left\langle \,j(x_2)j(x_2)\,\right\rangle \ \\&\quad -m^2\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )^2 \Bigg \}, \\&\underline{N=3} \qquad \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \Bigg \{\mathrm{im}\ \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \\&\quad \sum _{\begin{array}{c} i,r=1\\ i<r \end{array}}^3 \left\langle \,j(x_i)j(x_r)\,\right\rangle \\&\quad -\;\mathrm{im}^3\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )^3 \Bigg \}, \\&\underline{N=4} \qquad \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \Bigg \{ \left\langle \,j(x_1)j(x_2)j(x_3)j(x_4)\,\right\rangle \\&\quad -\; m^2\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )^2 \ \sum _{\begin{array}{c} i,r=1\\ i<j \end{array}}^4 \left\langle \,j(x_i)j(x_r)\,\right\rangle \\&\quad +\; m^4\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )^4 \Bigg \}, \\&\underline{N=5} \qquad \left\langle \,c(0)\ V_m(i\infty )\,\right\rangle \ \Bigg \{ \mathrm{im}\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )\\&\qquad \left\langle \,j(x_1)j(x_2)j(x_3)j(x_4)\,\right\rangle \\&\quad -\; \mathrm{im}^3\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )^3 \ \sum _{\begin{array}{c} i,r=1\\ i<r \end{array}}^5 \left\langle \,j(x_i)j(x_r)\,\right\rangle \\&\quad +\; \mathrm{im}^5\ \Big ( \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle \Big )^5 \Bigg \}, \end{aligned} \end{aligned}$$

(A.17)

where the only non-trivial correlator involving $\tilde{X}$ is given by

$$\begin{aligned} \left\langle \,\tilde{X}(i\infty ) j(0)\,\right\rangle _{C_N} \ = \ \frac{\sqrt{2}\ \pi }{N}. \end{aligned}$$

(A.18)