Structure constants of a single trace operator and determinant operators from hexagon

Abstract

We investigate the structure constant associated with a single trace operator and two determinant operators in $\mathcal{N}=4$ super Yang–Mills theory. In the holographic framework, this quantity corresponds to the interaction vertex between a closed string and two open strings attached to spherical D-branes. Based on diagrammatic intuition, we propose a conjecture that the structure constant at finite coupling can be elegantly expressed in terms of hexagon form factors. Specifically, this involves the preparation of two hexagon twist operators and the appropriate gluing of edges by integrating contributions from mirror particles and contracting boundary states. The gluing process yields the worldsheet for a closed string and two open strings attached to the D-branes. At weak coupling, the asymptotic expression simplifies to a sum over all possible partitions, not only for the edge related to the closed string but also for the edges representing half of the open string, taking into account reflection effects for the opposite open string edges. We validate this conjecture by directly computing various tree-level structure constants, and our results align well with the proposed conjecture. In particular, we find that there are multiple contractions in the tree-level problem as in four-point function problems. This is consistent with the fact that the worldsheet we are studying has one-dimensional moduli space.

Appendix A: review on open spin chain

Here, we review the open spin-chain which is holographically dual to the open string attached to the $Y=0$ brane [37]. The Hamiltonian is constructed by evaluating relevant boundary Feynman diagrams as well as bulk Feynman diagrams [37] and is explicitly given as

$$\begin{aligned} \begin{aligned} {\mathbb {H}}&=\frac{1}{2}\lambda \sum _{j=2}^{L-2}[ {{\mathbb {K}}}_{j,j+1}+2({{\mathbb {I}}}_{j,j+1}- {{\mathbb {P}}}_{j,j+1})]\\&\quad + \frac{1}{2} \lambda {{\mathbb {Q}}}_1^Y [ {{\mathbb {K}}}_{1,2}+2( {{\mathbb {I}}}_{1,2}- {{\mathbb {P}}}_{1,2})] {{\mathbb {Q}}}_1^Y\\&\quad + \frac{1}{2}\lambda {{\mathbb {Q}}}_L^Y [ {{\mathbb {K}}}_{L-1,L}+2( {{\mathbb {I}}}_{L-1,L}- {{\mathbb {P}}}_{L-1,L})] {{\mathbb {Q}}}_L^Y \\&\quad +\lambda ( {{\mathbb {I}}}- {{\mathbb {Q}}}_1^{{\bar{Y}}})+\lambda ( {{\mathbb {I}}}-{{\mathbb {Q}}}_L^{{\bar{Y}}}), \end{aligned} \end{aligned}$$

(A1)

where ${{\mathbb {I}}}_{j,j+1}$, ${{\mathbb {P}}}_{j,j+1}$, and ${{\mathbb {K}}}_{j,j+1}$ are, respectively, the identity operator, the permutation operator, and the trace operator acting on two adjacent sites $(j, j+1)$. The operator $Q^\phi _n$ is a projection operator defined as

$$\begin{aligned} \begin{aligned}&Q^\phi _n \mid \cdots {\mathop {\phi }\limits ^{{\mathop {\downarrow }\limits ^{n}}}} \cdots \rangle = 0 \\&Q^\phi _n \mid \cdots {\mathop {\psi }\limits ^{{\mathop {\downarrow }\limits ^{n}}}} \cdots \rangle = \mid \cdots {\mathop {\psi }\limits ^{{\mathop {\downarrow }\limits ^{n}}}} \cdots \rangle . \end{aligned} \end{aligned}$$

(A2)

Namely, if the field at the nth site is the same with $\phi$ in $Q^\phi _n$, the value vanishes. On the other hand, if the field is different, $Q^\phi _n$ just becomes the identity operator. One can notice that the projection operators are only related to Y-field or ${{{\bar{Y}}}}$-field.

Since the Hamiltonian is commuted with the total spin number operator, one can classify the whole Hilbert space in terms of numbers of excitations: $\mathcal{H}= \mathcal{H}_{1} \oplus \mathcal{H}_{2} \oplus \mathcal{H}_{3} \oplus \cdots \oplus \mathcal{H}_{L/2}$. Furthermore, we shall assume that the eigenstates can be written through the sum of plane waves with well-defined lattice momenta and their reflections with a negative sign of momenta. For example, the one-magnon eigenstate becomes

$$\begin{aligned} \mid \psi \rangle =\sum ^{L_{\textrm{eff}}}_{n=1_{\textrm{eff}}} (e^{ipn}+R(p)e^{-ipn})\mid n \rangle , \end{aligned}$$

(A3)

where we used $1_{\textrm{eff}}$ and $L_{\textrm{eff}}$, respectively, as the first site and the last site of the spin chain. This is because those can be differently defined for the type of excitations. It is not strange if we remind that the Y-excitation cannot be located at the first site.

Since Z and ${{{\bar{Z}}}}$ can be thought as sort of bound states, there would be in general four different scalar excitations: X, ${{{\bar{X}}}}$, Y, and ${{{\bar{Y}}}}$. However, we shall further exclude Y-excitation and ${{{\bar{Y}}}}$-excitation from our consideration, because those make interactions between D-brane parts of the determinants and strings nontrivial. For other scalar excitations, see the original paper [37]. For example, if we consider the following situation:

$$\begin{aligned} {{\mathcal {O}}}_{1}= & {} \epsilon _{i_{1}\ldots i_{N}}^{j_{1}\ldots j_{N}} Y_{j_{1}}^{i_{1}}\ldots Y_{j_{N-1}}^{i_{N-1}} \left( Z^{L_{1}-1} Y \right) _{j_{N}}^{i_{N}}, \qquad \nonumber \\ {{\mathcal {O}}}_{2}= & {} \epsilon _{i_{1}\ldots i_{N}}^{j_{1}\ldots j_{N}} {{{{\bar{Y}}}}}_{j_{1}}^{i_{1}}\ldots {{{{\bar{Y}}}}}_{j_{N-1}}^{i_{N-1}} \left( {{{\bar{Z}}}}^{L_{2}-1 }{{{\bar{Y}}}} \right) _{j_{N}}^{i_{N}}, \nonumber \\ {{\mathcal {O}}}_{3}= & {} = {\text {tr}}\left( {\tilde{Z}}^{L_3}\right) = {\text {tr}}\left( \left( Z+{{{\bar{Z}}}}+X-{{{\bar{X}}}}\right) ^{L_{3}}\right) , \end{aligned}$$

(A4)

the excitations Y and ${{{\bar{Y}}}}$ shall be contracted with ${{{\bar{Y}}}}$ and Y in the D-brane part. Thus, in our problem, it would be enough if we know the coordinate Bethe ansatz for X- and ${{{\bar{X}}}}$-excitations and the corresponding reflection factor.

Let us start with one-magnon problem. Note that there is actually no difference between X- and ${{{\bar{X}}}}$-excitations, since they are symmetric in Z-vacuum spin-chain of $Y=0$ brane. Thus, without loss of generality, we consider only X-excitation. For convenience, we further choose $\lambda =1$. We then have the much simpler Hamiltonian

$$\begin{aligned} {\mathbb {H}}=\sum _{l=1}^{L-1} (I_{l,l+1}-P_{l,l+1}). \end{aligned}$$

(A5)

If we assume the following one-magnon Bethe ansatz:

$$\begin{aligned} \mid \psi \rangle =\sum ^L_{n=1} (e^{ipn}+R(p) e^{-ipn})\mid n \rangle , \end{aligned}$$

(A6)

it is straightforward to check that the Hamiltonian can be diagonalized, such as

$$\begin{aligned} {\mathbb {H}} \mid \psi \rangle= & {} [(e^{ip}+R(p)e^{-ip})-(e^{2ip}+R(p)e^{-2ip})] \mid 1 \rangle \\{} & {} + [2(e^{2ip}+R(p)e^{-2ip})-(e^{ip}+R(p)e^{-ip})\\{} & {} -(e^{3ip}+R(p)e^{-3ip})]\mid 2 \rangle + \cdots \\{} & {} + [(e^{ipL}+R(p)e^{-ipL})-(e^{ip(L-1)}\\{} & {} +R(p)e^{-ip(L-1)})]\mid L \rangle = E\mid \psi \rangle \end{aligned}$$

if we demand the following identifications:

$$\begin{aligned} \begin{aligned}&[(e^{ip}+R(p)e^{-ip})-(e^{2ip}+R(p)e^{-2ip})]\mid 1\rangle \\&\quad =E(e^{ip}+R(p)e^{-ip})\mid 1 \rangle ,\\&\quad [2(e^{2ip}+R(p)e^{-2ip})-(e^{ip}+R(p)e^{-ip})\\&\qquad -(e^{3ip}+R(p)e^{-3ip})]\mid 2 \rangle =E(e^{2ip}+R(p)e^{-2ip})\mid 2 \rangle ,\\&\qquad \vdots \\&\quad [(e^{ipL}+R(p)e^{-ipL})-(e^{ip(L-1)}\\&\qquad +R(p)e^{-ip(L-1)})]\mid L \rangle =E(e^{ipL}+R(p)e^{-ipL})\mid L \rangle . \end{aligned} \end{aligned}$$

One can easily compute the Bethe equation, the reflection matrix and the energy eigenvalue from these equations. Let us summarize the results

$$\begin{aligned} e^{2ipL}=1, \quad R(p)=e^{ip},\quad E=4\sin ^2 \left( \frac{p}{2} \right) . \end{aligned}$$

(A7)

The two-magnon problem can be similarly studied. First, we need to remember there are only two cases: YY and ${{{\bar{Y}}}} {{{\bar{Y}}}}$. This is because $Y {{{\bar{Y}}}}$ can generate all other SO(6) ingredients, such as $X{{{\bar{X}}}}$ and $Z{{{\bar{Z}}}}$. One can start with the wavefunction such as

$$\begin{aligned} |\psi \rangle= & {} \sum _{n_{1} < n_{2} } f(n_1, n_2 ) |n_1,n_2 \rangle ,\quad \nonumber \\ |n_1, n_2 \rangle= & {} |Z \cdots Z {\mathop {X}\limits ^{{\mathop {\downarrow }\limits ^{n_{1}}}}} Z \cdots Z {\mathop {X}\limits ^{{\mathop {\downarrow }\limits ^{n_{2}}}}} Z \cdots Z \rangle . \end{aligned}$$

(A8)

We then need to solve $H |\psi \rangle = E |\psi \rangle$ with a general ansatz $f(n_1, n_2)$, such as

$$\begin{aligned} f(x_1,x_2 )= & {} A_{1} e^{i p_1 n_1 + i p_2 n_2 } + A_{2} e^{-i p_1 n_1 + i p_2 n_2 }\nonumber \\{} & {} + \cdots A_{8} e^{-i p_2 n_1 - i p_1 n_2 }. \end{aligned}$$

(A9)

It turned out that the correct wavefunction is

$$\begin{aligned} \begin{aligned} f(x_1,x_2 )&= e^{i p_1 n_1 + i p_2 n_2 } + R(p_{1})e^{-i p_1 n_1 + i p_2 n_2 } \\&\quad +R(p_{2})S(p_2,p_1 ) S(p_1, -p_2 )e^{i p_1 n_1 - i p_2 n_2 } \\&\quad + R(p_{1})R(p_{2}) S(p_2,p_1 ) S(p_1, -p_2 ) \\&\quad e^{-i p_1 n_1 - i p_2 n_2 } + S(p_2,p_1 ) e^{i p_2 n_1 + i p_1 n_2 } \\&\quad +R(p_{1}) S(p_1, -p_2 ) e^{i p_2 n_1 - i p_1 n_2 } \\&\quad + R(p_{2}) S(p_2,p_1 ) e^{-i p_2 n_1 + i p_1 n_2 } \\&\quad + R(p_1 )R(p_2 ) S(p_1, -p_2 )e^{-i p_2 n_1 - i p_1 n_2 }, \end{aligned} \end{aligned}$$

(A10)

where the boundary amplitude R(p) and the bulk scattering amplitude $S(p_1,p_2 )$ are given as

$$\begin{aligned} R(p) = e^{i p}, \qquad S(p_1,p_2 )= -\frac{1-2e^{i p_{1}}+e^{i p_{1} + i p_{2}}}{1-2e^{i p_{2}}+e^{i p_{1} + i p_{2}}}. \end{aligned}$$

(A11)

With the above wavefunction, we can really check that we solve the eigenvalue equation. The eigenvalue E is obtained as

$$\begin{aligned} E= & {} 4 - e^{i p_{1}}- e^{-i p_{1}}- e^{i p_{2}}- e^{-i p_{2}} \nonumber \\= & {} 4\sin ^2 \left( \frac{p_{1}}{2} \right) + 4\sin ^2 \left( \frac{p_{2}}{2} \right) . \end{aligned}$$

(A12)

Moreover, for consistency, the following BAEs should be satisfied:

$$\begin{aligned} e^{2 i p_{1} L}= & {} S(p_{1},p_{2}) R(p_1) S(p_{1},-p_{2}) R(-p_1 ) \end{aligned}$$

(A13)

$$\begin{aligned} e^{2 i p_{2} L}= & {} S(p_{2},p_{1}) R(p_2) S(p_{2},-p_{1}) R(-p_2 ). \end{aligned}$$

(A14)

All results for two-magnon are consistent with one-particle reduction. Although we shall not explicitly write down, the multi-magnon generalization can be straightforwardly performed.