-
In this section we review the spin chain model originated from ABJM theory [2, 3]. The quantum space in each site of the spin chain is a representation space of
$ \rm{SU(4)} $ group, alternating from fundamental representation "$ {\bf{4}} $ " to anti-fundamental one "$ \bar{{\bf{4}}} $ ".There are four kinds of R-matrices,
$ \begin{aligned} R_{ab}(u)=u\mathbb{I}_{ab}+\mathbb{P}_{ab},\quad R_{a\bar{b}}(u)=-(u+2)\mathbb{I}_{a\bar{b}}+\mathbb{K}_{a\bar{b}},\\ R_{\bar{a}\bar{b}}(u)=u\mathbb{I}_{\bar{a}\bar{b}}+\mathbb{P}_{\bar{a}\bar{b}},\quad R_{\bar{a}b}(u)=-(u+2)\mathbb{I}_{\bar{a}b}+\mathbb{K}_{\bar{a}b}, \end{aligned} $
(1) where the subscripts a(
$ \bar{a} $ ) or b($ \bar{b} $ ) indicate they belong to$ {\bf{4}} $ ($ \bar{{\bf{4}}} $ ) representation spaces, respectively.$ \mathbb{P} $ and$ \mathbb{K} $ are permutation and trace operators defined by means of the standard basis matrices$ \{e_{ij},i,j=1,2,3,4\} $ as$ \begin{aligned} \mathbb{P}=e_{ij}\otimes e_{ji},\quad \mathbb{K}=e_{ij}\otimes e_{ij}, \end{aligned} $
(2) where the repeated indices are summed implicitly. These R-matrices satisfy a total of eight Yang-Baxter equations which are expressed concisely as
$ \begin{aligned}[b] &R_{AB}(\lambda_A-\lambda_B)R_{AC}(\lambda_A-\lambda_C)R_{BC}(\lambda_B-\lambda_C)\\=\;& R_{BC}(\lambda_B-\lambda_C)R_{AC}(\lambda_A-\lambda_C)R_{AB}(\lambda_A-\lambda_B), \end{aligned} $
(3) where
$ A=\{a,\bar{a}\},\,B=\{b,\bar{b}\},\,C=\{c,\bar{c}\} $ .For the closed alternating spin chain with
$ 2L $ sites, we have the following two monodromy matrices,$ \begin{aligned} T_0(u)&=R_{01}(u)R_{0\bar{2}}(u)\cdots R_{0,2L-1}(u)R_{0,\overline{2L}}(u),\\ \bar{T}_{\bar{0}}(u)&=R_{\bar{0}1}(u)R_{\bar{0}\bar{2}}(u)\cdots R_{\bar{0},2L-1}(u)R_{\bar{0},\overline{2L}}(u), \end{aligned} $
(4) where
$ V_0 $ and$ V_{\bar{0}} $ are auxiliary spaces and the corresponding transfer matrices are$ \tau(u)= \rm{Tr}_{0}T_{0}(u) $ and$ \bar{\tau}(u)= \rm{Tr}_{\bar{0}}\bar{T}_{\bar{0}}(u) $ . Due to the Yang-Baxter relations (3), the transfer matrices commute with each other for arbitrary spectral parameters,$ \begin{aligned}[b]& \left[\tau(u),\tau(v)\right]=0,\quad \left[\bar{\tau}(u),\bar{\tau}(v)\right]=0,\\& \left[\tau(u),\bar{\tau}(v)\right]=0.\quad \forall u,v\in \mathbb{C} \end{aligned} $
(5) The Hamiltonian of the ABJM spin chain model is obtained from
$ \tau(u) $ and$ \bar{\tau}(u) $ as$ \begin{aligned} H_{\rm{ABJM}}=\frac{d}{du}\log \tau(u)\bigg|_{u=0}+\frac{d}{du}\log \bar{\tau}(u)\bigg|_{u=0}, \end{aligned} $
(6) and a direct computation gives its concrete expression,
$ \begin{aligned} H_{\rm{ABJM}}=\sum_{l=1}^{2L}\left(\mathbb{P}_{l,l+2}-\frac{1}{2}\mathbb{K}_{l,l+1}\mathbb{P}_{l,l+2}-\frac{1}{2}\mathbb{P}_{l,l+2}\mathbb{K}_{l,l+1}\right), \end{aligned} $
(7) which, up to an overall prefactor and a constant term, is exactly the anomalous dimension matrix of the dilatational operator in ABJM theory. Furthermore, we see that
$ H_{\rm{ABJM}} $ is a three-site interacting model with next-to-nearest local Hamiltonian density,$ \begin{aligned} h_{l,l+1,l+2}=\mathbb{P}_{l,l+2}-\frac{1}{2}\mathbb{K}_{l,l+1}\mathbb{P}_{l,l+2}-\frac{1}{2}\mathbb{P}_{l,l+2}\mathbb{K}_{l,l+1}. \end{aligned} $
(8) -
In this section we present the algebraic construction of the fused model for ABJM spin chain by introducing the Lax operator, the R-matrix and the boost operator.
-
In Eq.6, the Hamiltonian of ABJM spin chain is expressed as the sum of the conserved charges from two different transfer matrices. In the following, we will construct a fused model for ABJM spin chain in which the Hamiltonian
$ H_{\rm{ABJM}} $ is generated from a single transfer matrix. First we multiply the original two monodromy matrices$ T_0(u) $ and$ \bar{T}_{\bar{0}}(u) $ to get a new one$ {\bf{T}}_{0\bar{0}}(u)= T_0(u)\bar{T}_{\bar{0}}(u) $ . Then we rearrange the positions of R-matrices in$ {\bf{T}}_{0\bar{0}}(u) $ to get$ \begin{aligned}[b] {\bf{T}}_{0\bar{0}}(u)=&\left(R_{01}(u)R_{\bar{0}1}(u)R_{0\bar{2}}(u)R_{\bar{0}\bar{2}}(u)\right)\cdots\\ &\left(R_{0,2L-1}(u)R_{\bar{0},2L-1}(u)R_{0,\overline{2L}}(u)R_{\bar{0},\overline{2L}}(u)\right). \end{aligned} $
(9) Thus we find that if we treat the tensor product of two nearest quantum spaces as an enlarged new quantum space, such as
$ V_{2j-1,\overline{2j}}=V_{2j-1}\otimes V_{\overline{2j}} $ , and also introduce a new auxiliary space as$ V_{0\bar{0}}=V_0\otimes V_{\bar{0}} $ , we can define a new Lax operator on the tensor product space$ V_{0\bar{0}}\otimes V_{2j-1,\overline{2j}} $ as$ \begin{aligned} {\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)=R_{0,2j-1}(u)R_{0,\overline{2j}}(u)R_{\bar{0},2j-1}(u)R_{\bar{0},\overline{2j}}(u), \end{aligned} $
(10) or in more general form,
$ \begin{aligned} {\cal{L}}_{(a\bar{a}),(b\bar{b})}(u)=R_{ab}(u)R_{a\bar{b}}(u)R_{\bar{a}b}(u)R_{\bar{a}\bar{b}}(u), \end{aligned} $
(11) and then
$ {\bf{T}}_{0\bar{0}}(u) $ becomes$ \begin{aligned} {\bf{T}}_{0\bar{0}}(u)={\cal{L}}_{(0\bar{0}),(1\bar{2})}(u){\cal{L}}_{(0\bar{0}),(3\bar{4})}(u)\cdots {\cal{L}}_{(0\bar{0}),(2L-1,\overline{2L})}(u). \end{aligned} $
(12) Hence, we can see that
$ {\bf{T}}_{0\bar{0}}(u) $ represents a new spin chain of length L with isomorphic auxiliary space and quantum space in each site:$ \begin{aligned} V_{0\bar{0}}\cong V_{2j-1,\overline{2j}}={\bf{4}}\otimes \bar{{\bf{4}}},\quad j=1,2,\cdots,L. \end{aligned} $
(13) More importantly, the above spin chain is integrable since the following R-matrix
$ \begin{aligned} {\cal{R}}_{(a\bar{a}),(b\bar{b})}(u)=R_{a\bar{b}}(u)R_{ab}(u)R_{\bar{a}\bar{b}}(u)R_{\bar{a}b}(u) \end{aligned} $
(14) makes the "RLL" exchange relation holds:
$ \begin{aligned} {\cal{R}}_{(a\bar{a}),(b\bar{b})}(u-v){\cal{L}}_{(a\bar{a}),(c\bar{c})}(u){\cal{L}}_{(b\bar{b}),(c\bar{c})}(v)=&\\ {\cal{L}}_{(b\bar{b}),(c\bar{c})}(v){\cal{L}}_{(a\bar{a}),(c\bar{c})}(u)&{\cal{R}}_{(a\bar{a}),(b\bar{b})}(u-v). \end{aligned} $
(15) As a consistency condition, the R-matrix (14) itself should satisfy the Yang-Baxter relation:
$ \begin{aligned} {\cal{R}}_{(a\bar{a}),(b\bar{b})}(u-v){\cal{R}}_{(a\bar{a}),(c\bar{c})}(u){\cal{R}}_{(b\bar{b}),(c\bar{c})}(v)=&\\ {\cal{R}}_{(b\bar{b}),(c\bar{c})}(v){\cal{R}}_{(a\bar{a}),(c\bar{c})}(u)&{\cal{R}}_{(a\bar{a}),(b\bar{b})}(u-v). \end{aligned} $
(16) Using the Yang-Baxter equations (3), the above RLL relation (15) and Yang-Baxter relation (16) can be verified straightforwardly. Notice that the Lax operator
$ {\cal{L}}_{(a\bar{a}),(b\bar{b})}(u) $ , though quite similar to$ {\cal{R}}_{(a\bar{a}),(b\bar{b})}(u) $ , does not obey the intertwining relation:$ \begin{aligned} {\cal{L}}_{(a\bar{a}),(b\bar{b})}(u-v){\cal{L}}_{(a\bar{a}),(c\bar{c})}(u){\cal{L}}_{(b\bar{b}),(c\bar{c})}(v)\neq&\\ {\cal{L}}_{(b\bar{b}),(c\bar{c})}(v){\cal{L}}_{(a\bar{a}),(c\bar{c})}(u)&{\cal{L}}_{(a\bar{a}),(b\bar{b})}(u-v). \end{aligned} $
(17) We have so far established a new integrable model which will be called the fused model since both the auxiliary space and the quantum space are the fusion of two neighbouring representation spaces. The transfer matrix of the fused model is simply the multiplication of two original ones,
$ \begin{aligned} t(u)= \rm{Tr}_{0\bar{0}}{\bf{T}}_{0\bar{0}}(u)=\tau(u)\bar{\tau}(u), \end{aligned} $
(18) and the Hamiltonian generated from
$ t(u) $ is just$ H_{\rm{ABJM}} $ . However, from the viewpoint of the fused model,$ H_{\rm{ABJM}} $ turns out to be a nearest-neighbour interacting model$ \begin{aligned} H_{\rm{ABJM}}=\sum_{j=1}^L H_{(2j-1,\overline{2j}),(2j+1,\overline{2j+2})}, \end{aligned} $
(19) where the local Hamiltonian density is
$ \begin{aligned} H_{(2j-1,\overline{2j}),(2j+1,\overline{2j+2})}=h_{2j-1,\overline{2j},2j+1}+h_{\overline{2j},2j+1,\overline{2j+2}}. \end{aligned} $
(20) Now let us switch back to the original spin chain where the nearest two quantum spaces
$ {\bf{4}} $ and$ \bar{{\bf{4}}} $ are separate sites, then by construction the transfer matrix$ t(u) $ for fused model is two-site translation invariant. However we can easily find the Lax operator has the factorized form:$ \begin{aligned} {\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)={\cal{L}}_{(0\bar{0}),2j-1}(u){\cal{L}}_{(0\bar{0}),\overline{2j}}(u), \end{aligned} $
(21) where
$ \begin{aligned} {\cal{L}}_{(0\bar{0}),2j-1}(u)&=R_{0,2j-1}(u)R_{\bar{0},2j-1}(u),\\ {\cal{L}}_{(0\bar{0}),\overline{2j}}(u)&=R_{0,\overline{2j}}(u)R_{\bar{0},\overline{2j}}(u), \end{aligned} $
(22) and thus the transfer matrix can be written as
$ \begin{aligned} t(u)= \rm{Tr}_{0\bar{0}}{\cal{L}}_{(0\bar{0}),1}(u){\cal{L}}_{(0\bar{0}),\bar{2}}(u)\cdots {\cal{L}}_{(0\bar{0}),2L-1}(u){\cal{L}}_{(0\bar{0}),\overline{2L}}(u), \end{aligned} $
(23) which is obviously one-site shift invariant.
One last point we would like to mention: We could also use
$ {\cal{R}}_{(a\bar{a}),(b\bar{b})}(u) $ as the Lax operator to generate a new integrable spin chain, as shown below,$ \begin{aligned} \tilde{t}(u)= \rm{Tr}_{0\bar{0}}{\cal{R}}_{(0\bar{0}),(1\bar{2})}(u){\cal{R}}_{(0\bar{0}),(3\bar{4})}(u)\cdots {\cal{R}}_{(0\bar{0}),(2L-1,\overline{2L})}(u). \end{aligned} $
(24) The relation between
$ t(u) $ and$ \tilde{t}(u) $ can be found as follows: Notice that$ \begin{aligned} {\cal{R}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)=R_{2j-1,\overline{2j}}(0){\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)R^{-1}_{2j-1,\overline{2j}}(0), \end{aligned} $
(25) where the similarity transformation matrix
$ R_{2j-1,\overline{2j}}(0)= -2+\mathbb{K}_{2j-1,\overline{2j}} $ has the properties$ \begin{aligned} R_{2j-1,\overline{2j}}(0)^{T}=R_{2j-1,\overline{2j}}(0),\quad R_{2j-1,\overline{2j}}(0)^{T}R_{2j-1,\overline{2j}}(0)=4\mathbb{I}_{2j-1,\overline{2j}}, \end{aligned} $
(26) thus can be seen as a rotation in local quantum space
$ V_{2j-1,\overline{2j}}=V_{2j-1}\otimes V_{\overline{2j}} $ . Then we find$ t(u) $ and$ \tilde{t}(u) $ are related by a global rotation in the whole Hilbert space$ \otimes_{j=1}^L V_{2j-1,\overline{2j}} $ ,$ \begin{aligned} \tilde{t}(u)=\Lambda t(u) \Lambda^{-1}, \end{aligned} $
(27) where
$ \begin{aligned} \Lambda=R_{1\bar{2}}(0)R_{3\bar{4}}(0)\cdots R_{2L-1,\overline{2L}}(0), \end{aligned} $
(28) so do the Hamiltonian
$ \tilde{H}_{\rm{ABJM}} $ obtained from$ \tilde{t}(u) $ and$ H_{\rm{ABJM}} $ :$ \begin{aligned} \tilde{H}_{\rm{ABJM}}=\Lambda H_{\rm{ABJM}} \Lambda^{-1}. \end{aligned} $
(29) -
Now we proceed with an investigation of the boost operator for the fused model which can be used to generate higher conserved charges.
From the definition of the fused R-matrix in (14), we find
$ \begin{aligned} {\cal{R}}_{(a\bar{a}),(b\bar{b})}(0)=4\mathbb{P}_{ab}\mathbb{P}_{\bar{a}\bar{b}}. \end{aligned} $
(30) The R-matrix with the above condition is usually called regular. For the integrable model with regular R-matrix, the way to construct the boost operator is well-known in the literature [18] and applied to our fused model as follows: First it can be easily shown that the Lax operator is P-symmetric, that is
$ \begin{aligned} {\cal{L}}_{(a\bar{a}),(b\bar{b})}(u)={\cal{L}}_{(b\bar{b}),(a\bar{a})}(u), \end{aligned} $
(31) and thus the RLL relation in (15) can be written as
$ \begin{aligned} {\cal{R}}_{(1\bar{2}),(3\bar{4})}(v){\cal{L}}_{(0\bar{0}),(1\bar{2})}(u+v){\cal{L}}_{(0\bar{0}),(3\bar{4})}(u)\\ ={\cal{L}}_{(0\bar{0}),(3\bar{4})}(u){\cal{L}}_{(0\bar{0}),(1\bar{2})}(u+v){\cal{R}}_{(1\bar{2}),(3\bar{4})}(v). \end{aligned} $
(32) Then by taking the derivative with respect to v on both sides and setting
$ v=0 $ in the end, we get$ \begin{aligned}[b] &\dot{{\cal{R}}}_{(1\bar{2}),(3\bar{4})}(0){\cal{L}}_{(0\bar{0}),(1\bar{2})}(u){\cal{L}}_{(0\bar{0}),(3\bar{4})}(u)\\&+ {\cal{R}}_{(1\bar{2}),(3\bar{4})}(0)\dot{{\cal{L}}}_{(0\bar{0}),(1\bar{2})}(u){\cal{L}}_{(0\bar{0}),(3\bar{4})}(u) \\=\;&{\cal{L}}_{(0\bar{0}),(3\bar{4})}(u)\dot{{\cal{L}}}_{(0\bar{0}),(1\bar{2})}(u){\cal{R}}_{(1\bar{2}),(3\bar{4})}(0)\\&+ {\cal{L}}_{(0\bar{0}),(3\bar{4})}(u){\cal{L}}_{(0\bar{0}),(1\bar{2})}(u)\dot{{\cal{R}}}_{(1\bar{2}),(3\bar{4})}(0). \end{aligned} $
(33) Multiplying
$ {\cal{R}}_{(1\bar{2}),(3\bar{4})}(0) $ on both sides from the left and using the regularity condition$ {\cal{R}}_{(1\bar{2}),(3\bar{4})}(0) =4\mathbb{P}_{13}\mathbb{P}_{\bar{2}\bar{4}} $ gives$ \begin{aligned}[b] \left[\frac{1}{16}{\cal{R}}_{(1\bar{2}),(3\bar{4})}(0)\dot{{\cal{R}}}_{(1\bar{2}),(3\bar{4})}(0),{\cal{L}}_{(0\bar{0}),(1\bar{2})}(u){\cal{L}}_{(0\bar{0}),(3\bar{4})}(u)\right]\\ =-\dot{{\cal{L}}}_{(0\bar{0}),(1\bar{2})}(u){\cal{L}}_{(0\bar{0}),(3\bar{4})}(u)+{\cal{L}}_{(0\bar{0}),(1\bar{2})}(u)\dot{{\cal{L}}}_{(0\bar{0}),(3\bar{4})}(u), \end{aligned} $
(34) where
$ \dfrac{1}{16}{\cal{R}}_{(1\bar{2}),(3\bar{4})}(0)\dot{{\cal{R}}}_{1\bar{2},3\bar{4}}(0) $ is just the local Hamiltonian$ H_{(1\bar{2}),(3\bar{4})} $ plus an identity operator and thus can be replaced by$ H_{(1\bar{2}),(3\bar{4})} $ in the commutation relation. By the substitution of the indices:$ 1\rightarrow 2k-1,\, 2\rightarrow 2k,\, 3\rightarrow2k+1, \, 4\rightarrow 2k+2 $ , we have$ \begin{aligned}[b] &\left[{H}_{(2k-1,\overline{2k}),(2k+1,\overline{2k+2})},{\cal{L}}_{(0\bar{0}),(2k-1,\overline{2k})}(u){\cal{L}}_{(0\bar{0}),(2k+1,\overline{2k+2})}(u)\right] \\=\;&-\dot{{\cal{L}}}_{(0\bar{0}),{(2k-1,\overline{2k})}}(u){\cal{L}}_{(0\bar{0}),{(2k+1,\overline{2k+2})}}(u)+\\&{\cal{L}}_{(0\bar{0}),(2k-1,\overline{2k})}(u) \dot{{\cal{L}}}_{(0\bar{0}),(2k+1,\overline{2k+2})}(u). \end{aligned} $
(35) Then, by multiplying
$ \prod _{j=1}^{k-1}{\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u) $ from the left and$ \prod _{j=k+2}^{L}{\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u) $ from the right to both sides of the above equation, we get$ \begin{aligned}[b] &\left[{H}_{(2k-1,\overline{2k}),(2k+1,\overline{2k+2})},{\bf{T}}_{0\bar{0}}(u)\right]\\ =&-\left[\prod _{j=1}^{k-1}{\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)\right]\dot{{\cal{L}}}_{(0\bar{0}),{(2k-1,\overline{2k})}}(u) \left[\prod _{j=k+1}^{L}{\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)\right]\\ &+\left[\prod _{j=1}^{k}{\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)\right]\dot{{\cal{L}}}_{(0\bar{0}),{(2k+1,\overline{2k+2})}}(u)\\& \left[\prod _{j=k+2}^{L}{\cal{L}}_{(0\bar{0}),(2j-1,\overline{2j})}(u)\right]. \end{aligned} $
(36) Finally, by summing up the above equation for each
$ k=1,2,\cdots, L-1 $ , we find$ \begin{aligned}[b] &\left[\sum_{k=1}^{L-1}k{H}_{(2k-1,\overline{2k}),(2k+1,\overline{2k+2})},{{\bf{T}}}_{0\bar{0}}(u)\right]\\ =&-\frac{d{{\bf{T}}}_{0\bar{0}}(u)}{du}+L\left[\prod _{j=1}^{L-1}{\cal{L}}_{0\bar{0},(2j-1,\overline{2j})}(u)\right]\dot{{\cal{L}}}_{0\bar{0},(2L-1,\overline{2L})}(u). \end{aligned} $
(37) For an infinite spin chain or closed spin chain, we get
$ \begin{aligned} \frac{dt(u)}{du}=\left[{\cal{B}},t(u)\right], \end{aligned} $
(38) where
$ {\cal{B}} $ is the boost operator defined as$ \begin{aligned} {\cal{B}}=-\sum_k k {H}_{(2k-1,\overline{2k}),(2k+1,\overline{2k+2})}. \end{aligned} $
(39) The conserved charges are defined as the coefficients of the Taylor expansion of
$ \log t(u) $ at$ u=0 $ ,$ \begin{aligned} \log t(u)=i\sum_{n=0}Q_{n+1}u^n, \end{aligned} $
(40) then the first charge
$ Q_1 $ is simply$ -i\log t(0) $ which is non-local, since$ \begin{aligned} t(0)=4^L\mathbb{P}_{2L-3,2L-1}\cdots \mathbb{P}_{13}\mathbb{P}_{\overline{2L-2},\overline{2L}}\cdots \mathbb{P}_{\bar{2}\bar{4}} \end{aligned} $
(41) is the shift operator acting on the whole spin chain. The second charge
$ Q_2 $ corresponds to the spin chain Hamiltonian:$ \begin{aligned} iQ_2=\frac{d}{du}\log t(u)\bigg|_{u=0}=H_{\rm{ABJM}}. \end{aligned} $
(42) The rest of the higher charges can be derived by the boost operator
$ {\cal{B}} $ from the relation (38) as:$ \begin{aligned} Q_{n+1}=\frac{1}{n}\left[{\cal{B}},Q_n\right],\quad n=2,3,\cdots \end{aligned} $
(43) thus the next charge
$ Q_3 $ is found to be$ \begin{aligned}[b] 2Q_3&=-i\sum_j\left[H_{(2j-1,\overline{2j}),(2j+1,\overline{2j+2})},\,\,H_{(2j+1,\overline{2j+2}),(2j+3,\overline{2j+4})}\right]\\ &=-i\sum_j\Big\{\left[h_{\overline{2j},2j+1,\overline{2j+2}},\,\,h_{2j+1,\overline{2j+2},2j+3}+h_{\overline{2j+2},2j+3,\overline{2j+4}}\right]\\&+\Delta(j)\Big\}, \end{aligned} $
(44) where
$ \Delta(j) $ is a local operator with interaction range over five sites:$ \begin{aligned} \Delta(j)=\left[h_{2j-1,\overline{2j},2j+1},\,h_{2j+1,\overline{2j+2},2j+3}\right]. \end{aligned} $
(45) The form of the third charge
$ Q_3 $ shown above seemingly violates the generalized Reshetikhin condition proposed in [20] for integrable three-site model, in which$ \Delta(j) $ is conjectured to be a three-site operator. However, we would like the emphasize that our fused model is essentially a two-site model with auxiliary space and quantum space isomorphic to the tensor product space$ {\bf{4}}\otimes \bar{{\bf{4}}} $ , and thus even in its factorized form (23), we need two different reduced Lax operators$ {\cal{L}}_{(a\bar{a}),b}(u) $ and$ {\cal{L}}_{(a\bar{a}),\bar{b}}(u) $ , which are not regular either,$ \begin{aligned} {\cal{L}}_{(a\bar{a}),b}(0)=\mathbb{P}_{ab}(-2+\mathbb{K}_{\bar{a}b}),\quad {\cal{L}}_{(a\bar{a}),\bar{b}}(0)=(-2+\mathbb{K}_{a\bar{b}})\mathbb{P}_{\bar{a}\bar{b}}. \end{aligned} $
(46) These major differences indicate the fused model investigated in this work does not belong to the normal three-site interacting model considered in [20], and thus does not obey the conjecture.
As a final remark on the boost operator of ABJM spin chain, let us consider the possibility of the existence of boost operator in either of the two sub-chains,
$ \tau(u) $ or$ \bar{\tau}(u) $ . Suppose the boost operator we are looking for is a "matrix" type operator and is composed of the local inhomogeneous density, then we can write it in a very general form, for instance, the boost operator$ \mathfrak{b} $ for$ \tau(u) $ can be expressed as,$ \begin{aligned} \mathfrak{b}=\sum_j f(j) b_{j,j+1,\cdots,j+l-1}, \end{aligned} $
(47) where
$ f(j) $ is a function of the site position representing the inhomogeneity of the operator and$ b_{j,j+1,\cdots,j+l-1} $ is the local density with the interaction range over l sites starting at j-th site. The boost operator$ \mathfrak{b} $ should satisfy the condition:$ \begin{aligned} \dfrac{d\tau(u)}{du}=\left[\mathfrak{b},\tau(u)\right],\quad \forall u\in \mathbb{C}. \end{aligned} $
(48) At special point
$ u=0 $ , since$ \left[\tau^{-1}(0),\dot{\tau}(0)\right]=0 $ , the above equation will reduce to$ \begin{aligned} \tau^{-1}(0)\mathfrak{b}\tau(0)+\tau(0)\mathfrak{b}\tau^{-1}(0)=2\mathfrak{b}. \end{aligned} $
(49) Note that
$ \tau(0) $ is no longer a shift operator but has the following form$ \begin{aligned} \tau(0)=\left(-2+\mathbb{K}_{1\bar{2}}\right)\cdots \left(-2+\mathbb{K}_{2L-1,\overline{2L}}\right)\mathbb{P}_{2L-3,2L-1}\cdots \mathbb{P}_{35}\mathbb{P}_{13}, \end{aligned} $
(50) then it follows that
$ \tau(0) b_{j,j+1,\cdots,j+l-1}\tau^{-1}(0) $ not only shifts the sites that$ b_{j,j+1,\cdots,j+l-1} $ acts, but also increases its interaction range. As a result, the interaction range of the local density on both sides of (49) cannot be equal. So we conclude that there is no matrix type boost operator with inhomogeneous local density for integrable spin chain$ \tau(u) $ or$ \bar{\tau}(u) $ . -
In this section, we turn to the study of open spin chain Hamiltonian for fused model. First, let us review the construction of the Hamiltonian for ordinary
$ 2L $ -sites alternating spin chain with open boundaries. We need the following two so-called double row transfer matrices:$ \begin{aligned}[b]& \tau_b(u)= \rm{Tr}_0 K^+_{0}(u)T_0(u)K^-_0(u)T^{-1}_0(-u),\\& \bar{\tau}_b(u)= \rm{Tr}_{\bar{0}}\bar{K}^+_{\bar{0}}(u)\bar{T}_{\bar{0}}(u)\bar{K}_{\bar{0}}^-(u)\bar{T}^{-1}_{\bar{0}}(-u), \end{aligned} $
(51) where the symbol "b" is used to distinguish them from the closed spin chain transfer matrices defined in 4.
$ K_0^+(u)(\bar{K}^+_{\bar{0}}(u)) $ and$ K^-_0(u)(\bar{K}^-_{\bar{0}}(u)) $ are reflection matrices accounting for the left and right boundary local Hamiltonian, respectively. The two right boundary reflection matrices$ K^-_0(u) $ and$ \bar{K}^-_{\bar{0}}(u) $ satisfy four reflection equations (REs) given below:$ \begin{aligned}[b] R_{ab}(u-v)K^-_a(u)R_{ba}(u+v)K^-_b(v)=K^-_b(v)R_{ab}(u+v)K^-_a(u)R_{ba}(u-v),\\ R_{\bar{a}\bar{b}}(u-v)\bar{K}^-_{\bar{a}}(u)R_{\bar{b}\bar{a}}(u+v)\bar{K}^-_{\bar{b}}(v)=\bar{K}^-_{\bar{b}}(v)R_{\bar{a}\bar{b}}(u+v)\bar{K}^-_{\bar{a}}(u)R_{\bar{b}\bar{a}}(u-v),\\ R_{a\bar{b}}(u-v)K^-_a(u)R_{\bar{b}a}(u+v)\bar{K}^-_{\bar{b}}(v)=\bar{K}^-_{\bar{b}}(v)R_{a\bar{b}}(u+v)K^-_a(u)R_{\bar{b}a}(u-v),\\ R_{\bar{a}b}(u-v)\bar{K}^-_{\bar{a}}(u)R_{b\bar{a}}(u+v)K^-_b(v)=K^-_b(v)R_{\bar{a}b}(u+v)\bar{K}^-_{\bar{a}}(u)R_{b\bar{a}}(u-v), \end{aligned} $
(52) while the other two left boundary reflection matrices
$ K^+_0(u) $ and$ \bar{K}^+_{\bar{0}}(u) $ satisfy similar dual reflection equations and can be obtained from$ K^-_0(u) $ and$ \bar{K}^-_{\bar{0}}(u) $ by some isomorphism transformations. Due to these reflection relations, the transfer matrices form the commutating class,$ \begin{aligned}[b]& \left[\tau_b(u),\tau_b(v)\right]=0,\quad \left[\bar{\tau}_b(u),\bar{\tau}_b(v)\right]=0,\\& \left[\tau_b(u),\bar{\tau}_b(v)\right]=0,\quad \forall u,v\in \mathbb{C} \end{aligned} $
(53) showing the integrability of the open spin chain model. Then the boundary Hamiltonian from the original two open spin chains is given as
$ \begin{aligned} H_{b}=\frac{d}{du}\log \left(\tau_b(u)\bar{\tau}_b(u)\right)\bigg|_{u=0}. \end{aligned} $
(54) Now we consider the open spin chain from fused model. Since the quantum space and auxiliary space are both
$ {\bf{4}}\otimes\bar{{\bf{4}}} $ , we thus introduce two reflection matrices$ K^+_{0\bar{0}}(u) $ and$ K^-_{0\bar{0}}(u) $ defined on tensor product space$ V_0\otimes V_{\bar{0}} $ , of which$ K^-_{0\bar{0}}(u) $ satisfy the reflection equation,$ \begin{aligned}[b] &{\cal{R}}_{(a\bar{a}),(b\bar{b})}(u-v)K^-_{a\bar{a}}(u){\cal{R}}_{(b\bar{b}),(a\bar{a})}(u+v)K^-_{b\bar{b}}(v)\\=\;& K^-_{b\bar{b}}(v){\cal{R}}_{(a\bar{a}),(b\bar{b})}(u+v)K^-_{a\bar{a}}(u){\cal{R}}_{(b\bar{b}),(a\bar{a})}(u-v), \end{aligned} $
(55) and
$ K^+_{0\bar{0}}(u) $ satisfy similar dual reflection equation. The double row transfer matrix for our fused model is$ \begin{aligned} t_b(u)= \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(u){\bf{T}}_{0\bar{0}}(u)K^-_{0\bar{0}}(u){\bf{T}}^{-1}_{0\bar{0}}(-u), \end{aligned} $
(56) and the open spin chain Hamiltonian is then obtained as
$ \begin{aligned} \tilde{H}_b=\frac{d}{du}\log t_b(u)\bigg|_{u=0}. \end{aligned} $
(57) We can see that, unlike the closed spin chain case, the fused Hamiltonian
$ \tilde{H}_b $ is different from the Hamiltonian$ H_b $ composed of two original spin chains, hence representing two different open spin chain models. Though in very rare cases, with highly constrained reflection matrices, the two open spin chain models could be equivalent. For instance, if we assume that$ K^+_{0\bar{0}}(u)= K^+_0(u)\bar{K}^+_{\bar{0}}(u) $ and$ K^-_{0\bar{0}}(u)=K^-_0(u)\bar{K}^-_{\bar{0}}(u) $ , then the transfer matrix reduces to:$ \begin{aligned}[b] t_b(u)&= \rm{Tr}_{0\bar{0}} K^+_0(u)\bar{K}^+_{\bar{0}}(u)T_0(u)\bar{T}_{\bar{0}}(u)K^-_0(u)\bar{K}^-_{\bar{0}}(u)\bar{T}_{\bar{0}}^{-1}(-u)T^{-1}_0(-u)\\ &= \rm{Tr}_{0\bar{0}}K^+_0(u)T_0(u)K^-_0(u)\left[\bar{K}^+_{\bar{0}}(u)\bar{T}_{\bar{0}}(u)\bar{K}^-_{\bar{0}}(u)\bar{T}_{\bar{0}}^{-1}(-u)\right]T^{-1}_0(-u). \end{aligned} $
(58) Therefore, if we further impose the following condition
$ \begin{aligned} \left[\bar{K}^+_{\bar{0}}(u)\bar{T}_{\bar{0}}(u)\bar{K}^-_{\bar{0}}(u)\bar{T}_{\bar{0}}^{-1}(-u),\,T^{-1}_0(-u)\right]=0, \end{aligned} $
(59) then we will have
$ t_b(u)=\tau_b(u)\bar{\tau}_b(u) $ , which leads to the same Hamiltonian.We also note that, in order to deal with the open spin chain model with degrees of freedom on the boundary, the reflection K-matrices will also act on an additional internal space. By tracing over the auxiliary spaces
$ V_0\otimes V_{\bar{0}} $ , the interactions between the boundary and bulk can be achieved. Hence, in the most general settings, the reflection matrices should be treated as an operator-valued matrix on auxiliary space.Finally, let us discuss the concrete open spin chain Hamiltonian derived from the fused model transfer matrix (57). Since the whole calculation is straightforward but quite tedious, here we directly give the final results. For an open spin chain of length
$ 2L $ with internal degrees of freedom acting on the boundaries, the complete Hamiltonian$ \tilde{H}_b $ consists of three parts: the left boundary term$ H_l $ , the bulk Hamiltonian$ H_{\rm{in}} $ and the right boundary term$ H_r $ ,$ \begin{aligned} \tilde{H}_b=H_{l}+H_{\rm{in}}+H_r. \end{aligned} $
(60) The bulk part is just the ordinary closed spin chain Hamiltonian
$ \begin{aligned} H_{\rm{in}}=\sum_{l=1}^{2L-4}\left(2\mathbb{P}_{l,l+2}-\mathbb{P}_{l,l+2}\mathbb{K}_{l,l+1}-\mathbb{K}_{l,l+1}\mathbb{P}_{l,l+2}\right). \end{aligned} $
(61) The left boundary term can be further organized into three parts:
$ \begin{aligned} H_l=H_{l1}+H_{l2}+H_{l3}. \end{aligned} $
(62) Among them,
$ H_{l1} $ is a pure left boundary term, which acts trivially on the bulk Hilbert space, given as$ \begin{aligned}[b] H_{l1}=\;&\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\right]^{-1}\left[ \rm{Tr}_{0\bar{0}}\frac{dK^+_{0\bar{0}}(u)}{du}\bigg|_{u=0}\right]\\ &-\frac{1}{2}\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\right]^{-1}\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\left(-2+\mathbb{K}_{0\bar{0}}\right)\right]; \end{aligned} $
(63) $ H_{l2} $ is in fact an bulk term acting on the leftmost two sites,$ \begin{aligned} H_{l2}=-\frac{1}{2}\left(-2+\mathbb{K}_{12}\right); \end{aligned} $
(64) $ H_{l3} $ , representing the true bulk-boundary interaction, has the following form:$ \begin{aligned}[b] H_{l3}=\;&2\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\right]^{-1}\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\mathbb{P}_{01}\right]\\ &+\frac{1}{8}\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\right]^{-1}\left(-2+\mathbb{K}_{1\bar{2}}\right)\\&\times \left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\left(-2+\mathbb{K}_{0\bar{0}}\right)\mathbb{P}_{\bar{0}\bar{2}}\left(-2+\mathbb{K}_{0\bar{0}}\right)\right]\\ &\times\left(-2+\mathbb{K}_{1\bar{2}}\right). \end{aligned} $
(65) The right boundary term can be divided into two parts,
$ \begin{aligned} H_{r}=H_{r1}+H_{r2}, \end{aligned} $
(66) where
$ H_{r1} $ can be seen as the remaining part of the bulk Hamiltonian acting on the rightmost several sites, shown below,$ \begin{aligned}[b] H_{r1}= \;& \mathbb{P}_{\overline{2L-2},\overline{2L}}-\frac{1}{2}\mathbb{P}_{\overline{2L-2},\overline{2L}}\mathbb{K}_{2L-1,\overline{2L}}- \frac{1}{2}\mathbb{K}_{2L-1,\overline{2L}}\mathbb{P}_{\overline{2L-2},\overline{2L}}\\ &+\mathbb{P}_{2L - 3,2L - 1} - \frac{1}{2}\mathbb{P}_{2L-3,2L-1}\mathbb{K}_{2L-3,\overline{2L-2}}\\ &-\frac{1}{2}\mathbb{K}_{2L-3,\overline{2L-2}}\mathbb{P}_{2L-3,2L-1}-\frac{1}{4}\mathbb{K}_{2L-3,\overline{2L-2}}. \end{aligned} $
(67) $ H_{r2} $ includes the interaction between the bulk and the right boundary internal degrees of freedom. Since the expression of$ H_{r2} $ is quite lengthy, let us first define the following quantity,$ \begin{aligned} \Delta=\left(-2+\mathbb{K}_{2L-1,\overline{2L}}\right)K^-_{2L-1,\overline{2L}}(0)\left(-2+\mathbb{K}_{2L-1,\overline{2L}}\right), \end{aligned} $
(68) then
$ H_{r2} $ can be expressed in terms of following four parts,$ \begin{aligned} H_{r2}=\Delta_1+\Delta_2+\Delta_3+\Delta_4, \end{aligned} $
(69) where
$ \begin{aligned}[b] \Delta_1=&-\frac{1}{4}\Delta^{-1}\left(-2+\mathbb{K}_{2L-1,\overline{2L}}\right)\Delta ,\\ \Delta_2=&\left(-2+\mathbb{K}_{2L-1,\overline{2L}}\right)^{-1}\left[K^-_{2L-1,\overline{2L}}(0)\right]^{-1}\left[\frac{dK^-_{2L-1,\overline{2L}}(0)}{du}\bigg|_{u=0}\right]\\ &\times\left(-2+\mathbb{K}_{2L-1,\overline{2L}}\right),\\ \Delta_3=&\Delta^{-1}\left(\mathbb{P}_{2L-3,2L-1}-\frac{1}{2}\mathbb{P}_{2L-3,2L-1}\mathbb{K}_{2L-3,\overline{2L-2}}\right.\\&-\left.\frac{1}{2}\mathbb{K}_{2L-3,\overline{2L-2}}\mathbb{P}_{2L-3,2L-1} +\frac{1}{4}\mathbb{K}_{2L-3,\overline{2L-2}}\right)\Delta,\\ \Delta_4=&\Delta^{-1}\left(\mathbb{P}_{\overline{2L-2},\overline{2L}}-\frac{1}{2}\mathbb{P}_{\overline{2L-2},\overline{2L}}\mathbb{K}_{2L-1,\overline{2L}}- \frac{1}{2}\mathbb{K}_{2L-1,\overline{2L}}\mathbb{P}_{\overline{2L-2},\overline{2L}}\right.\\ &\left.+\frac{1}{4}\mathbb{K}_{2L-1,\overline{2L}}\right)\Delta. \end{aligned} $
(70) As shown above, the boundary Hamiltonians have very complicated forms, and by choosing different reflection K-matrices, we will have various boundary terms. However, we can find some common structures by analyzing the indices of the components of the boundary Hamiltonian. To be concrete, let us focus on the left boundary, especially the nontrivial bulk-boundary interaction term
$ H_{l3} $ , which acts on the boundary internal space$ V_{\rm{in}} $ and two leftmost quantum spaces$ V_1 $ and$ V_{\bar{2}} $ ,$ \begin{aligned} H_{l3}\in \rm{End}\left(V_{in}\otimes V_1\otimes V_{\bar{2}}\right). \end{aligned} $
(71) Note that the internal space
$ V_{\rm{in}} $ needn't to be isomorphic to$ V_1 $ or$ V_{\bar{2}} $ and thus can have different dimension. Now we look at the component of$ H_{l3} $ :$ \left[H_{l3}\right]_{i,J_1,I_2}^{j,I_1,J_2} $ , where$ \{i,j\}\in V_{\rm{in}},\,\{I_1,J_1\}\in V_1,\{I_2,J_2\}\in V_{\bar{2}} $ . There are several terms in$ H_{l3} $ and we will discuss them separately. First, for notational convenience, let us write$ \begin{aligned} B&=\left[ \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\right]^{-1}\in \rm{End} (V_{in}),\\ M&= \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\mathbb{P}_{01} \in\rm{End} (V_{in}\otimes V_1),\\ S&= \rm{Tr}_{0\bar{0}}K^+_{0\bar{0}}(0)\left(-2+\mathbb{K}_{0\bar{0}}\right)\mathbb{P}_{\bar{0}\bar{2}}\left(-2+\mathbb{K}_{0\bar{0}}\right) \in\rm{End} (V_{in}\otimes V_{\bar{2}}), \end{aligned} $
(72) then the component of each term in
$ H_{l3} $ reads:$ \begin{aligned}[b] \left(BM\right)_{i,j_1,I_2}^{j,I_1,J_2}&=\left(BM\right)_{i,j_1}^{j,I_1}\cdot \delta_{I_2}^{J_2},\\ \left(BS\right)_{i,j_1,i_2}^{j,i_1,J_2}&=\left(BS\right)_{i,I_2}^{j,J_2}\cdot \delta^{I_1}_{J_1},\\ \left(B\mathbb{K}_{1\bar{2}}S\right)_{i,J_1,I_2}^{j,I_1,J_2}&=\left(B_i^k S_{k,J_1}^{j,J_2}\right)\cdot \delta^{I_1}_{I_2},\\ \left(BS\mathbb{K}_{1\bar{2}}\right)_{i,J_1,I_2}^{j,I_1,J_2}&=\left(B_i^k S_{k,I_2}^{j,I_1}\right)\cdot \delta^{J_2}_{J_1},\\ \left(B\mathbb{K}_{1\bar{2}}S\mathbb{K}_{1\bar{2}}\right)_{i,J_1,I_2}^{j,I_1,J_2}&= \left(B_i^k S_{k,L}^{j,L}\right)\cdot \delta^{I_1}_{I_2}\delta^{J_2}_{J_1}, \end{aligned} $
(73) where we have used the component forms of
$ \mathbb{K} $ :$ \mathbb{K}_{J_1,I_2}^{I_1,J_2}=\delta^{I_1}_{I_2}\delta_{J_1}^{J_2} $ . For the other two left boundary terms, we can easily find$ \begin{aligned}[b] \left(H_{l1}\right)_{i,J_1,I_2}^{j,I_1,J_2}&=\left(H_{l1}\right)_i^j\cdot \delta^{I_1}_{J_1}\delta_{I_2}^{J_2},\\ \left(H_{l2}\right)_{i,J_1,I_2}^{j,I_1,J_2}&=\left(H_{l2}\right)^{I_1,J_2}_{J_1,I_2}\cdot \delta_i^j. \end{aligned} $
(74) Thus we find, for each of the left boundary terms, there exists a unique universal Kronecker delta factor, independent of the specific selection of the K-matrix. In other words, given an open spin chain Hamiltonian, if the components of the left boundary terms do not belong to the boundary types shown above, then such an open spin chain cannot be an integrable spin chain, at least not one originated from our fused model.
For a concrete example, we may consider the open spin chain Hamiltonian from flavored ABJM theory [13]. Due to the coupling between the bulk bi-fundamental fields and the boundary fundamental flavors, the bulk
$ SU(4) $ R-symmetry will break into a remaining$ SU(2)_R $ and a diagonal subgroup$ SU(2)_D $ :$ \begin{aligned} SU(4)_R \rightarrow SU(2)_R\times SU(2)_D \end{aligned} $
(75) In this case, the boundary internal space
$ V_{\rm{in}} $ is the two dimensional fundamental representation space of$ SU(2)_R $ , i.e.$ V_{\rm{in}}={\bf{2}}\cong \bar{{\bf{2}}} $ ,$ i,j\in\left\{1,2\right\} $ ; while$ V_{1} $ and$ V_{\bar{2}} $ are the four-dimensional representation spaces$ {\bf{2}}\times {\bf{2}} $ and$ \bar{{\bf{2}}}\times \bar{{\bf{2}}} $ of$ SU(2)_R\times SU(2)_D $ , respectively, whose component indices can be formulated by a pair of$ SU(2) $ indices, i.e.$ I=iA,J=jB, \,i,j\in\{1,2\},\,A,B\in\{1,2\} $ , and thus the delta function is simply given as:$ \delta^I_J=\delta^i_j\delta^A_B $ . Then it is a simple task to rewrite the boundary terms in (73) and (74) using the composite$ SU(2)_R\times SU(2)_D $ indices to replace$ SU(4)_R $ indices, i.e.$ I\rightarrow (iA) $ . The Hamiltonian of flavored ABJM spin chain has the following three types of left boundary terms [13, 22]:$ \begin{aligned} {\rm{type}}\,1: \qquad\quad\quad \delta^{A_1}_{A_2}\delta^{B_2}_{B_1}\delta^{j_2}_{i}\delta^{i_1}_{j_1}\delta^j_{i_2},\\ {\rm{type}}\,2: \qquad\quad\quad \delta^{A_1}_{B_1}\delta^{B_2}_{A_2}\delta^{j_2}_{i}\delta^{i_1}_{j_1}\delta^j_{i_2},\\ {\rm{type}}\,3: \qquad\quad\quad \delta^{A_1}_{B_1}\delta^{B_2}_{A_2}\delta^{i_1}_{i}\delta^{j}_{j_1}\delta^{j_2}_{i_2}. \end{aligned} $
(76) We can easily observe that the type 2 and type 3 terms have the factors
$ \delta^{I_{1}}_{J_1}=\delta^{A_1}_{B_1}\delta^{i_1}_{j_1} $ and$ \delta_{I_2}^{J_2}=\delta^{B_2}_{A_2}\delta^{j_2}_{i_2} $ , respectively, thus could be obtained from$ BM $ and$ BS $ terms in (73). As for type 1 term, it mixes the indices of all three spaces$ V_{\rm{in}}\otimes V_1\otimes V_{\bar{2}} $ and does not belong to any type of boundary terms in (73) and (74). So we see that the flavored ABJM spin chain cannot be generated from our fused model. Besides, by the same argument, we find the flavored ABJM spin chain cannot be obtained from another integrable boundary model$ H_b $ (54) either.
The Fused Model of Alternating Spin Chain from ABJM Theory
- Received Date: 2024-12-05
- Available Online: 2025-01-01
Abstract: In this paper we give an algebraic construction of the fused model for ABJM spin chain and find the corresponding boost operator. We also investigate the open spin Hamiltonian for fused model and point out the general common structures of the boundary terms.