1. Introduction
Quantum integrable system is an important topic in the modern research of mathematical physics, theoretical physics and condensed matter physics [1, 2]. The corresponding exact solutions can supply the benchmarks for the new numerical simulation methods and new physical concepts. Bethe ansatz is one of the most powerful methods to obtain the exact solutions of one-dimensional quantum many-body problems [3, 4].
The Heisenberg model is a typical quantum integrable system and has many applications in the theory of quantum magnetism and critical phenomena. By using the coordinate Bethe ansatz [5] or the quantum inverse scattering method [6–10], the Heisenberg model has been solved exactly. Based on the exact solutions, many interesting physical pictures such as the spinon and fractional charge have been found [11]. The Heisenberg model includes the nearest neighbor (NN) spin-exchanging interaction (i.e. the ${J}_{1}\mbox{--}{J}_{2}$ model [12]). Later, it was found that the next nearest neighbor (NNN) interaction in the strong correlated systems can also induce some new physics. Due to the competition between NN and NNN interactions, many novel magnetic ordered states and the Kosterlitz–Thouless-like quantum phase transitions arise in this kind of systems. We note that the ${J}_{1}\mbox{--}{J}_{2}$ model can not be solved by the Bethe ansatz. The chiral three-spin coupling is another typical spin-spin interaction which is used to describe the spin liquid phases in some low-dimensional systems [12–15].
The Yang–Baxter equation (YBE) plays a crucial role in the framework of Bethe ansatz [16, 17]. For the periodic boundary condition, the YBE can ensure the integrability of the system. While for the integrable open boundary conditions, besides the YBE, we also need the reflection equation, which can help us to prove the integrability and to obtain the exact solutions [18–20]. Recently, the integrable open boundary problems have attracted a lot of attentions because they can provide the believable physical mechanisms such as boundary bound states, topological order, edge state, surface energy and helical spinon excitations [21–24]. The corresponding results also have many important applications in modern theoretical physics such as open string theory [25] and non-equilibrium statistical physics [26, 27].
In this paper, we propose an exactly solvable quantum spin chain which has the NN, NNN and chiral three-spin couplings with the open boundary condition. We find that due to the existence of boundary fields, the spins at boundaries are polarized. The boundary fields can enhance the anisotropy of NN spin-spin interactions of the first and last bonds. We also find that the boundary fields can induce the Dzyloshinsky–Moriya (DM) interactions [28–30] at the boundaries. The DM interaction could lead to some interesting elementary excitations. In this system, the boundary fields are parallel thus the U(1) symmetry is held. By means of the Bethe ansatz, we obtain the exact solution of the system. The energy spectrum and the associated Bethe ansatz equations (BAEs) are given explicitly.
The paper is organized as follows. The model Hamiltonian is explained and its integrability of the system is proved in section 2 . In section 3 , using the algebraic Bethe ansatz method, we construct the corresponding Bethe states and the corresponding eigenvalue of the transfer matrix. The associated BAEs are also given. Section 4 is attributed to the concluding remarks.
2. Integrable Hamiltonian
The exactly solvable model studied in this paper is described by the Hamiltonian
$\begin{eqnarray}\begin{array}{rcl}H & = & \sum _{j=1}^{2N-1}\left\{{J}_{1}\left({\sigma }_{j}^{x}{\sigma }_{j+1}^{x}+{\sigma }_{j}^{y}{\sigma }_{j+1}^{y}\right)\right.\\ & & +{J}_{1}^{z}{\sigma }_{j}^{z}{\sigma }_{j\,+\,1}^{z}+{J}_{2}{\vec{\sigma }}_{j}\cdot {\vec{\sigma }}_{j+2}\\ & & +{J}_{3}{\vec{\sigma }}_{j+1}\cdot ({\vec{\sigma }}_{j}\times {\vec{\sigma }}_{j+2})+{J}_{3}^{z}{\sigma }_{j+1}^{z}\\ & & \left.\times ({\sigma }_{j}^{x}{\sigma }_{j+2}^{y}-{\sigma }_{j}^{y}{\sigma }_{j+2}^{x})\right\}\\ & & +{D}_{L}^{z}{\vec{n}}_{z}\cdot ({\vec{\sigma }}_{1}\times {\vec{\sigma }}_{2})+{D}_{R}^{z}{\vec{n}}_{z}\cdot ({\vec{\sigma }}_{2N-1}\times {\vec{\sigma }}_{2N})\\ & & +{h}_{L}{\sigma }_{1}^{z}+{h}_{R}{\sigma }_{2N}^{z}.\end{array}\end{eqnarray}$
Here ${\sigma }_{j}^{\alpha }$ $(\alpha =x,y,z)$ is the Pauli matrix along the α-direction on site j, $2N$ is the total number of sites and ${\vec{n}}_{z}$ is the unit vector along the z-direction. The interactions in the bulk include the anisotropic NN spin-exchanging, isotropic NNN interactions, and anisotropic chiral three-spin couplings. The boundary effects contain the boundary magnetic fields and DM interactions.Let us explain the Hamiltonian (2.1 ) more clearly. The boundary condition of the system is the open one, which is achieved by adding two magnetic fields at the boundaries. We consider the case that two boundary fields are parallel. Choose the direction of magnetic fields as the z-direction and denote the strengths as
$\begin{eqnarray}{h}_{L}=\displaystyle \frac{\sinh (2{\xi }_{-})\left[\cosh (4a)-\cosh (2\eta )\right]}{2\sinh \eta \left[\cosh (2a)-\cosh (2{\xi }_{-})\right]},\quad {h}_{R}={h}_{L}{| }_{{\xi }_{-}\to {\xi }_{+}},\end{eqnarray}$
where a is the model parameter, η is the anisotropic parameter, ${\xi }_{-}$ and ${\xi }_{+}$ are the boundary parameters quantifying the strengths of fields. Thus the boundary fields do not break the U(1) symmetry of the system.Due to the existence of magnetic fields, the spins at the boundaries are polarized, which leads to the anisotropy of two boundary bonds are enhanced. Then the NN spin-exchanging interactions read
$\begin{eqnarray}\begin{array}{rcl}{J}_{1} & = & \cosh (2a)+J{\delta }_{j,1}+J{| }_{{\xi }_{-}\to {\xi }_{+}}{\delta }_{j,2N-1},\\ {J}_{1}^{z} & = & \cosh \eta +{J}_{2}({\delta }_{j,1}+{\delta }_{j,2N-1}),\\ J & = & \displaystyle \frac{{\sinh }^{2}(2a)\left[\cosh (2a+2{\xi }_{-})+\cosh (2a-2{\xi }_{-})-1-\cosh (2\eta )\right]}{4{\sinh }^{2}\eta \left[\cosh (2a)-\cosh (2{\xi }_{-})\right]},\\ {J}_{2} & = & -\displaystyle \frac{{\sinh }^{2}(2a)\cosh \eta }{2{\sinh }^{2}\eta }.\end{array}\end{eqnarray}$
From above equation, we see that the NN couplings in the bulk keep unchanged except for the two boundary bonds. Meanwhile, because the NNN interactions are involved, the boundary fields can also induce the DM interactions of two boundary bonds along the z-direction and the corresponding strengths are $\begin{eqnarray}\begin{array}{rcl}{D}_{L}^{z} & = & -\displaystyle \frac{{\rm{i}}\left[{\rm{\cosh }}(4a)-{\rm{\cosh }}(2\eta )\right]\sinh (2a)\sinh (2{\xi }_{-})}{4{\sinh }^{2}\eta \left[{\rm{\cosh }}(2a)-{\rm{\cosh }}(2{\xi }_{-})\right]},\\ {D}_{R}^{z} & = & -{D}_{L}^{z}{| }_{{\xi }_{-}\to {\xi }_{+}}.\end{array}\end{eqnarray}$
The integrability of the system requires that the NNN spin-exchanging is isotropic and the strength is measured by J2. The chiral three-spin couplings are anisotropic and the corresponding strengths are2.1 ) requires that a is real if η, ${\xi }_{+}$ and ${\xi }_{-}$ are pure imaginary, while a is pure imaginary if η, ${\xi }_{+}$ and ${\xi }_{-}$ are real.
$\begin{eqnarray}\begin{array}{rcl}{J}_{3} & = & \displaystyle \frac{{(-1)}^{j}{\rm{i}}\sinh (2a)\cosh \eta }{2\sinh \eta },\\ {J}_{3}^{z} & = & {J}_{3}\displaystyle \frac{\cosh (2a)-\cosh \eta }{\cosh \eta }.\end{array}\end{eqnarray}$
It is clear that the chiral three-spin interactions are the staggered ones. Please note that the convention ${\vec{\sigma }}_{2N+1}=0$ has been used. The Hermitian of Hamiltonian (Next, we should show how to construct the integrable Hamiltonian (2.1 ). The integrability of the system is related with following R-matrix2.6 ) is defined in the direct product spaces ${V}_{0}\otimes {V}_{j}$ and has properties2.6 ) also satisfies the YBE
$\begin{eqnarray}\begin{array}{l}{R}_{0,j}(u)=\displaystyle \frac{1}{2}\left[\displaystyle \frac{\sinh (u+\eta )}{\sinh \eta }(1+{\sigma }_{j}^{z}{\sigma }_{0}^{z})\right.\\ \left.+\displaystyle \frac{\sinh u}{\sinh \eta }(1-{\sigma }_{j}^{z}{\sigma }_{0}^{z})\right]\\ +\displaystyle \frac{1}{2}({\sigma }_{j}^{x}{\sigma }_{0}^{x}+{\sigma }_{j}^{y}{\sigma }_{0}^{y})\\ =\displaystyle \frac{1}{\sinh \eta }\left(\begin{array}{cccc}\sinh (u+\eta ) & 0 & 0 & 0\\ 0 & \sinh u & \sinh \eta & 0\\ 0 & \sinh \eta & \sinh u & 0\\ 0 & 0 & 0 & \sinh (u+\eta )\end{array}\right),\end{array}\end{eqnarray}$
where u is the spectral parameter. The R-matrix ( $\begin{eqnarray}\begin{array}{l}{\rm{Initial}}\,{\rm{condition}}:\,{R}_{0,j}(0)={P}_{0,j},\\ {\rm{Unitary}}\,{\rm{relation}}:\,{R}_{0,j}(u){R}_{j,0}(-u)=\rho (u)\times {\rm{i}}{d},\\ {\rm{Crossing}}\,{\rm{relation}}:\,{R}_{0,j}(u)=-{\sigma }_{0}^{y}{R}_{0,j}^{{t}_{0}}(-u-\eta ){\sigma }_{0}^{y},\\ {\rm{PT-symmetry}}:\,{R}_{0,j}(u)={R}_{j,0}(u)={R}_{0,j}^{{t}_{0}{t}_{j}}(u),\end{array}\end{eqnarray}$
where $\rho (u)=-\sinh (u+\eta )\sinh (u-\eta ){\sinh }^{-2}\eta $ and ${P}_{\mathrm{1,2}}$ is the permutation operator possessing the property ${R}_{\mathrm{1,2}}(u)\,={P}_{\mathrm{0,1}}{R}_{\mathrm{0,2}}(u){P}_{\mathrm{0,1}}$. The R-matrix ( $\begin{eqnarray}\begin{array}{l}{R}_{\mathrm{1,2}}(u-v){R}_{\mathrm{1,3}}(u){R}_{\mathrm{2,3}}(v)\\ \quad =\ {R}_{\mathrm{2,3}}(v){R}_{\mathrm{1,3}}(u){R}_{\mathrm{1,2}}(u-v).\end{array}\end{eqnarray}$
Because we consider the integrable open boundary condition, we should introduce the reflection matrix ${K}^{-}(u)$ at one side [20, 31–33]
$\begin{eqnarray}\begin{array}{rcl}{K}^{-}(u) & = & \left(\begin{array}{cc}\sinh (u+{\xi }_{-}) & 0\\ 0 & -\sinh (u-{\xi }_{-})\end{array}\right)\\ & = & \left(\begin{array}{cc}{k}_{11}^{-}(u) & 0\\ 0 & {k}_{22}^{-}(u)\end{array}\right),\end{array}\end{eqnarray}$
which satisfies the reflection equation $\begin{eqnarray}\begin{array}{l}{R}_{\mathrm{1,2}}(u-v){K}_{1}^{-}(u){R}_{\mathrm{2,1}}(u+v){K}_{2}^{-}(v)\\ \quad =\ {K}_{2}^{-}(v){R}_{\mathrm{1,2}}(u+v){K}_{1}^{-}(u){R}_{\mathrm{2,1}}(u-v).\end{array}\end{eqnarray}$
The boundary magnetic field at the other side is quantified by the dual reflection matrix ${K}^{+}(u)$ with the form of $\begin{eqnarray}\begin{array}{rcl}{K}^{+}(u) & = & \left(\begin{array}{cc}\sinh (u+\eta +{\xi }_{+}) & 0\\ 0 & -\sinh (u+\eta -{\xi }_{+})\end{array}\right)\\ & = & \left(\begin{array}{cc}{k}_{11}^{+}(u) & 0\\ 0 & {k}_{22}^{+}(u)\end{array}\right),\end{array}\end{eqnarray}$
which satisfies the dual reflection equation $\begin{eqnarray}\begin{array}{l}{R}_{\mathrm{1,2}}(-u+v){K}_{1}^{+}(u){R}_{\mathrm{2,1}}(-u-v-2\eta ){K}_{2}^{+}(v)\\ \quad =\ {K}_{2}^{+}(v){R}_{\mathrm{1,2}}(-u-v-2\eta ){K}_{1}^{+}(u){R}_{\mathrm{2,1}}(-u+v).\end{array}\end{eqnarray}$
Starting from R-matrix (2.6 ), we construct the single-row monodromy matrix ${T}_{0}(u)$ and the reflecting one ${\hat{T}}_{0}(u)$ as2.7 ), we can prove that the transfer matrix t(u) has the crossing relation2.8 ), reflection equation (2.10 ) and the dual one (2.12 ), we find that the transfer matrices with different spectral parameter commutate with each other, i.e.2.1 ) is generated from the transfer matrix t(u) as
$\begin{eqnarray}\begin{array}{rcl}{T}_{0}(u) & = & {R}_{\mathrm{0,2}N}(u+a){R}_{\mathrm{0,2}N-1}(u-a)\\ & & \cdots {R}_{\mathrm{0,2}}(u+a){R}_{\mathrm{0,1}}(u-a),\\ {\hat{T}}_{0}(u) & = & {R}_{\mathrm{0,1}}(u+a){R}_{\mathrm{0,2}}(u-a)\\ & & \cdots {R}_{\mathrm{0,2}N-1}(u+a){R}_{\mathrm{0,2}N}(u-a),\end{array}\end{eqnarray}$
where the subscript 0 means the auxiliary space and the subscripts $\left\{1,2,\cdots ,2N\right\}$ denote the physical or quantum spaces. From the reflection matrix and the single-row monodromy matrices, we define the double-row monodromy matrix ${U}_{0}(u)$ as $\begin{eqnarray}{U}_{0}(u)={T}_{0}(u){K}_{0}^{-}(u){\hat{T}}_{0}(u),\end{eqnarray}$
which satisfies the reflection equation $\begin{eqnarray}\begin{array}{l}{R}_{\mathrm{1,2}}(u-v){U}_{1}(u){R}_{\mathrm{1,2}}(u+v){U}_{2}(v)\\ \quad =\,{U}_{2}(v){R}_{\mathrm{1,2}}(u+v){U}_{1}(u){R}_{\mathrm{1,2}}(u-v).\end{array}\end{eqnarray}$
By using the dual reflection matrix and the double-row monodromy matrix, we obtain the transfer matrix of the system $\begin{eqnarray}t(u)={{tr}}_{0}\left\{{K}_{0}^{+}(u){U}_{0}(u)\right\},\end{eqnarray}$
where tr0 means the partial trace in the auxiliary space. With the help of crossing symmetry ( $\begin{eqnarray}t(u)=t(-u-\eta ).\end{eqnarray}$
From the YBE ( $\begin{eqnarray}\left[t(u),t(v)\right]=0.\end{eqnarray}$
Thus the transfer matrix t(u) can generate all the conversed quantities and the system is integrable. The Hamiltonian ( $\begin{eqnarray}\begin{array}{l}H=\displaystyle \frac{{\sinh }^{3}\eta }{2{\rho }^{2N-2}(2a)\phi }\\ \left\{t(-a)\displaystyle \frac{\partial t(u)}{\partial u}{\left|{}_{u=a}+t(a)\displaystyle \frac{\partial t(u)}{\partial u}\right|}_{u=-a}\right\}-{l}_{0},\end{array}\end{eqnarray}$
where φ and l0 are the constants with the definitions $\begin{eqnarray}\begin{array}{l}\phi =-\sinh (2a+2\eta )\sinh (2a-2\eta )\sinh (a+{\xi }_{-})\\ \sinh (a-{\xi }_{-})\sinh (a+{\xi }_{+})\sinh (a-{\xi }_{+}),\\ {l}_{0}=-\displaystyle \frac{\cosh \eta }{4{\sinh }^{2}\eta \left[\cosh (4a)-\cosh (4\eta )\right]}\\ \left[-2-2\cosh (4a)+2\cosh (8a)+\cosh (4a-2\eta )\right.\\ -4\cosh (4a-4\eta )+9\cosh (2\eta )-6\cosh (4\eta )\\ +5\cosh (6\eta )-4\cosh (4a+4\eta )\\ \left.+\cosh (4a+2\eta )\right]-\displaystyle \frac{(N-2)\cosh \eta }{2{\sinh }^{2}\eta }\\ \left[1+\cosh (4a)-2\cosh (2\eta )\right].\end{array}\end{eqnarray}$
3. Exact solution
Now, we solve the Hamiltonian (2.1 ) by using the algebraic Bethe ansatz. The matrix form of single-row monodromy matrix ${T}_{0}(u)$ in the auxiliary space is2.7 ), the reflection single-row monodromy matrix reads
$\begin{eqnarray}{T}_{0}(u)=\left(\begin{array}{cc}A(u) & B(u)\\ C(u) & D(u)\end{array}\right),\end{eqnarray}$
where A(u), B(u), C(u) and D(u) are the operators defined in the quantum spaces. From the crossing symmetry ( $\begin{eqnarray}\begin{array}{rcl}{\hat{T}}_{0}(u) & = & \left(\begin{array}{cc}\hat{A}(u) & \hat{B}(u)\\ \hat{C}(u) & \hat{D}(u)\end{array}\right)\\ & = & \left(\begin{array}{cc}D(-u-\eta ) & -B(-u-\eta )\\ -C(-u-\eta ) & A(-u-\eta )\end{array}\right).\end{array}\end{eqnarray}$
Then we arrive at the matrix form of double-row monodromy matrix ${U}_{0}(u)$ $\begin{eqnarray}{U}_{0}(u)=\left(\begin{array}{cc}{ \mathcal A }(u) & { \mathcal B }(u)\\ { \mathcal C }(u) & { \mathcal D }(u)\end{array}\right).\end{eqnarray}$
We choose the vacuum state of the system (2.1 ) as2.8 ), we can prove that the following relation holds3.2 ), (3.5 ) and (3.7 ), we obtain the results of elements of double-row monodromy matrix ${U}_{0}(u)$ acting on the vacuum state
$\begin{eqnarray}| 0\rangle =| \uparrow {\rangle }_{1}\otimes | \uparrow {\rangle }_{2}\otimes \cdots \otimes | \uparrow {\rangle }_{2N-1}\otimes | \uparrow {\rangle }_{2N}.\end{eqnarray}$
The elements of monodromy matrix ${T}_{0}(u)$ acting on the vacuum state give $\begin{eqnarray}\begin{array}{rcl}A(u)| 0\rangle & = & \displaystyle \frac{{\sinh }^{N}(u+a+\eta ){\sinh }^{N}(u-a+\eta )}{{\sinh }^{2N}\eta }| 0\rangle ,\\ D(u)| 0\rangle & = & \displaystyle \frac{{\sinh }^{N}(u+a){\sinh }^{N}(u-a)}{{\sinh }^{2N}\eta }| 0\rangle ,\\ B(u)| 0\rangle & \ne & 0,\quad C(u)| 0\rangle =0.\end{array}\end{eqnarray}$
From the YBE ( $\begin{eqnarray}{T}_{1}(u){R}_{\mathrm{1,2}}(u+v){\hat{T}}_{2}(v)={\hat{T}}_{2}(v){R}_{\mathrm{1,2}}(u+v){T}_{0}(u).\end{eqnarray}$
Based on it, we obtain the commutative relation $\begin{eqnarray}\begin{array}{l}C(u)\hat{B}(v)=\displaystyle \frac{\sinh \eta }{\sinh (u+v+\eta )}\\ \quad \left[\hat{A}(v)A(u)-D(u)\hat{D}(v)\right]+\hat{B}(v)C(u).\end{array}\end{eqnarray}$
By using equations ( $\begin{eqnarray}\begin{array}{rcl}{ \mathcal A }(u)| 0\rangle & = & {k}_{11}^{-}(u)a(u)| 0\rangle ,\\ a(u) & = & \displaystyle \frac{{\sinh }^{2N}(u+a+\eta ){\sinh }^{2N}(u-a+\eta )}{{\sinh }^{4N}\eta },\\ { \mathcal D }(u)| 0\rangle & = & \displaystyle \frac{\sinh \eta {k}_{11}^{-}(u)}{\sinh (2u+\eta )}a(u)| 0\rangle \\ & & +\left[{k}_{22}^{-}(u)-\displaystyle \frac{{k}_{11}^{-}(u)\sinh \eta }{\sinh (2u+\eta )}\right]d(u)| 0\rangle ,\\ d(u) & = & a(u-\eta ),\\ & & { \mathcal B }(u)| 0\rangle \ne 0,\quad { \mathcal C }(u)| 0\rangle =0.\end{array}\end{eqnarray}$
We see that the vacuum state is the eigenstate of operators ${ \mathcal A }(u)$ and ${ \mathcal D }(u)$. The operator ${ \mathcal C }(u)$ acting on the vacuum gives zero and the operator ${ \mathcal B }(u)$ acting on the vacuum state generates other states thus can be regarded as the creation operator of eigenstates of the system. Assume the eigenstates are $\begin{eqnarray}| {\rm{\Phi }}\rangle ={ \mathcal B }({\mu }_{1}){ \mathcal B }({\mu }_{2})\cdots { \mathcal B }({\mu }_{M})| 0\rangle ,\end{eqnarray}$
where $\left\{{\mu }_{j}| j=1,2,\cdots ,M\right\}$ are the Bethe roots which will be determined later.According to equations (2.11 ) and (3.3 ), the transfer matrix can be expressed as
$\begin{eqnarray}t(u)={k}_{11}^{+}(u){ \mathcal A }(u)+{k}_{22}^{+}(u){ \mathcal D }(u).\end{eqnarray}$
If we define $\begin{eqnarray}\tilde{{ \mathcal D }}(u)={ \mathcal D }(u)-\displaystyle \frac{\sinh \eta }{\sinh (2u+\eta )}{ \mathcal A }(u),\end{eqnarray}$
the transfer matrix is rewritten as $\begin{eqnarray}t(u)=\left[{k}_{11}^{+}(u)+\displaystyle \frac{\sinh \eta {k}_{22}^{+}(u)}{\sinh (2u+\eta )}\right]{ \mathcal A }(u)+{k}_{22}^{+}\tilde{{ \mathcal D }}(u).\end{eqnarray}$
Next, we should act the transfer matrix t(u) on the assumed eigenstates (3.9 ). In order to obtain the detailed results, it is clear that we should exchange the orders between ${ \mathcal A }(u)$, $\tilde{{ \mathcal D }}(u)$ and ${ \mathcal B }({\mu }_{j})$. From the reflection equation (2.15 ), we have3.9 ) indeed are the eigenstates of the transfer matrix t(u). Because the system only has the U(1) symmetry, the number of Bethe roots could be $M\,=1,2,\cdots ,2N$.
$\begin{eqnarray}\begin{array}{rcl}{ \mathcal A }(u){ \mathcal B }(v) & = & \displaystyle \frac{\sinh (u+v)\sinh (u-v-\eta )}{\sinh (u-v)\sinh (u+v+\eta )}{ \mathcal B }(v){ \mathcal A }(u)\\ & & +\displaystyle \frac{\sinh (2v)\sinh \eta }{\sinh (u-v)\sinh (2v+\eta )}{ \mathcal B }(u){ \mathcal A }(v)\\ & & -\displaystyle \frac{\sinh \eta }{\sinh (u+v+\eta )}{ \mathcal B }(u)\tilde{{ \mathcal D }}(v),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\tilde{{ \mathcal D }}(u){ \mathcal B }(v)=\displaystyle \frac{\sinh (u-v+\eta )\sinh (u+v+2\eta )}{\sinh (u-v)\sinh (u+v+\eta )}{ \mathcal B }(v)\tilde{{ \mathcal D }}(u)\\ \quad -\displaystyle \frac{\sinh (2v)\sinh (2u+2\eta )\sinh \eta }{\sinh (2u+\eta )\sinh (2v+\eta )\sinh (u+v+\eta )}{ \mathcal B }(u){ \mathcal A }(v)\\ \quad -\displaystyle \frac{\sinh (2u+2\eta )\sinh \eta }{\sinh (u-v)\sinh (2u+\eta )}{ \mathcal B }(u)\tilde{{ \mathcal D }}(v).\end{array}\end{eqnarray}$
Repeatedly using above commutative relations, we obtain $\begin{eqnarray}t(u)| {\rm{\Phi }}\rangle ={\rm{\Lambda }}(u)| {\rm{\Phi }}\rangle +\mathrm{unwanted}\,\mathrm{terms},\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}{\rm{\Lambda }}(u)=\displaystyle \frac{\sinh (u+{\xi }_{+})\sinh (u+{\xi }_{-})\sinh (2u+2\eta )}{\sinh (2u+\eta )}\\ \quad \times \displaystyle \frac{{\sinh }^{2N}(u+a+\eta ){\sinh }^{2N}(u-a+\eta )}{{\sinh }^{4N}\eta }\\ \quad \prod _{j=1}^{M}\displaystyle \frac{\sinh (u+{\mu }_{j})\sinh (u-{\mu }_{j}-\eta )}{\sinh (u-{\mu }_{j})\sinh (u+{\mu }_{j}+\eta )}\\ \quad +\displaystyle \frac{\sinh (u+\eta -{\xi }_{+})\sinh (u+\eta -{\xi }_{-})\sinh (2u)}{\sinh (2u+\eta )}\\ \quad \times \displaystyle \frac{{\sinh }^{2N}(u+a){\sinh }^{2N}(u-a)}{{\sinh }^{4N}\eta }\\ \quad \prod _{j=1}^{M}\displaystyle \frac{\sinh (u-{\mu }_{j}+\eta )\sinh (u+{\mu }_{j}+2\eta )}{\sinh (u-{\mu }_{j})\sinh (u+{\mu }_{j}+\eta )}.\end{array}\end{eqnarray}$
If all the unwanted terms are zero, which gives that the Bethe roots $\left\{{\mu }_{j}| j=1,\cdots ,M\right\}$ must satisfy following BAEs, $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\sinh ({\mu }_{j}+{\xi }_{+})\sinh ({\mu }_{j}+{\xi }_{-})\sinh (2{\mu }_{j}+2\eta )}{\sinh ({\mu }_{j}+\eta -{\xi }_{+})\sinh ({\mu }_{j}+\eta -{\xi }_{-})\sinh (2{\mu }_{j})}\\ \quad =\ -{\left[\displaystyle \frac{\sinh ({\mu }_{j}+a)\sinh ({\mu }_{j}-a)}{\sinh ({\mu }_{j}+a+\eta )\sinh ({\mu }_{j}-a+\eta )}\right]}^{2N}\\ \quad \times \prod _{k=1}^{M}\displaystyle \frac{\sinh ({\mu }_{j}-{\mu }_{k}+\eta )\sinh ({\mu }_{j}+{\mu }_{k}+2\eta )}{\sinh ({\mu }_{j}+{\mu }_{k})\sinh ({\mu }_{j}-{\mu }_{k}-\eta )},\\ \quad j\,=\,1,2,\cdots ,M,\end{array}\end{eqnarray}$
the assumed states (According to (2.19 ), we obtain the eigenvalue of Hamiltonian (2.1 )3.17 ).
$\begin{eqnarray}\begin{array}{l}E=-\displaystyle \frac{\sinh (2a+\eta )\sinh (2a-\eta )}{2\sinh \eta }\\ \left\{\coth (a+{\xi }_{+})-\coth (a-{\xi }_{+})+\coth (a+{\xi }_{-})\right.\\ \quad -\coth (a-{\xi }_{-})+2\left[\coth (2a+2\eta )+\coth (2a-\eta )\right.\\ \quad \left.-\coth (2a-2\eta )-\coth (2a+\eta )\right]\\ \quad +2N\left[\coth (2a+\eta )-\coth (2a-\eta )+2\coth \eta \right]\\ \quad +2\sum _{j=1}^{M}\left[\coth (a+{\mu }_{j})\right.\\ \quad +\coth (a-{\mu }_{j}-\eta )-\coth (a-{\mu }_{j})\\ \quad \left.\left.-\coth (a+{\mu }_{j}+\eta )\right]\right\}-{l}_{0},\end{array}\end{eqnarray}$
where the Bethe roots must satisfy the BAEs (Next, we check above analytical results numerically. We first solve the BAEs (3.17 ) and obtain the values of Bethe roots. Substituting these values into (3.18 ), we obtain the eigenenergies of the Hamiltonian (2.1 ). The results are listed in table 1. After that, we numerically diagonalize the Hamiltonian (2.1 ) with same model parameters. We find that the eigenenergies of the Hamiltonian (2.1 ) obtained by solving the BAEs (3.17 ) are exactly the same as those obtained by the exact diagonalization. Meanwhile, the expression (3.18 ) gives the complete spectrum of the system.
Table 1. Numerical solutions of the BAEs ( |
| ${\mu }_{1}$ | ${\mu }_{2}$ | ${\mu }_{3}$ | ${\mu }_{4}$ | En | n |
|---|---|---|---|---|---|
| $-0.3000+1.1177{\rm{i}}$ | $-0.3000+1.3567{\rm{i}}$ | ⋯ | ⋯ | −9.3206 | 1 |
| $-0.6920-0.5799{\rm{i}}$ | $-0.6920+0.5799{\rm{i}}$ | $-0.3000+1.3595{\rm{i}}$ | ⋯ | −5.3779 | 2 |
| $-0.3000+0.5769{\rm{i}}$ | $-0.3000+1.3715{\rm{i}}$ | ⋯ | ⋯ | −4.6013 | 3 |
| $-0.6958-0.6407{\rm{i}}$ | $-0.6958+0.6407{\rm{i}}$ | $-0.3000-1.1392{\rm{i}}$ | ⋯ | −3.2736 | 4 |
| $-0.3000+0.5994{\rm{i}}$ | $-0.3000+1.1743{\rm{i}}$ | ⋯ | ⋯ | −3.1533 | 5 |
| $-0.3000-1.3867{\rm{i}}$ | ⋯ | ⋯ | ⋯ | −2.5986 | 6 |
| $-0.3000-1.2109{\rm{i}}$ | ⋯ | ⋯ | ⋯ | −1.7460 | 7 |
| $-0.6001-1.1974{\rm{i}}$ | $-0.6001+1.1974{\rm{i}}$ | ⋯ | ⋯ | 0.5545 | 8 |
| $-0.9231-0.7838{\rm{i}}$ | $-0.9231+0.7838{\rm{i}}$ | $-0.3000-0.8309{\rm{i}}$ | ⋯ | 2.1627 | 9 |
| $-0.3000-0.9609{\rm{i}}$ | ⋯ | ⋯ | ⋯ | 2.1688 | 10 |
| $-0.5752-0.8182{\rm{i}}$ | $-0.5752+0.8182{\rm{i}}$ | ⋯ | ⋯ | 2.1989 | 11 |
| −1.4562 | −0.9176 | $-0.3000+0.0712{\rm{i}}$ | 1.8613 | 3.7768 | 12 |
| −1.4094 | −0.9430 | $-0.3000+0.1085{\rm{i}}$ | ⋯ | 3.8209 | 13 |
| $-0.6644-0.2051{\rm{i}}$ | $-0.6644+0.2051{\rm{i}}$ | ⋯ | ⋯ | 4.1002 | 14 |
| $-0.3000-0.4181{\rm{i}}$ | ⋯ | ⋯ | ⋯ | 4.8439 | 15 |
| ⋯ | ⋯ | ⋯ | ⋯ | 6.4448 | 16 |
4. Conclusion
In this paper, we propose a method to construct new quantum integrable models. As an example, we construct an anisotropic quantum spin chain which includes the NN, NNN and chiral three-spin couplings with open boundary condition. We find that the boundary fields can enhance the anisotropy of boundary NN coupling bonds. Meanwhile, the boundary field can also induce a polarized boundary DM interaction along the z-direction. By using the algebraic Bethe ansatz, we obtain the exact solution of the system. The energy spectrum and the BAEs are give explicitly. We note that this method is universal and can be used to construct other integrable models with certain interesting interactions.
The boundary fields considered in this work are parallel. If the boundary fields are unparallel and do not break the integrability, which means that the reflection matrices (2.9 ) and (2.11 ) have non-diagonal elements, then the spins of quasi-particles may change after the boundary reflecting. Therefore, the total numbers of quasiparticles with fixed spin-components are not conserved. The U(1) symmetry of the system is broken and the algebraic Bethe ansatz does not work, because (3.4 ) is no longer the correct vacuum state. If the boundary fields satisfy some constraints [21, 22], we can use the generalized Bethe ansatz just like that has done for the XYZ spin chain [7]. For the most generic integrable non-diagonal boundary reflections, the corresponding system may be solved by the off-diagonal Bethe ansatz [34–36]. Moreover, the unparallel boundary fields can induce the completely anisotropic DM interactions at boundaries, then some novel magnetic ordered states such as the helical spinon excitations may arise.