Bethe ansatz approach for the steady state of the asymmetric simple exclusion process with open boundaries

The asymmetric simple exclusion process (ASEP) [1, 2], which describes the asymmetric diffusion of hard-core particles with anisotropic hopping rates, is a fundamental model in the study of non-equilibrium statistical mechanics and stochastic processes. The ASEP has been widely used in the research on mRNA translation [3], vehicular traffic [4] and motor-protein transport [5].

The ASEP is an integrable system [6, 7, 8]. Various techniques such as the matrix product ansatz [9, 10, 11, 12] and the Bethe ansatz [13, 14, 15, 16] have be employed to obtain its exact solutions. The ASEP with generic open boundaries can be mapped to an open XXZ chain with constrained non-diagonal boundary fields through a similarity transformation [6]. The non-diagonal boundaries break the $U(1)$ symmetry of both the open XXZ spin chain and the open ASEP, which prevents the use of conventional Bethe ansatz methods. However, the open XXZ model with the specific constraint exhibits an alternative symmetry. In Ref. [17], the authors mapped it to another open XXZ chain which is $U\mathfrak{{}_{q}sl}_{2}$-invariant using representation theory of the two-boundary Temperley-Lieb algebra. It has been shown that the spectrum of this specific constrained open XXZ chain (or open ASEP) can still be parameterized using Baxter’s homogeneous $T-Q$ relation [18, 19, 8, 14]. Consequently, a set of Bethe ansatz equations (BAEs) is derived, which enables the study of the model’s physical properties [8, 20, 21].

For the open ASEP, the eigenstates of the Markov transition matrix (or the transfer matrix) can be constructed using the modified algebraic Bethe ansatz method introduced in Ref. [19], the (modified) coordinate Bethe ansatz proposed in Refs. [14, 22], and the subsequent analogue — the chiral coordinate Bethe ansatz, presented in Refs. [23, 24]. However, due to the non-Hermitian nature of the system, constructing all the eigenstates of the Markov transition matrix is challenging. For example, the steady state is represented by a tensor product bra vector $\langle\Phi|$, which serves as the pseudo-vacuum in the modified Bethe ansatz method and corresponds to a trivial $T-Q$ relation without $Q$-function. In contrast, the ket vector $|\Phi\rangle$ is not factorized and has a notably complex expression.

In this paper, we study the open ASEP with a specific constraint (45). This constraint exhibits the following features: (1) it enables the construction of additional homogeneous $T-Q$ relations and BAEs; (2) it allows us to introduce a closed set of factorized chiral vectors; (3) it facilitates the parameterization of the eigenstates of the Markov transition matrix, including the right steady state $|\Phi\rangle$, using the solutions of the BAEs and the chiral basis. In this paper, we use the coordinate Bethe ansatz approach presented in Refs. [24, 23] to construct the steady state. We prove analytically that the steady state corresponds to a special string solution of BAEs and it allow us to get an elegant expression for the steady state. The current and density profile of the system under the steady state are also studied analytically and numerically in this paper.

The paper is organized as follows: Section I provides a brief introduction to the ASEP with open boundaries. In Section II, we demonstrate the system’s integrability. The $T-Q$ relations are presented in Section III. In Section IV, we introduce the constraint we are interested in and demonstrate another homogeneous $T-Q$ relations. A set of chiral vectors is constructed in Section V, where we also introduce the chiral coordinate Bethe ansatz method and a special string solution. The expression for the steady state is derived in Section VI. Finally, Section VII explores the ASEP with generic open boundaries. Some numerical results are provided in the appendix.

We study the ASEP with open boundaries which is described by the following Markov transition matrix [8]

$\displaystyle\mathcal{M}=\mathcal{M}_{1}+\sum_{k=1}^{N-1}\mathcal{M}_{k,k+1}+\mathcal{M}_{N},$

(1)

where

$\displaystyle\mathcal{M}_{k,k+1}=\left(\begin{array}[]{cccc}0&0&0&0\\[4.0pt] 0&-\frac{1}{q+1}&\frac{q}{q+1}&0\\[4.0pt] 0&\frac{1}{q+1}&-\frac{q}{q+1}&0\\[4.0pt] 0&0&0&0\end{array}\right),\qquad\mathcal{M}_{1}=\left(\begin{array}[]{cc}-\alpha&\gamma\\ \alpha&-\gamma\end{array}\right),\qquad\mathcal{M}_{N}=\left(\begin{array}[]{cc}-\delta&\beta\\ \delta&-\beta\end{array}\right).$

(10)

Here $\alpha,\,\beta,\,\gamma,\,\delta$ and $q$ are all non-negative real numbers, and the subscripts indicate on which sites the matrices $\mathcal{M}_{k,k+1}$ and $\mathcal{M}_{j}$ act non-trivially. The transition rates of open ASEP are illustrated in Figure 1.

The Markov transition matrix $\mathcal{M}$ can map to the Hamiltonian of an open spin-$\frac{1}{2}$ XXZ model with constrained boundary fields through a similarity transformation [6]

$\displaystyle\mathcal{M}=\frac{\sqrt{q}}{2(1+q)}\,G^{-1}HG,\quad G=\bigotimes_{n=1}^{N}\left(\begin{array}[]{cc}1&0\\ 0&\rho q^{\frac{1-n}{2}}\end{array}\right),$

(13)

where the gauge parameter $\rho$ can be an arbitrary non-zero complex number. The Hamiltonian $H$ in Eq. (13) reads

	$\displaystyle H=$	$\displaystyle\sum_{j=1}^{N-1}\left[\sigma^{x}_{j}\sigma_{j+1}^{x}+\sigma^{y}_{j}\sigma_{j+1}^{y}+\frac{q+1}{2\sqrt{q}}\left(\sigma_{j}^{z}\sigma_{j+1}^{z}-\mathbb{I}\right)\right]+\vec{h}_{1}\cdot\vec{\sigma}_{1}+\vec{h}_{N}\cdot\vec{\sigma}_{N}$
		$\displaystyle+\frac{q-1}{2\sqrt{q}}(\sigma_{1}^{z}-\sigma_{N}^{z})-(q+1)\left(\frac{\alpha+\gamma}{\sqrt{q}}\!+\!\frac{\beta+\delta}{\sqrt{q}}\right),$		(14)

where $\vec{h}_{1}$ and $\vec{h}_{N}$ are defined by

$\displaystyle\begin{aligned} &\vec{h}_{1}=(q+1)\left(\frac{\gamma+\alpha\rho^{2}}{\sqrt{q}\rho},\,{\rm i}\frac{\gamma-\alpha\rho^{2}}{\sqrt{q}\rho},\,\frac{\gamma-\alpha}{\sqrt{q}}\right),\\ &\vec{h}_{N}=(q+1)\left(\frac{\beta q^{\frac{N-1}{2}}+\delta\rho^{2}q^{\frac{1-N}{2}}}{\sqrt{q}\rho},\,{\rm i}\frac{\beta q^{\frac{N-1}{2}}-\delta\rho^{2}q^{\frac{1-N}{2}}}{\sqrt{q}\rho},\,\frac{\beta-\delta}{\sqrt{q}}\right).\end{aligned}$

(15)

As a stochastic process, the left steady state of Markov transition matrix is a simple factorized state [9]

$\displaystyle\langle\Phi|=\bigotimes_{n=1}^{N}(1,\,1).$

(16)

Both $\mathcal{M}$ in (1) and $H$ in (14) are non-Hermitian operators, and the conjugate transpose of $\langle\Phi|$ in Eq. (16) is not an eigenstate of $\mathcal{M}$. In this paper, we will use the Bethe ansatz method to study the right steady state of the open ASEP.

The ASEP is an integrable system and the corresponding $R$-matrix is [25]

$\displaystyle R(x)=\left(\begin{array}[]{cccc}a(x)&0&0&0\\ 0&b^{-}(x)&c^{+}(x)&0\\ 0&c^{-}(x)&b^{+}(x)&0\\ 0&0&0&a(u)\\ \end{array}\right),$

(21)

where $x$ is the spectral parameter and some functions in the $R$-matrix are defined as follows

	$\displaystyle a(x)=q-x,\quad b^{+}(x)=q(1-x),\quad b^{-}(x)=1-x,$
	$\displaystyle c^{+}(x)=(q-1)x,\quad c^{-}(x)=q-1.$		(22)

The $R$-matrix in (21) satisfies the Yang-Baxter equation (YBE) [26]

$\displaystyle R_{1,2}(x/y)R_{1,3}(x/z)R_{2,3}(y/z)=R_{2,3}(y/z)R_{1,3}(x/z)R_{1,2}(x/y).$

(23)

In order to present the reflection matrices more concisely, we introduce a new set of parameters $\{a,\,b,\,c,\,d\}$ where

$\displaystyle\begin{aligned} &\alpha=\frac{(q-1)ac}{(a+1)(c+1)(q+1)},\quad\gamma=\frac{(1-q)}{(a+1)(c+1)(q+1)},\\ &\beta=\frac{(q-1)bd}{(b+1)(d+1)(q+1)},\quad\delta=\frac{(1-q)}{(b+1)(d+1)(q+1)}.\end{aligned}$

(24)

The boundary $K$-matrices $K^{\pm}(x)$ are given by [25]

	$\displaystyle K^{-}(x)=\left(\begin{array}[]{cc}(ac+1)x^{2}+(a+c)x&x^{2}-1\\ ac(1-x^{2})&(ac+1)+(a+c)x\\ \end{array}\right),$		(27)
	$\displaystyle K^{+}(x)=\left(\begin{array}[]{cc}(bd+1)q^{2}+(b+d)qx&bd(q^{2}-x^{2})\\ q(x^{2}-q^{2})&(bd+1)qx^{2}+(b+d)q^{2}x\\ \end{array}\right),$		(30)

which satisfy the following reflection equation (RE) and dual RE respectively

	$\displaystyle R_{1,2}(xy^{-1})K_{1}^{-}(x)R_{2,1}(yx)K_{2}^{-}(y)=K_{2}^{-}(y)R_{1,2}(xy)K_{1}^{-}(x)R_{2,1}(y^{-1}x),$		(31)
	$\displaystyle R_{1,2}(xy^{-1})K_{1}^{+}(y)\widetilde{R}_{1,2}(yx)K_{2}^{+}(x)=K_{2}^{+}(x)\widetilde{R}_{2,1}(yx)K_{1}^{+}(y)R_{2,1}(xy^{-1}),$		(32)

where

$\displaystyle\widetilde{R}_{1,2}(x)=(R^{t_{1}}_{1,2}(x))^{-1})^{t_{1}}.$

(33)

Then, the transfer matrix can be constructed

$\displaystyle\tau(x)=\mathrm{tr}_{0}\left\{K_{0}^{+}(x)R_{0,N}(x)\cdots R_{0,1}(x)K_{0}^{-}(x)R_{1,0}(x)\cdots R_{N,0}(x)\right\}.$

(34)

Using the YBE (23), RE (31) and dual RE (32), we can prove that the transfer matrix $\tau(x)$ forms a commuting family, i.e., $[\tau(x),\,\tau(y)]=0$. The Markov transition matrix $\mathcal{M}$ can be obtained from the transfer matrix as follows

$\displaystyle\mathcal{M}=\left.\frac{1-q}{2(1+q)}\frac{\partial\ln\tau(x)}{\partial x}\right|_{x=1}-\mbox{const}.$

(35)

For the ASEP with generic open boundary conditions, the Hilbert space of the system splits into two invariant subspaces, i.e., $\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{2}$, whose dimensions are ${\rm dim}\,\mathcal{H}_{1}=2^{N}-1$ and ${\rm dim}\,\mathcal{H}_{2}=1$ respectively. The full spectrum of the quantum transfer matrix can be described by two homogeneous $T-Q$ relations, with the number of Bethe roots being $N-1$ and 0 respectively [19, 18, 8]. Let us recall these two $T-Q$ relations in this section.

First, we introduce some functions

	$\displaystyle a_{1}(x)=\frac{q^{3}-qx^{2}}{q-x^{2}}(ax+1)(cx+1)(bx+1)(dx+1)a^{2N}(x),$		(36)
	$\displaystyle d_{1}(x)=\frac{1-x^{2}}{q-x^{2}}(x+aq)(x+cq)(x+bq)(x+dq)\left[b^{-}(x)b^{+}(x)\right]^{N},$		(37)
	$\displaystyle a_{2}(x)=\frac{q^{3}-qx^{2}}{q-x^{2}}(x+a)(x+c)(x+b)(x+d)a^{2N}(x),$		(38)
	$\displaystyle d_{2}(x)=\frac{1-x^{2}}{q-x^{2}}(ax+q)(cx+q)(bx+q)(dx+q)\left[b^{-}(x)b^{+}(x)\right]^{N}.\qquad$		(39)

$T-Q$ relation I:     The eigenvalue of the transfer matrix $\tau(x)$ can be parameterized by the following $T-Q$ relation

$\displaystyle\Lambda_{1}(x)=a_{1}(x)\,\frac{q^{1-N}Q_{1}(qx)}{Q_{1}(x)}+d_{1}(x)\,\frac{q^{N-1}Q_{1}(q^{-1}x)}{Q_{1}(x)},$

(40)

where

$\displaystyle Q_{1}(x)=\prod_{k=1}^{N-1}(\lambda_{k}-x)(\lambda_{k}x-q).$

(41)

The Bethe roots $\{\lambda_{j}\}$ are given by the following BAEs

$\displaystyle\frac{d_{1}(\lambda_{j})}{a_{1}(\lambda_{j})}=-\prod_{k=1}^{N-1}\frac{(\lambda_{k}-q\lambda_{j})(\lambda_{k}\lambda_{j}-1)}{(\lambda_{k}-q^{-1}\lambda_{j})(\lambda_{k}\lambda_{j}-q^{2})},\quad j=1,\ldots,N-1.$

(42)

The eigenvalue of the Markov transition matrix ${\mathcal{M}}$ in terms of $\{\lambda_{j}\}$ is

$\displaystyle E=\frac{(1-q)^{2}}{q+1}\sum_{k=1}^{N-1}\frac{\lambda_{k}}{(\lambda_{k}-1)(\lambda_{k}-q)}-(\alpha+\beta+\gamma+\delta).$

(43)

The numerical results for small-size systems in Table 2 indicate that the solution of the BAEs in (42) yields $2^{N}-1$ eigenvalues of ${\mathcal{M}}$, leaving the energy level $E=0$ unaccounted for.

$T-Q$ relation II:     There exists second homogeneous $T-Q$ relation

$\displaystyle\Lambda_{2}(x)=a_{2}(x)+d_{2}(x),\quad$

(44)

which only gives the trivial eigenvalue of the Markov transition matrix $E=0$.

The functions $\Lambda_{1}(u)$ and $\Lambda_{2}(u)$ correspond to $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, respectively. The right eigenstates of $\tau(x)$ belonging to $\mathcal{H}_{1}$ and the left eigenstates belonging to $\mathcal{H}_{2}$ can be constructed via either the modified algebraic Bethe ansatz method [19] or the (chiral) coordinate Bethe ansatz method [14, 25, 23].

In the generic case, constructing the dual eigenstates, namely the left eigenstates belonging to $\mathcal{H}_{1}$ and the right eigenstates belonging to $\mathcal{H}_{2}$, is problematic. However, under certain constraint, the Hilbert space exhibits another subspaces, which allow us to use the Bethe ansatz method to construct the “missing” eigenstates of ${\mathcal{M}}$, such as the right steady state $|\Phi\rangle$. This inspiring discovery motivates our research.

In this paper, we study the open ASEP with the following constraint [22]

$\displaystyle abcdq^{N-1-M}=1,\quad\mbox{or equivalently}\,\,\alpha\beta q^{N-1-M}=\gamma\delta,\quad M=0,1,\ldots,N.$

(45)

Equation (45) has a two-fold meaning. First, it allows us to construct another homogeneous $T-Q$ relations and Bethe ansatz equations. Second, one can find a set of vectors which form an invariant subspace and can be used to expand certain eigenstates of the transfer matrix, with the expansion coefficients parameterizable by the Bethe roots.

$T-Q$ relation III:     Under constraint (45), using the off-diagonal Bethe ansatz method [27], another $T-Q$ relation can be constructed

$\displaystyle\Lambda_{3}(x)=a_{1}(x)\,\frac{q^{-\widetilde{M}}Q_{3}(qx)}{Q_{3}(x)}+d_{1}(x)\,\frac{q^{\widetilde{M}}Q_{3}(q^{-1}x)}{Q_{3}(x)},\qquad\widetilde{M}=N-1-M,$

(46)

where

$\displaystyle Q_{3}(x)=\prod_{k=1}^{\widetilde{M}}(\mu_{k}-x)(\mu_{k}x-q).$

(47)

The corresponding BAEs are

$\displaystyle\frac{d_{1}(\mu_{j})}{a_{1}(\mu_{j})}=-\prod_{k=1}^{\widetilde{M}}\frac{(\mu_{k}-q\mu_{j})(\mu_{k}\mu_{j}-1)}{(\mu_{k}-q^{-1}\mu_{j})(\mu_{k}\mu_{j}-q^{2})},\quad j=1,\ldots,\widetilde{M}.$

(48)

The eigenvalue of Markov transition matrix ${\mathcal{M}}$ is given by

$\displaystyle E=\frac{(1-q)^{2}}{q+1}\sum_{k=1}^{\widetilde{M}}\frac{\mu_{k}}{(\mu_{k}-1)(\mu_{k}-q)}-(\alpha+\beta+\gamma+\delta).$

(49)

$T-Q$ relation IV:     Another alternative $T-Q$ relation is

$\displaystyle\Lambda_{4}(x)=a_{2}(x)\frac{q^{-M}Q_{4}(qx)}{Q_{4}(x)}+d_{2}(x)\frac{q^{M}Q_{4}(q^{-1}x)}{Q_{4}(x)},\qquad$

(50)

where

$\displaystyle Q_{4}(x)=\prod_{k=1}^{M}(\nu_{k}-x)(\nu_{k}x-q).$

(51)

The corresponding BAEs read

$\displaystyle\frac{d_{2}(\nu_{j})}{a_{2}(\nu_{j})}=-\prod_{k=1}^{M}\frac{(\nu_{k}-q\nu_{j})(\nu_{k}\nu_{j}-1)}{(\nu_{k}-q^{-1}\nu_{j})(\nu_{k}\nu_{j}-q^{2})},\quad j=1,\ldots,M.$

(52)

The eigenvalue of Markov transition matrix ${\mathcal{M}}$ is given by

$\displaystyle E=\frac{(1-q)^{2}}{q+1}\sum_{k=1}^{M}\frac{\nu_{k}}{(\nu_{k}-1)(\nu_{k}-q)}.$

(53)

Based on the numerical results obtained from small-scale systems in Tables 3, 4,5, 6, we conjecture that $T-Q$ relation III in (46) and $T-Q$ relation IV in (50) also consitute the complete spectrum of the transfer matrix (the Markov transition matrix). It should be noted that the steady state always corresponds to a specific solution where all the Bethe roots go to infinity, as discussed in detail in Eq. (69). The functions $\Lambda_{3}(u)$ and $\Lambda_{4}(u)$ correspond to the subspaces $\mathcal{H}_{3}$ and $\mathcal{H}_{4}$, respectively. The left eigenstates belonging to $\mathcal{H}_{3}$ and the right eigenstates belonging to $\mathcal{H}_{4}$ can be constructed via the Bethe ansatz approach. In this paper, we focus on $T-Q$ relation IV in Eq. (50) in order to construct the right steady state. A summary of the invariant subspace and $T-Q$ relation is given in Table 1.

Factorized chiral vectors    Define the following chiral vectors

	$\displaystyle|n_{1},n_{2},\ldots,n_{m}\rangle=$	$\displaystyle\bigotimes_{k_{1}=1}^{n_{1}}\phi_{k_{1}}(k_{1}-1)\bigotimes_{k_{2}=n_{1}+1}^{n_{2}}\phi_{k_{2}}(k_{2}-2)\cdots\bigotimes_{k_{m}=n_{m-1}+1}^{n_{m}}\phi_{k_{m}}(k_{m}-m)$
		$\displaystyle\bigotimes_{k_{m+1}=n_{m}+1}^{N}\phi_{k_{m+1}}(k_{m+1}-m-1),\qquad\phi(x)=\binom{\gamma}{\alpha q^{x}}.$		(54)

The local vector $\phi(x)$ in Eq. (54) satisfies the following relations [24, 23]

	$\displaystyle{\mathcal{M}}_{n,n+1}\,\phi_{n}(x)\phi_{n+1}(x+1)=0,$		(55)
	$\displaystyle{\mathcal{M}}_{n,n+1}\,\phi_{n}(x)\phi_{n+1}(x)=\frac{q-1}{2(q+1)}\left(\sigma_{n}^{z}-\sigma_{n+1}^{z}\right)\phi_{n}(x)\phi_{n+1}(x),$		(56)
	$\displaystyle{\mathcal{M}}_{1}\,\phi_{1}(x)=\frac{(1-q^{-x})(\alpha q^{x}-\gamma)}{2}\,\phi_{1}(x)+\frac{(1-q^{-x})(\alpha q^{x}+\gamma)}{2}\,\sigma_{1}^{z}\,\phi_{1}(x),$		(57)
	$\displaystyle{\mathcal{M}}_{N}\,\phi_{N}(x)=\frac{\left(\alpha q^{x}-\gamma\right)\left(\alpha\beta-\gamma\delta q^{-x}\right)}{2\alpha\gamma}\phi_{N}(x)+\frac{\left(\alpha q^{x}+\gamma\right)\left(\alpha\beta-\gamma\delta q^{-x}\right)}{2\alpha\gamma}\,\sigma_{N}^{z}\,\phi_{N}(x),$		(58)
	$\displaystyle\sigma^{z}\phi(x)=\frac{q+1}{q-1}\,\phi(x)-\frac{2}{q-1}\,\phi(x+1)=\frac{1+q}{1-q}\,\phi(x)+\frac{2q}{q-1}\,\phi(x-1).$		(59)

Chiral basis of $\mathcal{H}_{4}$    Using Eqs. (55) - (59) repeatedly, one see that the following vectors

$\displaystyle|\!\underbrace{0,\ldots,0}_{m_{0}},\underbrace{j_{1},j_{2},\ldots,j_{k}}_{k},\underbrace{N,\ldots,N}_{m_{N}}\rangle,\quad 0<j_{1}<j_{2}\cdots<j_{k}<N,\quad m_{0}+k+m_{N}=M,\quad m_{0},k,m_{N}\geq 0,$

(60)

form an invariant subspace of the Hilbert space $\mathcal{H}_{4}$ under the constraint (45). It should be noted that the local state on the first site in $|\!\underbrace{0,\ldots,0}_{m_{0}},\underbrace{j_{1},j_{2},\ldots,j_{k}}_{k},\underbrace{N,\ldots,N}_{m_{N}}\rangle$ is $\phi_{1}(-m_{0})$.

To demonstrate the vectors in (60) more intuitively, we use a set of phase factor $\{z_{1},\ldots,z_{n}\}$ to represent the factorized state

$\displaystyle\bigotimes_{n=1}^{N}\binom{\gamma}{\alpha q^{z_{n}}}.$

(61)

For the chiral vectors in (60), we can easily get that

$\displaystyle z_{1}=0,-1,\ldots,-M,\quad z_{n+1}-z_{n}=0,1.$

(62)

The phases of the qubit increment by an amount 1 from site $n$ to site $n+1$ except at the points $n_{1},\ldots,n_{M}$ where kinks occur. The tensor product state $|n_{1},n_{2},\ldots,n_{M}\rangle$ is completely determined by the phase on the first site $z_{1}$ and the position of the kink and the integer $M$ is the maximum number of kinks. Our chiral vectors with kinks exhibit a structure similar to the Bernoulli product measures described in Refs. [28, 29, 30, 31]. The visualization of our chiral vectors in (60) is illustrated in Figure 2.

It should be noted that only $\sum_{n=0}^{M}\binom{N}{n}$ vectors in the set (60) are linearly independent, which matches the dimension of the subspace $\mathcal{H}_{4}$. For example, the linearly independent basis vectors can be selected as follows

$\displaystyle\begin{aligned} &|\!\underbrace{0,\ldots,0}_{m_{0}},\underbrace{j_{1},j_{2},\ldots,j_{k}}_{k}\rangle,\quad m_{0}+k=M,\quad m_{0}=0,\ldots,M,\quad 1\leq j_{1}<j_{2}\cdots<j_{k}\leq N,\\ {\rm or},\quad&|\underbrace{j_{1},j_{2},\ldots,j_{k}}_{k},\underbrace{N,\ldots,N}_{m_{N}}\rangle,\quad m_{N}+k=M,\quad m_{N}=0,\ldots,M,\quad 0\leq j_{1}<j_{2}\cdots<j_{k}\leq N-1.\end{aligned}$

(63)

Adding the auxiliary vectors makes the entire set in Eq. (60) symmetric and leads to explicit and symmetric forms for the expansion coefficients in the Bethe state [23].

The difference between our chiral vectors (54) and the tensor product states with excitations proposed in Ref. [14] needs to be clarified. Let us recall the tensor product state in Ref. [14] with our notation

	$\displaystyle|n_{1},n_{2},\ldots,n_{m}\rangle_{t}=$	$\displaystyle\bigotimes_{k_{1}=1}^{n_{1}-1}\phi_{k_{1}}(k_{1}-1)\,\varphi_{n_{1}}(t,n_{1}-1)\,\bigotimes_{k_{2}=n_{1}+1}^{n_{2}-1}\phi_{k_{2}}(k_{2}-2)\,\varphi_{n_{2}}(t,n_{2}-2)\,\bigotimes_{k_{3}=n_{2}+1}^{n_{3}-1}\phi_{k_{3}}(k_{3}-3)$
		$\displaystyle\cdots\bigotimes_{k_{m}=n_{m-1}+1}^{n_{m}-1}\phi_{k_{m}}(k_{m}-m)\,\varphi_{n_{m}}(t,n_{m}-m)\bigotimes_{k_{m+1}=n_{m}+1}^{N}\phi_{k_{m+1}}(k_{m+1}-m-1),$		(64)
	$\displaystyle\varphi(t,x)=$	$\displaystyle\,t\phi(x)+(1-t)\phi(x-1),\qquad t\in\mathbb{C}.$		(65)

It can be easily verified that $|n_{1},n_{2},\ldots,n_{m}\rangle_{t}$ is always a linear combination of the chiral vectors $\{|n^{\prime}_{1},n^{\prime}_{2},\ldots,n^{\prime}_{m}\rangle\}$. In particular, when $t=1$, we have $|n_{1},n_{2},\ldots,n_{m}\rangle_{t}=|n_{1},n_{2},\ldots,n_{m}\rangle$.

We see that the vectors constructed in Refs. [14, 22] to expand the Bethe vectors are all linearly independent, similar to those in Eq. (63). In contrast, auxiliary vectors are added in Eq. (60) to obtain an enlarged and symmetric set. This significant change of basis results in a different parameterization of the Bethe state. For large $M$, the number of auxiliary vectors is comparable to the number of independent vectors, and it is generally non-trivial to map the basis used in Refs. [14, 22] onto our enlarged basis, or vice versa.

Chiral coordinate Bethe ansatz    Define the quasi-momentum $\mathrm{p}_{j}$

$\displaystyle\mathrm{e}^{\mathrm{i}\mathrm{p}_{j}}=\frac{\nu_{j}-q}{\sqrt{q}(\nu_{j}-1)},$

(66)

where $\{\nu_{1},\ldots,\nu_{M}\}$ is the solution of BAEs (52).

One can thus use the chiral basis (60) and the quasi-momentum $\{\mathrm{p}_{j}\}$ to construct the Bethe-type eigenstate

	$\displaystyle|\Psi(\mathrm{p}_{1},\ldots,\mathrm{p}_{M})\rangle=$	$\displaystyle\sum_{n_{1},\ldots,n_{M}}\,\sum_{r_{1},\ldots,r_{M}}\,\sum_{\sigma_{1},\ldots,\sigma_{M}=\pm 1}Y_{n_{1},\ldots,n_{M}}A_{\sigma_{r_{1}},\ldots,\sigma_{r_{M}}}^{r_{1},\ldots,r_{M}}$
		$\displaystyle\times q^{-\frac{1}{2}\sum_{k}n_{k}}\exp\left[\mathrm{i}\sum_{k=1}^{M}\sigma_{r_{k}}n_{k}\mathrm{p}_{r_{k}}\right]|n_{1},n_{2},\ldots,n_{M}\rangle,$		(67)

where $r_{1},\ldots,r_{M}$ is the permutation of $1,\ldots,M$, $Y_{n_{1},\ldots,n_{M}}$ is a certain coefficient independent of $\{\mathrm{p}_{j}\}$ and the amplitudes $\Big{\{}A_{\sigma_{r_{1}},\ldots,\sigma_{r_{M}}}^{r_{1},\ldots,r_{M}}\Big{\}}$ are determined by the two-body scattering matrix $S_{j,k}$ and the boundary reflection matrices $S_{L}$ and $S_{R}$ as follows [23]

$\displaystyle\frac{A_{\ldots,\sigma_{k},\sigma_{j},\ldots}^{\ldots,k,j,\ldots}}{A_{\ldots,\sigma_{j},\sigma_{k},\ldots}^{\ldots,j,k,\ldots}}=S_{j,k}(\sigma_{j}\mathrm{p}_{j},\sigma_{k}\mathrm{p}_{k}),\quad\frac{A_{-1,\ldots}^{j,\ldots}}{A_{+1,\ldots}^{j,\ldots}}=S_{L}(\mathrm{p}_{j}),\quad\frac{A_{\ldots,-1}^{\ldots,k}}{A_{\ldots,+1}^{\ldots,k}}=\mathrm{e}^{2\mathrm{i}N\mathrm{p}_{k}}S_{R}(\mathrm{p}_{k}).$

(68)

This is the logic of the chiral coordinate Bethe ansatz method in Refs. [24, 23]. More details about the construction of Bethe state (67) are demonstrated in Ref. [23]. In this paper, we only focus on the right steady state.

Specific Bethe roots for the steady state    The steady state $|\Phi\rangle$ corresponds to a special solution of BAEs (52), i.e., all the Bethe roots $\{\nu_{1},\ldots,\nu_{M}\}$ go to infinity and form the following string

$\displaystyle\nu_{j}=\mathrm{e}^{\frac{2\mathrm{i}\pi(j-k)}{M}}\nu_{k},\quad\nu_{l}\to+\infty.$

(69)

For the steady state, the quasi-momentum tends to be nearly identical to each other

$\displaystyle\mathrm{e}^{\mathrm{i}\mathrm{p}_{j}}\to\frac{1}{\sqrt{q}}+\epsilon_{j},\quad j=1,\ldots,M,$

(72)

where $\{\epsilon_{j}\}$ are infinitesimal numbers. In this case, since $S_{j,k}(\mathrm{p}_{j},\mathrm{p}_{k})$ is now a constant not equal to $-1$ (see Eq. (71)), we no longer need to consider the bulk scattering $\pm\mathrm{p}_{j},\pm\mathrm{p}_{k}\to\pm\mathrm{p}_{k},\pm\mathrm{p}_{j}$ anymore, only the reflection $\pm\mathrm{p}_{j}\to\mp\mathrm{p}_{j}$ and the bulk scattering $\pm\mathrm{p}_{j},\mp\mathrm{p}_{k}\to\mp\mathrm{p}_{k},\pm\mathrm{p}_{j}$ contribute non-trivially. The wave-function (the expansion coefficient) in the Bethe-type steady state therefore has much simpler expansion than the generic one in (67). The explicit expression of the right steady state is shown in Section VI.