Bethe Ansatz and Q-operator for the open ASEP

The starting point for the results of this paper is to realize that the matrices d and e that are defined, for instance, in [47, 48], with the algebraic relation that they satisfy, de − qed = (1 − q), correspond to a special representation of the U_q(SU(2)) algebra (up to a simple gauge transformation that we present in section 2.6). In light of this, it seems natural to wonder whether a more general representation might be used, and produce different, and perhaps better, results.

Let us therefore define:

$$\begin{eqnarray*} X&=\left[\begin{array}{@{}cc@{}}n_0 & e\\ d & n_1\end{array}\right]= \left[\begin{array}{@{}cc@{}}1+x A & e\\ d & 1+y A\end{array}\right],\nonumber \\ \hat{X}&=\left[\begin{array}{@{}cc@{}}\hat{n}_0 & \hat{e}\\ \hat{d} & \hat{n}_1\end{array}\right]=\frac{(1-q)}{2}\left[\begin{array}{@{}cc@{}}1-x A & e\\ -d & -1+y A\end{array}\right]\nonumber \end{eqnarray*}$$

in basis {0, 1}, corresponding to the occupancy of one of the sites, and where matrices A, d and e satisfy:

For practical purposes, we will be using a specific solution to these equations, given by:

$$\begin{equation} A=\sum \limits _{n=0}^{\infty } q^n|\!|n\rangle \!\rangle \langle \!\langle n|\!| \end{equation} \tag{ 4 }$$

$$\begin{equation} d=\sum \limits _{n=1}^{\infty }(1-q^n)|\!|n-1\rangle \!\rangle \langle \!\langle n|\!|=S^-(1-A) \end{equation} \tag{ 5 }$$

and

$$\begin{equation} e=\sum \limits _{n=0}^{\infty }(1-x y q^n)|\!|n+1\rangle \!\rangle \langle \!\langle n|\!|=S^+(1-x y A) \end{equation} \tag{ 6 }$$

where S⁺ and S⁻ are simply operators increasing or decreasing n by 1 (not to be confused with spin operators). We recover the simpler versions of these matrices simply by taking x = y = 0.

A few remarks need to be made here. First of all, the matrix A that we have just defined plays an important role in building the matrix Ansatz for the multispecies periodic ASEP [48]. Secondly, we could have chosen for d and e their contragredient representation $\overline{e}={}^t\!d$ and $\overline{d}={}^t\!e$, which is equivalent to a gauge transformation on d and e. We will be using this fact abundantly in the rest of the chapter. Finally, we can actually define A, S⁺ and S⁻ over $\mathbb {Z}$ rather than $\mathbb {N}$, so that S⁺ and S⁻ are the inverse of one another: S⁺S⁻ = 1 (which would not work on $\mathbb {N}$ because of the cut at −1). Because of the term (1 − qⁿ) in d, which is 0 between states ||0〉〉 and || − 1〉〉, we are assured that, if starting from a state ||n〉〉 with n ⩾ 0, we can never go to one with n < 0 through any combination of d, e and A. We just need to make sure that those expressions are always applied to vectors that have non-zero coefficients only for n ⩾ 0, which is enforced by the matrix A_μ that we will define momentarily, on $\mathbb {N}$ alone. This fact will make many future calculations much easier.

All our matrices are now combinations of only A and S, which satisfy a simple algebra:

$$\begin{equation} A S=q\, S A, \end{equation} \tag{ 7 }$$

with S⁺ = S and S⁻ = (S)⁻¹.

We also need to define another diagonal matrix A_μ, given by

$$\begin{equation} A_\mu =\sum \limits _{n=0}^{\infty } {\rm e}^{-n\mu }|\!|n\rangle \!\rangle \langle \!\langle n|\!| \end{equation} \tag{ 8 }$$

such that

$$\begin{equation} A_\mu S={\rm e}^{-\mu } S A_\mu . \end{equation} \tag{ 9 }$$

Note that this does not have a well-defined trace for μ = 0. If we want to consider that limit, we will have to multiply A_μ by (1 − e^−μ) first.

Finally, we define the transfer matrix

$$\begin{equation} \fbox{$ T_\mu ^{{\rm per}}(x,y)={\rm Tr}\bigg[A_\mu \prod \limits _{i=1}^{L}X^{(i)}\bigg] $} \end{equation} \tag{ 10 }$$

where the product symbol refers to a matrix product in the auxiliary space (i.e. the internal space of matrices A and S⁺) and a tensor product in configuration space, and the trace is taken only on the auxiliary space. The superscript (i) only indicates to which physical site each matrix X corresponds to (and their place in the product). In other terms, the weight of that matrix between configurations $\mathcal{C}=\lbrace \tau _i\rbrace$ and $\mathcal{C}^{\prime }=\lbrace \tau _i^{\prime }\rbrace$, where τ_i is the number of particles on site i in $\mathcal{C}$, is given by

$$\begin{equation*} T_\mu ^{{\rm per}}(x,y)\big |_{\mathcal {C}^{\prime }\!,\mathcal {C}}={\rm Tr}\left[A_\mu \prod \limits _{i=1}^{L}X_{\tau _i^{\prime },\tau _i}\right].\nonumber \end{equation*}$$

Through a calculation which can be found in appendix A.1, we show that each of the local matrices in M_μ satisfy

$$\begin{equation} [M^{(i)}, X^{(i)}X^{(i+1)}]=\hat{X}^{(i)}X^{(i+1)}-X^{(i)}\hat{X}^{(i+1)} \end{equation} \tag{ 11 }$$

and, for M^(L), which contains the deformation,

$$\begin{equation} [M^{(L)}, X^{(L)}A_\mu X^{(1)}]=\hat{X}^{(i)}A_\mu X^{(i+1)}-X^{(i)}A_\mu \hat{X}^{(i+1)}. \end{equation} \tag{ 12 }$$

Note that these relations are in fact the infinitesimal equivalent of the so-called 'RLL equation' for the commutation of matrices X with different parameters, where $\hat{X}$ is the derivative of X for a well chosen variable.

We can now recover M_μ in its entirety by summing over i in (11) and (12). The hat matrices cancel out from one term to the next, and we are left with 0, so that:

$T_\mu ^{{\rm per}}(x,y)$ has therefore the same eigenvectors as M_μ.

Note that for a general set of fugacities {μ_i}, the corresponding transfer matrix is the same as the one we defined here, with matrices $A_{\mu _i}$ inserted at their appropriate place in the matrix product:

$$\begin{equation} T_{\lbrace \mu _i\rbrace }^{{\rm per}}(x,y)={\rm Tr}\left[A_{\mu _0}\prod \limits _{i=1}^{L}X^{(i)}A_{\mu _i}\right]. \end{equation} \tag{ 14 }$$

Considering the representation we chose for matrix e in (6), namely S(1 − xyA), an interesting case to consider is xy = q^{−k + 1} with $k\in \mathbb {N}^\star$ (which sets one coefficient to 0 in e).

Let us therefore impose y = 1/q^{k − 1}x. The four matrices in X become:

$$\begin{equation} d=\sum \limits _{n=1}^{\infty }(1-q^n)|\!|n-1\rangle \!\rangle \langle \!\langle n|\!| ,\quad e=\sum \limits _{n=0}^{\infty }(1- q^{n-k+1})|\!|n+1\rangle \!\rangle \langle \!\langle n|\!| \end{equation} \tag{ 15 }$$

$$\begin{equation} n_0=\sum \limits _{n=0}^{\infty }(1+q^{n}x)|\!|n\rangle \!\rangle \langle \!\langle n|\!| , \quad n_1=\sum \limits _{n=0}^{\infty }(1+q^{n-k+1}/x)|\!|n\rangle \!\rangle \langle \!\langle n|\!| \end{equation} \tag{ 16 }$$

and the coefficient of ||k〉〉〈〈k − 1|| in e vanishes. This makes all these matrices lower block-triangular (n₀ and n₁ obviously are, since they are diagonal, and d already was, but not e in general) with a block of size k (for n from 0 to k − 1 in the four series above) and one of infinite size (for n from k to ∞).

The coefficients of that second block happen to be the same as the coefficients of the whole matrix for x → q^kx and y → q/x and in the contragredient representation of d and e (i.e. exchanging and transposing them):

$$\begin{equation} ^t\!e=\sum \limits _{n=0}^{\infty }(1- q^{n+k})|\!|n-1\rangle \!\rangle \langle \!\langle n|\!| ,\quad {}^t\!d=\sum \limits _{n=1}^{\infty }(1-q^{n+1})|\!|n+1\rangle \!\rangle \langle \!\langle n|\!|, \end{equation} \tag{ 17 }$$

$$\begin{equation} n_0=\sum \limits _{n=0}^{\infty }(1+q^{n+k}x)|\!|n\rangle \!\rangle \langle \!\langle n|\!| ,\quad n_1=\sum \limits _{n=0}^{\infty }(1+q^{n+1}/x)|\!|n\rangle \!\rangle \langle \!\langle n|\!|. \end{equation} \tag{ 18 }$$

Indeed, taking n → n + k in (15) and (16), and removing the k negative indices, we get exactly (17) and (18).

Since the trace of a product of block-diagonal matrices is the sum of the traces of the products of the blocks, this gives us an equation for $T_\mu ^{{\rm per}}$, which is one of the two results essential to our derivation of the functional Bethe Ansatz:

where the factor e^−kμ is the first coefficient of A_μ on the second block. The transfer matrix t^[k] is the contribution coming from the first block, which can be written as:

$$\begin{equation} \fbox{$ t^{[k]}(x)={\rm Tr}\left[A_\mu \displaystyle\prod \limits _{i=1}^{L}X_k^{(i)}(x)\right] $} \end{equation} \tag{ 20 }$$

where X_k(x) contains the same entries as X(x, 1/q^{k − 1}x), but truncated at n = k − 1 (so that the auxiliary space is k-dimensional).

We saw that $T_\mu ^{{\rm per}}$ commutes with M_μ for any values of x and y, so it also commutes with another $T_\mu ^{{\rm per}}$ at different values of the parameters (this is in fact not certain, because any one of those matrices could have a degenerate eigenspace, but we will assume that it is true, for now; there is a better way to show that two matrices $T_\mu ^{{\rm per}}$ at different values of x and y commute, and we will come back to it in the next section). This tells us that those matrices also commute with t^[k](x) for any k, and that the t^[k](x)'s with different k's commute together. This matrix equation therefore implies the same relation for the eigenvalues Λ_i(x, y) of $T_\mu ^{{\rm per}}(x,y)$ and the eigenvalues $\Lambda ^{[k]}_i(x)$ of t^[k](x).

To go further, we need to examine the first two cases in this last equation. Note that, to be consistent with our notation for X, all the matrices will be written with the physical space as their outer space and the auxiliary space as their inner space (i.e. separated in 2 × 2 blocks of matrices acting on the auxiliary space), which is opposite to the standard convention in some cases.

For k = 1, the first block of X is given by:

$$\begin{equation*} X_1(x)=\left[\begin{array}{@{}c|c@{}}1+x & 0\\ \hline 0 & 1+\frac{\vphantom{a^{a^{2}}}1}{x}\end{array}\right].\nonumber \end{equation*}$$

The matrix t^[1](x), which is scalar inside of a given occupancy sector, is then given by:

$$\begin{equation} \fbox{$ t^{[1]}(x)=(1+x)^{L-N}(1+x^{-1})^{N}=h(x). $} \end{equation} \tag{ 21 }$$

This is the same as the function h(x) that is defined in [12], up to a sign.

For k = 2, the 2 × 2 blocks from n₀, e, n₁ and d are:

$$\begin{equation*} X_2(x)=\left[\begin{array}{@{}c c|c c@{}}1+x & 0 & 0 & 0 \\ 0 & 1+q x & 1-\frac{1}{q}\vphantom{a_{a_{2_{1}}}} & 0 \\ \hline 0 & 1-q & 1+\vphantom{a^{a^{2}}}\frac{1}{q x} & 0\\ 0 & 0 & 0 & 1+\frac{1}{x}\end{array}\right] \nonumber \end{equation*}$$

and

$$\begin{equation} \fbox{$ t^{[2]}(x)={\rm Tr}\bigg[A_\mu ^{[2]}\prod \limits _{i=1}^{L}X_2^{(i)}(x)\bigg] $} \end{equation} \tag{ 22 }$$

with $A_\mu ^{[2]}=\big[\begin{array}{@{}cc@{}}{ 1} & 0\\ { 0} & { {\rm e}^{-\mu }}\end{array}\big]$.

This matrix is, in fact, the standard transfer matrix for the periodic XXZ spin chain, with a two-dimensional auxiliary space. To write it in its usual form, we need to make a few transformations and change variables. To that effect, let us consider:

$$\begin{equation} \frac{1}{1+x} \left[\begin{array}{@{}c | c@{}}1 & 0 \\ \hline 0 & x\end{array}\right] \cdot\; X_2(x)=\left[\begin{array}{@{}c c|c c@{}}1 & 0 & 0 & 0 \\ 0 & \frac{1+q x}{1+x} & \frac{q-1}{q(1+x)} & 0 \\ \hline 0 &\frac{x(1-q)}{1+x} & \frac{1+q x}{q(1+x)} & 0\\ 0 & 0 & 0 & 1\end{array}\right] = \left[\begin{array}{@{}c c|c c@{}}1 & 0 & 0 & 0 \\ 0 & q \lambda &1-\lambda & 0 \\ \hline 0 &1-q \lambda & \lambda & 0\\ 0 & 0 & 0 & 1\end{array}\right] \end{equation} \tag{ 23 }$$

with $\lambda = \frac{1+q x}{q(1+x)}$, i.e. $x=-\frac{1-q \lambda }{q(1-\lambda )}$. We use the symbol · to signify a product in configuration space, so that it not be confused with a product in the auxiliary space (for which we use the usual product notation). This is the common form of the Lax matrix for the ASEP:

$$\begin{equation*} L^{(i)}(\lambda )= \left[\begin{array}{@{}c c|c c@{}}1 & 0 & 0 & 0 \\ 0 & q \lambda &1-\lambda & 0 \\ \hline 0 &1-q \lambda & \lambda & 0\\ 0 & 0 & 0 & 1\end{array}\right]=P^{(i)}(1+\lambda M^{(i)})\nonumber \end{equation*}$$

where P⁽ⁱ⁾ is a permutation matrix which has the effect of exchanging the physical space at site i with the auxiliary space (figure 1).

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The matrix we applied to X₂ from the left in (23) multiplies every entry by x for each occupied site (and since this number is conserved between the left and right entries of the transfer matrix, this operation actually commutes with t^[2], so that we have not modified its eigenvectors). All in all, this operation multiplies t^[2] by x^N/(1 + x)^L = 1/h(x). We therefore define:

$$\begin{equation} \hat{t}^{[2]}(\lambda )=\frac{t^{[2]}(x)}{h(x)}={\rm Tr}\left[A_\mu ^{[2]}\prod \limits _{i=1}^{L}L^{(i)}(\lambda )\right] \end{equation} \tag{ 24 }$$

which is the usual transfer matrix for the periodic ASEP with one marked bond.

Since L⁽ⁱ⁾(0) = P⁽ⁱ⁾ is a permutation matrix, and its derivative with respect to λ at 0 is $\frac{{\rm d}}{{\rm d}\lambda }L^{(i)}(0)=P^{(i)} M^{(i)}$, we find that $\hat{t}^{[2]}(0)$ is the matrix that transposes the whole system back one step, and that, for the whole transfer matrix $\hat{t}^{[2]}$ (figure 2):

$$\begin{equation} M_\mu =\bigl (\hat{t}^{[2]}(\lambda )\bigr )^{-1}\frac{{\rm d}}{{\rm d}\lambda }\hat{t}^{[2]}(\lambda )\bigl |_{\lambda =0}=\frac{{\rm d}}{{\rm d}\lambda }\log \bigl (\hat{t}^{[2]}(\lambda )\bigr )\bigl |_{\lambda =0} \end{equation} \tag{ 25 }$$

(we have not considered how A_μ comes into play, but we can easily check that it gives the correct term in M_μ).

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Written in terms of t^[2](x), this becomes:

Doing the same calculations at λ = ∞ instead of 0, after a few modifications (such as taking the contragredient representation for X and multiplying everything by h(x)/h(qx)) would have instead given us:

$$\begin{equation} M_\mu =(1-q)\frac{{\rm d}}{{\rm d}x}\log \biggl (\frac{t^{[2]}(x)}{h(qx)}\biggr )\biggl |_{x=-1} \end{equation} \tag{ 27 }$$

which is also what we would have obtained if we had considered a system with L − N particles, exchanging Q with P, and kept y = 1/qx as a variable. This identity is usually called the 'crossing symmetry' in the language of quantum spin chains.

The next step is to show that the eigenvalues Λ_i(x, y) of $T_\mu ^{{\rm per}}$ are in fact a product of a function of x and a function of y, which we will note as P_i(x) and Q_i(y). This factorization property is a well known fact for the periodic XXX [49, 50] and XXZ [34, 44] spin chains. This result will actually be derived in the next section, as there are a few preliminary calculations which need to be done first, mainly in order to find the R matrix of our system.

One way to go about this is through a method used in [51, 52] which consists in introducing two new Lax matrices, defined by:

$$\begin{eqnarray*} L_1&=L(a_1,b_1,c_1,d_1)=\left[\begin{array}{@{}cc@{}}a_1 A_1 & b_1 S_1\\ -c_1 S^{-1}_1 A_1& d_1\end{array}\right],\nonumber \\ \tilde{L}_2&=\tilde{L}(a_2,b_2,c_2,d_2)=\left[\begin{array}{@{}cc@{}}a_2 A_2 & -c_2 S_2 A_2\\ b_2 S^{-1}_2 & d_2\end{array}\right],\nonumber \end{eqnarray*}$$

where the operators A_i and S_j obey (9) for i = j (i.e. if they act on the same space), and commute otherwise.

We then take the product of those matrices (which is a matrix product on the physical two-dimensional space and a tensor product on the infinite-dimensional auxiliary spaces of L₁ and $\tilde{L}_2$):

$$\begin{equation*} L_1\tilde{L}_2=\left[\begin{array}{@{}cc@{}}a_1 a_2 A+b_1 b_2 S_1/S_2 & b_1 d_2 S_1-a_1 c_2 S_2 A\\ d_1 b_2 S_2^{-1}-c_1 a_2 S_1^{-1} A & c_1 c_2 S_2/ S_1 A + d_1 d_2\end{array}\right]\nonumber \end{equation*}$$

with A = A₁A₂. We will omit the notation · for the product between those new Lax matrices, since the indices are there to signify that their elements act on different spaces.

We now need to consider two special cases for the coefficients of L₁ and $\tilde{L}_2$. Let us first set them as follows: a₁ = x, c₂ = y and the rest is 1. We write the corresponding matrices as $L_1^+$ and $\tilde{L}_2^-$:

$$\begin{equation*} L_1^+(x) \tilde{L}_2^-(y)=\left[\begin{array}{@{}cc@{}}x A+ S_1 / S_2 & S_1-xy S_2 A\\ S_2^{-1}- S_1^{-1}A & y S_2/ S_1 A+1\end{array}\right] \nonumber \end{equation*}$$

and, by projecting each element on S₁ = S₂ = S (i.e. by applying ∑||i, j〉〉〈〈i + j|| to the right and its contragredient to the left), we get:

$$\begin{equation} \fbox{$ L_1^+(x) \tilde{L}_2^-(y)=\left[\begin{array}{@{}cc@{}}x A+ 1 & S(1-xy A) \\ S^{-1}(1- A) & y A+1\end{array}\right] =X(x,y). $} \end{equation} \tag{ 28 }$$

Naturally, we check that A and S satisfy the correct relations.

Let us now set a₂ = x, c₁ = y/q, c₂ = q and the rest to 1. We write the corresponding matrices as $L_1^-$ and $\tilde{L}_2^+$:

$$\begin{equation*} L_1^-(y)\tilde{L}_2^+(x)=\left[\begin{array}{@{}cc@{}}x A+ S_1/ S_2 & S_1- q S_2 A\\ S_2^{-1}- xy/q S_1^{-1}A & y S_2/ S_1 A+1\end{array}\right] \nonumber \end{equation*}$$

and, through the same operation as before, we get:

$$\begin{equation} \fbox{$ L_1^-(y)\tilde{L}_2^+(x)=\left[\begin{array}{@{}cc@{}}x A+ 1 & (1- A)S \\ (1- xy A)S^{-1} & y A+1\end{array}\right] =\overline{X}(x,y). $} \end{equation} \tag{ 29 }$$

In both of those special cases, one of the non-diagonal elements has a factor (1 − A) which allows us to truncate the representation at state ||0〉〉 and avoid some convergence issues. It would not be the case, however, if we were to construct $L_1^+(x) \tilde{L}_2^+(y)$ or $L_1^-(x) \tilde{L}_2^-(y)$, which we will therefore avoid at any cost.

We will now use this formalism in order to find the so-called 'R matrix' which is such that:

$$\begin{equation*} X(x,y)\cdot X(x^{\prime },y^{\prime }) R(x,y;x^{\prime },y^{\prime })=R(x,y;x^{\prime },y^{\prime }) X(x^{\prime },y^{\prime })\cdot X(x,y)\nonumber \end{equation*}$$

where R acts on the two auxiliary spaces of both X matrices. We will comment on the use of such a matrix at the end of this section.

Considering that $X(x,y)\cdot X(x^{\prime },y^{\prime })=L_1^+(x) \tilde{L}_2^-(y) L_3^+(x^{\prime }) \tilde{L}_4^-(y^{\prime })$, we will perform this exchange of parameters in steps, exchanging the parameters of only two L_i's at a time. The first thing we could try is to exchange y and x', but this would transform $\tilde{L}_2^-(y)$ into $\tilde{L}_2^+(x^{\prime })$, so we would have $L_1^+(x) \tilde{L}_2^+(x^{\prime })$ on the left, which we want to avoid. The solution is then to first exchange x' and y', then y' and y, and finally y and x', to obtain X(x, y') · X(x', y), and then do the same once more to exchange x and x'.

We first need to find f₁₂(x, y) such that $L_1^+(x) \tilde{L}_2^-(y) f_{1 2}(x,y) = f_{1 2}(x,y) L_1^-(y)\tilde{L}_2^+(x)$, which is to say:

$$\begin{equation*} X(x,y)f_{1 2}(x,y)=f_{1 2}(x,y)\overline{X}(x,y).\nonumber \end{equation*}$$

That f₁₂ may depend on A and S. In terms of the elements of X(x, y), this writes:

$$\begin{eqnarray*} &&{[}1+xA,f_{1 2}]=0,\nonumber \\ &&{[}1+yA,f_{1 2}]=0,\nonumber \\ &&S(1- xy A) f_{1 2}=f_{1 2}(1-A)S,\nonumber \\ &&S^{-1}(1- A) f_{1 2}=f_{1 2}(1-xyA)S^{-1}.\nonumber \end{eqnarray*}$$

The first and second equations tell us that f₁₂ commutes with A (i.e. it is diagonal), so it should be a function of A alone. The third or fourth equations then give us:

$$\begin{equation*} S(1- xy A) f_{1 2}[A]=f_{1 2}[A](1-A)S=S(1-qA)f_{1 2}[qA]\nonumber \end{equation*}$$

(where the second equality is due to the commutation of S with A), which we can rewrite as

$$\begin{equation*} \frac{f_{1 2}[A]}{f_{1 2}[q A]}=\frac{(1- q A)}{(1-xyA)}.\nonumber \end{equation*}$$

Iterating this last equation, we finally find:

$$\begin{equation} \fbox{$ f_{1 2}(xy)=\displaystyle\frac{(q A)_\infty }{(xyA)_\infty }, $} \end{equation} \tag{ 30 }$$

where (x)_∞ is the infinite q-Pochhammer symbol

$$\begin{equation*} (x)_\infty =\prod \limits _{k=0}^{\infty }(1-q^kx).\nonumber \end{equation*}$$

To exchange the parameters back, one simply has to apply $f_{1 2}^{-1}(xy)$.

This was for the exchange of parameters between the first and second or third and fourth matrices in $L_1^+(x) \tilde{L}_2^-(y) L_3^+(x^{\prime }) \tilde{L}_4^-(y^{\prime })$ (i.e. inside of a same X matrix). We now need to exchange parameters between the second and third matrices in that product.

Let us consider $\tilde{L}_2L_3$ with c₂ = y, c₃ = y'/q, and the rest set to 1:

$$\begin{equation*} \tilde{L}_2^-(y) L_3^-(y^{\prime })=\left[\begin{array}{@{}cc@{}}(1+ y y^{\prime }/q S_2/ S_3) A_2 A_3 & (S_3 -y S_2)A_2 \\ (S_2^{-1}-y^{\prime }/q S_3^{-1}) A_3 & S_3 /S_2+1\end{array}\right].\nonumber \end{equation*}$$

We need to find $g_{2 3}^-(y,y^{\prime })$ such that $\tilde{L}_2^-(y) L_3^-(y^{\prime })g_{2 3}^-(y,y^{\prime })=g_{2 3}^-(y,y^{\prime })\tilde{L}_2^-(y^{\prime }) L_3^-(y)$. As before, the commutation for each of the four elements of $\tilde{L}_2^-(y) L_3^-(y^{\prime })$ is:

$$\begin{eqnarray*} &&{[}(1+ y y^{\prime }/q S_2/ S_3) A_2 A_3,g_{2 3}^-]=0,\nonumber\\ &&{[}1+S_3/ S_2,g_{2 3}^-]=0,\nonumber \\ &&(S_3-y S_2)A_2 g_{2 3}^-=g_{2 3}^- (S_3 -y^{\prime } S_2)A_2 ,\nonumber \\ &&(S_2^{-1}-y^{\prime }/q S_3^{-1}) A_3 g_{2 3}^-=g_{2 3}^-(S_2^{-1}-y/q S_3^{-1}) A_3. \end{eqnarray*}$$

The first and second equations tell us that $g_{2 3}^-$ commutes with S₂/S₃ and with A₂A₃. the third and fourth suggest that $g_{2 3}^-$ keeps A₂ and A₃ separated, so it should only depend on S₂/S₃. The third equation then gives:

$$\begin{equation*} (S_3 -y S_2)A_2 g_{2 3}^-[S_2/ S_3]=g_{2 3}^-[q S_2/S_3] (S_3 -y S_2)A_2 =g_{2 3}^-[S_2/ S_3] (S_3 -y^{\prime } S_2)A_2 \nonumber \end{equation*}$$

which can be rewritten as:

$$\begin{equation*} \frac{g_{2 3}^-[S_2/S_3]}{g_{2 3}^-[q S_2/ S_3]}=\frac{(1 -y S_2/S_3)}{(1 -y^{\prime } S_2/S_3)}\nonumber \end{equation*}$$

and produces, through iteration:

$$\begin{equation} \fbox{$ g_{2 3}^-(y,y^{\prime })=\displaystyle\frac{(y S_2/S_3)_\infty }{(y^{\prime } S_2/S_3)_\infty }. $} \end{equation} \tag{ 31 }$$

Finally, we look for $g_{2 3}^+(x,x^{\prime })$ such that $\tilde{L}_2^+(x) L_3^+(x^{\prime })g_{2 3}^+(x,x^{\prime })=g_{2 3}^+(x,x^{\prime })\tilde{L}_2^+(x^{\prime }) L_3^+(x)$. Setting a₂ = x, c₂ = q, a₃ = x', and the rest to 1, in $\tilde{L}_2 L_3$, we find:

$$\begin{equation*} \tilde{L}_2^+(x) L_3^+(x^{\prime })=\left[\begin{array}{@{}cc@{}}(x x^{\prime }+ q S_2/ S_3) A_2 A_3 & (x S_3 -q S_2)A_2 \\ (x^{\prime } S_2^{-1}- S_3^{-1}) A_3 & S_3/S_2+1\end{array}\right] \nonumber \end{equation*}$$

and the exact same operations as before produce:

$$\begin{equation} \fbox{$ g_{2 3}^+(x,x^{\prime })=\displaystyle\frac{(x^{\prime } S_3/S_2)_\infty }{(x S_3/S_2)_\infty }. $} \end{equation} \tag{ 32 }$$

Let us note that g⁺ and g⁻ commute with the projection which we perform on $L_1\tilde{L}_2$ and $L_3\tilde{L}_4$ to get the X matrices.

We can finally forget everything about that alternative construction of X(x, y), and simply define those operators f and g^± as what we found them to be. After re-indexing them so that 1 refers to the auxiliary space of the first X matrix, and 2 to the second, we can write:

with

where we relabelled f₁₂ as f₁, f₃₄ as f₂, and $g^\pm _{23}$ as $g^\pm _{12}$, consistently with the indexes of the X matrices.

Applying those one after the other, we find the full R matrix:

$$\begin{equation} \fbox{$ X_1(x,y)\cdot X_2(x^{\prime },y^{\prime }) R(x,y;x^{\prime },y^{\prime })=R(x,y;x^{\prime },y^{\prime }) X_1(x^{\prime },y^{\prime })\cdot X_2(x,y) $} \end{equation} \tag{ 37 }$$

with

$$\begin{eqnarray} &&\fbox{$\begin{array}{l} R(x,y;x^{\prime },y^{\prime })= R_y(x,y;x^{\prime },y^{\prime })R_x(x,y^{\prime };x^{\prime },y) \\ \qquad\qquad\;\; = \displaystyle\frac{(q A_2)_\infty }{(x^{\prime }y^{\prime }A_2)_\infty }\frac{(y S_1/S_2)_\infty }{(y^{\prime } S_1/S_2)_\infty }\frac{(x^{\prime } y A_2)_\infty }{(q A_2)_\infty }\frac{(q A_1)_\infty }{(xy^{\prime }A_1)_\infty }\frac{(x^{\prime } S_2/S_1)_\infty }{(x S_2/S_1)_\infty }\frac{(x^{\prime }y^{\prime } A_1)_\infty }{(q A_1)_\infty } \vphantom{\frac{\frac{a}{b}} {\frac{a} {\frac{a}{b}}}}\end{array}$} \end{eqnarray} \tag{ 38 }$$

(cf figures 3–5), or an equivalent expression if we apply R_x to the left of R_y instead.

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There are many things to be said about that R matrix. First of all, it is the rigorous way to prove that $T^{{\rm per}}_\mu (x,y)$ commutes with $T^{{\rm per}}_\mu (x^{\prime },y^{\prime })$ for any value of those parameters: to do that, we insert R(x, y; x', y')R⁻¹(x, y; x', y') at any point in the matrix product expression of $T^{{\rm per}}_\mu (x,y)T^{{\rm per}}_\mu (x^{\prime },y^{\prime })$, and make R(x, y; x', y') commute to the left all the way around the trace (figure 6), exchanging parameters between the two rows along the way. When crossing the marked bond, we just need to note that R commutes with A_μ · A_μ.

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Equations (33) and (34) are also of interest by themselves: they tell us that $T^{{\rm per}}_\mu (x,y)$ and $T^{{\rm per}}_\mu (x^{\prime },y^{\prime })$ can actually exchange only one of their parameters and keep the other, instead of commuting altogether (by applying the procedure we just described, but with R_x or R_y instead of R). This decomposition of the R matrix is a well known property [49, 50], and will be extremely useful to us in the next section.

Moreover, considering the decomposition (19) which we obtained in the previous section for special values of the spectral parameters in $T^{{\rm per}}_\mu (x,y)$, by taking y = 1/q^{k − 1}x and y' = 1/q^{l − 1}x' in R(x, y; x', y'), we should be able to recover the R matrix between auxiliary dimensions k and l as an independent part of the whole matrix. In other words, this R matrix of twice infinite dimension should contain all the smaller R matrices for the XXZ chain. Taking y = 1/qx, in particular, should yield the Lax matrix X (possibly up to a permutation). Because of the complexity of equation (38), those facts are yet to be verified.

From the previous section, we now know that T^per satisfies:

$$\begin{equation} T^{{\rm per}}(x,y)T^{{\rm per}}(x^{\prime },y^{\prime })=T^{{\rm per}}(x,y^{\prime })T^{{\rm per}}(x^{\prime },y). \end{equation} \tag{ 39 }$$

Fixing x' = y' = 0 in that equation (although any other constants would do), we can therefore write:

with:

$$\begin{equation*} P(x)=\big[T_\mu ^{{\rm per}}(0,0)\big]^{-1}T_\mu ^{{\rm per}}(x,0) ,\quad Q(y)=T_\mu ^{{\rm per}}(0,y).\nonumber \end{equation*}$$

Since the matrices P and Q are combinations of the matrix $T_\mu ^{{\rm per}}$ taken at various values of its parameters, and considering equation (39), we immediately see that P(x) and Q(y) commute for any values of x and y.

This relation is crucial to our reasoning, and, put together with (19), will allow us to reach our conclusion in just a few more lines.

Using this, we can now rewrite (19) as:

$$\begin{equation} \fbox{$ P(x)Q(1/q^{k-1}x)=t^{[k]}(x)+{\rm e}^{-k\mu }P(q^k x)Q(q/x). $} \end{equation} \tag{ 41 }$$

The first and second orders of this equation give:

$$\begin{eqnarray} P(x)Q(1/x)&=h(x)+{\rm e}^{-\mu }P(q x)Q(q/x) , \end{eqnarray} \tag{ 42 }$$

$$\begin{eqnarray} P(q x)Q(1/q x)&=h(q x)+{\rm e}^{-\mu }P(q^2 x)Q(1/x) , \end{eqnarray} \tag{ 43 }$$

$$\begin{eqnarray} P(x)Q(1/q x)&=t^{[2]}(x)+{\rm e}^{-2\mu }P(q^2 x)Q(q/x) , \end{eqnarray} \tag{ 44 }$$

where we have written the first one twice (once at x, once at qx).

Combining those equations as Q(1/qx) ×(42) +e^−μQ(q/x) ×(43)−Q(1/x) ×(44) yields the T–Q relation:

$$\begin{equation} \fbox{$ t^{[2]}(x)Q(1/x)=h(x)Q(1/qx)+{\rm e}^{-\mu }h(qx)Q(q/x). $} \end{equation} \tag{ 45 }$$

Using this equation in conjunction with (27) then gives M_μ in terms of Q:

$$\begin{equation} M_\mu =(1-q)\frac{{\rm d}}{{\rm d}x}\log \biggl (\frac{Q(q/x)}{Q(1/x)}\biggr )\biggl |_{x=-1}. \end{equation} \tag{ 46 }$$

Let us note that using equation (41), we can obtain all the equations from the so-called 'fusion hierarchy' [35, 36], which gives equations on the decomposition of products of matrices t^[k], as well as the T–Q equations for any t^[k] (which involve products of k − 1 matrices Q). We find for instance that

$$\begin{equation*} t^{[2]}(q^2x)t^{[3]}(x)=h(q^2x)t^{[4]}(x)+{\rm e}^{-\mu }h(q^3x)t^{[2]}(x),\nonumber \end{equation*}$$

and

$$\begin{eqnarray*} &&t^{[3]}(x)Q(1/x)Q(1/qx)=h(x)Q(1/qx)Q(1/q^2x)+{\rm e}^{-\mu }h(qx)Q(q/x)Q(1/q^2x)\nonumber \\ &&+\,{\rm e}^{-2\mu }h(q^2x)Q(q/x)Q(x).\nonumber \end{eqnarray*}$$

See appendix C for more details on these relations.

At this point, in order to get an explicit result for the largest eigenvalue E(μ) of M_μ, we need some additional information, which we get by taking μ to 0.

We want to recover the expression for E(μ) which was found in [12]. Everything that we have shown so far is valid for all eigenspaces of M_μ, but we now need to look for any information that we might have specifically on the dominant eigenvalue or eigenvectors of M_μ, so that we can solve equation (46) explicitly in that eigenspace. Starting from the coordinate Bethe Ansatz, this was done, in [12], using prior knowledge on the behaviour of Bethe roots for μ → 0. We cannot use that same argument here, for two reasons: we did not start from the coordinate Bethe Ansatz, so that our Q-operator is not defined in terms of Bethe roots, and even though we can, in principle, argue that the two definitions of Q have to be consistent with one another, we do not expect to be able to do the same in the open case. Fortunately, we can get to the same result by purely algebraic calculations.

Let us go back to the definition of $T_\mu ^{{\rm per}}$ as given in (10), as well as that of X in terms of S and A:

$$\begin{equation*} T_\mu ^{{\rm per}}(x,y)={\rm Tr}\bigg[A_\mu \prod \limits _{i=1}^{L}X^{(i)}\bigg]\; {\rm with}\; X=\left[\begin{array}{@{}cc@{}}1+xA & S(1-xyA)\\ S^{-1}(1-A)& 1+yA\end{array}\right].\nonumber \end{equation*}$$

Expanding this expression in powers of A, we see that each weight in $T_\mu ^{{\rm per}}$ is a finite sum of traces of the form xⁱy^jq^kTr[A_μS^lA^m], where k, l, and m are integers, and the term q^k comes from ordering the powers of A and S. Because of the trace, those terms are 0 if l ≠ 0 (which accounts for the conservation of the total number of particles: the number of ds and es in the trace have to be the same). Otherwise, they are simply equal to xⁱy^jq^k/(1 − q^me^−μ). In particular, there is exactly one term with m = 0, which is equal to 1/(1 − e^−μ). In the non-deformed limit μ → 0, this term dominates the trace in each entry of the transfer matrix, as it is the only term that diverges, so that:

$$\begin{equation*} T_\mu ^{{\rm per}}(x,y)\sim \frac{1}{1-{\rm e}^{-\mu }}|1_N\rangle \langle 1_N|\nonumber \end{equation*}$$

where |1_N〉 is a vector with all entries equal to 1 if the total number of particles is N, and 0 otherwise. |1_N〉〈1_N| is the projector onto the dominant eigenspace (i.e. the steady state) of M in the sector with N particles.

From this, we draw two important conclusions. First, the prefactor in Q(y) diverges as 1/μ. Secondly, we see that this limit does not depend on x and y, which means that the roots of P(x) and Q(y), in this eigenspace, all go to infinity when μ goes to 0. This allows us to use a contour integral in (42) to separate the contribution from P(x) and P(qx) (whose roots all go to infinity, and are therefore all out of the unit circle for μ small enough) and those from Q(1/x) and Q(q/x) (whose roots, in terms of x, all go to 0, and are inside of the unit circle for μ small enough). We also find that the eigenvalue of t^[2] in that eigenspace is

$$\begin{equation*} \langle 1_N|t^{[2]}|1_N\rangle =h(x)+h(qx)\nonumber \end{equation*}$$

which gives E(0) = 0. Using this, we can solve (42) perturbatively in μ and find Q in terms of h. Putting the result in (46), we finally find E(μ). Since we will be doing the same for the open ASEP in the next section with only a few differences, we will give those final steps in detail only in section 2.5.

Let us first find out how the presence of boundaries make this case different from the previous one. Most of the results and calculations in this section are generalizations of those found in [16]. We define:

which has the same structure as the transfer matrix from [16], but with the X matrices replaced by their generalization. Note that, since we have two transfer matrices, we have, in principle, four free parameters (two in each matrix), but we must in fact put the same parameter twice in each matrix (so that U_μ depends only on x, and T_μ on y) if we want to be able to find suitable boundary vectors. To simplify notations, we will therefore rewrite X and $\hat{X}$ as:

$$\begin{equation*} X(x)=\left[\begin{array}{@{}cc@{}}n_x & e\\ d & n_x\end{array}\right] ,\quad \hat{X}=\frac{(1-q)}{2}\left[\begin{array}{@{}cc@{}}\tilde{n}_{x} & e\\ -d & -\tilde{n}_x\end{array}\right]\nonumber \end{equation*}$$

with n_x = 1 + xA and $\tilde{ n}_x=1-x A$.

Note that for a general set of fugacities, the generalization (14) holds, with the matrices $A_{\mu _i}$ being inserted in both U and T.

The conditions that these boundary vectors must satisfy is:

where we notice that the first two are the same as in [16, 17] with n_x replacing 1, and the next two also have an extra term (1 − q)yA. For x = y = 0, we naturally recover the conditions given in [16]. Note that those conditions were found by trial and error, so that they may not be unique.

2.1.1. Commutation relations

As in the periodic case, we will now see how the transfer matrices we have just defined commute with each of the individual jump matrices.

For the bulk matrices M⁽ⁱ⁾, we can simply use equation (11). Summing all of them, we get

$$\begin{eqnarray} \left[\sum _{i=1}^{L-1}M^{(i)},U_\mu \right]&=\langle \!\langle W|\!| A_\mu \hat{X}^{(1)}\prod _{i=2}^{L}X^{(i)} |\!|V\rangle \!\rangle - \langle \!\langle W|\!| A_\mu \prod _{i=1}^{L-1}X^{(i)} \hat{X}^{(L)} |\!|V\rangle \!\rangle , \end{eqnarray} \tag{ 53 }$$

$$\begin{eqnarray} \left[\sum _{i=1}^{L-1}M^{(i)},T_\mu \right]&= \langle \!\langle \tilde{W}|\!| A_\mu \hat{X}^{(1)}\prod _{i=2}^{L}X^{(i)} |\!|\tilde{V}\rangle \!\rangle - \langle \!\langle \tilde{W}|\!| A_\mu \prod _{i=1}^{L-1}X^{(i)} \hat{X}^{(L)} |\!|\tilde{V}\rangle \!\rangle . \end{eqnarray} \tag{ 54 }$$

Unlike in the periodic case, where the trace over the product of matrices ensures that all the $\hat{X}$ matrices cancel one another, we here need to consider the action of the boundary matrices m⁽⁰⁾ and m^(L) as well. We find that we cannot cancel the two boundary terms in (53), and those in (54), independently. Instead, we have to consider the commutation with the product U_μ(x)T_μ(y). We show, in appendix A.2, that

$$\begin{eqnarray} &&[m^{(0)}(\mu ),U_\mu T_\mu ]=- \langle \!\langle W|\!| A_\mu \hat{X}^{(1)}\prod _{i=2}^{L}X^{(i)} |\!|V\rangle \!\rangle \cdot T_\mu -U_\mu \cdot \langle \!\langle \tilde{W}|\!| A_\mu \hat{X}^{(1)}\prod _{i=2}^{L}X^{(i)} |\!|\tilde{V}\rangle \!\rangle , \end{eqnarray} \tag{ 55 }$$

$$\begin{eqnarray} &&[m^{(L)},U_\mu T_\mu ]= \langle \!\langle W|\!| A_\mu \prod _{i=1}^{L-1}X^{(i)}\hat{X}^{(L)} |\!|V\rangle \!\rangle \cdot T_\mu +U_\mu \cdot \langle \!\langle \tilde{W}|\!| A_\mu \prod _{i=1}^{L-1}X^{(i)}\hat{X}^{(L)} |\!|\tilde{V}\rangle \!\rangle , \end{eqnarray} \tag{ 56 }$$

which is exactly what is needed to compensate the terms coming from the bulk.

Putting those relations together, we can conclude that, for any values of x and y, we have:

2.1.2. Explicit expressions for the boundary vectors

We will need an expression for the boundary vectors in a few calculations later, so we determine them explicitly here.

For that, we need to define four new parameters in terms of the boundary rates:

$$\begin{eqnarray*} a&=\frac{1}{2\alpha }[(1-q-\alpha +\gamma )+\sqrt{(1-q-\alpha +\gamma )^2+4\alpha \gamma }] ,\nonumber \\ \tilde{a}&=\frac{1}{2\alpha }[(1-q-\alpha +\gamma )-\sqrt{(1-q-\alpha +\gamma )^2+4\alpha \gamma }],\nonumber \\ b&=\frac{1}{2\beta }[(1-q-\beta +\delta )+\sqrt{(1-q-\beta +\delta )^2+4\beta \delta }],\nonumber \\ \tilde{b}&=\frac{1}{2\beta }[(1-q-\beta +\delta )-\sqrt{(1-q-\beta +\delta )^2+4\beta \delta }] ,\nonumber \end{eqnarray*}$$

and reversely:

$$\begin{eqnarray*} \alpha &=\frac{(1-q)}{(1+a)(1+\tilde{a})}, \nonumber \\ \gamma &=-\frac{a \tilde{a}(1-q)}{(1+a)(1+\tilde{a})} ,\nonumber \\ \delta &=-\frac{b \tilde{b}(1-q)}{(1+b)(1+ \tilde{b})} ,\nonumber \\ \beta &=\frac{(1-q)}{(1+b)(1+ \tilde{b})}.\nonumber \end{eqnarray*}$$

In terms of these, equations (49)–(52) become:

$$\begin{eqnarray} &&[d+b\tilde{b}e+(1+b\tilde{b})xA-(b+\tilde{b})] |\!| V \rangle \!\rangle = 0 , \end{eqnarray} \tag{ 58 }$$

$$\begin{eqnarray} &&\langle \!\langle W|\!| [e+a\tilde{a}d+(1+a\tilde{a})xA-(a+\tilde{a})] = 0 , \end{eqnarray} \tag{ 59 }$$

$$\begin{eqnarray} &&[d+b\tilde{b}e+(b+\tilde{b})y A-(1+b \tilde{b})] |\!| \tilde{V} \rangle \!\rangle =0 , \end{eqnarray} \tag{ 60 }$$

$$\begin{eqnarray} &&\langle \!\langle \tilde{W}|\!| [e+a\tilde{a}d+(a+\tilde{a})y A-(1+a \tilde{a})] =0 . \end{eqnarray} \tag{ 61 }$$

We first focus on the right boundary. We saw that a possible representation for d and e is e = S₁(1 − x²A₁) in U_μ or e = S₂(1 − y²A₂) in T_μ, and $d=S_i^{-1}(1-A_i)$ (where indices 1 and 2 refer to the auxiliary spaces of U_μ and T_μ, respectively). Equations (58) and (59) become:

$$\begin{eqnarray*} {[}S_1^{-1}(1-A_1)+b\tilde{b}S_1(1-x^2 A_1)+(1+b\tilde{b})xA_1-(b+\tilde{b})] |\!| V \rangle \!\rangle &= 0 ,\nonumber \\ {[}S_2^{-1}(1-A_2)+b\tilde{b}S_2(1-y^2 A_2)+(b+\tilde{b})y A_2-(1+b \tilde{b})] |\!| \tilde{V} \rangle \!\rangle &=0,\end{eqnarray*}$$

which is to say, multiplying by S_i to the left:

$$\begin{eqnarray*} {[}(1-bS_1)(1-\tilde{b}S_1v)-(1-xS_1)(1-b\tilde{b}xS_1)A_1] |\!| V \rangle \!\rangle &= 0,\nonumber \\ {[}(1-S_2)(1-b\tilde{b}S_2)-(1-byS_2)(1-\tilde{b}yS_2)A_2] |\!| \tilde{V} \rangle \!\rangle &=0.\end{eqnarray*}$$

We can write those vectors as generating functions, which will make it easier for us to manipulate them. Let us then write $|\!| V \rangle \!\rangle =V(S_1) |\!|0\rangle \!\rangle =\sum \nolimits _{k=0}^\infty V_k S_1^k |\!|0\rangle \!\rangle$, which is such that A₁||V〉〉 = V(qS₁)||0〉〉. Those two last equations now become:

$$\begin{eqnarray*} \frac{V(S_1)}{V(qS_1)} |\!|0\rangle \!\rangle &=\frac{(1-xS_1)(1-b\tilde{b}xS_1)}{(1-bS_1)(1-\tilde{b}S_1)} |\!|0\rangle \!\rangle ,\nonumber \\ \frac{ \tilde{V} (S_2)}{ \tilde{V} (qS_2)} |\!|0\rangle \!\rangle &=\frac{(1-byS_2)(1-\tilde{b}yS_2)}{(1-S_2)(1-b\tilde{b}S_2)}|\!|0\rangle \!\rangle ,\nonumber \end{eqnarray*}$$

which we can iterate to get:

where we recall that (x)_∞ is the infinite q-Pochhammer symbol:

$$\begin{equation*} (x)_n=\prod \limits _{k=0}^{n-1}(1-q^kx),\qquad (x)_\infty =\prod \limits _{k=0}^{\infty }(1-q^kx).\nonumber \end{equation*}$$

We will sometimes write products of those in a more compact form, with only one symbol, such as (x, y)_∞ = (x)_∞(y)_∞.

As for the left boundary, it is simpler to treat it using the contragredient representation $\overline{X}$ of X on both vectors (which are then the exact symmetric of ||V〉〉 and $|\!| \tilde{V} \rangle \!\rangle$, but with a and $\tilde{a}$ replacing b and $\tilde{b}$), and then apply the operator f_i that we found in section 1.3 in order to go back to X. This gives us:

to which we could add factors $\frac{(q)_\infty }{(x^2)_\infty }$ and $\frac{(q)_\infty }{(y^2)_\infty }$ for normalization.

In fact, in most future calculations, we will use X for U_μ and $\overline{X}$ for T_μ, so that the factor $\frac{(y^2A_2)_\infty }{(qA_2)_\infty }$ goes to $|\!| \tilde{V} \rangle \!\rangle$ instead of $\langle \!\langle \tilde{W}|\!|$.

Note finally that scalar products involving those vectors (like, for instance, all the weights of U_μ and T_μ) are well defined only for |a|, |b|, $|\tilde{a}|$ and $|\tilde{b}|$ all smaller than 1. This corresponds to the ASEP being in its so-called 'maximal current' phase. We can assume this to be the case for now, but a simple analytic continuation can be performed in order to access more generic values of the parameters [47].

We now need to show two things: that the transfer matrix is a product of two commuting matrices, one depending on x and one on y, and that for special values of the parameters it decomposes into two independent blocks.

Unlike what we did for the periodic case, we will first show that U_μ(x)T_μ(y) factorizes into two commuting matrices P(x) and Q(y). Note that this decomposition is not trivial: U_μ and T_μ do not commute, so P is not simply equal to U_μ and Q to T_μ.

First, we need to show that U_μ(x)T_μ(y) commutes with U_μ(x')T_μ(y') for any values of x, y, x' and y'. If we want to be rigorous, the fact that both commute with M_μ is not enough (M_μ might have a degenerate eigenspace which is not degenerate for U_μ(x)T_μ(y)), so, as in the periodic case, we need to use R matrices to exchange spectral parameters. We cannot do as easily as in the periodic case, because of the boundaries: instead of taking R around a trace to exchange the top and bottom matrices, we now need to show that when applied to a boundary, R transfers the spectral parameters from a vector to another. We find the two following results (which are proven in appendix B.1 and illustrated on figure 7):

$$\begin{eqnarray*} &&R^{-1}(x,x;y,y) |\!| V(x) \rangle \!\rangle \cdot |\!| \tilde{V}(y) \rangle \!\rangle =|\!| V(y) \rangle \!\rangle \cdot |\!| \tilde{V}(x) \rangle \!\rangle ,\nonumber \\ &&R^{-1}(y,y;x,x) |\!| \tilde{V}(y) \rangle \!\rangle \cdot |\!| V(x)\rangle \!\rangle =|\!| \tilde{V}(x) \rangle \!\rangle \cdot |\!|V(y) \rangle \!\rangle ,\nonumber \end{eqnarray*}$$

where we use the same R matrix as defined in (38).

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These relations imply, at the level of the transfer matrices, that

which immediately gives that U_μ(x)T_μ(y) commutes with U_μ(x')T_μ(y'), but also that

$$\begin{equation*} U_\mu (x)T_\mu (y) U_\mu (0)T_\mu (0)=U_\mu (x)T_\mu (0) U_\mu (0)T_\mu (y)\nonumber \end{equation*}$$

so that we can write, dividing by U_μ(0)T_μ(0),

with

$$\begin{equation*} P(x)=U_\mu (x)\left[U_\mu (0)\right]^{-1} , Q(y)=U_\mu (0)T_\mu (y),\nonumber \end{equation*}$$

where P and Q commute with each other. Note that, in terms of P and Q and at y = 0, equation (66) becomes P(x)Q(0) = P(0)Q(x), so that P and Q are the same up to some constant (since P(0) is set to 1 but not Q(0)). We will not need to use this identity in our calculations, so we will keep using distinct names for those two objects, to have notations that are consistent with the periodic case.

In this section, we investigate whether anything happens for those special values of x and y that we considered in section 1.2. The calculations involved are slightly more complicated than they were then, not only because of the boundaries, but also because x and y are now separated, and the factors (1 − xyA) in e have been replaced by (1 − x²A₁) and (1 − y²A₂), which do not give anything useful for xy = q^{1 − k}.

The simple solution to this problem is to put x and y back together, using $g_{12}^-(x,y)$ (as defined in section 1.3) to go from $X(x,x)\cdot \overline{X}(y,y)$ to $X(x,y)\cdot \overline{X}(y,x)$. We choose to keep the second row of matrices in the contragredient representation to simplify future calculations. We recall that it corresponds to having f(xy) commute through X(x, y), so that

$$\begin{equation*} \overline{X}(x,y)=f(xy)X(x,y)f^{-1}(xy)=\left[\begin{array}{@{}cc@{}}n_x & {}^t\!d\\ {}^t\!e & n_y\end{array}\right].\nonumber \end{equation*}$$

This operation makes the boundary vectors more complicated: they are not a product of two vectors acting each on one row any more (see figure 8). We will write those vectors as K⁺ for the right boundary and K⁻ for the left one:

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We will denote by $K^+_{i,j}$ the coefficient of ||i, j〉〉 in K⁺ (i.e. the coefficient of $S_1^{i}S_2^{j}$ in the expansion of the ratios of q-Pochhammer symbols in K⁺), and by $K^-_{j,i}$ the coefficient of 〈〈i, j|| in K⁻. Note that, in this section, the normalization of the boundary vectors is important. The factors $\frac{(q)_\infty }{(x^2)_\infty }$ and $\frac{(q)_\infty }{(y^2)_\infty }$ in K⁺ and K⁻ are there so that $K^-_{0,0}=K^+_{0,0}=1$. Also note that the transpose of K⁻ can be obtained from K⁺ through the transformation $\lbrace x\leftrightarrow y,S_1\rightarrow aS_2,S_2\rightarrow aS_1, b\rightarrow 1/a,\tilde{b}\rightarrow \tilde{a}\rbrace$, so that we only need to do calculations on K⁺.

Now that x and y have been reunited in the bulk of the system, the same decomposition (into block triangular matrices) happens as did in the periodic case. We have to see whether the same happens to the boundary matrices K^±.

2.3.1. One-way boundaries

We first focus on the simpler case where $\tilde{a}=\tilde{b}=0$, which is to say γ = δ = 0. For this simpler case, we have:

$$\begin{equation*} K^+=\frac{(xS_1)_\infty }{(bS_1)_\infty }\frac{(y S_1/S_2)_\infty }{(x S_1/S_2)_\infty }\frac{(y^2A_2)_\infty }{(qA_2)_\infty }\frac{(byS_2)_\infty }{(S_2)_\infty } \frac{(q)_\infty }{(y^2)_\infty }\nonumber \end{equation*}$$

where the factor ||0₁, 0₂〉〉 on which this whole function acts is implicit.

This can be expanded into:

$$\begin{equation*} K^+=\frac{(xS_1)_\infty }{(bS_1)_\infty }\sum \limits _{k=0}^{\infty }\frac{(y/x)_k}{(q)_k}(x S_1/S_2)^k\sum \limits _{n=0}^{\infty }\frac{(b y)_n}{(y^2)_n}(S_2)^n\nonumber \end{equation*}$$

where the first sum comes from $\frac{(y S_1/S_2)_\infty }{(x S_1/S_2)_\infty }$, and the second from everything that is to the right of that.

We will shortly be in need of one of Heine's transformation formulae for basic hypergeometric series, which we give now. For a function ₂ϕ₁(a, b; c; z) defined as:

$$\begin{equation} _2\phi _1(a,b;c;z)=\sum \limits _{n=0}^{\infty }\frac{(a)_n(b)_n}{(q)_n(c)_n}z^n \end{equation} \tag{ 71 }$$

we have [53]:

$$\begin{equation} \fbox{$ _2\phi _1(a,b;c;z)=\displaystyle\frac{(z ab/c)_\infty }{(z)_\infty } {}_2\phi _1(c/a,c/b;c;zab/c). $} \end{equation} \tag{ 72 }$$

We will also need this simple identity on q-Pochhammer symbols:

$$\begin{equation} \fbox{$ (x)_{j+k}=(x)_j(x q^j)_{k}. $} \end{equation} \tag{ 73 }$$

Coming back to K⁺, we have:

$$\begin{eqnarray*} K^+&=\frac{(xS_1)_\infty }{(bS_1)_\infty }\sum \limits _{j=0}^{\infty }\frac{(b y)_j}{(y^2)_j}(S_2)^j\sum \limits _{k=0}^{\infty }\frac{(y/x)_k}{(q)_k}(x S_1/S_2)^k\frac{(b yq^j)_k}{(y^2q^j)_k}(S_2)^k\nonumber \\ &=\sum \limits _{j=0}^{\infty }\frac{(b y)_j}{(y^2)_j}(S_2)^j\frac{(xS_1)_\infty }{(bS_1)_\infty }{}_2\phi _1(y/x,byq^j;y^2q^j;xS_1)\nonumber \\ &=\sum \limits _{j=0}^{\infty }\frac{(b y)_j}{(y^2)_j}(S_2)^j{}_2\phi _1(xyq^j,y/b;y^2q^j,bS_1) \nonumber \end{eqnarray*}$$

where we used (73) between the first line and the second, and (72) between the second and the third. This finally gives us:

$$\begin{equation} \fbox{$ K^+_{i,j}(x,y)=\displaystyle\frac{(xyq^j)_i(y/b)_i b^i(b y)_j}{(q)_i(y^2)_{i+j}} $} \end{equation} \tag{ 74 }$$

and, at the left boundary:

$$\begin{equation} \fbox{$ K^-_{j,i}(x,y)=\displaystyle\frac{(xyq^i)_j(x/a)_i a^i(a x)_j}{(q)_j(x^2)_{i+j}}. $} \end{equation} \tag{ 75 }$$

We need to compare those values for (x, y) equal to (1/q^{k − 1}y, y) or (q/y, q^ky) (which are the same values as before, but keeping y as a variable instead of x). We get:

$$\begin{eqnarray*} &&K^+_{i,j}(1/q^{k-1}y,y)=\frac{(q^{j-k+1})_i(y/b)_i b^i(b y)_j}{(q)_i(y^2)_{i+j}},\nonumber \\ &&K^+_{i,j}(q/y,q^ky)=\frac{(q^{j+k+1})_i(yq^k/b)_i b^i(b yq^k)_j}{(q)_i(y^2q^{2k})_{i+j}}.\nonumber \end{eqnarray*}$$

Since (q^{j − k + 1})_i = 0 for j − k + 1 ⩽ 0 and i + j − k ⩾ 0, i.e. for j ⩽ k − 1 and i ⩾ k − j, the first matrix is, as we expected, block triangular.

We now consider, for j ⩾ k − 1, the ratio:

$$\begin{equation*} \frac{K^+_{i+k,j+k}(1/q^{k-1}y,y)}{K^+_{i,j}(q/y,q^ky)}=\frac{(y/b)_k b^k(b y)_k}{(y^2)_{2k}}\frac{(q^{j+1})_k}{(q^{i+1})_k}.\nonumber \end{equation*}$$

The term $\frac{(q^{j+1})_k}{(q^{i+1})_k}$ accounts for going to the contragredient representation on both lines (consistently with what happens to the X matrices), and can be transferred to the left boundary, where they compensate similar terms, emerging from the same calculation:

$$\begin{equation*} \frac{K^-_{j+k,i+k}(x,1/q^{k-1}x)}{K^-_{j,i}(q^k x,q/x)}=\frac{(x/a)_k a^k(a x)_k}{(x^2)_{2k}}\frac{(q^{i+1})_k}{(q^{j+1})_k}.\nonumber \end{equation*}$$

All the other terms depend only on k, and factor out of the matrix product. We will now put them together.

First, we need to make a few transformations on those factors. For y = 1/q^{k − 1}x, we have:

$$\begin{eqnarray*} &&(y/b)_k b^k=(-x)^{-k}q^{-k(k-1)/2}(bx)_k,\nonumber \\ &&(b y)_k=(bq^{-k+1}/x)_k,\nonumber \\ &&(x/a)_k a^k=(-x)^{k}q^{k(k-1)/2}(aq^{-k+1}/x)_k,\nonumber \end{eqnarray*}$$

which gives us:

$$\begin{eqnarray*} \frac{(x/a)_k a^k(a x)_k(y/b)_k b^k(b y)_k}{(x^2)_{2k}(y^2)_{2k}}&=\frac{(aq^{-k+1}/x)_k (bq^{-k+1}/x)_k (a x)_k (bx)_k }{(x^2)_{2k}(x^{-2}q^{-2k+2})_{2k}}\nonumber \\ &=\frac{h_b(q^k x)h_b(q/x)}{h_b(x)h_b(1/q^{k-1}x)}\nonumber \end{eqnarray*}$$

with

$$\begin{equation} \fbox{$ h_b(x)=\displaystyle\frac{(x^2)_{\infty }}{(ax,bx)_\infty }. $} \end{equation} \tag{ 76 }$$

We see that the boundary matrices, just as the bulk matrices, are upper block triangular, with a first block of auxiliary dimension k, and a second of infinite auxiliary dimension, which is the same as the full matrix at different values of the spectral parameters, up to a global factor which we have just computed. Put into equations, this becomes:

$$\begin{equation*} P(x)Q(1/q^{k-1}x)=t^{[2]}(x)+{\rm e}^{-2k\mu }\frac{h_b(q^k x)h_b(q/x)}{h_b(x)h_b(1/q^{k-1}x)}P(q^k x)Q(q/x)\nonumber \end{equation*}$$

where the factor e^−2kμ comes, as before, from the first coefficient of the matrices A_μ (of which there are now 2) in the second block.

To make things simpler, we can redefine all our matrices as:

$$\begin{equation} \tilde{P}(x)=h_b(x)P(x),\quad \tilde{Q}(x)=h_b(x)Q(x),\quad \tilde{t}^{[k]}(x)=h_b(x)h_b(1/q^{k-1}x)t^{[k]}(x) \end{equation} \tag{ 77 }$$

which is equivalent to choosing another normalization for U_μ and T_μ. Written in terms of those new transfer matrices, this last equation takes its definitive form:

This has the exact same form as equation (41), the only difference being a factor 2μ instead of μ in the right hand side. Note that, since $\tilde{P}$ and $\tilde{Q}$ are the same function up to a constant (as we saw at the end of section 2.2), we find that $\tilde{t}^{[k]}$ is symmetric under x↔1/q^{k − 1}x.

The rest of the reasoning is the same as in the periodic case. We first consider the case where k = 1. This gives us quite simply: t^[1](x) = (1 + x)^L(1 + 1/x)^L = h(x). From this, we get:

$$\begin{equation*} \tilde{t}^{[1]}(x)=h(x)h_b(x)h_b(x^{-1})=\frac{(1+x)^L(1+1/x)^L(x^2,x^{-2})_{\infty }}{(ax,bx,a/x,b/x)_\infty }=F(x).\nonumber \end{equation*}$$

This explains how the function F(x) replaces h(x) (from the periodic case) in all the expressions for the cumulants of the current that were found in [15].

For k = 2, through the same calculations as in the periodic case, we find the T–Q equation

$$\begin{equation*} t^{[2]}(x)Q(1/x)=h(x)Q(1/qx)+{\rm e}^{-2k\mu }\frac{h_b(qx)h_b(q/x)}{h_b(x)h_b(1/x)}h(qx)Q(q/x)\nonumber \end{equation*}$$

which translates into

$$\begin{equation} \tilde{t}^{\,[2]}(x)\tilde{Q}(1/x)=F(x)\tilde{Q}(1/qx)+{\rm e}^{-2k\mu }F(qx)\tilde{Q}(q/x). \end{equation} \tag{ 79 }$$

Note that although the equation for $\tilde{t}^{[2]}$ is more compact that that for t^[2], the former is not polynomial due to the presence of q-Pochhammer symbols in F and in the normalization of $\tilde{t}^{[2]}$, whereas the latter is.

We now have to verify that $\tilde{t}^{[2]}(x)$ is related to M_μ through some derivative. We will do this directly in the general case (with all four boundary rates), in a few pages. For now, we just give the two-dimensional boundary matrices that we find from K⁺ and K⁻:

$$\begin{equation*} K^+_2=\left[\begin{array}{@{}cc@{}}1 & \displaystyle\frac{qx(qx-b)}{(q^2x^2-1)} \\ \displaystyle\frac{x(1-qxb)}{(q^2x^2-1)} & 0\end{array}\right] ,\quad K^-_2=\left[\begin{array}{@{}cc@{}}1 & \displaystyle\frac{(a-x)}{(1-x^2)} \\ \displaystyle\frac{(ax-1)}{q(1-x^2)} & 0\end{array}\right].\nonumber \end{equation*}$$

2.3.2. Two-way boundaries

In the general case, where both boundaries have two non-zero rates, the calculations are much more involved, and can be found in appendix B.2. Starting from expression (70) for K⁺(x, y), and taking xy = q^{1 − k}, we find that

$$\begin{equation} \fbox{$ \displaystyle\frac{K^+_{i+k,j+k}(x,y)}{K^+_{i,j}(q^kx,q^ky)}=\displaystyle\frac{(x/b)_kb^k(xb)_k(y/\tilde{b})_k(\tilde{b})^k(y\tilde{b})_k}{(y^2 )_{2k}}(-y)^kq^{k(k-1)/2}\displaystyle\frac{(q^{j+1})_k}{(q^{i+1})_k} $} \end{equation} \tag{ 80 }$$

and the corresponding result for the left boundary:

$$\begin{equation} \fbox{$ \displaystyle\frac{K^-_{j+p,i+p}(x,y)}{K^-_{j,i}(q^px,q^py)}=\displaystyle\frac{(y/a)_pa^p(ya)_p(x/\tilde{a})_p(\tilde{a})^p(x\tilde{a})_p}{(x^2 )_{2p}}(-x)^pq^{p(p-1)/2}\displaystyle\frac{(q^{i+1})_p}{(q^{j+1})_p}. $} \end{equation} \tag{ 81 }$$

Using this, we do the exact same operations as in the previous case, replacing h_b with

$$\begin{equation} h_b(x)=\frac{(x^2)_{\infty }}{(ax,\tilde{a}x,bx,\tilde{b}x)_\infty }, \end{equation} \tag{ 82 }$$

and we get the completely general version of the function $F(x)=\tilde{t}^{[1]}(x)$:

We finally look at the transfer matrix $\tilde{t}^{[2]}(x)$ and try to relate it to M_μ, as we did in section 1.2 for the periodic case. By analogy with equation (24), we will try to rewrite $\tilde{t}^{[2]}(h)/F(x)$ as the standard two-dimensional transfer matrix.

The two-dimensional blocks from K⁺ and K⁻ for y = 1/qx are:

$$\begin{eqnarray*} K^+_2=\left[\begin{array}{@{}cc@{}}1 & \displaystyle\frac{qx(qx+qxb\tilde{b}-b-\tilde{b})}{(q^2x^2-1)} \\ \displaystyle\frac{x(1+b\tilde{b}-qxb-qx\tilde{b})}{(q^2x^2-1)} & -b\tilde{b}x\end{array}\right] ,\\ K^-_2=\left[\begin{array}{@{}cc@{}}1 & \displaystyle\frac{(a+\tilde{a}-x-a\tilde{a}x)}{(1-x^2)} \\ \displaystyle \frac{(ax+\tilde{a}x-1-a\tilde{a})}{q(1-x^2)} & -\displaystyle\frac{a\tilde{a}}{q x}\end{array}\right]\nonumber \end{eqnarray*}$$

and the corresponding blocks from the bulk are:

$$\begin{eqnarray*} X_2(x)&=\left[\begin{array}{@{}c c|c c@{}}1+x & 0 & 0 & 0 \\ 0 & 1+q x & 1-\frac{1}{q} & 0 \\ \hline 0 & 1-q & 1+\frac{1}{q x} & 0\\ 0 & 0 & 0 & 1+\frac{1}{x}\end{array}\right],\nonumber \\ \overline{X}_2(x)&=\left[\begin{array}{@{}c c|c c@{}}1+\frac{1}{q x}& 0 & 0 & 0 \\ 0 & 1+\frac{1}{x}& 1-q & 0 \\ \hline 0 & 1-\frac{1}{q} & 1+x & 0\\ 0 & 0 & 0 & 1+q x\end{array}\right].\nonumber \end{eqnarray*}$$

Consider these transformations, with $\lambda = \frac{1+q x}{q(1+x)}$, i.e. $x=-\frac{1-q \lambda }{q(1-\lambda )}$ :

$$\begin{equation*} L^{(i)}(\lambda )=\frac{1}{1+x} \left(\!\left[\begin{array}{@{}cc@{}}1 & 0\\ 0 & -q\end{array}\right]\! X_2(x)\!\left[\begin{array}{@{}cc@{}}1 & 0\\ 0 & -1/q\end{array}\right]\!\right) \!\cdot \!\left[\begin{array}{@{}c | c@{}}1 & 0 \\ \hline 0 & x\end{array}\right] = \left[\begin{array}{@{}c c|c c@{}}1 & 0 & 0 & 0 \\ 0 & q \lambda &1-q\lambda & 0 \\ \hline 0 &1- \lambda & \lambda & 0\\ 0 & 0 & 0 & 1\end{array}\right]\nonumber \end{equation*}$$

and

$$\begin{equation*} \overline{L}^{(i)}(\lambda )=\frac{1}{1+x} \left[\begin{array}{@{}c | c@{}}x & 0 \\ \hline 0 & 1\end{array}\right]\! \cdot \!\left(\!\left[\begin{array}{@{}cc@{}}0 & 1\\ 1 & 0\end{array}\right] \!\overline{X}_2(x)\!\left[\begin{array}{@{}cc@{}}0 & 1\\ 1 & 0\end{array}\right]\!\right) \! = \!\left[\begin{array}{@{}c c|c c@{}}1 & 0 & 0 & 1-q\lambda \\ 0 & \lambda & 0& 0 \\ \hline 0 & 0& q \lambda & 0\\ 1- \lambda & 0 & 0 & 1\end{array}\right]\nonumber \end{equation*}$$

where the matrix products inside the parentheses are done in the auxiliary space, on each element of X₂ or $\overline{X}_2$, and the third product is done on the physical space at each site. Notice that the inner products cancel out between one site and the next, and that the outer products are done between X₂ and $\overline{X}_2$ and amount to a global factor $\frac{x}{(1+x)^2}$ on each site. Taking a product of L matrices $X_2\cdot \overline{X}_2$, this transformation gives, apart from the inner products at each end of the chain, a global factor $\frac{x^L}{(1+x)^{2L}}$, which accounts for the bulk part h(x) of F(x). Also note that the matrices from $\overline{L}^{(i)}$ need to be transposed if multiplied from right to left.

Considering that $\tilde{t}^{[2]}$ has a factor h_b(x)h_b(1/qx), and that

$$\begin{equation*} h(x)h_b(x)h_b(1/qx)=F(x) \frac{(1-1/q^2x^2)(1-1/qx^2)}{(1-a/qx)(1-\tilde{a}/qx)(1-b/qx)(1-\tilde{b}/qx)}\nonumber \end{equation*}$$

the transformations we need to do on the boundary matrices are:

$$\begin{eqnarray*} \hat{K}^+_2(\lambda )&= \frac{(1-1/q^2x^2)}{(1-b/qx)(1-\tilde{b}/qx)}\left[\begin{array}{@{}cc@{}}1 & 0\\ 0 & -q\end{array}\right] K^+_2\left[\begin{array}{@{}cc@{}}0 & 1\\ 1 & 0\end{array}\right],\nonumber \\ \hat{K}^-_2(\lambda )&=\frac{(1-1/qx^2)}{(1-a/qx)(1-\tilde{a}/qx)}\left[\begin{array}{@{}cc@{}}0 & 1\\ 1 & 0\end{array}\right]K^-_2\left[\begin{array}{@{}cc@{}}1 & 0\\ 0 & -1/q\end{array}\right],\nonumber \end{eqnarray*}$$

of which we will not write the full expression in terms of λ. Their values and first derivatives at λ = 0, which is all we will need, are, in terms of the original boundary parameters:

$$\begin{equation} \hat{K}^+_2=\left[\begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right],\quad \frac{{\rm d}}{{\rm d}\lambda }\hat{K}^+_2=\left[\begin{array}{@{}cc@{}}-2\delta & 2\beta \\ 2\delta &1-q-2\beta \end{array}\right], \end{equation} \tag{ 84 }$$

$$\begin{eqnarray} \hat{K}^-_2=\left[\begin{array}{@{}c@{}c@{}}\displaystyle\frac{1\,{-}\,\alpha \,{+}\,\gamma }{1+q} &\displaystyle\frac{\gamma }{q} \\ \alpha & \displaystyle\frac{\alpha \,{-}\,\gamma \,{+}\,q}{1+q}\end{array}\right] ,\quad\!\!\! \displaystyle\frac{{\rm d}}{{\rm d}\lambda }\hat{K}^-_2=\left[\begin{array}{@{}c@{}c@{}}-\alpha \,{+}\,\gamma \,{+}\,A\,{-}\, B &\displaystyle\frac{\gamma (2q\,{-}\,\alpha \,{-}\,\gamma )}{q}\\ \alpha (1\,{+}\,q\,{-}\,\alpha \,{-}\,\gamma ) & -2\gamma \,{-}\,q\,{-}\,A\,{+}\, B\end{array}\right], \end{eqnarray} \tag{ 85 }$$

with $A=\frac{2+(\alpha -2)\alpha -\gamma ^2}{1+q}$ and $B=\frac{2(1-\alpha +\gamma )}{(1+q)^2}$.

We now put all these matrices together, obtaining $\tilde{t}^{[2]}(h)/F(x)$ as desired. The Lax matrices L⁽ⁱ⁾ are not exactly the same as those we had for the periodic case. Their values at λ = 0 are L⁽ⁱ⁾(0) = P⁽ⁱ⁾ and $\overline{L}^{(i)}(0)=\overline{P}^{(i)}$, where P⁽ⁱ⁾ is the permutation matrix exchanging the auxiliary space with the physical space when applied to the right in the matrix product, and $\overline{P}v$ is the same but when applied to the left. We see that the two boundaries do not play the same role. The right boundary matrix is simply the identity at λ = 0, and serves to connect the Lax matrices on the last site. The left boundary matrix is traced, at λ = 0, because of the Lax matrices on the first site, and we see that its trace is 1. All in all, at λ = 0, the whole transfer matrix is simply the identity (see figure 9(a)).

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As for its first derivative with respect to λ, we find that each pair of Lax matrices L⁽ⁱ⁾ and $\overline{L}^{(i)}$ have a total contribution of 2M⁽ⁱ⁾ + (1 − q)(τ_i − τ_{i + 1}), where τ_i is the number of particles on site i. The right boundary matrix, as can be seen in (84), gives a term 2m^(L) + (1 − q)τ_L. At the left boundary, we have three terms to consider, one involving the derivative of K⁻ and the first Lax matrices taken at 0, and two more where we take the derivative of the first Lax matrices, and the boundary matrix at 0. The sum of those terms gives a total contribution of 2m⁽⁰⁾ − (1 − q)τ₁. If we now sum all the terms that we have found, the parts proportional to (1 − q) all cancel out, and we are simply left with 2M_μ.

At the end of the day, we find that:

which is the same as equation (26), with h(x) replaced by F(x), and an extra factor $\frac{1}{2}$. We have not mentioned the matrices A_μ in those last calculations, because their behaviour is trivial, and it is left to the reader, as an exercise, to check that adding one between two sites gives the correct deformation for M_μ.

As for the periodic case, we could have done all those calculations around λ = ∞ instead of 0, which would have given an equivalent result:

$$\begin{equation} M_\mu =\frac{1}{2}(1-q)\frac{{\rm d}}{{\rm d} x} \log \biggl (\frac{\tilde{t}^{[2]}(x)}{F(qx)}\biggr )\bigg |_{x=-1}. \end{equation} \tag{ 87 }$$

In order to obtain the results found in [15], we need one last piece of information: how P and Q behave as μ goes to 0.

As in the periodic case, the final step is to consider the μ → 0 limit in our transfer matrices, in order to get some information specific to the dominant eigenspace of M_μ and the corresponding eigenvalue E(μ). Once more, this is made much more difficult by the presence of the boundaries, but the principle remains the same.

We first need to consider the behaviour of T_μ(y) alone. As we recall, it is defined by:

$$\begin{equation*} T_\mu (y)=\langle \!\langle \tilde{W}|\!| A_\mu \prod _{i=1}^{L}X^{(i)}|\!| \tilde{V} \rangle \!\rangle \nonumber \end{equation*}$$

where

$$\begin{equation*} X=\left[\begin{array}{@{}cc@{}}1+yA & S(1-y^2A)\\ S^{-1}(1-A)& 1+yA\end{array}\right]\nonumber \end{equation*}$$

and

$$\begin{equation*} \langle \!\langle \tilde{W}|\!|=\langle \!\langle 0|\!|\frac{(ay/S)_\infty (\tilde{a}y/S)_\infty }{(1/S)_\infty (a\tilde{a}/S)_\infty }\frac{(y^2A)_\infty }{(qA)_\infty } ,\qquad |\!| \tilde{V} \rangle \!\rangle =\frac{(byS)_\infty (\tilde{b}yS)_\infty }{(S)_\infty (b\tilde{b}S)_\infty } |\!|0\rangle \!\rangle .\nonumber \end{equation*}$$

As in the periodic case, we can expand each entry in T_μ as a finite sum of terms of the form $y^iq^k\langle \!\langle \tilde{W}|\!| A_\mu S^l A^m |\!| \tilde{V} \rangle \!\rangle$ (but this time, there is no constraint on l, as the number of particles is no longer conserved). In each entry, there is exactly one term with m = 0, equal to $\langle \!\langle \tilde{W}|\!| A_\mu S^l |\!| \tilde{V} \rangle \!\rangle$, where l is the difference between the number of particles in the initial and final configurations. We show, in appendix D, that for μ → 0, this term behaves as 1/(1 − e^−μ) times a prefactor which does not depend on l, and all the others remain finite, so that:

$$\begin{equation*} T_\mu (y)\sim \frac{1}{1-{\rm e}^{-\mu }}\frac{(ay,\tilde{a}y,by,\tilde{b}y)_\infty }{(y^2)_\infty (a\tilde{a},b\tilde{b},q)_\infty }|1\rangle \langle 1|.\nonumber \end{equation*}$$

Notice that the prefactor is equal to 1/h_b(y) as defined in (82), up to a constant.

We can do the same calculations for U_μ, and see that there is no divergence there, but it is not necessary. Instead, we just apply U_μ(x) on this last result. As we recall, U_μ(x) is defined by

$$\begin{equation*} U_\mu (x)= \langle \!\langle W|\!| A_\mu \prod _{i=1}^{L}X^{(i)} |\!| V \rangle \!\rangle \nonumber \end{equation*}$$

with

$$\begin{equation*} X=\left[\begin{array}{@{}cc@{}}n_x & e\\ d& n_x\end{array}\right]\nonumber \end{equation*}$$

and where 〈〈W|| and ||V〉〉 verify:

$$\begin{eqnarray*} [\beta (d+n_x) - \delta (e+n_x) -(1-q)] |\!| V \rangle \!\rangle &= 0 ,\nonumber \\ \langle \!\langle W|\!| [\alpha (e+n_x)- \gamma (d+n_x)-(1-q)] &= 0.\end{eqnarray*}$$

Since the vector |1〉 can be written as the tensor product of the constant vector |1_i〉 on each site i, applying each X to |1〉 gives the vector

$$\begin{equation*} X|1_i\rangle =\left[\begin{array}{@{}c@{}}e+n_x \\ d+n_x\end{array}\right]{\rm which\ we\ will\ note} \left[\begin{array}{@{}c@{}}E \\ D\end{array}\right].\nonumber \end{equation*}$$

Since, as we recall, d, e, and A satisfy

$$\begin{eqnarray*} &&de -q\, ed=(1-q)(1-x^2 A^2),\nonumber \\ &&A e=q\, e A , \nonumber \\ &&{\rm d} A=q\, A d,\nonumber \end{eqnarray*}$$

we find that D and E satisfy

$$\begin{equation} \fbox{$ DE-q\, ED=(1-q)(D+E) $} \end{equation} \tag{ 88 }$$

and the conditions on the boundary vectors simply write

The matrix U₀(x)T₀(y) becomes, up to a prefactor:

$$\begin{equation*} U_0(x)T_0(y)\sim |P^\star \rangle \langle 1|\nonumber \end{equation*}$$

where, for any configuration $\mathcal{C}=\lbrace \tau _i\rbrace$,

$$\begin{equation} \fbox{$ P^\star (\mathcal{C})=\langle \!\langle W|\!| \prod\limits_{i=1}^{L}[(1-\tau _i)E+\tau _i D] |\!| V \rangle \!\rangle $} \end{equation} \tag{ 91 }$$

(where we used the fact that A_μ = 1 for μ = 0), which does not depend on x.

Moreover, since we know that [M, U₀(x)T₀(y)] = 0 (by considering equation (57) at μ = 0), and that 〈1|M = 0 (because M is a stochastic matrix), we find that

$$\begin{equation} \fbox{$ M|P^\star \rangle =0 $} \end{equation} \tag{ 92 }$$

which is to say that |P^⋆〉 is the steady state of the open ASEP. We therefore recover the original 'matrix Ansatz', as presented in [17]. Note that a matrix Ansatz also exists for the steady state of the periodic multi-species ASEP [48], which one should be able to recover from an appropriate transfer matrix with spectral parameters and current-counting deformations, and which would be related to the U_q(SU(k)) algebra (for k different types of particles, counting holes as one of those types).

The normalization of U₀(x)T₀(y) still needs to be determined. Since we know that the only dependence in y is a prefactor 1/h_b(y), and that U_μ(x)T_μ(y) = U_μ(y)T_μ(x) (as shown in equation (66)), we conclude that, up to a constant term which we include in |P^⋆〉:

$$\begin{equation*} U_0(x)T_0(y)=\frac{1}{1-{\rm e}^{-\mu }} \frac{1}{h_b(x)h_b(y)}|P^\star \rangle \langle 1|\nonumber \end{equation*}$$

which is to say that, similarly to the periodic case, the prefactor in $\tilde{Q}(y)$ (as defined in (77)) diverges as 1/μ, and that h_b(x)h_b(y)U₀(x)T₀(y) is independent of x and y, so that the roots and poles of $\tilde{P}(x)$ and $\tilde{Q}(y)$ all go to infinity when μ goes to 0.

Since, in that limit, $\tilde{Q}(y)$ is a constant, we also find, from equation (79), that the eigenvalue of $\tilde{t}^{[2]}$ in that eigenspace is

$$\begin{equation*} \langle 1|\tilde{t}^{[2]}|P^\star \rangle =F(x)+F(qx).\nonumber \end{equation*}$$

It is straightforward to check that equation (88) then gives E(0) = 0, which we knew from the fact that M₀ is a stochastic matrix.

Starting from μ = 0, where, as we just saw, we know $\tilde{Q}$ explicitly, we can expand everything in series in μ, and find an explicit expression for E(μ) perturbatively in μ. This is what we do in the next section, where we put everything together and give a summary of the whole procedure.

In this section, we collect all the results we found for the open ASEP, and show how they lead to the expressions for the cumulants of the current obtained in [15].

The first step is to construct two transfer matrices U_μ(x) and T_μ(y):

$$\begin{eqnarray} U_\mu (x)&=h_b(x) \langle \!\langle W|\!| A_\mu \prod _{i=1}^{L}X^{(i)}(x,x) |\!| V \rangle \!\rangle , \end{eqnarray} \tag{ 93 }$$

$$\begin{eqnarray} T_\mu (y)&=h_b(y)\langle \!\langle \tilde{W}|\!| A_\mu \prod _{i=1}^{L}X^{(i)}(y,y)|\!| \tilde{V} \rangle \!\rangle \end{eqnarray} \tag{ 94 }$$

(involving the function h_b defined in equation (82)), such that, for any x and y, we have:

$$\begin{equation} [M_\mu , U_\mu (x)T_\mu (y)]=0. \end{equation} \tag{ 95 }$$

Using these, we can construct two commuting matrices P(x) and Q(y) as:

$$\begin{equation} P(x)=U_\mu (x)\left[U_\mu (0)\right]^{-1} , Q(y)=U_\mu (0)T_\mu (y) \end{equation} \tag{ 96 }$$

such that:

$$\begin{equation} U_\mu (x)T_\mu (y)=P(x)Q(y). \end{equation} \tag{ 97 }$$

We can show that those matrices verify, for any positive integer k:

$$\begin{equation} P(x)Q(1/q^{k-1}x)=t^{[k]}(x)+{\rm e}^{-2k\mu }P(q^k x)Q(q/x) \end{equation} \tag{ 98 }$$

(where we omit the tildes, since we have correctly normalized P and Q from the start). The matrix t^[k](x) is the one-parameter transfer matrix with a k-dimensional auxiliary space.

The first two of those relations write:

$$\begin{eqnarray} P(x)Q(1/x)&=F(x)+{\rm e}^{-2\mu }P(q x)Q(q/x) , \end{eqnarray} \tag{ 99 }$$

$$\begin{eqnarray} P(x)Q(1/qx)&=t^{[2]}(x)+{\rm e}^{-4\mu }P(q^2 x)Q(q/x) , \end{eqnarray} \tag{ 100 }$$

where F(x) is a scalar function given by

$$\begin{equation} F(x)=\frac{(1+x)^L(1+1/x)^L(x^2,x^{-2})_{\infty }}{(ax,\tilde{a}x,bx,\tilde{b}x,a/x,\tilde{a}/x,b/x,\tilde{b}/x)_\infty } \end{equation} \tag{ 101 }$$

and t^[2] is such that:

$$\begin{equation} M_\mu =\frac{1}{2}(1-q)\frac{{\rm d}}{{\rm d} x} \log \biggl (\frac{t^{[2]}(x)}{F(qx)}\biggr )\bigg |_{x=-1}. \end{equation} \tag{ 102 }$$

Using equation (99) at x and qx, and equation (100), we can find the T-Q equation:

$$\begin{equation} t^{[2]}(x)Q(1/x)=F(x)Q(1/qx)+{\rm e}^{-2\mu }F(qx)Q(q/x) \end{equation} \tag{ 103 }$$

which allows us to express M_μ in terms of Q instead:

$$\begin{equation} M_\mu =\frac{1}{2}(1-q)\frac{{\rm d}}{{\rm d} x} \log \biggl (\frac{Q(q/x)}{Q(1/x)}\biggr )\bigg |_{x=-1}. \end{equation} \tag{ 104 }$$

We now consider:

$$\begin{equation} B=-{\rm e}^{2\mu }\bigl (Q(0)\bigr )^{-1}=-{\rm e}^{2\mu }\bigl (U_\mu (0)T_\mu (0)\bigr )^{-1}. \end{equation} \tag{ 105 }$$

As we saw in section 2.4, the first eigenvalue of B goes to 0 for μ → 0, which is not a priori the case for the others. Moreover, all the roots and poles of P(x) and Q(y) go to infinity in that eigenspace.

From here on, we restrict ourselves to that specific eigenspace, so that P, Q and B refer to functions rather than matrices.

The next step is to rewrite equation (99) in a different way, and see that all the information we have about P and Q makes it solvable. This was done in [12] for the periodic case, and we reproduce it here, under a slightly different form, for the open case. Let us therefore define a function W as:

$$\begin{equation} W(x)=-\frac{1}{2}\log \biggl (\frac{P(x)Q(1/x)}{{\rm e}^{-2\mu }P(q x)Q(q/x)}\biggr ), \end{equation} \tag{ 106 }$$

and a convolution kernel K, as:

$$\begin{equation} K(x,\tilde{x})=2\sum _{k=1}^{\infty }\frac{q^k}{1-q^k}((x/\tilde{x})^k+(x/\tilde{x})^{-k}) \end{equation} \tag{ 107 }$$

along with the associated convolution operator X:

$$\begin{equation} X[f](x)=\oint _{c_1}\frac{{\rm d}\tilde{x}}{\imath 2\pi \tilde{x}}f(\tilde{x})K(x,\tilde{x}). \end{equation} \tag{ 108 }$$

Using those, as we show in appendix E, one can rewrite equation (99) in terms of only one unknown function W:

which is the same as equation (59) in [12].

The last step is to take equations (104) and (106) at x = 0, to find:

$$\begin{equation} E(\mu )=\frac{1}{2}(1-q)\frac{{\rm d}}{{\rm d}x}\log \biggl (\frac{Q(q/x)}{Q(1/x)}\biggr )\biggl |_{x=-1} , \mu =-W(0). \end{equation} \tag{ 110 }$$

Considering what we said before about the roots and poles of P(x) being outside of the unit circle, and those of Q(1/x) being inside, we can replace $\frac{1}{2}\log \bigl (\frac{Q(q/x)}{Q(1/x)}\bigr )$ by −W(x) when expressing E(μ) as a contour integral over the unit circle (since P will not contribute), and obtain:

$$\begin{equation} \fbox{$ \mu =\displaystyle-\oint _{c_1}\frac{{\rm d}z}{\imath 2\pi z}W(z) $} \end{equation} \tag{ 111 }$$

and

$$\begin{equation} \fbox{$ E(\mu )=-(1-q)\displaystyle\oint _{c_1}\frac{{\rm d}z}{\imath 2\pi (1+z)^2}W(z), $} \end{equation} \tag{ 112 }$$

in which we recognize (13) and (14) from [15]. All this is done for a < 1 and b < 1, but can then be generalized to any a and b through the same reasoning as in [47] for the mean current, replacing the unit circle c₁ by small contours around $\Gamma =\lbrace 0,q^k a,q^k \tilde{a},q^k b,q^k \tilde{b}\rbrace$.

In this last result, E(μ) is expressed as an implicit function of μ, through the variable B. In order to get an explicit expression, one has to invert (111) to get B in terms of μ, and inject the result into (112). Since B is of order μ around μ = 0, this can be done perturbatively in B and μ, to yield the coefficients of E(μ) expanded in powers of μ. Those are, up to a factorial, the cumulants of the current in our system. In the general case, they can be expressed as combinations of multiple contour integrals around powers of F convolved through K [15]. In the simpler case of the TASEP, where q = 0, that convolution kernel vanishes, and (111) and (112) simplify to

$$\begin{equation*} \mu =-\frac{1}{2}\sum \limits _{k=1}^{\infty }\oint _{\Gamma }\frac{{\rm d}z}{\imath 2\pi z}F(z)^k\frac{B^k}{k}\nonumber \end{equation*}$$

and

$$\begin{equation*} E(\mu )=-\frac{(1-q)}{2}\sum \limits _{k=1}^{\infty }\oint _{\Gamma }\frac{{\rm d}z}{\imath 2\pi (1+z)^2}F(z)^k\frac{B^k}{k},\nonumber \end{equation*}$$

where F(z) is also much simpler, and has at most nine poles. This makes calculating the cumulants of the current much easier, especially in the case where a = b = 0 [14].

In this section, we explain how our construction for the open ASEP can be translated for the spin-1/2 XXZ chain with non-diagonal boundary conditions [6].

Let us first define the bulk Hamiltonian of the XXZ spin chain of length L:

$$\begin{eqnarray*} &&H_{b}=\frac{1}{2}\sum \limits _{k=1}^{L-1}h^{(i)}\end{eqnarray*}$$

with h⁽ⁱ⁾ acting as:

$$\begin{eqnarray*} h^{(i)}=\left[\begin{array}{@{}cccc@{}}\Delta & 0 & 0 & 0 \\ 0 & -\Delta & 1 & 0 \\ 0 & 1& -\Delta & 0 \\ 0 & 0 & 0 & \Delta \end{array}\right]\nonumber\end{eqnarray*}$$

on sites i and i + 1 (in basis {00, 01, 10, 11}, as usual), and as the identity on the rest of the chain. We define Δ as $\frac{1}{2}(q^{-1/2}-q^{1/2})$, which is not the usual definition for the XXZ chain (that can be obtained simply by replacing q by q²).

Let us also write the deformed Markov matrix $M_{\lbrace \!\mu _i\!\rbrace }$ for the special choice of weights defined by:

$$\begin{equation*} \left\lbrace \mu _0=\frac{1}{2}\log {\biggl (\frac{\gamma }{\alpha }\biggr )+\nu _0}, \mu _i=\frac{1}{2}\log {(q)}, \mu _L=\frac{1}{2}\log {\biggl (\frac{\delta }{\beta }\biggr )+\nu _L} \right\rbrace \nonumber \end{equation*}$$

which is on the line $\mu =\frac{1}{2} \log {\bigl (\frac{\gamma \delta }{\alpha \beta }q^{L-1}\bigr )}+\imath \mathbb {R}$ if ν₀ and ν_L are imaginary numbers (in which case $M_{\lbrace \!\mu _i\!\rbrace }$ is Hermitian). The deformed local matrices become:

$$\begin{eqnarray*} m^{(0)}(\mu _0)&=\left[\begin{array}{@{}cc@{}}-\alpha & \sqrt{\alpha \gamma } {\rm e}^{-\nu _0} \\ \sqrt{\alpha \gamma } {\rm e}^{\nu _0} & -\gamma \end{array}\right],\nonumber \\ M^{(i)}(\mu _i)&= \left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 0 \\ 0 & -q &\sqrt{q} & 0 \\ 0 & \sqrt{q}& -1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right],\nonumber \\ m^{(L)}(\mu _l)&=\left[\begin{array}{@{}cc@{}}-\delta & \sqrt{\beta \delta }{\rm e}^{\nu _L} \\ \sqrt{\beta \delta }{\rm e}^{-\nu _L} & -\beta \end{array}\right].\nonumber \end{eqnarray*}$$

It is straightforward to check that in this case, we have $M_{\lbrace \!\mu _i\!\rbrace }=\sqrt{q}H+\epsilon$, where is a constant, with the boundary matrices being equal to:

$$\begin{eqnarray*} h^{(0)}&=\frac{1}{2\sqrt{q}}\left[\begin{array}{@{}cc@{}}(1-q-\alpha +\gamma ) & 2\sqrt{\alpha \gamma }\, {\rm e}^{-\nu _0} \\ 2\sqrt{\alpha \gamma }\, {\rm e}^{\nu _0} & (-1+q+\alpha -\gamma )\end{array}\right]\nonumber \\ &=\frac{q^{-1/2}-q^{1/2}}{2}\frac{1}{(1+a)(1+\tilde{a})}\left[\begin{array}{@{}cc@{}}a+\tilde{a}& 2\sqrt{-a\tilde{a}}\,{\rm e}^{\nu _0} \\ 2\sqrt{-a\tilde{a}}\,{\rm e}^{-\nu _0} &-a-\tilde{a}\end{array}\right],\nonumber \\ h^{(L)}&=\frac{1}{2\sqrt{q}}\left[\begin{array}{@{}cc@{}}(-1+q+\beta -\delta ) & 2\sqrt{\beta \delta }\,{\rm e}^{\nu _L} \\ 2\sqrt{\beta \delta }\,{\rm e}^{-\nu _L} & (1-q-\beta +\delta )\end{array}\right]\nonumber \\ &=\frac{q^{-1/2}-q^{1/2}}{2}\frac{1}{(1+b)(1+\tilde{b})}\left[\begin{array}{@{}cc@{}}-b-\tilde{b}& 2\sqrt{-b\tilde{b}}\,{\rm e}^{-\nu _L} \\ 2\sqrt{-b\tilde{b}}\,{\rm e}^{\nu _L} &b+\tilde{b}\end{array}\right].\nonumber \end{eqnarray*}$$

Since we have three nontrivial parameters in each of those matrices, they are completely general: we can write (without restricting ourselves to Hermitian matrices)

$$\begin{eqnarray*} h^{(0)}=a_z\sigma _z+a_+\sigma ^++a_-\sigma ^-&=\left[\begin{array}{@{}cc@{}}a_z& a_- \\ a_+ &-a_z\end{array}\right],\nonumber \\ h^{(L)}=b_z\sigma _z+b_+\sigma ^++b_-\sigma ^-&=\left[\begin{array}{@{}cc@{}}b_z& b_- \\ b_+ &-b_z\end{array}\right],\nonumber \end{eqnarray*}$$

with

$$\begin{eqnarray*} &&a_z=(1-q-\alpha +\gamma )/2\sqrt{q} ,\qquad a_+=\sqrt{\alpha \gamma /q} \,{\rm e}^{\nu _0} ,\qquad a_-=\sqrt{\alpha \gamma /q} \,{\rm e}^{-\nu _0},\nonumber \\ &&b_z=(-1+q+\beta -\delta )/2\sqrt{q} ,\qquad b_+=\sqrt{\beta \delta /q} \,{\rm e}^{-\nu _L} ,\qquad b_-=\sqrt{\beta \delta /q} \,{\rm e}^{\nu _L},\nonumber \end{eqnarray*}$$

which is to say

$$\begin{eqnarray*} \nu _0&=-2\log (a_+/a_-),\nonumber \\ \alpha &=\sqrt{(\sqrt{q}a_z-(1-q)/2)^2+a_+a_-}-\sqrt{q}a_z+(1-q)/2,\nonumber \\ \gamma &=\sqrt{(\sqrt{q}a_z-(1-q)/2)^2+a_+a_-}+\sqrt{q}a_z-(1-q)/2,\nonumber \end{eqnarray*}$$

and

$$\begin{eqnarray*} \nu _L&=-2\log (b_-/b_+),\nonumber \\ \beta &=\sqrt{(\sqrt{q}b_z+(1-q)/2)^2+b_+b_-}+\sqrt{q}b_z+(1-q)/2,\nonumber \\ \delta &=\sqrt{(\sqrt{q}a_z+(1-q)/2)^2+b_+b_-}-\sqrt{q}b_z-(1-q)/2.\nonumber \end{eqnarray*}$$

Those are well defined for any values of a_z, a₊, a₋, b_z, b₊ and b₋.

Considering expression (14), or its equivalent for an open chain, and noting that $A_{\frac{\mu _i}{2}}=A^{-1/4}$, we can rewrite U(x) and T(y) in a way better suited to this situation:

$$\begin{eqnarray*} U(x)&= \langle \!\langle \phi |\!| \prod _{i=1}^{L}Y^{(i)}(x) |\!| \psi \rangle \!\rangle ,\nonumber \\ T(y)&= \langle \! \langle \tilde{\phi }|\!| \prod _{i=1}^{L}Y^{(i)}(y) |\!| \tilde{\psi } \rangle \!\rangle ,\nonumber \end{eqnarray*}$$

with

$$\begin{eqnarray*} Y(x)=\left[\begin{array}{@{}cc@{}}N_x & \Sigma _+\\ \Sigma _- & N_x\end{array}\right]\nonumber \end{eqnarray*}$$

where

$$\begin{eqnarray*} N_x&=A^{-1/2}-xA^{1/2}\nonumber \\ \Sigma _+&=A^{-1/4}e A^{-1/4}=q^{-1/4}S^+(A^{-1/2}-x^2A^{1/2}),\nonumber \\ \Sigma _-&=A^{-1/4}{\rm d} A^{-1/4}=q^{1/4}S^-(A^{-1/2}- A^{1/2}).\nonumber \end{eqnarray*}$$

The boundary vectors become:

$$\begin{eqnarray*} \langle \!\langle \phi |\!| &= \langle \!\langle W|\!| A_{\mu _0}A^{1/4},\nonumber \\ |\!| \psi \rangle \!\rangle &=A_{\mu _L}A^{1/4}|\!| V\rangle \!\rangle ,\nonumber \\ \langle \! \langle \tilde{\phi }|\!| &= \langle \!\langle \tilde{W}|\!| A_{\mu _0}A^{1/4},\nonumber \\ |\!| \tilde{\psi } \rangle \!\rangle &=A_{\mu _L}A^{1/4}|\!| \tilde{V} \rangle \!\rangle .\nonumber \end{eqnarray*}$$

Matrices N_x, Σ₊ and Σ₋ satisfy the U_q[SU(2)] algebra [45]:

$$\begin{eqnarray*} &&[\Sigma_+,\Sigma_-]=(q^{-1/2}-q^{1/2})(A^{-1}-x^2A),\nonumber \\ &&\Sigma _- A=q A \Sigma _-,\nonumber \\ &&A \Sigma _+=q \Sigma _+ A,\nonumber\end{eqnarray*}$$

and the conditions on the boundary vectors become:

$$\begin{eqnarray*} \langle \!\langle \phi |\!| &[a_+\Sigma _+-a_-\Sigma _--2a_zN_x-(q^{-1/2}-q^{1/2})xA^{1/2}]=0,\nonumber \\ & [b_-\Sigma _--b_+\Sigma _++2b_zN_x-(q^{-1/2}-q^{1/2})xA^{1/2}] |\!| \psi \rangle \!\rangle =0,\nonumber \\ \langle \! \langle \tilde{\phi }|\!| &[a_+\Sigma _+-a_-\Sigma _--2a_zN_y+(q^{-1/2}-q^{1/2})A^{-1/2}]=0,\nonumber \\ & [b_-\Sigma _--b_+\Sigma _++2b_zN_y+(q^{-1/2}-q^{1/2})A^{-1/2}] |\!| \tilde{\psi } \rangle \!\rangle =0.\nonumber \end{eqnarray*}$$

We may note that the structure of this solution bears a strong resemblance to that of the Lindblad master equation found in [54, 55]. In that case, the algebraic relations satisfied by the boundary vectors are different from ours, as there is only one vector per boundary but two equations per vector, which may constrain the values of the boundary parameters. The connection between integrable spin chains and certain boundary-driven Lindblad models is also investigated in [56, 57], and a matrix product structure of the density matrix was observed in [58] for an open Hubbard chain, but the precise conditions for such systems to be integrable have yet to be understood.

The commutation relation for the periodic case or for the bulk of the open case involves a calculation almost identical to that which can be found in the appendices of [16].

Let us recall:

$$\begin{eqnarray*} X&=\left[\begin{array}{@{}cc@{}}n_0 & e\\ d & n_1\end{array}\right]= \left[\begin{array}{@{}cc@{}}1+x A & e\\ d & 1+y A\end{array}\right],\nonumber \\ \hat{X}&=\left[\begin{array}{@{}cc@{}}\hat{n}_0 & \hat{e}\\ \hat{d} & \hat{n}_1\end{array}\right]=\frac{(1-q)}{2}\left[\begin{array}{@{}cc@{}}1-x A & e\\ -d & -1+y A\end{array}\right].\nonumber \end{eqnarray*}$$

The parts of the transfer matrix and of the Markov matrix involving sites i and i + 1 are:

$$\begin{equation*} X^{(i)}X^{(i+1)}=\left[\begin{array}{@{}cccc@{}}n_0 n_0 & n_0 e & e n_0 & e e \\ n_0 d & n_0 n_1 & e d & e n_1 \\ d n_0 & d e & n_1 n_0 & n_1 e \\ d d & d n_1 & n_1 d & n_1 n_1\end{array}\right] , M_{i}=\left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 0 \\ 0 & -q & 1 & 0 \\ 0 & q & -1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right].\nonumber \end{equation*}$$

The commutator of those two gives:

$$\begin{eqnarray} [M^{(i)}, X^{(i)}X^{(i+1)}]&=&\left[\begin{array}{@{}cccc@{}} 0 & q(n_0e-en_0) & en_0-n_0e & 0 \\ dn_0-q n_0d & de-q ed & (1-q)ed & n_1e-q en_1 \\ q n_0d-dn_0 & (q-1)de & q ed-de & q en_1-n_1e \\ 0 & q(dn_1-n_1d) & n_1d-dn_1 & 0 \end{array}\right]\nonumber \\ &&=\left[\begin{array}{@{}cccc@{}} 0 & \hat{n}_0e-n_0\hat{e} & \hat{e}n_0-e\hat{n}_0 & 0 \\ \hat{n}_0d-n_0\hat{d} & \hat{n}_0n_1-n_0\hat{n}_1 & \hat{e}d-e\hat{d} & \hat{e}n_1-e\hat{n}_1 \\ \hat{d}n_0-d\hat{n}_0& \hat{d}e-d\hat{e} & \hat{n}_1n_0-n_1\hat{n}_0 & \hat{n}_1e-n_1\hat{e} \\ 0 & \hat{d}n_1-d\hat{n}_1 & \hat{n}_1d-n_1\hat{d} & 0\end{array}\right]\nonumber \\ &=&\fbox{$\hat{X}^{(i)}X^{(i+1)}-X^{(i)}\hat{X}^{(i+1)}, $} \end{eqnarray} \tag{ A.1 }$$

where we got from the second to the third line using:

$$\begin{eqnarray} &&de -q ed=(1-q)(1-x y A^2) =\hat{n}_0n_1-n_0\hat{n}_1\nonumber ,\\ &&q(n_0e-en_0)=(q-1)xA e=\hat{n}_0e-n_0\hat{e}\nonumber ,\\ &&en_0-n_0e=(1-q)x e A =\hat{e}n_0-e\hat{n}_0\nonumber ,\\ &&n_1e-q en_1=(1-q)e=\hat{e}n_1-e\hat{n}_1\nonumber ,\\ &&q en_1-n_1e=(q-1)e=\hat{n}_1e-n_1\hat{e}\nonumber ,\\ &&q(dn_1-n_1d)=(q-1)y\,{\rm d} A =\hat{d}n_1-d\hat{n}_1\nonumber ,\\ &&n_1d-dn_1=(1-q) yA d=\hat{n}_1d-n_1\hat{d}\nonumber ,\\ &&dn_0-q n_0d=(1-q)d=\hat{n}_0d-n_0\hat{d}\nonumber ,\\ &&q n_0d-dn_0=(q-1)d=\hat{d}n_0-d\hat{n}_0. \end{eqnarray} \tag{ A.2 }$$

We recall, for the left boundary vectors:

$$\begin{eqnarray*} &&\langle \!\langle W|\!| [\alpha (e+n_x)- \gamma (d+n_x)-(1-q)] = 0,\nonumber \\ &&\langle \!\langle \tilde{W}|\!| [\alpha (e-n_y) - \gamma (d-n_y)+(1-q)y A] =0.\nonumber \end{eqnarray*}$$

We consider the commutator of the left boundary matrix from M_μ with the relevant part of U_μ and T_μ (omitting the matrix A_μ here, because, as in the periodic case, its action is trivial), and use those relations. We get:

$$\begin{eqnarray*} \left[m^{(0)},\langle \!\langle W|\!| X^{(1)}\right]&=\langle \!\langle W|\!|\left[\begin{array}{@{}cc@{}}\gamma d-\alpha e & -(\alpha -\gamma )e \\ (\alpha -\gamma )d & \alpha e-\gamma d\end{array}\right]\nonumber \\ &=(\alpha -\gamma )\langle \!\langle W|\!|\left[\begin{array}{@{}cc@{}}n_x & -e \\ d & -n_x\end{array}\right]+(1-q)\langle \!\langle W|\!|\left[\begin{array}{@{}cc@{}}-1 & 0 \\ 0 & 1\end{array}\right]\nonumber \\ \left[m^{(0)},\langle \!\langle \tilde{W}|\!| X^{(1)}\right]&=\langle \!\langle \tilde{W}|\!|\left[\begin{array}{@{}cc@{}}\gamma d-\alpha e & -(\alpha -\gamma )e \\ (\alpha -\gamma )d & \alpha e-\gamma d\end{array}\right]\nonumber \\ &=(\alpha -\gamma )\langle \!\langle \tilde{W}|\!|\left[\begin{array}{@{}cc@{}}-n_y & -e \\ d & n_y\end{array}\right]+(1-q)\langle \!\langle \tilde{W}|\!|\left[\begin{array}{@{}cc@{}}y A & 0\\ 0 & -y A\end{array}\right].\nonumber \end{eqnarray*}$$

The first terms in each of those two equations cancel one another:

$$\begin{equation*} \left[\begin{array}{@{}cc@{}}n_x & -e \\ d & -n_x\end{array}\right]\cdot X(y)+X(x)\cdot \left[\begin{array}{@{}cc@{}}-n_y & -e \\ d & n_y\end{array}\right]=0.\nonumber \end{equation*}$$

We note that if we had had two different spectral parameters for the two diagonal terms in X(x) or X(y), this relation would not have been possible (or we would have needed two equations on each boundary vector instead of one).

As for the second terms, a straightforward calculation gives:

$$\begin{equation*} (1-q)\left(\left[\begin{array}{@{}cc@{}} -1 & 0\\ 0& 1\end{array}\right]\cdot X(y)+ X(x)\cdot \left[\begin{array}{@{}cc@{}} yA & 0\\ 0 & -yA\end{array}\right]\right)=-\hat{X}(x)\cdot X(y)-X(x)\cdot \hat{X}(y),\nonumber \end{equation*}$$

so that

$$\begin{equation*} [m^{(0)}(\mu ),U_\mu T_\mu ]=- \langle \!\langle W|\!| A_\mu \hat{X}^{(1)}\prod _{i=2}^{L}X^{(i)} |\!|V\rangle \!\rangle \cdot T_\mu -U_\mu \cdot \langle \!\langle \tilde{W}|\!| A_\mu \hat{X}^{(1)}\prod _{i=2}^{L}X^{(i)} |\!|\tilde{V}\rangle \!\rangle .\nonumber \end{equation*}$$

The exact same calculations can be done at the right boundary, and yields:

$$\begin{equation*} [m^{(L)},U_\mu T_\mu ]= \langle \!\langle W|\!| A_\mu \prod _{i=1}^{L-1}X^{(i)}\hat{X}^{(L)} |\!|V\rangle \!\rangle \cdot T_\mu +U_\mu \cdot \langle \!\langle \tilde{W}|\!| A_\mu \prod _{i=1}^{L-1}X^{(i)}\hat{X}^{(L)} |\!|\tilde{V}\rangle \!\rangle .\nonumber \end{equation*}$$

We start from $I_0=R^{-1}(x,x;y,y) |\!| V(x) \rangle \!\rangle \cdot |\!| \tilde{V}(y) \rangle \!\rangle =\frac{(x y)_\infty }{(q)_\infty }\frac{(q A_2)_\infty }{(x y A_2)_\infty } K^+$, as defined in ...

$$\begin{equation} \fbox{$ I_0=\displaystyle\frac{(xS_1)_\infty (b\tilde{b}xS_1)_\infty }{(bS_1)_\infty (\tilde{b}S_1)_\infty } \frac{(q A_2)_\infty }{(x y A_2)_\infty }\frac{(y S_1/S_2)_\infty }{(x S_1/S_2)_\infty }\frac{(y^2A_2)_\infty }{(qA_2)_\infty } \frac{(byS_2)_\infty (\tilde{b}yS_2)_\infty }{(S_2)_\infty (b\tilde{b}S_2)_\infty }\frac{(x y)_\infty }{(y^2)_\infty } $} \end{equation} \tag{ B.12 }$$

where the term ||0₁, 0₂〉〉 is implicit.

We expand all but the leftmost two ratios:

$$\begin{equation*} I_0=\frac{(xS_1)_\infty (b\tilde{b}xS_1)_\infty }{(bS_1)_\infty (\tilde{b}S_1)_\infty }\sum \limits _{n,m,k}\frac{(x y)_k(y/x)_{n+m-k}}{(q)_k(q)_{n+m-k}}(x S_1)^{n+m-k}S_2^k\frac{(q)_{n+m}}{(y^2)_{n+m}}\frac{(y b)_n(y/b)_{m}}{(q)_n(q)_{m}}(b \tilde{b})^m.\nonumber \end{equation*}$$

Using equation (B.1), and relabelling n − r and m − s as n and m, we get:

$$\begin{eqnarray*} I_0&=\frac{(xS_1)_\infty (b\tilde{b}xS_1)_\infty }{(bS_1)_\infty (\tilde{b}S_1)_\infty }\sum \limits _{n,m,r,s}\frac{(x y)_{r+s}(y/x)_{n+m}}{(y^2)_{n+m+r+s}}\frac{(y b)_{n+r}(y/b)_{m+s}}{(q)_n(q)_{m}(q)_r(q)_{s}}(q^sx S_1)^{n}(b \tilde{b}xS_1)^{m}S_2^{r}(b \tilde{b}S_2)^{s}\nonumber \\ &=\frac{(xS_1)_\infty (b\tilde{b}xS_1)_\infty }{(bS_1)_\infty (\tilde{b}S_1)_\infty }\sum \limits _{r,s}\frac{(y b)_{r}(y/b)_{s}}{(q)_r(q)_{s}}\frac{(x y)_{r+s}}{(y^2)_{r+s}}S_2^{r}(b \tilde{b}S_2)^{s}\Phi _1\big({ y/x,}{{{\begin{array}{@{}c@{}} { q^s y/b}\\ { q^ryb}\end{array}}}}{ ;q^{r+s}y^2;}{{{\begin{array}{@{}c@{}} { b \tilde{b}xS_1}\\ { q^sx S_1}\end{array}}}}\big).\nonumber \end{eqnarray*}$$

We then use equation (B.7):

$$\begin{eqnarray*} I_0&=\frac{(xS_1)_\infty }{(\tilde{b}S_1)_\infty }\sum \limits _{r,s}\frac{(y b)_{r}(y/b)_{s}}{(q)_r(q)_{s}}\frac{(x y)_{r+s}}{(y^2)_{r+s}}S_2^{r}(b \tilde{b}S_2)^{s}\frac{(q^{r+s}xy)_n(q^sy/b)_n(y\tilde{b})_n}{(q^{r+s}y^2)_n(q)_n}(bS_1)^n\frac{(q^{n+s}xy\tilde{b}S_1)_\infty }{(q^sxS_1)_\infty }\nonumber \\ &=\frac{(xS_1)_\infty }{(\tilde{b}S_1)_\infty }\sum \limits _{n,s}\frac{(y/b)_{s+n}}{(q)_{s}}\frac{(x y)_{s+n}}{(y^2)_{s+n}}(b \tilde{b}S_2)^{s}\frac{(y\tilde{b})_n}{(q)_n}(bS_1)^n\frac{(q^{n+s}xy\tilde{b}S_1)_\infty }{(q^sxS_1)_\infty }{}_2\phi _1\big({{{\begin{array}{@{}c@{}} { yb,xyq^{s+n}}\\ { y^2q^{s+n}}\end{array}}}}{ ;S_2}\big).\nonumber \end{eqnarray*}$$

Then we use (B.5) on ₂ϕ₁:

$$\begin{eqnarray*} &&I_0=\frac{(xS_1)_\infty (xbS_2)_\infty }{(\tilde{b}S_1)_\infty (S_2)_\infty }\sum \limits _{n,r,s}\frac{(y/b)_{r+s+n}}{(y^2)_{r+s+n}}\frac{(y/x)_{r}}{(q)_{r}}\frac{(x y)_{s+n}(y\tilde{b})_n}{(q)_n(q)_{s}}\frac{(q^{n+s}xy\tilde{b}S_1)_\infty }{(q^sxS_1)_\infty }(xbS_2)^r(b\tilde{b}S_2)^s(bS_1)^n\nonumber \\ &&\hphantom{ I_0}=\frac{(xS_1)_\infty (xbS_2)_\infty }{(\tilde{b}S_1)_\infty (S_2)_\infty }\sum \limits _{r,s}\frac{(y/b)_{r+s}}{(y^2)_{r+s}}\frac{(y/x)_{r}}{(q)_{r}}\frac{(x y)_{s}}{(q)_{s}}(xbS_2)^r(b\tilde{b}S_2)^s\nonumber\\ &&\hphantom{ I_0=}\times\,\frac{(q^{s}xy\tilde{b}S_1)_\infty }{(q^sxS_1)_\infty }{}_3\phi _2\big(\begin{array}{@{}c@{}} { y/bq^{r+s},xyq^s,y\tilde{b}}\\ { y^2q^{r+s},q^{s}xy\tilde{b}S_1}\end{array}{ ;bS_1}\big) \end{eqnarray*}$$

and (B.8) on ₃ϕ₂:

$$\begin{equation*} I_0=\frac{(xS_1)_\infty (xbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\sum \limits _{n,m,r,s}\frac{(y/b)_{r+s}(yb)_{n+m}}{(y^2)_{n+m+r+s}}\frac{(xy)_{n+s}(y/x)_{m+r}}{(q)_n(q)_{m}(q)_r(q)_{s}}(\tilde{b} S_1)^{n}(q^s xS_1)^{m}(xbS_2)^{r}(b \tilde{b}S_2)^{s}.\nonumber \end{equation*}$$

We then use (B.1) in reverse (and exchange k and n + m − k for convenience) to find:

$$\begin{eqnarray} &&\fbox{$ I_0=\displaystyle\frac{(xS_1)_\infty (xbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\sum \limits _{n,m,k}\frac{(y/x)_k(xy)_{n+m-k}}{(q)_k(q)_{n+m-k}}\tilde{b}^{n+m-k} x^{k}\frac{(q)_{n+m}}{(y^2)_{n+m}}\frac{(y/b)_n(yb)_{m}}{(q)_n(q)_{m}}(xbS_2)^n(xS_1)^m. $}\nonumber\\ \end{eqnarray} \tag{ B.13 }$$

In this expression, we can see that, after rescaling S₁ to S₁/b, we have the symmetry {S₁↔S₂, b↔1/b}, so that

$$\begin{eqnarray*} &&I_0=\frac{(xbS_2)_\infty (x\tilde{b}S_2)_\infty }{(S_2)_\infty (b\tilde{b}S_2)_\infty } \frac{(q A_1)_\infty }{(x y A_1)_\infty }\frac{(y S_2/S_1)_\infty }{(x S_2/S_1)_\infty }\frac{(y^2A_1)_\infty }{(qA_1)_\infty } \frac{(yS_1)_\infty (b\tilde{b}yS_1)_\infty }{(bS_1)_\infty (\tilde{b}S_1)_\infty }\frac{(x y)_\infty }{(y^2)_\infty }\end{eqnarray*}$$

which is to say, after rearranging a few terms:

$$\begin{equation*} R^{-1}(x,x;y,y) |\!| V(x) \rangle \!\rangle \cdot |\!| \tilde{V}(y) \rangle \!\rangle =|\!| V(y) \rangle \!\rangle \cdot |\!| \tilde{V}(x) \rangle \!\rangle .\nonumber \end{equation*}$$

Through the exact same calculations, we can also show that

$$\begin{equation*} R^{-1}(y,y;x,x) |\!| \tilde{V}(y) \rangle \!\rangle \cdot |\!| V(x)\rangle \!\rangle =|\!| \tilde{V}(x) \rangle \!\rangle \cdot |\!|V(y) \rangle \!\rangle .\nonumber \end{equation*}$$

What we want to prove here, taking xy = q^{1 − p}, is twofold. First, we need to show that $K^+_{i,j}(x,y)=0$ for j ⩽ p − 1 and i ⩾ p. Then, we need to find the relation between $K^+_{i+p,j+p}(x,y)$ and $K^+_{i,j}(q^px,q^py)$, which we will do by calculating $\overline{I_0}=D_{S_1}^p D_{S_2}^p I_0$.

The first part is rather straightforward. Considering equation (B.14), and since $K^+=\frac{(q)_\infty }{(xy)_\infty }\frac{(xy A_2)_\infty }{(q A_2)_\infty } I_0$, we see that K⁺ has a factor $\frac{(xy A_1)_\infty }{(x y A_2)_\infty }$ which results in a factor $\frac{(xy)_i}{(x y)_j}$ in $K^+_{i,j}$. Taking xy = q^{1 − p}, this factor is 0 precisely when j ⩽ p − 1 and i ⩾ p, and since the rest of the expression for K⁺ has no pole for xy = q^{1 − p}, we conclude that the corresponding terms in K⁺ are equal to 0. Note that, even though that prefactor is infinite for i ⩽ p − 1 and j ⩾ p, the absence of poles in K⁺ tells us that it is compensated by the rest of the expression.

We now need to calculate $\overline{I_0}=D_{S_1}^p D_{S_2}^p I_0$. In order to do this, we first notice that

$$\begin{equation*} \frac{(q)_{n+m}}{(q)_{n+m-k}}\tilde{b}^{n+m-k}=D_{\tilde{b}}^k \tilde{b}^{n+m}\nonumber \end{equation*}$$

and we use equation (B.9) on $\frac{(y/x)_k(xy)_{n+m-k}}{(y^2)_{n+m}}$ in (B.13) to write:

$$\begin{eqnarray*} I_1&=&\frac{(xS_1)_\infty (xbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\sum \limits _{n,m,k}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^kt)_\infty }t^{\log _q(xy)+n+m-k}\frac{x^k}{(q)_k}D_{\tilde{b}}^k \tilde{b}^{n+m}\nonumber\\ &&\times \frac{(y/b)_n(yb)_{m}}{(q)_n(q)_{m}}(bS_2)^n(S_1)^m\nonumber \\ &=&\sum \limits _{k}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^kt)_\infty }t^{\log _q(xy)}\frac{(x/t)^k}{(q)_k}D_{\tilde{b}}^k\frac{(xS_1)_\infty (y b \tilde{b}tS_1)_\infty }{(bS_1)_\infty (\tilde{b}tS_1)_\infty }\frac{(xbS_2)_\infty (y \tilde{b}tS_2)_\infty }{(S_2)_\infty (b \tilde{b}tS_2)_\infty }\nonumber \end{eqnarray*}$$

where $I_0=I_1 \frac{(xy)_\infty (y/x)_\infty }{(y^2)_\infty (q)_\infty }$. That prefactor will be put back in later when we re-sum the integral. Note that the term (xy)_∞ gives 0 when we take xy = q^{1 − p}, but since that term is part of I₀ and not of K⁺, it merely serves to simplify the calculation, and will be taken away at the end. For that reason, we will not apply xy = q^{1 − p} on that term, but keep it in its generic form.

We will now differentiate with respect to S₁ and S₂. Consider the following relation, which is easy to prove by recursion: if ab/cd = q^{1 − p}, then

$$\begin{equation*} D_{X}^p\frac{(a)_\infty (b)_\infty }{(c)_\infty (d)_\infty }=(a/c)_p(b/c)_pc^p\frac{(q^pa)_\infty (q^pb)_\infty }{(c)_\infty (d)_\infty }.\nonumber \end{equation*}$$

This relation applies to both groups of q-Pochhammer symbols at the right in the integral (remember that xy = q^{1 − p}), and gives:

$$\begin{eqnarray*} D_{S_1}^p\frac{(xS_1)_\infty (y b \tilde{b}tS_1)_\infty }{(bS_1)_\infty (\tilde{b}tS_1)_\infty }&=(x/b)_p(y\tilde{b}t)_pb^p\frac{(q^pxS_1)_\infty (q^py b \tilde{b}tS_1)_\infty }{(bS_1)_\infty (\tilde{b}tS_1)_\infty }\nonumber \\ D_{S_2}^p\frac{(xbS_2)_\infty (y \tilde{b}tS_2)_\infty }{(S_2)_\infty (b \tilde{b}tS_2)_\infty }&=(xb)_p(y\tilde{b}t)_p\frac{(q^pxbS_2)_\infty (q^py\tilde{b}tS_2)_\infty }{(S_2)_\infty (b \tilde{b}tS_2)_\infty }.\nonumber \end{eqnarray*}$$

We then have:

$$\begin{eqnarray*} &&\overline{I_1}=(x/b)_pb^p(xb)_p\frac{(q^pxS_1)_\infty (q^pxbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\nonumber \\ &&\hphantom{ \overline{I_1}=}\times\,\sum \limits _{k}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^kt)_\infty }t^{\log _q(xy)}\frac{(x/t)^k}{(q)_k}D_{\tilde{b}}^k(q^py\tilde{b}t)_p(q^py\tilde{b}t)_p\frac{(y b \tilde{b}tS_1)_\infty (y \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }.\nonumber \end{eqnarray*}$$

We use the formula for the q-derivative of a product (B.10) on the derivative with respect to $\tilde{b}$, with $f(\tilde{b})=(y\tilde{b}t)_p$ and the rest in g. Considering that $D_{\tilde{b}}^l(y\tilde{b}t)_p=(q^pyt)^l(q^{-p})_l(y\tilde{b}tq^l)_{p-l}$, we find:

$$\begin{eqnarray*} &&D_{\tilde{b}}^k(y\tilde{b}t)_p(y\tilde{b}t)_p\frac{(y b \tilde{b}tS_1)_\infty (y \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }\nonumber \\ &&= \sum \limits _{l=0}^{k}\frac{(q)_k}{(q)_l(q)_{k-l}}(q^pyt)^l(q^{-p})_l(y\tilde{b}tq^k)_{p-l}D_{\tilde{b}}^{k-l}(y\tilde{b}t)_p\frac{(q^py b \tilde{b}tS_1)_\infty (q^py \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }\nonumber \end{eqnarray*}$$

so that, renaming k − l as k, we have:

$$\begin{eqnarray*} &&\overline{I_1}=(x/b)_pb^p(xb)_p\frac{(q^pxS_1)_\infty (q^pxbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\sum \limits _{k,l}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^{k+l}t)_\infty }t^{\log _q(xy)}(q^pxy)^l\nonumber\\ &&\times\,\frac{(q^{-p})_l}{(q)_l}(y\tilde{b}tq^{k+l})_{p-l}\frac{(x/t)^k}{(q)_k}D_{\tilde{b}}^k(y\tilde{b}t)_p\frac{(q^py b \tilde{b}tS_1)_\infty (q^py \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }.\nonumber \end{eqnarray*}$$

Consider the part that depends on l (writing $\frac{1}{(y/xq^{k+l}t)_\infty }=\frac{(y/xq^{k}t)_l}{(y/xq^{k}t)_\infty }$):

$$\begin{eqnarray*} &&\sum \limits _{l}(y/xq^{k}t)_l(q^pxy)^l\frac{(q^{-p})_l}{(q)_l}(y\tilde{b}tq^{k+l})_{p-l}\nonumber \\ &&=\sum \limits _{l}(y/xq^{k}t)_l((y\tilde{b}tq^{k+p-1})^{-1})_{p-l}\frac{(q)_p}{(q)_l(q)_{p-l}}(xy)^l(y\tilde{b}tq^{k})^{p-l}(-1)^pq^{p(p-1)/2}\nonumber \end{eqnarray*}$$

which, through (B.11), is equal to:

$$\begin{eqnarray*} &&\sum \limits _{l}\frac{(y/\tilde{b})_l(q^{1-p}/xy)_{p-l}}{(q)_l(q)_{p-l}}(y\tilde{b}tq^{k})^l(xy)^{p-l}(-1)^pq^{p(p-1)/2}(q)_p\nonumber \\ &&=\sum \limits _{l}\frac{(y/\tilde{b})_l(xyq^l)_{p-l}}{(q)_l(q)_{p-l}}(y\tilde{b}tq^{k})^l(-1)^lq^{l(l-1)/2}(q)_p.\nonumber \end{eqnarray*}$$

Using that xy = q^{1 − p}, only the term for l = p survives, namely $(y/\tilde{b})_p(y\tilde{b}tq^{k})^p(-1)^pq^{p(p-1)/2}$, so that:

$$\begin{eqnarray*} &&\overline{I_1}=(x/b)_pb^p(xb)_p(y/\tilde{b})_p(y\tilde{b})^p(-1)^pq^{p(p-1)/2}\frac{(q^pxS_1)_\infty (q^pxbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\nonumber \\ &&\hphantom{ \overline{I_1}=}\times\,\sum \limits _{k,l}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^{k}t)_\infty }t^{\log _q(xy)+p}\frac{(q^px/t)^k}{(q)_k}D_{\tilde{b}}^k(y\tilde{b}t)_p\frac{(q^py b \tilde{b}tS_1)_\infty (q^py \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }.\nonumber \end{eqnarray*}$$

Using again the formula for the derivative of a product and relabelling k − l as k again, we find:

$$\begin{eqnarray*} &&\overline{I_1}=(x/b)_pb^p(xb)_p(y/\tilde{b})_p(y\tilde{b})^p(-1)^pq^{p(p-1)/2}\frac{(q^pxS_1)_\infty (q^pxbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\nonumber \\ &&\times\,\sum \limits _{k,l}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^{k+l}t)_\infty }t^{\log _q(xy)+p}(q^{2p}xy)^l\frac{(q^{-p})_l}{(q)_l}(y\tilde{b}tq^{k+l})_{p-l}\nonumber \\ &&\times\,\frac{(q^px/t)^k}{(q)_k}D_{\tilde{b}}^k\frac{(q^py b \tilde{b}tS_1)_\infty (q^py \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }.\nonumber \end{eqnarray*}$$

We have almost the same sum as before over l, only with an extra term q^pl which changes the result to:

$$\begin{eqnarray*} &&\sum \limits _{l}(y/xq^{k}t)_l(q^{2p}xy)^l\frac{(q^{-p})_l}{(q)_l}(y\tilde{b}tq^{k+l})_{p-l}\nonumber \\ &&=\sum \limits _{l}\frac{(q^py/\tilde{b})_l(xyq^{p+l})_{p-l}}{(q)_l(q)_{p-l}}(y\tilde{b}tq^{k})^l(-1)^lq^{l(l-1)/2}(q)_p\nonumber \end{eqnarray*}$$

so that, this time, the whole sum survives to taking xy = q^{1 − p}.

We now have to re-sum the integral over t. First, we expand the two ratios of q-Pochhammer symbols on the right, and apply the derivative with respect to $\tilde{b}$:

$$\begin{equation*} D_{\tilde{b}}^k\frac{(q^py b \tilde{b}tS_1)_\infty (q^py \tilde{b}tS_2)_\infty }{(\tilde{b}tS_1)_\infty (b \tilde{b}tS_2)_\infty }=t^{n+m}\sum \limits _{n,m}\frac{(q)_{n+m}}{(q)_{n+m-k}}\tilde{b}^{n+m-k}\frac{(q^py/b)_n(q^pyb)_{m}}{(q)_n(q)_{m}}(bS_2)^n(S_1)^m\nonumber \end{equation*}$$

so that

$$\begin{eqnarray*} &&\overline{I_1}=(x/b)_pb^p(xb)_p(y/\tilde{b})_p(y\tilde{b})^p(-1)^pq^{p(p-1)/2}(q)_p\frac{(q^pxS_1)_\infty (q^pxbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\nonumber \\ &&\hphantom{ \overline{I_1}=}\times\,\sum \limits _{n,m,k,l}\int \nolimits _0^1\frac{d_qt}{t}\frac{(qt)_\infty }{(y/xq^{k}t)_\infty }t^{\log _q(xy)+p+n+m+l-k}\frac{(q^py/\tilde{b})_l(xyq^{p+l})_{p-l}}{(q)_l(q)_{p-l}}(y\tilde{b}q^{k})^l(-1)^lq^{l(l-1)/2}\nonumber \\ &&\hphantom{ \overline{I_1}=}\times\,\frac{(q^px)^k}{(q)_k}\frac{(q)_{n+m}}{(q)_{n+m-k}}\tilde{b}^{n+m-k}\frac{(q^py/b)_n(q^pyb)_{m}}{(q)_n(q)_{m}}(bS_2)^n(S_1)^m.\nonumber \end{eqnarray*}$$

We only have to consider the part that depends on k and l, as we need to keep the terms that depend on n and m intact. We re-sum the integral, going back from I₁ to I₀:

$$\begin{equation*} \sum \limits _{k,l}\frac{(y/x)_k(x y )_{n+m-k+l+p}}{(y^2)_{n+m+l+p}}\frac{(q^py/\tilde{b})_l(xyq^{p+l})_{p-l}}{(q)_l(q)_{p-l}}(y\tilde{b}q^{k})^l(-1)^lq^{l(l-1)/2}\frac{(q^px)^k}{(q)_k}\frac{(q)_{n+m}}{(q)_{n+m-k}}\tilde{b}^{n+m-k}.\nonumber \end{equation*}$$

Noticing that (xy)_{n + m − k + l + p}(xyq^{p + l})_{p − l} = (xyq^{p + l})_{n + m − k}(xy)_2p, we find:

$$\begin{equation*} \tilde{b}^{n+m}(q)_{n+m}(xy)_{2p}\sum \limits _{k,l}\frac{(y/x)_k(x y q^{p+l})_{n+m-k}}{(q)_k(q)_{n+m-k}}(q^{p+l}x/\tilde{b})^k\frac{(q^py/\tilde{b})_l}{(q)_l(q)_{p-l}}\frac{(y\tilde{b})^l}{(y^2)_{n+m+l+p}}(-1)^lq^{l(l-1)/2}.\nonumber \end{equation*}$$

Using (B.11) on the sum over k, we get:

$$\begin{eqnarray} &&\tilde{b}^{n+m}(q)_{n+m}(xy)_{2p}\sum \limits _{k,l}\frac{(y\tilde{b})_k(q^{p+l}y/\tilde{b})_{n+m-k}}{(q)_k(q)_{n+m-k}}(q^{p+l}x/\tilde{b})^k\frac{(q^py/\tilde{b})_l}{(q)_l(q)_{p-l}}\frac{(y\tilde{b})^l}{(y^2)_{n+m+l+p}}(-1)^lq^{l(l-1)/2}.\nonumber\\ \end{eqnarray} \tag{ B.14 }$$

We now write:

$$\begin{equation*} (q^{p+l}y/\tilde{b})_{n+m-k}\frac{(q^py/\tilde{b})_l}{(y^2)_{n+m+l+p}}=\frac{(q^py/\tilde{b})_{n+m-k+l}}{(y^2)_{n+m+l+p}}=\frac{(q^py/\tilde{b})_\infty }{(y^2)_\infty }\frac{(y^2q^{n+m+l+p})_\infty }{(q^{n+m-k+l+p}y/\tilde{b})_\infty }\nonumber \end{equation*}$$

and

$$\begin{equation*} \frac{(y^2q^{n+m+l+p})_\infty }{(q^{n+m-k+l+p}y/\tilde{b})_\infty }=\sum \limits _{r}\frac{(y \tilde{b}q^k)_r}{(q)_r}(q^{p+n+m-k+l}y/\tilde{b})^r.\nonumber \end{equation*}$$

Putting this into (B.14), we get:

$$\begin{eqnarray*} &&\tilde{b}^{n+m}(q)_{n+m}\frac{(xy)_{2p}}{(q)_p}\frac{(q^py/\tilde{b})_\infty }{(y^2)_\infty }\sum \limits _{k,l,r}\frac{(y\tilde{b})_{k+r}}{(q)_k(q)_{n+m-k}(q)_r}\nonumber \\ &&\times\,(q^{p}x/\tilde{b})^k(q^{p+n+m-k}y/\tilde{b})^r\frac{(q)_p}{(q)_l(q)_{p-l}}q^{l(l-1)/2}(-q^{k+r}y\tilde{b})^l.\nonumber \end{eqnarray*}$$

Since $(a)_p=\sum \limits _{l}\frac{(q)_p}{(q)_l(q)_{p-l}}q^{l(l-1)/2}(-a)^l$, this simplifies to

$$\begin{eqnarray*} &&\tilde{b}^{n+m}(q)_{n+m}\frac{(xy)_{2p}}{(q)_p}\frac{(q^py/\tilde{b})_\infty }{(y^2)_\infty }\sum \limits _{k,r}\frac{(y\tilde{b})_{k+r}}{(q)_k(q)_{n+m-k}(q)_r}(q^{p}x/\tilde{b})^k(q^{p+n+m-k}y/\tilde{b})^r(q^{k+r}y\tilde{b})_p\nonumber \\ &&=\tilde{b}^{n+m}(q)_{n+m}\frac{(xy)_{2p}}{(q)_p}\frac{(q^py/\tilde{b})_\infty }{(y^2)_\infty }\sum \limits _{k,r}\frac{(q^py\tilde{b})_{k}(q^{p+k}y\tilde{b})_{r}}{(q)_k(q)_{n+m-k}(q)_r}\nonumber\\ &&\times\,(q^{p}x/\tilde{b})^k(q^{p+n+m-k}y/\tilde{b})^r(y\tilde{b})_p.\nonumber \end{eqnarray*}$$

We can now get rid of the sum over r. We have:

$$\begin{equation*} \frac{(q^py/\tilde{b})_\infty }{(y^2)_\infty }\sum \limits _{r}\frac{(q^{p+k}y\tilde{b})_{r}}{(q)_r}(q^{p+n+m-k}y/\tilde{b})^r=\frac{(q^py/\tilde{b})_{n+m-k}}{(y^2)_{n+m+2p}}\nonumber \end{equation*}$$

which gives

$$\begin{equation*} (y\tilde{b})_p\tilde{b}^{n+m}\frac{(xy)_{2p}}{(q)_p}\frac{(q)_{n+m}}{(y^2)_{n+m+2p}}\sum \limits _{k}\frac{(q^py\tilde{b})_{k}(q^py/\tilde{b})_{n+m-k}}{(q)_k(q)_{n+m-k}}(q^{p}x/\tilde{b})^k.\nonumber \end{equation*}$$

Using (B.11) one final time on the sum over k gives:

$$\begin{equation*} (y\tilde{b})_p\tilde{b}^{n+m}\frac{(xy)_{2p}}{(q)_p}\frac{(q)_{n+m}}{(y^2)_{n+m+2p}}\sum \limits _{k}\frac{(y/x)_{k}(q^{2p}xy)_{n+m-k}}{(q)_k(q)_{n+m-k}}(q^{p}x/\tilde{b})^k.\nonumber \end{equation*}$$

We can at last put this back into $\overline{K^+}$. We get:

$$\begin{eqnarray*} &&\overline{I_0}=(x/b)_pb^p(xb)_p(y/\tilde{b})_p(\tilde{b})^p(y\tilde{b})_p(-y)^pq^{p(p-1)/2}\frac{(xy)_{2p}}{(y^2 )_{2p}}\nonumber \\ &&\times\,\frac{(q^pxS_1)_\infty (q^pxbS_2)_\infty }{(bS_1)_\infty (S_2)_\infty }\sum \limits _{n,m,k}\frac{(y/x)_{k}(q^{2p}xy)_{n+m-k}}{(q)_k(q)_{n+m-k}}(q^{p}x)^k\tilde{b}^{n+m-k}\nonumber \\ &&\times\,\frac{(q)_{n+m}}{(y^2q^{2p})_{n+m}}\frac{(q^py/b)_n(q^pyb)_{m}}{(q)_n(q)_{m}}(bS_2)^n(S_1)^m\nonumber \end{eqnarray*}$$

which is to say

$$\begin{equation*} \overline{I_0}(x,y)=\frac{(x/b)_pb^p(xb)_p(y/\tilde{b})_p(\tilde{b})^p(y\tilde{b})_p}{(y^2)_{2p}}(-y)^pq^{p(p-1)/2}(xy)_{2p} I_0(q^px,q^py).\nonumber \end{equation*}$$

In terms of K⁺, this gives us, for xy = q^{1 − p}:

$$\begin{equation} \fbox{$ \displaystyle\frac{K^+_{i+p,j+p}(x,y)}{K^+_{i,j}(q^px,q^py)}=\displaystyle\frac{(x/b)_pb^p(xb)_p(y/\tilde{b})_p(\tilde{b})^p(y\tilde{b})_p}{(y^2 )_{2p}}(-y)^pq^{p(p-1)/2}\displaystyle\frac{(q^{j+1})_p}{(q^{i+1})_p}. $} \end{equation} \tag{ B.15 }$$

The equivalent result for K⁻ is:

$$\begin{equation} \fbox{$ \displaystyle\frac{K^-_{j+p,i+p}(x,y)}{K^-_{j,i}(q^px,q^py)}=\displaystyle\frac{(y/a)_pa^p(ya)_p(x/\tilde{a})_p(\tilde{a})^p(x\tilde{a})_p}{(x^2 )_{2p}}(-x)^pq^{p(p-1)/2}\displaystyle\frac{(q^{i+1})_p}{(q^{j+1})_p}. $} \end{equation} \tag{ B.16 }$$

This first calculation is rather straightforward. Consider the PQ of order 1

$$\begin{equation} P_0Q_0=h_0+{\rm e}^{-k\mu }P_1Q_{-1} \end{equation} \tag{ C.1 }$$

and multiply the PQ equation of order k

$$\begin{equation} P_0Q_{k-1}=t^{[k]}_0+{\rm e}^{-k\mu }P_kQ_{-1} \end{equation} \tag{ C.2 }$$

by the product of all Q_l's for l between 0 and k − 2:

$$\begin{equation*} t^{[k]}_0\prod \limits _{l=0}^{k-2}Q_l=P_0\prod \limits _{l=0}^{k-1}Q_l-{\rm e}^{-k\mu }P_k\prod \limits _{l=-1}^{k-2}Q_l.\nonumber \end{equation*}$$

The right-hand side of the equation can then be rewritten to give

$$\begin{equation*} t^{[k]}_0\prod \limits _{l=0}^{k-2}Q_l=\sum \limits _{n=0}^{k-1}{\rm e}^{-n\mu }(P_nQ_n-{\rm e}^{-\mu }P_{n+1}Q_{n-1})\prod \limits _{l=-1}^{n-2}Q_l\prod \limits _{l=n+1}^{k-1}Q_l\nonumber \end{equation*}$$

in which we can simply use equation (C.1) shifted by n to obtain the T–Q equation:

$$\begin{equation} \fbox{$ t^{[k]}_0\prod \limits _{l=0}^{k-2}Q_l=\sum \limits _{n=0}^{k-1}\,{\rm e}^{-n\mu }h_n\prod \limits _{l=-1}^{n-2}Q_l\prod \limits _{l=n+1}^{k-1}Q_l. $} \end{equation} \tag{ C.3 }$$

As we see, this equation involves products of k − 1 matrices Q. By transferring all the Qs to the right-hand side, we find an alternative expression:

$$\begin{equation} t^{[k]}_0=\sum \limits _{n=0}^{k-1}{\rm e}^{-n\mu }h_n\frac{Q_{-1}Q_{k-1}}{Q_{n-1}Q_n}. \end{equation} \tag{ C.4 }$$

Let us consider $t^{[2]}_0t^{[k]}_1$. Using the PQ equations of order 2 and k, we find:

$$\begin{eqnarray*} &&t^{[2]}_0t^{[k]}_1=(P_0Q_1-{\rm e}^{-2\mu }P_2Q_{-1})(P_1Q_k-{\rm e}^{-k\mu }P_{k+1}Q_0)\nonumber \\ &&\hphantom{ t^{[2]}_0t^{[k]}_1}=P_0Q_kP_1Q_1+{\rm e}^{-(k+2)\mu }P_2Q_0P_{k+1}Q_{-1}-{\rm e}^{-2\mu }P_2Q_kP_1Q_{-1}-{\rm e}^{-k\mu }P_0Q_0P_{k+1}Q_1.\nonumber \end{eqnarray*}$$

The first two terms of this sum can also be obtained by considering $t^{[k+1]}_0h_1$:

$$\begin{eqnarray*} &&t^{[k+1]}_0h_1=(P_0Q_k-{\rm e}^{-(k+1)\mu }P_{k+1}Q_{-1})(P_1Q_1-{\rm e}^{-\mu }P_{2}Q_0)\nonumber \\ &&\hphantom{ t^{[k+1]}_0h_1}=P_0Q_kP_1Q_1+{\rm e}^{-(k+2)\mu }P_2Q_0P_{k+1}Q_{-1}-{\rm e}^{-(k+1)\mu }P_{k+1}Q_1P_1Q_{1}-{\rm e}^{-\mu }P_0Q_kP_{2}Q_{-1}\nonumber \end{eqnarray*}$$

so that

$$\begin{eqnarray*} &&t^{[2]}_0t^{[k]}_1-t^{[k+1]}_0h_1={\rm e}^{-(k+1)\mu }P_{k+1}Q_1P_1Q_{1}+{\rm e}^{-\mu }P_0Q_kP_{2}Q_{-1}\nonumber\\ &&\hphantom{ t^{[2]}_0t^{[k]}_1-t^{[k+1]}_0h_1=}-{\rm e}^{-2\mu }P_2Q_kP_1Q_{-1}-{\rm e}^{-k\mu }P_0Q_0P_{k+1}Q_1\nonumber \\ &&\hphantom{ t^{[2]}_0t^{[k]}_1-t^{[k+1]}_0h_1}={\rm e}^{-\mu }(P_0Q_0-{\rm e}^{-\mu }P_1Q_{-1})(P_2Q_k-{\rm e}^{-(k-1)\mu }P_{k+1}Q_1)\nonumber \\ &&\hphantom{ t^{[2]}_0t^{[k]}_1-t^{[k+1]}_0h_1}={\rm e}^{-\mu }t^{[k-1]}_2h_0.\nonumber \end{eqnarray*}$$

We obtain the fusion equations:

$$\begin{equation} \fbox{$ t^{[2]}_0t^{[k]}_1=h_1t^{[k+1]}_0+{\rm e}^{-\mu }h_0t^{[k-1]}_2 $} \end{equation} \tag{ C.5 }$$

which, for k → ∞, allows to recover the T–Q equation (45).

Through the same reasoning, we also find

$$\begin{equation} t^{[k]}_0t^{[2]}_{k-1}=h_{k-1}t^{[k+1]}_0+{\rm e}^{-\mu }h_kt^{[k-1]}_0, \end{equation} \tag{ C.6 }$$

and combining the two, we find

$$\begin{equation} t^{[2]}_0t^{[k]}_1t^{[2]}_{k}=h_1h_kt^{[k+2]}_0+{\rm e}^{-\mu }h_0h_kt^{[k]}_2+{\rm e}^{-\mu }h_1h_{k+1}t^{[k]}_0+{\rm e}^{-2\mu }h_0h_{k+1}t^{[k-2]}_2. \end{equation} \tag{ C.7 }$$