Time-Dependent Matrix Product Ansatz for Interacting Reversible Dynamics

Abstract

We present an explicit time-dependent matrix product ansatz (tMPA), which describes the time-evolution of any local observable in an interacting and deterministic lattice gas, specifically for the rule 54 reversible cellular automaton of Bobenko et al. (Commun Math Phys 158(1):127, 1993. https://doi.org/10.1007/BF02097234). Our construction is based on an explicit solution of real-space real-time inverse scattering problem. We consider two applications of this tMPA. Firstly, we provide the first exact and explicit computation of the dynamic structure factor in an interacting deterministic model, and secondly, we solve the extremal case of the inhomogeneous quench problem, where a semi-infinite lattice in the maximum entropy state is joined with an empty semi-infinite lattice. Both of these exact results rigorously demonstrate a coexistence of ballistic and diffusive transport behaviour in the model, as expected for normal fluids.

Appendices

The Action of the Matrices T, $T^{\prime }$, ${\overline{T}}$ and ${\overline{T}}^{\prime }$

In this appendix we explicitly compute the powers of the matrices T and ${\overline{T}}$ ($T^{\prime }$ and ${\overline{T}}^{\prime }$) acting onto the left (right), with the matrices defined as

$$\begin{aligned} \begin{aligned} T^{\phantom {\prime }}&=(V_0+V_1)(W_0+W_1),\qquad&{\overline{T}}^{\phantom {\prime }}&=(W_0+W_1)(V_0+V_1),\\ T^{\prime }&= (W_0^{\prime }+W_1^{\prime })(V_0^{\prime }+V_1^{\prime }),&{\overline{T}}^{\prime }&= (V_0^{\prime }+V_1^{\prime })(W_0^{\prime }+W_1^{\prime }), \end{aligned} \end{aligned}$$

(69)

but first let us discuss the general structure of the matrices $M\in \{\alpha V_0+\beta V_1,\alpha W_0+\beta W_1;\,\alpha ,\beta \in {\mathbb {R}}\}$ and $M^{\prime }\in \{\alpha V^{\prime }_0+\beta V^{\prime }_1,\alpha W^{\prime }_0+\beta W^{\prime }_1;\,\alpha ,\beta \in {\mathbb {R}}\}$. We start by defining the projectors to the subspace of unactivated vectors, i.e. the subspace defined by $a=0$, the subspace of activated vectors with width 0 ($a=1$ and $w=0$) and to the subspace of vectors with $a=1$ and $w\le 1$,

$$\begin{aligned} \begin{aligned} P_0 \mathinner {|{c,w,n,a}\rangle }&= \delta _{a,0}\mathinner {|{c,w,n,0}\rangle },\\ Q^{\phantom {\prime }}\mathinner {|{c,w,n,a}\rangle }&= \delta _{a,1}\delta _{w,0} \mathinner {|{c,0,n,1}\rangle },\\ Q^{\prime }\mathinner {|{c,w,n,a}\rangle }&= \delta _{a,1}\delta _{w,0} \mathinner {|{c,0,n,1}\rangle } +\delta _{a,1}\delta _{w,1} \mathinner {|{c,1,n,1}\rangle }. \end{aligned} \end{aligned}$$

(70)

The subspace with $a=1$ is invariant to multiplication by matrices $M^{T}$ and $M^{\prime }$,

$$\begin{aligned} P_0 M^T P_0 = P_0 M^T,\qquad P_0 M^{\prime } P_0 = P_0 M^{\prime }, \end{aligned}$$

(71)

and the value of w inside the $a=1$ subspace cannot increase, which implies the following,

$$\begin{aligned} \begin{aligned} Q M^T Q&= M^T Q ,\qquad&Q M^{\prime } Q&= M^{\prime } Q ,\\ Q^{\prime } M^T Q^{\prime }&= M^T Q^{\prime } ,\qquad&Q^{\prime } M^{\prime } Q^{\prime }&= M^{\prime } Q^{\prime }. \end{aligned} \end{aligned}$$

(72)

Additionally, the matrices M ($M^{\prime }$) commute with the raising/lowering operators defined in the main text Eq. (23) as long as $w\ge 1$ ($w\ge 2$). Explicitly,

$$\begin{aligned} \begin{aligned} (1-Q)\left[ {\mathbf {c}}^{\pm },M^T\right]&=0, \qquad&(1-Q)\left[ {\mathbf {w}}^{\pm },M^T\right]&=0,\\ (1-Q^{\prime })\left[ {\mathbf {c}}^{\pm },M^{\prime }\right]&=0, \qquad&(1-Q^{\prime })\left[ {\mathbf {w}}^{\pm },M^{\prime }\right]&=0. \end{aligned} \end{aligned}$$

(73)

We wish to obtain $\mathinner {\langle {v}|}T^x$, $\mathinner {\langle {v}|}{\overline{T}}^{x}$ (or $T^{\prime \,x}\mathinner {|{v}\rangle }$, ${\overline{T}}^{\prime \,x}\mathinner {|{v}\rangle }$) for an arbitrary vector $\mathinner {|{v}\rangle }\in {\mathcal {V}}$. Due to the mentioned properties, it is convenient to first express the $w\ge 1$ (or $w\ge 2$) projections,

$$\begin{aligned} \begin{gathered} \mathinner {\langle {v}|}T^x(1-Q),\qquad \mathinner {\langle {v}|}{\overline{T}}^x(1-Q),\\ (1-Q^{\prime })T^{\prime \,x}\mathinner {|{v}\rangle },\qquad (1-Q^{\prime }){\overline{T}}^{\prime \,x}\mathinner {|{v}\rangle }, \end{gathered} \end{aligned}$$

(74)

and compute the relevant overlaps using these vectors [for specific examples see Eqs. (96), (109), (121) and (122)].

Furthermore, the matrices $V_s^T$, $W_s^T$ and $V_s^{\prime }$, $W_s^{\prime }$ differ only in the boundary terms; explicitly,

$$\begin{aligned} (1-Q^{\prime })(V_s^{\prime }-V_s^{T})=0,\qquad (1-Q^{\prime })(W_s^{\prime }-W_s^{T})=0, \end{aligned}$$

(75)

therefore we can express the products of right-soliton matrices by projecting the corresponding left-soliton products to the subspace with $w\ge 2$ and transpose them,

$$\begin{aligned} \begin{aligned} (1-Q^{\prime })T^{\prime \, x}\mathinner {|{v}\rangle }&= \Big (\mathinner {\langle {v}|}T^x (1-Q)(1-Q^{\prime })\Big )^T,\\ (1-Q^{\prime }){\overline{T}}^{\prime \, x}\mathinner {|{v}\rangle }&= \Big (\mathinner {\langle {v}|}{\overline{T}}^x (1-Q)(1-Q^{\prime })\Big )^T. \end{aligned} \end{aligned}$$

(76)

Thus, it suffices to express $\mathinner {\langle {v}|} T^x(1-Q)$ and $\mathinner {\langle {v}|} {\overline{T}}^x(1-Q)$.

1.1 The powers $T^m$

The matrices $W_s$, $V_s$ restricted to the subspace with $a=0$ are simple, as are T and ${\overline{T}}$,

$$\begin{aligned} \mathinner {\langle {c,w,n,0}|} T P_0&=\mathinner {\langle {c,w,n,0}|} {\overline{T}} P_0 \nonumber \\&=2 \mathinner {\langle {c+1,w-1,0,0}|} + \mathinner {\langle {c+1,w-1,1,0}|}+\mathinner {\langle {c+1,w-1,2,0}|}. \end{aligned}$$

(77)

Since the subspace with $a=1$ is an invariant subspace of the left action of matrices $V_s$, $W_s$, the following holds

$$\begin{aligned}&\mathinner {\langle {c,w,n,0}|} T^x P_0 = \mathinner {\langle {c,w,n,0}|} {\overline{T}}^x P_0 \nonumber \\&\qquad =4^{x-1}\left( 2 \mathinner {\langle {c+x,w-x,0,0}|} + \mathinner {\langle {c+x,w-x,1,0}|}+\mathinner {\langle {c+x,w-x,2,0}|}\right) , \end{aligned}$$

(78)

as long as $x\le w$, otherwise the r.h.s. is 0.

Now let us focus on the subspace spanned by $\{\mathinner {|{c,w,n,1}\rangle }; c\ge 0, w>0, n\in \{0,1,2\}\}$. Due to the value of w and c decreasing, it is convenient to express the left action of $T^x$ to the basis vectors $\mathinner {\langle {c,w,n,1}|}$ in the following form

$$\begin{aligned} \begin{aligned} \mathinner {\langle {c,w,n,1}|} T^x (1-Q)&= \sum _{m,p} f_{x}^n(m,p) \mathinner {\langle {c-m,w-x-p,0,1}|}\\&\quad + \sum _{m,p} g_{x}^n(m,p) \mathinner {\langle {c-m,w-x-p,1,1}|}\\&\quad + \sum _{m,p} h_{x}^n(m,p) \mathinner {\langle {c-m,w-x-p,2,1}|}, \end{aligned} \end{aligned}$$

(79)

where $f_x^n$, $g_x^n$, $h_x^n$ are some unknown coefficients that have to satisfy the following recurrence relation

$$\begin{aligned} \begin{aligned} f^n_{x+1}(m,p)&=f^n_x(m,p)+f^n_x(m-1,p-1)+g^n_x(m,p-1)\\&\quad +g^n_x(m-1,p-1)+h^n_x(m,p)+h^n_x(m,p-1),\\ g^n_{x+1}(m,p)&=f^n_x(m,p)+g^n_x(m-1,p-1)+h^n_x(m,p),\\ h^n_{x+1}(m,p)&=f^n_x(m-1,p)+g^n_x(m-1,p-1)+h^n_x(m-1,p-1). \end{aligned} \end{aligned}$$

(80)

A family of solutions is parametrized by 4 parameters, $\alpha$, $\beta$, $\gamma$ and $\delta$,

$$\begin{aligned} \begin{aligned} f_x(m,p)&=\left( {\begin{array}{c}x-m+p+\alpha \\ m+\beta \end{array}}\right) \left( {\begin{array}{c}x+m-p+\gamma \\ p+\delta \end{array}}\right) ,\\ g_x(m,p)&=\left( {\begin{array}{c}x-m+p+\alpha \\ m+\beta \end{array}}\right) \left( {\begin{array}{c}x+m-p-1+\gamma \\ p+\delta \end{array}}\right) ,\\ h_x(m,p)&=\left( {\begin{array}{c}x-m+p+\alpha \\ m-1+\beta \end{array}}\right) \left( {\begin{array}{c}x+m-p-1+\gamma \\ p+\delta \end{array}}\right) . \end{aligned} \end{aligned}$$

(81)

Taking into account the appropriate initial conditions, it is possible to express the coefficients $f_x^n$, $g_x^n$, $h_x^n$ in terms of this solution with the following parametrization,

$$\begin{aligned} \begin{aligned} n&=0:\qquad&(\alpha ,\beta ,\gamma ,\delta )&=(0,0,0,0),\\ n&=1:&(\alpha ,\beta ,\gamma ,\delta )&=(0,0,0,-1),\\ n&=2:&(\alpha ,\beta ,\gamma ,\delta )&=(-1,0,1,0). \end{aligned} \end{aligned}$$

(82)

Now we are almost able to express the whole $\mathinner {\langle {v}|}T^{x}(1-Q)$ for any vector $\mathinner {\langle {v}|}$. The last remaining property is

$$\begin{aligned} \begin{aligned} \mathinner {\langle {c,w,0,0}|} T (1-P_0)&=\mathinner {\langle {c,w-1,0,1}|},\\ \mathinner {\langle {c,w,1,0}|} T (1-P_0)&=\mathinner {\langle {c,w-1,0,1}|}+\mathinner {\langle {c-1,w-1,0,1}|}\\&\quad +\mathinner {\langle {c-1,w-1,1,1}|}+\mathinner {\langle {c-1,w-1,2,1}|},\\ \mathinner {\langle {c,w,2,0}|} T (1-P_0)&=\mathinner {\langle {c,w-1,0,1}|}+\mathinner {\langle {c-1,w-1,2,1}|}. \end{aligned} \end{aligned}$$

(83)

Combining the Eqs. (78) and (83) with the expressions (81), and (82), we can explicitly obtain the coefficients in the basis expansion of $\mathinner {\langle {v}|}T^x(1-Q)$ in terms of sums of coefficients (81). For sufficiently simple $\mathinner {\langle {v}|}$ they simplify, as for example in the case $\mathinner {\langle {v}|}=\mathinner {\langle {0,t,0,0}|}=\mathinner {\langle {l(t)}|}$,

(84)

with the coefficients ${\mathcal {A}}_x^{n}(m,p)$ given by

$$\begin{aligned} \begin{aligned} {\mathcal {A}}_x^0(m,p)&=2^{2x+p-m-1}\left( {\begin{array}{c}m-p-1\\ p\end{array}}\right) , \\ {\mathcal {A}}_x^{1,2}(m,p)&=2^{2x+p-m-1}\left( {\begin{array}{c}m-p-2\\ p\end{array}}\right) . \end{aligned} \end{aligned}$$

(85)

1.2 The powers ${\overline{T}}^m$

Similarly, the left action of ${\overline{T}}^x$ on $\mathinner {\langle {c,w,n,1}|}$ can be expressed in terms of basis vectors via coefficients ${\bar{f}}_x^n$, ${\bar{g}}_x^n$, ${\bar{h}}_x^n$ as

$$\begin{aligned} \begin{aligned} \mathinner {\langle {c,w,n,1}|} T^x (1-Q)&= \sum _{m,p} {\bar{f}}_{x}^n(m,p) \mathinner {\langle {c-m,w-x-p,0,1}|}\\&\quad + \sum _{m,p} {\bar{g}}_{x}^n(m,p) \mathinner {\langle {c-m,w-x-p,1,1}|}\\&\quad + \sum _{m,p} {\bar{h}}_{x}^n(m,p) \mathinner {\langle {c-m,w-x-p,2,1}|}, \end{aligned} \end{aligned}$$

(86)

with the coefficients satisfying a recurrence relation similar to (80),

$$\begin{aligned} \begin{aligned} {\bar{f}}^n_{x+1}(m,p)&={\bar{f}}^n_x(m,p)+{\bar{f}}^n_x(m-1,p-1)+{\bar{g}}^n_x(m-1,p)\\&\quad +{\bar{g}}^n_x(m-1,p-1)+{\bar{h}}^n_x(m,p)+{\bar{h}}^n_x(m-1,p),\\ {\bar{g}}^n_{x+1}(m,p)&={\bar{f}}^n_x(m,p)+{\bar{g}}^n_x(m-1,p-1)+{\bar{h}}^n_x(m,p),\\ {\bar{h}}^n_{x+1}(m,p)&={\bar{f}}^n_x(m,p-1)+{\bar{g}}^n_x(m-1,p-1)+{\bar{h}}^n_x(m-1,p-1). \end{aligned} \end{aligned}$$

(87)

Again, a family of solutions is parametrized by 4 parameters,

$$\begin{aligned} \begin{aligned} {\bar{f}}_x(m,p)&=\left( {\begin{array}{c}x-m+p+\alpha \\ m+\beta \end{array}}\right) \left( {\begin{array}{c}x+m-p+\gamma \\ p+\delta \end{array}}\right) ,\\ {\bar{g}}_x(m,p)&=\left( {\begin{array}{c}x-m+p-1+\alpha \\ m+\beta \end{array}}\right) \left( {\begin{array}{c}x+m-p+\gamma \\ p+\delta \end{array}}\right) ,\\ {\bar{h}}_x(m,p)&=\left( {\begin{array}{c}x-m+p-1+\alpha \\ m+\beta \end{array}}\right) \left( {\begin{array}{c}x+m-p+\gamma \\ p-1+\delta \end{array}}\right) , \end{aligned} \end{aligned}$$

(88)

and the values of parameters corresponding to particular solutions ${\bar{f}}^n_x$, ${\bar{g}}^n_x$, ${\bar{h}}^n_x$ are

$$\begin{aligned} \begin{aligned} n&=0:\qquad&(\alpha ,\beta ,\gamma ,\delta )&=(0,0,0,0),\\ n&=1:&(\alpha ,\beta ,\gamma ,\delta )&=(0,-1,0,0),\\ n&=2:&(\alpha ,\beta ,\gamma ,\delta )&=(1,0,-1,0). \end{aligned} \end{aligned}$$

(89)

The relation equivalent to (83) is

$$\begin{aligned} \mathinner {\langle {c,w,0,0}|} {\overline{T}} (1-P_0)&=\mathinner {\langle {c,w-1,0,1}|}+\mathinner {\langle {c+1,w-1,2,1}|},\nonumber \\ \mathinner {\langle {c,w,1,0}|} {\overline{T}} (1-P_0)&= \mathinner {\langle {c,w-1,0,1}|}+\mathinner {\langle {c,w-1,1,1}|}+\mathinner {\langle {c+1,w-1,2,1}|},\nonumber \\ \mathinner {\langle {c,w,2,0}|} {\overline{T}} (1-P_0)&=\mathinner {\langle {c+1,w-1,2,1}|}. \end{aligned}$$

(90)

As before, it is possible to explicitly express $\mathinner {\langle {v}|} {\overline{T}}^x$ in terms of sums of coefficients $f^n_x$, $g^n_x$, $h^n_x$ for any vector $\mathinner {\langle {v}|}$. For some special vectors, the expressions are simple. For example,

(91)

with the coefficients $\bar{{\mathcal {A}}}_x^n$ defined as

$$\begin{aligned} \begin{aligned} \bar{{\mathcal {A}}}_x^0(m,p)&=\left( {\begin{array}{c}m-p-1\\ p\end{array}}\right) \sum _{y=m-x}^{2x+p-m-1}\left( {\begin{array}{c}2x+p-m-1\\ y\end{array}}\right) ,\\ \bar{{\mathcal {A}}}_x^1(m,p)&=\left( {\begin{array}{c}m-p-1\\ p\end{array}}\right) \sum _{y=m-x}^{2x+p-m-2}\left( {\begin{array}{c}2x+p-m-2\\ y\end{array}}\right) ,\\ \bar{{\mathcal {A}}}_x^2(m,p)&=\left( {\begin{array}{c}m-p-1\\ p-1\end{array}}\right) \sum _{y=m-x}^{2x+p-m-2}\left( {\begin{array}{c}2x+p-m-2\\ y\end{array}}\right) . \end{aligned} \end{aligned}$$

(92)

Note that we assumed $w\ge x$.

The Inhomogeneous Quench

We wish to explicitly obtain the overlaps

$$\begin{aligned} \begin{aligned} L(x,t)&=\mathinner {\langle {l(t)}|} T^x V_0 U^{t-x}\mathinner {|{r}\rangle },\\ R(x,t)&=\mathinner {\langle {l^{\prime }}|}{\overline{T}}^{\prime \, x} V_0^{\prime } U^{\prime \, t-x}\mathinner {|{r^{\prime }(t)}\rangle }, \end{aligned} \end{aligned}$$

(93)

with $0\le x \le t$ and

$$\begin{aligned} \begin{aligned} T^{\phantom {\prime }}&=(V_0+V_1)(W_0+W_1),\qquad&U^{\phantom {\prime }}&=W_0 V_0,\\ {\overline{T}}^{\prime }&= (V_0^{\prime }+V_1^{\prime })(W_0^{\prime }+W_1^{\prime }),&U^{\prime }&=W_0^{\prime }V_0^{\prime }. \end{aligned} \end{aligned}$$

(94)

Let us start with the overlap that corresponds to the left moving solitons.

1.1 Expressing the overlap L(x, t)

The matrices $V_s$, $W_s$ act trivially on the vectors $\mathinner {\langle {0,0,n,1}|}$,

$$\begin{aligned} \begin{aligned} \mathinner {\langle {0,0,n,1}|}V_s&=\mathinner {\langle {0,0,n,1}|}W_s=\mathinner {\langle {0,0,s\cdot \max \{n+1,2\},1}|},\\ \mathinner {\langle {0,0,n,1}|}T&=2\mathinner {\langle {0,0,0,1}|}+\mathinner {\langle {0,0,1,1}|}+\mathinner {\langle {0,0,2,1}|}, \end{aligned} \end{aligned}$$

(95)

therefore we can treat these vectors separately. Since they are the only vectors with the nonzero overlap with $\mathinner {|{r}\rangle }$, computing L(x, t) is equivalent to summing up the contributions of $\mathinner {\langle {0,0,n,1}|}$ vectors that are created at different steps. Explicitly,

$$\begin{aligned} \begin{aligned} L(x,t)&=\underbrace{(1-\delta _{x,0})\mathinner {\langle {l(t)}|}T\mathinner {|{r}\rangle }}_{\equiv L_1(x,t)}+ \underbrace{\sum _{y=1}^{x} 4^{x-y} \mathinner {\langle {l(t)}|}T^{y-1}(1-Q) T \mathinner {|{r}\rangle }}_{\equiv L_2(x,t)}\\&\quad + \underbrace{\mathinner {\langle {l(t)}|}T^x(1-Q)V_0 \mathinner {|{r}\rangle } + \sum _{z=1}^{t-x-1} \mathinner {\langle {l(t)}|}T^x V_0 U^{z-1} (1-Q) U \mathinner {|{r}\rangle }}_{\equiv L_3(x,t)}. \end{aligned} \end{aligned}$$

(96)

Since $L(0,t)=0$, let us from now on assume $x>0$ to simplify the notation. The first contribution is easy; if $t\ne 0$, the only nonzero overlap occurs for $x=t=1$,

$$\begin{aligned} L_1(x,t)=\delta _{x,1}\delta _{t,1}. \end{aligned}$$

(97)

The second contribution $L_2(x,t)$ is obtained from (84) as

(98)

Taking into account the form of the coefficients from (84) the first two contributions combine into

(99)

The function u(m, n) satisfies the following recurrence relation,

$$\begin{aligned} u(m,n) = \frac{1}{2} u(m-1,n) + \frac{1}{2} u(m-3,n) + \theta _{m-2x-1} 2^{2n-m} \left( {\begin{array}{c}m-2n\\ n-1\end{array}}\right) , \end{aligned}$$

(100)

which implies

$$\begin{aligned} 2 u(m,n) + u(m-1,n) + u(m-2,n) = {\left\{ \begin{array}{ll} 2; &{} n=0,\,m\ge 0,\\ 0; &{} n=0,\,m<0,\\ \sum _{y=0}^{m-2n-1} 2^{-y}\left( {\begin{array}{c}y\\ n-1\end{array}}\right) ;&n>0. \end{array}\right. }\nonumber \\ \end{aligned}$$

(101)

Therefore, the expression (99) simplifies into

(102)

where $\theta _x$ is a discrete Heaviside function,

$$\begin{aligned} \theta _x={\left\{ \begin{array}{ll}1;&{} x\ge 0,\\ 0; &{} x<0.\end{array}\right. } \end{aligned}$$

(103)

The other part is obtained by observing

$$\begin{aligned} \mathinner {\langle {c,w,0,1}|}U^z(1-Q)U\mathinner {|{r}\rangle }=\delta _{c,0}\delta _{w,z+1}, \end{aligned}$$

(104)

which implies

(105)

Inserting the explicit forms of the coefficients ${\mathcal {A}}_x^n$ and simplifying the expression we obtain

$$\begin{aligned} \begin{aligned} L_3(x,t)&=\delta _{x,1}\delta _{t,1}+\theta _{2x-t-2} 2^{t-1}\left( {\begin{array}{c}2x-t-2\\ t-x-1\end{array}}\right) \\&\quad +2^x\bigg ( \underbrace{\sum _{y=0}^{x-1} 2^y \left( {\begin{array}{c}x-1-y\\ y\end{array}}\right) }_{\frac{2^x}{3}\left( {1-\left( \frac{1}{2}\right) ^x}\right) } -\sum _{y=t-x}^{x-1}2^y \left( {\begin{array}{c}x-1-y\\ y\end{array}}\right) \bigg ). \end{aligned} \end{aligned}$$

(106)

Finally, the whole contribution of the left MPA is

$$\begin{aligned} L(x,t)= & {} 3\cdot 2^{2x-3} \delta _{t,x}(1-\delta _{t,1}) +2\delta _{x,1}\delta _{t,1} +16\delta _{x,3}\delta _{t,4}\nonumber \\&\quad + 2^{2x} \sum _{y=0}^{2x-t-4} 2^{-y}\left( {\begin{array}{c}y\\ t-x-1\end{array}}\right) \nonumber \\&\quad + 2^{2x}\frac{{1-\left( \frac{1}{2}\right) ^x}}{3} -2^x \sum _{y=t-x}^{x-1}2^y \left( {\begin{array}{c}x-1-y\\ y\end{array}}\right) \nonumber \\&\quad + \theta _{2x-t-3} 2^{t-1} \left( 3\left( {\begin{array}{c}2x-t-3\\ t-x-1\end{array}}\right) +2\left( {\begin{array}{c}2x-t-3\\ t-x-2\end{array}}\right) \right) . \end{aligned}$$

(107)

1.2 Overlap R(x, t)

We start by observing

$$\begin{aligned} V_0^{\prime } U^{\prime \, t-x}\mathinner {|{r^{\prime }(t)}\rangle } =V_0^{\prime }U^{\prime \, t-x}\mathinner {|{0,t+1,0,0}\rangle }=\mathinner {|{t-x,x+1,0,0}\rangle }, \end{aligned}$$

(108)

therefore the contribution of right moving solitons to the density profile is

$$\begin{aligned} R(x,t)=\mathinner {\langle {l^{\prime }}|} {\overline{T}}^{\prime \, x}V_0^{\prime }U^{\prime \, t-x}\mathinner {|{r^{\prime }(t)}\rangle } =\mathinner {\langle {l^{\prime }}|}{\overline{T}}^{\prime \, x}\mathinner {|{t-x,x+1,0,0}\rangle }. \end{aligned}$$

(109)

We use the same approach as before; as soon as vectors $\mathinner {|{0,0,n,1}\rangle }$ are created, we compute their overlap with the left boundary vector, while we keep propagating the other vectors,

$$\begin{aligned} \begin{aligned} R(x,t)&=\mathinner {\langle {l^{\prime }}|}{\overline{T}}^{\prime } (1-Q) {\overline{T}}^{\prime \, x-1}\mathinner {|{t-x,x+1,0,0}\rangle }\\&\quad +\sum _{y=1}^{x-2}4^{x-y-1}\mathinner {\langle {r}|} {\overline{T}}^{\prime }(1-Q){\overline{T}}^{\prime \, y}\mathinner {|{t-x,x+1,0,0}\rangle }+ \mathinner {\langle {r}|}{\overline{T}}^{\prime }\mathinner {|{t-x,x+1,0,0}\rangle }, \end{aligned}\nonumber \\ \end{aligned}$$

(110)

where we used the fact that $\mathinner {\langle {r}|}=(\mathinner {|{r}\rangle })^T$ is the $w=0$ part of $\mathinner {\langle {l^{\prime }}|}$, i.e.

$$\begin{aligned} \mathinner {\langle {r}|}=\mathinner {\langle {l^{\prime }}|}-\left( \mathinner {\langle {0,1,0,1}|}+\mathinner {\langle {0,1,2,1}|}\right) =\mathinner {\langle {l^{\prime }}|}Q. \end{aligned}$$

(111)

The expression for $(1-Q^{\prime }){\overline{T}}^{\prime \, y}\mathinner {|{t-x,x+1,0,0}\rangle }$ is straightforwardly obtained from Eq. (91) by transposing it and removing the vectors with $w=1$, therefore the right overlap reads

(112)

Due to the coefficients inside the sum vanishing for almost all values of y, the overlap can be equivalently expressed as

(113)

Inserting the explicit values of $\bar{{\mathcal {A}}}_y^n$ and simplifying the whole expression yields

$$\begin{aligned} R(x,t)&=2\delta _{x,2}\delta _{t,2}+2\delta _{x,3}\delta _{t,3} \nonumber \\&\quad + 2 \left( s(t-2)+ s (t-3) + 2 s(t-4)\right) \sum _{z=t-x}^{2x-t-3}\left( {\begin{array}{c}2x-t-3\\ z\end{array}}\right) \nonumber \\&\quad + 2 (1-\delta _{t,2})\left( s(t-2)+ 3 s (t-4) + 4 s(t-6)\right) \sum _{z=t-x-1}^{2x-t-3}\left( {\begin{array}{c}2x-t-3\\ z\end{array}}\right) ;\nonumber \\ s(m)&=\sum _{z=0}^{\lfloor \frac{m}{2}\rfloor } 4^{z} \left( {\begin{array}{c}m-2z\\ z\end{array}}\right) . \end{aligned}$$

(114)

The function s(m) satisfies the following recurrence relation,

$$\begin{aligned} s(m+3)=s(m+2)+4 s(m), \end{aligned}$$

(115)

which together with the initial condition $s(0)=s(1)=s(2)=1$ implies²

$$\begin{aligned} s(m)+3 s(m-2) + 4 s(m-4) = s(m)+2(m-1)+2s(m-2)=2^m, \end{aligned}$$

(116)

and the whole contribution from the right moving solitons is

$$\begin{aligned} R(x,t)=2\delta _{x,2}\delta _{t,2} + \theta _{2x-t-3}2^{t-1}\left( {\begin{array}{c}2x-t-3\\ t-x-1\end{array}}\right) +2^t\sum _{z=t-x}^{2x-t-3}\left( {\begin{array}{c}2x-t-3\\ z\end{array}}\right) .\qquad \end{aligned}$$

(117)

The Free Regime of the Inhomogeneous Quench

In this appendix we present an alternative derivation of the expression for the density profile (54), which also provides some physical insights into the result. Before the quench, there are no solitons in the right half-infinite chain. When we join the two half-chains, the right moving solitons from the left that reach the boundary continue moving to the right unperturbed with the velocity 1, since there are no left moving solitons to slow them down. Therefore an intermediate area with only right moving solitons is established between the vacuum and the part that contains both types of solitons. This can be seen on an example in Fig. 10. Due to the maximal velocity of the solitons being $v_{\text {max}}=1$, this area is limited to the right by $x=t$.

Fig. 10

One realization of the inhomogeneous quench up to $t=74$. Between the empty space on the right and the area filled with left and right moving solitons, there is a section, where all the particles move to the right and do not scatter, corresponding to the free regime

Fig. 11

A path of a soliton (red bordered squares) that moves to the left with the effective velocity 1 / 3. Due to propagation rules, the solitons can not scatter more frequently, therefore this is the slowest possible effective speed

The left border is determined by the right most possible position of the left moving solitons, which is $x=-t/3$ due to the effective soliton speed being bounded from bellow by 1 / 3 (see Fig. 11). The ballistic part of the profile is therefore described by the $-t/3+1\le x \le t$ part of the profile in (54), which can also be derived by assuming that solitons enter this area randomly with uniform probability.

Let us look at the intermediate area of the chain at some fixed time t and let us join two consecutive sites together, so that sites $t-(2k-1)$ and $t-2k$ constitute a supersite labeled by $n=k$,

(118)

All the sites with $x\ge t$ are empty, while a site with $x\le t-1$ is occupied if there is a right moving soliton going through it, in which case the whole supersite has to be occupied and the neighbouring supersites have to be empty. Since the solitons enter this area randomly, the site $n=1$ is occupied with probability 1 / 2. If site $n-1$ is full, site n has to be empty and if $n-1$ is empty, site n is occupied with probability 1 / 2. This can be expressed in a matrix form as

(119)

where $p_n$ is the probability of the site n being full. Taking into account $p_1=\frac{1}{2}$, we obtain

$$\begin{aligned} p_n=\frac{1}{3}\left( 1-\left( -\frac{1}{2}\right) ^n\right) , \end{aligned}$$

(120)

which matches the ballistic part of the profile (54).

The Dynamic Structure Factor

As already discussed in Appendices A and B , the matrices act trivially on the vectors $\mathinner {\langle {0,0,n,1}|}$, which implies that a general overlap, $\mathinner {\langle {l(t)}|}M_1M_2\cdots M_{2t+1}\mathinner {|{r}\rangle }+\mathinner {\langle {l^{\prime }}|}M_1^{\prime }M_2^{\prime }\cdots M_{2t+1}^{\prime }\mathinner {|{r^{\prime }(t)}\rangle }$, can be straightforwardly determined using the projections $\mathinner {\langle {l(t)}|}M_1 M_2 \cdots M_j(1-Q)$, with $j=0,1,\ldots 2t+1$. Explicitly, the overlaps (62) can be expressed as

(121)

and

(122)

where we introduced D, $D^{\prime }$ to denote the difference of the matrices,

$$\begin{aligned} D=T V_1-V_1 {\overline{T}},\qquad D^{\prime }=V_1^{\prime } {\overline{T}}^{\prime }-T^{\prime } V_1^{\prime }. \end{aligned}$$

(123)

Therefore to obtain $\varDelta C_{l}(x,t)$ it suffices to express the projections $\mathinner {\langle {l(t)}|}T^x (1-Q)$, $\mathinner {\langle {l(t)}|}T^x D (1-Q)$ and $\mathinner {\langle {l(t)}|}T^{x}D {\overline{T}}^y(1-Q)$ and then compute the relevant overlaps with the right vector $\mathinner {|{r}\rangle }$ as shown in (121). The right moving soliton counterpart is very similar; since the matrices $W_s^{\prime }$, $V_s^{\prime }$ are the same as $W_s^{T}$ and $V_s^T$ in the $w\ge 2$ subspace, we can just take the corresponding left moving soliton vectors, transpose them, remove the terms with $w \le 1$ [similarly as in (76)] and compute the overlaps from (122).

The procedure is straightforward but lengthy, therefore we split it into multiple parts. In Appendix D.1 we use the relations from Appendix A to explicitly write the vectors $\mathinner {\langle {l(t)}|}T^x(1-Q)$, $\mathinner {\langle {l(t)}|}T^x D (1-Q)$ and $\mathinner {\langle {l(t)}|}T^x(1-Q){\overline{T}}^y$ in terms of basis vectors $\mathinner {\langle {c,w,n,a}|}$ by introducing the coefficients ${\mathcal {A}}_x^n$, ${\mathcal {B}}_x^n$, ${\mathcal {C}}_{x,y}^n$ and ${\mathcal {D}}_{x,y}^n$. In Appendix D.2 we proceed to express the overlaps $\varDelta C_{l,r}(x,t)$. We split the overlaps into multiple parts corresponding to different coefficients and we simplify the contributions. They are expressed in terms of single binomial coefficients, their single sums and triple sums. The contributions consisting of triple sums are simplified in Appendix D.3, where also the whole overlaps $\varDelta C_{l,r}(x,t)$ are expressed. Additionally, another subsection is included at the end, Appendix D.4, where we show the equivalence of expressions (64) and (66) from the main text.

1.1 The explicit form of different contributions to the overlaps

We start by expressing the vectors $\mathinner {\langle {l(t)}|}T^x(1-Q)$, $\mathinner {\langle {l(t)}|}T^x D (1-Q)$ and $\mathinner {\langle {l(t)}|}T^xD{\overline{T}}^y(1-Q)$. The first one can be expressed in terms of the basis vectors $\mathinner {\langle {c,w,n,a}|}$ with the coefficients ${\mathcal {A}}_x^n(m,p)$, as introduced in (84) and (85). Acting on it with $T V_1 - V_1 {\overline{T}}$ we straightforwardly obtain

$$\begin{aligned} \begin{aligned} \mathinner {\langle {l(t)}|}&T^x D(1-Q)\\&\left. \begin{aligned}&= 4^x \bigg (-2\mathinner {\langle {x+1,t-(x+1);0;0}|} +\mathinner {\langle {x+1,t-(x+1);0;1}|} \\&\quad -\mathinner {\langle {x+1,t-(x+1);0;2}|} +\mathinner {\langle {x+1,t-(x+1);1;2}|}\bigg ) \end{aligned}\right\} \mathinner {\langle {s(x,t)}|}\\&\left. \begin{aligned}&\quad - \sum _{p}\sum _{m}{\mathcal {B}}^0_x(m,p)\mathinner {\langle {x-m,t-(x+1)-p;1;0}|}\\&\quad + \sum _{p}\sum _{m}{\mathcal {B}}^1_x(m,p)\mathinner {\langle {x-m,t-(x+1)-p;1;1}|}\\&\quad + \sum _{p}\sum _{m}{\mathcal {B}}^2_x(m,p)\mathinner {\langle {x-m,t-(x+1)-p;1;2}|} \end{aligned}\right\} \mathinner {\langle {c(x,t)}|}, \end{aligned}\nonumber \\ \end{aligned}$$

(124)

with the following coefficients

$$\begin{aligned} \begin{aligned} {\mathcal {B}}_x^0(m,p)&=2^{2x+p-m}\left( {\begin{array}{c}m-p\\ p\end{array}}\right) ,\\ {\mathcal {B}}_x^1(m,p)&=2^{2x+p-m}\left( {\begin{array}{c}m-p-1\\ p\end{array}}\right) ,\\ {\mathcal {B}}_x^2(m,p)&=2^{2x+p-m-1}\left( {\begin{array}{c}m-p\\ p-1\end{array}}\right) . \end{aligned} \end{aligned}$$

(125)

At this point it is convenient to split $\mathinner {\langle {l(t)}|}T^x D{\overline{T}}^y(1-Q)$ into two parts; the first part corresponds to acting with ${\overline{T}}^y$ from right to the first two lines from (124),

$$\begin{aligned} \begin{aligned} \mathinner {\langle {s(x,t)}|}{\overline{T}}^{y}(1-Q)&= \sum _{m,p}{\mathcal {C}}_{x,y}^0(m,p)\mathinner {\langle {x-m,t-(x+y+1)-p;1;0}|}\\&\quad +\sum _{m,p}{\mathcal {C}}_{x,y}^1(m,p)\mathinner {\langle {x-m,t-(x+y+1)-p;1;1}|}\\&\quad +\sum _{m,p}{\mathcal {C}}_{x,y}^2(m,p)\mathinner {\langle {x-m,t-(x+y+1)-p;1;2}|}, \end{aligned} \end{aligned}$$

(126)

where the coefficients ${\mathcal {C}}_{x,y}^n(m,p)$ are expressed in terms of ${\bar{f}}_y^n$, ${\bar{g}}_y^n$ and ${\bar{h}}_y^n$ as introduced in (89) and (88),

(127)

Similarly, the second part is

$$\begin{aligned} \begin{aligned} \mathinner {\langle {c(x,t)}|}{\overline{T}}^{y}(1-Q)&= \sum _{m,p}{\mathcal {D}}_{x,y}^0(m,p)\mathinner {\langle {x-m,t-(x+y+1)-p;1;0}|}\\&\quad +\sum _{m,p}{\mathcal {D}}_{x,y}^1(m,p)\mathinner {\langle {x-m,t-(x+y+1)-p;1;1}|}\\&\quad +\sum _{m,p}{\mathcal {D}}_{x,y}^2(m,p)\mathinner {\langle {x-m,t-(x+y+1)-p;1;2}|}, \end{aligned} \end{aligned}$$

(128)

with

(129)

1.2 The explicit overlaps $\varDelta C_{l,r}(x,t)$

To express the overlaps (121) and (122), we group the contributions from the different coefficients into separate groups,

$$\begin{aligned} \begin{aligned} \varDelta C_l(x,t)&\equiv \varDelta a(x,t) +\varDelta b(x,t) +\varDelta c(x,t) +\varDelta d(x,t),\\ \varDelta C_r(x,t)&\equiv \varDelta a^{\prime }(x,t) +\varDelta b^{\prime }(x,t) +\varDelta c^{\prime }(x,t) +\varDelta d^{\prime }(x,t), \end{aligned} \end{aligned}$$

(130)

where $\varDelta a(x,t)$ and $\varDelta a^{\prime }(x,t)$ include all the contributions from ${\mathcal {A}}_x^n$ coefficients,

$$\begin{aligned} \begin{aligned} \varDelta a(x,t)&=4^{t-x-1}\bigg (-{\mathcal {A}}_{x}^1(x-2,t-x-3)+{\mathcal {A}}_x^1(x-1,t-x-2)\\&\quad +2{\mathcal {A}}_x^1(x-1,t-x-3)-2{\mathcal {A}}_x^1(x-2,t-x-2)\bigg ),\\ \varDelta a^{\prime }(x,t)&=4^{t-x-1}\bigg ( {\mathcal {A}}_x^0(x,t-x-1)+\frac{1}{2}{\mathcal {A}}_x^0(x,t-x-2) -{\mathcal {A}}_x^0(x-1,t-x-2)\\&\quad -{\mathcal {A}}_x^1(x,t-x-1)+2{\mathcal {A}}_x^1(x-1,t-x-1) -\frac{3}{2}{\mathcal {A}}_x^1(x,t-x-2) \bigg ), \end{aligned}\nonumber \\ \end{aligned}$$

(131)

$\varDelta b(x,t)$ and $\varDelta b^{\prime }(x,t)$ include the contributions of ${\mathcal {B}}_{x}^n$,

$$\begin{aligned} \varDelta b(x,t)&=4^{t{-}x{-}2}\bigg ( {-}4{\mathcal {B}}^0_x(x,t-x-2){-}{\mathcal {B}}_x^0(x,t{-}x{-}3)\nonumber \\&\quad {-}{\mathcal {B}}_x^0(x{-}1,t{-}x{-}3){+}4{\mathcal {B}}_x^1(x{-}1,t{-}x{-}2){+}3{\mathcal {B}}^1_x(x{-}1,t{-}x{-}3)\nonumber \\&\quad {+}2{\mathcal {B}}_x^2(x,t{-}x{-}2){+}2{\mathcal {B}}_x^2(x{-}1,t{-}x{-}2){+}{\mathcal {B}}_x^2(x{-}1,t{-}x{-}3) \bigg ),\nonumber \\ \varDelta b^{\prime }(x,t)&=4^{t{-}x{-}2}\bigg ( {-}2{\mathcal {B}}_x^0(x,t{-}x{-}2){-}\frac{1}{2}{\mathcal {B}}_x^0(x,t{-}x{-}3)\nonumber \\&\quad {+}\frac{3}{2}{\mathcal {B}}_x^1(x{-}1,t{-}x{-}3) {+}{\mathcal {B}}_x^2(x,t{-}x{-}2){+}\frac{1}{2}{\mathcal {B}}_x^2(x{-}1,t{-}x{-}3) \bigg ), \end{aligned}$$

(132)

$\varDelta c(x,t)$ and $\varDelta c^{\prime }(x,t)$ contain the contributions from ${\mathcal {C}}_{x,y}^n$,

$$\begin{aligned} \varDelta c(x,t)&= \sum _{y=1}^{t-x-2} 4^{t-x-y-2}\bigg ( 4{\mathcal {C}}_{x,y}^0(x,t-x-y-2)+{\mathcal {C}}_{x,y}^0(x,t-x-y-3) \nonumber \\&\quad +{\mathcal {C}}_{x,y}^0(x-1,t-x-y-3) +4{\mathcal {C}}_{x,y}^1(x-1,t-x-y-2) \nonumber \\&\quad +3{\mathcal {C}}_{x,y}^1(x-1,t-x-y-3)+ 2{\mathcal {C}}_{x,y}^1(x,t-x-y-2) \nonumber \\&\quad +2{\mathcal {C}}_{x,y}^2(x-1,t-x-y-2)+{\mathcal {C}}_{x,y}^2(x-1,t-x-y-3) \bigg ),\nonumber \\ \varDelta c^{\prime }(x,t)&= \sum _{y=1}^{t-x-2} 4^{t - x - y - 2} \bigg ( 2 {\mathcal {C}}_{x,y}^0\left( x, t - x - y - 2\right) + \frac{1}{2} {\mathcal {C}}_{x,y}^0\left( x, t - x - y - 3\right) \nonumber \\&\quad + \frac{3}{2} {\mathcal {C}}_{x,y}^1\left( x - 1, t - x - y - 3\right) + {\mathcal {C}}_{x,y}^2\left( x, t - x - y - 2\right) \nonumber \\&\quad + \frac{1}{2} {\mathcal {C}}_{x,y}^2\left( x - 1, t - x - y - 3\right) \bigg ), \end{aligned}$$

(133)

and $\varDelta d(x,t)$, $\varDelta d^{\prime }(x,t)$ contain the contributions from ${\mathcal {D}}_{x,y}^n$,

$$\begin{aligned} \varDelta d(x,t)&= \sum _{y=1}^{t-x-2} 4^{t-x-y-2}\bigg ( 4{\mathcal {D}}_{x,y}^0(x,t-x-y-2)+{\mathcal {D}}_{x,y}^0(x,t-x-y-3)\nonumber \\&\quad +{\mathcal {D}}_{x,y}^0(x-1,t-x-y-3) +4{\mathcal {D}}_{x,y}^1(x-1,t-x-y-2)\nonumber \\&\quad +3{\mathcal {D}}_{x,y}^1(x-1,t-x-y-3) + 2{\mathcal {D}}_{x,y}^2(x,t-x-y-2) \nonumber \\&\quad +2{\mathcal {D}}_{x,y}^2(x-1,t-x-y-2)+{\mathcal {D}}_{x,y}^2(x-1,t-x-y-3) \bigg ),\nonumber \\ \varDelta d^{\prime }(x,t)&= \sum _{y=1}^{t-x-2} 4^{t - x - y - 2} \bigg ( 2 {\mathcal {D}}_{x,y}^0\left( x, t - x - y - 2\right) + \frac{1}{2} {\mathcal {D}}_{x,y}^0\left( x, t - x - y - 3\right) \nonumber \\&\quad + \frac{3}{2} {\mathcal {D}}_{x,y}^1\left( x - 1, t - x - y - 3\right) + {\mathcal {D}}_{x,y}^2\left( x, t - x - y - 2\right) \nonumber \\&\quad + \frac{1}{2}{\mathcal {D}}_{x,y}^2\left( x - 1, t - x - y - 3\right) \bigg ). \end{aligned}$$

(134)

The contributions $\varDelta a(x,t)$, $\varDelta b(x,t)$, $\varDelta a^{\prime }(x,t)$ and $\varDelta b^{\prime }(x,t)$ can be expressed in terms of simple binomial coefficients by plugging the coefficients ${\mathcal {A}}_x^n$, ${\mathcal {B}}_x^n$ into Eqs. (131) and (132),

$$\begin{aligned} \varDelta a(x,t)+\varDelta b(x,t)= & {} \theta _{2x-t-2} 2^{3t-2x-3} \left( \left( {\begin{array}{c}2x-t-2\\ t-x-3\end{array}}\right) -\left( {\begin{array}{c}2x-t-2\\ t-x-2\end{array}}\right) \right) \nonumber \\&\quad -\theta _{2x-t+1} 2^{3t-2x-7} \left( \left( {\begin{array}{c}2x-t+1\\ t-x-4\end{array}}\right) -\left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) \right) ,\nonumber \\ \varDelta a^{\prime }(x,t)+\varDelta b^{\prime }(x,t)= & {} \theta _{2x-t-2} \left( 2\left( {\begin{array}{c}2x-t-2\\ t-x-1\end{array}}\right) -\left( {\begin{array}{c}2x-t-1\\ t-x-3\end{array}}\right) \right) \nonumber \\&\quad -\theta _{2x-t+1} 2^{3t-2x-8}\left( \left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) -\left( {\begin{array}{c}2x-t+1\\ t-x-4\end{array}}\right) \right) .\qquad \end{aligned}$$

(135)

Similarly, the sums (133) can be simplified into the following form,

$$\begin{aligned} \varDelta c(x,t)= & {} \theta _{t-2x-4} 2^{t+2x-1}\left( 6\left( {\begin{array}{c}t-2x-4\\ x\end{array}}\right) +5\left( {\begin{array}{c}t-2x-4\\ x-1\end{array}}\right) +\left( {\begin{array}{c}t-2x-4\\ x-2\end{array}}\right) \right) ,\nonumber \\ \varDelta c'(x,t)= & {} \theta _{t-2x-4} 2^{t+2x-1}\left( 3\left( {\begin{array}{c}t-2x-4\\ x\end{array}}\right) +\left( {\begin{array}{c}t-2x-4\\ x-1\end{array}}\right) \right) , \end{aligned}$$

(136)

by observing that for any $u\ge 0$ the following holds,

$$\begin{aligned} \sum _{m=0}^{\lfloor \frac{u}{2}\rfloor } 4^m\left( {\begin{array}{c}u-2m\\ m\end{array}}\right) =2^{u-1} +\frac{1-\frac{ \mathrm{i} }{\sqrt{7}}}{4}\left( \frac{-1+ \mathrm{i} \sqrt{7}}{2}\right) ^{\!u}\!\! +\frac{1+\frac{ \mathrm{i} }{\sqrt{7}}}{4}\left( \frac{-1- \mathrm{i} \sqrt{7}}{2}\right) ^{\!u}\!\!\equiv a_u.\nonumber \\ \end{aligned}$$

(137)

However, simplifying the contributions $\varDelta d(x,t)$ and $\varDelta d^{\prime }(x,t)$ requires a bit more work.

1.3 Contributions $\varDelta d(x,t)$ and $\varDelta d^{\prime }(x,t)$

We start by noting that both the remaining contributions can be expressed in terms of the following triple sum,

(138)

as

$$\begin{aligned} \begin{aligned} \varDelta d(x,t) - 2\varDelta d^{\prime }(x,t)&= 2^{3t-2x-6}\left( {\begin{array}{c}2x-t+2\\ t-x-3\end{array}}\right) \\ {}&\quad +2^{2t-5} \bigg ( s_{x,t}( -3, -4, -3) + s_{x,t}( -3, -4, -2) + 2 s_{x,t}( -3, -4, -1) \\ {}&\quad + s_{x,t}( -2, -3, -6) + s_{x,t}( -2, -3, -5) + 2 s_{x,t}( -2, -3, -4) \\ {}&\quad - 2 s_{x,t}( -1, -3, -4) - 2 s_{x,t}( -1, -3, -3) - 4 s_{x,t}( -1, -3, -2) \bigg ), \end{aligned}\qquad \end{aligned}$$

(139)

and

$$\begin{aligned} \begin{aligned} \varDelta d^{\prime }(x,t)&= 2^{3t-2x-8} \left( 8 \left( {\begin{array}{c}2 x - t + 2\\ t - x - 2\end{array}}\right) + \left( {\begin{array}{c}2 x - t + 3\\ t - x - 3\end{array}}\right) \right) \\&\quad +2^{2 t-7} \bigg ( s_{x,t}(-3,-5,-4) + 3 s_{x,t}(-3,-5,-2) + s_{x,t}(-2,-4,-7) \\ {}&\quad + 3 s_{x,t}(-2,-4,-5) + 2 s_{x,t}(-2,-4,-3) + 2 s_{x,t}(-2,-4,-2) \\ {}&\quad + 8 s_{x,t}(-2,-3,-1) - 2 s_{x,t}(-1,-4,-5) - 6 s_{x,t}(-1,-4,-3) \\ {}&\quad + 2 s_{x,t}(-1,-3,-6) + 2 s_{x,t}(-1,-3,-5) + 8 s_{x,t}(-1,-2,-4) \\ {}&\quad - 4 s_{x,t}(0,-3,-4) - 4 s_{x,t}(0,-3,-3) -16 s_{x,t}(0,-2,-2) \bigg ). \end{aligned} \end{aligned}$$

(140)

Note that instead of explicitly expressing $\varDelta d(x,t)$ we simplified the expressions a bit by introducing $\varDelta d(x,t)-2\varDelta d^{\prime }(x,t)$.

We first observe that the inner-most sum in (138) can be evaluated,

$$\begin{aligned} \sum _{w=0}^{\frac{u}{2}}2^{-u+2w}\left( {\begin{array}{c}u-2w\\ w\end{array}}\right) = 2^{-u}a_{u}, \end{aligned}$$

(141)

with $a_u$ defined in (137). If $u\ge 0$, the coefficients $a_u$ satisfy the following recurrence relation,

$$\begin{aligned} a_{u+1}=a_u+4 a_{u-2}. \end{aligned}$$

(142)

This enables us to rewrite the expression (139) in terms of simpler double sums,

(143)

by grouping together the terms $s_{x,t}(\alpha ,\beta ,\gamma )$ with the same $\alpha$, $\beta$. Explicitly,

$$\begin{aligned} s_{x,t}(\alpha ,\beta ,\gamma )+s_{x,t}(\alpha ,\beta ,\gamma +1)+2 s_{x,t}(\alpha ,\beta ,\gamma +2) \mapsto 2^{-(t+\alpha +\beta )}{\bar{s}}(x+\alpha ,t-x+\beta ).\nonumber \\ \end{aligned}$$

(144)

Furthermore, we have to subtract the terms that contain $a_{n}$ with $n<0$, since the relation does not hold for them. Taking care of these corner cases, the contribution $\varDelta d(x,t)-2\varDelta d^{\prime }(x,t)$ can be rewritten as

$$\begin{aligned} \begin{aligned}&\varDelta d(x,t)-2\varDelta d^{\prime }(x,t)=2^{3t-2x-6}\left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) -2^{3t-2x-5}\left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) \\&+ 2^{t+3} {\bar{s}}(x-3,t-x-4)+2^{t+1}{\bar{s}}(x-2,t-x-3)-2^{t+1}{\bar{s}}(x-1,t-x-3). \end{aligned}\nonumber \\ \end{aligned}$$

(145)

It is possible to further simplify this result by noting that the functions ${\bar{s}}(m,n)$ satisfy the following two relations,

$$\begin{aligned} \begin{aligned} {\bar{s}}(m+1,n)&={\bar{s}}(m,n)+4{\bar{s}}(m-1,n-1)+\varepsilon (m,n),\\ {\bar{s}}(m,n+1)&={\bar{s}}(m,n)+4{\bar{s}}(m-1,n-1)+\eta (m,n),\\ \varepsilon (m,n)&=\theta _{n-m-1} 4^{m+1}\left( {\begin{array}{c}n-m-1\\ m+1\end{array}}\right) ,\\ \eta (m,n)&=\theta _{m-n-1} 4^{n+1}\left( {\begin{array}{c}m-n-1\\ n+1\end{array}}\right) , \end{aligned} \end{aligned}$$

(146)

which enable us to obtain

$$\begin{aligned} \varDelta d(x,t)-2\varDelta d^{\prime }(x,t)= {\left\{ \begin{array}{ll} -2^{2x+t-1}\left( {\begin{array}{c}t-2x-2\\ x-1\end{array}}\right) ; &{} x\le \frac{t-2}{2},\\ -2^{3t-2x-6}\left( \left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) -\left( {\begin{array}{c}2x-t+1\\ t-x-4\end{array}}\right) \right) ;&x \ge \frac{t-1}{2}. \end{array}\right. } \end{aligned}$$

(147)

The contribution (140) is a bit more complicated, since the sums $s_{x,t}(\alpha ,\beta ,\gamma )$ with the same $\alpha$, $\beta$ do not simplify as before. Therefore we split the coefficients $a_n$ into two parts,

$$\begin{aligned} 2^{-n}a_n=\frac{1}{2}+b_n, \end{aligned}$$

(148)

and we treat the different contributions to $\varDelta d^{\prime }(x,t)$ separately,

$$\begin{aligned} \varDelta d^{\prime }(x,t) = \varDelta d_c^{\prime }(x,t) + \varDelta d_r^{\prime }(x,t) + \varDelta d_i^{\prime }(x,t), \end{aligned}$$

(149)

where $\varDelta d_c^{\prime }(x,t)$ includes all the constant terms,

$$\begin{aligned} \varDelta d_c^{\prime }(x,t) = 2^{3t-2x-8}\left( 8\left( {\begin{array}{c}2x-t+2\\ t-x-2\end{array}}\right) -\left( {\begin{array}{c}2x-t+2\\ t-x-3\end{array}}\right) \right) , \end{aligned}$$

(150)

and $\varDelta d_{i,r}^{\prime }(x,t)$ correspond to different parts of the coefficients $a_n$. Explicitly,

$$\begin{aligned} \varDelta d^{\prime }_r(x,t)&=2^{t-2}\bigg ( 16{\bar{s}}(x-3,t-x-5)+8{\bar{s}}(x-2,t-x-4) \nonumber \\&\quad +4{\bar{s}}(x-2,t-x-3) -4{\bar{s}}(x-1,t-x-4) +{\bar{s}}(x-1,t-x-3) \nonumber \\ {}&\quad +{\bar{s}}(x-1,t-x-2) -{\bar{s}}(x,t-x-3)-{\bar{s}}(x,t-x-2) \bigg ), \end{aligned}$$

(151)

and

$$\begin{aligned} \varDelta d^{\prime }_i(x,t)&=-2^{t-1} \bigg ( 4{\bar{s}}_2(x-3,t-x-5) - 4{\bar{s}}_3(x-3,t-x-5) \nonumber \\ {}&\quad + 2{\bar{s}}_1(x-2,t-x-4) - 2{\bar{s}}_2(x-2,t-x-4) - 4{\bar{s}}_2(x-2,t-x-3) \nonumber \\ {}&\quad - {\bar{s}}_0(x-1,t-x-4) + {\bar{s}}_1(x-1,t-x-4) - {\bar{s}}_0(x-1,t-x-3) \nonumber \\ {}&\quad - 2{\bar{s}}_1(x-1,t-x-3) + {\bar{s}}_0(x-1,t-x-2) - 2{\bar{s}}_1(x-1,t-x-2) \nonumber \\ {}&\quad - {\bar{s}}_1(x,t-x-3) + {\bar{s}}_0(x,t-x-2) \bigg ), \end{aligned}$$

(152)

with the generalized sums ${\bar{s}}_\gamma (m,n)$ defined as

(153)

Similarly as before, the contribution (151) reduces into

$$\begin{aligned} \varDelta d^{\prime }_r(x,t)&=2^{t-2}\bigg ( -4\eta (x-2,t-x-4)-\eta (x-1,t-x-3)+\eta (x,t-x-3) \nonumber \\&\quad -2\varepsilon (x-1,t-x-2) \bigg )\nonumber \\ {}&=-\theta _{t-2x-2}2^{2x+t-1}\left( {\begin{array}{c}t-2x-2\\ x\end{array}}\right) . \end{aligned}$$

(154)

In order to simplify the last part (152), we first observe that the following relations hold,

$$\begin{aligned} \begin{aligned} {\bar{s}}_\gamma (m+1,n)&= {\bar{s}}_{\gamma }(m,n) + 4{\bar{s}}_{\gamma }(m-1,n-1) + \varepsilon _{{\gamma }}(m,n),\\ {\bar{s}}_{\gamma }(m,n+1)&= {\bar{s}}_{\gamma +1}(m,n) + 4 {\bar{s}}_{\gamma +3}(m-1,n-1) + \eta _{{\gamma }}(m,n),\\ \varepsilon _{{\gamma }}(m,n)&= \theta _{n-m-1} 4^{m+1}\left( {\begin{array}{c}n-(m+1)\\ m+1\end{array}}\right) b_{n+m+1+\gamma },\\ \eta _{{\gamma }}(m,n)&=\theta _{m-n-1} 4^{n+1}\left( {\begin{array}{c}m-(n+1)\\ n+1\end{array}}\right) b_{\gamma }. \end{aligned} \end{aligned}$$

(155)

Using them, we obtain,

$$\begin{aligned} \begin{aligned} \varDelta d^{\prime }_{i}(x,t)&=-2^{t-1}\bigg ( \eta _{0}(x,t-x-3)-\eta _1(x-1,t-x-3) \\ {}&\quad -4\eta _2(x-2,t-x-4) +4\eta _3(x-1,t-x-3)\bigg ) \\ {}&=-\theta _{2x-t+1} 2^{3t-2x-7}\left( 3\left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) +4\left( {\begin{array}{c}2x-t+1\\ t-x-2\end{array}}\right) \right) , \end{aligned} \end{aligned}$$

(156)

which yields

$$\begin{aligned} \varDelta d^{\prime }(x,t)={\left\{ \begin{array}{ll} -2^{2x+t-1}\left( {\begin{array}{c}t-2x-2\\ x\end{array}}\right) ; &{} x\le \frac{t-2}{2},\\ 2^{3t-2x-9}\left( \left( {\begin{array}{c}2x-t+1\\ t-x-3\end{array}}\right) -\left( {\begin{array}{c}2x-t+1\\ t-x-4\end{array}}\right) \right) ;&x\ge \frac{t-1}{2}. \end{array}\right. } \end{aligned}$$

(157)

By combining the Eqs. (135), (136), (147) and (157), we can finally express the left and right overlap as

$$\begin{aligned} \varDelta C_l(x,t)= & {} {\left\{ \begin{array}{ll} 2^{2x+t-1}\left( 4\left( {\begin{array}{c}t-2x-4\\ x\end{array}}\right) -3\left( {\begin{array}{c}t-2x-4\\ x-2\end{array}}\right) -\left( {\begin{array}{c}t-2x-4\\ x-3\end{array}}\right) \right) ;&{} x\le \frac{t-4}{2},\\ 0;&{} \frac{t-3}{2}\le x \le \frac{t+1}{2},\\ 2^{3t-2x-3}\left( \left( {\begin{array}{c}2x-t-2\\ t-x-3\end{array}}\right) -\left( {\begin{array}{c}2x-t-2\\ t-x-2\end{array}}\right) \right) ;&\frac{t+2}{2}\le x, \end{array}\right. }\nonumber \\ \varDelta C_r(x,t)= & {} {\left\{ \begin{array}{ll} 2^{2x+t-1}\left( 2\left( {\begin{array}{c}t-2x-4\\ x\end{array}}\right) -\left( {\begin{array}{c}t-2x-4\\ x-1\end{array}}\right) -\left( {\begin{array}{c}t-2x-4\\ x-2\end{array}}\right) \right) ;&{} x\le \frac{t-4}{2},\\ 0;&{} \frac{t-3}{2}\le x \le \frac{t+1}{2},\\ 2^{3t-2x-3}\left( 2\left( {\begin{array}{c}2x-t-2\\ t-x-1\end{array}}\right) -\left( {\begin{array}{c}2x-t-2\\ t-x-3\end{array}}\right) -\left( {\begin{array}{c}2x-t-2\\ t-x-4\end{array}}\right) \right) ;&\frac{t+2}{2}\le x. \end{array}\right. }\nonumber \\ \end{aligned}$$

(158)

1.4 Equation (66)

To show that (66) is equivalent to (64), it suffices to prove that the polynomial ${\tilde{p}}(u,x)$, defined as

$$\begin{aligned} {\tilde{p}}(u,x) = \sum _{n=2x}^{3x}\left( \prod _{\begin{array}{c} j=2x\\ j\ne n \end{array}}^{3x}\frac{u-j}{n-j}\right) \left( 2 s(n)-s(n+1)\right) , \end{aligned}$$

(159)

is equivalent to the sum p(u, x),

$$\begin{aligned} p(u,x) = \sum _{m=0}^{x}4^m\left( 2\left( {\begin{array}{c}u-2m\\ m\end{array}}\right) -\left( {\begin{array}{c}u+1-2m\\ m\end{array}}\right) \right) , \end{aligned}$$

(160)

where $s(u)=p(u,\lfloor \frac{u}{2}\rfloor )$ was defined in the main text³ and $u\ge 0$, $2x \le u$. Clearly, if $\frac{u}{2}\ge x \ge \frac{u+1}{3}$, both expressions coincide, therefore it is sufficient to show that ${\tilde{p}}(u,x)$ satisfies the same relation as p(u, x),

$$\begin{aligned} p(u,x)=p(u-1,x)+4p(u-2,x)-4^{x+1}\left( 2\left( {\begin{array}{c}u-2x-3\\ x\end{array}}\right) -\left( {\begin{array}{c}u-2x-2\\ x\end{array}}\right) \right) .\nonumber \\ \end{aligned}$$

(161)

After some straightforward manipulation of the sums, we obtain the following

(162)

Expressing it in terms of the sums $r(x,\alpha )=\sum _{n=0}^{x+1}(-1)^n\left( {\begin{array}{c}x+1\\ n\end{array}}\right) s(n+\alpha )$ and taking into account the following properties,

$$\begin{aligned} \left. \begin{aligned} r(x+1,\alpha )&=-4 r(x,\alpha -2)\\ r(0,\alpha )&=-4 s(\alpha -2) \end{aligned}\right\} \, r(x,\alpha )=(-4)^{x+1} s(\alpha -2x-2), \end{aligned}$$

(163)

yields

$$\begin{aligned} {\tilde{p}}(u,x)\!-\!{\tilde{p}}(u-1,x)\!-\!4{\tilde{p}}(u-2,x)= \!-4^{x+1}\!\left( \! 2\left( {\begin{array}{c}u-2x-3\\ x\end{array}}\right) -\left( {\begin{array}{c}u-2x-2\\ x\end{array}}\right) \!\right) .\nonumber \\ \end{aligned}$$

(164)