Non-compact Quantum Spin Chains as Integrable Stochastic Particle Processes

Abstract

In this paper we discuss a family of models of particle and energy diffusion on a one-dimensional lattice, related to those studied previously in Sasamoto and Wadati (Phys Rev E 58:4181–4190, 1998), Barraquand and Corwin (Probab Theory Relat Fields 167(3–4):1057–1116, 2017) and Povolotsky (J Phys A 46(46):465205, 2013) in the context of KPZ universality class. We show that they may be mapped onto an integrable ${\mathfrak {sl}}(2)$ Heisenberg spin chain whose Hamiltonian density in the bulk has been already studied in the AdS/CFT and the integrable system literature. Using the quantum inverse scattering method, we study various new aspects, in particular we identify boundary terms, modeling reservoirs in non-equilibrium statistical mechanics models, for which the spin chain (and thus also the stochastic process) continues to be integrable. We also show how the construction of a “dual model” of probability theory is possible and useful. The fluctuating hydrodynamics of our stochastic model corresponds to the semiclassical evolution of a string that derives from correlation functions of local gauge invariant operators of $\mathcal {N}=4$ super Yang–Mills theory (SYM), in imaginary-time. As any stochastic system, it has a supersymmetric completion that encodes for the thermal equilibrium theorems: we show that in this case it is equivalent to the ${\mathfrak {sl}}(2|1)$ superstring that has been derived directly from $\mathcal {N}=4$ SYM.

Appendices

Taylor Expansion

1.1 Derivation (2.15)

The action of the bulk generator $-\mathcal {H}^t_{i,i+1}$ on a function of the rescaled variables $(x_i^{(M)}, x_{i+1}^{(M)})$ reads

$$\begin{aligned} -\mathcal {H}^t_{i,i+1} f\big (x_i^{(M)}, x_{i+1}^{(M)}\big )= & {} \sum _{k=1}^{x_i M} \; \frac{1}{k} \Big [ f\Big (x_i^{(M)} - \frac{k}{M}, x_{i+1}^{(M)} + \frac{k}{M}\Big ) - f\Big (x_i^{(M)} , x_{i+1}^{(M)}\Big ) \Big ] \nonumber \\&+ \sum _{k=1}^{x_i M} \; \frac{1}{k} \Big [ f\Big (x_i^{(M)} + \frac{k}{M}, x_{i+1}^{(M)} - \frac{k}{M}\Big ) - f\Big (x_i^{(M)} , x_{i+1}^{(M)}\Big ) \Big ].\nonumber \\ \end{aligned}$$

(A.1)

We only consider the scaling limit of the right jumps, the proof is analogous for the left jumps. On one hand we notice that, by a bivariate Taylor expansion, we may write

$$\begin{aligned}&\sum _{k=1}^{x_i M} \; \frac{1}{k} \Big [ f\Big (x_i^{(M)} - \frac{k}{M}, x_{i+1}^{(M)} + \frac{k}{M}\Big ) - f\Big (x_i^{(M)} , x_{i+1}^{(M)}\Big ) \Big ] \nonumber \\&\quad = \sum _{k=1}^{x_i M} \; \frac{1}{k} \sum _{a=1}^{\infty } \frac{1}{a!} \sum _{b=0}^a {a \atopwithdelims ()b} \Big [\Big (\frac{\partial }{\partial x_i}\Big )^b \Big ( \frac{\partial }{\partial x_{i+1}} \Big )^{a-b} f\Big (x_i^{(M)} , x_{i+1}^{(M)}\Big )\Big ] \Big (-\frac{k}{M} \Big )^b \Big ( \frac{k}{M} \Big )^{a-b}.\nonumber \\ \end{aligned}$$

(A.2)

This implies

$$\begin{aligned}&\lim _{M\rightarrow \infty } \sum _{k=1}^{x_i M} \; \frac{1}{k} \Big [ f\Big (x_i^{(M)} - \frac{k}{M}, x_{i+1}^{(M)} + \frac{k}{M}\Big ) - f\Big (x_i^{(M)} , x_{i+1}^{(M)}\Big ) \Big ] \nonumber \\&\quad =\sum _{a=1}^{\infty } \frac{1}{a!} \sum _{b=0}^a {a \atopwithdelims ()b} \Big [\Big (\frac{\partial }{\partial x_i}\Big )^b \Big ( \frac{\partial }{\partial x_{i+1}} \Big )^{a-b} f(x_i , x_{i+1})\Big ] (-1)^b \int _{0}^{x_i} \alpha ^{a-1} d\alpha \end{aligned}$$

(A.3)

where the convergence of the Riemann sum

$$\begin{aligned} \sum _{k=1}^{x_iM} \frac{1}{k} \Big (\frac{k}{M} \Big )^a \longrightarrow \int _{0}^{x_i} \alpha ^{a-1} d\alpha \qquad \text {as} \quad M\rightarrow \infty \end{aligned}$$

(A.4)

has been exploited. On the other hand we may also write

$$\begin{aligned}&\int _{0}^{x_i} \frac{d \alpha }{\alpha }\Big \{ f(x_i - \alpha , x_{i+1} + \alpha ) - f(x_i,x_{i+1})\Big \} \nonumber \\&\quad =\int _{0}^{x_i} \frac{d \alpha }{\alpha } \sum _{a=1}^{\infty } \frac{1}{a!} \sum _{b=0}^a {a \atopwithdelims ()b} \Big (\frac{\partial }{\partial x_i}\Big )^b \Big ( \frac{\partial }{\partial x_{i+1}} \Big )^{a-b} f(x_i, x_{i+1}) (-\alpha )^b (\alpha )^{a-b} \nonumber \\&\quad =\int _{0}^{x_i} \frac{d \alpha }{\alpha } \sum _{a=1}^{\infty } \frac{1}{a!} \sum _{b=0}^a {a \atopwithdelims ()b} \Big (\frac{\partial }{\partial x_i}\Big )^b \Big ( \frac{\partial }{\partial x_{i+1}} \Big )^{a-b} f(x_i, x_{i+1}) (-1)^b \alpha ^{a}. \end{aligned}$$

(A.5)

The right hand sides of (A.3) and (A.5) do coincide, thus concluding the proof.

1.2 Derivation (2.16)

The action of the bulk generator $-\mathcal {H}^t_{1}$ on a function of the rescaled variables $x_1^{(M)}$ reads

$$\begin{aligned} -\mathcal {H}^t_{1} f\big (x_1^{(M)}\big )= & {} \sum _{k=1}^{x_1 M} \; \frac{1}{k} \Big [f\Big (x_1^{(M)} - \frac{k}{M}\Big ) - f\Big (x_1^{(M)}\Big ) \Big ] \nonumber \\&+ \sum _{k=1}^{\infty } \; \frac{[\beta _1^{(M)}]^k}{k} \Big [f\Big (x_1^{(M)} + \frac{k}{M}\Big ) - f\Big (x_1^{(M)}\Big ) \Big ] . \end{aligned}$$

(A.6)

By Taylor expansion, we may write

$$\begin{aligned} -\mathcal {H}^t_{1} f\big (x_1^{(M)}\big )= & {} \sum _{k=1}^{x_1 M} \; \frac{1}{k} \sum _{n=1}^\infty \frac{1}{n!} f^{(n)}\Big (x_1^{(M)}\Big ) \Big (-\frac{k}{M}\Big )^n \nonumber \\&+ \sum _{k=1}^{\infty } \; \frac{[\beta _1^{(M)}]^k}{k} \sum _{n=1}^\infty \frac{1}{n!} f^{(n)}\Big (x_1^{(M)}\Big ) \Big (\frac{k}{M}\Big )^n . \end{aligned}$$

(A.7)

This yields in the limit

$$\begin{aligned} - \lim _{M\rightarrow \infty } \mathcal {H}^t_{1} f\big (x_1^{(M)}\big )= & {} \sum _{n=1}^\infty \frac{1}{n!} f^{(n)}(x_1) (-1)^n \int _{0}^{x_1} \alpha ^{n-1} d\alpha \nonumber \\&+ \sum _{n=1}^\infty \frac{1}{n!} f^{(n)}(x_1) \int _{0}^{\infty } \alpha ^{n-1} e^{-\lambda _1 \alpha } d\alpha . \end{aligned}$$

(A.8)

where, besides (A.4), it has been used the convergence

$$\begin{aligned} \sum _{k=1}^{\infty }\frac{[\beta _1^{(M)}]^k}{k} \Big (\frac{k}{M} \Big )^n = \sum _{k=1}^{\infty }\frac{[1-\frac{\lambda _1}{M}]^k}{k} \Big (\frac{k}{M} \Big )^n \longrightarrow \int _{0}^{\infty } \alpha ^{n-1} e^{-\lambda _1 \alpha } d\alpha , \qquad \text {as} \quad M\rightarrow \infty . \end{aligned}$$

(A.9)

Clearly, the right hand sides of (A.8) coincides with the right hand side of (2.16).

Infinite Sums

1.1 Derivation (3.39)

It is straightforward to see that (3.39) is true for $k\le l$. In the case $k>l$ the sum reduces to

$$\begin{aligned} \sum _{m=l}^k \left( {\begin{array}{c}k\\ m\end{array}}\right) \left( {\begin{array}{c}m\\ l\end{array}}\right) \beta ^{k-m}(-\beta )^{m-l}\psi (m+1)=\sum _{m=l}^k \sum _{r=1}^m\left( {\begin{array}{c}k\\ m\end{array}}\right) \left( {\begin{array}{c}m\\ l\end{array}}\right) \frac{(-1)^{m-l}\beta ^{k-l}}{m+1-r} =-\frac{\beta ^{k-l}}{k-l} \end{aligned}$$

(B.1)

where we first used

$$\begin{aligned} \psi (m+1)=\psi (1)+\sum _{r=1}^m\frac{1}{m+1-r} \end{aligned}$$

(B.2)

and exchanged the sums in the second step.

1.2 Derivation (3.41)

After noting that the first sum truncates and shifting the second sum we rewrite (3.41) as

$$\begin{aligned} \begin{aligned} \sum _{m_2=1}^\infty \frac{(-1)^{l+m_2} }{m_2}\alpha ^{m_2}\beta ^{k-l+m_2}&\left( {\begin{array}{c}m_2\\ l\end{array}}\right) \, _2F_1(-k,m_2+1;-l+m_2+1;1) \\&= \sum _{m_2=1}^\infty (-1)^{l+m_2}\alpha ^{m_2}\beta ^{k-l+m_2} \frac{\Gamma (m_2)\Gamma (k-l)}{\Gamma (l+1)\Gamma (-l) \Gamma (k-l+m_2+1)}\\ \end{aligned} \end{aligned}$$

(B.3)

The case $k>l$ in (3.41) is straightforward, the denominator is finite while the numerator diverges. For $k\le l$ one has to be more careful. One gets

$$\begin{aligned} \begin{aligned} \sum _{m_2=1}^\infty (-1)^{l+m_2}\alpha ^{m_2}\beta ^{k-l+m_2}&\frac{\Gamma (m_2)\Gamma (k-l)}{\Gamma (l+1)\Gamma (-l) \Gamma (k-l+m_2+1)}\\&=\frac{(-1)^{k+l}}{(l-k)!}\sum _{m_2=1}^\infty (-1)^{m_2}\alpha ^{m_2}\beta ^{k-l+m_2} \frac{\Gamma (m_2)}{\Gamma (k-l+m_2+1)}\\ \end{aligned} \end{aligned}$$

(B.4)

To show the remaining relations we substitute

$$\begin{aligned} \alpha =\frac{1}{1-\beta }=\sum _{p=0}^\infty \beta ^p \end{aligned}$$

(B.5)

Using the binomial series

$$\begin{aligned} \alpha ^{m_2}= \frac{1}{(1-\beta )^{m_2}}=\sum _{n=0}^\infty \frac{\Gamma (m_2+n)}{\Gamma (m_2) \Gamma (1+n)}\beta ^n \end{aligned}$$

(B.6)

we get for $k=l$

$$\begin{aligned} \begin{aligned} \sum _{m_2=1}^\infty \frac{(-1)^{m_2} \alpha ^{m_2} \beta ^{m_2}}{m_2}&=\sum _{n=0}^\infty \sum _{m_2=1}^\infty \frac{(-1)^{m_2} \beta ^{n+m_2}(m_2+n-1)!}{n!m_2!}\\&=\sum _{q=1}^\infty \sum _{m_2=1}^q\frac{(-1)^{m_2} \beta ^{q}(q-1)!}{(q-m_2)!m_2!}=-\sum _{q=1}^\infty \frac{\beta ^{q}}{q}\\ \end{aligned} \end{aligned}$$

(B.7)

For $k<l$ we get

$$\begin{aligned} \begin{aligned}&\frac{(-1)^{k+l}\beta ^{k-l}}{(l-k)!}\sum _{m_2=l-k}^\infty \sum _{n=0}^\infty \frac{(-1)^{m_2} \beta ^{n+m_2}(m_2+n-1)!}{n!(k-l+m_2)!}\\&\quad = \frac{(-1)^{k+l}\beta ^{k-l}}{(l-k)!}\sum _{m_2=l-k}^\infty \sum _{q=m_2}^\infty \frac{(-1)^{m_2} \beta ^{q}(q-1)!}{(q-m_2)!(k-l+m_2)!}\\&\quad = \frac{(-1)^{k+l}\beta ^{k-l}}{(l-k)!}\sum _{q=l-k}^\infty \sum _{m_2=l-k}^q \frac{(-1)^{m_2} \beta ^{q}(q-1)!}{(q-m_2)!(k-l+m_2)!} \end{aligned} \end{aligned}$$

(B.8)

Finally noting that

$$\begin{aligned} \sum _{m=0}^p\frac{(-1)^{m}}{(p-m)!m!}=\delta _{p,0} \end{aligned}$$

(B.9)

we obtain the final result.