Think of a Young tableau $R$ as collection of rows $y_1 \geq ... \geq y_d > y_{d+1}=0$ and all others zero, with $\ell(Y):= \sum_j y_j$ and for a box $s=(i,j)\in R$ we have $a_Y(s):=y_i-j$ and $l_Y(s):=y_j^D-i$, where $y^D_j$ corresponds to the transpose $R^D$ of $R$.

How do we prove that

$$ Z_k=\sum_{\ell_1+\ell_2=k} \prod_{i,j} \prod_{(s,s')\in R_i\times R_j} \frac{1}{1-\alpha_i \alpha_j^{-1} t_1^{-l_{R_j}(s)} t_2^{1+a_{R_i}(s)}} \frac{1}{1-\alpha_i \alpha_j^{-1} t_1^{1+l_{R_i}(s')} t_2^{-a_{R_j}(s')}} $$

(where the sum means sum over all Young tableaux $R_1R_2$ such that $\ell(R_1)+\ell(R_2)=k$)

can be expanded, in the limit $t_1t_2=1$, and after defining $t_i:=e^{i2\epsilon_i}$, $\alpha_i:=e^{i2a_i}$, $\epsilon:=\epsilon_1$ as

$$Z_k = \sum_{\ell_1+\ell_2=k} \prod_{(i,m)\neq(j,n)} \frac{\sinh (a_i-a_j +\epsilon(y_{i,n}-y_{j,m}+m-n))} {\sinh (a_i-a_j +\epsilon(m-n))}$$

where $y_{i,n}$ denotes the $n$-th row length of diargam $Y_i$

What I've been able to do: reduce it to proving the following.

Define $f_R(q)=\sum_{i=1}^d \sum_{k=1}^{\mu_i} q^{k-i}$, $\sum_kC_k(R_1,R_2)q^k=(q+q^{-1}-2)f_{R_1}(q)f_{R_2}(q)+f_{R_1}(q)+f_{R_2}(q)$, and $2a=a_1-a_2$.

Then I need to prove that

$$ \prod_k \frac1{(\sinh 2a +\epsilon k)^{C_k(R_1,R_2^t)}}= \prod_{s\in R_1}\prod_{s'\in R_2} \frac1{\sinh 2a -\epsilon(l_{R_2}(s)+1+a_{R_1}(s))} \frac1{\sinh 2a+\epsilon(1+l_{R_1}(s')-a_{R_2}(s'))} $$

Refs: