Given a semisimple complex Lie algebra $\frak{g}$ of rank $r$, with Chevally generators $E_i,F_i,K_i$. Let $V$ be a finite dimensional representation of $\mathfrak{g}$ such that each weight space of $V$ is $1$-dimensional. Let $(i_1,\dots,i_k)$ be an ordered set of elements of $\{1,\dots,r\}$ (allowing repeats), and let $\{j_1,\dots,j_k\}$ be some permutation of $\{1,\dots,r\}$. For $v$ a highest weight of $V$, the elements $$ F_{i_1} F_{i_2} \cdots F_{i_k}(v),\quad \text { and } \quad F_{j_1} F_{j_2} \cdots F_{j_k}(v), $$ must have the same weight. Thus by our assumption they must differ by a scalar multiple. Will this scalar multiple always be an element of $\mathbb{Q}$?

Let me flesh out the answer given in the comments. Any simple Lie algebra can be defined over the field of rational numbers. Then its irreducible finite dimensional representations can be constructed as quotients of Verma modules. That is $$ V_\lambda = \mathfrak{U}(\mathfrak{g}_\mathbb{Q}) \otimes_{ \mathfrak{U}(\mathfrak{b}_\mathbb{Q})} \mathbb{Q}_\lambda / \text{maximal submodule}, $$ where $\mathbb{Q}_\lambda$ is the one dimensional representation on which the split Cartan subalgebra of $\mathfrak{g}_\mathbb{Q}$ acts by character $\lambda.$

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