Let $k$ be a field of characteristic zero, let $k[x,y]$ be the polynomial ring in two variables $x$ and $y$ and let $A_1(k)$ be the first Weyl algebra, namely, the $k$-algebra generated by non-commuting variables $X$ and $Y$ subject to the relation $YX-XY=1$.
When one has an equation in $k[x,y]$ one sometimes substitutes $(x,y)$ by $(a,b) \in k^2$.
If I am not wrong, it is not possible to substitute $(X,Y)$ by $(a,b) \in k^2$, since in the non-commutative case the substitution mapping is not a well-defined $k$-algebra morphism, for example, if $f: A_1 \to k$, $f(X):=1,f(Y):=1$, then $f(YX-XY)=f(Y)f(X)-f(X)f(Y)=1\cdot 1-1\cdot 1=0 \neq 1=f(1)=f(YX-XY)$.
Now, in the commutative case, if one knows that for given $p,q \in k[x,y]$, $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x=1$, then one sometimes considers $\operatorname{Jac}(p,q)|(a,b)$, for some $(a,b) \in k^2$ (or one considers the Jacobian at a point $(x,b)$ or $(a,y)$).
Can we replace $(X,Y)$ by $(a,b) \in k^2$, when we are given $P,Q \in A_1(k)$ such that $[P,Q]:=PQ-QP=1$?
I guess that the answer is no, and is connected to the above discussion. But perhaps something else interesting can be said in the non-commutative case?
Thank you very much!