Are Weyl algebra $A_1$ and it's opposite algebra isomorphic?

Question

Let $A$ be a noncommutative ring and $ab=c$ in $A$. $A'$ is it's opposite ring if $ba=c$ in $A'$. If $A$ is a Weyl algebra $A_1$, are $A$ and $A'$ isomorphic? I have an idea, but I think it's wrong. Let $p, q$ be generators of $A$. So in $A$ we have $pq=qp+1$. In $A': qp=qp+1 \implies 0=1$, so there can't be an isomorphism if $0 \neq 1$ in $A$. I think it's a bullshit, but no other ideas... I need your help! It's a task from my research seminar "Geometry and dynamics" for first-year undergraduates in HSE, Moscow.

Answer 1

The Weyl algebra $A$ is given by two generators $p$ and $q$ subject to the single relation $p q - q p = 1$. The opposite algebra $A^{\mathrm{op}}$ is therefore given by two elements $p'$ and $q'$ subject to the relation $q' p' - p' q' = 1$. Consequently, we have an isomorphism of algebras $A ≅ A^{\mathrm{op}}$ given on generators by $p \mapsto q'$ and $q \mapsto p'$.

Alternatively, let $$ be the three-dimensional Heisenberg Lie algebra, which given by the vector space basis $p, q, c$ subject to the commutator relation $[p, q] = c$ and $c$ central in $$. We have $A ≅ \mathrm{U}() / (c - 1)$ where $\mathrm{U}()$ is the universal enveloping algebra of $$. We have $ ≅ ^{\mathrm{op}}$ via $p \mapsto q$, $q \mapsto p$ and $c \mapsto c$. This isomorphism of Lie algebras induces isomorphisms of algebras $$ A ≅ \mathrm{U}() / (c - 1) ≅ \mathrm{U}(^{\mathrm{op}}) / (c - 1) ≅ \mathrm{U}()^{\mathrm{op}} / (c - 1) ≅ ( \mathrm{U}() / (c - 1) )^{\mathrm{op}} ≅ A^{\mathrm{op}} \,. $$ (This is the same isomorphism as in the first argumentation.)