Weyl algebra confusion I am following http://planetmath.org/weylalgebra
I don't understand why the the more concrete definition satisfies $\partial_i x_j - x_j 
\partial_i = \delta_{ij}$.
Indeed, if $i=j$ then we would have $$\partial_i x_i - x_i \partial_i = 1 - x_i \partial_i = 1$$ and hence $$ -x_i \partial_i = 0$$ which is nonsense, no?
 A: You're confusing things, since $x_i$ is not a variable, it is an operator. In particular, $$\partial_i(x_i p(x_1,\ldots,x_n))=p(x_1,\ldots,x_n)+x_i (\partial_i p)(x_1,\ldots,x_n)$$  is not $p(x_1,\ldots,x_n)$ unless $\partial_i p=0$. 
A: The $\partial_i$ doesn't act on $x_i$ as a derivative within the algebra, so $\partial_ix_i \not= 1$ is the problem. Instead you are supposed to view the $x_i$ as an operator as well.
The confusion is that $x_i$ is an element of the algebra, but it is also a function which the algebra acts on. I will then denote $X_i$ for the function (not an element of the Weyl algebra).
For example we could apply the operator $x_i$ (which means multiplication by $X_i$) to the function $X_i$ to get $x_i(X_i) = X_i^2$. The operator $\partial_i$ then says to derive $X_i^2$ to get $\partial_i(X_i^2) = 2X_i$. So, $\partial_ix_i(X_i) = 2X_i$ which is not the same as the identity operator $1X_i = X_i$.
If we also act by $x_i\partial_i$ on $X_i$, we first derive to get $\partial_iX_i = 1$, then act by $x_i$ to get $x_i(1) = X_i$. Since $2X_i - X_i = X_i$, we see that $(\partial_ix_i - x_i\partial_i)(X_i) = X_i$, i.e. it acts as identity and $\partial_ix_i - x_i\partial_i = 1$.
Also, if $j\not= i$ then $x_j$ is a scalar relative to $\partial_i$, so of course the derivative commutes with multiplication and $\partial_ix_j - x_j\partial_i = 0$.
