I'm having trouble understanding how the canonical commutator works on differential operators: in this article, page 2, they define the multiplication-by-$x$ operator, that is a linear operator acting on some space of functions of the variable $x$, that simply multiplies any function by $x$. Then they extend it into the following time dependent operator: $$\hat{x}(\tau)=e^{H\tau}xe^{-H\tau}$$ with $H=(a-bx)\partial_x+\frac{\sigma^2}{2}x\partial^2_x$. Then they claim that in order to get the equation of motion (the derivative of $\hat{x}(\tau)$) they use the canonical commutator: $$[\partial_x,x]=\partial_xx-x\partial_x=1$$ so that $$[H,x]=a-bx+\sigma^2x\partial_x$$ and the equation of motion becomes $$\frac{d}{d\tau}\hat{x}(\tau)=a-b\hat{x}(\tau)+\sigma^2\hat{x}(\tau)\hat{p}(\tau) \tag{1}$$ where $\hat{p}(\tau)=e^{H\tau}\partial_xe^{-H\tau}$.
I tryed to do the explicit calculation: $[H,x]=((a-bx)\partial_x+\frac{\sigma^2}{2}x\partial^2_x)x-x((a-bx)\partial_x+\frac{\sigma^2}{2}x\partial^2_x)=$ $(a-bx)\partial_xx-x(a-bx)\partial_x+\frac{\sigma^2}{2}x\partial^2_xx-x\frac{\sigma^2}{2}x\partial^2_x=(a-bx)(\partial_xx-x\partial_x)+\frac{\sigma^2}{2}x\partial_x(\partial_xx-x\partial_x)=$ $a-bx+\frac{\sigma^2}{2}x\partial_x$ but I have a $\sigma^2/2$ here and not just $\sigma^2$ as they claim. Also I don't understand then how to get to the equation of motion since the right hand side seems to be just their $[H,x]$ multiplied by $e^{H\tau}$ on the left and $e^{-H\tau}$ on the right, but why should $e^{H\tau}[H,x]e^{-H\tau}$ equal $\frac{d}{d\tau}\hat{x}(\tau)$?
