Sign wrong in angular momentum (Quantum Mechanics)

Question

For small angles $\theta$ the rotation along a particular axis $n$ is given by

$R(n,\theta)(r)=Id+ \theta (n \times r)+ o(\epsilon)$. Now, the rotation operator in Quantum Mechanics is given by $R(n,\theta)(r)=r-\frac{i}{\hbar} \theta \langle n , L \rangle r + o(\epsilon)$

But if I check this for $n=e_z$ we have: $R(n,\theta)(r)=r+ \theta (e_z \times r)+ o(\epsilon)=r+ \theta (x e_y - y e_x)+ o(\epsilon)$ and $R(n,\theta)(r)=r-\frac{i}{\hbar} \theta l_z r + o(\epsilon)=r-\theta (x \partial_y - y \partial_x)r + o(\epsilon)=r-\theta (x e_y - y e_x) + o(\epsilon)$

so obviously the last two expressions of the last two rows differ in a sign and I do not see why.

Answer 1

Your expression are mathematically correct, but I think that the problem is that the "$\vec r$" you are using in the $2$ kind of equations is not the same, so you cannot compare.

The "$\vec r$" you use with $\vec n \wedge \vec r$, represents clearly coordinates, and coordinates change under an infinitesimal rotation.

In the second kind of equations, you are using differential operators, applying on "physical" functions of $\vec r$. For instance, we can take the example of the temperature $T(\vec r)$. This is a physical quantity depending on the physical space point. The change of coordinates $\vec r \to \vec {r'}$ does not change the physical space-point and its temperature, it corresponds only to a change of coordinates system, so the new temperature $T'(\vec {r'})$ must verify $T'(\vec {r'}) = T(\vec r)$ and the variation of the temperature, in function of the initial coordinates $\vec r$, goes like the inverse of the variation of the coordinates :

$\vec r \to \vec r' = \vec r + \delta \vec r, \quad \quad T(\vec r) \to T'(\vec r)=T(\vec r - \delta \vec r), \quad \quad T'(\vec {r'}) = T(\vec r)$

Now, you may consider $\vec r_f$, as a physical function of $\vec r$, with $ \vec r_f(\vec r) = \vec r$, in the initial system of coordinates , so we have the following variation for the function $\vec r_f$ :

$\vec r = \vec r_f(\vec r) \to \vec r'_f(\vec r) =\vec r_f(\vec r - \delta \vec r) = \vec r - \delta \vec r\quad \quad \vec {r'}_f(\vec {r'}) = r_f(\vec r)$

So, considering $\vec r$ as coordinates, or $\vec r$ as a function, give $2$ different kind of variations. As coordinates, you have $\vec r \to \vec r + \delta \vec r$, and as a physical function, you have $\vec r \to \vec r - \delta \vec r$