RUber said:
How do you define the "action" of an operator?
You can quickly break up the operator as:
##e^{\vec a \cdot \vec \nabla } ( f(\theta, \phi) g(r) ) = e^{\vec a \cdot [g' , \, \frac{\partial f}{r \partial \theta } , \, \frac{\partial f}{r\sin \theta \partial \phi }] }##
##= e^{a_r g' + a_\theta \frac{\partial f}{r \partial \theta } +a_\phi \frac{\partial f}{r\sin \theta \partial \phi }}##
Thanks for your reply, unfortunately, it is not that simple. Remembering that when you decompose the gradient operator into ## \overrightarrow{\nabla} = \underbrace{\frac{\partial}{\partial r}}_{\hat{A}} + \underbrace{\frac{1}{r}\frac{\partial}{\partial \theta}}_{\hat{B}} + \underbrace{\frac{1}{r \sin\theta}\frac{\partial}{\partial \phi}}_{\hat{C}} ## and then take exponential, two remarks:
a. ## e^{\vec a \cdot \vec \nabla } \ne e^{\hat{A}}.e^{\hat{B}}.e^{\hat{C}} \ne e^{\hat{B}}.e^{\hat{A}}.e^{\hat{C}} \ne etc..## since the definition of the exponential of an operator is: ## e^{A} = \mathbb{1} + \hat{A} + \frac{\hat{A}^2}{2!} + \dots##.
b. Hence the successive applications of ## e^{\hat{A}}, e^{\hat{B}}, e^{\hat{C}}## onto the function (as in your case) are not allowed since it obviously yields different results. One has to bring the whole gradient operator ## \overrightarrow{\nabla}## into the definition above. And this will be very complicated.
Furthermore, if we want to recover your expression, there is a condition: ## \hat{A}^n(f.g) = \big(\hat{A}(f.g)\big)^n ##, which is not satisfied with the gradient in general.
An interesting thing is that if we decompose the gradient into cartesian coordinates, i.e ## \overrightarrow{\nabla} = \vec{i}\frac{\partial}{\partial x} + \vec{j}\frac{\partial}{\partial y} + \vec{k}\frac{\partial}{\partial z} ##, as the differential over x, y, z commute, we can perform the successive applications. In spherical coordinates, this property does not hold...
