Action of gradient exponential operator

[duc]

Homework Statement

Find the action of the operator $e^{\vec{a} . \vec{\nabla}} \big( f(\theta,\phi) . g(r) \big)$ where \nabla is the gradient operator given in spherical coordinates, f and g are respectively scalar functions of the angular part $( \theta, \phi)$ and the radial part $( r = |\vec{r}| )$. $\vec a$ is a constant vector.

Homework Equations

Is there any textbook which gives an explicit formula for such expression, if anyone knows, please tell me.
Thanks very much!

duc

 

[RUber]

How do you define the "action" of an operator?
You can quickly break up the operator as:
$e^{\vec a \cdot \vec \nabla } ( f(\theta, \phi) g(r) ) = e^{\vec a \cdot [g' , \, \frac{\partial f}{r \partial \theta } , \, \frac{\partial f}{r\sin \theta \partial \phi }] }$
$= e^{a_r g' + a_\theta \frac{\partial f}{r \partial \theta } +a_\phi \frac{\partial f}{r\sin \theta \partial \phi }}$

 

[duc]

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...

 

[mfb]

I'm not sure, but I would expect that at least the radial derivative commutes with the angular ones as your function can be factorized. This could help to get rid of some r-dependent things.

 

[duc]

I'm not sure, but I would expect that at least the radial derivative commutes with the angular ones as your function can be factorized. This could help to get rid of some r-dependent things.

The radial derivative $\frac{\partial}{\partial r}$ will commute with $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \phi}$ when the function to be derived is factorizable.

Ps: Sorry, I'm wrong about the commutations of $\frac{\partial}{\partial x_i}, (x_i = x,y,z)$. It in fact depends on the property of the function $f(x,y,z)$ for which one takes the derivative.

 

[mfb]

The radial derivative $\frac{\partial}{\partial r}$ will commute with $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \phi}$ when the function to be derived is factorizable.

That's what I meant - and you know the function is factorizable.

 

[duc]

That's what I meant - and you know the function is factorizable.

Yes i know. What matters is the factor in front of each derivative, e.g $\frac{1}{r \sin\theta}$. Hence, only if the derivatives $\frac{\partial}{\partial r}, \frac{\partial}{\partial \phi}$ involves, it follows that: $e^{\frac{\partial}{\partial r} + \frac{\partial}{\partial \phi}} = e^{\frac{\partial}{\partial \phi}} \cdot e^{\frac{\partial}{\partial r}} = e^{\frac{\partial}{\partial r}} \cdot e^{\frac{\partial}{\partial \phi}} \quad (1)$ for factorizable functions. It means that when the factor $\frac{1}{r\sin\theta}$ appears, one has to find the kernel of the commutator generated by the operators: $\overrightarrow{e}_r \, \frac{\partial}{\partial r}$ and $\overrightarrow{e}_{\phi} \, \frac{1}{r\sin\theta} \, \frac{\partial}{\partial \phi}$. The expression (1) is realized for functions belonging to kernel of the commutator. And once again the unit vector also depends on angles $(\theta,\phi)$

 

[mfb]

It allows to sort all expressions like $$(\alpha (\partial_r + \frac{1}{r} \partial_\phi))^2$$
$$=(\alpha \partial_r)^2 + \alpha \partial_r \alpha \frac{1}{r} \partial_\phi + \alpha \frac{1}{r} \partial_\phi \alpha \partial_r + (\alpha \frac{1}{r} \partial_\phi)^2 \\
= (\alpha \partial_r)^2 + \alpha (\partial_r \frac{1}{r} + \frac{1}{r} \partial_r) \alpha \partial_\phi + (\alpha \frac{1}{r} \partial_\phi)^2$$
Not sure if that helps.