Action of gradient exponential operator

In summary, the homework statement is that there is not a straightforward formula for the action of an operator, and it requires factorization of the function to be derived.
  • #1
duc
9
0

Homework Statement



Find the action of the operator ## e^{\vec{a} . \vec{\nabla}} \big( f(\theta,\phi) . g(r) \big) ## where [itex]\nabla[/itex] 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
 
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  • #2
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 }}##
 
  • #3
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...
 
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  • #4
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.
 
  • #5
mfb said:
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.
 
  • #6
duc said:
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.
 
  • #7
mfb said:
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) ##
 
  • #8
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.
 

1. What is the gradient exponential operator?

The gradient exponential operator is a mathematical tool used in vector calculus and differential equations to describe the change of a scalar or vector quantity in multiple dimensions. It is represented by the symbol ∇ex and is also known as the exponential gradient operator or the exponential gradient.

2. What is the purpose of the gradient exponential operator?

The gradient exponential operator is used to describe the rate of change of a quantity in a given direction. It is particularly useful in solving differential equations and understanding the behavior of physical systems.

3. How is the gradient exponential operator related to the gradient operator?

The gradient exponential operator is closely related to the gradient operator (represented by the symbol ∇). While the gradient operator gives the direction and magnitude of the steepest increase of a scalar or vector quantity, the gradient exponential operator gives the direction and rate of change of the quantity in a given direction.

4. What is the formula for the action of gradient exponential operator?

The action of the gradient exponential operator on a function f(x) in multiple dimensions is given by the formula ∇ex f(x) = (∂/∂x1 + ∂/∂x2 + ∂/∂x3 + ...) f(x).

5. How is the gradient exponential operator used in physics?

In physics, the gradient exponential operator is used to describe the change of physical quantities such as temperature, pressure, and velocity in multiple dimensions. It is used in various fields including fluid dynamics, electromagnetism, and quantum mechanics to model and analyze physical systems.

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