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Action of gradient exponential operator

  1. Aug 24, 2015 #1

    duc

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    1. The problem statement, all variables and given/known data

    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.

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

    duc
     
  2. jcsd
  3. Aug 24, 2015 #2

    RUber

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    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 }}##
     
  4. Aug 24, 2015 #3

    duc

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    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...
     
    Last edited: Aug 24, 2015
  5. Aug 24, 2015 #4

    mfb

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    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.
     
  6. Aug 24, 2015 #5

    duc

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    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.
     
  7. Aug 25, 2015 #6

    mfb

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    That's what I meant - and you know the function is factorizable.
     
  8. Aug 26, 2015 #7

    duc

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    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) ##
     
  9. Aug 26, 2015 #8

    mfb

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    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.
     
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