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Figuring symmetries of a differential operator from its eigenfunctions

  1. Oct 21, 2013 #1
    So, I understand that the derivative operator, [itex]D=\frac{d}{dx}[/itex] has translational invariance, that is: [itex]x \rightarrow x - x_0[/itex], and its eigenfunctions are [itex]e^{\lambda t}[/itex]. Analogously, the theta operator [itex]\theta=x\frac{d}{dx}[/itex] is invariant under scalings, that is [itex]x \rightarrow \alpha x[/itex], and its eigenfunctions are [itex]x^\lambda[/itex]. Taking logarithms and exponentials, I have constructed a sequence of operators and their respective eigenfunctions, all with the property that [itex]\{L(\frac{d}{dx})\}f^\lambda(x)=\lambda f^\lambda(x)[/itex]. I've taken a picture and attached it to this post.

    My guess is that associated with every single one of these operators is some symmetry, some sort of coordinate transformation [itex]x \rightarrow f(x)[/itex] under which the operator is invariant. For the [itex]x\log x \frac{d}{dx}[/itex] operator, its invariant under [itex]x \rightarrow x^k[/itex], by inspection. How can I figure out what sort of symmetry a given operator has, given its eigenfunctions?

    Physically, symmetries are associated with conservation laws. For a system whose differential equations are governed by this sort of differential operators, what sort of conserved quantities should I expect?
     

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  2. jcsd
  3. Oct 24, 2013 #2
    ##\frac{d}{dx}## is associated with translational symmetry. An infinitesimal translation is produced by acting on a function with the operator ##1 + \epsilon \frac{d}{dx}##, with ##\epsilon## infinitesimal.

    ##x \frac{d}{dx}## is associated with scale invariance. An infinitesimal rescaling is produced by acting on a function with the operator ##1 + \epsilon x \frac{d}{dx}##, with ##\epsilon## infinitesimal.

    Presumably your other operators ##O## can be associated with symmetry transformations with an infinitesimal transformation being implemented by ##1 + \epsilon O##?
     
  4. Oct 31, 2013 #3
    Ok, I agree with you. How can I figure out the "finite" version of the transformation from its infinitesimal counterpart?
     
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