Meaning of this operator

  • Thread starter Karlisbad
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From the "Lie Group" theory point of view we know that:

[tex] p [/tex] := is the generator for traslation (if the Lagrangian is invariant under traslation then p is conserved)

[tex] L [/tex]:= s the generator for rotation (if the Lagrangian is invariant under traslation then L is conserved)

(I'm referring to momentum p and Angular momentum L, although the notation is obvious :blushing: )

My question is if we take the "Lie derivative" and "covariant derivative" as a generalization of derivative for curved spaces.. if we suppose they're Lie operators..what's their meaning?..if the momentum operator acts like this:

[tex] pf(x)\rightarrow \frac{df}{dx} [/tex] derivative of the function..could the same holds for Lie and covariant derivative (covariant derivative is just a generalization to gradient, and i think that Lie derivatives can be expressed in some cases as Covariant derivatives, in QM the momentum vector applied over the wave function is just the gradient of the [tex] \psi [/tex]
 

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You confuse several levels here. What you call Lie operator is a left (or right) invariant vector field, an element of a Lie algebra. The example you gave for ##p## is just a possible representation, better realization of a Lie algebra. If we come from a group of smooth functions we will get a natural operation of the Lie algebra elements as Lie derivatives on these functions. Your example looks like the Poincaré group (algebra). For a general context of Lie derivatives see:
https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/ and following parts

And here is an example of a realization of ##\mathfrak{sl}(2) \cong \mathfrak{su}(2)## as differential operators on ##\mathcal{C}^\infty(\mathbb{R})## (sec. 6.2 and 7.3):
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
 

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