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Are Noether charges a rep of the generators on the Hilbert space

by a2009
Tags: charges, hilbert, noether, symmetry
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Feb11-11, 05:04 AM
P: 25
I'm trying to understand the relationship between conserved charges and how operators transform. I know that we can find conserved charges from Noether's theorem. If (for internal symmetries) I call them [tex] Q^a = \int d^3x \frac{\partial L}{\partial \partial_0 \phi_i} \Delta \phi_i^a [/tex] then is it always the case that operators transform like

[tex] \hat O \rightarrow e^{i t_a Q^a} \hat O e^{-i t_a Q^a} [/tex]

i.e. are the conserved charges the rep of the generators on the Hilbert space?

Thanks for any help!
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Feb11-11, 03:22 PM
Sci Advisor
P: 882
Yes, they do generate the correct transformation on the fields AND satisfy the Lie algebra of the symmetry group. More importantly, they ( in the internal case) DONíT need to be CONSERVED to do the job.


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