The text I am reading claims that if there exists a symmetry generated by $Q$, and under this transformation an operator transforms by$$f\rightarrow f +\delta f,$$ then by Noether's theorem $$\delta f = i\epsilon[Q, f].$$

However, this is not the Noether's theorem I am used to, which I see to be similar to this statement, but only for time translations (with generator $H$). How does one go about showing this generalisation of Noether's theorem?


2 Answers 2


This seems not to be precisely the statement of Noether's theorem, but the consequence of the Liouville equation of motion: $$ \frac{df}{dt} = \frac{\partial f}{\partial t} + i \{Q, f\} $$ Where $\{,\}$ -denotes Poisson bracket. Usually, you will see Hamiltonian $H$ instead of $Q$. However, any charge can be treated in the same way as Hamiltionian. And for small $\epsilon$, and $f$ not depending explicitly on time, it gives: $$ \delta f = i \epsilon \{Q, f\} $$ As claimed


I think you might want to have a look at

Baez, J. C. (2020). Getting to the Bottom of Noether's Theorem. arXiv preprint arXiv:2006.14741.

where I believe he discusses a version of Noether's theorem phrased this way in Section 2.

On page 15 specifically, it discusses a version of Noether's theorem which is expressed as the Poisson bracket of two things being zero. You have not defined $\epsilon$ above, but I think perhaps it is part of the calculus of notation you are using (and as of yet I'm still ignorant of it), and might be equivalent to what's written in Baez's article.

In fact the difference between the two reminds me of something I read recently at this solution regarding thinking of derivations (the last line right before the Proof of [1] line), which might explain the transition.


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