I'm trying to prove the conserved quantities for $N$-vortices but having a difficult time doing so. I know it's something to do with Noether's theorem but not sure how to go about it. The Hamiltonian is $$ H = \frac{1}{4\pi}\sum_{\beta \neq \alpha}^N\Gamma_{\alpha}\Gamma_{\beta}logl_{\alpha\beta} $$ Where l is the distance between vortices. I need to use the fact that H is translational invariant and rotational invariant which I can show. The translational invariance gives

$$ X = \sum_{\alpha}^N\Gamma_{\alpha}z_{\alpha} $$

and the rotational invariance should give

$$ I = \sum_{\alpha}^N\Gamma_{\alpha}|z_{\alpha}|^2 $$

But again, I'm not sure how to show that X and I are conserved. Not necessarily looking a solution but just to point me in the right direction further than just 'use Noether's theorem'. Any help is appreciated!!


1 Answer 1


Assuming no explicit time dependence, the Heisenberg equation of motion relates the time derivative of an operator to its commutator with the Hamiltonian as follows: $$\frac{d}{dt}\mathcal{O} = \frac{i}{\hbar} [H,\mathcal{O}]$$ From this we see that, in such cases, an operator that commutes with the Hamiltonian corresponds to a conserved quantity. The objects $\Gamma$ in your system probably come with some (anti)commutation relations, which you could exploit to test if $X$ and $I$ are conserved.

  • $\begingroup$ Thanks for the reply! I have used this relation prior in a separate proof using the Poisson bracket I believe. So you are saying that with this relationship proposed, using invariance to translations and such should give the bracket = 0? $\endgroup$
    – D.Cecchi
    Apr 17, 2021 at 17:36
  • $\begingroup$ Yes, if you're dealing with a classical system, the corresponding result holds at the level of Poisson brackets. See, for example, here: en.wikipedia.org/wiki/Conserved_quantity#Hamiltonian_mechanics $\endgroup$ Apr 18, 2021 at 4:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.