Looking around I see one version of Noether's Theorem that creates conserved quantities from symmetries that preserve the Lagrangian (e.g. http://math.ucr.edu/home/baez/noether.html), and another theorem also called Noether's Theorem that finds conserved quantities by seeing if they Poisson commute with the Hamiltonian (e.g. page 29 of http://www.math.ucla.edu/~tao/preprints/chapter.pdf).

Are these two results actually the same results disguised?

For example, Terry Tao in http://www.math.ucla.edu/~tao/preprints/chapter.pdf on page 83 derives the total charge $\int |u|^2 \, dx$ as an invariant of the free Schroedinger equation, He says this is a consequence of the map $u \mapsto e^{i\phi} u$. I tried writing the Schroedinger equation in Lagrangian form, by deciding that the real part of $u$ would be "position", and the imaginary part of $u$ would be "momentum." but then the map $u \mapsto e^{i\phi} u$ ended up being a map that didn't map position to a new position, but rather mixed position and velocity in a rather unpleasant manner. Would there be some other smart choice of deciding which variables are "position" and "momentum" that would cause the total charge to arise as the conserved quantity given by http://math.ucr.edu/home/baez/noether.html?

  • 1
    $\begingroup$ Also asked here: physics.stackexchange.com/questions/210886/… $\endgroup$ – Stephen Montgomery-Smith Oct 6 '15 at 0:55
  • $\begingroup$ Yes they are the same. When bracket of a quantity is zero with Hamiltonian, it means that quantity is an integral of motion (doesn't change, is invariant). This quantity is a conserved charge with respect to some symmetry (I don't remember how we construct this symmetry exactly but I know it can be done). Conversely obviously if you have a conserved charge (coming from Lagrangian consideration) clear it is an integral of motion (since it is conserved). So its Poisson bracket with Hamiltonian will be zero. $\endgroup$ – Hamed Oct 6 '15 at 1:01
  • $\begingroup$ Why would real part be position and imaginary time momentum? $\endgroup$ – Hamed Oct 6 '15 at 1:04
  • $\begingroup$ @hamedp To your first comment, the symmetry would presumably be the flow generated by treating the conserved quantity as a Hamiltonian. But how would you identify which variables should be used as momentum and which should be chosen as position in order to make the Lagrangian form of Noether's Theorem apply? $\endgroup$ – Stephen Montgomery-Smith Oct 6 '15 at 1:07
  • $\begingroup$ @hamedp To your second comment - it was an arbitrary choice, and all I could think of. $\endgroup$ – Stephen Montgomery-Smith Oct 6 '15 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.