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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?

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Stephen Montgomery-Smith
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1 Answers1

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I) As a purist, I disapprove of the common praxis to call the implication $$ \tag{1} \{Q,H\}+\frac{\partial Q}{\partial t}~=~0 \qquad\Rightarrow\qquad \frac{dQ}{dt}~\approx~0.$$ for a 'Hamiltonian version of Noether's theorem', cf. my Phys.SE answer here.

Moreover, the implication (1) is not equivalent to the full Noether's theorem for various reasons. Firstly, the issue of possible singular Legendre transformation may render a comparison of Lagrangian and Hamiltonian formulations difficult. Secondly, Noether's theorem also works for so-called horizontal variations of the form $\delta t=\ldots$ not accounted for in (1).

Rather, the implication (1) is just a trivial consequence of Hamilton's equations. See also e.g. nLab.

II) Concerning OP's example of the free Schrödinger field, the Hamiltonian formulation is e.g. discussed in this Phys.SE post. OP's suggestion to split in real and imaginary parts $$\tag{1} \phi~=~(\phi^1+i\phi^2)/\sqrt{2} $$ is strictly speaking not necessary, cf. e.g. this Phys.SE post, but the logic is somewhat simpler if one does split. The non-zero fundamental Poisson bracket reads

$$\tag{2} \{\phi^1({\bf x},t),\phi^2({\bf y},t)\}~=~\delta^3 ({\bf x}-{\bf y}).$$

The infinitesimal global(=${\bf x}$-independent) phase symmetry

$$\tag{3} \delta \phi~=~-i\epsilon \phi, $$

where $\epsilon$ is an infinitesimal parameter, reads in components

$$\tag{4} \delta \phi^1~=~\epsilon \phi^2\qquad\text{and}\qquad \delta \phi^2~=~-\epsilon \phi^1.$$

The zero-component of the Noether 4-current is then

$$\tag{5} j^0~:=~\phi^2\frac{\partial {\cal L}}{\partial \dot{\phi}^1} -\phi^1\frac{\partial {\cal L}}{\partial \dot{\phi}^2}~=~|\phi|^2. $$

Hence the Noether charge is

$$\tag{6} Q(t)~:=~ \int d^3x~ j^0({\bf x},t)~=~ \int d^3x|\phi({\bf x},t)|^2.$$

It is straightforward to see that this Noether charge (6) generates the infinitesimal global phase symmetry (4)

$$\tag{7} \delta \phi^1~=~\epsilon \{\phi^1,Q\} \qquad\text{and}\qquad \delta \phi^2~=~\epsilon\{\phi^2,Q\} ;$$

that it Poisson commutes with the Hamiltonian density

$$ \tag{8} {\cal H}~:=~\frac{1}{2m}|\nabla\phi|^2 ~=~\frac{1}{4m}(\nabla\phi^1)^2 +\frac{1}{4m}(\nabla\phi^2)^2;$$

and that it is conserved on-shell.

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