Suppose you have two charged particles that interact by e.m. potential ##V(\vec{r_1},\vec{r_2})##, the total charge is conserved. Since there's a conserved quantity, it must exist a transformation for which the hamiltonian is invariant (Noether theorem). Let's be the operator ##U## ##(U^{\dagger}=U^{-1})## the generator of the aforementioned trasformation, you have that:
##H'=U^{\dagger}HU##
and
##H'=H##
so that
##[H,U]=0##

Now the questions are:
What is ##U##?
Is ##U## a continuous or discrete transformation?
Is ##U## an observable (##U=U^{\dagger}##)?

Don't know if such a thing exists in NRQM.
But in QFT charge conservation comes from a U(1) symmetry of the field.

samalkhaiat
Suppose you have two charged particles that interact by e.m. potential ##V(\vec{r_1},\vec{r_2})##, the total charge is conserved. Since there's a conserved quantity, it must exist a transformation for which the hamiltonian is invariant (Noether theorem). Let's be the operator ##U## ##(U^{\dagger}=U^{-1})## the generator of the aforementioned trasformation, you have that:
##H'=U^{\dagger}HU##
and
##H'=H##
so that
##[H,U]=0##

Now the questions are:
What is ##U##?
Unitary operator on Hilbert space, see next post.
Is ##U## a continuous or discrete transformation?
Well you mentioned Noether theorem! Does Noether theorem applies to discrete symmetry?
Is ##U## an observable (##U=U^{\dagger}##)?
See the next post.

samalkhaiat
Don't know if such a thing exists in NRQM.
But in QFT charge conservation comes from a U(1) symmetry of the field.
You can treat QM as non-relativistic field theory and apply Noether theorem. Indeed, the Schrodinger equations for $\Psi$ and $\Psi^{ \dagger }$ are the Euler-Lagrange equations of the following Lagrangian $$\mathcal{ L } = \frac{ \hbar }{ i } ( \Psi^{ \dagger } \partial_{ t } \Psi - \Psi \partial_{ t } \Psi^{ \dagger } ) + \frac{ \hbar^{ 2 } }{ 2 m } \nabla \Psi^{ \dagger } \cdot \nabla \Psi + V ( r ) \Psi^{ \dagger } \Psi .$$ This Lagrangian is invariant under $U(1)$ phase transformation, i.e. $$\delta \mathcal{ L } = 0 , \ \ \mbox{ when } , \ \ \delta \Psi = i \alpha \Psi , \ \delta \Psi^{ \dagger } = - i \alpha \Psi^{ \dagger } .$$ Now do the algebra and you will find (when the fields satisfy the E-L equation) the following continuity equation $$\partial_{ t } \left( \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ t } \Psi ) } \delta \Psi + C.C \right) = \nabla \cdot \left( \frac{ \partial \mathcal{ L } }{ \partial ( \nabla \Psi ) } \delta \Psi + C.C \right) .$$ This is nothing but the familiar QM equation $$\partial_{ t } ( \Psi^{ \dagger } \Psi ) = - \frac{ \hbar }{ 2 m i } \nabla \cdot ( \Psi^{ \dagger } \nabla \Psi - \Psi \nabla \Psi^{ \dagger } ) .$$ Integrating this over the whole volume of the field and using the divergence theorem, we find the following conserved charge (operator) $$\frac{ d Q }{ d t} = \frac{ d }{ d t } \int d^{ 3 } x \ \Psi^{ \dagger } ( x ) \ \Psi ( x ) = 0 .$$ You can show that $[ Q , H ] = 0$ and that the unitary operator $U( \alpha ) = \exp ( i \alpha Q )$ generates the correct transformations on the fields through $$\delta \Psi ( x ) = [ i \alpha Q , \Psi ( x ) ] .$$

Sam

Thank you very much for the help!

Does Noether theorem applies to discrete symmetry?
Professor told me that Noether theorem does not apply to discrete transformation (like parity), he told this 100 times, but i continue to forget every time! My apologies

You can treat QM as non-relativistic field theory and apply Noether theorem. Indeed, the Schrodinger equations for $\Psi$ and $\Psi^{ \dagger }$ are the Euler-Lagrange equations of the following Lagrangian $$\mathcal{ L } = \frac{ \hbar }{ i } ( \Psi^{ \dagger } \partial_{ t } \Psi - \Psi \partial_{ t } \Psi^{ \dagger } ) + \frac{ \hbar^{ 2 } }{ 2 m } \nabla \Psi^{ \dagger } \cdot \nabla \Psi + V ( r ) \Psi^{ \dagger } \Psi .$$ This Lagrangian is invariant under $U(1)$ phase transformation, i.e. $$\delta \mathcal{ L } = 0 , \ \ \mbox{ when } , \ \ \delta \Psi = i \alpha \Psi , \ \delta \Psi^{ \dagger } = - i \alpha \Psi^{ \dagger } .$$ Now do the algebra and you will find (when the fields satisfy the E-L equation) the following continuity equation $$\partial_{ t } \left( \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ t } \Psi ) } \delta \Psi + C.C \right) = \nabla \cdot \left( \frac{ \partial \mathcal{ L } }{ \partial ( \nabla \Psi ) } \delta \Psi + C.C \right) .$$ This is nothing but the familiar QM equation $$\partial_{ t } ( \Psi^{ \dagger } \Psi ) = - \frac{ \hbar }{ 2 m i } \nabla \cdot ( \Psi^{ \dagger } \nabla \Psi - \Psi \nabla \Psi^{ \dagger } ) .$$ Integrating this over the whole volume of the field and using the divergence theorem, we find the following conserved charge (operator) $$\frac{ d Q }{ d t} = \frac{ d }{ d t } \int d^{ 3 } x \ \Psi^{ \dagger } ( x ) \ \Psi ( x ) = 0 .$$ You can show that $[ Q , H ] = 0$ and that the unitary operator $U( \alpha ) = \exp ( i \alpha Q )$ generates the correct transformations on the fields through $$\delta \Psi ( x ) = [ i \alpha Q , \Psi ( x ) ] .$$

Sam
That's still a whole different affair though. I mean U(1) invariance of the position state representation is not at all the same as U(1) invariance of the dynamical variable itself.

samalkhaiat
That's still a whole different affair though. I mean U(1) invariance of the position state representation is not at all the same as U(1) invariance of the dynamical variable itself.
First of all, the “position state representation” i.e. the wavefunction IS NOT invariant under U(1)-transformation. Second, under $U(1)$, $\Psi (x)$ (whether it is regarded as a wavefunction or dynamical field variable) transforms according to $$\Psi ( x ) \to e^{ i \alpha} \Psi ( x ) .$$ Infinitesimally, we write this as $$\delta \Psi ( x ) = i \alpha \Psi ( x ) .$$

I think it's fairly obvious that by U(1) invariance I meant invariance of the lagrangian.

samalkhaiat