- #1

- 38

- 3

##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}##)?

You should upgrade or use an alternative browser.

- Thread starter Clear Mind
- Start date

- #1

- 38

- 3

##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}##)?

- #2

- 737

- 210

But in QFT charge conservation comes from a U(1) symmetry of the field.

- #3

Science Advisor

- 1,793

- 1,192

Unitary operator on Hilbert space, see next post.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##?

Well you mentioned Noether theorem! Does Noether theorem applies to discrete symmetry?Is ##U## a continuous or discrete transformation?

See the next post.Is ##U## an observable (##U=U^{\dagger}##)?

- #4

Science Advisor

- 1,793

- 1,192

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

But in QFT charge conservation comes from a U(1) symmetry of the field.

Sam

- #5

- 38

- 3

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 apologiesDoes Noether theorem applies to discrete symmetry?

- #6

- 737

- 210

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

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.

- #7

Science Advisor

- 1,793

- 1,192

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 [itex]U(1)[/itex], [itex]\Psi (x)[/itex] (

- #8

- 737

- 210

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

- #9

Science Advisor

- 1,793

- 1,192

Right, you still need to tell us what exactly you meant byI think it's fairly obvious that by U(1) invariance I meant invariance of the lagrangian.

Share:

- Replies
- 3

- Views
- 505

- Replies
- 22

- Views
- 473

- Replies
- 13

- Views
- 1K

- Replies
- 12

- Views
- 764

- Replies
- 3

- Views
- 822

- Replies
- 7

- Views
- 605

- Replies
- 3

- Views
- 528

- Replies
- 3

- Views
- 550

- Replies
- 5

- Views
- 506

- Replies
- 16

- Views
- 814