- #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}##)?

- 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

- 692

- 142

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

- #3

samalkhaiat

Science Advisor

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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

samalkhaiat

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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

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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.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

- #7

samalkhaiat

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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] (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.

- #8

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

- #9

samalkhaiat

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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.