How Does the Pauli-Lujanski Tensor Relate to Gauge Invariance?

[thalisjg]

I want to proof that
[M_{[itex]\mu[/itex][itex]\nu[/itex]},W_{[itex]\sigma[/itex]}]=i(g_{[itex]\nu[/itex][itex]\sigma[/itex]}W_{[itex]\mu[/itex]}-g_{[itex]\mu[/itex][itex]\sigma[/itex]}W_{[itex]\nu[/itex]})
I can reduce this expression but I can't find the correctly answer.
Thanks!

 

[Bill_K]

You mean the Pauli-Lubanski vector. You can grind through a lot of algebra to show this, but actually there's nothing to prove. It's an identity!

M_μν is the 4-dimensional rotation operator, and consequently [M_μν, V_σ] = i(g_νσV_μ - g_μσV_ν) for any 4-vector V_μ.

 

[thalisjg]

Sorry about the mistakes.
Thanks!

 

[samalkhaiat]

I want to proof that
[M_{[itex]\mu[/itex][itex]\nu[/itex]},W_{[itex]\sigma[/itex]}]=i(g_{[itex]\nu[/itex][itex]\sigma[/itex]}W_{[itex]\mu[/itex]}-g_{[itex]\mu[/itex][itex]\sigma[/itex]}W_{[itex]\nu[/itex]})
I can reduce this expression but I can't find the correctly answer.
Thanks!

Sandwitch [itex]W_{ \mu }[/itex] between [itex]U = \exp ( - i \omega_{ \mu \nu } J^{ \mu \nu } / 2 )[/itex] and [itex]U^{ \dagger }[/itex]:
[tex]
U^{ \dagger } W_{ \mu } U = \frac{ 1 }{ 2 } \epsilon_{ \mu \nu \rho \sigma } U^{ \dagger } J^{ \nu \rho } U U^{ \dagger } P^{ \sigma } U .
[/tex]
Now, use the transformation rules
[tex]
U^{ \dagger } J^{ \nu \rho } U = \Lambda^{ \nu }{}_{ \lambda } \Lambda^{ \rho }{}_{ \eta } J^{ \lambda \eta } ,
[/tex]
[tex]U^{ \dagger } P^{ \sigma } U = \Lambda^{ \sigma }{}_{ \delta } P^{ \delta } ,[/tex]
and the identity
[tex]
\epsilon_{ \mu \nu \rho \sigma } \Lambda^{ \nu }{}_{ \lambda } \Lambda^{ \rho }{}_{ \eta } \Lambda^{ \sigma }{}_{ \delta } = \Lambda_{ \mu }{}^{ \gamma } \epsilon_{ \gamma \lambda \eta \delta } ,
[/tex]
you get
[tex]U^{ \dagger } W_{ \mu } U = \Lambda_{ \mu }{}^{ \nu } W_{ \nu } .[/tex]
Write the infinitesimal version of this.

Sam

 

[Bill_K]

you get
[tex]U^{ \dagger } W_{ \mu } U = \Lambda_{ \mu }{}^{ \nu } W_{ \nu } .[/tex]
Write the infinitesimal version of this.

Again, this simply states the obvious fact that W_μ is a vector.

 

[samalkhaiat]

Again, this simply states the obvious fact that W_μ is a vector.

I believe that proving that “obvious fact” is not trivial at all. Bellow is my reasons why:

i) Having single space-time index does not guarantee the vector nature of an object.

ii) The presence of the [itex]\epsilon[/itex] symbol in the definition of [itex]W_{ \mu }[/itex].

iii) In QFT, both [itex]P_{ \mu }[/itex] and [itex]J_{ \mu \nu }[/itex] will have contributions from the gauge potential which itself is not a genuine vector.

iv) Young researchers should learn about the tricks of the trade and use them to prove as many “obvious facts” as they possibly can.

Sam

 

[Avodyne]

I believe that proving that “obvious fact” is not trivial at all. Bellow is my reasons why:

i) Having single space-time index does not guarantee the vector nature of an object.

Well, most of the time it does. The vector potential [itex]A_{ \mu }[/itex] in a gauge theory is the only exception I can think of, and even then only gauge-dependent results are sensitive to this issue.

ii) The presence of the [itex]\epsilon[/itex] symbol in the definition of [itex]W_{ \mu }[/itex].

The [itex]\epsilon[/itex] symbol is invariant under any Lorentz transformation that does not involve time reversal.

iii) In QFT, both [itex]P_{ \mu }[/itex] and [itex]J_{ \mu \nu }[/itex] will have contributions from the gauge potential which itself is not a genuine vector.

[itex]P_{ \mu }[/itex] and [itex]J_{ \mu \nu }[/itex] depend only on the field strength [itex]F_{ \mu\nu }[/itex], which is a genuine tensor.

iv) Young researchers should learn about the tricks of the trade and use them to prove as many “obvious facts” as they possibly can.

I fully agree with this one! :)

 

[samalkhaiat]

Well, most of the time it does. The vector potential [itex]A_{ \mu }[/itex] in a gauge theory is the only exception I can think of

One exception is enough to make the proof of the statement non-trivial.

and even then only gauge-dependent results are sensitive to this issue.

Do you think that [itex]P^{ \mu }[/itex] and [itex]J^{ \mu \nu }[/itex] are gauge invariant operators? :)

The [itex]\epsilon[/itex] symbol is invariant under any Lorentz transformation that does not involve time reversal.

I expressed this fact by writing the explicit transformation law for [itex]\epsilon[/itex] symbol. Students need to know this guy is invariant, don’t they?

[itex]P_{ \mu }[/itex] and [itex]J_{ \mu \nu }[/itex] depend only on the field strength [itex]F_{ \mu\nu }[/itex], which is a genuine tensor.

Without the use of the field equations ( off-shell), the canonical [itex]P^{ \mu }[/itex] & [itex]J^{ \mu \nu }[/itex] both depend on the gauge potential.

Sam

 

[Avodyne]

One exception is enough to make the proof of the statement non-trivial.

Agreed!

Do you think that [itex]P^{ \mu }[/itex] and [itex]J^{ \mu \nu }[/itex] are gauge invariant operators? :)

The "improved", Belinfante versions are indeed gauge invariant.

 

[samalkhaiat]

The "improved", Belinfante versions are indeed gauge invariant.

As a matter of fact, Bellinfante procedure is possible because it does not affect the Poincare’ charges [itex]( P^{ \mu }, J^{ \mu \nu } )[/itex]. It only add a total divergence to the Poincare’ (canonical) currents, [itex]( T^{ \mu \nu }, J^{ \mu \nu \rho } )[/itex], leaving [itex]P^{ \mu }[/itex] and [itex]J^{ \mu \nu }[/itex] unchanged.
I can state (and prove on general grounds) the following claim:
“Even in a gauge invariant theory, the energy-momentum vector and the angular momentum tensor cannot be invariant under the c-number gauge transformations of the theory”.

Sam