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

[thalisjg]

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

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

Sam

 

[Bill_K]

you get
$$U^{ \dagger } W_{ \mu } U = \Lambda_{ \mu }{}^{ \nu } W_{ \nu } .$$
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 $\epsilon$ symbol in the definition of $W_{ \mu }$.

iii) In QFT, both $P_{ \mu }$ and $J_{ \mu \nu }$ 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 $A_{ \mu }$ 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 $\epsilon$ symbol in the definition of $W_{ \mu }$.

The $\epsilon$ symbol is invariant under any Lorentz transformation that does not involve time reversal.

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

$P_{ \mu }$ and $J_{ \mu \nu }$ depend only on the field strength $F_{ \mu\nu }$, 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 $A_{ \mu }$ 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 $P^{ \mu }$ and $J^{ \mu \nu }$ are gauge invariant operators? :)

The $\epsilon$ symbol is invariant under any Lorentz transformation that does not involve time reversal.

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

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

Without the use of the field equations ( off-shell), the canonical $P^{ \mu }$ & $J^{ \mu \nu }$ 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 $P^{ \mu }$ and $J^{ \mu \nu }$ 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 $( P^{ \mu }, J^{ \mu \nu } )$. It only add a total divergence to the Poincare’ (canonical) currents, $( T^{ \mu \nu }, J^{ \mu \nu \rho } )$, leaving $P^{ \mu }$ and $J^{ \mu \nu }$ 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