# Proof: Pauli-Lujanski tensor

1. Jul 9, 2013

### 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!

2. Jul 9, 2013

### 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!

Mμν is the 4-dimensional rotation operator, and consequently [Mμν, Vσ] = i(gνσVμ - gμσVν) for any 4-vector Vμ.

3. Jul 9, 2013

Thanks!

4. Jul 9, 2013

### samalkhaiat

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

5. Jul 9, 2013

### Bill_K

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

6. Jul 10, 2013

### samalkhaiat

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

7. Jul 10, 2013

### Avodyne

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.

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

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

I fully agree with this one! :)

8. Jul 11, 2013

### samalkhaiat

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

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

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

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

Sam

9. Jul 11, 2013

### Avodyne

Agreed!

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

10. Jul 11, 2013

### samalkhaiat

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