# Pauli-Lubanski pseudovector commutation relations

1. Feb 17, 2012

### Factvince

1. The problem statement, all variables and given/known data
Hi. This is not a homework question per se, but more of a personal question, but I thought I'd post it here.
I'm trying to prove the commutation relations of the Pauli-Lubanski pseudovector

W_\mu\equiv-\frac{1}{2} \epsilon_{\mu\nu\rho\sigma}J^{\nu\rho}P^\sigma

with the Lorentz transformation generators $$J^{\mu\nu}.$$

I'm supposed to find

\left[J_{\mu\nu},W_\rho\right]= \mathrm{i}\left( \eta_{\nu\rho}W_\mu-\eta_{\mu\rho}W_\nu\right)

but I simply can't.

2. Relevant equations
Obviously I have to use
\begin{align}
\left[P_\mu,P_\nu\right]&= 0,\\
\left[P_\mu,J_{\nu\rho}\right]&= \mathrm{i}\left(\eta_{\mu\rho}P_\nu-\eta_{\mu\nu}P_\rho\right),\\
\left[J_{\mu\nu},J_{\rho\sigma}\right]&= \mathrm{i}\left(\eta_{\mu\rho}J_{\sigma\nu}- \eta_{\nu\rho}J_{\sigma\mu}- \eta_{\mu\sigma}J_{\rho\nu}+ \eta_{\nu\sigma}J_{\rho\mu}\right)
\end{align}

3. The attempt at a solution

My calculations gave

\left[J_{\kappa\lambda},W_\mu\right]= -i\left(\eta_{\tau\lambda} \epsilon_{\kappa\rho\mu\sigma} J^{\rho\tau}P^\sigma- \frac{1}{2} \eta_{\tau\lambda} \epsilon_{\kappa\rho\mu\sigma} J ^{\sigma\rho}P^\tau- \eta_{\tau\kappa} \epsilon_{\lambda\rho\mu\sigma} J^{\rho\tau}P^\sigma+ \frac{1}{2}\eta_{\tau\kappa} \epsilon_{\lambda\rho\mu\sigma} J^{\sigma\rho}P^\tau\right).

I'm pretty confident this is correct, but in the meantime I don't see where to go when I get here.

Any help very much appreciated !

Last edited: Feb 17, 2012
2. Feb 17, 2012

### tiny-tim

Welcome to PF!

Hi Factvince! Welcome to PF!

You can restore that missing equation by pressing the "EDIT" button and typing a space at least every 50 characters (it's a PF bug).

3. Feb 17, 2012

### Factvince

Thanks very much tiny-tim !

4. Feb 17, 2012

### Factvince

I went a little further and I can sense I could do something with antisymmetry arguments here :

\begin{align*}
\left[W_\mu,J_{\kappa\lambda}\right]&= \frac{\mathrm{i}}{2}\left\{\eta_{\tau\lambda}\left( \epsilon_{\mu\kappa\rho\sigma} J^{\rho\tau}P^\sigma+ \epsilon_{\mu\sigma\kappa\rho} J^{\tau\sigma}P^\rho+ \epsilon_{\mu\rho\kappa\sigma} J^{\sigma\rho}P^\tau\right)\right\}\\
&- \frac{\mathrm{i}}{2}\left\{\eta_{\tau\kappa}\left( \epsilon_{\mu\lambda\rho\sigma} J^{\rho\tau}P^\sigma+ \epsilon_{\mu\sigma\lambda\rho}J ^{\tau\sigma}P^\rho+ \epsilon_{\mu\rho\lambda\sigma}J ^{\sigma\rho}P^\tau\right)\right\}.
\end{align*}

Being able to "switch" \mu and \tau would solve the problem, but I don't see any good way to justify such a move here.

5. Aug 28, 2012

### sfn17

Some Fierz reshuffling of indices seems to be useful.

I found for myself the following identity (not confirmed!!):

$\epsilon_{\kappa\lambda\mu\nu}g_{\sigma\tau} -\epsilon_{\kappa\lambda\tau\nu}g_{\sigma\mu}= \epsilon_{\mu\tau\lambda\nu}g_{\sigma\kappa} -\epsilon_{\mu\tau\kappa\nu}g_{\sigma\lambda}$

This formula makes possible the exchange of the antisymmetric pair κλ with μτ.
(I know that the ordering of the indices in my formula is not perfect.)

For disentangling all those indices in the actual problem, I tried to use the pictorial notation of Penrose. But I didn't come to a solution either... :-/

6. Sep 10, 2012

### Factvince

My teacher gave me the answer months ago and I forgot about this thread: since Minkwoski space is four-dimensional, any expression which is antisymmetrised over five indices is identically zero. So one can write

\eta_{\tau\lambda}\epsilon_{\mu\varkappa\rho\sigma} + \eta_{\mu\lambda}\epsilon_{\varkappa\rho\sigma\tau} + \eta_{\varkappa\lambda}\epsilon_{\rho\sigma\tau\mu} + \eta_{\rho\lambda}\epsilon_{\sigma\tau\mu\varkappa} + \eta_{\sigma\lambda}\epsilon_{\tau\mu\varkappa\rho} = 0

and that pretty much wraps it up. By the way, there were a couple of mistakes in the last result I gave due to the fact that I had the wrong sign for the

\left[P_\mu,J_{\nu\rho}\right]

commutator.