The Pauli-Lubanski vector and angular momentum

[redtree]

TL;DR Summary

I have seen it stated but not proven that in the formula for the Pauli-Lubanski vector, the orbital angular momentum drops out from $M_{\rho \sigma}$ because of the anti-symmetry with the momentum operator, such that it includes only the spin part of the angular momentum. Where can I find an explicit proof of this statement?

Given $$M_{\rho \sigma} = i (x^{\rho} \partial_{\sigma} - x^{\sigma} \partial_{\rho})$$
and $$W^{\mu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}$$

Why does $W^{\mu}$ pick up only the spin part of the total angular momentum?

 

[Haborix]

Looks to me like you can just plug in $M_{\rho \sigma}$ and the derivative for $P_{\nu}$ and just compute that $W_\mu$ is zero, as you have defined it (without the spin component). You will just need to appeal to the anti-symmetry of the Levi-Cevita symbol. I see a couple terms with a repeated index in the Levi-Cevita symbol and couple where it gets contracted with symmetric terms. It is about 3 lines of index algebra/calculus.

 

[redtree]

I apologize, but I don't see it.

 

[haushofer]

I apologize, but I don't see it.

You only see it if you perform the calculation explicitly. You probably need to use that partial derivatives commute. Also, don't forget to add a test function on which the operators act for clarity.

If you have problems, you need to be more explicit than "I don't see it". Stating your problem very specifically enforces you to think more clearly, bringing you to a solution more closely.

Last note: "I've seen it stated" requires an explicit source here on PF. We need context to help.

 

[Haborix]

Just to get things started:

$W^{\mu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}=\frac{i}{2} \epsilon^{\mu \nu \rho \sigma} \partial_{\nu} (x_{\rho} \partial_{\sigma} - x_{\sigma} \partial_{\rho})$

or with test function:

$W^{\mu}f = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}f=\frac{i}{2} \epsilon^{\mu \nu \rho \sigma} \partial_{\nu} \left( (x_{\rho}\partial_{\sigma} - x_{\sigma} \partial_{\rho})f\right)$

 

[haushofer]

Now work it out. The derivative on the coordinates gives a (symmetric!) metric, and partial derivatives commute. Conclusion?

 

[redtree]

Is this correct? I apologize if the Latex syntax is wrong for Physics Forums. If it is, maybe you can cut and paste into another Latex editor....? (or you can tell me what I did wrong and I can repost)

$$\begin{equation}
\begin{split}
W^{\mu} &= \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}=\frac{i}{2} \epsilon^{\mu \nu \rho \sigma} \partial_{\nu} (x_{\rho} \partial_{\sigma} - x_{\sigma} \partial_{\rho})
\end{split}
\end{equation}$$
where
$$\begin{equation}
\begin{split}
\epsilon^{\mu \nu \rho \sigma} \partial_{\nu} (\epsilon^{\mu \nu \rho \sigma} x_{\rho} \partial_{\sigma} - x_{\sigma} \partial_{\rho}) &= \epsilon^{\mu \nu}\partial_{\nu} (\epsilon^{\rho \sigma}x_{\rho} \partial_{\sigma} - \epsilon^{\sigma \rho} x_{\sigma} \partial_{\rho})
\\
&= \epsilon^{\mu \nu}\partial_{\nu} (\epsilon^{\rho \sigma}x_{\rho} \partial_{\sigma} + \epsilon^{\rho \sigma} x_{\rho} \partial_{\sigma})
\\
&= 2 \epsilon^{\mu \nu \rho \sigma} \partial_{\nu} x_{\rho} \partial_{\sigma}
\\
&= -2 \epsilon^{\mu \rho \nu \sigma} x_{\rho} \partial_{\nu} \partial_{\sigma}
\end{split}
\end{equation}$$
such that
$$\begin{equation}
\begin{split}
W^{\mu} &=-i \epsilon^{\mu \rho \nu \sigma} x_{\rho} \partial_{\nu} \partial_{\sigma}
\end{split}
\end{equation}$$

 

[haushofer]

Is this correct? I apologize if the Latex syntax is wrong for Physics Forums. If it is, maybe you can cut and paste into another Latex editor....? (or you can tell me what I did wrong and I can repost)

$$\begin{equation}
\begin{split}
W^{\mu} &= \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}=\frac{i}{2} \epsilon^{\mu \nu \rho \sigma} \partial_{\nu} (x_{\rho} \partial_{\sigma} - x_{\sigma} \partial_{\rho})
\end{split}
\end{equation}$$
where
$$\begin{equation}
\begin{split}
\epsilon^{\mu \nu \rho \sigma} \partial_{\nu} (\epsilon^{\mu \nu \rho \sigma} x_{\rho} \partial_{\sigma} - x_{\sigma} \partial_{\rho}) &= \epsilon^{\mu \nu}\partial_{\nu} (\epsilon^{\rho \sigma}x_{\rho} \partial_{\sigma} - \epsilon^{\sigma \rho} x_{\sigma} \partial_{\rho})
\\
&= \epsilon^{\mu \nu}\partial_{\nu} (\epsilon^{\rho \sigma}x_{\rho} \partial_{\sigma} + \epsilon^{\rho \sigma} x_{\rho} \partial_{\sigma})
\\
&= 2 \epsilon^{\mu \nu \rho \sigma} \partial_{\nu} x_{\rho} \partial_{\sigma}
\\
&= -2 \epsilon^{\mu \rho \nu \sigma} x_{\rho} \partial_{\nu} \partial_{\sigma}
\end{split}
\end{equation}$$
such that
$$\begin{equation}
\begin{split}
W^{\mu} &=-i \epsilon^{\mu \rho \nu \sigma} x_{\rho} \partial_{\nu} \partial_{\sigma}
\end{split}
\end{equation}$$

I'm not sure why you split up the epsilon symbol, but I think you're looking in the right direction. Take a look at the expression

$\epsilon^{\mu \nu \rho \sigma} \partial_{\nu} (x_{\rho} \partial_{\sigma}f - x_{\sigma} \partial_{\rho}f)$

with $f$ a test function. If the derivative $\partial_{\nu}$ hits e.g. $x_{\rho}$ you get $\eta_{\nu\rho}$ which is symmetric; if it hits $\partial_{\sigma}$ you get $\partial_{\nu}\partial_{\sigma}$ which is symmetric. Same for the other two terms. Contracting these symmetric terms with the epsilon symbol gives you zero. So you get

$\epsilon^{\mu \nu \rho \sigma}(\eta_{\nu\rho} \partial_{\sigma}f + x_{\rho}\partial_{\nu}\partial_{\sigma}f) - [\sigma \leftrightarrow \rho] = 0 + 0 - (0+0) = 0$

It could be even done quicker if you realise that the epsilon symbol is antisymmetric in $\sigma$ and $\rho$, so it becomes

$2\epsilon^{\mu \nu \rho \sigma}(\eta_{\nu\rho} \partial_{\sigma}f + x_{\rho}\partial_{\nu}\partial_{\sigma}f)$

If you don't understand why the contraction of an antisymmetric and symmetric tensor gives zero, you should work that out first.

 

[redtree]

I understand why symmetric and anti-symmetric tensors contract to zero.

I apologize, but then I don't see how spin come out as non-zero.

 

[haushofer]

Ok, let's define the total angular momentum

$J_{\rho\sigma} = M_{\rho\sigma} + S_{\rho\sigma}$

with $S_{\rho\sigma}$ the spin-operator. Now calculate

$\epsilon^{\mu\nu\rho\sigma}P_{\nu}J_{\rho\sigma}$

 

[redtree]

Clearly
$$\begin{equation}
\begin{split}
\epsilon^{\mu\nu\rho\sigma} P_{\nu} J_{\rho\sigma} &= \epsilon^{\mu\nu\rho\sigma} P_{\nu} (M_{\rho \sigma} + S_{\rho \sigma})\\
&= \epsilon^{\mu\nu\rho\sigma} P_{\nu} M_{\rho \sigma} + \epsilon^{\mu\nu\rho\sigma} P_{\nu} S_{\rho \sigma}\\
\end{split}
\end{equation}$$
but don't orbital and spin angular momentum have the similar commutation relations? So why the difference?

Sorry if I'm being daft.

 

[PeterDonis]

don't orbital and spin angular momentum have the similar commutation relations?

With themselves and with the total angular momentum $J$, yes. With the linear momentum $P$, no.

 

[redtree]

Why is that?

 

[Haborix]

Because spin and angular momentum operators act on two completely different subspaces of the space of states. Angular momentum and linear momentum act on the same subspace, i.e., the $L^2(\mathbb{R}^3)$ part.

 

[redtree]

Can you prove that mathematically in a simple and intuitive way? The only proof I've found is https://www-user.rhrk.uni-kl.de/~apelster/Vorlesungen/WS2021/v6.pdf.

 

[redtree]

For anyone coming across this thread in the future, it seems the above hyperlink has been taken down.

The best mathematical proof I've found for the decomposition of total angular momentum into orbital and spin angular momentum can be found at: https://repository.tudelft.nl/file/File_9e551b4b-b048-4b6e-9707-b4f6dc956d46?preview=1.