Angular momentum operator derived from Lorentz invariance

[Gene Naden]

I am working through Lessons in Particle Physics by Luis Anchordoqui and Francis Halzen; the link is https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf. I am on page 11, equation 1.3.20. The authors have defined an operator $L_{\mu\nu} = i( x_\mu \partial \nu - x_\nu \partial \mu)$. They state that $L_{23}$ is the x component of angular momentum. Reviewing the Schrödinger equation, I see that the momentum operator $\vec{p} = -i \nabla$, from which I get

$\partial_1=i p_{1}$ and $\partial_2 = i p_{2}$.

Substituting these in the definition for L, I get $L_1 = L_{23} = i(x_2 i p_3 - x_3 i p_2) = - (\vec{r} \times \vec{p})_1$

So I get $L_{1}$ equal to minus the x component of the angular momentum. I am wondering where I went wrong. I have worked through about 20 equations in this reference and have never found an error!

 

[eys_physics]

The contravariant components of the four vector $x^{\mu}$ are
$$x_\mu=g_{\mu\nu}x^\nu$$ where $g_{\mu\nu}=\rm{diag}(1,-1,-1,-1)$, which explains the minus sign.

 

[Gene Naden]

Thank you. There doesn't seem to be a "like" button, or I would "like" your response.

 

[eys_physics]

You are welcome. There is a like button to the bottom right of each post. But, you need to have your mouse point over the relevant post. Please, see the attached screenshot.