How Do Lorentz Group Commutation Relations Apply to Spin Matrices?

[Markus Kahn]

Homework Statement

Prove that the sets $(S_{\mu\nu})_L$ and $(S_{kl})_R$, where
$$
\left( S _ { k \ell } \right) _ { L } = \frac { 1 } { 2 } \varepsilon _ { j k \ell } \sigma _ { j } = \left( S _ { k \ell } \right) _ { R } \quad\text{and}\quad \left( S _ { 0 k } \right) _ { L } = \frac { 1 } { 2 } i \sigma _ { k } = ( S ^ { 0 k }) _ { R }
$$
satisfy the commutation relation of the Lorentz group, namely
$$
\left[M_{\mu \nu}, M_{\rho \sigma}\right]=-\eta_{\mu \rho} M_{\nu \sigma}+\eta_{\mu \sigma} M_{\nu \rho}-\eta_{\nu \sigma} M_{\mu \rho}+\eta_{\nu \rho} M_{\mu \sigma}.
$$

The Attempt at a Solution

My attempt was straight forward
$$
\begin{align*}
[(S_{kl})_L, (S_{bc})_L]
&= \frac{1}{4}\varepsilon_{jkl}\varepsilon_{abc}[\sigma_j,\sigma_a] = \frac{1}{4}\varepsilon_{jkl}\varepsilon_{abc} (2i \varepsilon_{jau}\sigma_u) = \frac{i}{2}\varepsilon_{jkl}\varepsilon_{abc} \varepsilon_{jau}\sigma_u\\
&=\frac{i}{2} (\delta_{ka}\delta_{lu}-\delta_{ku}\delta_{al})\varepsilon_{abc}\sigma_u = \frac{i}{2}\varepsilon_{kbc}\sigma_l -\frac{i}{2} \varepsilon_{lbc}\sigma_k
\end{align*}
$$
but this seems to lead to nowhere. One of my problems here is that $(S_{kl})_L$ is only defined for $k,l\in\{1,2,3\}$ but $M_{\mu\nu}$ is defined for $\mu,\nu\in\{0,\dots,3\}$... I'm not sure how to make sense of this but I honestly also don't know where I made a mistake in the above calculation...

 

[stevendaryl]

My attempt was straight forward
$$
\begin{align*}
[(S_{kl})_L, (S_{bc})_L]
&= \frac{1}{4}\varepsilon_{jkl}\varepsilon_{abc}[\sigma_j,\sigma_a] = \frac{1}{4}\varepsilon_{jkl}\varepsilon_{abc} (2i \varepsilon_{jau}\sigma_u) = \frac{i}{2}\varepsilon_{jkl}\varepsilon_{abc} \varepsilon_{jau}\sigma_u\\
&=\frac{i}{2} (\delta_{ka}\delta_{lu}-\delta_{ku}\delta_{al})\varepsilon_{abc}\sigma_u = \frac{i}{2}\varepsilon_{kbc}\sigma_l -\frac{i}{2} \varepsilon_{lbc}\sigma_k
\end{align*}
$$

I find working with the Levi-Civita symbols extremely tedious and error-prone, but I think that's correct. You can check for a couple of special cases:

$[(S_{xy})_L, (S_{yz})_L] = [i/2 \sigma_z, i/2 \sigma_x] = (-1/4) (- 2 i ) \sigma_y = (i/2) \sigma_y$

Your formula gives:
$(i/2) \varepsilon_{xyz} \sigma_y - (i/2) \varepsilon_{yyz} \sigma_x$

which simplifies to the same thing.

but this seems to lead to nowhere. One of my problems here is that $(S_{kl})_L$ is only defined for $k,l\in\{1,2,3\}$ but $M_{\mu\nu}$ is defined for $\mu,\nu\in\{0,\dots,3\}$... I'm not sure how to make sense of this but I honestly also don't know where I made a mistake in the above calculation...

$S_{\mu \nu}$ is defined for all $\mu$ and $\nu$. If $\mu = \nu = 0$, then it's zero. If $\mu = j$ and $\nu = k$ are both 1,2 or 3, then it's $S_{jk}$. If $\mu = 0$ and $\nu = j$, then it's $S_{0j} = (i/2) \sigma_j$