Proving commutation relation

  • #1
104
12

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...
 

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,581
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##
 

Related Threads on Proving commutation relation

Replies
2
Views
3K
Replies
5
Views
5K
Replies
2
Views
2K
Replies
6
Views
941
  • Last Post
Replies
19
Views
2K
  • Last Post
Replies
7
Views
786
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
907
  • Last Post
Replies
1
Views
862
Top