I need to prove a commutator relation, but I'm getting stuck at the definition of the matrices.

$(L_{ab})_{cd} = \delta_{ac} \delta_{bd} - \delta_{ad} \delta_{bc}$ with $a<b$ and $a, b \in 1,2,3,4$. What does this definition of the matrix mean? Can someone explain this to me?


Let me illustrate in the case $2\times2$.

The matrix $L_{ab}$ would be $$ \left[\begin{array}{ccc} \delta_{a1}\delta_{b1}-\delta_{a1}\delta_{b1} &\delta_{a1}\delta_{b2}-\delta_{a2}\delta_{b1}\\ \delta_{a2}\delta_{b1}-\delta_{a1}\delta_{b2} &\delta_{a2}\delta_{b2}-\delta_{a2}\delta_{b2} \end{array}\right] .$$

Then for $L_{11}$ we have $$ \left[\begin{array}{ccc} \delta_{11}\delta_{11}-\delta_{11}\delta_{11} &\delta_{11}\delta_{12}-\delta_{12}\delta_{11}\\ \delta_{12}\delta_{11}-\delta_{11}\delta_{12} &\delta_{12}\delta_{12}-\delta_{12}\delta_{12} \end{array}\right] = \left[\begin{array}{ccc} 0&0\\ 0&0 \end{array}\right] $$

But for $L_{12}$ $$ \left[\begin{array}{ccc} \delta_{11}\delta_{21}-\delta_{11}\delta_{21} &\delta_{11}\delta_{22}-\delta_{12}\delta_{21}\\ \delta_{12}\delta_{21}-\delta_{11}\delta_{22} &\delta_{12}\delta_{22}-\delta_{12}\delta_{22} \end{array}\right] = \left[\begin{array}{ccc} 0&1\\ -1&0 \end{array}\right] $$

  • $\begingroup$ Thank you! But what do the second indices 'cd' mean. So, I'd understand it if it was only $L_{ab}$, but I'm very confused about the $(L_{ab})_{cd}$ notation. $\endgroup$ – Darius Nov 15 '15 at 17:47
  • $\begingroup$ the $cd$ indexes determines the entry in the $c$-row and $d$-column of the matrix $L_{ab}$ $\endgroup$ – janmarqz Nov 15 '15 at 17:51
  • $\begingroup$ you oughtta use that $\delta_{ij}=1$ when $i=j$ and $\delta_{ij}=0$ when $i\neq j$... this is the famous delta Kronecker $\endgroup$ – janmarqz Nov 15 '15 at 17:55
  • $\begingroup$ Thanks, now I got it! So basically I could build this matrix in 4x4 by following your idea. Now I have to prove this commutator relation: $\left[L_{ab}, L_{cd} \right] = \delta_{ad}L_{bc} + \delta_{bc}L_{ab} - \delta_{ac} L_{bd} - \delta_{bd} L_{ac}$. DO you have any hint on proving this? I guess writing out the matrices and doing the multiplication is pretty straight forward and not so nice. $\endgroup$ – Darius Nov 15 '15 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.