This is how I began,
Proof.
Using
$$ \epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} – \delta_{jm}\delta_{kl} \tag{1}\label{eq1} $$
I wrote it as
$ \epsilon_{ijk}\epsilon_{ijk} = \delta_{jj}\delta_{kk} – \delta_{jk}\delta_{kj} \tag{2}\label{eq2}$
and
$ \delta_{jj} = 3 \tag{3}\label{eq3} $
$\therefore \delta_{jj} \delta_{kk} = 9 $ and $$ \delta_{jk}\delta_{kj} = ? \tag{4}\label{eq4}$$
I'm confused on how to simplify $\ref{eq4}$. My first thoughts would be that it's $\delta_{jk}\delta_{kj} = \delta_{kk}\delta_{kk} = 3*3 = 9$ since it only has a value when $j=k$ but after explicitly writing out the terms I can see that
If $j=1$ then
$$\begin{equation}\begin{aligned}
\delta_{1k}\delta_{1k} = 1 \text{ when k=1, but 0 otherwise} \\
\delta_{2k}\delta_{2k} = 1 \text{ when k=2, but 0 otherwise} \\
etc …
\end{aligned}\end{equation}\tag{5}\label{eq5}$$
$\therefore \delta_{jk}\delta_{kj} = 3 $
But I don't quite understand why $ \delta_{jk}\delta_{kj} = 3 $ without explicitly writing out the terms. Can you help me rationalize this or look at it a different way so that I can understand it more intuitively?
Best Answer
Use the Kronecker delta contraction
$\delta_{ik}\delta_{kj}=\delta_{ij}$
where we get terms of $1=\delta_{jj}$ iff $i=j$. Then in three dimensions summing these unities over three values of $j$ gives $3$.