Simplify the following in index notation

Question

Simplify the following in index notation

$I_{s,t}\delta_{s,n}\delta_{n,t}$

Since both $\delta$ 's contain an $n$ index does it simplify to

$I_{s,t}\delta_{s,t}$

Then can you simplify further since $I$ and $\delta$ contain $t$ and $s$ indices?

Answer 1

Yes it does simplify to $I_{st}\delta_{st}$. Summing over $n$ (we use Einsteins summation convention) we have $I_{st}\delta_{sn}\delta_{nt} = I_{st}\delta_{st}$ since in order for $\delta_{sn}\delta_{nt}$ to be non-zero we need $s=n=t$.

In the sum $I_{st}\delta_{st}$ the only non-zero terms comes from $s=t=1,2,3\ldots, N$ so the sum is simply

$$I_{st}\delta_{st} = \sum_{n=1}^N I_{nn}$$

We can also write this in matrix notation as $I_{st}\delta_{st} = \text{tr}(I)$ where tr is the trace.