Sifting Property

It might help to see it on a simple problem:

i={1,2,3,4,5}

j=3

$\displaystyle \sum_{i=1}^5 \delta_j^i a_i = (0)(2) + (0)(4) + (1)(6) + (0)(8) = 6$

$\displaystyle \sum_{i=1}^5 \delta_j^i a_i = a_j$

If we are allowed to use Einstein Notation then the summation symbol is dropped:

$\delta_j^i a_i = a_j$

For this problem we just get one term for the answer.

the problem could have been complicated in a way that would have led to several terms in the answer, and we would like to show an example of this.

All indices run 1 to 4:

$a^j \delta^i_j b_i$