Need help understanding product of Kronecker delta function with an outer product

Question

I need help solving the the below expression which is from the book Modern Quantum Mechanics by J. J. Sakurai and Jim Napolitano.

$$A=\sum_{a''}\sum_{a'}|a''\rangle a'\delta_{a'a''}\langle a'| =~?$$

I think that kronecker delta function will go with the ket $|a''\rangle$

$A=\sum_{a''}\sum_{a'}\delta_{a'a''}|a''\rangle a'\langle a'|$

and will change the term $\delta_{a'a''}|a''\rangle$ to $|a'\rangle$ and the expression will become

$$A=\sum_{a'}|a'\rangle a'\langle a'|,$$

but I'm not sure. Please help.

Answer 1

The Kronecker delta tells you that when you sum over $a''$, there is only one term will be zero except for the term in which $a''=a'$. As a result, you can simply set $a''=a'$ and drop the sum over $a''$ to obtain $$A = \sum_{a''} \sum_{a'} |a''\rangle a' \delta_{a' a''}\langle a'| = \sum_{a'} |a'\rangle a' \langle a'|$$

As an example, let $a',a''\in\{1,2\}$. Then

$$A = \sum_{a''=1}^2 \sum_{a'=1}^2 |a''\rangle a' \delta_{a' a''} \langle a'|$$ $$= |1\rangle\big(1 \cdot \delta_{11}\big)\langle 1| + |2\rangle\big(1 \cdot \delta_{12}\big)\langle 1| +|1\rangle\big(2 \cdot \delta_{21}\big)\langle 2| +|2\rangle\big(2\cdot \delta_{22}\big)\langle 2| $$ $$= |1\rangle\langle 1| + |2\rangle 2 \langle 2| = \sum_{a'=1}^2 |a'\rangle a' \langle a'|$$