# Inner products with orthonormal bases

Probably a stupid question here - I think it's a case of me not having sufficient mathematical background to follow this through. In Leonard Susskind's Theoretical Minimum book, he represents the ket-vector A as such: $$|A\rangle = \displaystyle\sum_{i} a_{i}|i\rangle$$

Which I follow, as I do when he takes the inner product with $\langle j|$ and collapses $\langle j | i \rangle$ to $\delta_{ij}$. However, could somebody please explain to me how he collapses the summation

$$\langle j|A\rangle = \displaystyle\sum_{i} a_{i} \delta_{ij}$$

to

$$\langle j | A\rangle = a_{j}$$

It's probably a really stupid question, so apologies for that, but is it possible that someone on here could step me through it?

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Hint: use the definition of $\delta_{ij}$ ;-) it is different from zero only when... – yuggib Sep 1 '14 at 9:17
@yuggib Seems so simple now I realise it - thanks for the hint! – user58309 Sep 1 '14 at 10:01

Note that: \begin{aligned} \langle j | A \rangle & = \langle j |\left( \sum\limits_{i} a_i |i \rangle \right) \\& = \langle j| \left(a_1 |1 \rangle + a_2 |2 \rangle + a_3 |3 \rangle + \cdots + a_j |j \rangle + \cdots \right) \\& = a_1 \underbrace{\langle j | 1 \rangle}_{=\delta_{j1} = 0} + a_2 \underbrace{\langle j | 2 \rangle}_{=\delta_{j2} = 0} + a_3 \underbrace{\langle j | 3 \rangle}_{=\delta_{j3} = 0} + \cdots + a_j \underbrace{\langle j | j \rangle}_{=\delta_{jj} = 1} + \cdots \\& = a_j \end{aligned}
If you think in terms of vectors and matrices with indices, $a_i$ is a vector and $\delta_{ij}$ is the unit matrix. The summation on the right hand side of your equation represents nothing but a multiplication of the vector $a_i$ with the latter. Since free indices are preserved, the remaining object carries the index $j$. You can write down an arbitrary vector and multiply it with the unit matrix to see that it is true.
You can also just take the basic property of the Kronecker delta $\delta_{ij}$, namely that it is only nonzero when $i$ and $j$ are equal. The consequence this has for the sum is now that only those terms with $i=j$ remain, i.e. $a_j$.