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I'm working through the article Quantum Computing for the Very Curious and am stuck on one aspect of the proof it gives for unitary matrices being length preserving. I've included an annotated screenshot as I can't link to this specific part of the article: screenshot of proof

My question is: at the place I've marked with the red arrow, why did the author introduce the index k?

The sum is over the product of the jth entry of $U\psi$ and $(U\psi)^*$ i.e. the same entry in both vectors. If $(U\psi)_j$ expands to $\sum_{l}U_{jl}\psi_l$ (as explained above), then couldn't we write $(U\psi)^*_j$ as $\sum_{l}U^*_{jl}\psi^*_l$ and eliminate the need for the k index?

I get that it ultimately works because lower down $\delta_{kl}$ is zero whenever k != l. But it seems like the proof would be simpler without introducing k:

$$\|U|\psi \rangle \|^2 = \sum_j(U\psi)^*_j(U\psi)_j$$

$$= \sum_{jl}U^*_{jl}\psi^*_lU_{jl}\psi_l$$

$$= \sum_{jl}U^*_{jl}U_{jl}\psi^*_l\psi_l$$

$$= \sum_{jl}U^\dagger_{lj}U_{jl}\psi^*_l\psi_l$$

$$= \sum_{l}I_{ll}\psi^*_l\psi_l$$

$$= \sum_{l}\psi^*_l\psi_l$$

$$= \|\psi\|^2$$

Overall, the proof in the article (and others like https://math.stackexchange.com/a/2850254/748472) make sense to me but I was surprised / thrown off by the k above and since this is all new to me, I can't shake the feeling that I'm missing something.

Thanks in advance for indulging me.

triangle_man
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    The thing is that you expand two different terms, so you need two different indices. For instance, when expanding $(a+b)(c+d+e)$, the final sum has $6=2 \times 3$ terms. In general, if you’re expanding $AB$, where $A$ and $B$ are sums of $n$ terms, there will be $n^2$ terms in the expansion, requiring two indices. It’s merely a coincidence in this case that terms with $k \neq l$ ultimately contribute nothing. – Aphelli Feb 06 '20 at 22:33
  • Thank you, @Mindlack for your explanation. It took me a while to grok what you're saying and when I did, it revealed a deeper misunderstanding on my part, so I'm glad I pushed on this (I had wondered if I was just splitting hairs when I wrote the post). I think what confused me was this: I saw we're multiplying the jth element of $U\psi$ and $(U\psi)^*$ i.e. the same element index in both vectors, but didn't see that each of those terms is itself a sum of many terms and we have to multiply their sums together, resulting in the $n^2$ expansion, as you say. – triangle_man Feb 07 '20 at 21:42
  • I think having all the sums squished into the single $\sum$ rather than seeing the intermediate step of $\sum_j(\sum_{k}U^_{jk}\psi^k)(\sum_jU{jl}\psi_l)$ confused me. Thanks again for taking the time to answer. – triangle_man Feb 07 '20 at 21:42

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