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Here in this video from 15:14 Arvin Ash demonstrates a quantum-full-adder circuit, he goes on further to illustrate how it can perform multiple operations simultaneously via superposition. I took some screenshots asper:
$a)$quantum full adder
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quantum full adder
I have no issues with the concept of multiple operations being performed simultaneously but how would measurement make sense here (since we have a superposition of all states of the simultaneously performed additions).

Here's what I mean, for example with Grover's search algorithm we are interested in isolating a single state from a superposition of multiple states, from the video it seems to me we are interested in isolating multiple states from a superposition of multiple states. Since qubits collapse only to a single state after measurement, How would measurement here yield multiple states or is there a work around?

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Fair warning: I would consider this video typical of Arvin Ash's content, in that it is a mix of correct details and incorrect assertions. Eg. at 16:13 he claims a quantum computer can search 8 items in 1 step which is just wrong.

It's misleading to say there are multiple additions being done at the same time. Suppose I told you I have two numbers, A and B. I tell you there's a 20% chance that A=1 and B=2, a 30% chance that A=3 and B=0 and a 50% chance that A=2 and B=2. Now I tell you that I computed C=B+A. Do you jump into the air and say "WOW! There were 3 cases and you updated all of them! You computed 1+2=3, and 3+0=3, and 2+2=4! You just did 3 additions at the same time!"?

It's a mistake to assume that the amount of computational work you'd need to do to track the effects of a change is the same as the amount of computational work that can actually be extracted from applying that change. Yes, a quantum computer can apply one addition and force you to do a million additions to update your state vector tracking its state. But that's very different from claiming if you have a million additions to do then you can get a quantum computer to do them in one addition. And anyone who is mixing up those two things, without calling it out, is doing a disservice to your understanding.

So my answer to your question is: it doesn't do multiple additions at the same time. There's a sense in which it does, but going down that road leads to misunderstanding.

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  • $\begingroup$ thought as much! $\endgroup$
    – LiNKeR
    Oct 8 '21 at 23:16

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