# Isn't measuring the results the main bottleneck of quantum computers?

I have researched a bit about quantum computers, and hopefully understood that the main advantage does lie in the utilisation of superpositions in order to compute (manipulate qubits) in parallel instead of sequential. While conventional bits can only hold a single information at a time, qubits are capable of holding two states simultaneously - let say i want to add two 4-bit numbers - not only I don't need 8 bits (4 bits for each number) for just a single addition, but with just 4 qubits I am able to "encode" for example 8 additions ((2^4)/2) and compute them all at once.

To visualize my idea: 4 qubits, therfore 2^4 states:

01.) 0000
02.) 0001
03.) 0010
04.) 0011
05.) 0100
06.) 0101
07.) 0110
08.) 0111
09.) 1000
10.) 1001
11.) 1010
12.) 1011
13.) 1100
14.) 1101
15.) 1110
16.) 1111


Let's assume we have circuit that adds two consecutive "lines" (=states) together and stores the value in the latter one, i.e. we add 1.) and 2.) and store the sum in 2.), we add 3.) and 4.) and store the sum in 4.) and so on and so forth.

But

... how am I able now to "read" (measure) two results at the same time? What do I mean by that? What if I not only want to measure the result saved in line 1.), but also line 15.)?

I mean, the moment I measure - let's say line (=state) 1.) - the qubits act like a bit, and only one single information is given while the other informations are loss.

It may be true that quantum computers compute in parallel, but at the end, I am only able to measure a single outcome-scenario, thus losing the advantages of parallel-computation.

I know that I misunderstood something, or even everything with regards to quantum-computers. That's why I am asking.

Tldr: What advantages do we gain from quantum-computer's parallel calculations if we are able to measure only a single outcome while losing every other possible outcome? It feels like sequential computing with extra steps.

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• – Yuval Filmus Dec 22 '20 at 21:41