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Let's say you want to do an experiment with 14+ qubits. You apply some arbitrary unitary operator $U \in (\mathbb{C}^2)^{\otimes n} \times (\mathbb{C}^2)^{\otimes n}$ to the state $|\psi\rangle \in (\mathbb{C}^2)^{\otimes n} $. That is

$$ U|\psi \rangle = |\phi \rangle $$

We can take $|\psi \rangle = |0\rangle^{\otimes n} $ to fits with current quantum computing setting. Now, if we do this experiment with $2^{13} = 8192$ shots, how is this enough to build up the statistical distribution as we have more than $2^{14}$ slots to distribute them to. If your output state $|\phi\rangle$ is particular eigenstate, says $|0110\cdots 1 \rangle$, then this many shots is more than enough. But if $|\phi\rangle$ is in a $2^n$ superposition state, then how is this enough? We wouldn't have enough experimental data to build up an accurate statistical distribution. Of course, I can repeat my experiment/job with 8192 shots a bunch of times and average out the results but even then it would still take a huge amount of experiments to have enough number of shots to get meaningful results... especially for variational type quantum algorithms... where one would need millions of shots to get within chemical precision.

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Reading out all the probabilities for all the possible output bit strings isn't common in quantum computing. The ideal case is to induce an interference effect that will allow your result to be read out with just one shot. Though that isn't something most algorithms achieve, they nevertheless use only $O(1)$ shots, or some other complexity that is far less than $O(2^n)$.

Nevertheless, if you want more than 8192 you could submit several jobs and combine results.

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  • $\begingroup$ James, Thanks for taking your time to answer. I understand that you can use more than one job to add up the result to built up the statistical distribution if needed as I mentioned in the question. However, even this is not reasonable for a 53 qubit device, don't you think? That is 2^53 eigenstates and so the amount of job you need to perform is enormous. I understand algorithm like, Shor's factoring and many other, end up with spitting out only of of the eigenstate, hence you don't need to do so many samples/shots to retrieve the answer. $\endgroup$ – KAJ226 Sep 10 at 14:52
  • $\begingroup$ (...cont...) let's say i want to benchmark the performance of my 2^53 or whatever number of qubit system. I want to see if it can generate the right quantum states, and I have enough classical resources to read through all those probabilities. I would like to do some arbitrary unitary operation on the qubit state and read out its output state, and making sure it gave me the right answer. In such case, I would need to see the entire probabilities. $\endgroup$ – KAJ226 Sep 10 at 14:58

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