How can we only use 8192 shots for an experiment with 14 or more qubits?

Question

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.

Answer 1

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.