4
$\begingroup$

For a given unitary, I want to know whether this unitary gate is correctly evolved in the circuit. In the simulator, I can use "statevector" to get the state vector to check the correctness of the evolution.

But I don't know whether such a unitary can still evolve correctly on a real quantum computer. Is there a corresponding way to check its correctness?

In more detail, I give a Hamiltonian H, then I get unitary U=expm(-1t*H). Then I add this unitary gate to the circuit usingqc.unitary(U, [0,1]). So how can I check whether it evolves correctly in the real quantum computer?

Improve this question
$\endgroup$
2
  • $\begingroup$ What do mean by "U gate evolves correctly" and by "problem with U-gate itself"? Please try to be a little more specific and edit your question accordingly. $\endgroup$
    – SimoneGasperini
    Jan 20 at 10:31
  • $\begingroup$ To reconstruct quantum state you can use quantum tomography. However, bear in mind that the tomography is exponentially complex in number of qubits. $\endgroup$
    – Martin Vesely
    Jan 20 at 12:42

2 Answers 2

Reset to default
3
$\begingroup$

There is no way to get a "perfectly correct" state evolution on a real quantum computer since the hardware is affected by different types of noise (e.g. coherent gate error, state decoherence, etc.) and it is not error-corrected.

What you can do is to first run your quantum circuit implementing the unitary $U$ on a perfect simulator, in order to check it is doing what it is supposed to do, and then set it up to run on your real device. In this case, you can be pretty sure that the deviation between the simulation and the actual execution is due to noise effects only.

Improve this answer
$\endgroup$
3
$\begingroup$

To characterize the nature of your unitary transformation you can use quantum process tomography. Qiskit experiments supports this experiment. For an example see the similar tutorial on state tomography.

Note that with tomography you need to characterize the unitary by applying it to all of the input and measurement basis so there is an overhead to the characterization.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.