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Let's say I've got a qubit, being in one of two states:

either [ 0.995+0j, 0.098+0j ]
or [ 0.995+0j, -0.098+0j ],

that is the state I get when applying a RY-gate with θ = ±π/x to |0>. In the both states above, x would be 16, but let's just say it is just a large number, so the probability to measure |1> is nearly 0%. Is there any way to determine which of these cases is true, so whether the phase angle now is 0 or π?

I am not very familiar with either quantum computers or the english language, so I hope that I managed to somehow express what I mean and that you can forgive me the mistakes I have most probably made.

JRE F
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1 Answers1

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One can determine the phase angle by looking at the sign and the initialized state. In your case Qubit was initialized to |0> thus when when applied ry pi, it's phase will change to -1 for |1> and just 1 for rz O.

the result you're seeing up there is not perfect due to randomness and error generated by the simulator (if you got that answer from a simulator) or actual error due to it being performed on actual quantum computer( if you got that answer from a real quantum computer). So the theoretical answer should be just |0> but due to noise you got the following, which is both true and part of the process.

Samar Singh
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