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I am trying to vary the "quantumness" of this simple circuit by inserting some cnot gates into the circuit with an RY gate to control the "quantumness" My Circuit

When I change RY the probabilities will change, when I set RY to pi/2 it will go back to roughly 5050 for states 00000 and 01111.

My question is; why once I add measurements at the end of this circuit for qubits 0,1,2,3 does my circuit act as if RY = (pi/2) or as if the cnot gates I have implemented to vary "quantumness" aren't there.

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  • $\begingroup$ I have not looked at this too closely yet, but remember that a CNOT gate does not affect the control qubit, just the target, so the only gates affecting $q[0], q[1], q[2]$ and $q[3]$ are the H, and you first, second and last CNOT's. RY is not affecting any of your measured qubits. $\endgroup$
    – PGibbon
    Nov 24, 2022 at 16:24
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    $\begingroup$ Your question is not clear, what's the issue? What do you mean by "quantumness"? $\endgroup$
    – Dani007
    Nov 25, 2022 at 0:17
  • $\begingroup$ @PGibbon Hello, thanks for the reply, I am aware of what you mentioned but I am messing around with the fact that changing my RY affects the probabilities from q[0],q[1],q[2] and q[3]. $\endgroup$ Nov 27, 2022 at 18:26
  • $\begingroup$ @Dani007 Hello, thanks for the reply, I couldn't think of a word so I used "quantumness" what I am talking about is that when I change my RY from pi/2 the probabilities of the state of qubits 0,1,2 and 3 change. I thought this indicated that changing RY creates noise within the circuit. This is what I meant by "quantumness" adding more of this variable noise into the circuit. $\endgroup$ Nov 27, 2022 at 18:28
  • $\begingroup$ The reason for this change in the probabilities is not "noise", it's entanglement. You're adding correlation between the first 4 qubits and the last qubit. The probability of measuring 0000 and 1111 for the first 4 qubits is 1/2 each but the probability of measuring 0000 and 1111 conditioned on the last qubit being 0 is $cos(\theta /2)^2/2$ and $sin(\theta /2)^2/2$ hence they won't generally be the same. $\endgroup$
    – Dani007
    Nov 27, 2022 at 19:44

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When you add a measurement to the composer, it will act as a simulation with only one shot, hence the behaviour you are observing.

If you want to see the statevector directly computed on the page then I suggest you just remove the measure. Moreover if you want to study a little bit more its elevation then you can click on the button "Inspect" right above the drawing of the circuit (I think).
And if you want to simulate real computation or send on real hardware, then don't forget to put back measures, then all you need to do is click on "Setup and run" (blue button at the top right of the screen).

If you have more questions I'll be happy to help! :)

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  • $\begingroup$ Hello, thanks for the reply, I have another question, what is the reason that adding those CNOT gates inbetween those barriers into the circuit alters the probabilities of the qubits 0,1,2 and 3 when they are just being the control and not the target. $\endgroup$ Nov 27, 2022 at 18:32

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