# Why does taking a measurement completely change my probabilities of states

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" 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.

• 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 $q, q, q$ and $q$ are the H, and you first, second and last CNOT's. RY is not affecting any of your measured qubits. Nov 24, 2022 at 16:24
• Your question is not clear, what's the issue? What do you mean by "quantumness"? Nov 25, 2022 at 0:17
• @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,q,q and q. Nov 27, 2022 at 18:26
• @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. Nov 27, 2022 at 18:28
• 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. Nov 27, 2022 at 19:44