Preferably using the gates found in QASM/Qiskit and the qubit stays in the superposition and no measurement is made, i.e. the output of if it is in a superposition or not is a binary answer in another qubit.

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    $\begingroup$ The short answer for a general superposition is no, you can‘t. To obtain information you have to measure and that will affect your qubit. $\endgroup$ – Cryoris May 28 at 8:26

The easiest thing to do would be to check the circuit or just run several qubits through the circuit and measure these qubits. If checking the circuit or running several qubits through the circuit is not possible, then I think applying a CNOT gate to said qubit and a predetermined qubit should allow us to know. Set said qubit to be the control qubit and the predetermined qubit as the target. If the control qubit is in superposition, the final state of these qubit will be in a Bell state.


After the edit to the question, assuming there is no measurement at all, then the answer would be no. However, if what you mean is that we cannot measure or alter only said qubit, then I think we can try the following circuit:

![the circuit

Let the first qubit be the qubit in question. The last qubit will be measured later to determine whether the first qubit is in superposition or not. If the last qubit is always |1⟩ or |0⟩, then the first qubit is not in superposition. This is because the final state of the circuit is:

$|\psi⟩ = \frac{\alpha}{\sqrt{2}}|0⟩(|00⟩+|10⟩) + \frac{\beta}{\sqrt{2}}|1⟩(|11⟩+|01⟩)$

Please let me know if I make any mistake as I am also new to this.


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