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This is a picture from Wiki(https://en.wikipedia.org/wiki/Quantum_logic_gate). Can someone give me a simple example by using two qubits?

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    $\begingroup$ If you apply a unitary $F$ to the first qubit (q[0]), then your state would be the same as if you applied the conjugate transpose $F^\dagger$ to the second qubit (q[1]). Can you be more specific? $\endgroup$ – Mark S Apr 12 at 2:05
  • $\begingroup$ could you please edit the title to something that reflects what is being asked? $\endgroup$ – glS Apr 13 at 8:01
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Since Fourier transform and inverse Fourier transform for one qubit is only Hadamard gate, for two qubit case following two circuits are equivalent.

First circuit (Fourier transform applied on qubit $q_0$)

Circuit 1

First circuit (inverse Fourier transform applied on qubit $q_1$)

Circuit 2

Both circuits return state

$$ |\psi\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle - |11\rangle). $$


EDIT: I have just realized that the gate $F$ is general unitary transformation and not the QFT (I was missleaded by F = Fourier). However, my example is also valid. It is a particular case for two qubits asked for in the question.

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