## Could someone give an example of this pic?

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1If you apply a unitary $F$ to the first qubit (q), then your state would be the same as if you applied the conjugate transpose $F^\dagger$ to the second qubit (q). Can you be more specific? – Mark S – 2020-04-12T02:05:26.383

could you please edit the title to something that reflects what is being asked? – glS – 2020-04-13T08:01:12.580

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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 $$q_0$$) First circuit (inverse Fourier transform applied on qubit $$q_1$$) 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.