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When implementing the inverse quantum Fourier transform, in addition to reversing the circuit, does one need to take the conjugate transpose of the phase shift gates in the circuit as well?

Sanchayan Dutta
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Jack Ceroni
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1 Answers1

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Yes.

You have been given a factorization $QFT=U_1 \cdots U_n$ where each $U_i$ is an individual gate.

$$ QFT^{-1} = U_n^{-1} \cdots U_1^{-1}\\ = U_n^{\dagger} \cdots U_1^{\dagger}\\ $$

A lot of the individual gates will have the property that $U_i = U_i^\dagger = U_i^{-1}$. These are the involutions like NOT, CNOT, etc. In those cases you are lucky and don't have to worry about conjugating.

In other gates, you do have to change parameters in order to acheive the conjugate transpose. Like $R_Z (\beta)$ vs $R_Z (-\beta)$. In these sorts of basic gates, the conjugate transpose is straightforward to see how parameters like Euler angles change.

AHusain
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