6
$\begingroup$

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?

$\endgroup$

1 Answer 1

10
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ And this is done in addition to reversing the order of the gates in the QFT circuit, correct? $\endgroup$ Commented Dec 5, 2018 at 23:16
  • $\begingroup$ See the order is 1 through n before and then n through 1 after. $\endgroup$
    – AHusain
    Commented Dec 5, 2018 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.