[Measured Quantum Fourier Transform]

I've recently learned the Quantum Fourier Transform, and was shown its circuit. The circuit I've seen is composed of Hadamard gates and controlled Rotation gates.

I saw in some exercises (Nielsen & Chuang; problem 5.2) that it's possible to modify the QFT circuit to use only 1-qubit gates, if we want to measure it in the computational basis.

I tried to think how to do it, but I'm having trouble to understand how it's possible. Isn't any conversion of the controlled rotation gate result in an un-reversible gate?

Help would be appreciated.

  • $\begingroup$ Do you have any reference for this result? $\endgroup$
    – Tristan Nemoz
    Feb 28 at 9:09
  • 3
    $\begingroup$ please add a reference do where you saw this statement. The way you're stating it, it's not possible: a product of single-qubit unitaries gives a product unitary, which the QFT is not $\endgroup$
    – glS
    Feb 28 at 11:20
  • $\begingroup$ The full statement was "Show that if we want to measure the output of the QFT in the computational basis then we can modify the circuit to use only 1-qubit gates". I might have miss written it $\endgroup$
    – Gabi G
    Feb 28 at 13:04
  • 1
    $\begingroup$ it's hard to tell, but it might refer to the fact that the output of the QFT is writable as a tensor product for computational basis inputs. Ie $\operatorname{QFT}_n|x\rangle=\bigotimes_k (|0\rangle+\omega^{2^k x}|1\rangle)$, $\omega\equiv\exp(2\pi i/N)$. So for fixed input $|x\rangle$, and measuring in the computational basis, the output probabilities are uncorrelated, hence can be simulated with only local operations. But that only works for a fixed input $|x\rangle$ $\endgroup$
    – glS
    Feb 28 at 16:25
  • $\begingroup$ Could you elaborate? It's supposed to be QFT on the group $\mathbb{Z}_N$ $\endgroup$
    – Gabi G
    Feb 28 at 19:28

1 Answer 1


This is called qubit recycling (when combined with introducing the qubits-to-QFT one by one). All you have to do is take the normal circuit, measure each qubit immediately after it gets Hadamard'ed, and use the deferred measurement principle to turn the CPHASE gates into classically-controlled single-qubit phase gates. Like this:

enter image description here


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