# Is it possible to modify the QFT circuit to use only 1-qubit gates?

[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.

• Do you have any reference for this result? Feb 28 at 9:09
• 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
– glS
Feb 28 at 11:20
• 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 Feb 28 at 13:04
• 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 $|x\rangle$
– glS
Feb 28 at 16:25
• Could you elaborate? It's supposed to be QFT on the group $\mathbb{Z}_N$ Feb 28 at 19:28