It is well known that entanglement in a quantum state is not affected when you perform a combination of 1-qubit unitary transformations.

I have seen that the QFT can be decomposed into product of 1-qubit unitary transformations (for example in Wiki). However, it is not very clear that QFT preserves entanglement or not. I have seen sources suggesting that it does, doesn't, and unclear.

Does anyone have a definite answer for this question?

  • 1
    $\begingroup$ Those aren't 1-qubit unitaries. There are controls so they are 2-qubit unitaries. $\endgroup$ – AHusain Feb 19 '20 at 22:58

TLDR: the Fourier transform is entangling.

We can immediately agree on two things:

  • if you input a computational basis state (separable) to the Fourier transform, it outputs a separable state
  • the circuit involves entangling gates

Neither of these actually resolves the question. Could there be another separable basis which is converted to entangled states? Could entangling gates combine to give an overall separable operation (example: swap gate being composed of 3 controlled-nots)?

The best way to resolve this is with an example. So, take the simplest example: the Fourier transform on two qubits. The circuit looks like this:

enter image description here

This is easy to analyse because there's only one entangling gate: the controlled-$S$. Any single-qubit gates that come after it do not affect entanglement. Now, if controlled-$S$ acted on an input of $|+\rangle|+\rangle$, it would create entanglement. So, we can work backwards: if I input $(H\otimes I)|+\rangle|+\rangle=|0+\rangle$ to the circuit, the output must be entangled. Separable input yields entangled output, so the circuit is entangling.


I expect you're referring to this image:

from Wikipedia

Where it says that $R_m$ is given by a particular 2x2 matrix. In this notation, $R_m$ is a 1-qubit unitary, but the circuit is applying a different gate, what we might call $CR_m$: a controlled version of $R_m$. This is a 2-qubit gate, given by a 4x4 matrix, with the definition

$$ CR_m (|0\rangle |\psi\rangle) = |0\rangle |\psi\rangle$$ $$ CR_m (|1\rangle |\psi\rangle) = |1\rangle (R_m |\psi\rangle)$$

This is part of the circuit notation, but not particularly clear in the article; I'll edit it now. But anyway, this is a 2-qubit gate, so it does create entanglement.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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