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?
TLDR: the Fourier transform is entangling.
We can immediately agree on two things:
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:
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:
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.
The OP was asking about preserving entanglement that no one has answered yet. Here is my attempt to answer and initiate a discussion.
In my opinion, QFT does not preserve entanglement (no proof),inferred via a numerical test. Create a Bell state and compute EOF (entanglement of formation) which should be 1 since it is maximally entangled. Apply QFT to this state and measure new EOF. EOF < 1 implies loss of entanglement.
