Does QFT exploit entanglement?

Question

I was studying the quantum circuit for the Quantum Fourier Transform (QFT) on the Mike & Ike, and they write the result of the transformation as a product state. More precisely they wrote the output of the QFT in the form $$\frac{1}{\sqrt{2^n}} (\left| 0 \right> + e^{2\pi i 0.j_n} \left| 1\right>) (\left| 0\right> + e^{2\pi i 0.j_{n-1}j_n} \left| 1\right>)...(\left| 0\right> + e^{2\pi i 0.j_1j_{2}...j_n} \left| 1\right>).$$

Is it an entangled state? It seems just the (tensor) product of single-qubit states to me, but I think I am missing something here.

Answer 1

Yes, the formula you have shows that applying QFT to a given computational basis state $|j\rangle = |j_1 j_2 \dots j_n\rangle$ results in an unentangled output state. However when applied to superposition states the output can certainly be entangled.

Note that the same effect occurs for gates like CNOT and CZ: The result of applying these entangling gates to computational basis states is another (unentangled) computational basis state but the gates are still capable of producing entanglement in general.