The (classical) Fourier Transform is famous for the reversible switch between the time-domain and frequency-domain of time dependent functions. Likewise there would be a change in domain associated with the QFT. I’d like to say between the computational basis and the (what?). I have never seen it described in this way.
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It exchanges the computational basis and the frequency basis, just like it does classically. It conjugates clock matrices into shift matrices and vice versa.
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$\begingroup$ @Craig, Totally agree. A bit dynamic oriented interpretation would be position basis to momentum basis change. (or, vice versa; depending on how the transformation is defined. ) Although this interpretation might not give direct insight into the problem being solved via QFT, most of the time. $\endgroup$ Commented Jun 12 at 19:36
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The QFT is just a classical Fourier transform, specifically a discrete Fourier transform (DFT): The DFT transforms an $N$-component vector $(a_x)_{x=1,\dots,N}$ as $$ (a_x)\mapsto b_p = \frac{1}{\sqrt{N}} \sum e^{i n x p} a_x\ . $$ This transformation is unitary, and it is precisely this unitary which the QFT implements.
(The point, of course, is that is (i) does so on a vector in Hilbert space, which describes the state of a quantum mechanical system, and (ii) that it can be implemented efficiently if the state is stored in a register of qubits.)
If you want an interpretation in the quantum mechanical context, I would say the best perspective is to exchange $\hat x$ and $\hat p$ (i.e., position and momentum) in the finite dimensional Heisenberg-Weyl group (i.e., the shift and the phase operation). In that sense, it is better to think of the "conventional" Fourier transformation as exchanging position and momentum, rather than time and frequency. (This operation also makes sense in infinite dimensions, indeed, the "classical" Fourier trafo, when applied to a single-particle wavefunction, does precisely that.)
answered Jun 12 at 21:42