# Can a QFT be implemented as a physical change of the measurement basis?

Suppose you had an "analog" quantum computer, where a register would store a wavefunction $$\psi(x)$$ where $$x$$ is a continuous variable (with $$0\leq x\leq 1$$, say). Instead of gates, you would somehow apply unitary operators to the wavefunction. In this context, it seems to me that the QFT would essentially be equivalent to changing to the momentum basis, $$\psi(k)$$. This could be done by using a different physical apparatus to measure the state of the register, and wouldn't require the actual application of any unitary operators to the wavefunction. Could there be a similar result for a qubit register which could be measured in two different non-commuting bases? If not, why?

• When you say measure the state of the register? What are you measuring? $\hat{x}$ or $\hat{k}$? It sounds like you are thinking of the function $\psi (x)$ as classically stored by the way this is asked. – AHusain Nov 16 '19 at 21:13
• Yes, classically stored. So instead of a register you might have a sort of 'quantum abacus' which is well-enough isolated from the environment that the beads can be in superpositions of different positions on the rods. You could then use some kind of physical unitary evolution to transform the wavefunction for a single bead $\psi(x)$ into its fourier transform $\tilde{\psi}(x)$ and then measure $\hat{x}$. But alternatively you could use a different measurement apparatus to measure the momentum of the bead (measure $\hat{k}$). – Maxwell Aifer Nov 16 '19 at 21:49
• My question is then whether you could use the same trick for actual qubits (e.g. superconducting loops or ion traps), where the wavefunction is defined on a finite set of values ($\psi[n]$, $0\leq n < 2^N$, where $N$ is the number of cubits). Could you then represent the QFT mapping $\psi[n] \to \psi[k]$ as a physical measurement whose eigenstates are states with definite $k$, and if so, what would the operator be that corresponds to this measurement? – Maxwell Aifer Nov 16 '19 at 21:58
• You can edit your question with these clarifying comments. – AHusain Nov 16 '19 at 22:06