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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?

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  • $\begingroup$ 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. $\endgroup$ – AHusain Nov 16 '19 at 21:13
  • $\begingroup$ 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 $\hat{k}$). $\endgroup$ – Maxwell Aifer Nov 16 '19 at 21:49
  • $\begingroup$ 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? $\endgroup$ – Maxwell Aifer Nov 16 '19 at 21:58
  • $\begingroup$ You can edit your question with these clarifying comments. $\endgroup$ – AHusain Nov 16 '19 at 22:06
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Yes, you can in principle do any measurement as a direct physical operation. The real obstacle is to come up with a way of actually doing it in practice.

The basis states of QFT-ish measurement span every qubit in the register. This means that whatever physical mechanism you are using must be merging together information as large as the register. But physics is local, and your measurement is not, implying you need to build up your big information-merging measurement out of many local interactions. That set of local interactions will probably be isomorphic to quantum gates.

For example, you can do a QFT-ish measurement on photons by putting a diffraction grating between them and the detectors... but isn't putting down the grating just a ultimately roundabout way of applying a gate before measuring?

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  • $\begingroup$ Thanks for your response, this is quite helpful. I see how the measurement might well end up being exactly like the gates it was meant to replace. I guess the question is really about where you draw the line between the "hardware" and "software" of a quantum computer. That said, can we still say there is a lower bound on the complexity of the QFT or similar operations if there is, in principle, a hypothetical machine which performs it in O(1)? $\endgroup$ – Maxwell Aifer Nov 15 '19 at 22:19
  • $\begingroup$ @MaxwellAifer That seems very unlikely, because in order to verify that you are in a particularly frequency eigenstate you need to fold information from all N qubits. Using fixed-size arbitrary-distance interactions this takes at least O(log N) depth due to standard circuit bounds. Using fixed-distance interactions it takes at least O(N^(1/3)) time due to the light speed barrier in 3d space. $\endgroup$ – Craig Gidney Nov 15 '19 at 22:32
  • $\begingroup$ Wow, I had not thought of the light speed barrier, that's fascinating. I will have to read up on these circuit bounds, thanks again $\endgroup$ – Maxwell Aifer Nov 15 '19 at 22:38

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