Timeline for How to compute the energy of a non-eigenstate in a Heisenberg limited form?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 13 at 14:23 | vote | accept | Pablo | ||
| Aug 9 at 11:07 | answer | added | Pablo | timeline score: 1 | |
| Aug 8 at 14:37 | history | became hot network question | |||
| Aug 8 at 10:51 | answer | added | Norbert Schuch | timeline score: 4 | |
| Aug 8 at 10:38 | comment | added | Norbert Schuch | One naive attempt could be to take $|\psi\rangle^{\otimes N}$ and do phase estimation for $\mathcal H = N^{-1}\sum H_i$ ($H_i$ is $H$ acting on the $i$th copy.) For each eigenstate of $\mathcal H$, this will give Heisenberg-limited accuracy, and my feeling is that a central limit type argument should tell that the error from averaging the different eigenstates concentrates rapidly. | |
| Aug 8 at 10:27 | comment | added | Norbert Schuch | BTW, is your question really "What is the best way to achieve a Heisenberg-limited measurement", or rather "Is it possible to achieve ... "? | |
| Aug 8 at 10:24 | comment | added | Norbert Schuch | It seems that if you want to get a better accuracy than $1/\epsilon^3$ you will need protocols which use several copies of $|\psi\rangle$ at once, since otherwise there is nothing which distinguishes $|\psi\rangle$ from a mixture of eigenstates, so repeated phase estimation and averaging should be optimal in that case. --- Does the Grover method give you $1/\epsilon^2$? | |
| Aug 7 at 23:32 | answer | added | Mark Spinelli | timeline score: 1 | |
| Aug 7 at 21:11 | comment | added | Pablo | I think it is! In Quantum Phase Estimation you need to call the unitary $U = e^{-iH\tau}$ $O(1/\epsilon)$ times to achieve precision $\epsilon$ in the measurement. The thing is, when you do QPE, you will measure the energy of one eigenstate in the superposition, so you'd need to sample $O(1/\epsilon^2)$ many points to get an expectation value of the average of the distribution. So overall the cost would be $O(1/\epsilon^3)$. The question is improving this to inverse linear. | |
| Aug 7 at 19:35 | comment | added | Mark Spinelli | Thanks! My poor-man's "computer-science" mind is wondering what you mean by the Heisenberg limit? There's a time-energy uncertainty in the precision you'd get from the Quantum Phase Estimation even if $|\psi\rangle$ where promised to be an eigenstate of $H$, because you'd need to actually simulate $\exp(-iHt)$ many times to avoid conflict with the No-Fast-Forwarding Theorem. But I don't think that's what you mean by the Heisenberg limit, is it? | |
| Aug 7 at 19:19 | comment | added | Pablo | Not dumb, it's just that the scaling would be $O(1/\epsilon^2)$, right? And we'd prefer to achieve $O(1/\epsilon)$. | |
| Aug 7 at 17:57 | comment | added | Mark Spinelli | Dumb question but what's wrong with sampling? | |
| Aug 7 at 16:12 | history | asked | Pablo | CC BY-SA 4.0 |