4 ms·

Grid of atoms is both a quantum computer and an optimization solver

58 points by isaacfrond 4y ago

amrb 4y ago
Still find it nuts you need to operate the system at 0 kelvin. I assume if we had super efficient cooling equipment (not LN2) we could also do superconductors on a small scale?
- ta988 4y ago
  LN2 is not enough you need LHe for most superconductors. It is usually cooled by LHe that is itself shieled by LN2 to limit losses. LN2 is relatively cheap not LHe. That's why there are recylcing systems to recover the boiling off He in MRIs, NMRs and other supraconducting systems. There are systems that allow cooling without wasting LHe/LN2 but they are pretty limited in size or costly: https://hexus.net/tech/news/cooling/131414-worlds-first-absolute-zero-magnetic-cooler-developed/ https://hexus.net/tech/news/cooling/131414-worlds-first-abso... https://www.wired.co.uk/article/mri-magnet-cooling https://www.wired.co.uk/article/mri-magnet-cooling https://en.wikipedia.org/wiki/Dilution_refrigerator https://en.wikipedia.org/wiki/Dilution_refrigerator edit: s/LH2/LHe
  - carterschonwald 4y ago
    Liquid He right? Liquid hydrogen is relatively cheap
    - ta988 4y ago
      Thanks will edit
- packetlost 4y ago
  You use lasers to optically trap and cool the atoms, you can get them insanely cold very reliably with methods like this: https://en.wikipedia.org/wiki/Optical_molasses https://en.wikipedia.org/wiki/Optical_molasses https://en.wikipedia.org/wiki/Polarization_gradient_cooling https://en.wikipedia.org/wiki/Polarization_gradient_cooling https://en.wikipedia.org/wiki/Doppler_cooling https://en.wikipedia.org/wiki/Doppler_cooling The components in a system like this are probably operating at room temperature.
- phonebucket 4y ago
  > Still find it nuts you need to operate the system at 0 kelvin. I assume if we had super efficient cooling equipment (not LN2) we could also do superconductors on a small scale? Neutral atom quantum computers (the type outlined in the article) work at room temperature. Superconducting qubit based quantum computers do indeed require cooling to these levels, though.
Gravityloss 4y ago
If we want to simulate molecular systems that have quantum effects, with a quantum computer, is it a good fit? Say we want to create designer molecules for flow batteries. A few benzene rings and some "arms" with a few carbons and hydroxyl group at the end. The potential design space is absolutely humongous. Can a quantum computer simulate something like that relatively directly, and then see how that reacts in various situations, oxidizing and reducing? If you have a system with 20 atoms, how many qubits would you need, and so on? Right now researchers make a best guess, then synthesize the material. That can take months. Then they make experiments, and usually there are some surprises. Machine learning has been proposed to help but probably has limitations...
- jlokier 4y ago
  Little known fact: Simulating quantum physical systems was the original motivation for building quantum computers, because those simulations are infeasible without one, and someone figured out it would be, in principle, feasible to run those simulations using a computer based on quantum effects. There's a neat symmetry to it. We can't be certain quantum computers work at useful scales, until we've built one, because there's no feasible way to simulate the physics of a useful size quantum computer without using one.
  - jacobr1 4y ago
    > there's no feasible way to simulate the physics of a useful size quantum computer without using one. What do the diverging growth curves look like? We probably could brute force simulate the exponential states within a small-n using conventional computing, just by paying for more parallelism and time. At some point that doesn't scale (more computers needed than atoms on earth, for longer than the heat-death of the universe or whatever) but surely there are interesting problems we can model before whatever inflection point exists? If so - what kind of scale is that?
- ta988 4y ago
  In theory yes, but we can't get one with enough qbits (you would need more qbits than your molecule because you need to encode positions and interactions of electrons at the very least), enough configurability (current q-computers are somewhat limited to which qbit can interact with which through which mean) and that is able to keep decoherence away (which limits how much it can scale). It seems that we are really hitting a wall every time we want to scale them to a useful size and keep the number of possible interactions between qbits high and noise/decoherence level low. So for now we don't have the technology (if ever), it is not just a problem of "just scaling by adding more qbits" we need a new approach.
cwillu 4y ago
Smells like d-wave nonsense. ‹finishes reading› Yeah, it's the same nonsense. Reminder: quantum computing is not expected to be able to improve the solutions of NP-complete problems by more than a quadratic factor, unless some commonly accepted conjectures can be proven false. If that happens, you'll be hearing about it from more than just a analog computer company's marketing materials.
whitten 4y ago
So you can make a grid with less than 40 quantum atom pairs, each of which is coupled in some way to another atom, and since measuring one atom guarantees the other atom is in the opposite state, you can bypass Heisenberg uncertainty. But wait. To solve an NP problem of size n, you need square(n) atoms. But 40 atoms is barely more than 36, so you can solve a NP problem of size 6, which is easy and possible. I think brute force on an NP problem is factorial(n) So factorial(6) == 6 ! == 6 * 5 * 4 * 3 * 2 * 1 == 720. Am I misunderstanding something here ?
- karmakaze 4y ago
  The other atom is not in the 'opposite state' for all properties, rather it has one property which is entangled so whatever accuracy you just measured, you know to the same accuracy of the other in the entangled property only. Also 40 bits in superposition can represent 2^40 possibilities.
  - aaplok 4y ago
    > 40 bits in superposition can represent 2^40 possibilities. In classical QC yes, is this the case any longer in the Rydberg blockade configuration? From what I understood from the article, it's what GP said: the atom configuration implements the independent set problem using entanglement. So the number of qubits is the order of the graph, which is the size of the problem. I'm not sure where the quadratic bound is derived from, and I didn't understand where the solution to the maximum independent set comes from so maybe I didn't understand the article... There are so many other gray areas in the article that reading it feels a bit like riding on the hype train.
- petters 4y ago
  > I think brute force on an NP problem is factorial(n) No, NP ⊆ EXPTIME, which means that exponential time should be enough.

58 points by isaacfrond 4y ago

amrb 4y ago

Still find it nuts you need to operate the system at 0 kelvin. I assume if we had super efficient cooling equipment (not LN2) we could also do superconductors on a small scale?

ta988 4y ago
LN2 is not enough you need LHe for most superconductors. It is usually cooled by LHe that is itself shieled by LN2 to limit losses. LN2 is relatively cheap not LHe. That's why there are recylcing systems to recover the boiling off He in MRIs, NMRs and other supraconducting systems. There are systems that allow cooling without wasting LHe/LN2 but they are pretty limited in size or costly: https://hexus.net/tech/news/cooling/131414-worlds-first-absolute-zero-magnetic-cooler-developed/ https://hexus.net/tech/news/cooling/131414-worlds-first-abso... https://www.wired.co.uk/article/mri-magnet-cooling https://www.wired.co.uk/article/mri-magnet-cooling https://en.wikipedia.org/wiki/Dilution_refrigerator https://en.wikipedia.org/wiki/Dilution_refrigerator edit: s/LH2/LHe
- carterschonwald 4y ago
  Liquid He right? Liquid hydrogen is relatively cheap
  - ta988 4y ago
    Thanks will edit
packetlost 4y ago
You use lasers to optically trap and cool the atoms, you can get them insanely cold very reliably with methods like this: https://en.wikipedia.org/wiki/Optical_molasses https://en.wikipedia.org/wiki/Optical_molasses https://en.wikipedia.org/wiki/Polarization_gradient_cooling https://en.wikipedia.org/wiki/Polarization_gradient_cooling https://en.wikipedia.org/wiki/Doppler_cooling https://en.wikipedia.org/wiki/Doppler_cooling The components in a system like this are probably operating at room temperature.
phonebucket 4y ago
> Still find it nuts you need to operate the system at 0 kelvin. I assume if we had super efficient cooling equipment (not LN2) we could also do superconductors on a small scale? Neutral atom quantum computers (the type outlined in the article) work at room temperature. Superconducting qubit based quantum computers do indeed require cooling to these levels, though.

Gravityloss 4y ago

If we want to simulate molecular systems that have quantum effects, with a quantum computer, is it a good fit? Say we want to create designer molecules for flow batteries. A few benzene rings and some "arms" with a few carbons and hydroxyl group at the end. The potential design space is absolutely humongous. Can a quantum computer simulate something like that relatively directly, and then see how that reacts in various situations, oxidizing and reducing? If you have a system with 20 atoms, how many qubits would you need, and so on? Right now researchers make a best guess, then synthesize the material. That can take months. Then they make experiments, and usually there are some surprises. Machine learning has been proposed to help but probably has limitations...

jlokier 4y ago
Little known fact: Simulating quantum physical systems was the original motivation for building quantum computers, because those simulations are infeasible without one, and someone figured out it would be, in principle, feasible to run those simulations using a computer based on quantum effects. There's a neat symmetry to it. We can't be certain quantum computers work at useful scales, until we've built one, because there's no feasible way to simulate the physics of a useful size quantum computer without using one.
- jacobr1 4y ago
  > there's no feasible way to simulate the physics of a useful size quantum computer without using one. What do the diverging growth curves look like? We probably could brute force simulate the exponential states within a small-n using conventional computing, just by paying for more parallelism and time. At some point that doesn't scale (more computers needed than atoms on earth, for longer than the heat-death of the universe or whatever) but surely there are interesting problems we can model before whatever inflection point exists? If so - what kind of scale is that?
ta988 4y ago
In theory yes, but we can't get one with enough qbits (you would need more qbits than your molecule because you need to encode positions and interactions of electrons at the very least), enough configurability (current q-computers are somewhat limited to which qbit can interact with which through which mean) and that is able to keep decoherence away (which limits how much it can scale). It seems that we are really hitting a wall every time we want to scale them to a useful size and keep the number of possible interactions between qbits high and noise/decoherence level low. So for now we don't have the technology (if ever), it is not just a problem of "just scaling by adding more qbits" we need a new approach.

cwillu 4y ago

Smells like d-wave nonsense. ‹finishes reading› Yeah, it's the same nonsense. Reminder: quantum computing is not expected to be able to improve the solutions of NP-complete problems by more than a quadratic factor, unless some commonly accepted conjectures can be proven false. If that happens, you'll be hearing about it from more than just a analog computer company's marketing materials.

whitten 4y ago

So you can make a grid with less than 40 quantum atom pairs, each of which is coupled in some way to another atom, and since measuring one atom guarantees the other atom is in the opposite state, you can bypass Heisenberg uncertainty. But wait. To solve an NP problem of size n, you need square(n) atoms. But 40 atoms is barely more than 36, so you can solve a NP problem of size 6, which is easy and possible. I think brute force on an NP problem is factorial(n) So factorial(6) == 6 ! == 6 * 5 * 4 * 3 * 2 * 1 == 720. Am I misunderstanding something here ?

karmakaze 4y ago
The other atom is not in the 'opposite state' for all properties, rather it has one property which is entangled so whatever accuracy you just measured, you know to the same accuracy of the other in the entangled property only. Also 40 bits in superposition can represent 2^40 possibilities.
- aaplok 4y ago
  > 40 bits in superposition can represent 2^40 possibilities. In classical QC yes, is this the case any longer in the Rydberg blockade configuration? From what I understood from the article, it's what GP said: the atom configuration implements the independent set problem using entanglement. So the number of qubits is the order of the graph, which is the size of the problem. I'm not sure where the quadratic bound is derived from, and I didn't understand where the solution to the maximum independent set comes from so maybe I didn't understand the article... There are so many other gray areas in the article that reading it feels a bit like riding on the hype train.
petters 4y ago
> I think brute force on an NP problem is factorial(n) No, NP ⊆ EXPTIME, which means that exponential time should be enough.