4 ms·

I'm not sure where you're getting your information: everyone in this thread seems to be parroting the same popsci headline that it's not a "real" quantum comput

I'm not sure where you're getting your information: everyone in this thread seems to be parroting the same popsci headline that it's not a "real" quantum computer, but that completely misses the point. Regardless of whether or not it's a general purpose computing machine with programmable logic, it can still perform mathematical optimization of an objective function over a superposition of bits. Saying that it's not quantum because it's not general purpose is a complete distortion of the facts.

kirrent 10y ago
Whether or not you agree that D-wave has proper superposition and is properly performing quantum annealing your original statement was different from this one. You wrote: "it can test the superposition of the entire search space instead of picking at points here and there." That's similar to common misconceptions on how quantum computers work which might attract some of your downvotes, but more importantly you're also misrepresenting why quantum annealing is faster than thermal annealing.
krastanov 10y ago
Here is my claim in precise terms: superposition of all combinatorial states is a very non-classical state. Preparing such a state is still an active area of research, and nobody has yet made one in a usable fashion. The D-wave device does not have a superposition of states, rather it is tracking a single classical state as far as I know. No need for harsh language in your response. Edit: removed an incorrect remark about entanglement.
- vhfnfnncs 10y ago
  >The D-wave device does not have a superposition of states, rather it is tracking a single classical state I'm going to need a citation on that. Even if only some small fraction of the qubits in the 2^2000 search space are fully entangled, that's still a massive improvement over the current state of the art using conventional computers because of how absolutely huge the space is.
  - johncolanduoni 10y ago
    Not if the entanglement is partitioned into very small groups of qubits. That's the only thing we have even circumstantial evidence of, and it doesn't appear to gain us much over classical computers. This is especially bad for the "it'll be better with more qubits" hypothesis D-Wave is chasing, since everything we know about quantum control suggests that if you can't solve this problem with a 1000-qubit system, doing it with a 2000-qubit system is going to be even harder. We're still waiting for one problem, no matter how arbitrary, where the D-Wave beats basic consumer computers, let alone a few million dollars worth of computing power. And if D-Wave (or anyone else) had found one for their 2000-qubit computer, it would be in this press release. So no, any gain in performance form the D-Wave hardware is still in the "theoretical but without any theoretical backing" stage.
    - logicallee 10y ago
      >We're still waiting for one problem, no matter how arbitrary, where the D-Wave beats basic consumer computers This is like saying that NASA has never gone to space, because there is not ONE thing they have EVER done in space, no matter how arbitrary, (including going to the moon) that I care about more than literature. Let's look at the parts of my analogy: NASA has not gone to space BECAUSE it is not better than something irrelevant. DWAVE does not have a quantum computer BECAUSE it is not better than something irrelevant. What on Earth do classical computers have to do with whether DWAVE is using quantum effects? This is very irrelevant. I can't believe so many people use this to "prove" that it's not a quantum computer. This is a complete and utter non sequitur. This is very basic logic. It's like saying that until the minute a machine beat Kasparov at chess, computers couldn't play chess. In a very narrow sense of cheating, sure, a computer could fake playing chess if a person is hidden and playing for the computer. Maybe the comouter isn't even powered on. Likewise, in a narrow sense a classical computer could be hidden in the DWAVE's cabinet and producing the output with a python script. Maybe the DWAVE isn't even powered on. (and has a cheater inside, a macbook.) But short of this scenario it is a completely inappropriate point to make. It doesn't say anything about how dwave works. -> Are you accusing them of not even powering it on and hiding a macbook in the case, which produces all its results? (if not, how is your contention different.)
johncolanduoni 10y ago
I'm parroting the information I've gotten from hearing people who work with the D-Wave describe their research, reading their papers, and friends who work in the area. So no, I didn't just read a popsci headline that tickled my contrarian bone.
chriswarbo 10y ago
> Saying that it's not quantum because it's not general purpose is a complete distortion of the facts. Being not-general-purpose means it's not a computer. If it's not a computer, it can't be a quantum computer. As far as I can tell, D-wave's machines do appear to use quantum effects, but that doesn't make them quantum computers. They're quantum calculators.
- johncolanduoni 10y ago
  A non-general quantum computer could still be considered to be a "quantum computer" in the way that physicists use the term. We'd definitely consider a computer that could "only" run Shor's algorithm to be one, so that's not a reason to write off the D-Wave.

kirrent 10y ago

Whether or not you agree that D-wave has proper superposition and is properly performing quantum annealing your original statement was different from this one. You wrote: "it can test the superposition of the entire search space instead of picking at points here and there." That's similar to common misconceptions on how quantum computers work which might attract some of your downvotes, but more importantly you're also misrepresenting why quantum annealing is faster than thermal annealing.

krastanov 10y ago

Here is my claim in precise terms: superposition of all combinatorial states is a very non-classical state. Preparing such a state is still an active area of research, and nobody has yet made one in a usable fashion. The D-wave device does not have a superposition of states, rather it is tracking a single classical state as far as I know. No need for harsh language in your response. Edit: removed an incorrect remark about entanglement.

vhfnfnncs 10y ago
>The D-wave device does not have a superposition of states, rather it is tracking a single classical state I'm going to need a citation on that. Even if only some small fraction of the qubits in the 2^2000 search space are fully entangled, that's still a massive improvement over the current state of the art using conventional computers because of how absolutely huge the space is.
- johncolanduoni 10y ago
  Not if the entanglement is partitioned into very small groups of qubits. That's the only thing we have even circumstantial evidence of, and it doesn't appear to gain us much over classical computers. This is especially bad for the "it'll be better with more qubits" hypothesis D-Wave is chasing, since everything we know about quantum control suggests that if you can't solve this problem with a 1000-qubit system, doing it with a 2000-qubit system is going to be even harder. We're still waiting for one problem, no matter how arbitrary, where the D-Wave beats basic consumer computers, let alone a few million dollars worth of computing power. And if D-Wave (or anyone else) had found one for their 2000-qubit computer, it would be in this press release. So no, any gain in performance form the D-Wave hardware is still in the "theoretical but without any theoretical backing" stage.
  - logicallee 10y ago
    >We're still waiting for one problem, no matter how arbitrary, where the D-Wave beats basic consumer computers This is like saying that NASA has never gone to space, because there is not ONE thing they have EVER done in space, no matter how arbitrary, (including going to the moon) that I care about more than literature. Let's look at the parts of my analogy: NASA has not gone to space BECAUSE it is not better than something irrelevant. DWAVE does not have a quantum computer BECAUSE it is not better than something irrelevant. What on Earth do classical computers have to do with whether DWAVE is using quantum effects? This is very irrelevant. I can't believe so many people use this to "prove" that it's not a quantum computer. This is a complete and utter non sequitur. This is very basic logic. It's like saying that until the minute a machine beat Kasparov at chess, computers couldn't play chess. In a very narrow sense of cheating, sure, a computer could fake playing chess if a person is hidden and playing for the computer. Maybe the comouter isn't even powered on. Likewise, in a narrow sense a classical computer could be hidden in the DWAVE's cabinet and producing the output with a python script. Maybe the DWAVE isn't even powered on. (and has a cheater inside, a macbook.) But short of this scenario it is a completely inappropriate point to make. It doesn't say anything about how dwave works. -> Are you accusing them of not even powering it on and hiding a macbook in the case, which produces all its results? (if not, how is your contention different.)

johncolanduoni 10y ago

I'm parroting the information I've gotten from hearing people who work with the D-Wave describe their research, reading their papers, and friends who work in the area. So no, I didn't just read a popsci headline that tickled my contrarian bone.

chriswarbo 10y ago

> Saying that it's not quantum because it's not general purpose is a complete distortion of the facts. Being not-general-purpose means it's not a computer. If it's not a computer, it can't be a quantum computer. As far as I can tell, D-wave's machines do appear to use quantum effects, but that doesn't make them quantum computers. They're quantum calculators.

johncolanduoni 10y ago
A non-general quantum computer could still be considered to be a "quantum computer" in the way that physicists use the term. We'd definitely consider a computer that could "only" run Shor's algorithm to be one, so that's not a reason to write off the D-Wave.