14 ms·

Chinese Researchers Achieve Quantum Entanglement Record

305 points by 0xbxd 8y ago

foxes 8y ago
I visited USTC several years ago, (2014) and in this time they have basically doubled the number of qubits they can entangle, so this is a great step forward. At around ~100 qubits quantum computing becomes very useful, so maybe we won't have to wait too long.
- ssdd 8y ago
  That's great. Now next thing that needs to be solved is to make this tech available for general public use just like computers are these days. Problems that are waiting to be solved by it are very large and of huge impact.
  - reaperducer 8y ago
    Problems that are waiting to be solved by it are very large and of huge impact I know virtually nothing about quantum computing, so can you give some examples? Whenever I've asked anyone who seemed to know anything about the field, all they can come up with is weather forecasting and simulating nuclear explosions. Not exactly "general use," as you put it. In the back of my mind, I know that quantum computing is a big deal, and we're in ENIAC days with it. But I don't have a good understanding of where it goes or why.
    - adrianN 8y ago
      Quantum computers are extremely useful for simulating chemistry. They can do it exponentially faster than classical computers. Cheap useful quantum computers can revolutionize biology and materials science.
      - deleted 8y ago
        [deleted]
    - chrisweekly 8y ago
      ENIAC?
      - SimbaOnSteroids 8y ago
        One of the first computers. https://en.wikipedia.org/wiki/ENIAC https://en.wikipedia.org/wiki/ENIAC
      - deleted 8y ago
        [deleted]
    - techbio 8y ago
      Protein folding and gene research. Cure malaria, cancer, feed the world etc.
    - foxes 8y ago
      There are already lots of 'practical' applications in the literature. A famous one is Shor's factorisation algorithm. But there are others - everything from doing pagerank to simulating other quantum systems. Many classical algorithms, which run in ~O(poly(n)) can have an 'equivalent' quantum algorithm in ~O(log(n)) - an exponential speedup. There is still debate as to how the complexity class of problems which are efficient on a quantum computer (BQP) relate to other complexity classes. Its suspected P lies entirely within BQP. However, at least initially, I think quantum computers will be a specialized piece of equipment. Classical computing is pretty good for your average person. Quantum computers will be used for more heavy compute tasks.
      - bdamm 8y ago
        Is Shor's an application of entanglement though?
        teraflop 8y ago
        Yes, all non-trivial quantum computing is based on entanglement.
      - zjaffee 8y ago
        I'd also add that the need for localized personal computing services goes down as network bandwidth increases, if there's anything that people really need quantum computers for, it's not unreasonable to suggest that they will be handled on the backend with aggregated results sent to clients as is the case for most computation today.
      - moefh 8y ago
        > Its suspected P lies entirely within BQP. It's actually known that BQP contains P[1]. It also contains BPP. What's not known is the relationship between BQP and NP (most experts suspect there's no containment in either direction). [1] See this for an easy proof: https://people.eecs.berkeley.edu/~vazirani/f04quantum/notes/lec5/lec5.pdf https://people.eecs.berkeley.edu/~vazirani/f04quantum/notes/...
        8y ago
    - kart23 8y ago
      Doesn't basically all encryption become meaningless with quantum computers? Remember how they always said 'it would take a billion years to crack this encryption with a computer'? Well soon we are going to have computers that don't play by the rules.
  - jblow 8y ago
    We all have these things in our pockets that are massive supercomputers by the standards of my college days. And we all use them just to rant at each other about political things that we in reality are totally misinformed about ... and otherwise just waste our lives. I am not sure why quantum computing is supposed to change this. Even software that is supposed to be useful is so terribly slow. Computers are between 2 and 4 orders of magnitude faster than programmers today experientially believe they are, because today’s culture of programming has rotted so thoroughly. Do you really need a quantum computer when 3 orders of magnitude are just sitting there on the table waiting to be picked up?
    - ahelwer 8y ago
      Quantum computers won't change anything about people posting misinformed political opinions. Three orders of magnitude speedup is a constant factor. Quantum computers provide between sqrt(n) and log(n) speedup depending on the problem - for example, biochemistry simulation for improved drug designs and such. That seems worthwhile.
    - Erlich_Bachman 8y ago
      > And we all use them just to rant at each other about political things that we in reality are totally misinformed about ... and otherwise just waste our lives. So if I decide to use the device that I have bought and invested in, to "rant" about something (to connect to other people through it, essentially), or even just connect to my loved ones sending "meaningless" information (as many I'm sure would call it), or if I choose to use it for anything else, who is to say that it's "wasting my life"? You? Why do you decide what is wasteful and what is not? Maybe all those people want to do those things? You can't say that this technology is wasted or is not used properly without also implicitly assuming moral and philosophical authority about what people should choose and not to choose with their time and other resources, including the money used to buy such devices and resources spend on developing them. Why do you assume that you can be such an authority? Of course, looking at this in another way, you definitely have the ultimate authority in this area in this one regard: where it applies to your own life. Which I guess is another way of saying that what you wrote says more about your outlook on life than on the underlying technology and its social ramifications.
  - Cobord 8y ago
    Because of the expense it will have to start off as QCaaS. You send the really hard part of your problem to whichever person actually has the hardware and they'll return the answer. If you are concerned with them not knowing your data, there are workarounds to that as well. So available to general public in a weaker sense.
- bscphil 8y ago
  What's the size needed for cracking large-primes based encryption? If not at 100 qubits, what are some useful calculations that we can do at that size?
  - deepGem 8y ago
    So the number of raw qubits are in the order of a million for large prime factorization or to make Shor's algorithm a reality. This is because the physical qubits are highly unstable and prone to errors. The error correction methodologies will result in a fault-tolerant logical qubit. Depending on the error correction algorithm, the ratio of logical: data/physical qubit is 1:1000 at the minimum. You need 4000 logical qubits at least for large prime factorization give or take 4 million raw qubits.
    - amelius 8y ago
      How fast do you think the number of qubits will grow over the coming years? Will it be exponential, like Moore's law?
      - misterprime 8y ago
        There was an AMA not long ago with someone in the quantum computer industry and iirc he stated that it isn't certain if we even can go that high. Guessing how fast that will happen is not reasonable at this point.
    - bscphil 8y ago
      This is probably a dumb question, but could you spread out the error correction in time? I know it's hard to create very large entangled systems, so would it be possible/easier to do the calculation with 4000 real qubits 1000 times and use the repetition for error correction? (Checking for correct factorization is easy, after all.)
      - simcop2387 8y ago
        That's been one strategy I've seen mentioned in regards to Shor's algorithm many times. Since there's the probability of an error, you still have to check it and rerun it a few times until you confirm the right answer. That's still many times faster than you'd get conventionally, so it's a reasonable trade off. But it doesn't address the possibility that you can't entangle enough qubits to begin with, or keep them stable long enough to actually perform the calculations, which is where the error correction stuff comes in. I don't know enough about the challenges there to comment on how many you'd need (though 1000:1 seems high to my naive intuition, i'd guess maybe 10-100 for magnitude myself).
downrightmike 8y ago
Even neater is that they are using quantum entanglement in radar to detect stealth planes: https://www.popsci.com/china-quantum-radar-detects-stealth-planes-missiles#page-2 https://www.popsci.com/china-quantum-radar-detects-stealth-p...
- misterprime 8y ago
  From the article: they are planning to do this, and hopeful that it will work. Not actually using it yet.
  - seanmcdirmid 8y ago
    Also, if they got it to work they would surely keep it secret.
    - JackFr 8y ago
      Also, if they didn't get it to work they would surely keep it secret.
      - seanmcdirmid 8y ago
        Yes, and if the military was involved at all, they wouldn’t even talk about the possibility.
- barbegal 8y ago
  I understand the core concept of quantum radar but does it actually work in practice as a radar? i.e. you can resolve targets with a small range and angular resolution.
  - mike00632 8y ago
    Theoretically you would combine the new tech with the old tech. Quantum state would be another measurement to make on the incoming wave on top of measurements like frequency and amplitude.
angrygoat 8y ago
Here's a pre-print of the USTC team's paper: 18-qubit entanglement with photon's three degrees of freedom https://arxiv.org/abs/1801.04043 https://arxiv.org/abs/1801.04043
- deleted 8y ago
  [deleted]
ArtWomb 8y ago
Congrats! US House science committee also set to pass national initiative in quantum tech with on order of $1B funding: National Quantum Initiative - Action Plan https://www.lightourfuture.org/getattachment/85484dca-465a-46f4-8c8c-090aeb845d09/FINAL-Action-Plan-for-a-NQI-Apr-3-2018.pdf https://www.lightourfuture.org/getattachment/85484dca-465a-4...
ahelwer 8y ago
Crash course in entanglement: The state of each qbit is represented by a state vector of two complex numbers [a, b] where |a|^2 + |b|^2 = 1. There are two special qbit values called the classical basis: [1, 0] which is the classical bit 0, and [0, 1] which is the classical bit 1. If a qbit is not in one of the two classical states, we say it is in superposition. When a qbit is in superposition, we can measure it[0] and it will collapse probabilistically to 0 or 1; for a qbit [a, b], the probability that it collapses to 0 is |a|^2 and the probability that it collapses to 1 is |b|^2. Things get more interesting when we have multiple qbits. If we have two qbits [a, b] and [c, d], we define their product state as their tensor product [ac, ad, bc, bd]. For example, if we have two qbits both in state [1/sqrt(2), 1/sqrt(2)], their product state would be [1/2, 1/2, 1/2, 1/2]. We use the product state to calculate the action of a quantum logic gate that operates on multiple qbits - for a gate which operates on two qbits, we can always represent its action as a 4x4 matrix. Usually we can move back and forth between the product state representation and writing out the individual qbits states. However, in certain scenarios something very special happens: we cannot factor the product state back into the individual state representation! Consider the product state [1/sqrt(2), 0, 0, 1/sqrt(2)]. If you try to write this as a tensor product of two states [a, b] and [c, d], you cannot! It cannot be factored; the qbits have no individual value, and we say they are entangled. Well, what does this mean? It means when you measure one qbit, even if the qbits are very far apart, you instantly know the value of the other qbit. So if I entangled two qbits in the state [1/sqrt(2), 0, 0, 1/sqrt(2)], give you one, then we go to opposite ends of the universe, if I measure my qbit and see a 0 I'll know your qbit instantly also collapsed to 0 (or collapsed to 1 if I measured 1). This phenomenon has been experimentally-verified to occur faster than light. It is instantaneous, as far as we can tell. So, local realism is wrong! Spooky action at a distance is real! There is an important caveat: while the qbits seem to coordinate in some faster-than-light way, you cannot use this to communicate in a faster-than-light way. All we have is a shared random number generator. I can't send some chosen bit from my reference frame to yours. This is called the no-communication theorem. If you found this interesting, I have a full video on quantum computing for computer scientists here: https://youtu.be/F_Riqjdh2oM https://youtu.be/F_Riqjdh2oM [0] IDGAF about your chosen quantum mechanics interpretation, don't @ me
- bdamm 8y ago
xpuente 8y ago
uhmmm ... https://www.quantamagazine.org/gil-kalais-argument-against-quantum-computers-20180207/ https://www.quantamagazine.org/gil-kalais-argument-against-q...
- Cobord 8y ago
  Take that more as a challenge and a caution to avoid overhype.
nabla9 8y ago
Question: Why superfluid 4He is not quantum-entangled? What differentiates macroscopic many-body quantum states from many-body quantum entanglement?
- barbegal 8y ago
  Superfluid 4He is quantum-entangled and non-local effects are observable. https://www.sciencedaily.com/releases/2017/03/170321110344.htm https://www.sciencedaily.com/releases/2017/03/170321110344.h...
  - nabla9 8y ago
    So why it's not taken into account when quantum-entanglement records are broken? You can have flasks of superfluid 4He with thousands of moles of entangled atoms clearly visible.
    - barbegal 8y ago
      Quite correct but usually quantum-entanglement records are for qubits rather than generic atoms. And qubits must be able to be measured in isolation e.g. the polarisation of a single photon can be measured. With superfluid 4He you can't isolate the qubits.
      - nabla9 8y ago
        Ah, that makes sense. Thank you.
tzakrajs 8y ago
> “It’s as though you took six bits in your computer, but each bit tripled in how much information it could hold,” Schreppler said "Each bit is tripled in how much information it could hold"? Normally bits have two states, but Shreppler is claiming these qubits can have six states? What am I missing?
- albertgoeswoof 8y ago
  I thought qubits had an infinite range of states?
  - tzakrajs 8y ago
    Oh yeah, thats right.. Isn't it based on orientation in a spherical sense? I guess you could have an infinite number of those orientations.
    - whatshisface 8y ago
      A pure quantum state is a basis vector, written like |n>. It's called a basis vector because it shares a list of mathematical properties with the x̂,ŷ,ẑ coordinate basis vectors you might be familiar with. They can be combined, (like coordinate vectors) by writing coefficients before them. For example, 1/√2 |a> + 1/√2 |b> would be a combination of state a and state b. Combinations of states obey the condition that the sum of the squares of each coefficient equals one. This is the unitary constraint, and above you can see that 1/√2^2 + 1/√2^2 = 1/2 + 1/2 = 1. Coefficients can be complex numbers, which consist of a real part and an imaginary part. So, putting all of this together, a qbit that may be in the state |1> or the state |0> may also be in the state (a+bi)|1> + (c+di)|0>. With the unitary constraint, we can cut out a degree of freedom by introducing the equation |a+bi|^2 + |c+di|^2 = 1. This leaves three degrees of freedom, which can be mapped in to a sphere by a change of coordinates. Three spherical coordinates plus the unitary constraint specifies the four-number state combination.
      - tzakrajs 8y ago
        Very insightful, thank you!
xtrapolate 8y ago
Is entanglement an actual physical phenomenon? What exactly needs to happen in-order for a photon to "become" entangled?
- dvh 8y ago
  Physical? As in everyday Newtonian physical? It doesn't work on quantum levels, just like Bohr's model of atom doesn't work during big bang because gravity we were successfully ignoring suddenly is on the same scale or bigger then electrostatic force. We will never be able to intuitively (physically) grasp quantum world because our intuition was trained on trees, rocks and cats.
- mundo 8y ago
  Not sure what "actual physical phenomenon" means, but probably no. If you mean can you look at a photon and observe something about it that tells you it's entangled to some other photon, no. If you mean do we know what physical process underlies entanglement, also no. If you mean does entanglement really happen as opposed to being theoretical, yes. The link from the first paragraph of the article may help: https://www.scientificamerican.com/article/chinese-researchers-achieve-stunning-quantum-entanglement-record/ https://www.scientificamerican.com/article/chinese-researche...
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kelvinsilva 8y ago
https://www.reddit.com/r/QRL/ https://www.reddit.com/r/QRL/
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Want to raise funding for your science field? Contact the opposing partys propaganda office immediately! You are one cold war announcement away from getting funded without due diligence! What are you waiting for?

305 points by 0xbxd 8y ago

I visited USTC several years ago, (2014) and in this time they have basically doubled the number of qubits they can entangle, so this is a great step forward. At around ~100 qubits quantum computing becomes very useful, so maybe we won't have to wait too long.

ssdd 8y ago
That's great. Now next thing that needs to be solved is to make this tech available for general public use just like computers are these days. Problems that are waiting to be solved by it are very large and of huge impact.
- reaperducer 8y ago
  Problems that are waiting to be solved by it are very large and of huge impact I know virtually nothing about quantum computing, so can you give some examples? Whenever I've asked anyone who seemed to know anything about the field, all they can come up with is weather forecasting and simulating nuclear explosions. Not exactly "general use," as you put it. In the back of my mind, I know that quantum computing is a big deal, and we're in ENIAC days with it. But I don't have a good understanding of where it goes or why.
  - adrianN 8y ago
    Quantum computers are extremely useful for simulating chemistry. They can do it exponentially faster than classical computers. Cheap useful quantum computers can revolutionize biology and materials science.
    - deleted 8y ago
      [deleted]
  - chrisweekly 8y ago
    ENIAC?
    - SimbaOnSteroids 8y ago
      One of the first computers. https://en.wikipedia.org/wiki/ENIAC https://en.wikipedia.org/wiki/ENIAC
    - deleted 8y ago
      [deleted]
  - techbio 8y ago
    Protein folding and gene research. Cure malaria, cancer, feed the world etc.
  - foxes 8y ago
    There are already lots of 'practical' applications in the literature. A famous one is Shor's factorisation algorithm. But there are others - everything from doing pagerank to simulating other quantum systems. Many classical algorithms, which run in ~O(poly(n)) can have an 'equivalent' quantum algorithm in ~O(log(n)) - an exponential speedup. There is still debate as to how the complexity class of problems which are efficient on a quantum computer (BQP) relate to other complexity classes. Its suspected P lies entirely within BQP. However, at least initially, I think quantum computers will be a specialized piece of equipment. Classical computing is pretty good for your average person. Quantum computers will be used for more heavy compute tasks.
    - bdamm 8y ago
      Is Shor's an application of entanglement though?
      - teraflop 8y ago
        Yes, all non-trivial quantum computing is based on entanglement.
    - zjaffee 8y ago
      I'd also add that the need for localized personal computing services goes down as network bandwidth increases, if there's anything that people really need quantum computers for, it's not unreasonable to suggest that they will be handled on the backend with aggregated results sent to clients as is the case for most computation today.
    - moefh 8y ago
      > Its suspected P lies entirely within BQP. It's actually known that BQP contains P[1]. It also contains BPP. What's not known is the relationship between BQP and NP (most experts suspect there's no containment in either direction). [1] See this for an easy proof: https://people.eecs.berkeley.edu/~vazirani/f04quantum/notes/lec5/lec5.pdf https://people.eecs.berkeley.edu/~vazirani/f04quantum/notes/...
      - 8y ago
  - kart23 8y ago
    Doesn't basically all encryption become meaningless with quantum computers? Remember how they always said 'it would take a billion years to crack this encryption with a computer'? Well soon we are going to have computers that don't play by the rules.
- jblow 8y ago
  We all have these things in our pockets that are massive supercomputers by the standards of my college days. And we all use them just to rant at each other about political things that we in reality are totally misinformed about ... and otherwise just waste our lives. I am not sure why quantum computing is supposed to change this. Even software that is supposed to be useful is so terribly slow. Computers are between 2 and 4 orders of magnitude faster than programmers today experientially believe they are, because today’s culture of programming has rotted so thoroughly. Do you really need a quantum computer when 3 orders of magnitude are just sitting there on the table waiting to be picked up?
  - ahelwer 8y ago
    Quantum computers won't change anything about people posting misinformed political opinions. Three orders of magnitude speedup is a constant factor. Quantum computers provide between sqrt(n) and log(n) speedup depending on the problem - for example, biochemistry simulation for improved drug designs and such. That seems worthwhile.
  - Erlich_Bachman 8y ago
    > And we all use them just to rant at each other about political things that we in reality are totally misinformed about ... and otherwise just waste our lives. So if I decide to use the device that I have bought and invested in, to "rant" about something (to connect to other people through it, essentially), or even just connect to my loved ones sending "meaningless" information (as many I'm sure would call it), or if I choose to use it for anything else, who is to say that it's "wasting my life"? You? Why do you decide what is wasteful and what is not? Maybe all those people want to do those things? You can't say that this technology is wasted or is not used properly without also implicitly assuming moral and philosophical authority about what people should choose and not to choose with their time and other resources, including the money used to buy such devices and resources spend on developing them. Why do you assume that you can be such an authority? Of course, looking at this in another way, you definitely have the ultimate authority in this area in this one regard: where it applies to your own life. Which I guess is another way of saying that what you wrote says more about your outlook on life than on the underlying technology and its social ramifications.
- Cobord 8y ago
  Because of the expense it will have to start off as QCaaS. You send the really hard part of your problem to whichever person actually has the hardware and they'll return the answer. If you are concerned with them not knowing your data, there are workarounds to that as well. So available to general public in a weaker sense.
bscphil 8y ago
What's the size needed for cracking large-primes based encryption? If not at 100 qubits, what are some useful calculations that we can do at that size?
- deepGem 8y ago
  So the number of raw qubits are in the order of a million for large prime factorization or to make Shor's algorithm a reality. This is because the physical qubits are highly unstable and prone to errors. The error correction methodologies will result in a fault-tolerant logical qubit. Depending on the error correction algorithm, the ratio of logical: data/physical qubit is 1:1000 at the minimum. You need 4000 logical qubits at least for large prime factorization give or take 4 million raw qubits.
  - amelius 8y ago
    How fast do you think the number of qubits will grow over the coming years? Will it be exponential, like Moore's law?
    - misterprime 8y ago
      There was an AMA not long ago with someone in the quantum computer industry and iirc he stated that it isn't certain if we even can go that high. Guessing how fast that will happen is not reasonable at this point.
  - bscphil 8y ago
    This is probably a dumb question, but could you spread out the error correction in time? I know it's hard to create very large entangled systems, so would it be possible/easier to do the calculation with 4000 real qubits 1000 times and use the repetition for error correction? (Checking for correct factorization is easy, after all.)
    - simcop2387 8y ago
      That's been one strategy I've seen mentioned in regards to Shor's algorithm many times. Since there's the probability of an error, you still have to check it and rerun it a few times until you confirm the right answer. That's still many times faster than you'd get conventionally, so it's a reasonable trade off. But it doesn't address the possibility that you can't entangle enough qubits to begin with, or keep them stable long enough to actually perform the calculations, which is where the error correction stuff comes in. I don't know enough about the challenges there to comment on how many you'd need (though 1000:1 seems high to my naive intuition, i'd guess maybe 10-100 for magnitude myself).

downrightmike 8y ago

Even neater is that they are using quantum entanglement in radar to detect stealth planes: https://www.popsci.com/china-quantum-radar-detects-stealth-planes-missiles#page-2 https://www.popsci.com/china-quantum-radar-detects-stealth-p...

misterprime 8y ago
From the article: they are planning to do this, and hopeful that it will work. Not actually using it yet.
- seanmcdirmid 8y ago
  Also, if they got it to work they would surely keep it secret.
  - JackFr 8y ago
    Also, if they didn't get it to work they would surely keep it secret.
    - seanmcdirmid 8y ago
      Yes, and if the military was involved at all, they wouldn’t even talk about the possibility.
barbegal 8y ago
I understand the core concept of quantum radar but does it actually work in practice as a radar? i.e. you can resolve targets with a small range and angular resolution.
- mike00632 8y ago
  Theoretically you would combine the new tech with the old tech. Quantum state would be another measurement to make on the incoming wave on top of measurements like frequency and amplitude.

angrygoat 8y ago

Here's a pre-print of the USTC team's paper: 18-qubit entanglement with photon's three degrees of freedom https://arxiv.org/abs/1801.04043 https://arxiv.org/abs/1801.04043

deleted 8y ago
[deleted]

ArtWomb 8y ago

Congrats! US House science committee also set to pass national initiative in quantum tech with on order of $1B funding: National Quantum Initiative - Action Plan https://www.lightourfuture.org/getattachment/85484dca-465a-46f4-8c8c-090aeb845d09/FINAL-Action-Plan-for-a-NQI-Apr-3-2018.pdf https://www.lightourfuture.org/getattachment/85484dca-465a-4...

ahelwer 8y ago

Crash course in entanglement: The state of each qbit is represented by a state vector of two complex numbers [a, b] where |a|^2 + |b|^2 = 1. There are two special qbit values called the classical basis: [1, 0] which is the classical bit 0, and [0, 1] which is the classical bit 1. If a qbit is not in one of the two classical states, we say it is in superposition. When a qbit is in superposition, we can measure it[0] and it will collapse probabilistically to 0 or 1; for a qbit [a, b], the probability that it collapses to 0 is |a|^2 and the probability that it collapses to 1 is |b|^2. Things get more interesting when we have multiple qbits. If we have two qbits [a, b] and [c, d], we define their product state as their tensor product [ac, ad, bc, bd]. For example, if we have two qbits both in state [1/sqrt(2), 1/sqrt(2)], their product state would be [1/2, 1/2, 1/2, 1/2]. We use the product state to calculate the action of a quantum logic gate that operates on multiple qbits - for a gate which operates on two qbits, we can always represent its action as a 4x4 matrix. Usually we can move back and forth between the product state representation and writing out the individual qbits states. However, in certain scenarios something very special happens: we cannot factor the product state back into the individual state representation! Consider the product state [1/sqrt(2), 0, 0, 1/sqrt(2)]. If you try to write this as a tensor product of two states [a, b] and [c, d], you cannot! It cannot be factored; the qbits have no individual value, and we say they are entangled. Well, what does this mean? It means when you measure one qbit, even if the qbits are very far apart, you instantly know the value of the other qbit. So if I entangled two qbits in the state [1/sqrt(2), 0, 0, 1/sqrt(2)], give you one, then we go to opposite ends of the universe, if I measure my qbit and see a 0 I'll know your qbit instantly also collapsed to 0 (or collapsed to 1 if I measured 1). This phenomenon has been experimentally-verified to occur faster than light. It is instantaneous, as far as we can tell. So, local realism is wrong! Spooky action at a distance is real! There is an important caveat: while the qbits seem to coordinate in some faster-than-light way, you cannot use this to communicate in a faster-than-light way. All we have is a shared random number generator. I can't send some chosen bit from my reference frame to yours. This is called the no-communication theorem. If you found this interesting, I have a full video on quantum computing for computer scientists here: https://youtu.be/F_Riqjdh2oM https://youtu.be/F_Riqjdh2oM [0] IDGAF about your chosen quantum mechanics interpretation, don't @ me

bdamm 8y ago

xpuente 8y ago

uhmmm ... https://www.quantamagazine.org/gil-kalais-argument-against-quantum-computers-20180207/ https://www.quantamagazine.org/gil-kalais-argument-against-q...

Cobord 8y ago
Take that more as a challenge and a caution to avoid overhype.

nabla9 8y ago

Question: Why superfluid 4He is not quantum-entangled? What differentiates macroscopic many-body quantum states from many-body quantum entanglement?

barbegal 8y ago
Superfluid 4He is quantum-entangled and non-local effects are observable. https://www.sciencedaily.com/releases/2017/03/170321110344.htm https://www.sciencedaily.com/releases/2017/03/170321110344.h...
- nabla9 8y ago
  So why it's not taken into account when quantum-entanglement records are broken? You can have flasks of superfluid 4He with thousands of moles of entangled atoms clearly visible.
  - barbegal 8y ago
    Quite correct but usually quantum-entanglement records are for qubits rather than generic atoms. And qubits must be able to be measured in isolation e.g. the polarisation of a single photon can be measured. With superfluid 4He you can't isolate the qubits.
    - nabla9 8y ago
      Ah, that makes sense. Thank you.

tzakrajs 8y ago

> “It’s as though you took six bits in your computer, but each bit tripled in how much information it could hold,” Schreppler said "Each bit is tripled in how much information it could hold"? Normally bits have two states, but Shreppler is claiming these qubits can have six states? What am I missing?

albertgoeswoof 8y ago
I thought qubits had an infinite range of states?
- tzakrajs 8y ago
  Oh yeah, thats right.. Isn't it based on orientation in a spherical sense? I guess you could have an infinite number of those orientations.
  - whatshisface 8y ago
    A pure quantum state is a basis vector, written like |n>. It's called a basis vector because it shares a list of mathematical properties with the x̂,ŷ,ẑ coordinate basis vectors you might be familiar with. They can be combined, (like coordinate vectors) by writing coefficients before them. For example, 1/√2 |a> + 1/√2 |b> would be a combination of state a and state b. Combinations of states obey the condition that the sum of the squares of each coefficient equals one. This is the unitary constraint, and above you can see that 1/√2^2 + 1/√2^2 = 1/2 + 1/2 = 1. Coefficients can be complex numbers, which consist of a real part and an imaginary part. So, putting all of this together, a qbit that may be in the state |1> or the state |0> may also be in the state (a+bi)|1> + (c+di)|0>. With the unitary constraint, we can cut out a degree of freedom by introducing the equation |a+bi|^2 + |c+di|^2 = 1. This leaves three degrees of freedom, which can be mapped in to a sphere by a change of coordinates. Three spherical coordinates plus the unitary constraint specifies the four-number state combination.
    - tzakrajs 8y ago
      Very insightful, thank you!

xtrapolate 8y ago

Is entanglement an actual physical phenomenon? What exactly needs to happen in-order for a photon to "become" entangled?

dvh 8y ago
Physical? As in everyday Newtonian physical? It doesn't work on quantum levels, just like Bohr's model of atom doesn't work during big bang because gravity we were successfully ignoring suddenly is on the same scale or bigger then electrostatic force. We will never be able to intuitively (physically) grasp quantum world because our intuition was trained on trees, rocks and cats.
mundo 8y ago
Not sure what "actual physical phenomenon" means, but probably no. If you mean can you look at a photon and observe something about it that tells you it's entangled to some other photon, no. If you mean do we know what physical process underlies entanglement, also no. If you mean does entanglement really happen as opposed to being theoretical, yes. The link from the first paragraph of the article may help: https://www.scientificamerican.com/article/chinese-researchers-achieve-stunning-quantum-entanglement-record/ https://www.scientificamerican.com/article/chinese-researche...

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