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Breakthrough for Quantum Measurement

said_captain_said_wo writes to tell us that PhysicsWeb is reporting that two teams of physicists have developed a new method for measuring the state of quantum bits in a quantum computer without disturbing the state. From the article: "In the future, the Josephson capacitance could be used for operations in a large-scale quantum computer," says Mika Sillanpaa of Helsinki University. "The Josephson inductance and Josephson capacitance together would also allow us to build new types of quantum 'band engineered' electronic devices, such as low-noise parametric amplifiers."

19 of 201 comments (clear)

  1. Re:Heisenberg Uncertainty Principle? by arrrrg · · Score: 5, Informative

    > Wouldnt this violate the Heisenberg Uncertainty Principle?

    Reading a qubit doesn't violate the Uncertainty Principle by itself; if qubits couldn't be read or written, they'd be worthless. The issue you are probably thinking of is that entanglements between qubits will be destroyed by the reading process (and there is no way to "read" such entanglements without destroying the individual qubit values).

  2. Re:Heisenberg Uncertainty Principle? by maxzilla · · Score: 3, Informative

    Not neccisarily, by one quantum bit changing can be measured because it influences a number of other bits in a uniform way. I think you could imagine one gear spinning that drives a set of other gears, as the quantum bit changes it changes the whole setup of the capacitor so it is evident which way the bit was moving without changing its spin by measuring it. you change the spin of one of the bits in the capacitor, not all of the bits or the original bits

  3. Re:Heisenberg Uncertainty Principle? by BlkItlStl · · Score: 3, Informative

    I don't think so. I'm pretty sure that has to do with limits on measuring momentum and location. http://en.wikipedia.org/wiki/Uncertainty_principle

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    Nothing succeeds like the appearance of success
  4. Re:Heisenberg Uncertainty Principle? by arrrrg · · Score: 5, Informative

    I should clarify: reading the qubit will destroy all quantum effects (superposition as well as entanglement), effectively making the qubits look like ordinary bits (when you open the box, the cat's either dead or alive, not both). However, quantum computers are designed with this in mind; reading the output destroys any quantum properties it may have, but a computation can be repeated many times to get an idea of what uncertainty was present in the output.

  5. Credit where it's due. by kimmop · · Score: 5, Informative
    The article isn't totally clear about it but the Finnish university in question is the Helsinki University of Technology (in the city of Espoo) and not the University of Helsinki. These are the largest two universities in Finland and both have Physics departments so the distinction is important.

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    Binaries may die but source code lives forever

    1. Re:Credit where it's due. by grimJester · · Score: 2, Informative

      I just realized I've studied at both but graduated from neither. How did you collapse those wavefunctions again?

  6. Re:Heisenberg Uncertainty Principle? by Anonymous Coward · · Score: 5, Informative

    The uncertainty principle just states that you cannot know both
    the position and momentum of a particle at the same time ... in
    other words you can know both but not precisely, if you know one
    precisely you do not know the other ... because you have interacted
    with it.

    If you know that the particle is in a certain band, you do not need
    to know its location ... that IS its location ... or it is essentially
    trapped .... you do not care.

    If you have your cow in a pasture, you do not care where it is, as
    long as it is eating grass or hay in the pasture and how not escaped. ... best I can do.

  7. Re:Unchanged State by quoll · · Score: 5, Informative
    The article seems misleading in its wording. It says "read the value of a qubit without changing its value." This can't mean that it doesn't change the original quantum value, as this makes no sense. Reading a quantum value (a qubit) collapses the probability to the value read, by definition. This means that the value is no longer quantum. The original probability function cannot be read (though it can be calculated).

    The statement without changing its value must refer to reading the value reliably. When reading the state of an individual subatomic particle it is extremely easy to have the result perturbed by noise. Given that there is a probability of reading an alternative value, then it is not normally possible to tell when the wrong value was read. It appears that this makes the process much more reliable.

    IAAQP (I Am A Quantum Physicist). Though I could still learn a thing or two about subatomic physics.

  8. Re:Implications in reverse order by insignificant1 · · Score: 2, Informative

    I concur. NIST Boulder, as an example that I am familiar with, is developing certain techniques that can be used for quantum computing. (http://tf.nist.gov/ion/index.htm)

    But the reason why the Time and Frequency division at NIST cares is because these techniques may yield better clocks in the future. (In fact, many breakthroughs in fundamental theoretical/experimental physics are applicable to clocks.) Meantime, however, the project gets mainstream-media publicity for quantum computing implications, gets funding from NSA, QuIST/DARPA, etc.

    I'm sure it's a windfall for the physicists to do fundamental research, though, so "Hoorah Quantum Computers!!! (and cut me that check...)"

    At least one other group (http://qdev.boulder.nist.gov/) is working on research similar to that published on PhysicsWeb, specifically using Josephson junctions in creating quantum bits & logic. That group was just recently setting up an honest phase measurement system, though, so are probably a bit behind in the research mentioned.

    The reason that group plays with Josephson junction devices (how they justify it under the NIST banner)? Voltage standards. The reason they tell other funding agencies? Quantum computing, communication, and good-old-Alice-and-Bob.

  9. Re:But is it a flux capacitor? by Anonymous Coward · · Score: 1, Informative

    I'm guessing here, but it's probably something like this:

    Consider if you had a single cooper pair in the system. From a non-quantum point of view, the cooper pair must either be in the top or bottom superconducting loop. However, the cooper pair can tunnel through the insulator to the other loop so the system exists as a superposition of the two states.

    Of course if you start with the cooper pair on one side, the chance of it being found on the other side should increase with time. I'm guessing that they're cooling down the system to the point where the state of the system changes slowly enough that it can serve as a relatively stable qubit. Then if you want to prepare a particular state, you can start with the cooper pair on one side and then lower the temperature to "freeze" it at the right moment so that it's in the desired state.

    So can any physicists tell me if I got this right? I've studied a little quantum computing but we didn't cover implementation that much.

  10. Some corrections by Catullus · · Score: 3, Informative

    Quantum computers are not known to be very good at solving NP-complete problems, and in fact it is considered very unlikely that they will be able to solve such problems efficiently. Grover's algorithm provides a square-root speed-up in solving any problem in NP; however, this is not enough to make an unfeasible problem feasible, and for any given NP-complete problem, there is likely to be a classical algorithm that outperforms this "brute force" approach.

    Grover's algorithm is only the provably best implementation in a "black box" setting, which is unrealistic for many problems.

    Finally, quantum computers are not known to be able to do anything useful for protein folding - this would be an application of an efficient quantum algorithm for the graph isomorphism problem, which nobody has come up with yet...

  11. Re:spooky action at a distance by gauge+boson · · Score: 5, Informative

    presumably, given entanglement [wikipedia.org], measurement of qbit state allows potentially for instant communication ?

    No, it doesn't. The closest you can come is instant synchronization of states, but you don't get to choose what state that is. For example, you can have two particles entangled to have the same (or opposite, as in the EPR thought experiment) spin orientation, but you can't send a signal from one to the other by choosing the orientation. Instead, it's random whether each one is spin up or spin down - the only guarantee is the relationship between the measurements. This would be great for things like cryptographic key exchange, since you can't have a man-in-the-middle attack if there is no middle, but it's useless for sending information. See: The No-Communication Theorem (warning: requires crazy math skills to avoid the MEGO effect)

    nothing can travel faster than light.

    I call bullshit. Relativity prohibits* local superluminal motion; non-locally, it's fair game. See, for example, the Alcubierre Warp Drive - the only question of whether it's possible or not (aside from new physics) rests on whether there's any local superluminal energy propagation at the edge of the spacetime bubble. Plus, QM allows for lots more in the way of non-local effects (even if you assume hidden variables, since Bell's Theorem rules out local hidden variables based on current experimental results), though, as I noted above, it still doesn't allow for superluminal communication (or teleportation, for that matter).

    * Minor caveat: this is not counting tachyons, since nobody knows if they exist.

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    This is sqrt(not) a sig.
  12. Helsinki U. of Tech., not Helsinki U. by msmikkol · · Score: 4, Informative

    Just a minor correction to the linked article: Mika Sillanpää worked at the Helsinki University of Technology, not at the Helsinki University when he wrote the paper in question.

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    The aim of science is not to open the door to infinite wisdom, but to set a limit to infinite error.
    -Bertolt Brecht
  13. Re:Quantum cryptographic links? by Anonymous Coward · · Score: 2, Informative
    Sounds to me like the security of quantum fiber-optic links are now in question. This isn't directly applicable to taping one, but it's a start.

    The security of quantum key distribution (QKD) does not depend on the technology of the eavesdropper: it is assumed she can do any attack allowed by quantum mechanics. The security only depends on Alice and Bob's (the legitimate users) ability to actually produce and measure the quantum states used by the protocol. Finally, there are *proofs* of the security of QKD. The only way it becomes insecure is if we learn that quantum mechanics is an incorrect theory.

    (Not a quantum physicist, but I can play one on slashdot can't I?)

    Sure. Just try to play a better one next time! ;)

  14. Because it isn't an insulator, of course by Flying+pig · · Score: 4, Informative

    The description of the Josephson Junction is aimed at all the non-physicists out there. The "insulating layer" is a bandgap layer. The point is that cooper paired electrons can tunnel through it, i.e. it acts as a superconductor itself. It is an insulator for ordinary electrons only. And the definition of capacitor is nothing at all to do with physical conductors or insulators. It is a region of space where a potential gradient can be created, and the capacitance is the measure of how much energy has to be pumped into the region in order to create a given potential gradient. "Empty space" requires the lowest energy and has the lowest capacitance per unit volume, while certain ceramics with relatively mobile but limited electrons have very high values. If you cannot create a potential difference across your region of space, you have no capacitance - and at first sight, if that region is superconducting you cannot have a potential difference.

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    Pining for the fjords
  15. Re:Heisenberg Uncertainty Principle? by mumrah · · Score: 2, Informative

    The Heisenberg Uncertainty Principle states that the more precisly you measure momentum, the less precisly you can measure position. This is a specific case of a more general uncertainty principle that states if two observables (momentum, position, speed, energy) do not commute, then they cannot be measured to close percision.

    The reason that x (position) and p (momentum) do not commute comes from their operators. In x-space (what you're used to), the operator for x is just x, the operator for p is -i*hbar*d/dt. Long story short, x and p do not commute because of the time derivative in the p operator. This physically means that these two particular observables cannot both be measured with high precision, and as i turns out the product of the uncertainty in x with the uncertainty in p must be less than or equal to hbar/2 (which is the Heisenberg Uncertainty Principle). Not sure how relavent all of this is.

  16. Re:Shroedinger's cat? by slavemowgli · · Score: 2, Informative

    I don't think you understand. It's not about someone being *aware* of whether the metaphorical cat is dead or alive - it's about the quantum state being disturbed. Measuring it can disturb it; whether the measurement is presented to a human mind afterwards or whether it's thrown away, for example, is irrelevant. So, yes, if an ant crawls into the box, the cat will either be dead or alive.

    As for QM not making any testable hypotheses - that's also not true. Quite the opposite, in fact; QM works exceptionally well so far, and our findings continue to match its predictions. Do read up on it a bit - it's a very fascinating topic.

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    quidquid latine dictum sit altum videtur.
  17. Probabilistic computing? by Urusai · · Score: 3, Informative

    Or aleatory computing? I realize there are certain problems that are deterministically intractable but with feasible probabilistic "solutions", but statistics based computing is just...dirty. I don't think a lot of people understand that quantum computing doesn't actually provide hard answers, that you have to run the same "algorithm" a lot of times to get an approximation.

  18. Re:Shroedinger's cat? by tomhudson · · Score: 2, Informative

    I always had a problem with that experiment - it implies that human intellegence has some link with the state of the universe.

    Sounds weird, but its already been proven to be the case - look for two-slit diffraction experiments if yo really want to warp your brain. And no, it doesn't mean that humans are special - its just one case where takeing a measurement alters the state of an object.

    Think on a macro scale. You take a cold thermometer and put it in a big bucket - the bucket's temperature doesn't change much, the thermometer does, and registers the change.

    Now, substitute a drop of water for the bucket. Doesn't matter how hot that water is, sticking it on the cold thermometer is going to change its temperature by a significant amount.

    Now go to the sub-atomic level. There is no way that ANY measurement can't help but affect the thing being measured. Its like the drop of water - you tried to measure its temperature, and in doing so, changed its temperature.

    Objects affect each other when they interact - or, as the saying goes, "shit happens"