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Science Technology

Breakthrough for Quantum Measurement 201

Posted by ScuttleMonkey from the states-from-your-stateless-news dept.
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."
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Breakthrough for Quantum Measurement More

Breakthrough for Quantum Measurement

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  • Heisenberg Uncertainty Principle? (Score:4, Insightful)

    by SRA8 ( 859587 ) on Wednesday November 23, 2005 @03:32AM (#14098696)
    Wouldnt this violate the Heisenberg Uncertainty Principle?

    • Re:Heisenberg Uncertainty Principle? (Score:5, Funny)

      by Anonymous Coward on Wednesday November 23, 2005 @03:35AM (#14098702)
      I'm not sure
    • Maybe. Maybe not, who can tell?

    • Re:Heisenberg Uncertainty Principle? (Score:5, Informative)

      by arrrrg ( 902404 ) on Wednesday November 23, 2005 @03:39AM (#14098719)
      > 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).

      • Re:Heisenberg Uncertainty Principle? (Score:5, Informative)

        by arrrrg ( 902404 ) on Wednesday November 23, 2005 @03:50AM (#14098744)
        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.
        • Actually, some algorithms depend on this behavior. For example, the Quantum Factorization algorithm uses this behavior to resolve one of the registers in the calculation, the other register contains only probability waveforms that are consistent with the value read in from the first register.
        • 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.
      • But don't worry, they are not certain it is working yet.

        They are still uncertain, until they are measured. Does measuring one make the other certain?

        Quantum cryptography? Wassup with that now?

        o.0
    • 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

      • If I understod it right, you just described quantum entanglement. So, looking for that single gear will change the spin of all the other gears.

        The GP had a very good point, because what the submiter says is known to be impossible. That is also why I'm reading the comments, to discover what the researches really did. But it seems that I'll need to RTFA.

    • 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 [wikipedia.org]
      • The Heisenberg uncertainty principle can be generalized to "conjugate" quantities.

        For instance, in condensed matter physics, the wave functions that have to do with superfluidity have an uncertainty relationship between the number of particles participating in the condensate and the quantum phase of the condensate.
      • >http://en.wikipedia.org/wiki/Uncertainty_principl e [wikipedia.org]

        It should be noted that the uncertainity principle also applies to Wikipedia itself, insofar that you can never be entirely sure if the information posted there is legit ;-)

    • Re:Heisenberg Uncertainty Principle? (Score:4, Funny)

      by Obvius ( 779709 ) on Wednesday November 23, 2005 @03:42AM (#14098728)
      Heisenburg's Uncertainty Principle - DxDp>=hbar/2 Or as my old prof used to say "When you've got energy, you don't have the time. And when you've got the time, you don't have the energy."

    • Re:Heisenberg Uncertainty Principle? (Score:5, Informative)

      by Anonymous Coward on Wednesday November 23, 2005 @04:37AM (#14098868)
      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.
      • 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
    • Heisenberg's Uncertainty Principle places no limit on measuring just the position, or just the momentum; it only limits measuring both with precision. Measuring the states of quantum bits won't violate similar principles provided you're measuring commutating properties, and not non-commutating properties like position and momentum.

    • Wouldnt this violate the Heisenberg Uncertainty Principle?

      "You keep using that word. I do not think it means what you think it means."

      There are certain pairs of quantities ("conjugate variables") which cannot both be measured to arbitrary precision. If you measure, for example, the position of a particle along the x-axis (in your choice of coordinate system) very precisely, you cannot measure the momentum along that same axis very well. The product of the two imprecisions (or uncertainties) is a sma

  • Shroedinger's cat? (Score:5, Funny)

    by mcrbids ( 148650 ) on Wednesday November 23, 2005 @03:38AM (#14098714) Journal
    Does this mean that we can find out if the cat is dead without opening the box? Sure sounds like it.

    IANASPP (I Am Not A Sub-atomic Particle Physicist) but this seems to be quite a breakthrough that might save millions of subatomic cats from untimely deaths...

    Anybody with some actual knowledge care to elucidate?

    • Re:Shroedinger's cat? (Score:3, Interesting)

      by Anonymous Coward
      Does this mean that we can find out if the cat is dead without opening the box? Sure sounds like it.


      That's exatcly what it means, the way the headline presents it, which would mean that QP is disproved, which again leads me to beleive that the poster has misunderstood something.
      • Given that TFA makes exactly the same statement, it's not really the poster who's misunderstanding you're seeing.
        I guess what they managed is to do a measurement which behaves like an ideal quantum measurement: After you measured the system to be in state x, it really is in state x (unlike e.g. a typical photon measurement where after measuring the state of the photon, it isn't actually in any state because it doesn't exist any more).
        Which means that if the system is already in an eigenstate of the measurem
    • It's proven. Shroedinger's cat is dead.
    • Lots of ways... it stops moving about, for example. Also the cat itself knows whether it's dead or not.

      I always had a problem with that experiment - it implies that human intellegence has some link with the state of the universe. I don't buy that for one minute. What happens if an ant crawls into the box for example? Because it's not 'really' an observer the cat is still half alive?? Plus since it's by definition unobservable it's also unprovable - so the whole experiment relies on philosophy rather th

      • Re:Shroedinger's cat? (Score:2, Interesting)

        by tendays ( 890391 )

        What happens if an ant crawls into the box for example? Because it's not 'really' an observer the cat is still half alive??

        Yes the cat is still half alive, and the ant is half seeing a dead cat and half seeing a cat alive. What happens when the ant walks in the box is that its state gets correlated to that of the cat.

        Note that the either state of the ant is unaware of the other.

        When you open the box you will either see a dead cat and an ant that has been seeing a dead cat all along, or a living cat and

        • Re:Shroedinger's cat? (Score:3, Insightful)

          by Y2 ( 733949 )

          What happens if an ant crawls into the box for example? Because it's not 'really' an observer the cat is still half alive??

          Yes the cat is still half alive, and the ant is half seeing a dead cat and half seeing a cat alive. What happens when the ant walks in the box is that its state gets correlated to that of the cat.

          Note that the either state of the ant is unaware of the other.

          When you open the box you will either see a dead cat and an ant that has been seeing a dead cat all along, or a living cat and

          • Because this whole problem is about what the human observes. You could choose some other human standing outside of the room in which this takes place, who would walk into this room eventually and either: a human that has opened the box and seen a dead cat and an ant that has seen a dead cat; or a human that has opened the box and seen a live cat and an ant that has seen a live cat.

            And that room could be inside a much larger house, and that house inside one of those pretentious "gated communities", which is
      • 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
      • If we send a robot into the forest to cut down a tree, the tree falls even if a human intelligence does not witness or perceive it.

        Likewise if we use a machine to measure the position of a particle, the superposition collapses whether or not a human intelligence ever reads the results. In QM measurement is an act with consequences, just like swinging an axe at a tree. But it doesn't require a human intelligence.
      • ..it's not 'really' an observer..

        I think this is a common misunderstanding. The whole buisiness with the "Cat" is a rather over-stretched metaphor. When you read "observation" think "interaction". In this sense anything else is an observer if it interacts with the "Cat" (probes its state). This gets more complex of course when one realises that "everything is connected to everything else" [*]

        ...it's also unprovable..

        I realise the following is probably more detailed than necessary, however it's wort

      • Plus since it's by definition unobservable it's also unprovable - so the whole experiment relies on philosophy rather than science.

        The cat is a thought experiment that's useful for getting across the general idea. As far as I know, it's never been argued as something that would actually be performed, because cats are macroscopic.

        You can use the concept on the quantum scale to predict lab-testable results like the two-slit experiment [wikipedia.org]. If you put detectors (e.g. observers) in the slits, the effect disappears.
    • Does this mean that we can find out if the cat is dead without opening the box?

      Sure, just ask the cat.
    • Its closer to "we can find out if the cat is dead without accidentally killing him in the process" (or, even freakier, "we can find out if the cat is dead without accidentally bringing him back to life in the process").
    • "IANASPP (I Am Not A Sub-atomic Particle Physicist)"
      Probably just as well. If you were, you would be too small to use a keyboard or to set up your experiments.

      Sorry. Well, half-sorry. I just couldn't resist.
    • Does this mean that we can find out if the cat is dead without opening the box?

      No, it means that you can find out that the cat is alive without killing it in the process.

      Most quantum measurements require you to kill the cat in order to determine that it was alive when the measurement started. For instance, drop a large weight on the box and listen for screams coming out. Think of measuring photons: the usual way to measure a photon is to allow it to be absorbed by something such as a photomultiplier tub

  • Is quantum computing useful beyond decryption? (Score:3, Interesting)

    by jimmyhat3939 ( 931746 ) on Wednesday November 23, 2005 @03:44AM (#14098733) Homepage
    The problem with quantum computing, as I understand it, is there are very very few applications.

    Essentially, it's only useful in a situation where you need to repeatedly run the same computation over and over again with different input values to see which of those values produces a valid output.

    I have a friend who has suggested repeatedly that eventually computers will contain some sort of quantum processor that helps with such tasks as gaming. I don't think this is realistic because of the serialness of the tasks that quantum computing tackles. In particular, something like rendering an environment in real-time won't be helped because there's an unpredictable input (the human).
    --
    Free 411! 1-800-411-SAVE [1800411save.com]

    • Re:Is quantum computing useful beyond decryption? (Score:5, Funny)

      by smeek ( 617646 ) on Wednesday November 23, 2005 @03:55AM (#14098764)
      Quantum computing is also good for solving problems in quantum mechanics. No, really.

    • Re:Is quantum computing useful beyond decryption? (Score:5, Insightful)

      by Jace of Fuse! ( 72042 ) on Wednesday November 23, 2005 @04:07AM (#14098790) Homepage
      In particular, something like rendering an environment in real-time won't be helped because there's an unpredictable input (the human).

      Durring the 1/60th (or less) of a second that your system is rendering a single frame in that game, the state of the scene and all objects (as well as light positions, textures, and overlays) is very static. It just doesn't seem like it to you, because you are very slow compared to your computer.

      There could be hundreds of applications of a Quantum Co-Processor in a game, from testing for occlusion in a 3D scene, to making AI decisions in computer controlled characters.

      Quantum Computing may very well not be immediately useful in many traditional computation tasks ("While this value is true then do that") but it will open up whole new ways of tackling processes that are time consuming with today's methods ("do any of these things give us this, that, or something in between?").

      Just thinking about it gives me that Fuzzy Logic Feeling...
    • Maybe not on your PC, but a box made by these guys: http://hardware.slashdot.org/article.pl?sid=05/06/ 22/0610220&tid=126&tid=137 [slashdot.org]
      attached to an MMORG server...

      Full disclosure: I work for them...

      Paul B.

  • juz wondering.. would this result mean anything to the already available systems whereby quantum properties are used to securely send data from point to point??

  • no fair (Score:4, Funny)

    by FidelCatsro ( 861135 ) * <fidelcatsro@gmaDALIil.com minus painter> on Wednesday November 23, 2005 @03:51AM (#14098747) Journal
    You changed the state of the quantum bits by measuring them

  • Implications in reverse order (Score:5, Interesting)

    by Flying pig ( 925874 ) on Wednesday November 23, 2005 @03:53AM (#14098754)
    Talk about looking for grant funding! Problem is, scientific illiterates in Government etc. think they understand what a quantum computer can do (application a long way in the future if at all) but not what you can do with very low noise parametric amplifiers (which might be relatively near term applications.) In terms of exciting progress in studies of brain function, small scale biochemistry, remote sensing and signal processing, very low noise amplifiers are critical components, whereas quantum computers don't yet exist, and by the time they do conventional computers should be adequate to deal with a lot of the data processing.

    Not to knock the discovery, which is very interesting, but it's a pity quantum computers have to be dragged into everything to justify research. I doubt that Tom's Hardware will be reviewing millikelvin coolers for your qubit box any time in the next 20 years (though I'd like to be proved wrong)

    • 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 [nist.gov])

      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, ge

    • I thought the main point of quantum computing is that it could solve problems that can't be tackled by digital computers. So, for example, whereas factoring a 64-bit prime number might take a fairly hefty digital computer a decent chunk of time, all a quantum computer would need is 64 qbits and the problem would be effectively solved.

      As a mathematician in training, my biggest worry is that all the interesting cryptography jobs will have been obsoleted by the time I get that far.
      • So, for example, whereas factoring a 64-bit prime number might take a fairly hefty digital computer a decent chunk of time

        I can factor even the largest Mersenne primes in under two seconds in my head. Maybe I can help these scientists out a little bit with factoring primes... :)
  • presumably, given entanglement [wikipedia.org], measurement of qbit state allows potentially for instant communication ? (which would be really spooky [wikipedia.org]!).

  • Unchanged State (Score:3, Interesting)

    by squoozer ( 730327 ) on Wednesday November 23, 2005 @03:58AM (#14098771)

    I thought the state had to be changed to measure it or am I confusing a technique used in quantum crytography with this technique in quantum computing. As an ex-chemist my understanding of things quantum was never that good anyway but I seem to remember someone saying that in order to measure something you had to change it. Any physicists in the house?

    • Re:Unchanged State (Score:5, Informative)

      by quoll ( 3717 ) on Wednesday November 23, 2005 @05:16AM (#14098950)
      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.

  • Physicist in the House (Score:5, Interesting)

    by anomalousman ( 316636 ) on Wednesday November 23, 2005 @04:09AM (#14098797)
    They can do what they say, but it's a lot more trivial than measuring the entire quantum state of the system, which is, as others have suggested above, impossible.

    The Heisenberg Unccertainty principle implies that measuring a quantity must add noise in the conjugate quantity. For example, measuring the momentum of an object spreads out the wavefunction. Another example, measuring the state of a qubit (whether it is a zero or a one) destroys the relative phase between the zero and the one.

    So the "non-destructive" measurement they are talking about means that they aren't changing it from a zero to a one or vice-versa. But they are (and must) destroy the information about the phase of the qubit state during the measurement. For a more in-depth discussion, look up "quantum nondemolition measurements".
    • How much experimental evidence do we have to prove the Heisenberg Uncertainty Principle? I couldn't find anything in the Wikipedia article providing any verification of the principle. If Einstein was right in his arguments, then it's possible that there are missing variables underlying quantum states that could be measured, correct?
    • I'm a newb when it comes to physics. I like to read about it, but I don't quite get how measuring something creates such a change.

      Is it our processes for trying to detect and measure states that create a change or do we not know why?

  • Credit where it's due. (Score:5, Informative)

    by kimmop ( 121096 ) on Wednesday November 23, 2005 @04:14AM (#14098805) Homepage
    The article isn't totally clear about it but the Finnish university in question is the Helsinki University of Technology [www.hut.fi] (in the city of Espoo) and not the University of Helsinki [helsinki.fi]. These are the largest two universities in Finland and both have Physics departments so the distinction is important.

  • Crap! (Score:5, Funny)

    by werewolf1031 ( 869837 ) on Wednesday November 23, 2005 @04:16AM (#14098810)
    I changed the article by reading it! Someone tell me what it says now...?

  • mis-statement (I think) (Score:3, Interesting)

    by sdedeo ( 683762 ) on Wednesday November 23, 2005 @04:17AM (#14098812) Homepage Journal
    It's been awhile; I do GR now, not QM (much simpler.) But any measurement will change the state; this is the famous "collapse of the wavefunction" (in the Copenhagen interpretation.) What they mean is that the measurement will collapse the wavefunction as usual, but that it will not then alter the system being measured so that the state changes. i.e., if the amplitude is 0.1 A and 0.9 B, and the measurement collapses the wf to B 90% of the time as it will, then when the measurement is done the system will be B 90% of the time as expected, and it will be B "the right" times.
    • You've just done a very good job of reducing several paragraphs from the article to a nice, simple form without oversimplifying them. Your last clause, clarifying that the results have to be right, not just in the aggregate, but seperately, for this research to be significant, is an especially good example of how not to oversimplify a scientific subject.
      One more pedantic point though - I've never liked the use of the word 'collapse' in the Copenhagen interpretation. It has a
    • any measurement will change the state

      Not quite. If the system is originally in an eigenstate of the operator being measured, then it should be possible to perform a measurement without changing the state. For instance, if a photon is in a pure momentum eigenstate, then it should be possible to measure the photon's momentum without changing its state.

      The problem is that most quantum measurements destroy the state, even in such a case. For instance, with the photon, we usually absorb the photon in order to

  • Quantum cryptographic links? (Score:3, Interesting)

    by evilviper ( 135110 ) on Wednesday November 23, 2005 @04:19AM (#14098821) Journal
    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.

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

    • Re:Quantum cryptographic links? (Score:2, Informative)

      by Anonymous Coward
      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 o

  • Let's see, a "Josephson junction consists of two superconducting layers separated by a thin insulating layer." Two conductors separated by an insulator. Can anybody tell me why they wouldn't have predicted that it would behave like a capacitor seeing as how two conductors separated by an insulator *is* the definition of a capacitor?

    • Because it isn't an insulator, of course (Score:5, Informative)

      by Flying pig ( 925874 ) on Wednesday November 23, 2005 @07:55AM (#14099379)
      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.

  • Helsinki U. of Tech., not Helsinki U. (Score:4, Informative)

    by msmikkol ( 155023 ) on Wednesday November 23, 2005 @06:39AM (#14099161)
    Just a minor correction to the linked article: Mika Sillanpää worked at the Helsinki University of Technology [www.tkk.fi], not at the Helsinki University [helsinki.fi] when he wrote the paper in question.

  • Which is going to be more important for us?

    Holographic storage or quantum computing?

  • Shor's Algorithm (Score:2, Interesting)

    by stelmach ( 894192 )
    I'm trying to grasp what the implications are of this. Let's take Shor's algorithm as an example. It is my understanding that the Quantum Fourier Transform (QFT) is applied to the result of the algorithm to peak the probability amplitudes, which will help the result to collapse into the correct state when measured. So does this mean that the QFT will not need to be applied, and the result of Shor's algorithm can be read with 100% accuracy?
  • Quantum Computers will usher in a golden age in computing. There are all sorts of applications that they could be used for. For a time they'll serve a role that most super-computers today serve and that's for engineering computations and scientific experiments that require massive number crunching.

    For long term space travel like the proposed mission to Mars a quantum computer would be invaluable. It would be able to monitor the crew and spacecraft faster than today's computers and will be able to react t

  • thank goodness (Score:3, Funny)

    by revery ( 456516 ) <<ten.2cac> <ta> <selrahc>> on Wednesday November 23, 2005 @09:36AM (#14100087) Homepage
    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.

    I'm very glad, as I have a current-model parametric amplifier and man is it LOUD....
    I should have figured as much, seeing as how it goes up to 11.

  • Useful? (Score:3, Interesting)

    by necro81 ( 917438 ) on Wednesday November 23, 2005 @09:37AM (#14100099) Journal
    From RTFA, I am wondering if this new discovery will actually be of much use to anyone. The apparatus involves cooling down to a few millikelvin. I am guessing that this is so that the thermal noise in the circuit is greatly reduced, and also because the superconducting threshold of whatever their Johnson capacitor is made from might in fact be that cold. Pure copper becomes a superconductor, but not until several degrees Kelvin, I believe.

    In any event, cooling down to such temperatures implies a couple of things: lots and lots of very expensive equipment to cool down a tiny tiny volume of space. Even the first transistors didn't require such great lengths.

    The article also makes reference to the capacitance of the Johnson capacitor changing signs depending on the state of the qubit, which is part of how the whole thing works. Does this mean that someone has discovered negative capacitance? Whoa! What would that mean?

    • "... [T]he superconducting threshold of whatever their Johnson capacitor is made from might in fact be that cold."

      As a Canadian, I can vouch for this. In extremely cold temperatures, my Johnson shrinks significantly and becomes much easier to measure.

      ... But what exactly is a 'Johnson capacitor'? Is that like Viagra or something?

  • They're so silly (Score:3, Funny)

    by jav1231 ( 539129 ) on Wednesday November 23, 2005 @10:41AM (#14100647)
    Hey, it could be worse. Quantum Theorists could be using their imagination to sit around and dream about getting laid.
  • It's funny to read Josephson this and Josephson that in the summary. Brian Josephson [cam.ac.uk] is of course a very famous Nobel prize winner, but he has become a pariah in the scientific community.

    Look at his web page [cam.ac.uk]! He's pushing cold fusion, ESP and other paranormal powers, and all kinds of bizarre theories. He's gotten into fights with the highly respected archiv.org [archiv.org] physics publication site over their habit of removing crackpot papers. In short Josephson is an embarrassment to the scientific community, someone w
    • Agreed that he's rather nutty. I got to see him speak in Lindau a little over a year ago, and even got to ask him some questions later on. An interesting guy, to say the least, but he's a nice person and pretty much harmless.

      Besides, you never know, he might even be right about one of those things. If you throw enough darts you might just hit something.

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