Could you please explain what a measurement is, since clearly interaction with something isn't enough or it would be measured by the air.
I'd like to help you with your question but I've got to pop over to Stockholm to collect my Nobel prize.
Seriously, nobody really knows what a measurement is. That's the whole point of Schroedinger's Cat.
Here's a real experiment:
Well there was a nice ascii art diagram but I can't get it through the lameness filter. Filter error: Please use fewer 'junk' characters.
Basically H polarized photon source that goes into a +/- measurer then the two paths are joined and the H/V polarization measured again with H photons going to detector 1 and V photons going to detector 2
Every photon ends up at detector1 if we don't actually look which path the photon took through the +/- branches. As soon as we actually look which path the photon went then the H/V measurement gives 50/50 results to each detector.
|HH> + |VV> = |++> + |-->, so Alice and Bob get the same result if they measure H/V or +/- |HH> - |VV> = |+-> + |-+>, so Alice and Bob get the same result if they measure H/V but opposite results if they measure +/-.
You let them go there on their own. Photons have this innate ability to go places at the speed of light without having to put them in a box and carry them there.
Yes, there is a problem with them interacting with the environment and decohering but that's an engineering problem rather than a physics problem. Provided you're not sending them too far then most will get through without issue.
In any formal experiment you have to allow for the probability of decohering - and so you don't expect 100% correlation. The trick is to get that probability low enough that the errors don't swamp the signal.
Alice can't tell if Bob modulates his angle. Alice and Bob can't even agree who made the measurement first which makes it kind of hard to imagine how they can use it to communicate.
Alice measures her photon and gets one of two results completely at random. Bob measures his photon and gets one of two results completely at random.
When they get together they discover that, although their individual result were random, their combined results display a correlation related to the relative angle of the polarizations they measured.
Bob can't be in a superposition of the two results because Alice's measurement has already determined what result he must have got. If Alice sends a polarized photon (that she's created and knows the polarization) to Bob then she knows what result Bob will get if he measures the same polarization even if Bob doesn't know what measurement to make and just makes a random guess.
The speed of light prohibits Alice from knowing whether Bob actually bothered to do the experiment until later but Alice knows in advance, provided Bob hadn't taken a coffee break, what result Bob got, even though, from Bob's POV, his result was completely random.
i.e. Alice and Bob can have a random number generator that can simultaneously generate the same stream of random numbers at opposite ends of the universe and yet nobody can predict what the next random number will be until Alice and Bob makes their measurement.
(Of course, in the simple random number example, the results could be explained by a local hidden variable but we can design experiments where a local hidden variable cannot explain the results)
That is, unless I'm missing something fundamental.
Yes, you're missing something fundamental.
Going back to one photon.
We'll have four polarization states H, V (the normal horizontal and vertical polarization) and +, - the 45 degree polarizations.
Now Alice produces a stream of H photons and sends them to Bob. Now if Bob measures to see if they're H/V then he will always get H.
But if Bob measures if they're +/- he'll get 50/50 + and -, with each individual photon being + or - at random.
After measuring +/-, if Bob then remeasures H/V he'll again get 50/50 H and V. The measuring of +/- destroys the knowledge about H/V
If Bob measures at an angle other than 45 degrees then he'll get different proportions but he'll get sin^2 theta with one polarization and cos^2 theta with the other polarization.
Now lets consider entangled photons that will always give the same result for Alice and Bob. Initially we'll assume that Alice will always measure the horizontal polarization (0 degrees) Now lets consider that the photon "knows in advance" whether it will go through a horizontal polarizer i.e. it has (an infinite number) of hidden variables. Regardless of what measurement Bob does, an ensemble of photons can distribute values amongst these hidden variables so that Bob gets the expected correlations relative to Alice and the angle of his measurement.
But now let Alice vary her angle as well. Now the correlation depends on the difference in angle between Alice and Bob. But that angle isn't known (and hasn't even been decided) at the point the photon has been created. It could have a big "look up table" saying "If Alice angle is n and Bob angle is m then do/don't go through Alice's filter and do/don't go through Bob's filter BUT the photon that arrives at Bob's detector has to know what measurement Alice will/has done and the photon that arrives at Alice's detector has to know what measurement Bob will/has done.
But because Alice and Bob independently randomly decide what angle to measure "long" after the photon was created and their independent decisions are made so close together in time that neither can know what the other has/will do when they make their measurement due to the speed of light limit then there is no way for the photon to use its "lookup table" and get the correct statistical results.
It doesn't matter how you construct that "lookup table", unless you allow some sort of faster than light communication, using the lookup table will give different results to QM.
If you want the formal maths for that bit of hand waving then lookup Bell's inequality. He actually deduced the inequality that could be tested to prove no local hidden variable theorem was consistent with the results of QM based on measuring particle spins while most of the tests that have been done have used polarization of photons but the underlying theory is the same.
These experiments have already been done, and Bell's inequality has come down on the side of QM. Because Alice and Bob make their measurements so close together in time, not all observers will agree which one is first but (perhaps unfortunately) Alice and Bob will agree who was first and who was second. What this experiment does is close even that loophole - even Alice and Bob will be unable to agree who made the first measurement and who made the second.
What's already been done is to ensure that Alice and Bob decide what measurement to do, and make the measurement, so close to the same time that it's impossible for there to be any way for Bob's equipment or photon to "know" what Alice is going to do (or vice-versa) except at superluminal velocities.
Just to clarify this paragraph because I've realized it's confusing.
Alice and Bob both randomly decide to measure the H/V polarization or the +/- (45 degree) polarization.
Then they get together and compare results. Where one has measured H/V and the other +/- then they throw the results away because they don't tell them anything useful, but where they've made the same measurement they find they always get the same (or opposite) results.
It's when they make the measurement that neither they, nor their equipment or photon can "know" what the other is doing.
There's also something called Bell's inequality that basically proves that the results of all of Alice and Bob's possible cannot be "known" by the photon ahead of time. (no local hidden variables).
I'm not sure what you are trying to suggest, but you can't use entangled photons to communicate faster than light.
I've not RTFA - it's down - but basically the EPR effect allows someone to create two photons and then measure if their polarization is H or V. The result is completely random BUT, both photons will always give the same result.
Now Alice measures her photon first and lets say we get H, then Bob's photon must instantaneously turn into H (previously it was a mixture of H and V - the dead and alive cat) so that when he measures his photon he also gets H.
What's already been done is to ensure that Alice and Bob decide what measurement to do, and make the measurement, so close to the same time that it's impossible for there to be any way for Bob's equipment or photon to "know" what Alice is going to do (or vice-versa) except at superluminal velocities.
But because Alice and Bob are in the same inertial frame there's still, at least in theory, a concept of who did the measurement first and who did it second. (Alice and Bob can have synchronized clocks and record the time they did the experiment. Then they can, using normal communication, tell each other what time they did the experiment and they'll both agree who was first and who was second.)
What this experiment will do is mean that Alice and Bob won't agree about who was first and who was second. Alice and Bob's clocks cannot remain synchronized, so that according to Alice, and people sitting next to her, she did the measurement first, but according to Bob, and people sitting next to him, he did the experiment first. And BOTH will be correct because the two measurements are space like rather than time like.
Asus got an early lead but if they want to throw their advantage away and return to differentiating their product only by price and color that is their right. There are more than enough other mfrs eager to push the mindshare across the threshold and bring about the unchained era of personal computing.
Actually I think the eeepc is another example of the fact that manufacturers "just don't get it" when it comes to "free" software.
The eeepc has a significant number of closed source bits and pieces. Most of them don't work quite the way people want, so people learn how to hack things in order to get it to do what they really want - an example is the popup when you plug in an SD card or USB stick. And then ASUS releases an update that breaks that previous hack and people get annoyed.
Also, because there are pieces that aren't open, nobody can easily support the factory install. So openssl is still broken and, last time I looked, no fixed package is available. So ASUS are probably finding that it's quite expensive supporting Linux.
Had they just made the whole thing open then there would rapidly have grown up a "support team" around the factory install. Security issue - within 24 hours the sourceforge project would have been updated. Asus then just need to have someone monitoring the sourceforge project and "backport" any fixes into their official distribution.
A little more work would be involved if they also want to pull in "improvements" from the sourceforge project. But not that much work.
"This study also enabled us to explore Prediction 3, that incompetent individuals fail to gain insight into their own incompetence by observing the behaviour of other people."
"[After seeing the answers of others] If anything, bottom-quartile participants tended to raise their already inflated self-estimates, although not to a significant degree"
The fundamental problem is that, even with the right answers in front of them, the incompetent are unable to distinguish the right from the wrong answers. What the internet brings to the incompetent is AN answer so now they THINK they know.
The competent can, of course, filter the wheat from the chaff.
I especially like the concluding remarks from that paper. "That worry is that this article may contain faulty logic, methodological errors, or poor communication. Let us assure our readers that to the extent this article is imperfect, it is not a sin we have committed knowingly."
Yes, corners first. I don't know about the same 8 move combination but IMO there are only 8 pieces that aren't obvious how to position.
First you do four corners on a face - anybody can do this without any "method". (at least I could)
Then you position the second four corners and rotate them as necessary - this is more tricky.
Then you position the eight edges on opposite faces (leaving the middle slice unsolved). (How to do this was completely obvious to me, YMMV)
Then you need to position and rotate the remaining four edges - again more tricky.
Finally, you might need to reposition the centres (By default I ignore the four centres on the middle slice and minimize the moves for the edges, sometimes leaving me with a four spot pattern to "solve" at the end)
If you are saying there is no way to rotate a centre face without moving any of the other pieces then you are wrong.
Proof:
F1U1 (front 90, Upper 90) repeated 105 times returns the cube to its original state.
But 105 is 1 mod 4. Therefore the Front and Upper centres have been rotated by 90 degrees at the end of this move.
R1F1 has similar properties, as does R1U1.
Therefore F1U1^105 R1F1^105 R1U1^315 will rotate F centre through 180 and R and U centers through 360 - i.e. one centre will be rotated by 180 degrees.
Funnily enough, only last week I got a cube that had polarization of the centres. I very quickly found a move to rotate one centre 90 clockwise and an adjacent 90 anticlockwise but the move to rotate a single centre through 180 degrees took me a very long time to find. (The above 525 step move is not original to me and was only pointed out to me once I'd found a 12 step move to do the same thing) Also the cube I had actually had one centre that could be in any orientation when viewed on its own. Only when comparing with the other centres was it possible to deduce that the centre was rotated from it's factory state.
I think you're getting confused with CPT invariance. Reverse the charge, reverse the parity and reverse time and everything looks the same.
IIRC CP symmetry (reverse the charge, reverse the parity) is broken with the decay of the Kzero and anti-Kzero. (There was an article about this some years or decades ago in SciAm but I'm too lazy to go and look it up now). Given that then I don't think, in general, an anti-particle can be a particle moving backwards in time.
Actually, the message telling which bits were send and/or measured at a 45% angle is only exchanged after Bob has measured the quantum bits. So, even can't just store them, measure them and re-inject them after the fact.
So, rather than just eavesdropping the message from Bob to Alice, she would actually need to destructively intercept it and change it.
Eve isn't intercepting the bits. She's creating a pair of correlated photons without actually making any measurement. She sends one on to Bob and stores the other for measuring later.
a|H> + b|V> => a|HH> + b|VV>. I'm pretty sure this transformation is allowed for all values of a and b. Note that she is NOT cloning (a|H> + b|V>)(a|H> + b|V>) which would be prohibited.
Having said that, I think my quick thoughts were incorrect. I was assuming |HH> = k|++> + k|--> when I think it should have been 1/2|++> + 1/2|+-> + 1/2|-+> + 1/2|--> (k=1/sqrt2) now that I think about it properly.
Or better yet, what would happen if some new device could record without observing?
"record without observing" doesn't make sense. In the words of Pauli, "That's not right. That's not even wrong." But trying to guess that you mean "observe without affecting in any way".
Then you can violate causality. Give me a device that can "record without observing" and I'll build you a device to communicate faster than the speed of light.
You don't transmit the message using QC, you transmit a OTP. So if Eve does intercept it then all she gets is a bunch of random bits, Alice and Bob detect the interception, throw the OTP away and start again.
Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?
Lookup quantum cloning and the "no cloning theorem".
But basically (and this is a naive implementation that won't actually work), Alice transmits to Bob using linearly polarized photons. Now, if you remember from your school days, if you shine a light though a polarization filter and then through another filter at the same angle, all the light that gets through the first filter gets through the second filter as well.
So, let Alice transmit a horizontally polarized photon |H> if she wants to send a 1 and |V> if she wants to send a 0.
Bob uses a horizontally polarized filter <H| to measure Alice's photon. <H|H> = 1 (the photon gets through and he detects it, <H|V> = 0. The photon gets stopped and he doesn't detect it.
So far, so good but... Eve does exactly what you suggested and measures the photon and then regenerates it - so Bob doesn't see any difference.
Now it starts getting clever... Again, hopefully you remember from your school days that, if you send that polarized light though a second filter at an angle to the first, a proportion of the light gets through (cos^2 theta). But, QM says that you cannot predict which photons will make it through the second filter, it's entirely probabilistic.
So as well as using |H>,|V> to transmit 1 and 0, Alice also uses |+>,|-> where these are 45 degree polarizations. Alice uses one or the other completely at random.
Bob, when he measures at his end also choses whether to measure the horizontal polarization <H| or diagonal polarization <+| completely at random. There are eight cases:
<H|H> = Alice and Bob use the same polarization angle so Bob detects the photon <H|V> = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon <H|+> = Alice and Bob's filters are at 45 degrees so Bob may or may not detect a photon <H|-> = ditto <+|H> = ditto <+|V> = ditto <+|+> = Alice and Bob use the same polarization angle so Bob detects the photon <+|-> = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
Once Alice and Bob have done this, Bob tells Alice which measurement he's done (over a classical channel, they don't care who might eavesdrop.) If Alice and Bob have used the same basis - i.e. Alice used |H>,|V> to transmit her bit and Bob used <H| to measure it or Alice used |+>,|-> to transmit her bit and Bob used <+| to measure it then Alice says "correct" and Bob knows what bit Alice sent. If Bob used the wrong measurement then Alice says "incorrect" and they both throw that bit away.
So, on average, for each two bits that Alice tries to send to Bob, one will be thrown away and the other will be a good value.
Now Eve tries to eavesdrop. If she measures <H| then she'll detect |H> or |V>. She can retransmit that value but, if Alice sent |+> or |-> instead then she'll have corrupted the bit. If she measures <+| instead then she can retransmit |+> or |-> but if Alice sent |H> or |V> then she'll corrupt that bit instead. Infact, on average, regardless of which measurement she makes, she'll end up corrupting 1/2 of the values that Alice and Bob have "successfully" exchanged.
Now Eve can get really clever. Instead of measuring the photon, she can clone it and then measure her clone. Now it turns out that there is a limit to how good her cloning machine can be so, although it won't corrupt half of the bits that Alice and Bob transmit, it will corrupt at least 1/6.
Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?
Lookup quantum cloning and the "no cloning theorem".
But basically (and this is a naive implementation that won't actually work), Alice transmits to Bob using linearly polarized photons. Now, if you remember from your school days, if you shine a light though a polarization filter and then through another filter at the same angle, all the light that gets through the first filter gets through the second filter as well.
So, let Alice transmit a horizontally polarized photon |H> if she wants to send a 1 and |V> if she wants to send a 0.
Bob uses a horizontally polarized filter = 1 (the photon gets through and he detects it, = 0. The photon gets stopped and he doesn't detect it.
So far, so good but... Eve does exactly what you suggested and measures the photon and then regenerates it - so Bob doesn't see any difference.
Now it starts getting clever... Again, hopefully you remember from your school days that, if you send that polarized light though a second filter at an angle to the first, a proportion of the light gets through (cos^2 theta). But, QM says that you cannot predict which photons will make it through the second filter, it's entirely probabilistic.
So as well as using |H>,|V> to transmit 1 and 0, Alice also uses |+>,|-> where these are 45 degree polarizations. Alice uses one or the other completely at random.
Bob, when he measures at his end also choses whether to measure the horizontal polarization = Alice and Bob use the same polarization angle so Bob detects the photon
= Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
= Alice and Bob's filters are at 45 degrees so Bob may or may not detect a photon
= ditto
= ditto
= ditto
= Alice and Bob use the same polarization angle so Bob detects the photon
= Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
Once Alice and Bob have done this, Bob tells Alice which measurement he's done (over a classical channel, they don't care who might eavesdrop.) If Alice and Bob have used the same basis - i.e. Alice used |H>,|V> to transmit her bit and Bob used,|-> to transmit her bit and Bob used or |V>. She can retransmit that value but, if Alice sent |+> or |-> instead then she'll have corrupted the bit. If she measures or |-> but if Alice sent |H> or |V> then she'll corrupt that bit instead. Infact, on average, regardless of which measurement she makes, she'll end up corrupting 1/2 of the values that Alice and Bob have "successfully" exchanged.
Now Eve can get really clever. Instead of measuring the photon, she can clone it and then measure her clone. Now it turns out that there is a limit to how good her cloning machine can be so, although it won't corrupt half of the bits that Alice and Bob transmit, it will corrupt at least 1/6.
(Actually, in the naive scheme outlined above I think Eve can do: a|H> + b|V> => a|HH> + b|VV>, store her photon, wait for Bob to measure, eavesdrop the message from Bob to Alice and then make the same measurement on her stored photon. But this only works because the only possible values for a,b in the naive scheme are (0,1), (1,0), (1/sqrt2,1/sqrt2), (1/sqrt2, -1/sqrt2) but I'm right on the limits of my understanding of QM and entangled photons now so I could be completely wrong)
What is keeping NASA and the ESA from working together to create a tiny habitat to send to mars? I'm not talking anything fancy. How about sending a plant to Mars and keeping it alive? You have all the challenges of putting a living organism into space, getting it to mars, landing it on mars, and getting a habitat inflated, powered up, and surviving, all without having to risk the life of a human being.
Because there's nothing to be gained from doing that that we can't do on Earth.
With the exception of gravity, we can create an environment on Earth identical to that on Mars. So if you want to practice raising plants in a Martian environment then you can do it much more cheaply on Earth.
Instead, use that rocket launch to get more experiments to Mars to investigate things that actually need a presence on the planet.
Also, the unnamed "scientists" who were lucky enough to find this are Alicia Soderberg of Princeton University & her colleagues, just so we give credit where credit is due.
Completely unrelated, but I had to go back and reread the first name after seeing that surname:
Whenever I have an interview for a job I always take a copy of my CV to give to the person interviewing me. The CV that the interviewer is holding varies from close to what I sent to the agency to bearing absolutely no resemblance to what I wrote. New skills, missing skills, sometimes I think agencies must cut and paste from several different peoples CVs.
Very, very occasionally (happened to me just once) I got a call from the agency - "Is it alright if we change this to say this instead. This client is looking for a particular skill and it's not immediately obvious to a non technical person from what you've written that you've been doing this for the last three years"
CVs can often be dramatically improved without lying. If you know a good salesman then it's worth asking them to help with the writing of your CV. Once you get to the technical people it probably doesn't matter too much but you've probably got to get your CV through HR screening first. I'd have no problems with agencies reworking my CV if they came back to me first to make sure it was accurate.
Slashdot Mirror
That might allow communication.
Never!
Faster than light communication violates causality - A causes B but B happens before A. It's just absurd.
If you think you've found a way around the speed of light limit then there's something you don't understand yet.
Tim.
Could you please explain what a measurement is, since clearly interaction with something isn't enough or it would be measured by the air.
I'd like to help you with your question but I've got to pop over to Stockholm to collect my Nobel prize.
Seriously, nobody really knows what a measurement is. That's the whole point of Schroedinger's Cat.
Here's a real experiment:
Well there was a nice ascii art diagram but I can't get it through the lameness filter.
Filter error: Please use fewer 'junk' characters.
Basically H polarized photon source that goes into a +/- measurer then the two paths are joined and the H/V polarization measured again with H photons going to detector 1 and V photons going to detector 2
Every photon ends up at detector1 if we don't actually look which path the photon took through the +/- branches. As soon as we actually look which path the photon went then the H/V measurement gives 50/50 results to each detector.
Tim.
Even without any knowledge of quantum mechanics, I could have told them they'd always get the same or opposite results.
Ha Ha!
What I was trying to say was:
(dropping factors of sqrt 2)
|H> = |+> + |->
|V> = |+> - |->
|HH> + |VV> = |++> + |-->, so Alice and Bob get the same result if they measure H/V or +/-
|HH> - |VV> = |+-> + |-+>, so Alice and Bob get the same result if they measure H/V but opposite results if they measure +/-.
etc.
Tim.
You let them go there on their own. Photons have this innate ability to go places at the speed of light without having to put them in a box and carry them there.
Yes, there is a problem with them interacting with the environment and decohering but that's an engineering problem rather than a physics problem. Provided you're not sending them too far then most will get through without issue.
In any formal experiment you have to allow for the probability of decohering - and so you don't expect 100% correlation. The trick is to get that probability low enough that the errors don't swamp the signal.
Tim.
No.
They can't communicate using this.
Alice can't tell if Bob modulates his angle. Alice and Bob can't even agree who made the measurement first which makes it kind of hard to imagine how they can use it to communicate.
Alice measures her photon and gets one of two results completely at random.
Bob measures his photon and gets one of two results completely at random.
When they get together they discover that, although their individual result were random, their combined results display a correlation related to the relative angle of the polarizations they measured.
Tim.
Realise I've not really answered this.
Bob can't be in a superposition of the two results because Alice's measurement has already determined what result he must have got. If Alice sends a polarized photon (that she's created and knows the polarization) to Bob then she knows what result Bob will get if he measures the same polarization even if Bob doesn't know what measurement to make and just makes a random guess.
The speed of light prohibits Alice from knowing whether Bob actually bothered to do the experiment until later but Alice knows in advance, provided Bob hadn't taken a coffee break, what result Bob got, even though, from Bob's POV, his result was completely random.
i.e. Alice and Bob can have a random number generator that can simultaneously generate the same stream of random numbers at opposite ends of the universe and yet nobody can predict what the next random number will be until Alice and Bob makes their measurement.
(Of course, in the simple random number example, the results could be explained by a local hidden variable but we can design experiments where a local hidden variable cannot explain the results)
Tim.
That is, unless I'm missing something fundamental.
Yes, you're missing something fundamental.
Going back to one photon.
We'll have four polarization states H, V (the normal horizontal and vertical polarization) and +, - the 45 degree polarizations.
Now Alice produces a stream of H photons and sends them to Bob. Now if Bob measures to see if they're H/V then he will always get H.
But if Bob measures if they're +/- he'll get 50/50 + and -, with each individual photon being + or - at random.
After measuring +/-, if Bob then remeasures H/V he'll again get 50/50 H and V. The measuring of +/- destroys the knowledge about H/V
If Bob measures at an angle other than 45 degrees then he'll get different proportions but he'll get sin^2 theta with one polarization and cos^2 theta with the other polarization.
Now lets consider entangled photons that will always give the same result for Alice and Bob. Initially we'll assume that Alice will always measure the horizontal polarization (0 degrees) Now lets consider that the photon "knows in advance" whether it will go through a horizontal polarizer i.e. it has (an infinite number) of hidden variables. Regardless of what measurement Bob does, an ensemble of photons can distribute values amongst these hidden variables so that Bob gets the expected correlations relative to Alice and the angle of his measurement.
But now let Alice vary her angle as well. Now the correlation depends on the difference in angle between Alice and Bob. But that angle isn't known (and hasn't even been decided) at the point the photon has been created. It could have a big "look up table" saying "If Alice angle is n and Bob angle is m then do/don't go through Alice's filter and do/don't go through Bob's filter BUT the photon that arrives at Bob's detector has to know what measurement Alice will/has done and the photon that arrives at Alice's detector has to know what measurement Bob will/has done.
But because Alice and Bob independently randomly decide what angle to measure "long" after the photon was created and their independent decisions are made so close together in time that neither can know what the other has/will do when they make their measurement due to the speed of light limit then there is no way for the photon to use its "lookup table" and get the correct statistical results.
It doesn't matter how you construct that "lookup table", unless you allow some sort of faster than light communication, using the lookup table will give different results to QM.
If you want the formal maths for that bit of hand waving then lookup Bell's inequality. He actually deduced the inequality that could be tested to prove no local hidden variable theorem was consistent with the results of QM based on measuring particle spins while most of the tests that have been done have used polarization of photons but the underlying theory is the same.
These experiments have already been done, and Bell's inequality has come down on the side of QM. Because Alice and Bob make their measurements so close together in time, not all observers will agree which one is first but (perhaps unfortunately) Alice and Bob will agree who was first and who was second. What this experiment does is close even that loophole - even Alice and Bob will be unable to agree who made the first measurement and who made the second.
Tim
What's already been done is to ensure that Alice and Bob decide what measurement to do, and make the measurement, so close to the same time that it's impossible for there to be any way for Bob's equipment or photon to "know" what Alice is going to do (or vice-versa) except at superluminal velocities.
Just to clarify this paragraph because I've realized it's confusing.
Alice and Bob both randomly decide to measure the H/V polarization or the +/- (45 degree) polarization.
Then they get together and compare results. Where one has measured H/V and the other +/- then they throw the results away because they don't tell them anything useful, but where they've made the same measurement they find they always get the same (or opposite) results.
It's when they make the measurement that neither they, nor their equipment or photon can "know" what the other is doing.
There's also something called Bell's inequality that basically proves that the results of all of Alice and Bob's possible cannot be "known" by the photon ahead of time. (no local hidden variables).
Tim.
I'm not sure what you are trying to suggest, but you can't use entangled photons to communicate faster than light.
I've not RTFA - it's down - but basically the EPR effect allows someone to create two photons and then measure if their polarization is H or V. The result is completely random BUT, both photons will always give the same result.
Now Alice measures her photon first and lets say we get H, then Bob's photon must instantaneously turn into H (previously it was a mixture of H and V - the dead and alive cat) so that when he measures his photon he also gets H.
What's already been done is to ensure that Alice and Bob decide what measurement to do, and make the measurement, so close to the same time that it's impossible for there to be any way for Bob's equipment or photon to "know" what Alice is going to do (or vice-versa) except at superluminal velocities.
But because Alice and Bob are in the same inertial frame there's still, at least in theory, a concept of who did the measurement first and who did it second. (Alice and Bob can have synchronized clocks and record the time they did the experiment. Then they can, using normal communication, tell each other what time they did the experiment and they'll both agree who was first and who was second.)
What this experiment will do is mean that Alice and Bob won't agree about who was first and who was second. Alice and Bob's clocks cannot remain synchronized, so that according to Alice, and people sitting next to her, she did the measurement first, but according to Bob, and people sitting next to him, he did the experiment first. And BOTH will be correct because the two measurements are space like rather than time like.
Tim.
Asus got an early lead but if they want to throw their advantage away and return to differentiating their product only by price and color that is their right. There are more than enough other mfrs eager to push the mindshare across the threshold and bring about the unchained era of personal computing.
Actually I think the eeepc is another example of the fact that manufacturers "just don't get it" when it comes to "free" software.
The eeepc has a significant number of closed source bits and pieces. Most of them don't work quite the way people want, so people learn how to hack things in order to get it to do what they really want - an example is the popup when you plug in an SD card or USB stick. And then ASUS releases an update that breaks that previous hack and people get annoyed.
Also, because there are pieces that aren't open, nobody can easily support the factory install. So openssl is still broken and, last time I looked, no fixed package is available. So ASUS are probably finding that it's quite expensive supporting Linux.
Had they just made the whole thing open then there would rapidly have grown up a "support team" around the factory install. Security issue - within 24 hours the sourceforge project would have been updated. Asus then just need to have someone monitoring the sourceforge project and "backport" any fixes into their official distribution.
A little more work would be involved if they also want to pull in "improvements" from the sourceforge project. But not that much work.
Tim.
Here's the old adage: You know how stupid the average person is? Statistically, half the people are more stupid than that.
Unskilled and Unaware of It: How Difficulties in Recognizing One's Own Incompetence Lead to Inflated Self-Assessments
http://www.apa.org/journals/features/psp7761121.pdf
"This study also enabled us to explore Prediction 3, that incompetent individuals fail to gain insight into their own incompetence by observing the behaviour of other people."
"[After seeing the answers of others] If anything, bottom-quartile participants tended to raise their already inflated self-estimates, although not to a significant degree"
The fundamental problem is that, even with the right answers in front of them, the incompetent are unable to distinguish the right from the wrong answers. What the internet brings to the incompetent is AN answer so now they THINK they know.
The competent can, of course, filter the wheat from the chaff.
I especially like the concluding remarks from that paper. "That worry is that this article may contain faulty logic, methodological errors, or poor communication. Let us assure our readers that to the extent this article is imperfect, it is not a sin we have committed knowingly."
Tim.
Yes, corners first. I don't know about the same 8 move combination but IMO there are only 8 pieces that aren't obvious how to position.
First you do four corners on a face - anybody can do this without any "method". (at least I could)
Then you position the second four corners and rotate them as necessary - this is more tricky.
Then you position the eight edges on opposite faces (leaving the middle slice unsolved). (How to do this was completely obvious to me, YMMV)
Then you need to position and rotate the remaining four edges - again more tricky.
Finally, you might need to reposition the centres (By default I ignore the four centres on the middle slice and minimize the moves for the edges, sometimes leaving me with a four spot pattern to "solve" at the end)
Tim.
If you are saying there is no way to rotate a centre face without moving any of the other pieces then you are wrong.
Proof:
F1U1 (front 90, Upper 90) repeated 105 times returns the cube to its original state.
But 105 is 1 mod 4. Therefore the Front and Upper centres have been rotated by 90 degrees at the end of this move.
R1F1 has similar properties, as does R1U1.
Therefore F1U1^105 R1F1^105 R1U1^315 will rotate F centre through 180 and R and U centers through 360 - i.e. one centre will be rotated by 180 degrees.
Funnily enough, only last week I got a cube that had polarization of the centres. I very quickly found a move to rotate one centre 90 clockwise and an adjacent 90 anticlockwise but the move to rotate a single centre through 180 degrees took me a very long time to find. (The above 525 step move is not original to me and was only pointed out to me once I'd found a 12 step move to do the same thing)
Also the cube I had actually had one centre that could be in any orientation when viewed on its own. Only when comparing with the other centres was it possible to deduce that the centre was rotated from it's factory state.
Tim.
However, they're becoming irrelevant anyway.
http://news.bbc.co.uk/1/hi/education/7434463.stm
Tim.
I think you're getting confused with CPT invariance. Reverse the charge, reverse the parity and reverse time and everything looks the same.
IIRC CP symmetry (reverse the charge, reverse the parity) is broken with the decay of the Kzero and anti-Kzero. (There was an article about this some years or decades ago in SciAm but I'm too lazy to go and look it up now). Given that then I don't think, in general, an anti-particle can be a particle moving backwards in time.
Tim.
Actually, the message telling which bits were send and/or measured at a 45% angle is only exchanged after Bob has measured the quantum bits. So, even can't just store them, measure them and re-inject them after the fact.
So, rather than just eavesdropping the message from Bob to Alice, she would actually need to destructively intercept it and change it.
Eve isn't intercepting the bits. She's creating a pair of correlated photons without actually making any measurement. She sends one on to Bob and stores the other for measuring later.
a|H> + b|V> => a|HH> + b|VV>. I'm pretty sure this transformation is allowed for all values of a and b. Note that she is NOT cloning (a|H> + b|V>)(a|H> + b|V>) which would be prohibited.
Having said that, I think my quick thoughts were incorrect. I was assuming
|HH> = k|++> + k|--> when I think it should have been 1/2|++> + 1/2|+-> + 1/2|-+> + 1/2|-->
(k=1/sqrt2) now that I think about it properly.
Tim.
Or better yet, what would happen if some new device could record without observing?
"record without observing" doesn't make sense. In the words of Pauli, "That's not right. That's not even wrong." But trying to guess that you mean "observe without affecting in any way".
Then you can violate causality. Give me a device that can "record without observing" and I'll build you a device to communicate faster than the speed of light.
Tim.
You don't transmit the message using QC, you transmit a OTP. So if Eve does intercept it then all she gets is a bunch of random bits, Alice and Bob detect the interception, throw the OTP away and start again.
The best that Eve can do is a DOS attack.
Tim.
Grrr. < needed :-(
... Again, hopefully you remember from your school days that, if you send that polarized light though a second filter at an angle to the first, a proportion of the light gets through (cos^2 theta). But, QM says that you cannot predict which photons will make it through the second filter, it's entirely probabilistic.
Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?
Lookup quantum cloning and the "no cloning theorem".
But basically (and this is a naive implementation that won't actually work), Alice transmits to Bob using linearly polarized photons. Now, if you remember from your school days, if you shine a light though a polarization filter and then through another filter at the same angle, all the light that gets through the first filter gets through the second filter as well.
So, let Alice transmit a horizontally polarized photon |H> if she wants to send a 1 and |V> if she wants to send a 0.
Bob uses a horizontally polarized filter <H| to measure Alice's photon. <H|H> = 1 (the photon gets through and he detects it, <H|V> = 0. The photon gets stopped and he doesn't detect it.
So far, so good but... Eve does exactly what you suggested and measures the photon and then regenerates it - so Bob doesn't see any difference.
Now it starts getting clever
So as well as using |H>,|V> to transmit 1 and 0, Alice also uses |+>,|-> where these are 45 degree polarizations. Alice uses one or the other completely at random.
Bob, when he measures at his end also choses whether to measure the horizontal polarization <H| or diagonal polarization <+| completely at random. There are eight cases:
<H|H> = Alice and Bob use the same polarization angle so Bob detects the photon
<H|V> = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
<H|+> = Alice and Bob's filters are at 45 degrees so Bob may or may not detect a photon
<H|-> = ditto
<+|H> = ditto
<+|V> = ditto
<+|+> = Alice and Bob use the same polarization angle so Bob detects the photon
<+|-> = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
Once Alice and Bob have done this, Bob tells Alice which measurement he's done (over a classical channel, they don't care who might eavesdrop.) If Alice and Bob have used the same basis - i.e. Alice used |H>,|V> to transmit her bit and Bob used <H| to measure it or Alice used |+>,|-> to transmit her bit and Bob used <+| to measure it then Alice says "correct" and Bob knows what bit Alice sent. If Bob used the wrong measurement then Alice says "incorrect" and they both throw that bit away.
So, on average, for each two bits that Alice tries to send to Bob, one will be thrown away and the other will be a good value.
Now Eve tries to eavesdrop. If she measures <H| then she'll detect |H> or |V>. She can retransmit that value but, if Alice sent |+> or |-> instead then she'll have corrupted the bit. If she measures <+| instead then she can retransmit |+> or |-> but if Alice sent |H> or |V> then she'll corrupt that bit instead. Infact, on average, regardless of which measurement she makes, she'll end up corrupting 1/2 of the values that Alice and Bob have "successfully" exchanged.
Now Eve can get really clever. Instead of measuring the photon, she can clone it and then measure her clone. Now it turns out that there is a limit to how good her cloning machine can be so, although it won't corrupt half of the bits that Alice and Bob transmit, it will corrupt at least 1/6.
(Actually, in the naive scheme outlined above
Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?
... Again, hopefully you remember from your school days that, if you send that polarized light though a second filter at an angle to the first, a proportion of the light gets through (cos^2 theta). But, QM says that you cannot predict which photons will make it through the second filter, it's entirely probabilistic.
,|-> to transmit her bit and Bob used or |V>. She can retransmit that value but, if Alice sent |+> or |-> instead then she'll have corrupted the bit. If she measures or |-> but if Alice sent |H> or |V> then she'll corrupt that bit instead. Infact, on average, regardless of which measurement she makes, she'll end up corrupting 1/2 of the values that Alice and Bob have "successfully" exchanged.
Lookup quantum cloning and the "no cloning theorem".
But basically (and this is a naive implementation that won't actually work), Alice transmits to Bob using linearly polarized photons. Now, if you remember from your school days, if you shine a light though a polarization filter and then through another filter at the same angle, all the light that gets through the first filter gets through the second filter as well.
So, let Alice transmit a horizontally polarized photon |H> if she wants to send a 1 and |V> if she wants to send a 0.
Bob uses a horizontally polarized filter = 1 (the photon gets through and he detects it, = 0. The photon gets stopped and he doesn't detect it.
So far, so good but... Eve does exactly what you suggested and measures the photon and then regenerates it - so Bob doesn't see any difference.
Now it starts getting clever
So as well as using |H>,|V> to transmit 1 and 0, Alice also uses |+>,|-> where these are 45 degree polarizations. Alice uses one or the other completely at random.
Bob, when he measures at his end also choses whether to measure the horizontal polarization = Alice and Bob use the same polarization angle so Bob detects the photon
= Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
= Alice and Bob's filters are at 45 degrees so Bob may or may not detect a photon
= ditto
= ditto
= ditto
= Alice and Bob use the same polarization angle so Bob detects the photon
= Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
Once Alice and Bob have done this, Bob tells Alice which measurement he's done (over a classical channel, they don't care who might eavesdrop.) If Alice and Bob have used the same basis - i.e. Alice used |H>,|V> to transmit her bit and Bob used
Now Eve can get really clever. Instead of measuring the photon, she can clone it and then measure her clone. Now it turns out that there is a limit to how good her cloning machine can be so, although it won't corrupt half of the bits that Alice and Bob transmit, it will corrupt at least 1/6.
(Actually, in the naive scheme outlined above I think Eve can do:
a|H> + b|V> => a|HH> + b|VV>, store her photon, wait for Bob to measure, eavesdrop the message from Bob to Alice and then make the same measurement on her stored photon. But this only works because the only possible values for a,b in the naive scheme are (0,1), (1,0), (1/sqrt2,1/sqrt2), (1/sqrt2, -1/sqrt2) but I'm right on the limits of my understanding of QM and entangled photons now so I could be completely wrong)
Tim.
QC allows you to securely exchange a OTP. And it doesn't depend on using entangled particles at all.
What is keeping NASA and the ESA from working together to create a tiny habitat to send to mars? I'm not talking anything fancy. How about sending a plant to Mars and keeping it alive? You have all the challenges of putting a living organism into space, getting it to mars, landing it on mars, and getting a habitat inflated, powered up, and surviving, all without having to risk the life of a human being.
Because there's nothing to be gained from doing that that we can't do on Earth.
With the exception of gravity, we can create an environment on Earth identical to that on Mars. So if you want to practice raising plants in a Martian environment then you can do it much more cheaply on Earth.
Instead, use that rocket launch to get more experiments to Mars to investigate things that actually need a presence on the planet.
Tim.
This is just plain wrong.
Muon catalyzed fusion is documented and reproduceable. It can also occur at room temperature or lower.
It's probably not viable as a power source though.
http://en.wikipedia.org/wiki/Muon-catalyzed_fusion
Tim.
Also, the unnamed "scientists" who were lucky enough to find this are Alicia Soderberg of Princeton University & her colleagues, just so we give credit where credit is due.
Completely unrelated, but I had to go back and reread the first name after seeing that surname:
http://en.wikipedia.org/wiki/Lenna
is another Soderberg that geeks might be familiar with.
Tim.
It's not just brokers.
Whenever I have an interview for a job I always take a copy of my CV to give to the person interviewing me. The CV that the interviewer is holding varies from close to what I sent to the agency to bearing absolutely no resemblance to what I wrote. New skills, missing skills, sometimes I think agencies must cut and paste from several different peoples CVs.
Very, very occasionally (happened to me just once) I got a call from the agency - "Is it alright if we change this to say this instead. This client is looking for a particular skill and it's not immediately obvious to a non technical person from what you've written that you've been doing this for the last three years"
CVs can often be dramatically improved without lying. If you know a good salesman then it's worth asking them to help with the writing of your CV. Once you get to the technical people it probably doesn't matter too much but you've probably got to get your CV through HR screening first. I'd have no problems with agencies reworking my CV if they came back to me first to make sure it was accurate.
Tim.