Well, you are missing the barn, but that's not really your fault. After all, the power of quantum computing doesn't come from the properties of one qubit (which is what I described above), but from the behaviour of several of them (of course understanding one of them is a key ingredient of understanding several; however other than with classical bits, it's not sufficient). The magic thing which happens there (and which simply doesn't exist in a single qubit) is something called entanglement. It's entanglement which makes quantum computers special. And it's exactly the controlled not gate which creates entanglement (well, there are of course other ways to create entanglement, but CNOT is the method of choice in quantum computing). The problem with entanglement is, however, that it cannot be such easily depicted as the behaviour of a single qubit. Let me try anyway:
First, let's look at what a COT does on classical bits. There's one control bit, and one target bit. If the control bit is 1, then the target bit is flipped (i.e. a NOT operation is done on it), and if the control bit is 0, the target bit is left unchanged (that's why it is called controlled not: The control bit controls if a "not" operation is applied on the target bit, or if it is not applied).
Ok, now what if we replace the classical bits with quantum bits? Replacing the target bit is trivial: As explained above, a NOT is just a 180 degree rotation around an axis through the equator. However what about replacing the control bit? Now, let's look at the behaviour which results:
First the obvious cases: If the control qubit is exactly 1 (i.e. at the north pole), the target qubit is simply rotated. On the other hand, if the control qubit is exactly 0 (i.e. at the south pole), the target bit is left alone. This is nothing but the straightforward translation of the classical behaviour.
But what happens if the control qubit is in a superposition of 0 and 1 (i.e. somewhere else on the sphere)? Note that we don't want to do a measurement during the process (otherwise it would be simple: Measuring the control qubit gives a definite 0 or 1, and then we could apply the corresponding transformation). On the other hand, if we measure the control qubit after applying the CNOT, we still find that the target qubit has rotated exactly if we measure an 1. But remember that it's completely unpredictable if we will measure 0 or 1, thus we cannot predict if the target qubit was rotated or not. That is, the target qubit is in some sense rotated and unrotated at the same time! However, if looked at in isolation, the target qubit cannot be distinguished from the one we would have gotten by just measuring the control qubit, applying or not applying the rotation depending on the result, and then simply forgetting the measurement result. However, one can distinguish both cases if one can measure both qubits: You can measure the control qubit along another axis (or equivalently, rotate it before measuring it), and you'll get (at least statistically) different results in both cases. Indeed, if you look at it from the rotation axis, but in both qubits, you'll even find that the roles of the control and target qubit are reversed, i.e. if the target qubit happens to be in one of the two eigenstates of the rotation, the control qubit is rotated around the z axis, and for the other eigenstate it's not changed. This shows that the CNOT operation really affects not only the "target" qubit, but also the "control" qubit. Of course for a pure 0 or 1 state of the control qubit, this rotation doesn't have any effect, because those states are exactly the eigenstates of the z-axis rotation.
Now where do those "both-rotated-and-not-rotated" states lie on the bloch sphere? Well, if you look at the single qubit in isolation, you can describe its state by some point inside the sphere, but that's not the whole picture (indeed, being in the inside always tells you that there
If a qubit is both 0 and 1 at the same time, what the hell does an inverter do? Make it 1 and 0 at the same time instead?
Yes. If it consists of "the same amount" 0 and 1, then it will simply not change. That's called an eigenstate of the operator. However, you might be in a superposition of 0 and 1 where you have a bit more 0 than 1, and then you'll get something which is a bit more 1 than 0. Actually, it's still a bit more complicated than that. There are many ways to have a state which is 0 and 1 at the same time, and in most cases, the not gate will change one such state int another one.
Actually the easiest way to understand a single qubit is to look at the so-called bloch sphere. If you take a sphere (i.e. the surface of a ball), then the states 0 and 1 are on opposite ends of the sphere, say the south pole and the north pole. Then every single point on the sphere represents a state of the qubit. The equator represents those states which are 0 and 1 to the same amount, the northern hemisphere contains all states which are more 1 than 0, and the southern hemisphere contains all the states which are more 0 than 1. Now a not operation means turning the sphere around an axis through the equator by 180 degrees. This obviously moves the south pole to the north pole, and vice versa (i.e. 0 gets 1, and 1 gets 0, so it's obviously a not gate). Anything on the equator (i.e. anything which is "half 0 and half 1") gets back to the equator; however except for the two points on the turning axis (the eigenstates of the operation), every point ends up at a different place on the equator.
Now if you measure the qubit, you only get a 0 or 1 (you can in principle also measure along other axes, but since you can get any axis onto the north-south axis by simply rotating the sphere, it suffices to look at the one axis). If the state is the 1 state (the north pole), you are sure to get an 1, and if it's the 0 state (the south pole), you are sure to get a 0. For other states, you may get either 0 or 1 (in that sense they are both 0 and 1 at the same time), but the probability of getting 1 is the higher, the more northern you are. At the equator you are half way between south pole and north pole, so you get 0 and 1 with equal probability. In the southern hemisphere, you're more likely to get 0 as result, and on the northern hemisphere, you're more likely to get 1 as a result.
Of course a quantum floating point exception, which stops your quantum program. Now the interesting part is if you divide by a number that is both 0 and not 0. Then you both trigger and don't trigger an exception, which means your program both will terminate and not terminate at the same time. And you thought the classical halting problem was tough...:-)
In Soviet Russia, quantum dots teleport YOU... oh wait, that wasn't Soviet Russia, it was USS Enterprise. And it wasn't quantum dots, but Scotty. And it also wasn't YOU, but Captain Kirk. But except for those unimportant details, it's correct!:-)
Or similar, what if you created a clone of yourself, and then the clone killed the original? Now change that to a process where the clone-creation itself destroyed the original, and you'd be right at the teleportation scenario.
Could it be possible to organise things such that information sent from a third party, equidistant between the sender and reciever, was used as the classical communication channel?
No. The classical information sent is generated by the sending process itself, at the place of the sender.
Actually the speed is determined by the speed of the classical information you need to transmit in the process. So if you send some messenger who carries it the traditional way, quantum teleportation speed is limited by how fast the messenger can run (assuming you manage to preserve the quantum state for so long, of course).
The fact that nothing you do to the one part of the particle can directly influence any measurement outcome on the other side. You need additional (classical) information from the sending side in order to distinguish any measurement results from completely random garbage. One way to think of it is to think of the quantum information as encrypted by an automatically generated one-time pad during transmission, and you cannot read it until you get the pad, which is only revealed to the sender and thus has to be transmitted classically.
While this is quantum mechanics, you also have to define "moving matter" quantum mechanically. That is, for true matter transport, one of the teleported quantum properties would have to be the particle number. In other words, the quantum teleportation would have to change the expectation value of the particle number operator depending on its expectation value on the sender's side before teleportation.
Not every interaction between a particle and its surrounding is a measurement. But the point is that in quantum teleportation you don't measure the state you want to teleport. There is a measurement involved (and it's that measurement which destroys the original), but the measurement's outcome doesn't give you any information about the transmitted state. The outcome is, however, needed to reconstruct that state on the receiving side (that's why you can't transmit the state faster than light; you just need that classical information).
Of course you are free to measure the state after receiving it.
If you would "teleport" position, then the position of the "teleported" particle would obviously be at the same position as the original. Of course that type of teleportation is also possible; it's usually called "doing nothing":-)
im forbidden to use gpl code in a closed application, i MUST release all my source code if i use gpl code in my application.
Yes. That's because by using the code in a closed source application you'd not give to those receiving the closed source application the rights you got for the GPLed code. In short, you don't get the right to not pass on the rights you did get for the code you received.
In effect there are two things you have to differentiate. The first is things you actually do, like reading the code, changing it, compiling it, running it, distributing it, etc. The GPL allows you to do all those things. The second is things you permit or forbid, that is, telling others what they may do. That's a freedom the GPL does very explicitly not give you.
The BSD license obviously gives you personally more freedom (because you are not only allowed to do something, but you are also allowed to deny those same things from others). But that's because the GPL is designed to increase overall freedom. That's comparable to a law which disallows slavery: That law obviously reduces your freedom (by taking away your freedom to have slaves), but it does so in order to increase overall freedom (by denying you from restricting the freedom of others).
But this isn't an OSI approved license: Open CASCADE Technology Public License (version 6.2) This company has been abusing the definition of "open source" for years. A shame really, as the software is very nice.
I'm too lazy to go into the details of the license. Which of the ten rules does it violate? Or is it just that the license doesn't have the "approval stamp"?
Since acceleration is equivalent to gravity, this might be solved by having regular acceleration phases long enough for women to ovulate again. Or even frequent enough so they don't stop ovulating to begin with. Given the other health issues caused by zero-G, it's probably a good idea to have regular acceleration phases anyway.
Slashdot Mirror
Well, maybe penguins can afford smaller brains because they use them more efficiently ... :-)
But did that shark have a frigging laser attached to its head?
The researchers didn't find windows in prehistoric Peru.
Well, you are missing the barn, but that's not really your fault. After all, the power of quantum computing doesn't come from the properties of one qubit (which is what I described above), but from the behaviour of several of them (of course understanding one of them is a key ingredient of understanding several; however other than with classical bits, it's not sufficient). The magic thing which happens there (and which simply doesn't exist in a single qubit) is something called entanglement. It's entanglement which makes quantum computers special. And it's exactly the controlled not gate which creates entanglement (well, there are of course other ways to create entanglement, but CNOT is the method of choice in quantum computing). The problem with entanglement is, however, that it cannot be such easily depicted as the behaviour of a single qubit. Let me try anyway:
First, let's look at what a COT does on classical bits. There's one control bit, and one target bit. If the control bit is 1, then the target bit is flipped (i.e. a NOT operation is done on it), and if the control bit is 0, the target bit is left unchanged (that's why it is called controlled not: The control bit controls if a "not" operation is applied on the target bit, or if it is not applied).
Ok, now what if we replace the classical bits with quantum bits? Replacing the target bit is trivial: As explained above, a NOT is just a 180 degree rotation around an axis through the equator. However what about replacing the control bit? Now, let's look at the behaviour which results:
First the obvious cases: If the control qubit is exactly 1 (i.e. at the north pole), the target qubit is simply rotated. On the other hand, if the control qubit is exactly 0 (i.e. at the south pole), the target bit is left alone. This is nothing but the straightforward translation of the classical behaviour.
But what happens if the control qubit is in a superposition of 0 and 1 (i.e. somewhere else on the sphere)? Note that we don't want to do a measurement during the process (otherwise it would be simple: Measuring the control qubit gives a definite 0 or 1, and then we could apply the corresponding transformation). On the other hand, if we measure the control qubit after applying the CNOT, we still find that the target qubit has rotated exactly if we measure an 1. But remember that it's completely unpredictable if we will measure 0 or 1, thus we cannot predict if the target qubit was rotated or not. That is, the target qubit is in some sense rotated and unrotated at the same time! However, if looked at in isolation, the target qubit cannot be distinguished from the one we would have gotten by just measuring the control qubit, applying or not applying the rotation depending on the result, and then simply forgetting the measurement result. However, one can distinguish both cases if one can measure both qubits: You can measure the control qubit along another axis (or equivalently, rotate it before measuring it), and you'll get (at least statistically) different results in both cases. Indeed, if you look at it from the rotation axis, but in both qubits, you'll even find that the roles of the control and target qubit are reversed, i.e. if the target qubit happens to be in one of the two eigenstates of the rotation, the control qubit is rotated around the z axis, and for the other eigenstate it's not changed. This shows that the CNOT operation really affects not only the "target" qubit, but also the "control" qubit. Of course for a pure 0 or 1 state of the control qubit, this rotation doesn't have any effect, because those states are exactly the eigenstates of the z-axis rotation.
Now where do those "both-rotated-and-not-rotated" states lie on the bloch sphere? Well, if you look at the single qubit in isolation, you can describe its state by some point inside the sphere, but that's not the whole picture (indeed, being in the inside always tells you that there
Yes. If it consists of "the same amount" 0 and 1, then it will simply not change. That's called an eigenstate of the operator. However, you might be in a superposition of 0 and 1 where you have a bit more 0 than 1, and then you'll get something which is a bit more 1 than 0.
Actually, it's still a bit more complicated than that. There are many ways to have a state which is 0 and 1 at the same time, and in most cases, the not gate will change one such state int another one.
Actually the easiest way to understand a single qubit is to look at the so-called bloch sphere. If you take a sphere (i.e. the surface of a ball), then the states 0 and 1 are on opposite ends of the sphere, say the south pole and the north pole. Then every single point on the sphere represents a state of the qubit. The equator represents those states which are 0 and 1 to the same amount, the northern hemisphere contains all states which are more 1 than 0, and the southern hemisphere contains all the states which are more 0 than 1. Now a not operation means turning the sphere around an axis through the equator by 180 degrees. This obviously moves the south pole to the north pole, and vice versa (i.e. 0 gets 1, and 1 gets 0, so it's obviously a not gate). Anything on the equator (i.e. anything which is "half 0 and half 1") gets back to the equator; however except for the two points on the turning axis (the eigenstates of the operation), every point ends up at a different place on the equator.
Now if you measure the qubit, you only get a 0 or 1 (you can in principle also measure along other axes, but since you can get any axis onto the north-south axis by simply rotating the sphere, it suffices to look at the one axis). If the state is the 1 state (the north pole), you are sure to get an 1, and if it's the 0 state (the south pole), you are sure to get a 0. For other states, you may get either 0 or 1 (in that sense they are both 0 and 1 at the same time), but the probability of getting 1 is the higher, the more northern you are. At the equator you are half way between south pole and north pole, so you get 0 and 1 with equal probability. In the southern hemisphere, you're more likely to get 0 as result, and on the northern hemisphere, you're more likely to get 1 as a result.
Well, as long as it's not quantum Gates ... :-)
Of course a quantum floating point exception, which stops your quantum program. Now the interesting part is if you divide by a number that is both 0 and not 0. Then you both trigger and don't trigger an exception, which means your program both will terminate and not terminate at the same time. And you thought the classical halting problem was tough ... :-)
And that helps? Yes. It will kill all the bugs.
In Soviet Russia, quantum dots teleport YOU ... oh wait, that wasn't Soviet Russia, it was USS Enterprise. And it wasn't quantum dots, but Scotty. And it also wasn't YOU, but Captain Kirk. But except for those unimportant details, it's correct! :-)
Or similar, what if you created a clone of yourself, and then the clone killed the original?
Now change that to a process where the clone-creation itself destroyed the original, and you'd be right at the teleportation scenario.
No. The classical information sent is generated by the sending process itself, at the place of the sender.
It's effectively a one-time pad. It's unbreakable, even in principle.
Any amount of redundancy will not help with one-time pads.
Actually the speed is determined by the speed of the classical information you need to transmit in the process. So if you send some messenger who carries it the traditional way, quantum teleportation speed is limited by how fast the messenger can run (assuming you manage to preserve the quantum state for so long, of course).
The fact that nothing you do to the one part of the particle can directly influence any measurement outcome on the other side. You need additional (classical) information from the sending side in order to distinguish any measurement results from completely random garbage. One way to think of it is to think of the quantum information as encrypted by an automatically generated one-time pad during transmission, and you cannot read it until you get the pad, which is only revealed to the sender and thus has to be transmitted classically.
While this is quantum mechanics, you also have to define "moving matter" quantum mechanically. That is, for true matter transport, one of the teleported quantum properties would have to be the particle number. In other words, the quantum teleportation would have to change the expectation value of the particle number operator depending on its expectation value on the sender's side before teleportation.
Not every interaction between a particle and its surrounding is a measurement. But the point is that in quantum teleportation you don't measure the state you want to teleport. There is a measurement involved (and it's that measurement which destroys the original), but the measurement's outcome doesn't give you any information about the transmitted state. The outcome is, however, needed to reconstruct that state on the receiving side (that's why you can't transmit the state faster than light; you just need that classical information).
Of course you are free to measure the state after receiving it.
If you would "teleport" position, then the position of the "teleported" particle would obviously be at the same position as the original. Of course that type of teleportation is also possible; it's usually called "doing nothing" :-)
Why? Is there a better explanation for why the WMDs in Iraq were not found? They were simply teleported away!
SCNR
Yes. That's because by using the code in a closed source application you'd not give to those receiving the closed source application the rights you got for the GPLed code. In short, you don't get the right to not pass on the rights you did get for the code you received.
In effect there are two things you have to differentiate. The first is things you actually do, like reading the code, changing it, compiling it, running it, distributing it, etc. The GPL allows you to do all those things. The second is things you permit or forbid, that is, telling others what they may do. That's a freedom the GPL does very explicitly not give you.
The BSD license obviously gives you personally more freedom (because you are not only allowed to do something, but you are also allowed to deny those same things from others). But that's because the GPL is designed to increase overall freedom. That's comparable to a law which disallows slavery: That law obviously reduces your freedom (by taking away your freedom to have slaves), but it does so in order to increase overall freedom (by denying you from restricting the freedom of others).
I'm too lazy to go into the details of the license. Which of the ten rules does it violate?
Or is it just that the license doesn't have the "approval stamp"?
Well, getting out is easy. It's also very final ...
Since acceleration is equivalent to gravity, this might be solved by having regular acceleration phases long enough for women to ovulate again. Or even frequent enough so they don't stop ovulating to begin with. Given the other health issues caused by zero-G, it's probably a good idea to have regular acceleration phases anyway.
It's its store's stores. :-)
Well, maybe next they'll offer their experience in finding copyright violators to hunt down violators of the copyright on GPLed software? :-)