Alice sends a photon to Bob. If Eve tries to measure the polarization, and measures it on the wrong axis, there’s a chance Bob won’t get the result Alice sent. From what I understand, if Eve copies the photon, using a laser or some other method of getting entangled photons, and she measures the copied photon, the same result will happen to Bob. What happens if Eve copies the photon, and waits until Bob reads it before she does?
Also, you referred to virtual particles as a convenient fiction when responding to someone else. I assumed that they were akin to a particle being in a place with more potential energy than there is energy in a system during quantum tunneling. The particle is real. It’s just that due to the fact that the kinetic energy is negative, it behaves in a way that makes the waveform small at any real distance. Was I completely off base?
Also, should I have just edited my old post instead of adding a new one?
What happens if Eve copies the photon, and waits until Bob reads it before she does?
Not my field, but it seems to me that it should be the same thing that happens if Bob tries to read the photon after Eve has already done so. You can only read the quantum information off once. Now, an interesting question is, what happens if Eve goes off into space at near lightspeed, and reads the photon at a time such that the information “Bob has read the photon” hasn’t had time to get to her spaceship? If I understand correctly, it doesn’t matter! This scenario is just a variant of the Bell’s-inequality experiment.
Also, you referred to virtual particles as a convenient fiction when responding to someone else. I assumed that they were akin to a particle being in a place with more potential energy than there is energy in a system during quantum tunneling. The particle is real. It’s just that due to the fact that the kinetic energy is negative, it behaves in a way that makes the waveform small at any real distance. Was I completely off base?
So firstly, in quantum tunneling the particle never occupies the forbidden area. It goes from one allowed area to another without occupying the space between; hence the phrase “quantum leap”. Of course this is not so difficult to imagine when you think of a probability cloud rather than a particle; if you think of a system with parts ABC, where B is forbidden but A and C are allowed, then there is at any time a zero probability of finding the particle in B, but a nonzero probability to find it in A and C. This is true even if at some earlier time you find it in A, because, so to speak, the wave function can go where the particle can’t. So, yes, if you ever found the particle in B its kinetic energy would be negative, but in fact that doesn’t happen. So now we come to matters of taste: The wave function does exist within B; is this a mathematical fiction, because no experiment will find the particle there, or is it real since it explains how you can find the particle at C?
Then, back to virtual particles. The mass of a virtual particle can be negative; it is really unclear to me what it would even mean to observe such a thing. Therefore I think of them as a convenient fiction. But they are certainly a very helpful fiction, so, you know, take your choice.
Also, should I have just edited my old post instead of adding a new one?
I don’t think so, the number of comments here is so large that it would be very easy to miss an edit.
You can only read the quantum information off once.
Bob knows the right way to polarize it, though. If Eve tries to read it but polarizes it wrong, it would mess with the polarization of Bob’s particle, so there’s a chance he’d notice. If Bob polarizes it the way Alice did, and then Eve polarizes it wrong when she reads it, will Bob notice? If Bob notices, he just predicted the future. If he does not, then he can tell whether or not when Eve reads it constitutes “future”, violating relativity of simultaneity.
So firstly, in quantum tunneling the particle never occupies the forbidden area.
If you solve Schroedinger’s time-independent equation for a finite well, there is non-zero amplitude outside the well. If you calculate kinetic energy on that part of the waveform, it will come out negative. You obviously wouldn’t be able to observe it outside the well, in the sense of getting it to decohere to a state where it’s mostly outside the well, without giving it enough energy to be in that state. That’s just a statement about how the system evolves when you put a sensor in it. If you trust the Born probabilites and calculate the probability of being in a configuration space with a particle mid-quantum tunnel, it will come out finite.
… it is really unclear to me what it would even mean to observe such a thing.
I don’t really care about observation. It’s just a special case of how the system evolves when there’s a sensor in it. I want to know how virtual particles act on their own. Do they evolve in a way fundamentally different from particles with positive kinetic energy, or are they just what you get when you set up a waveform to have negative energy, and watch it evolve?
Bob knows the right way to polarize it, though. If Eve tries to read it but polarizes it wrong, it would mess with the polarization of Bob’s particle, so there’s a chance he’d notice. If Bob polarizes it the way Alice did, and then Eve polarizes it wrong when she reads it, will Bob notice? If Bob notices, he just predicted the future. If he does not, then he can tell whether or not when Eve reads it constitutes “future”, violating relativity of simultaneity.
Good point. My initial answer wasn’t fully thought through; I again have to note that this isn’t really my area of expertise. There is apparently something called the no-cloning theorem, which states that there is no way to copy arbitrary quantum states with perfect fidelity and without changing the state you want to copy. So the answer appears to be that Eve can’t make a copy for later reading without alerting Bob that his message is compromised. However, it seems to be possible to copy imperfectly without changing the original; so Eve can get a corrupted copy.
There is presumably some tradeoff between the corruption of your copy, and the disturbance in the original message. You want to keep the latter below the expected noise level, so for a given noise level there is some upper limit on the fidelity of your copying. To understand whether this is actually a viable way of acquiring keys, you’d have to run the actual numbers. For example, if you can get 1024-bit keys with one expected error, you’re golden: Just try the key with each bit flipped and each combination of two bits flipped, and see if you get a legible message. This is about a million tries, trivial. (Even so, Alice can make things arbitrarily difficult by increasing the size of the key.) If we expected corruption in half the bits, that’s something else again.
I don’t know what the limits on copying fidelity actually are, so I can’t tell you which scenario is more realistic.
As I say, this is a bit out of my expertise; please consider that we are discussing this as equals rather than me having the higher status. :)
If you solve Schroedinger’s time-independent equation for a finite well, there is non-zero amplitude outside the well. If you calculate kinetic energy on that part of the waveform, it will come out negative. You obviously wouldn’t be able to observe it outside the well, in the sense of getting it to decohere to a state where it’s mostly outside the well, without giving it enough energy to be in that state. That’s just a statement about how the system evolves when you put a sensor in it. If you trust the Born probabilites and calculate the probability of being in a configuration space with a particle mid-quantum tunnel, it will come out finite.
You are correct. It seems to me, however, that you would not actually observe a negative energy; you would instead be seeing the Heisenberg relation between energy and time, \Delta E \Delta t >= hbar/2; in other words, the particle energy has a fundamental uncertainty in it and this allows it to occupy the classically forbidden region for short periods of time.
I don’t really care about observation. It’s just a special case of how the system evolves when there’s a sensor in it. I want to know how virtual particles act on their own. Do they evolve in a way fundamentally different from particles with positive kinetic energy, or are they just what you get when you set up a waveform to have negative energy, and watch it evolve?
Your original question was whether virtual particles are real; perhaps I should ask you, first, to define the term. :) However, they are at least as real as the different paths taken by the electron in the two-slit experiment; if you set things up so that particular virtual-particle energies are impossible, the observed probabilities change, jsut like blocking one of the slits.
Well, as they can have negative mass you have to assume that their gravitational interactions are, to coin a phrase, counterintuitive. (That is, even for quantum physicists! :) ) But, of course, we don’t have any sort of theory for that. As far as interactions that we actually know something about go, they are the same, modulo the different mass in the propagator. (That is, the squiggly line in the Feynman diagram, which has its own term in the actual path integral; you have to integrate over the masses.)
I’m confused about part of quantum encryption.
Alice sends a photon to Bob. If Eve tries to measure the polarization, and measures it on the wrong axis, there’s a chance Bob won’t get the result Alice sent. From what I understand, if Eve copies the photon, using a laser or some other method of getting entangled photons, and she measures the copied photon, the same result will happen to Bob. What happens if Eve copies the photon, and waits until Bob reads it before she does?
Also, you referred to virtual particles as a convenient fiction when responding to someone else. I assumed that they were akin to a particle being in a place with more potential energy than there is energy in a system during quantum tunneling. The particle is real. It’s just that due to the fact that the kinetic energy is negative, it behaves in a way that makes the waveform small at any real distance. Was I completely off base?
Also, should I have just edited my old post instead of adding a new one?
Not my field, but it seems to me that it should be the same thing that happens if Bob tries to read the photon after Eve has already done so. You can only read the quantum information off once. Now, an interesting question is, what happens if Eve goes off into space at near lightspeed, and reads the photon at a time such that the information “Bob has read the photon” hasn’t had time to get to her spaceship? If I understand correctly, it doesn’t matter! This scenario is just a variant of the Bell’s-inequality experiment.
So firstly, in quantum tunneling the particle never occupies the forbidden area. It goes from one allowed area to another without occupying the space between; hence the phrase “quantum leap”. Of course this is not so difficult to imagine when you think of a probability cloud rather than a particle; if you think of a system with parts ABC, where B is forbidden but A and C are allowed, then there is at any time a zero probability of finding the particle in B, but a nonzero probability to find it in A and C. This is true even if at some earlier time you find it in A, because, so to speak, the wave function can go where the particle can’t. So, yes, if you ever found the particle in B its kinetic energy would be negative, but in fact that doesn’t happen. So now we come to matters of taste: The wave function does exist within B; is this a mathematical fiction, because no experiment will find the particle there, or is it real since it explains how you can find the particle at C?
Then, back to virtual particles. The mass of a virtual particle can be negative; it is really unclear to me what it would even mean to observe such a thing. Therefore I think of them as a convenient fiction. But they are certainly a very helpful fiction, so, you know, take your choice.
I don’t think so, the number of comments here is so large that it would be very easy to miss an edit.
Bob knows the right way to polarize it, though. If Eve tries to read it but polarizes it wrong, it would mess with the polarization of Bob’s particle, so there’s a chance he’d notice. If Bob polarizes it the way Alice did, and then Eve polarizes it wrong when she reads it, will Bob notice? If Bob notices, he just predicted the future. If he does not, then he can tell whether or not when Eve reads it constitutes “future”, violating relativity of simultaneity.
If you solve Schroedinger’s time-independent equation for a finite well, there is non-zero amplitude outside the well. If you calculate kinetic energy on that part of the waveform, it will come out negative. You obviously wouldn’t be able to observe it outside the well, in the sense of getting it to decohere to a state where it’s mostly outside the well, without giving it enough energy to be in that state. That’s just a statement about how the system evolves when you put a sensor in it. If you trust the Born probabilites and calculate the probability of being in a configuration space with a particle mid-quantum tunnel, it will come out finite.
I don’t really care about observation. It’s just a special case of how the system evolves when there’s a sensor in it. I want to know how virtual particles act on their own. Do they evolve in a way fundamentally different from particles with positive kinetic energy, or are they just what you get when you set up a waveform to have negative energy, and watch it evolve?
Good point. My initial answer wasn’t fully thought through; I again have to note that this isn’t really my area of expertise. There is apparently something called the no-cloning theorem, which states that there is no way to copy arbitrary quantum states with perfect fidelity and without changing the state you want to copy. So the answer appears to be that Eve can’t make a copy for later reading without alerting Bob that his message is compromised. However, it seems to be possible to copy imperfectly without changing the original; so Eve can get a corrupted copy.
There is presumably some tradeoff between the corruption of your copy, and the disturbance in the original message. You want to keep the latter below the expected noise level, so for a given noise level there is some upper limit on the fidelity of your copying. To understand whether this is actually a viable way of acquiring keys, you’d have to run the actual numbers. For example, if you can get 1024-bit keys with one expected error, you’re golden: Just try the key with each bit flipped and each combination of two bits flipped, and see if you get a legible message. This is about a million tries, trivial. (Even so, Alice can make things arbitrarily difficult by increasing the size of the key.) If we expected corruption in half the bits, that’s something else again.
I don’t know what the limits on copying fidelity actually are, so I can’t tell you which scenario is more realistic.
As I say, this is a bit out of my expertise; please consider that we are discussing this as equals rather than me having the higher status. :)
You are correct. It seems to me, however, that you would not actually observe a negative energy; you would instead be seeing the Heisenberg relation between energy and time, \Delta E \Delta t >= hbar/2; in other words, the particle energy has a fundamental uncertainty in it and this allows it to occupy the classically forbidden region for short periods of time.
Your original question was whether virtual particles are real; perhaps I should ask you, first, to define the term. :) However, they are at least as real as the different paths taken by the electron in the two-slit experiment; if you set things up so that particular virtual-particle energies are impossible, the observed probabilities change, jsut like blocking one of the slits.
Well, as they can have negative mass you have to assume that their gravitational interactions are, to coin a phrase, counterintuitive. (That is, even for quantum physicists! :) ) But, of course, we don’t have any sort of theory for that. As far as interactions that we actually know something about go, they are the same, modulo the different mass in the propagator. (That is, the squiggly line in the Feynman diagram, which has its own term in the actual path integral; you have to integrate over the masses.)