In response to falenas108's "Ask an X" thread. I have a PhD in experimental particle physics; I'm currently working as a postdoc at the University of Cincinnati. Ask me anything, as the saying goes.
This is an experiment. There's nothing I like better than talking about what I do; but I usually find that even quite well-informed people don't know enough to ask questions sufficiently specific that I can answer any better than the next guy. What goes through most people's heads when they hear "particle physics" is, judging by experience, string theory. Well, I dunno nuffin' about string theory - at least not any more than the average layman who has read Brian Greene's book. (Admittedly, neither do string theorists.) I'm equally ignorant about quantum gravity, dark energy, quantum computing, and the Higgs boson - in other words, the big theory stuff that shows up in popular-science articles. For that sort of thing you want a theorist, and not just any theorist at that, but one who works specifically on that problem. On the other hand I'm reasonably well informed about production, decay, and mixing of the charm quark and charmed mesons, but who has heard of that? (Well, now you have.) I know a little about CP violation, a bit about detectors, something about reconstructing and simulating events, a fair amount about how we extract signal from background, and quite a lot about fitting distributions in multiple dimensions.
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.)