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.)