Here's a brief description of how one scheme of quantum encryption actually works. Hopefully that helps.

Alice sends a bunch of quantum bits (qubits) to Bob. These qubits are binary information encoded in one of four ways. Either they're encoded as spin up/down (up being 1 and down being 0) or they're encoded in left/right (left being 1 and right being 0). (The choice of 1 and 0 is arbitrary once you've chosen either up/down or left/right.) Alice flips a coin each time she sends a qubit of data to determine if she sends up/down or left/right.

Now here's the tricky part with these up/down left/right states. Bob has to choose whether to align his measurements with up/down or left/right. If he chooses the right alignment, he can tell exactly what Alice sent, i.e. measuring an up with up/down always gives up. If Bob chooses the wrong alignment, however, he will measure incorrectly, getting each incorrect state 50% of the time, i.e. if Bob measures an up state with left/right, he'll get left 50% of the time and right 50% of the time. So what Bob actually does is randomly align his detector.

After Bob is done measuring, he calls Alice up on the phone (this, by the way, is why "faster-than-light" communication fails here), and Alice tells Bob which states were up/down and which were left/right. This doesn't actually give an eavesdropper any useful information, since the states have already been measured and thrown out. Bob then knows the 50% of cases where he improperly aligned his detector are garbage and discards them. The remaining 50% of measurements are proper results.

Now here's the trick to making it eavesdropper proof. Let's say Eve the eavesdropper is catching qubits along the way. She has no way of knowing if her measurement is properly aligned or not, so if she sees an up, she doesn't know if it's an actually up, or a left or a right. Since measuring an up forces it to become an up (one of the rules of quantum mechanics), she has a choice of either sending an up along to Bob, or creating and sending a left or right. If Eve guesses right, she won't be caught, but what if she guesses wrong?

Let's say she sends a left when it was actually an up. If that qubit is one of the 50% that Bob didn't discard when talking to Alice, Bob knows that his measurement of it should be 100% accurate (because he and Alice agreed that they sent and measured with up/down alignment). But since Bob is actually measuring a left from Eve, he has a 50% chance of measuring up and 50% of measuring down. So all Alice has to do is encode a bunch of extra qubits in her data that don't contain the encrypted information. Bob has discarded 50% of them, but she finds those he didn't discard and tells him (over the phone), "By the way, bits x,y,z,... are extra data and should have values of A,B,C,..." Bob then checks his measured values for those qubits. If he finds errors, then he knows someone has tampered with the qubits along the way. If not, he trusts that the remaining qubits are safe.

You've probably noticed that all of this has to do with probabilities. So if Eve gets really lucky and happens to send out exact copies of the original states, she won't be detected. The chance of this happening can be made arbitrarily small by sending more of those extra qubits.