That sounds a little crazy. And how it terms are there? And here's what I mean by doing nothing else. We've never seen a blue electron. It's true. And the process of doing a classical computation is, build a machine which is governed by classical mechanics, OK, that takes this initial configuration and replaces it with a new one, f of 0, 1. So I need to build that computer out of objects available to me. And this gives me 0, and this gives me 0 plus 1, which is 1. And then I impose U-NOT, and it gives me an out state. No. This is the fourth, and presumably final time that I will be teaching this class. If I send a random pile of electrons into a color box, useful thing to know, they come out about half and half. How many people think it's not? So the question was, what's the difference between this experiment and the last one. Consider the single electron. It's sort of like calling mice, mice, and calling all the other mammals non-mice. And that may or may not be a good model the real world. Matt provides this function for us, U. You cannot possibly. Lecture 24: Entanglement: QComputing, EPR, and Bell's Theorem. So again, think about your prediction your head, come to a conclusion, raise your hand when you have an idea. That's too minor. In fact, this could be 0. So remember at the very beginning, we talked about Bell's experiment. So this violates linearity and a unitarity rather badly. And these are great questions. I don't know the details of the history, but the lore is that this was really from his observation. When I say that the barrier is out, what I mean is it's not in the way. » So all this does, it's just a set of mirrors. And I want to run this experiment using the barriers to tease out how the electrons transit through our apparatus. And I take those which come out the white aperture. Have we learned anything about the state of the second particle? ALLAN ADAMS: Oh, sure, you can. So does chalk. Two complex numbers. And here's the reason. So here's a prediction. Neuroscience is much harder. Good. I'm wearing a mic. So how many numbers do I need to specify this state of n qubits? This is so sad. It's correct. No, so 0, 0. So now we take that 50% of electrons that comes out the soft aperture, which had previously been observed to be white and soft. We turn on a magnetic field in z, and what happens is the spin precesses around the z-axis. We turn a magnetic field with a known amplitude with a known amount of time, details here don't matter so much. So again, as usual, the way we take our input to our output is we build a machine that implements an algorithm of our choice by arranging the physical evolution of the system under time. Similarly, with the Hadamard, I take an initial state n, and I've apply U Hadamard. And doesn't seem to have anything to do with f. In fact, we haven't measured f. Two. It was 0, 1/2. OK. We leave it alone. Here's the basic idea. Read full story → Newly observed phenomenon could lead to new quantum devices. Lecture 9: Operator Methods for the Harmonic Oscillator. Now is this the separable state? So this is now the state 1 upon root 2. And the utility of this box is that the color can be inferred from the position. And the empirical fact is that every electron, every electron that's ever been observed is either black or white and no other color. So experiment four. Right? Just a single electron at a time going through this apparatus. All right, now that everyone has had a quick second to think through this one, let me just talk through what I'd expect from the point of these experiments. So how does this differ? Some books are aimed towards people who are interested in materials science, some books that are aimed towards people interested in philosophy. Hold on a sec. We get precession. So this is an i and a minus i. I can pull out the i so that the phi after is i, a phase, upon root 2 times up z, minus down z. Yeah. Well, if they're the same exactly, then either it's 0 plus 0, in which case we get 0, or it's 1 plus 1, in which case we get 0. But it's a superposition, and you've lost certainty that you'll measure up in the z-direction. Lecture 13: More on Scattering. We call this correlation entanglement. I need 0, 1. AUDIENCE: Well, earlier we showed that [INAUDIBLE] so it'll take those paths equally--. Yeah. So cosine pi over 2 is 0. So let me show you entanglement. So actually, this problem was there already in the experiment we did. So the input and output are superpositions of the values we'll measure. Can you do this? But you get these interference terms. And that's quantum mechanics. The answer is you always, always see one electron on one of the paths. OK, so here's something you can't do. Lecture 24. The universe around you is inescapably probabilistic. No one has ever found a fluorescent electron. It introduces the basic features of quantum mechanics. Psi is equal to-- well, that's 0, and we know we've lost this, so this particular subsystem, this particular qubit has been collapsed to the state 0, so we have 0 times c, 0, minus d, 1 for the second particle. That's not random. We tested them over and over. Well, let's think about the logic here. And this is easy to check. So that's the upper end. I think it's going to happen at the end of the semester. So the coefficient bd must be-- both of those must be non-zero. That's what I mean by the interference effects. OK, so let's make all this much more explicit. These are the eigenstates. I don't know whether that's true or not. Suppose I send some random collection of electrons into a color box. So if it took both paths, here's what you should be able to do. So the question is, what do you do if the checking process is probabilistic? Lecture 8: Quantum Harmonic Oscillator. So here is-- I don't know how he came up with this, but he's clever. It's going to be an introduction to a basic idea then is going to haunt, plague, and charm us through the rest of the semester. And you can compute the operator representing spin along a particular direction by taking a unit vector in that direction and taking the dot product. That there's an exactly analogous problem for n qubits. 0 plus 1 times 0, also known as 1 over root 2, 0, 0 plus 1, 0. And we did all of this with a single evaluation of our function f, right here.