University of California - Axioms of QM, Two Qubits, and Quantum Entanglement lyrics

SPEAKER 1: In this lecture, we'll talk about systems of two qubits. These quantum systems exhibit a remarkable property called entanglement. And this property of entitlement plays a critical role in quantumcomputation. So in this and the next couple of lectures, we'll try to get an intuition for entanglement and what this property's all about. But before we do that, in this video, I want to formalize what we spoke about last time about qubits and super positions. OK. So let's get started. So recall that the energy of an electron in an atom is quantized. What this means is that the electron cannot take one any arbitrary energy, but the energy must be in one of a discrete number of energy levels.So now imagine that we've limited the energy of the electron so that it's in the ground state, or the first excited state, and so on, up through the k minus first excited state. So if this electron were a cla**ical system, then it could store one of K pieces of information, which we might denote by 0, 1, through k minus 1. Now the superposition principle, which is one of the basic axioms of quantum mechanics says that, in general, the state of the system is going to be any linear superposition of these allowable states. So what the superposition principle says is that the general state is a linear superposition of 0 through k minus 1, each with its amplitude, alpha sub j, which is a complex number. And these amplitudes are normalized so that summation alpha j magnitude squared is equal to 1 for j equal to 0 to k minus 1. OK so as we saw last time, interpreting what the state means is not very easy. Because it's hard to make sense of what it means when we say the electron is in the ground state with amplitude minus 1 over 2 or 1 over 2 plus I over 2. ------------------------------------------------------------------------------------------- OK so now of course, one way we can interpret this is through there measurement axiom, which says that when we measure the system, or when we look at it, then the probability that the outcome is j is the magnitude of alpha j squared. So the fact that the state is normalized means that we get, with probability 1, we see some outcome j between 0 and k minus 1. Moreover, we also saw that the measurement disturbs the system and the new state, which we'll denote by psi prime, is the j-th excited state if the measurement outcome was j. So we can do a quick example. So if k was 3, so we have a three-state system, we might have our state as 0 with aptitude 1/2 plus I over 2, 1 with amplitude minus 1 over 2, 2 with amplitude I over 2. And now if we measure, the probability vc0 is 1/2, and the new state is 0. The probability vc1 is 1/4, new state is 1. And the probability vc2 is 1/4, and the new state would be just ket 2. Here's a nice cartoon which depicts how, in the world that we are used to, what a superposition might look like. A superposition of taking the left route around the tree and the right route around the tree. -------------------------------------------------------------------------------------------- OK. So now, let's look at a geometric interpretation of the quantum state. So geometrically, here's what the superposition principle says. So what it says is that the state of a quantum system, a K-level quantum system, is a unit vector in a K-dimensional complex vector space. So what it says is that state is given by a unit vector in a k-dimensional complex vector space. And this vector space is also called a Hilbert space. OK. So what do we mean by this? So we have a k-dimensional complex vector space, which has an orthonormal basis consisting of the states ket 0, ket 1, ket 2, through ket k minus 1. And the state psi is a unit vector. And if we write psi as alpha 0,0 plus alpha 1, 1 plus alpha ket 2. We could also have written it as alpha 0, alpha 1, alpha 2, in standard vector notation, where alpha 0, alpha 1, alpha 2 are complex numbers. So it belongs to a three-dimensional complex vector space. We can also talk about what happens when we measure the system. So what happens when we measure is the state vector psi gets projected onto one of the basis states. So if you're measuring in the standard basis 0, 1, 2, then psi gets projected onto the state zero with probability which is cosine squared of the angle theta zero it makes with the state zero. So the probability that the outcome is zero is cosine squared theta 0. And if that's the outcome, then psi is projected onto the zero state, it becomes the zero state. Now this is true, in general, that the state vector is projected onto each of the vectors in the orthonormal bases in which we are measuring, with probability cosine squared theta, where theta is the angle which it makes with that particular vector. But how do we define theta in general? So if you are given two different vectors, and maybe they are complex vectors, how to we define theta or cosine theta? So remember that the way we define cosine theta --so if you have two vectors-- so if you have this vector psi, which is given by alpha 0, alpha 1, alpha 2. And phi, which is given by beta 0, beta 1, beta 2, And these are complex numbers. Then the way we define the cosine of angle between them, is we add the inner product between alpha 0--between psi and phi. Which you get by taking the sum of the corresponding products. And if these are complex numbers, then you take convex conjugates of one side. So the inner product is just alpha 0 complex conjugate times beta 0 plus alpha 1 complex conjugate times beta 1 plus alpha 2 complex conjugate times beta 2. Which is also given by taking the row vector with complex conjugate of psi, and then the column vector, phi. And we can define the angle between these two vectors by saying that cosine theta is the absolute value of the inner product. Sorry, and if these are complex numbers, then it's not the absolute value. It's the magnitude of the complex number. ----------------------------------------------------------------------------------------- OK. Let's work out an example of all this. So let's say that we're working in a two-dimensional space. So it has our standard bases, ket 0, ket 1. So we are working with a qubit. These are the standard basis states, zero and one. These are the states plus and minus that we talked about last time. Plus is an equal superposition of zero and one. Minus is the state which is 1 over square root 2, 0 minus 1 over square root 2, 1. And now, let's take a state psi, which looks like this. It's one half 0 plus square root 3 over 2, 1. So you know that this, in a plus of course, makes a 45 degree angle. And what about psi? It makes a 60 degree angle. And so the angle between psi and plus is 15 degrees. So if you were measuring psi in the plus-minus basis, the probability you would see plus is, of course, cosine squared 15. But how would you work this out as a vector? Well what you would say is that the probability that the outcome of the measurement is plus is just given by the inner product between these two vectors. So what's the inner product? Well you can just multiply corresponding coordinates. Or you can write it like 1/2 square root 3 over 2 times 1 over square root 2, 1 over square root 2. Of course, you've got to take this product squared, the magnitude squared. But everything is real, so you just take the square. And that's 1 over 2 square root 2 plus square root 3 over 2 square root 2, whole thing squared. Which is 1 plus square root 3 over 2 square root 2. all squared. Which is 1 plus 3 plus 2 times square root 3 over 8. Which is 2 plus square root 3 over 4. Now of course, you can work out something similar for probability of minus. ---------------------------------------------------------------------------------------- But the other way you can do this is, of course you can realize that probability of plus plus probability of minus is 1. So the probability of minus is just 2 minus square root 3 over 4. Now let me show you one way you can do this calculation which, of course, should give you the same value. But it might be worth doing this other calculation just to give you intuition. OK, so here's what we're going to do. What we'll do is, we want to measure psi in the plus minus basis. So what we'll do is we'll first rewrite psi as alpha plus plus beta minus. And now, once we write it as alpha plus plus beta minus, what's the probability of seeing plus when we measured it in the past minus basis? It's just alpha magnitude squared. So how do you rewrite this in the plus minus basis? Well let's rewrite zero as a linear combination of plus and minus. And you can see here that zero is 1 over square root 2 plus plus 1 over square root 2 minus. Because if you add plus and minus, the one vector cancels and you get a multiple of zero. Similarly, one is 1 over square root 2 plus minus 1 over square root 2 minus. So what you can do is write psi as 1/2 zero, which is 1 over square root 2 plus plus 1 over square root 2 minus, plus 3 over 2 minus, which is 1 over square root 2 plus, minus 1 over square root 2 minus. Which is, we can gather terms now, so we have over 2 square root 2. And then for plus we get 1 plus square root 3 times plus. And then for minus we have, again, 2 square root 2 in the denominator. And we get 1 minus square root 3. And so if you were to figure out what's the probability that we'll see plus when we do a measurement, we just take the square of this. Which is something you'd recognize from the previous calculation. So the probability of plus is just the square of this, which is 1 plus 3 plus twice square root 3 over 8. Which is what we saw before, which was 2 plus square root 3 over 4.