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PROFESSOR: So we go back to the integral. We think of k. We'll write it as k naught plus k tilde. And then we have psi of x0 equal 1 over square root of 2pi e to the ik naught x-- that part goes out-- integral dk tilde phi of k naught plus k tilde e to the ik tilde x dk. OK. So we're doing this integral. And now we're focusing on the integration near k naught, where the contribution is large. So we write k as k naught plus a little fluctuation. dk will be dk tilde. Wherever you see a k, you must put k naught plus k tilde. And that's it. And why do we have to worry? Well, we basically have now this peak over here, k naught. And we're going to be integrating k tilde, which is the fluctuation, all over the width of this profile. So the relevant region of integration for k tilde is the range from delta k over 2 to minus delta k over 2. So maybe I'll make this picture a little bigger. Here is k naught. And here we're going to be going and integrate ...

PROFESSOR: Today's lecture continues the thing we're doing with scattering states. We send in a scattering state. That is an energy eigenstate that cannot be normalized into a step barrier. And we looked at what could happen. And we saw all kinds of interesting things happening. There was reflection and transmission when the energy was higher than the barrier. And there was just reflection and a little exponential decay in the forbidden region if the energy was lower than the energy of the barrier. We also observed when we did the packet analysis that a wave packet sent in would have a delay in coming back out. It doesn't come out immediately. And that's the property of those complex numbers that entered into the reflection coefficient. Those complex numbers were a phase that had an energy dependence. And by the time you're done with analyzing how the wave packet is moving, there was a delay. So today we're going to see another effect that is somet...

PROFESSOR: In order to learn more about this subject, we must do the wave packets. So this is the place where you really connect this need solution of Schrodinger's equation, the energy eigenstates, to a physical problem. So we'll do our wave packets. So we've been dealing with packets for a while, so I think it's not going to be that difficult. We've also been talking about stationary phase and you've practiced that, so you have the math ready. We should not have a great difficulty. So let's new wave packets with-- I'm going to use A equals 1 in the solution. Now, I've erased every solution, and we'll work with E greater than v naught to begin with. The reason I want to work with E greater than v naught is because there is a transmitted wave. So that's kind of nice. So what am I going to do? I'm going to write it this following way. So here is a solution with A equals to 1. e to the i kx plus k minus k bar over k plus k bar...

PROFESSOR: We'll begin by discussing the wave packets and uncertainty. So it's our first look into this Heisenberg uncertainty relationships. And to begin with, let's focus it as fixed time, t equals zero. So we'll work with packets at t equals zero. And I will write a particular wave function that you may have at t equals 0, and it's a superposition of plane waves. So it would be e to the ikx. You sum over many of them, so you're going to sum over k, but you're going to do it with a weight, and that's 5k. And there's a lot to learn about this, but the physics that is encoded here is that any wave at time equals 0, this psi of x at time equals 0, can be written as a superposition of states with momentum h bar k. You remember e to the ikx represents a particle or a wave that carries momentum h bar k. So this whole idea here of a general wave function being written in this way carries physical meaning for us. It's a quantum mechanical...

PROFESSOR: Welcome back to recitation. In this video we're going to do another solid of revolution problem. So what I'd like to do in this problem is to find the volume of the solid generated by rotating the region bounded by the following curves: y equals 0, x equal 4, and y equals square root of x, around the line x equals 6. And you can choose your favorite method to do this. What I would like first, when you're doing this kind of problem, is get a rough sketch of the region. So you have some picture of what's actually going to happen. You don't necessarily need a three-dimensional picture. But at least have the two-dimensional region and understand the where the rotation line is, with respect to that region. So I will give you a little bit of time to work on that problem. And when I come back I'll show you how I do it. OK, welcome back. So again, what we're doing in this video is we're going to be looking, finding the volume of a solid ...

PROFESSOR: Welcome back. So in this session, we're going to look at unit step and impulse responses. So in this question, we ask you to find the unit impulse response to these two equations, x dot plus 2x equals f of t, and 2 x dot dot plus 27 x dot-- oops, it should be a 7-- plus 7 x dot plus 3x equals f of t. In the second part, you're asked to find the unit step response for the first equation. So here, the key points are really to remember what do we mean by unit impulse and unit step response. Which initial condition correspond to these responses? And what functions of f of t do you choose in each case? So why don't you pause the video and work through this problem? And I'll be right back. Welcome back. So let's look at the equation A. So the unit impulse response is simply-- I'm going to write this down, unit impulse response-- is simply the solution to the following problem, to our differential equation, x dot plus 2x that we're given, w...

PROFESSOR: Hi everyone. Welcome back. So today, I'd like to tackle a problem in undetermined coefficients, specifically find a particular solution to each of the following equations using undetermined coefficients. So for part A, we have x dot plus 3x equals t squared plus t. And for B, we have x dot dot plus x dot equals t to the four. So I'll let you work this problem out. And I'll come back in a minute. Hi Everyone. Welcome back. So we're asked to solve this problem using the method of undetermined coefficients. And specifically, the observation is if we have a differential equation with constant coefficients, and we have a forcing on the right-hand side which is a polynomial, then there's always going to be a particular solution, which is a polynomial, that has the form of some constant times t to the power of r plus constant t^(r-1) plus a constant c_(r-2) t^(r-2) plus dot, dot, dot, plus c_1*t and then possibly plus c_0. And typically, the proble...

PROFESSOR: This definition in which the uncertainty of the permission operator Q in the state psi. It's always important to have a state associated with measuring the uncertainty. Because the uncertainty will be different in different states. So the state should always be there. Sometimes we write it, sometimes we get a little tired of writing it and we don't write it. But it's always implicit. So here it is. From the analogous discussion of random variables, we were led to this definition, in which we would have the expectation value of the square of the operator minus the square of the expectation value. This was always-- well, this is always a positive quantity. Because, as claim 1 goes, it can be rewritten as the expectation value of the square of the difference between the operator and its expectation value. This may seem a little strange. You're subtracting from an operator a number, but we know that numbers can be thought as operators as well. Opera...

[MUSIC PLAYING] PROFESSOR: Hello, everyone. Today we'll be taking a look at how light interacts with the surface of a solar cell. Right now I'm standing next to a solar module made up of individual silicon solar cells. If you look closely, these cells actually appear black. And they appear black for a very important reason. Solar engineers work very hard to make their solar cells as efficient as possible. Reflected light is lost energy, so good engineers will want to minimize the total amount of reflected light. To make solar cells absorb as much light as possible, and appear black, solar engineers do two things. First, they grow this very thin film of a dielectric layer on the surface. This layer is aptly called an anti-reflection coating. Second, they texture the wafer. And today we'll demonstrate how texturing is performed, and quantify its enhancement for reducing light reflection. Silicon wafers don't start out black. In fact, they appear gray. Polish...

[MUSIC PLAYING] PROFESSOR: Hello everyone, today we're going to learn how a Solar Cell is able to turn light generated mobile charges into electricity. Today's lesson will use everything we've learned in the past videos to understand this effect. So, make sure you understand the material from the previous videos before watching. First, let's go over the structure of a Solar Cell. Here's a cell that I made. And we can see that a metal ribbon is connected to the top metal contacts, which form a grid. The spaces between the grid lines allow light to enter the cell. If we flip over the cell we see the entire back surface is coated with metal, which allows easy extraction of charge from the back surface. Additionally, we have another metal ribbon that's connected to the backside. Now, let's hook up our Solar Cell to an ammeter to measure the current. So, here we have an ammeter connected to our Solar Cell and our light source which will simulate the...

[MUSIC PLAYING] PROFESSOR: Hello, everyone. Today we'll talk about doping, which is the process of intentionally adding impurities to a semiconductor in order to change its electrical properties. Doping is a critical process in the tech world. It's used in manufacturing almost all semiconductor technologies today. Without doping, the solar industry would not exist, but even though doping is common today, the effects of impurities confused semiconductor physicists in the 1950s, who had trouble reproducing results. Eventually, they realized that contamination levels, as low as 1 in a billion, were vastly changing the electrical properties of their samples. Today, we'll show you how this works with a very simple experiment. We'll be measuring the electrical conductivity of two silicon slabs using an ohmmeter. One is doped with impurities, phosphorus in our case, and the other is ultra-pure, or what we call intrinsic. Let's go over our experiment. We'l...

PROFESSOR: The fact is that angular momentum is an observable, and as such it deserves attention. There is an active way of thinking of observables, and we have not developed it that much in this course. But for example, with a momentum operator you've learned that the momentum operator can give you the differential operator. It's a derivative, and derivatives tell you how to move, how a function varies. So with the momentum operator, for example, you have the momentum operator p hat, which is h bar over i d dx. And you could ask the question of, OK, so the momentum operator moves or takes a derivative, does the momentum operator move a function? Does it generate a translation? And the answer is, yes. That's another way of thinking of the momentum operator as a generator of translations. But how does it do it? This is a Hermitian operator, and it takes a derivative. It doesn't translate the function. But there is a universal trick that if you exponentiate ...

PROFESSOR: Welcome to another help session on recombinant DNA. Today, we're going to be discussing about transformation and protein expression. As you can imagine, there are often many times we will need a large amount of protein. But it can be difficult to get it from the original source. For example, you need insulin to treat diabetes, but it's not exactly practical to get a lot of insulin from humans. In order to get a lot of the desired protein, often other organisms will be used to express this protein. But it's a multi-step process. For example, let's say we want to express our human insulin in bacteria. Well, the human gene has both introns and exons, as you remember from lecture. Exons are what are actually cut together in order to produce the final mature mRNA, which is later used to express the protein. Bacteria, on the other hand, don't have introns. They just have exons. So they are only capable of reading a gene that just has the exons and...