You are here
PHYS 201: Fundamentals of Physics II
Lecture 23
 Quantum Mechanics V: Particle in a Box
Overview
The allowed energy states of a free particle on a ring and a particle in a box are revisited. A scattering problem is studied to expose more quantum wonders: a particle can tunnel into the classically forbidden regions where kinetic energy is negative, and a particle incident on a barrier with enough kinetic energy to go over it has a nonzero probability to bounce back.
Lecture Chapters

0

528

1153

3243
Transcript  Audio  Low Bandwidth Video  High Bandwidth Video 

htmlFundamentals of Physics IIPHYS 201  Lecture 23  Quantum Mechanics V: Particle in a BoxChapter 1: Review of Wave Functions [00:00:00]Professor Ramamurti Shankar: Okay, so I’m going to begin with something I did last time, because it was not completed, then we’ll take it from there. And we had a lot more questions last time, which is good, very helpful, so I know what may be bothering you. So don’t hesitate to stop me. The new thing we did last time was, if I give you some wave function, ψ(x), and I ask you what happens if I measured energy, the recipe is the same as in momentum. You’ve got to find functions called ψ_{E}(x), which are supposed to be states in which, if the electron was there, it’ll have a definite energy. Now, not worrying for a moment on how you get them, the recipe for what happens when you measure energy is that if you’re given some arbitrary function ψ, that’s a state the particle is in. And you’re asking, “What happens if I measure energy? What will I get?” The instruction is to write that function as a sum over these functions, each associated with a definite energy. And once you have done that, the probability that you will get a certain energy E = the absolute value squared of that particular coefficient. Not different from ψ_{p} , same rule. And the rule for A_{E} is also the same, or similar. A_{E} is the integral of complex conjugate of this function associated with definite energy, times the function you’re thinking about, integrated over space. It’s the same rule. So it can happen, the problem can take many forms. Sometimes one can give you a function ψ(x). Then you take all these functions, do all these integrals, find all the coefficients and you square them. Other times the coefficients may be given to you; you’re just asked, “What’s the odd for this energy or that energy?” But now the question is, what are these functions, ψ_{E}(x). Whereas functions of definite momentum are given once and for all by this formula, e^{ipx/ℏ}, I wish I could give you a function of definite energy and say, “These are functions of definite energy. Maybe instead of p, there’ll be some E on the top.” No, it’s not like that. You’ve got to do a lot more work, because the reason is that the functions of definite energy depend on what the energy is in classical mechanics. And in classical mechanics, the energy depends on the momentum and on the potential energy. That means on the position. Every problem has a different formula for E. If it’s a harmonic oscillator, this will be ½kx^{2}. If it’s a particle moving up and down in gravity, it could be mgy. If it’s two particles connected by a spring, it will be something else. So in each problem, our electron in Hydrogen atom, is the coulomb force, coulomb potential, e_{1}e_{2}/r^{2} times some 4Π’s. So in each problem, there’s a different V and for each V, you’ve got to find the E — I mean, you’ve got to find the functions. And what is the rule? The rule is the following. You have to solve the following equation, called the Schrödinger equation, d^{2}ψ/dx^{2} + V(x)ψ(x) = E times — you just have to solve this equation. So let’s not worry about how you solve it. That’s a problem in mathematics, but the point is that, given a potential V, you come to this equation, and you go to the math department and say, “Give me all the solutions of this equation.” And they’ll give you all the solutions of the equation. They will find, in fact, only for certain special values of E that the equation has acceptable solutions. And for each one, they’ll give you the solution and you start with that solution and do all of this. You follow that? That’s the program. So I did a very simple example last time, which was two examples. First I said, take the problem with no potential, no force, particle is free of all forces. Then the equation is very simple, ℏ^{2}/2m d^{2}ψ — I’m sorry, d^{2}ψ/dx^{2}. There is no Vψ. That is just Eψ. So let’s rearrange this to the following form: d^{2}ψ/dx^{2} + k^{2}ψ = 0, where k^{2} is just 2mE/ℏ^{2}. You know if you multiply by 2m/ℏ^{2}, you’ll get this equation. Since this combination appears all the time, I want to give it a name, k^{2}. But you must understand, what’s the relation of k to anything else you know will be clear in a minute. So what’s the solution to this one? You can write them in terms of sines and cosines, but I want to write them in terms of exponentials and the solution was — there are two solutions. One is e^{ikx}, where k is defined that way. Other is e^{ikx}. You can check. If you took this function and take two derivatives of either this term of that term, both will give you −k^{2} times the function itself. Every time you take a derivative, you bring down an ik or a ik. If you do it twice, it’s −k^{2}. And A and B are arbitrary right now. There is no restriction coming from that equation. And what is k? You know that functions of definite momentum look like e^{ipx/ℏ}, and I have something that says e^{ikx}. It’s very clear that p is just ℏk, or k is p/ℏ. So if you like, this equation I got, we can write as ℏ^{2}k^{2} is 2mE. That means p^{2} is 2mE. That means E = p^{2}/2m. So let’s understand what that means. Yes. Student: Just jumping back for a second. At the bottom of the top board, you wrote the E for classical mechanics. Did the equation to the right of that come from that? Professor Ramamurti Shankar: In a way, it does. That’s correct. I will tell you how it does. But right now, I’m just giving you a recipe. If in classical mechanics, you know what the kinetic and potential energy of the particle are. Kinetic is always p^{2}/2m. Potential is V(x) where V depends on the context. Then in the quantum theory, the functions of definite energy obey an equation which looks a lot like that in many ways, right? So I’ve given you some clues or some more discretion in my notes on what is the universal recipe for guessing the equation in every context. I’ll talk about it a little bit. But right now, you can take the view that this guy just tells me that if you want states of definite energy, solve that equation. Look, so what is this saying? Think about this. It’s saying, if you want a state of definite energy, it is really made up of states of definite momentum, and the momenta that appear here are not arbitrary. They are simply the momenta that you will have, even in classical mechanics, for that definite energy, because if E is p^{2}/2m, p would be ±√2mE. That’s exactly what we have. If k = here,… this says k = √2mE/ℏ, and ℏk is just √2mE. So that looks like classical mechanics, right? Chapter 2: Particle on a Ring [00:08:48]Even before you took the course, if I said there’s a particle on a ring. Its energy is E, what is its momentum? You will say since energy is p^{2}/2m, p must be ±√2mE and they mean going one way, and going the other way. These are the only two options. That part is very much like classical mechanics. The k is related to E exactly the way it would be. What is novel here is that this state, simultaneously can yield either the clockwise momentum or the counterclockwise momentum. That’s the sense in which it’s a different thing, you understand? A classical particle of energy E will have one of these two momenta, but on any given day, on any given trial, it will have only one of these two. It cannot have both the signs. Either it’s going like this or going like that. In quantum theory, it’s in the state of limbo. Not this alone, not that alone, both. The probability of going clockwise is proportional to A^{2}, and the probability of going counterclockwise is proportional to B^{2}. If you want, you can put the square root of L’s and so on, but they’re all common to everything, so the relative probability is just in the ratio of A^{2} is to B^{2}. It’s an admixture. And there is no restriction on how much of A and how much of B you should put into your answer. You can pick the extreme case where only A is nonzero. That’s fine. That’s a particle who has a definite energy and a definite momentum. You can pick this one, a definite energy in the opposite momentum, but in general, both can be nonzero. It’s a state of definite energy whose momentum is known in magnitude but not in sign, because both allowed signs appear here. The other result from quantum mechanics, which is surprising, is whereas a particle in a ring can have any momentum, in quantum theory, the demand that when you go around the ring, if you add an L to the x, you should come back to the same function, makes the requirement that kL is some multiple of 2Π.That means k = 2Πn/L, or p is 2Πℏ/L times n. The allowed values of k or p are restricted in this way, and the allowed values of energy then, which is p^{2}/2m, are restricted to be 4Π^{2}ℏ^{2} over 2m L^{2} times an integer n^{2}. And this is a problem where energy is quantized. It’s really quantized because momentum is quantized and the energy in this simple example is simply p^{2}/2m. That means if you shine light on the system, it will absorb light only at certain frequencies, and it will emit light only at certain frequencies. And what are those frequencies? Well, all of this there’s a bunch of numbers. ℏ, m, L they’re not changing. n can be 0 + or  1, + or  2 etc. For every n you put in, you’ll get this basic number times 0 squared or 1 squared or 2 squared. So you will find that will be the state E_{0}, which is n=0. Then above that, the energy of that will be 2Π^{2}ℏ^{2} over mL squared. Then there’ll be n=1, and there’ll be n=1. Both will have the same energy, because energy depends only on n^{2}. The energy of that will be — I’m sorry. I made a mistake. E_{0} is simply 0, because that corresponds to n=0. E_{1} will be what I wrote here, with n=1, which will be 2Π^{2}ℏ^{2} over mL squared. So what I’m saying is, at a given energy, you can have a state of right moving clockwise or anticlockwise momentum, or a crazy mixture of the two. But normally, if you wish, you may select from these mixtures, two special cases, one which is all A and one which is all B. We like those because now we can visualize them as states having a definite momentum clockwise, a definite momentum counterclockwise. That’s what one really means by drawing two lines at that energy. So if you absorb a photon, you’ve got to give an energy equal to the difference of these two numbers, which is that. And that will be the ℏω of your photon. And if you emit a photon and come from there down here, that photon frequency will be given by that. So this is how a quantum system is usually probed. We cannot actually look in and see anything. We deal with it by shining light and we see what light it absorbs, what light it emits. The frequency of light is a clue as to energy level differences, you understand? Not to the energy themselves, but to the energy level difference that turns into photon energy. But you can put together the levels by putting together all the differences, so you measure all the differences. So one famous line is the hydrogen 21 centimeter line. Hydrogen atom has two levels which are close, and the wavelength of that is 21 centimeters. There’s a standard fingerprint of hydrogen anywhere in the universe. Sometimes, if the wavelength is not 21 but 22, you can say, “Why is it 22? The wavelength is longer. Maybe it’s not hydrogen. Maybe it’s somebody else.” But the answer is that it is hydrogen, but that galaxy is moving away from you, so that its light is Doppler shifted into the red. If something was coming towards you, it would be blue shifted. So you can tell, if you believe that hydrogen atoms all over the universe are the same hydrogen atoms, and the frequency shift is only due to the motion of the galaxy, you find two things. First, what is it made of and secondly, what’s the speed of the galaxy? So the red shift is what one uses to find the whole story about galaxies. Yes? Student: In the context of a photon, or an atom being ___________ by a photon, what does the L correspond to? Professor Ramamurti Shankar: What is L for an atom, right? That’s your question. Now, this is a fake atom. This fake atom is an electron moving on a ring, correct? That L is actually the size of that ring. You can go to a lab and you can make a ring of certain micron, then the mass of the electron is known. That L would be the length of the ring. For an atom, what you have to do is that you must solve it, solve this equation, where V(x) = what? In three dimensions, it will be the charge of the nucleus, which is Z times e, charge of the electron, and it’s attractive 4Πε_{0}r, which is the separation between them. So you’ll have to solve a similar equation, but that equation will have d/dx and d/dy and d/dz and so on, right? It’s in three dimensions. But this V(r) will determine the allowed functions in that problem. That won’t be an L, but a certain size will emerge even from that problem. That size will be the size of the orbit. For example, if you look at the function, the lowest energy state will look like e to the r/a_{0} times some number, where a_{0} is called the Bohr radius. That will be roughly 10^{8} centimeters. That a_{0} will be a length — you can say where is the length and the height of an atom, right? It’s no box. The length is the length made out of eℏm. You can make a length out of those things. That’s what will come out. So if you want, you can take any textbook that gives you the formula for hydrogen’s Bohr radius, that’s supposedly the radius of the orbit, and you will find its length a_{0} made out of the parameters in the problem, which is the charge of the nucleus, the charge of the electron, the mass of the electron. Okay, so this is, if you want, a simple atom that we have manufactured, because we can see quantization of energy. Yes? Student: When you go to the potential of 0, is it possible to measure both momentum and position, in that if you measure the energy and a position, don’t you know something about the momentum? Professor Ramamurti Shankar: But you cannot measure energy and position at the same time either, because energy — what you can really measure at the same time is energy and momentum. If you can measure p, you can measure p^{2}, or p^{3} or any function of p. Because knowing a state of well defined momentum, in this problem, doesn’t do anything to the energy, because once you know that momentum is p, p^{2}/2m is the energy. So it turns out, this is a fortunate case where the two variables, energy and momentum, are simply related to each other in the fact that one is essentially the square of the other. But if you have two variables, one involving x and p, at the same time, then you cannot measure them simultaneously with either x or with p. Yes? Student: What is a_{0} again? Professor Ramamurti Shankar: a_{0}. Gee, I don’t know. The Bohr radius. I think it’s ℏ squared over me squared, but you have to check. I don’t guarantee that answer, but I think it’s something like that. You can see if it’s got at least units of length. I’m too lazy to try that. I think it’s something like this. Anyway, suppose it is something like this. My main point is, it is a length made out of Planck’s constant, the mass of the electron and the electric charge. Together they make a length, and that’s the length that controls the size of the orbit. Chapter 3: Particle in a Box [00:19:13]All right, so this was the simpler problem. Now I want to take a more complicated problem of what I call a particle in a box. So that is a potential. See, normally, if you make a hole in the ground, it may take this shape, and this is the location and this is the height, but the height is just the potential energy directly, because it’s mg times the height. So imagine a particle somewhere here. In classical mechanics — by the way, the rule for doing quantum mechanics is, in classical theory, what’s the potential experienced by the particle? That’s what you ask. And it is that potential that enters this crazy Schrödinger equation. So you need to know classical mechanics before you can do quantum mechanics. You should know what potential the particle experiences in classical theory, and stick that into this equation to find the allowed energies in quantum theory. So if it’s a vibrating system, there is some omega for vibrations. That ω is √k/m, and you can get k from that one. Or if it’s a system like an atom, where there’s an electric charge between them, electric force between them, you should know the electric potential. So you should know the potential in classical mechanics, and then you do the quantum theory from there. So imagine a particle in this potential in classical theory, and you ask yourself, what does it do in quantum theory? Well, classical theory says, let’s plot the energy of the particle, right? It’s some number that’s not supposed to change as the particle moves. That means here, if the energy is kinetic + potential, here the potential is 0, it’s all kinetic. But as you start going near the edge, there are two parts here. That’s the potential and that’s the kinetic. And right here, all the energy is potential, and E = potential. There is no K. What does that mean? That means the particle comes here, slows down, comes to a rest, then goes back. It can never be found here if it had this energy, because it’s got more — let’s see. It’s got more potential energy than total energy. That means it’s got negative kinetic energy. That’s not possible. So every classical particle will execute bounded oscillations at any energy, between what are called the turning points. The left turning point and the right turning point is as far as you can go at that energy. If you’ve got more energy, of course you can go further out, and if you’ve got even more energy, you can just roll up to the top of the hill and escape. That’s classical mechanics. Now we want to ask ourselves, in quantum mechanics, what does the particle do? That’s our goal. So I’m going to take a following simpler problem, because it’s mathematically easier. I’m going to take a potential that looks like this. It is V_{0} to the right of L. It’s V_{0} to the left of the origin, and it is 0 here. It’s got very sharp walls, because that makes it easy to solve the equation. In real life, potentials are rounded off. My fake potential has got sharp corners. I want to take a particle of some energy and ask, what are the allowed energies in this problem? What do the corresponding functions look like? That’s what we’re asking. So the rule is to go back to that equation, which now takes the following form: d^{2}ψ/dx^{2}. I’m going to stop writing the subscript e. From now on, you should understand, I’m trying to find states of definite energy E. 2m times E  V ψ = 0. I’ve just rearranged the Schrödinger equation, so that it looks like second derivative + all kinds of stuff. So how do we solve this problem? Well, V varies with position, so there are three regions in the problem. There’s region I inside the well. Inside this well, V is 0. Outside this well in region II, it = V_{0}, some constant. Likewise in region III to the right of the well, it’s also V_{0}. So we’ve got to solve this equation in 3 different regions, middle, left and right. First let’s solve it in the middle because it’s very easy. What is the middle defined as? Middle, region I, V = 0. So the equation is d^{2}ψ/dx^{2} + 2mE/ℏ^{2} is 0. So let me write it as d^{2}ψ/dx^{2} + k^{2}ψ = 0. Just like before. This k is the same as before, k = √2mE/ℏ, in region I. So what’s the answer to that one? Again, it’s exponentials, but I’m purposely going to write it in region 1 as follows: as Acoskx + Bsinkx. I hope you all know that this has exponentials, our old friends the trigonometric functions, are also the solution to this equation, because the second derivative of cosine is roughly  the cosine. The second derivative of sine will also give you −k^{2} times the sine. So rather than using exponentials, I’m going to use sine and cosine, because I have an ultimate motive. You realize that exponentials and sines and cosines are all connected by Euler’s great formula, so if I want, I can take e^{ikx} and write it as cos + i sine. e^{ikx}, write that as cos  i sine. Combine the cosines, and call the coefficient A. Combine the coefficients for sine and call it B, and get this form. There are two alternative ways. You just write this, you will see why I picked that. That’s the answer in region I. We’re all set. Right now, A is arbitrary, B is arbitrary and k is arbitrary. There is no restriction at all on any of them. Do you understand that? Because if you take this function, it satisfies that equation absolutely. So the free parameters are A, B and k, are all arbitrary. That means it can have any k. But the energy is ℏ^{2}k^{2}/2m, so it can have any energy. But we’re not done yet, because we only looked at region I. Have to look at region II and region III, and then you will find out what happens. So let’s go to region II So in region II, the equation obeyed by the ψ, is ψ (let me call it region II) over dx squared = 2m/ℏ^{2} times V_{0}  E ψ. I just put it on the other side now. Now this is all a constant. That’s why we picked the potential, which is piecewise constant, so it’s some constant. You’re asking yourself, what ψ has the property that when I take 2 derivatives, it looks like this number times ψ? So let me call this whole combination kappa squared. Then the answer is ψ of 2 = C times e^{κx} + D times e^{κx}. Why? Because two derivatives of this guy will give me kappa squared times itself; likewise here. And kappa has been cooked up so that kappa squared is what I want. Yes, now so κ = 2m/ℏ^{2} times V_{0}  E under root. So these are now real exponentials and not imaginary exponentials. The reason that used to be sines and cosines, or imaginary exponentials, is that inside the well, E  V is a positive number. Outside the well, I’m assuming that E  V is a negative number. In other words, I’m going to consider the well to be, in the end, infinitely deep. And the energy of the particle, whatever it is, is some finite number. I want to send the walls to infinity. That’s called particle in a box. So I’m taking V_{0} finite but very, very large. Then it’s very clear that V_{0}  E is a positive number. This kappa is what I have here. Now look at these two functions. This function grows exponentially as you go to the right, to + infinity. That means the wave function describes a particle that would rather be at infinity than anywhere near this well. So it’s not the problem we’re describing where the particle we expect is somewhere near the well. Furthermore, the square of ψ can never be normalized to 1, because an exponentially growing function, the area under that is exponentially large. There’s no way to fix that. So on physical grounds, we drop the solution, because we don’t have to pick C to be nonzero. We have the freedom, and the physical condition tells you, no one wants a function growing at infinity. We want functions falling at infinity. This is the only function that falls at infinity. And how does it fall at infinity, how does it fall off? It looks like e to the √2m/ℏ^{2} times V_{0}(x). I’m ignoring e compared to this very large V_{0}. I’m taking V_{0} much, much bigger than any energy I’m looking at. So it’s falling exponentially, and the number on top is like square root of V_{0}. So what do you think happens if V_{0} becomes very, very large? Well, if you plot the function that’s exponentially falling, it falls like that. But as you increase V_{0}, it falls more and more quickly. You follow that? And eventually, even V_{0} goes to infinity; it just doesn’t have a life outside. It just vanishes immediately. Whatever you start with goes to 0 immediately. That basically means ψ_{II}(x) is just 0 in region III, if V_{0} goes to infinity. So let me emphasize what we are doing here. We have a particle in some kind of well, which is very, very tall, not yet infinite, and you give it some energy E that cannot change. In this region, in classical mechanics, that would be the entire kinetic energy, because potential energy is 0. What about here? This is the total energy available to you. Potential energy is all of that, so kinetic energy has to be minus some number. That can never be. That means that particle will never be found in the region outside this well. Do you understand? As long as the energy is less than it takes to go to the top of the barrier and climb out, a classical particle — imagine rounding it off. You can come rolling with any energy. If you don’t have enough energy to go over the top, you cannot be found in this region. But in quantum theory, as long as the well is not infinitely large, there’s an exponentially small decaying wave function. So a classical particle will turn around at the classical turning points. A quantum particle will go slightly outside the forbidden region, into the forbidden region in both directions of the well. That is a purely quantum phenomenon. It is forbidden, meaning it’s discouraged from going there, but it’s not absolutely forbidden. That’s called going over the — well, it’s going into the classical forbidden region, but as the barrier gets higher and higher, the excursion into the forbidden region becomes less and less, because the function falls very rapidly. And I’m considering, for simplicity, a barrier which is infinitely tall, in which case it doesn’t go there at all, even in quantum theory. Therefore, by a similar argument, ψ is 0 here and ψ is 0 here, and in region I, I’ve shown you, ψ = Acoskx + Bsinkx, where the energy is related to k by ℏ^{2}k^{2}/2m. So first when we did only region I, you get the impression that A and B and k can be whatever you like. But now we have an extra restriction. The restriction is that if you draw the function in this well, it can do whatever it wants, but it must vanish at the two ends, because it’s 0 here and it’s 0 here. Just by continuity of ψ, every ψ that’s allowed must vanish at the two ends. That’s a new restriction, because the function that I have here in general will not vanish at the two ends, but I’m going to demand that it vanish. That will tell me something about the allowed values of A, B and k. You will see what it is. First I demand that ψ should vanish on the left end of the wall, which is the wall at x = 0. This is the wall at x = L. ψ of 0 should be 0. I take the function I have. When this is 0, x is 0, that guy is gone. I get A times cosine 0 which is just A. But that has to vanish. That means the coefficient A has to vanish. It is not allowed. So even though it looked like, before we went to the other regions, you can have any A and any B. If you’re only in the middle region, that’s correct. But if you want to connect your answer in the middle region to a 0 on either side, then the coefficient has to be such that the ψ you have in the middle vanishes at the left end. So A is gone. So I’m now left with only the following thing: ψ(x) = Bsinkx. We did not have to kill B because that x = 0 sine vanished, so it is allowed. But now I demand furthermore it should vanish at x = L. Well, at x = L, what do I find? I want 0 to = BsinkL. That should be 0. Now one simple way to get there is to kill B, say B = 0, but then you’ve got no solution. You killed A and you killed B. It’s certainly true that ψ = 0 is the solution to this equation, but it doesn’t describe anything. So you don’t want to kill B, so you want to blame the 0 on the sine, and the sine, as you know, vanishes for any multiple of Π. You can have 0 times Π, 2Π, 3Π, 4Π, 6Π. They all give you 0 for sine. Therefore k is quantized to be one of these numbers. Therefore k has got only values nΠ/L. I’m going to call it k_{n}. And the energy, which is really ℏ^{2}k^{2}/2m L squared — sorry, k^{2}/2m, becomes now ℏ^{2}Π^{2}/2m L^{2} times n^{2}. And ψ looks like B — ψ’s of n looks like Bsin(nΠx/L). You guys see that? These are the allowed function. The sine function is very nice. It vanishes on the left end, because when x is 0, it’s gone. When x=L, the argument of the sine is a multiple of Π, it vanishes. So it looks like this when n = 1. It looks like that when n=2, and so on. So it’s got more and more wiggles in the region in between, but they are designed so that they vanish precisely at the edges. So it’s like it’s exactly the case of a violin string that’s clamped to two ends, which has to vanish at the two ends, because you don’t let it move there. In between, it can do this and it can do that. It can have more and more wiggles. So what you find is, of the three things that we thought were arbitrary, A is gone, and k is not arbitrary. k is again quantized to be nΠ/L, so that the function does n half oscillations between one end and the other end. And therefore there’s a corresponding quantization of energy. Yes? Student: What does that tell you? If you square ψ_{n}, does that give the probability of finding it in each one, the energy levels? Professor Ramamurti Shankar: No. Let’s talk about that now. So let’s write down the function. The ψ sub energy that we were looking for, we have finally found out, so let’s have a picture in our mind. Here is the well. It’s going all the way to infinity. There is the state n=1, and the function looks like this. Then there is the state — let me draw it right there, it looks like that, n=2 and so on. In the ψ_{n} (let me label the energy by n) equals some constant B times sin(nΠx/L). That means if you want a particle in the well to have a definite energy, the wave function must look like one of these sine functions. It must look like this or it must look like that, or the higher harmonics. They are the only allowed functions for states of definite energy. Now before I answer any further questions, the allowed values of n here you might think are 0, + or  1, + or  2 etc., but I claim they’re just 1, 2, 3 and so on. You should think about why. What’s wrong with n=0? Is that a possible value? Can anybody tells me what happens to this function if — yes? Student: It identically vanishes Professor Ramamurti Shankar: If n=0, the function identically vanishes, right? Because sine of 0 times Πx/L is identical. That’s not what you want. You want a nontrivial solution. So we drop this guy. But how about n=1? Why don’t I keep that? Yes? Oh, I thought I imagined a hand going up. Yes? Student: It’s the same thing as B sine of the positive value. Professor Ramamurti Shankar: That’s correct. If you put n=1, the function ψ of 1 is B sin(Πx/L), but that = 1 times B sin(Πx/L). This guy is just ψ_{1}. So it’s just a multiple of the ψ_{1} function, because 1 times the state is the same state. You don’t get a new function; you just get a multiple of the old function, and in quantum mechanics, they’re all the same. But be very careful. When I did a particle on a ring, I did e^{ipx/ℏ}, where I said p = 2Πℏ/L times any integer n. There I let n = 0, +  1, +  2 etc. Why did I allow that? First of all, in that case, e to the i times 0 is not 0. It’s in fact constant. So for a particle going in a ring, the 0 momentum state is actually allowed. And +1 and 1 are not the same, because e^{ipx} and e^{ipx}, forget the ℏ, they are not proportional to each other. You understand that? You can never write e^{ipx} as a multiple of e^{ipx}. You see that? If there were such a multiple, let me call the multiple as N, then N is really e^{2ipx} by bringing it the other side. That’s certainly not a constant. So e^{ipx} can never be converted to e^{ipx} by multiplying by any number, so it’s an independent solution. On the other hand, sinx and sin(x) are related to each other by a single factor of 1. So that’s the reason why in that case of a particle going in a ring, the solutions are labeled by all integers, positive, negative and 0, but a particle in a box, we label only by positive integers. Negative integers give you the same function and 0 kills the function, so that’s not allowed. Okay, so what about B? What shall I do with B? What equation’s going to determine B? Yes, any idea? Yes? Student: Normalize right? Professor Ramamurti Shankar: Right. So first of all, the equation will never tell you what B is. That’s what I want you to understand. The equation that I’ve written here for the energy functions on the top, if size of B is a solution, you can multiply both sides by any number, like 15. Then 15 times ψ will also be a solution. That’s a property of a linear equation that the answer is not given in overall scale. Give me one answer; I rescale it, it’s still an answer. So linear equation will never tell me the overall size of the function. That is picked by the extra convenient restriction that ψ^{2}dx = 1. That will determine B and I’m claiming the answer is the √(2/L) is what you want B to be. The answer is again simple. The average value of sine squared over 1 half period is 1 half. If you square or integrate or whatever, you’ll get L/2, so B^{2} times L/2 should be 1, and B should therefore be the √(2/L). So let me write down now the final result of this problem, which I think is very, very instructive. If you can follow this much, you’ve really got 80 percent of the quantum mechanics I want to teach you. If you’re comfortable with this example, because we’re trying to find out, in quantum theory, you’ve got a particle in a box, which is like a hole in the floor, which is so deep that it’s infinitely deep. But even if it’s not infinite but very deep, the answer’s good enough. You want to know what it will do. In classical mechanics, what will it do? If you’re a particle in a box, what’s your lowest energy state? Yes? Student: Floor of the box. Professor Ramamurti Shankar: If you’re in prison with infinitely tall walls, you just sit on the floor and mope, right? That’s the lowest energy state. But that’s not allowed in quantum mechanics, because you have a definite x, which is wherever you’re sitting, a definite p, you’re not moving. Such a state is not allowed. So in quantum mechanics, a particle cannot simply sit still. Therefore, it will necessarily have an uncertainty in position, and it will try to spread itself over the box. And the typical uncertainty is of order. If I say, “Where is this particle?” I don’t know anything. It’s got to be somewhere in the box. That’s the estimate for Δx. So Δp has got to be bigger than ℏ/L. Forget all the 2Π’s. So it cannot have 0 momentum. It’s momentum has to be fluctuating with that range. And the kinetic energy, which is in fact the total energy inside the box, which is p^{2}/2m, the size of that is ℏ^{2}/2m L^{2}. So that is your estimate for what should be the energy of a particle in a box of size L. So what happens is, the box of size L forces it to have a spread in momentum, because you’ve squeezed it to a position known to within L. The spread in momentum translates into spread in kinetic energy, and if I take the spread in kinetic energy to be a rough estimate of the lowest kinetic energy I could have, you get a number which except for factors of Π is in fact the right answer. So people always estimate energies of particles by doing this calculation. For example, take a nucleus, which is 10^{13} centimeters across. I put a proton there. What’s the minimum kinetic energy of that proton? You find by the following method: Δx is 10^{13}, because it’s somewhere in the nucleus. Δp is that, and you do that, you will get some energy. That will be the typical energy of a particle inside the nucleus. So particles inside the nucleus cannot sit still. More you cram them, it’s like children. You put them in a tighter room, they’re jiggling more and more. That’s called zero point motion. You cannot just nail them in x without letting them spread in p. That means you cannot get the lowest energy So lowest energy is some number you can estimate by the uncertainty principle. So the next question was, let’s take one of these functions, say ψ’s of 3, the square root of (2/L)sin(3Πx/L). If you plot that guy, that function will look like this. Now what was your question? If I measure a particle’s position, what’s the probability? You’ve just got to square this function. It will look like this. That’s the probability. If you measure the energy, what will you get? You will get exactly one energy corresponding to n=3. It’s in a state of definite energy, but not definite position. Position has got some ups and downs. The probability for x looks like that. So everyone with me? This is like knowing the spectrum of this artificial atom. In a real atom, the electron is forced to stay near the nucleus, because the 1 over r potential, if you plot it as a function of r, looks like a well. And the particle is somewhere here. In this problem, the particle is confined by the box. Whenever you have a confining thing that keeps the particle in a certain region, the energy gets quantized. And we found out in our first example the quantized energy levels of a particle in a box. And you can probe the particle by shining light and the energy levels are plotted like this. This is n=1, this is n=2, which is n=3. Zero is here. The zero state is not allowed. You’ve got to have n=1 and 4 times that will be n=2, and 9 times that will be n=3. Energy grows like n^{2}. And then you can come from there to there emitting a photon, or you can go from there to there emitting a photon, or you can go from here to here absorbing a photon. You can do all kinds of things. That’s how if you want you can test quantum mechanics. Put a known particle inside the well and shine light and see what light it absorbs. Yes. Student: So the ψ of 3, is that the ψ of the energy function _________? Professor Ramamurti Shankar: Yes. Exactly right. So what I should really write, he’s absolutely right, I should write E = ℏ^{2}Π^{2}/2m L squared 3 squared, right? The third energy level. So I’m short circuiting all of that and just calling it ψ_{3}. Normally, if you put a label 3, it’s not clear what we’re talking about. Maybe it’s the position x=3. Maybe it’s the momentum p=3. But in this lecture, since we’re talking only about energy, I thought I will just call it ψ_{3}, but at least I should call it ψ_{E3}. Student: [inaudible] Professor Ramamurti Shankar: Yeah, I just did that. Student: [inaudible] Professor Ramamurti Shankar: You can take the complex conjugate, but for a real function, you can just square it, right? You understand? If ψ is real, ψ*is the same as ψ. So ψ*is also (2/L)sin(3Πx/L). So ψ*ψ is just a square. Student: On the first board, you have A_{E} = _________ ψ*E ψ _______. I’m wondering why there’s no _______________. Professor Ramamurti Shankar: No. I think I know what the problem is you’re having now. That formula is not relevant now. I’ll tell you when that formula will come into play. It will come into play in the following situation. I have a particle in a box whose wave function looks like this. Somebody put it in that state, okay? It’s an allowed wave function because it vanishes at the ends like it should. Beyond that, it does anything it wants. That is my ψ(x), okay? If I measure the energy of that guy, now you tell me, if I measure the energy in that state, what answer will I get? Student: [inaudible] Professor Ramamurti Shankar: Pardon me? Yes. I’ll get one of those quantized values, do you understand? A particle in a box under energy measurement can give only one of those numbers. Now this guy is not one of those functions, you understand. This is a randomly chosen function. But when I measure energy, so what will I do? Maybe you can guess now. Student: Write it as a sum of energy functions Professor Ramamurti Shankar: That’s correct. You will have to write this guy as a sum of A — let me call it n for energy now. A √(2/L) sin(nΠx/L). I mean, this is just A_{E} ψ’s of E of x. Except E is labeled by an integer, so I’m writing it this way. Then I have to find A_{n}. And if I want to find A_{n}, that = integral of (2/L)sin(nΠx/L). If you want, you can put a conjugate, but it doesn’t matter, because sine is real. Multiply it by Mr. Crazy here, whatever it is, and do their integral. It will give you some number for every n. A_{1}, A_{2}, A_{3}, A_{4}, and the square of every one of them will give you the odds that you will find this energy or that energy or some other energy. So what quantum mechanics tells you is, an energy measurement of a particle in a box cannot give a random answer. It can only give one of these answers. That’s the quantization of energy. It does not mean every particle in a box is in a state of definite energy. A particle in a box can be any wave function, provided it vanishes at the ends and doesn’t spill out of the infinitely tall wall. So for every one of them, you can ask the usual questions: what happens when I measure position? Square the function. What happens when I measure momentum? Write it in terms of e to the ipx functions. What happens when I measure energy? Write it in terms of these functions. Okay? Chapter 4: Scattering [00:54:03]All right, so I want to do one last topic here, which is called scattering, but I don’t want to proceed till you guys have understood this part. This is covered in all the books and maybe you can read the books or you can read my notes. You can talk to everybody. But we have solved — from this, all quantum problems that we all do are exactly isomorphic to this problem. It’s the same thing we do over and over again, except the equations you solve are more difficult. They can be in higher dimensions. They can involve more than one particle. The potential could be complicated, but the philosophy is always the same. There’ll be some Schrödinger equation, and you want to find all the functions that satisfy it, find the allowed energies and the allowed functions. We will see that states of definite energy have a very privileged role in quantum mechanics. That will be clear in the next two lectures. But right now, here’s what I want to do. This is a very small variation of what we have done, so most of the hard work has been done. I’m just going to cash in on that. Take the following problem in classical mechanics: here is the level ground. Then I make a little change in the landscape so that this has got certain height and therefore certain potential, V_{0}. And from this end, I roll a marble and see what happens. The marble is given some energy E, which because there is no potential here, it’s just p^{2}/2m. Question is, what will happen? Can you tell me what will happen in this problem, to this marble? Student: Don’t we have to use the Schrödinger equations? Professor Ramamurti Shankar: No. Classical mechanics. Student: Oh, the marble will only go up as far as it has the amount of energy that that would equal the potential energy at that point. Professor Ramamurti Shankar: That’s correct. So he said it will go as far as energy equals potential, so let p^{2}/2m have this value. That’s my energy. And it does not change. That’s the law of conservation of energy. So here, since energy is kinetic + potential, it’s all kinetic. Somewhere here, that part of it is potential; the rest of it is kinetic. Here kinetic vanishes. Here you’re not allowed. So you will go, you will climb that far and you’ll come back. But if your initial energy was that, then you will come here, going very fast. You will slow down, but you still make it on the other side, but with lower velocity. That’s very simple. There’s only one answer you can get. You throw something. If it comes back, it means the barrier is taller than what you have. Suppose you cannot see this barrier. All you have is a marble, and you want to know how tall the barrier is. Very easy. Throw the marble at a known speed. If it comes back, then you haven’t reached the top of the barrier. Keep on increasing the kinetic energy till one day it doesn’t come back. That’s when you’re sitting right here. That’s one way to tell how tall this hill is without actually going there. But now what I want to do is do a quantum mechanical experiment in exactly the same situation. I want to send a particle from the left and I want to see what happens. So now, to simplify the problem, I’m going to take the potential to again have only 2 values. At x = 0, it jumps to a value V_{0}. So this problem is just the Schrödinger equation in the left region, Schrödinger equation in the right region. So I’m going to spare you some of the calculations, because it’s so familiar. One will look like Ae^{ikx} + Be^{ikx}, where k^{2} is just 2mE/ℏ^{2}, here. First, let’s take a case where E is bigger than V_{0}, okay? Then I go to region II. Region II I’ve got to solve the Schrödinger equation with a V_{0} in it. Then you will find in region II, ψ II looks like Ce^{ik’x} + De^{−ik’x}, where k’^{2} will be 2m times E  V_{0} over ℏ^{2}. Because when you have a potential, you must always have — look at the Schrödinger equation here — 2m/ℏ^{2} at E  V. So E is not 0. The k’ now, so what does this mean? You can see very simply, it’s got some momentum here. It’s got a smaller momentum here, because k’^{2} is going to be smaller than k^{2}, because V_{0} is nonzero here. That is just the statement that when the particle’s moving here, it’s slower than when it’s here, because it has to climb up the hill. So the wave functions in this region has got an incoming, right moving wave of momentum, ℏk. It’s got to reflect that wave of momentum ℏk. And in this region again, it’s got a right moving and a left moving, because this k can be of either sign. So you can pick the coefficients, A, B, C, D arbitrarily, but I’m going to choose D = 0, because what I have in mind is a problem where the particle actually came in from the left. If it came in from the left, I’m prepared for some of it coming back reflected, and some of it getting transmitted. There’s nobody coming in from the right at + infinity, because I’m not sending anything from there. I’m sending my particle from the left. So I pick a solution in which D, which is anything, I pick that to be 0. So here are my two solutions. I hope you understand that k and k’ are not the same number. k’ is smaller because the momentum on the right hand side is going to be less, because some of it’s eaten up by potential energy. So what people do is they draw a picture like this. They draw a wave that’s going at some speed, and it comes in this region and it slows down. That’s what the function will look like. The real function’s made of exponentials, but the real and imaginary parts will have rapid oscillations on the left and slow oscillations on the right, because k is bigger than k’. Very good. So now I have the following condition: at the interface where the two regions meet, the ψ must match from the left and from the right. So x = 0, coming from the left, I want ψ_{1} at 0 should be the same at ψ_{2} at 0. That will tell me that A + B = C. Because if you put x = 0, all the exponentials become one, and A + B must = C. There’s the condition that if you come from the two sides, the wave function has to match. But in the Schrödinger theory, you make a second requirement that the derivatives also match. The two functions coming from the left and right cannot join like this. They must join with the same slope. The reason the slope must be the same is that if the slope changed abruptly at one point, the rate of change of slope will be infinite there, because d^{2}ψ/dx^{2} is the rate of change of the slope. But in the Schrödinger equation, the second derivative ψ, if it’s infinite, there is no compensating terms in the equation, because E and V are all finite. So that infinity will not be balanced, so you can never have that. So second derivative must be finite; first derivative must be continuous. So I make the requirement that dψ/dx in region I at x = 0 must match dψ II dx at x = 0. What does that give me? Well, take the derivative and equate them at x = 0, you’ll find ik times A  B = ik’ times C. Take the derivative and then equate the derivatives at x = 0. Derivatives brings down ik for this, ik for that. Blah, blah, you put them together. So these are the two conditions. Now what causes mild panic is that I have three unknowns and two equations. I’ve got A, B and C, and I’ve got only two equations. So what do you think is going on? Anybody have an idea? It’s that even in the particle in a box, you’ll remember the overall height of the wave function is never determined by the Schrödinger equation. It’s whatever you like. So what people do is they say, “Let’s pick A = 1.” That sets the overall height arbitrarily. Then I get 1 + B = C and ik times 1  B = ik’ [C}. Now of course you can solve for B and you can solve for C. I’m not going to do that here; it’s very easy. You’re going to find B = k − k’ over k + k’. You’re going to find C = 2k over k + k’. That’s simple algebra of simultaneous equations. I want you to look at the answer. You remember now what A, B and C are. A comes in, B goes back and C goes to the right. What do you find strange about the answer? It’s that if you came in from the left with an energy bigger than the barrier, there’s a certain probability that you will go backwards. In classical mechanics, if your energy is more than the barrier, you will never come back. The probability is 0. In quantum theory, even though your energy is more than the barrier, there’s a certain probability you’ll get reflected and certain probability to get transmitted. The square of this is like the odds of going to the right. The square of this is like the odds of getting reflected. The point is the reflection probability does not vanish even if the energy is bigger than the barrier height. That’s because this is quantum mechanics and you’re really dealing with waves and not particles. And waves have the property that when the medium changes, there is some reflection. And to the quantum mechanical wave ψ, a barrier looks like a change in refractive index, so it will get reflected and some will get transmitted. But the interesting thing is, in the quantum problem, you can send your billiard ball to find how tall the mountain is. And even if the energy was more than the mountain, sometimes it’ll come back. So that’s not a reliable way to measure that. The last example is, what if your energy is less than this? What if your energy is like here, lower than the barrier? If you’re lower than the barrier, you will find that k’ will equal i times some κ, because it will become imaginary, and you will have then a function that looks like this. The function will oscillate, then it will die exponentially in this region. It’s exactly what I did earlier, like particle in a box. In the classically forbidden region, it goes a little distance but not very far. So it does get reflected if the energy is less, but you can actually find it in a region where it’s not supposed to be. And you can say that’s going to be very embarrassing for the particle to be found here, because its kinetic energy should be negative here. So what’s the particle say to you when you catch it there? It will say, “In order to locate me here, you shine some light on me, and the energy I got from the photon made me legitimate in that region.” It will come back to positive energy. So the act of measurement will prevent it from ever being caught in that region, but it’s got an amplitude, a probability to be caught there. Final thing I want to mention is, if this barrier, instead of going on forever, terminated there and became 0 again, this will leak out here and start going out with the same wavelength as that one. That’s called barrier penetration. That means if you send a particle here with an energy not enough to overcome the barrier in classical mechanics, in quantum mechanics, it’s got a small chance of being found on the other side. On the other side, it can go scot free. It’s not allowed in this region, because there the kinetic energy is negative. Here kinetic energy is the same as before, except the height may be very small. So this means no barrier is completely safe in quantum theory. So I’m going to give you my final survival tip. I’ve told you so many situations you can be in where this course will help you. So here’s another one. You are in a prison and it doesn’t have infinite walls, but it’s got some walls. What’s your strategy? I say go ram yourself into that wall as often as you can, because there’s a small probability that you can suddenly find yourself on the other side. This is what happens in alpha decay. In alpha decay, there’s helium nuclei sitting inside a nucleus. There’s a barrier that keeps it from coming out of the nucleus, but the barrier goes up and comes down. So if you go a certain distance from the nucleus, you’re again free. So the alpha particle does exactly what I told you. It goes rattling back and forth inside the nucleus and once in a while, it penetrates and comes to the other side, and that’s the alpha decay of the nucleus. So this really happens. It may not happen to you in the situation I described, but the probability is not zero. [end of transcript] Back to Top 
mp3  mov [100MB]  mov [500MB] 