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PHYS 201: Fundamentals of Physics II
Lecture 21
 Quantum Mechanics III
Overview
The fact that the wave function provides the complete description of a particle’s location and momentum is emphasized. Measurement collapses the wave function into a spike located at the measured value. The quantization of momentum for a particle on a ring is deduced.
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htmlFundamentals of Physics IIPHYS 201  Lecture 21  Quantum Mechanics IIIChapter 1: Review of the Particle Wave Function [00:00:00]Professor Ramamurti Shankar: All right. Shall we begin now? So as usual, and as promised, I will tell you what we have done so far, and that’s all you really need to know, to follow what’s going to happen next. The first thing we learned is if you’re studying a particle living in one dimension, that’s all I’m going to do the whole time because it’s mathematically easier and there is not many new things you gain by going to higher dimensions. There’s a particle somewhere in this onedimensional universe. Everything you need to know about that particle is contained in a function called the wave function, and is denoted by the symbol ψ. By the way, everything I’m doing now is called kinematics. In other words, kinematics is a study of how to describe a system completely at a given time. For example, in classical mechanics for a single particle the complete description of that particle is given by two things, where is it, and what its momentum is. Dynamics is the question of how this changes with time. If you knew all you could know about the particle now can you predict the future? By predict the future we mean can you tell me what x will be at a later time and p will be at a later time. That’s Newtonian mechanics. So the kinematics is just how much do you need to know at a given time, just x and p. Once you’ve got x and p everything follows. As I mentioned to you, the kinetic energy, for example, which you write as ½mv^{2}, is given just in terms of p, or in higher dimensions the angular momentum is some cross product of position and momentum, so you can get everything of interest just by giving the position and momentum. I claim now the equivalent of this pair of numbers in this quantum world is one function, ψ(x). So it’s a lot more information than you had in classical physics. In classical physics two numbers tell you the whole story. Quantum theory says “give me a whole function”, and we all know a function is really infinite amount of information because at every point x the function has a height and you’ve got to give me all that. Only then you have told me everything. That’s the definition. And the point is ψ can be real, ψ can be complex, and sometimes ψ is complex. So we can ask, you’ve got this function, you say it tells me everything I can know, well, what can I find out from this function? The first thing is that if you took the absolute square of this function, that is the probability density to find it at the point x. By that I mean if you multiply both by some infinitesimal Δx that is the probability that the particle will be between x and x + dx. That means that you take this ψ and you square it. So you will get something that will go to 0 here, go up, go to zero, do something like that. This is ψ^{2}, and that’s your probability density and what we mean by density is, if the function p(x) has an area with the xaxis like p(x)dx that’s the actual probability that if you look for this guy you’ll find him or her or it in this interval, okay? So we will make the requirement that the total probability to find it anywhere add up to 1. That is a convention because — well, in some sense. It’s up to you how you want to define probability. You say, “What are the odds I will get through this course, 50/50?” That doesn’t add up to 1 that adds up to 100, but it gives you the impression the relative odds are equal. So you can always give odds. Jimmy the Greek may tell you something, 7 is to 4 something’s going to happen. They don’t add up to 1 either. I mean, 7 divided by 11 is one thing and 4 divided by 11 is the absolute probability. So in quantum theory the wave function you’re given need not necessarily have the property that its square integral is 1, but you can rescale it by a suitable number, I mean, if it’s not 1, but if it’s 100 then you divide it by 10 and that function will have a square integral of 1. That’s the convention and it’s a convenience, and I will generally assume that we have done that. And I also pointed out to you that the function ψ and the function 3 times ψ stand for the same situation in quantum mechanics. So this ψ is not like any other ψ. And if ψ is a water wave 3 inches versus 30 inches are not the same situation, they describe completely different things. But in quantum mechanics ψ and a multiple of ψ have the same physics because the same relative odds are contained in them. So given one ψ you’re free to multiply it by any number, in fact, real or complex and that doesn’t change any prediction, so normally you multiply it by that number which makes the square integral in all of space equal to 1. Such a function is said to be normalized. If it’s normalized the advantage is the square directly gives you the absolute probability density and integral of that will give you 1. That’s one thing we learned. You understand now? What are the possible functions I can ascribe with the particle? Whatever you like within reason; it’s got to be a single value and it cannot have discontinuous jumps. Beyond that anything you write down is fine. That’s like saying what are the allowed positions, or allowed momentum for a particle in classical mechanics? Anything, there are no restrictions except x should be real and p should be real. You can do what you want. Similarly all possible functions describe possible quantum states. It’s called a quantum state. It’s this crazy situation where you don’t know where it is and you give the odds by squaring ψ. That’s called a quantum state and it’s given by a function ψ. All right, now I also said there is one case where I know what’s going on. So let me give you one other case. Maybe I will ask you to give me one case. The particle is known to be very close to x = 5 because I just saw it there and ε later I know it’s still got to be there because I just saw it. Now what function will describe that situation? You guys know this. Want to guess? What will ψ look like so that the particle is almost certainly near x = 5? Student: [inaudible] Professor Ramamurti Shankar: Yeah? Centered where? Student: Centered at 5. Professor Ramamurti Shankar: At 5, everybody agree with that? I mean the exact shape we don’t know. Maybe that’s why you’re hesitating, but here is the possible function that describes a particle that’s location isn’t known to within some accuracy. So one look at it, it tells you, “Hey, this guy’s close to 5.” I agree that you can put a few wiggles on it, or you can make it taller or shorter if you change the shape a bit, but roughly speaking here is what functions, describing particles of reasonably well know location, look like. They’re centered at the point which is the wellknown location. On the other hand, I’m going to call is ψ x = 5. That is a function, and the subscript you put on the function is a name you give the function. We don’t go to a party and say, “Hi, I am human.” You say, “I’m so and so,” because that tells you a little more than whatever species you belong to. Similarly these are all normalizable wave functions, but x = 5 is one member of the family, which means I’m peaked at x = 5. Another function I mentioned is the function ψ_{p}(x). That’s the function that describes a particle of momentum p. We sort of inferred that by doing the double slit experiment. That function looks like this. Some number A times e^{ipx/ℏ}. Now you can no longer tell me you have no feeling for these exponentials because it’s going to be all about the exponential. I’ve been warning you the whole term. Get used to those complex exponentials. It’s got a real part. It’s got an imaginary part, but more natural to think of a complex number as having a modulus and a phase, and I’m telling you it’s a constant modulus. I don’t know what it is. But the phase factor should look like ipx/ℏ. So if I wrote a function e to the i times 96x/ℏ, and I said, “What’s going on?” Well, that’s a particle whose momentum is 96. So the momentum is hidden in the function right in the exponential. It’s everything x of the i, the x and the ℏ. Whatever is sitting there that’s the momentum. I am going to study such states pretty much all of today. So let’s say someone says, “Look, I produced a particle in a state of momentum p. Here it is. Let’s normalize this guy.” To normalize the guy you’ve got to take ψ^{2} and you’ve got to take the dx and you’ve got to get it equal to 1. If you take the absolute square of this, A absolute square is some fixed number. I hope you all know the absolute value of that is 1 because that times its conjugate, which is e^{−ipx/ℏ}will just give you e^{0} which is 1. I want 1 times dx over all of space to be equal to 1. That’s a hopeless task because you cannot pick an A to make that happen, because all of space, the integral of dx over all of space is the length of the universe you’re living in, and if that’s infinite no finite A will do it. So that poses a mathematical challenge, and people circumvent it in many ways. One is to say, “Let’s pretend our universe is large and finite.” It may even be the case because we don’t know. And I’m doing quantum mechanics in which I’m fooling around in a tiny region like atoms and molecules, and it really doesn’t matter if the universe even goes beyond this room. It goes beyond this room. It goes beyond the planet. It goes beyond the solar system. I grant you all that, but I say allow me to believe that if it goes sufficiently far enough it’s a closed universe. So a closed universe is like this. A closed onedimensional universe is a circle. In that universe if you throw a rock it’ll come back and hit you from behind. In fact you can see the back of your head in this universe because everything goes around in a circle. All right, so that’s the world you take. Chapter 2: Particle on a Ring [00:11:21]Now that looks kind of artificial for the real world, which we all agree seems to be miles and miles long, but I don’t care for this purpose what L is as long as it’s finite. If L is finite, any number you like, then the ψ of p of x will look like 1/√L e^{ipx/ℏ}. Do you agree? If you take the absolute square of this you’ll get a 1/L. This’ll become 1. The integral of 1 of L over the length of space is just 1. So that’s the normalized wave function. Now this is also a very realistic thing if in practice your particle is restricted to live in a circle. Again, there are a lot of experiments being done, including at Yale, where there’s a tiny metallic ring, a nano scale object, and the electrons are forced to lived in that ring. So the ring has a radius and this L is just 2ΠR where R is the radius. There L is very real. It’s not something you cooked up. It’s the size of the ring, but sometimes even if you’re not doing particle in a ring, if you’re doing particle in a line you just pretend that the line closes in on itself. That if you start at the origin and you go on the two sides it closes in. If this is x = 0 you go right and you go left they all meet at the back and it’s a closed ring. So let’s imagine what life looks like for a particle forced to live in a ring of circumference L. The normalized wave function looks like this of momentum p. This is 1/√L e^{ipx/ℏ}. The probability to find this guy at some point x which is the absolute value of ψ^{2} is just 1/L. That means the probability is constant over the entire ring. We don’t know where it is. You can find it anywhere with equal probability. It’s always true that if you know the momentum you don’t know where it is. Now let’s ask one other question. Here is the circle in which I’m living. If you look at the real part and imaginary part with their sines and cosines they kind of oscillate. Do you understand that? When the x varies they oscillate. And one requirement you make that if you go all the way around and come back it’s got to close in on itself. Because if you increase x from anywhere by an amount L, which is going all the way around the circle, you’ve got to come back to where you start. The function has to come back where you started meaning it’s a single valued function. If you say what’s ψ here you’ve got to get one number. That means if you start at some point, you go on moving, you follow the ψ, you go all the way around and you come back you shouldn’t have a mismatch. It should agree. So the only allowed functions are those obeying the condition ψ(x+L) = ψ(x). You understand that? Take any point you like, follow the function for a distance L. That means you come back and the function better come back. So a function like e to the −x^{2} is not a good function to go around a circle and come back. If you go in real space, if you go on all the way it doesn’t come back to where you are. So you’ve got the right functions which have the property that when you add L to them you come back where you are. That’s a condition of single valuedness. So that means in this function if I take it at some point e^{ipx}at some point x then I add to it L, namely I’m getting e^{ipx/ℏ}times e^{ipL/ℏ}. That has to agree with the starting value which is e^{ipx/ℏ}. In other words I’m comparing the wave function at the point x and the point x + L. See, if your world is infinite this is x, this is x + L, this is x + 2L, they’re all different points. What the function does here has nothing to do with what it does here, but if you wrapped out the region back into itself what it does when you go a distance L is no longer independent of what it does at the starting point. It has to be the same thing. Therefore, I’m just applying the test. I’m saying take the function at x + L, factorize it this way, demand that it be equal to a function of x. These cancel and you learn then that e^{ipL/ℏ}should be 1. Now how can that be 1? Well, one way is p = 0, but I hope you know that it’s not the only way. Do you know how else it can be 1 should p be 0? Student: [inaudible] Professor Ramamurti Shankar: Pardon me? Student: [inaudible] Professor Ramamurti Shankar: I didn’t hear that, sorry. Student: It’s infinite. Professor Ramamurti Shankar: I didn’t hear that again. Student: It’s infinite [inaudible]. Professor Ramamurti Shankar: You still have to say it louder. Student: If it is infinite [inaudible]. Professor Ramamurti Shankar: Any other answers? Yeah? Student: If it’s a multiple of 2Π. Professor Ramamurti Shankar: Is that what you said multiple of 2Π? Student: Sure. Professor Ramamurti Shankar: Good. I’m glad you didn’t say it loud enough. Okay good. It’s a multiple of 2Π because any trigonometry function if you add 2Π, or 4Π, or 6Π whose argument doesn’t change then you should never forget the fact that this exponential is the sum of two trigonometric functions cosine + i sine, and they all come back. Therefore, it is too strong to say p should be 0. p should be such that pL/ℏ is 2Π times any integer when the integer can be 0, plus minus 1, plus minus 2 etcetera. That means p has to be equal to 2Πℏ/L times m where m is an integer, positive, negative, 0. Now this is a very big moment in your life. Why is it a big moment? Yes? You don’t know? Well if you haven’t done it before, this is the first time you’re able to deduce the quantization of a dynamical variable. This is the first time you realize this is the quantum of quantum theory. The allowed momenta for this particle living in there, you might think it can zip around at any speed it likes, it cannot, especially in a ring of nano proportions these values of p are all discrete. p times.. 2Πℏ/L times 0 is here, times 1 is here, times 2 is here, 3 is here, 1, 2. These are the only allowed values of p. So that’s the case of quantization, and the quantization came from demanding that the wave function had a certain behavior that’s mathematically required. The behavior in question is single valuedness. Now I can say this in another way. Let’s write the same relationship in another way. So let me go here. So I had the allowed values of p are 2Πℏ/L times m. I can write it as p times L/2Π is equal to mℏ. L/2Π is the radius of the circle. So I find that p times R is equal to mℏ. You guys know what p times R is for a particle moving in a circle at momentum p? What is the momentum times the radial distance to the center? Student: Angular momentum. Professor Ramamurti Shankar: Angular momentum. That’s usually denoted by a quantity L in quantum mechanics. So angular momentum is an integral multiple of ℏ. That’s something you will find even in high school people will tell you angular momentum is an integral multiple of ℏ. Where does it come from and how does it come from quantum mechanics? Here is one simple context in which you can see the angular momentum is quantized to these values. Now what I will do quite often is to write the state ψ as ψ of p this way, e^{2Π}, I’m sorry, e^{ipx/ℏ}, but to remind myself that p is quantized to be integer multiples of some basic quantum where the multiple is just m, I may also write the very same function the ψ’s of m is 1/√L e^{ix/ℏ}, but for p you put the allowed value which is 2Πℏ/L times m. And what do you get? You get 1/√L times e^{2Πimx/L}. They’re all oscillating exponentials, but you realize that the label p and the label m are equally good. If I tell you what m is you know what the momentum is because you just multiply by this number. So quite often I’ll refer to this wave function by this label m which is as good as the label p. The nice thing about the label m is that m ranges over all integers. p is a little more complicated. p is also quantized, but the allowed values are not integers, but integers times this funny number. In the limit in which L is very, very large, compared to ℏ, the spacing between the allowed values becomes very, very close, and you may not even realize that p is taking only discrete values. So when you do a microscopic problem where L is 1 meter the spacing between one allowed p and another allowed p will be so small you won’t notice it. So even if you lived in a real ring of circumference 1 meter the momenta that you’ll find in the particle will look like any momentum is possible. That’s because the allowed values of p of so densely close, packed, that you don’t know whether you’ve got this one, or that one or something in between because you cannot measure it that well. So it’ll smoothly go into the classical world of all allowed momenta if L becomes macroscopic. And the notion of macroscopic is ” how big is big”. Well, it should be comparable to ℏ. I mean, it should be much bigger than ℏ then it’ll become continuous. So now I’m going to ask the following question. If I have a particle in this world, in this onedimensional ring and I plot the wave function, some function ψ(x), suppose it’s not one of these functions? It’s not e^{ipx/ℏ}. It’s some random thing I wrote here. Of course it meets itself when you go around a circle. That’s a periodic function. Be very careful. A periodic function doesn’t mean it oscillates with a period. In this case periodic means when I go around the loop it comes back to a starting value. It doesn’t do something like this where the two don’t match. That’s what I mean by periodic. It doesn’t mean it’s a nice oscillatory function. These guys are periodic and oscillatory with the period. These are periodic only in the sense that if you go around the ring you come back to the starting value of the function. So I give you some function like this and I ask you what’s going on with this guy. What can you say about the particle? So can you tell me anything now given this function ψ(x)? Does it tell you any information? You mean if I draw the function like that, you get no information from it. Is that what you’re saying? You must know — yes? Student: You square it you get the probability Professor Ramamurti Shankar: Yes, look. If you knew that you’ve got to say that because if I think that you didn’t realize that I know we are both in serious trouble. That’s correct. I want to reinforce the notion over and over again. Wave functions should tell you something. Square it you get the probability. If you don’t square it don’t think it’s the probability because it can be negative. It can even be complex. So don’t forget if you square it you get the probability. Define a certain position. That means if you went around with this little Heisenberg microscope all over the ring and you catch it and you say, “Good, I found it here,” and you do that many, many times. By many, many times I mean you take a million particles in a million rings each in exactly this quantum state and make measurements, then your histogram will look like the square of this function. But there is more to life than just saying, “Where is the particle?,” because in classical mechanics you also ask, “What is its momentum.” The question I’m asking you is, “What is the momentum you will get when you measure the momentum of a particle in this quantum state? You know the answer only on a few special cases. If by luck your function happened to look like one of these functions — right now I’ve not told you what ψ is. It’s whatever you like, but if it looked like one of these you’re in good shape because then the momentum is whatever p you find up here, but it may not look like that. This guy doesn’t look like that. So then the question is if you measure momentum what answer will you get. Now that is something — anybody know what the answer is? Now I accept your silence because you’re not supposed to know this. This is a postulate in quantum mechanics. It’s not logic. It’s not mathematics. No one could have told you 300 years ago this is the right answer. So this is another postulate just like saying, ψ^{2} is probability defined at x. There is a new postulate. It addresses the following question. If I shift my attention from position, to momentum, and I ask, “What are the odds for getting different answers for momentum?” That answer’s actually contained in the same function ψ of x in the following fashion. Take this function ψ(x) and write it as a sum over p of these functions, ψ_{p} of x with some coefficient A_{p}. I can also write it equally well as the same function labeled by integer m and I want to call the coefficient m. They’re exactly the same thing, p and m are synonymous. The same function. This function is called ψ_{m} because they contain the same information on the momentum. So either you can write it this way if you want to see the momentum highlighted, or you can write the function this way if you want to see the quantum number m highlighted, but they stand for the same physics. If m is equal to 4 it means the cosine and the sine contained in the complex exponential finish 4 complete oscillations as they go around the cycle. All right, so here’s what you’re told. Take the arbitrary periodic function. Write it as a sum of these functions, each of definite momentum, with some coefficient. Then the probability that you will get a momentum p when you measure it, which is the same as the probability you will get the corresponding m, is nothing but the absolute square of this coefficient. In other words, anyway, this is a postulate. Let me repeat the postulate. Somebody gives you a function. You write the function as a sum of all these periodic functions, each with the index m, multiply each with a suitable number so that they add up to give you the function that’s provided to you. Once you’ve done that the coefficient squared with the particular value m is the probability you will get that value for m or the corresponding momentum. So there are two questions one can ask at this point. First question is what makes you think that you can write any function I give you as a sum of these functions. Realize what this means. I’m saying I can write any function as e^{2Πimx/L}√L times A_{m}. I’m writing out this function explicitly for you. Now that is a mathematical result I will not prove here, it’s called Fourier series and it tells you that every periodic function, namely that which comes back to itself when you go around a period length L, can be written as a sum of these periodic functions with suitably chosen coefficients. It can always be done. That is analogous to the statement that if you are say living in three dimensions and there is a vector V, and you pick for yourself three orthonormal vectors, i, j and k, then any vector V can be written as V_{x} times i + V_{y} times j+ V_{z} times k. In other words, I challenge you to write any arrow in three dimensions starting from the origin and pointing in any direction of any finite length. I’ll build it up for you using some multiple of i, some of j, and some of k. We know that can be done in fact. That’s if you want the technical definition of threedimension. Yes? Student: With the Fourier case would there be one unique way of writing ________? Professor Ramamurti Shankar: Good point. That’s correct. There will be a unique way. Student: There is? Professor Ramamurti Shankar: There is. You agree in three dimensions there is no other mixture except this one. If somebody comes with the second way of writing it you can show that the second way will coincide with the first way. So the expansion of a function, in what are called these trigonometric or exponential functions and it’s called the Fourier series of the function, has unique coefficients. And I’ll tell you right now what the formula for the coefficient is, but first I’m telling you that just like it’s natural to build a vector out of some building blocks i, j and k, it’s natural to build up periodic functions with these building blocks the ψ_{m}. The only difference is there you need only three guys. Here you need an infinite number of them because a range of m goes from minus to plus infinity, all integers. But it’s still remarkable that given all of them you can build any function you like, including this thing I just wrote down arbitrarily, can be built. The fact that you can prove it, that you can do it, I don’t want to prove because it’s kind of tricky, but I will prove the second part of it which is given that such an expansion exists how do I find these coefficients given a function? So let’s ask a similar question in the usual case of vectors. How do I find the coefficients? Suppose I write the vector V as e_{1} times V_{1} + e_{2} times V_{2} + e_{3} times V_{3}. Don’t worry e_{1}, e_{2} and e_{3} are the usual guys, e_{1} is i, e_{2} is j and e_{3} is k. People like to do that because in mathematics you may want to go to 96 dimensions, but we’ve only got 26 letters, so if you stuck to i, j and k you’re going to have trouble, but with numbers you never run out of numbers. So you label all the dimensions by some number, which in this case happens to go from 1 to 3. You also know that these vectors i and j have some very interesting properties, i⋅iis 1. That’s the same as j⋅j. That’s the same as k⋅k. And that i⋅k and i⋅j are 0 and so on. Namely the dot product of one guy with himself is 1, and any one with anything else is 0. That just tells you they all have unit length and they’re mutually perpendicular. I want to write this as e_{i}⋅e_{j}, but this could be 1, 2 or 3. That could be 1, 2 or 3. I want to say this is equal to 1 if i is equal to j. This is equal to 0 if i is not equal to j. This is a usual vector analysis. I’m just saying the dot product of basis vectors has this property, 1 if they match, 2 if they’re different, so there’s a shortcut for this and that’s write the symbol δ_{ij}. δ_{ij} is called Kronecker’s delta. Kronecker’s in a lot of things, but this is one place where his name has been immortalized. He just said instead of saying this all the time, 1 if they’re equal, 0 if they’re different, why don’t you call it my symbol, the Kronecker’s symbol, δ_{ij}. It’s understood that this whole thing simply says δ_{ij}. This is shorthand. That means if on the left hand side there are two guys with indices i and j if the indices are equal the right hand side is 1, indices are unequal right hand side is 0. Do you understand that this gives you the fact that each vector is of length 1 and that each is perpendicular to the other two. Now that is what we can use now to find out. So let me write the vector V in this notation as e_{i} times V_{ii} from 1 to 3. You’re all familiar with this way to write the sum? So I come with a certain vector. The vector is not defined in any axis. It’s just an arrow pointing in some direction. It’s got a magnitude and direction. And I say, “Can you write the vector in terms of i, j and k?” And the answer is, “Yes, of course I can.” So here’s your vector V. Here is, if you like, e_{1}, e_{2} and e_{3}. The claim is some mixture of e_{1}, e_{2}, e_{3} will add up to this V. That’s granted, but how much? How much e_{1} do I need? How much e_{2} do I need? There’s a very simple trick for that. Anybody know what that trick is? You might know that trick. You’ve seen it anywhere? Here is the trick. Suppose you want to find V_{2}? You take the dot product of both these things with e_{2}. Take e_{2} dot this, and take dot product with e_{2}. What happens is you will get — the dot product can go inside. You’ll get e_{i}⋅e_{2}times V_{i} where i goes from 1 to 3. What is e_{i}⋅e_{2}? e_{i}⋅e_{2}is δ_{i2}. That means if this index size is equal to 2 you get 1. If not equal to 0 you strike out. You get 0. So of all the three terms in this only one will survive. That’s the one when i is equal to 2 in which case you will get 1. That’s multiplying V_{2} so it’ll give you V_{2}. So to find the component number 3 you take the dot product of the given vector with e_{3}, and that will give you, you can see, that will give you V_{3} or V_{2} or whatever you like. I’m going to use a similar trick now in our problem. The trick I’m going to use is the following. So the analogy is just like you had V = sum over i e_{i} times V_{i}, I have ψ of x is equal to sum over m of some A_{m} times the function ψ_{m} of x. You understand that? ψ_{m} is a particular function which I don’t want to write every time, but if you insist it is 2Πimx/L divided by √L. So I’m trying to find this guy. How much is it? That’s the question. Now here we had a nice rule. The rule says e_{i}⋅e_{j} is δ_{ij}. That was helpful in finding the coefficients. There’s a similar rule on the right hand side which I will show you and we can all verify it together. The claim is ψ_{m}*(x) times ψ_{n}(x) dx from 0 to L is in fact δ_{mn}. So the basis vectors ψ_{m} are like e_{i}. The dot product of two basis vectors being δ_{ij} is the same here, like same integral, but one of them star times the other one is 1 if it’s the same function and 0 if it’s not the same function. Let’s see if this is true. If m is equal to n, can you do this in your head? If m is equal to n what do we have? This number times this conjugate is just 1. You just get 1/L. An integral of 1/L dx, so if you want I will write it here, it’s 2Πi times n −mx/L dx from 0 to L. Do you understand that? You take the conjugate of the first function. That where’s it’s a m here, and take the second function which is e^{2Πin}^{/L} times x. I’ll wait until you have time to digest that. The product of ψ_{m}* with ψ_{n} you combine the two exponentials, but the thing that had m in it has a m here because you conjugated everything. So I’m saying, “What is the integral going to be?” If n is equal to m you can see that in your head. This is e^{0}. That is just 1. The integral of dx is L. That cancels the L. You get 1. That’s certainly true if m is equal to n. If m is not equal to n, suppose it is 6? It doesn’t matter. This will complete 6 full oscillations in the period, the sine and the cosine, but whenever you integrate a sine or a cosine over some number of full periods you get 0. So this exponential, when integrated over a full cycle, if it’s got a nonzero exponent integer exponent will give you 0. So you see the remarkable similarity between usual vector analysis and these functions. This is an arbitrary vector. This is an arbitrary function. e_{1}, e_{2}, e_{3} are basis vectors; ψ_{m} are if you want basis function. I can build any vector out of these unit vectors. I can build any function out of these basis functions. And finally, if I want to find out a coefficient, what should I do? You want to find the coefficient A_{n}. What did I do here to find V_{j}? You take V⋅e_{j} where j could be whatever number you picked. Because if you take the dot product of the two sides when you take dot product of e_{j} the only term who survives is i = j. That’ll give you the V_{j}. So similarly here I claim the following is true. If you do the integral of ψ_{n}* x times the given function dx from 0 to L you will get A_{n}. So once I show this I’m done. Now if you don’t have the stomach for this proof you don’t have to remember this proof. That’s up to you. See, I don’t know how much you guys want to know. I’m trying to keep the stuff I just tell you without proof to a minimum. I felt bad telling you that every function can be expanded this way, but the coefficients being given by the formula is not too far away, so I want to show you how it’s done. You may not know the details why this is working, but you should certainly know that if you want coefficient number 13 you’ve got to take ψ_{13}* and multiply with the given function and integrate. That you’re supposed to know, so why does this work? So let’s see what this does. We are trying to take ψ_{n}*(x). The given function looks like a sum A_{m}ψ_{m}(x) dx 0 to L summed over m. So this summation you see you can bring the ψ in here if you like. It doesn’t matter. Then do the integral over x then you will find this is giving me — maybe I’ll write it here. That is going to be equal to sum over m A_{m} of ψ_{m} of ψ_{n}* x ψ_{m}(x) dx, and that is going to be δ_{mn}. That means I will vanish unless m equals n, and when m equals n I will give you 1. That means the only term that survives from all these terms is the one where m matches n, so the thing that comes out is A_{n}. Once again, you will see this in my notes, but do you have any idea of what I did or where this is going? In quantum theory if you want to know what’ll happen if I measure momentum for a particle living in a ring you have to write the given function in terms of these special functions each item defined as a definite momentum with suitable coefficients. The rule for finding coefficient A_{n} is to do this integral of ψ_{n}*. This one, this is the rule. A_{n} is the integral of ψ_{n}*ψ dx. And once you found the coefficients for all possible n then the probability that you will have some momentum corresponding to m is just A_{m}^{2}. That is a recipe. What the recipe tells you is if your function is made up as a special function with a definite momentum of course you will get that momentum as the answer when you measure it. If your function is a sum over many different momentum functions then you can get any of the answers in the sum, but if it had a big coefficient in the expansion is more likely to be that answer. If it had a small coefficient it’s less likely. If it had no coefficient you won’t get that momentum at all. That’s like saying if you had ψ(x) it’s likely where it’s big, unlikely where it’s small and impossible where it is 0. So that’s your job. Anytime someone gives you a function you have to find these coefficients A_{n} then look at them. They’ll tell you what the answer is. So that’s what I’m going to do now. I’m going to take some trial functions and go through this machinery of finding the coefficients and reading off the answers. So maybe if I do an example you’ll see where this is going. So let’s take an example where I’m going to pick first of all a very benign function then maybe a more difficult function. The function I want to pick is this. Some number n cosine 6Πx/L, somebody gives you that function. That is not a state of definite momentum because it is not e to the i something x. So we already know when you measure momentum you won’t get a unique answer. You’ll get many answers, but what are the many answers? What are the many odds is what we’re asking. I forgot to mention one thing in my postulate. What I forgot to mention is that for all this to be true it is important that the functions, the momentum functions, are all normalized and the given function is also normalized. You should first normalize your function then expand it in terms of these normalized functions of definite momentum. Only then the squares of the coefficient are the absolute probabilities. By that I mean if you do this calculation, and you then went and added all these A_{m}squares, you will find amazingly it adds up to 1. Because you can show mathematically that if ψ*ψ dx is 1 then the coefficients of expansion squared will also be 1. So if you took a normalized ψ then the probabilities you get are absolute probabilities because they will add up to 1. There’s also a mathematical result which I am not showing. All right, so let’s go to this problem. So the first job is, normalize your ψ. So how to normalize this function I’m going to demand that if you square ψ and you integrate it, so I say N^{2} times integral of cosine square as 6Πx/L dx from 0 to L should be 1. Now you don’t have to look up the table of integrals because when you take a cosine squared or a sine squared over any number of full cycles the average value is a half. That means over the length L this integral will be L/2. So you want N^{2}L/2 to be 1 or you want n to be square root of 2/L. So the normalized function looks like square root of 2/L cosine 6Πx/L. So that’s the first job. Second is you can ask now what are the coefficients A_{n}. So A_{n} is going to be integral 1/√L e^{2Πinx/L} times this function square root of 2/L cosine 6Πx/L dx. Are you with me? That’s the rule. Take the function, multiply it by ψ_{n}*, which is this guy here, this is the ψ_{n}*, this is the ψ that’s given to you, and you’re integrating the product in the integral. That’ll give you A_{n}. But I claim, yeah, if you want you can do this integrals, but I think there’s a quicker way to do that. Have any idea of what the quicker way just by looking at it? In other words, if you can guess the expansion I’ll say I will take it. Can you guess what the answer looks like without doing the work? In other words I want you to write this function as a sum over exponentials just by looking at it. Can you tell me what it is? I want you to write it in the form sum over m A_{m} e^{2Πimx/L}. You can find the A_{m} by doing all this nasty work, but I’m saying in this problem there’s a much quicker way. You see it anywhere? What’s the relation between trigonometric functions and exponential functions? Yep? Student: Euler’s formula? Professor Ramamurti Shankar: And what does it say? Student: e^{ix} = cosine x + i sine x. Professor Ramamurti Shankar: Yes, but now I want to go the other way. It is certainly true that cosine θ, I’m sorry, e^{iθ}is cosine θ + I sine θ. I told you, you forget the formula at your own peril. The conjugate of that is e^{iθ} is cosine θ i sine θ. If you add these two you’ll find cosine θ is (e^{iθ} + e^{iθ})/2, and if you subtract them and divide by 2i you’ll find sine θ is e to the minus over 2i. I wanted you to be familiar with complex numbers enough so that when you see a cosine the two exponentials jump out at you. Otherwise you will be doing all these hard integrals that you don’t have to do. That’s like saying sine squared + cosine squared is 1, but you don’t look it up. That’s something you know. What you look up is your Social Security number or mother’s maiden name. You’re allowed to forget those things. You look in a book. That’s the name. But this you have to know because you cannot go anywhere without this because if you don’t know it all the time you must have done these trigonometry calculations in school. You’ve got to plug in the right stuff at the right time for things to all cancel out. You cannot say, “I will look it up,” because you don’t know what to look up. So everything should be in your head, and the minimum you should know is this, the minimum. Therefore, I can come to this function here and I can write it in terms of — what I’m telling you is ψ(x), which was given to us, is equal to the square root of 2/L times cosine 6Πx/L, but I’m going to write it as square root of 2/L times ½e^{6Πix/L}+ e^{6Πix/L}. So that I will write it very explicitly as 1/√2 times e^{6Πix/L}/√L + (1/√2)e^{6Πix/L}divided by √L. In other words, what I’ve done is rather than doing integrals I just massaged the given function and managed to write it as a sum over normalized functions, as I said, with definite momentum with some coefficients. In other words, here it is in the form that we want. A_{m}e^{2Πimx/L}. You agree that I have got it to the form I want? You see that? You take the cosine, write it as sum of exponentials, then put factors of √L so that that’s a normalized function, and that’s a normalized function, and everybody else is A. So what do you find here? What are the A’s for this problem? By comparing the two what do you find are the A’s in this problem? What is A_{14}? Let me ask you this. If you compare this to this one what m are you getting? Do that in your head. If you compare this to this one what is m? Student: 3. Professor Ramamurti Shankar: Pardon me? Student: 3. Professor Ramamurti Shankar: m is 3 for this guy, so this is really ψ_{3} and this is ψ_{3} because you can see there are two momenta in the problem. And you can stare at them, you can see right away. Therefore, A_{3} is 1/√2, and A_{3}is also 1/√2. Therefore, the probability for m = 3 is ½, and the probability for m = 3 is ½ and nothing else. All other A’s are 0 because they don’t have a role. So you might think every term must appear. It need not. So this is a good example that tells you only two of the m’s make it to the final summation. All the other m’s are 0 and they happen to come with equal coefficients, 1 over root 2. The square of that is 1 over 2, and you can see these probabilities nicely add up to 1. I told you if you normalize the initial function the probability for everything will add up to 1. So this is a particle. If a particle is in a wave function cosine 6Πx/L, which is a real wave function, and you can plot that guy going around the circle, it describes a particle whose momentum, if you measure, will give you only 1 of 2 answers, m = 3 is p = 2Πℏ/L times 3 or  3 plus or minus 6Πℏ/L. You can see that too. I mean, just look at this function. If you call it e^{ipx} p is 6Πℏ/L, so this particle has only two possible answers when you measure momentum. So it is not as good as the single exponential which has only one momentum in it. This is made up of two possible momenta, but you won’t get anything else. You will not get any other momentum if you measure this. So if you got m = 14 as a momentum there’s something wrong. It’s still probabilities, but it tells you that there is non0 probability only for these two. Now in a minute I will take on a more difficult problem where you cannot look at the answer, you cannot look at the function ψ and just by fiddling with it bring it to this form. You understand? It is very fortunate for you that I gave you cosine which is just the sum of two exponentials so the two what are called plane wave exponentials are staring at you and you pick them up. I can write other functions, crazy functions for which you will have to do the integral to find the coefficient. Chapter 3: The Measurement Postulate [00:56:26]But I’m going to tell you one other postulate of quantum mechanics. In the end I will assemble all the postulates for you. I’m going to tell you one more postulate. It’s called the measurement postulate. The measurement postulate says that if you make a measurement, and you found the particle to be at a certain location x, then right after the measurement the wave function of the particle will be a spike at the point x because you know that you found it at x it means if that has any meaning at all if you repeat the measurement immediately afterwards you’ve got to get the same answer. Therefore, your initial function could have looked like that, but after the measurement it collapses to a function at the point where you found it. This is called the collapse of the wave function. It goes from being able to be found anywhere to being able to be found only where you found it just now. It won’t stay that way for long, but right after measurement that’s what it’ll be. But the state of the system changes following the measurement. And if you measured x it’ll turn into a wave function with well defined x which we know is a spike at x equal to wherever you found it. Similarly, if you measured momentum and you found it at — if you measure momentum, first of all you’ll get only one of two answers, m = 3 or m = 3. If you got m = 3 the state after the measurement will reduce to this. This guy will be gone. This will be the state. So from being able to have two momenta which are equal and opposite the act of measurement will force it to be one or the other. It can give you either one, but once you’ve got it that’s the answer. It’s like the double slit. It can be here and it can be there. It is not anywhere in particular, but if you shine light and you catch it, right after the measurement it is in front of one slit or the other. Think of this as double slit. There’s some probability for 3 and some for 3, but if you catch it at 3 it’ll collapse to this one. So what happens is in the sum over many terms the answer will correspond to one of them, and whichever one you got only that one term will remain. Everything will be deleted from the wave function. So the act of measurement filters out from the sum, the one term corresponding to the one answer that you got, this is called the collapse of the wave function. So in classical mechanics when you measure the position of a particle nothing much happens. It doesn’t even know you measured it because there are ways to measure it without affecting it in any way. So if had a momentum p at a position x before the measurement it’s the answer right after the measurement because you can do noninvasive measurements. In quantum theory there are no such measurements in general. In general the measurement will change the wave function from being in one of many options to the one option you got, but the answer depends on what you measure. With the same wave function on a ring, let’s say, if you measure position and you found it here that’ll be the answer. If you measured momentum and you got 5 it’ll be something with 5 oscillations in it. So it’ll collapse to that particular function and what it collapses it depends on what you measure. And for a long time I’m going to focus only on x and p. Of course there are other things you can measure and I don’t want to go there right now, but if you measure x it collapses to a spike at that location. If you measure p it collapses to the one term wherever that was, that one e^{ipx} in the sum. So another interesting thing is the only measurement, only answers you will get in the measurement are the allowed values of momentum. You’ll never get a momentum that’s not allowed. And once you get one of the allowed answers there are two things I’m telling you. One is the probability you will get that answer is proportional to the square of that coefficient in the expansion of the given function. In our problem there are only two nonzero coefficients. Both happen to be 1/√2. It happened to be an equal mixture, but you can easily imagine some other problem where this is 1/√6, and something that is 1/√6, and something that is 1/√3, they should all add up to 1 when you square and add them. Then you can get all those answers with those probabilities. So the last thing I’m going to do is just one more example of this where you actually have to do an integral and you cannot just read off the answer by looking at it. Then we’re done with this whole momentum thing. I will write the postulates later, so I don’t want to write it in my handwriting, but one final time. Measurement of x collapses the function ψ(x) to a spike at x. Measurement of p collapses ψ to that particular plane wave with that particular p or m on the exponent. And after that that’s what’s taken to be. People always ask, “How do we know what state the quantum system is in?” Who tells you what ψ(x) is? It’s the act of measurement. If you measure the guy and you found him at x = 5 the answer is ψ is a big spike there. If you measure momentum and you’ve got m = 3 the answer is that particular wave function with 3 on the exponent. So measurements are a way to prepare states. By the way, it’s very important that if you had a system like this the probability for getting a certain x, say at this x, is proportional to the square of that number. Once you took the state and didn’t measure x, but measured momentum it’ll become one of these oscillatory functions with a definite wavelength. It’s a complex exponential, but I’m showing you the real part. Now if you measure position, in fact the square of this will be flat. Remember e^{ipx} is flat. So in classical mechanics if I measure position, and I measure momentum, and I measure position again I’ll keep getting the same answers. I see where it is. I see how fast it’s moving. Again, I see where it is. If I do all of this in rapid succession you will get the same answer xp xp xp. In quantum theory once you measure x it’ll become a big spike at that point. You can get all kinds of answers p. If you measure p you’ve got an answer. That answer is a new wave function which is completely flat. That means if you measure x you’ll no longer get the old x. In fact you can get any x. That’s why you can never filter out a state with well defined x and p because states of well defined x are very spiky, and states of welldefined p are very broad, so you cannot have it both ways. And that’s what I want to show you in this final example. I’m going to take the following wave function on this ring. The function I’m going to take is, ψ is some ne^{−αx}, mod x [ x ]means this is x = 0. To the right it falls exponentially and to the left it falls exponentially. Is it clear what I’m saying here? This is my ring and I’m trying to plot the function. It is highest at the origin and falls exponentially equally for positive and negative x. So that’s the meaning of mod x. So how far can you go before ψ becomes negligible? Well, that’s when e^{x} is a big number. So I’m going to assume that when you go all the way around half the circle, I’m going to assume that αL is much bigger than 1. That means its function dies very quickly spreading negligible beyond some distance. How far does a function live? Roughly speaking you can go a distance Δx so that α times Δx is roughly 1. Because if you plot this exponential you ask, when does it come to say half its value or one fourth of its value, you’ll find it’s some number of order 1/α. So this is a particle whose position has an uncertainty of order 1/α. So you can make it very narrow in space so you know pretty much where it is or you can make it broad, but I want to consider only those problems, where even if it’s broad it’s dead by the time you go to the other side to the back. That’s for mathematical convenience. So this is the state and I want to ask myself any question we can ask. First question I can ask is if I look for its position what will I find? I think we have done it many, many times. You square the guy you get n^{2}e^{2αx}. So that’s the shape. That shape looks the same. If you square an exponential you get another exponential. But now let me demand that this integral dx be equal to 1. So what is that integral? I hope you can see that this function is an even function of x because it’s mod x. So it’s double the answer I get for positive x. Oh yeah, sorry, not minus infinity. It’s −L/2 to +L/2. You can go this way L/2 and we can go that way L/2. So it’s n^{2}is 2 times e to the α x 2α x dx. Student: Should it be L over 2? Professor Ramamurti Shankar: Pardon me? Student: Should it be L over 2. Professor Ramamurti Shankar: Yes, thank you. So I’m going to now assume that in the upper limit of the integration instead of going up to L/2 I go to infinity. It doesn’t matter because this guy is dead long before that. That’s why I made that choice. So that I’m going to write this 2 times N^{2} times 0 to infinity e^{2αx} dx. That’s a pretty trivial integral. It’s just 1/2α. If you want I will write it out. It’s e^{2αx}/2α from 0 to infinity. At the upper limit when you put infinity you get 0. At the lower limit when you put x = 0 you get 1. There’s a minus sign from this and you get N^{2}/α. It should be 1, or N is equal to the √α. So my normalized wave function, this is the first order of business, is square root of αe^{−αx}^{}. I just squared it and I integrated it. The only funny business I did was instead of cutting off the upper integral at L/2 I cut if off at infinity because e^{−L/2} is e to the minus nine million, let’s say. So I don’t care if it’s nine million or infinity. To make the life simpler I just did it that way. Now I want to ask you, what is A(p)? A(p), the coefficient of the expansion remember is e^{−ipx/ℏ}/√L times ψ(x) dx from −L/2 to +L/2. Now I’m going to work with p rather than m. I will go back and forth. You should use to the notion that the momentum can be either labeled by the actual momentum or the quantum number m which tells you how much momentum you have in multiples of 2Πℏ/m. So you’ve got to do this integral. So let’s write this integral. This looks like −L/2 to L/2. In fact I’m going to change this integral to minus infinity to plus infinity because this function e^{ipx/ℏ}times e^{−αx} dx. Do you understand that these limits can be made to be plus and minus infinity because area under a graph that’s falling so rapidly, whether it’s between minus and plus L/2 or minus and plus infinity, is going to be the same. It’s just that this integral is so much easier to do. Now you cannot jump out and do this integral because it’s a mod x here. So mod x is not x, it is x when it’s positive and it’s x when it’s negative. So you’ve got to break this integral into two parts. One part where x is positive from 0 to infinity, I’m sorry, I also forgot a root, α and A√L. Do you see that? So this is really square root of α/L times e^{αx} e^{−ipx/ℏ} dx plus another integral from minus infinity to 0 e^{+αx}times e^{−ipx/ℏ} dx. I split the integral into two parts. So I didn’t make a mistake here. This really is e^{+αx}because x is negative. So what one does in such situations, I’m going to do it quickly and you can go home and check it, just calculus, change the variable from x to x everywhere. In the terms of new variable this will make a x. That’ll become +x. dx will become minus of the new variable. The limits will be plus infinity to 0, and you can flip that for another change of sign 0 to infinity. So that was a very rapid slightofhand, but I don’t want to delay that. This is just — you go home and if you want check that if x goes to x you get that. So you notice this is the complex conjugate of this one. Whatever function I’m integrating here is the conjugate of it because this is real, αx, and ipx has become +ipx. So if I find the first part of the integral I just take that times its conjugate and I’m done. So what do I get for that? I get square root of α/L times — remember integral e^{αx} dx from 0 to infinity is 1/α, but what I have here is dx e^{α} + ip/ℏ x_{0}to infinity plus the complex conjugate. Now you may be very nervous about doing this integral with a complex number in it because real we all know the integral is just 1/α e^{α}. It turns out that it’s true even if it’s got an imaginary part as long as you have a positive real part. In other words, the answer here doesn’t depend on this guy being real. So it’s really α over L times 1 over α + ip/ℏ plus a complex conjugate which is α − ip/ℏ. Now you should be able to combine these two denominators and you get square root of α/L divided by α^{2} plus p^{2} over ℏ^{2} times 2α. Again, this is something you can go and check, but I don’t want to wait until everyone can do this thing because I want to tell you the punch line. So this is what it looks like. A(p) looks like a whole bunch of numbers I’m not going to worry about, but look at the denominator. It’s α^{2} + p^{2} over ℏ^{2}, or if you want multiply by ℏ^{2}. There are some other numbers I’m not interested in. The numbers are not important. How does it vary with p is all I’m asking you to think about. I’m sorry, A(p), this is just A(p), but I want the A(p) squared that looks like the square of this. All I want you to notice is that this function is peaked at p = 0 and falls very rapidly as p increases. When p is 0 you’ve got the biggest height. When does it become half as big or one fourth as big? Roughly when p^{2} is equal to h^{2}α^{2} because that’s when these numbers become comparable, therefore when p is ℏα this function will have a denominator which is twice what it had here or maybe one fourth. I’m not worried about factors like 1 and 2. The point is in momentum space, in momentum you can get all kinds of values of p, but the odds decrease very rapidly for p bigger than ℏα. So the most likely value is 0, but that’s the spread. And Δp is ℏα, or α times Δp is ℏ, and that is the uncertainty principle because α is just Δx. By the way, I will publish these notes too, so you don’t have to worry if you didn’t write everything down. I suggest you — yes? Student: Can you go over again what cc is? Professor Ramamurti Shankar: Complex conjugate. That’s what I did. Whatever number this one is the other guy is obtained by changing i to i. Look, all I want you to notice is this. I took a function whose width is roughly 1/α. Then I looked at what kind of momenta I can get. Then I find that the narrower the function bigger the spread in the possible momenta you can get. So squeezing it in x broadens it out in p. And that’s the origin of the uncertainty principle. It’s simply a mathematical result. The functions which are narrow in x have a Fourier series which is very broad in p. The quantum mechanics relates p to momentum and therefore the uncertainty principle. So anyway, I’ll give you some homework on this, and you can also fill in the blanks of this derivation, which I think is very useful. 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