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PHYS 201: Fundamentals of Physics II
Lecture 22
 Quantum Mechanics IV: Measurement Theory, States of Definite Energy
Overview
It is shown how to extract the odds for getting different values of momentum from a generic wave function by writing it as a sum over functions of definite momentum. A recipe is given for finding states of definite energy, which requires solving a differential equation that depends on what potential the particle is experiencing. The particle in a box is considered and the allowed energies derived.
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htmlFundamentals of Physics IIPHYS 201  Lecture 22  Quantum Mechanics IV: Measurement Theory, States of Definite EnergyChapter 1: Review of Wave Functions [00:00:00]Professor Ramamurti Shankar: So I got an interesting email this weekend from two of your buddies, Jerry Wang and Emma Alexander. The title of the email is this. It says Casaplanka, so I already know these guys are up to no good. And the question is the following. They said, “You’ve written a wave function ψ(x) and A(p). There seems to be no reference of time, so where is time in this,” and they couldn’t stop there. That would be a good place to stop but they went on to say, “Are you saying the ψ is just a ψ as time goes by?” So that’s the kind of stuff that appeals to me, so I don’t care if you don’t learn any quantum mechanics, but if you can do this kind of stuff I’m not worried about you. But now I have to give a serious answer to that serious question, which I sort of mentioned before, which is that everything I’ve done so far is for one instant in time. I hope I made it very clear. So here’s the analogy. If someone comes to you and says, “Tell me all about Newtonian mechanics. How does it work? What’s the scheme like?” You say, “At any instant someone has to give you the x and p of that particle at that time,” and that’s all you need to know. That’s all you can ask and that’s all you need to know. Everything about that one particle is completely given by this x and p. That is called a state. That’s the complete description of any system is called at state, and a state in classical mechanics for one particle in one dimension is just a pair of numbers. It’s a state because given that I can predict the future for you. I can predict the future because if I knew the initial momentum, namely initial velocity. I know how fast the guy is moving, and I know where it is at the initial x. A tiny amount of time later it’ll be at x + vt, but t is very small. Then since I know the acceleration from Newton’s Laws, which is rate of change of velocity, if I knew the initial velocity or initial momentum I can get the momentum a little later. Then in that manner I can inch forward in time. So you need Newton’s Laws, and Newton’s Laws tell you, if you like, rate of change of momentum is the force. So if you like you can write this mdx/dt squared is the force. So this part is called dynamics and this part is called kinematics. So dynamics is how do things change with time? Kinematics is what do you need to know about the particle? In the quantum version of this if you say, “What constitutes complete knowledge of the particle?,” the answer is this function ψ(x). If you knew this function you know all that there is to know about the particle. We’re not saying it’s real. It can be complex. Then you can ask, “If you knew the ψ at one time what’s it going to be later on?” That’s the analog of this one. So I have not come to that yet. So this is a pretty long story. It takes so long just to tell you what to do at one time. Then I’m going to tell you how to go from now to later. In Newtonian mechanics it’s just a oneword answer, x and p now is the whole story. So anyway, since we’ve got four lectures including today I will, of course, tell you before it’s all over what is the analog of this. Namely I’ll tell you a formula for dψ/dt. How does ψ change with time? So let’s summarize what we know. I want to do this every class so that it gets into your system. If it is true that ψ(x) tells me everything I need to know let’s ask questions of this ψ. We can say, “Where is the particle?” So there is some ψ. You’re told then the probability that you will find it somewhere is the absolute value. Probability density is proportional to ψ^{2}. So it doesn’t tell you where the particle is. It gives you the odds, and the odds are this. Then you can say, “Okay, x is just one of the variables of interest to me. How about momentum? If I measure momentum what answers will I get, and what are the odds for the different answers?” That question has a longer answer and it goes as follows. Take the ψ that’s given to you, and write is as a sum with some coefficients of functions that I call ψ_{p}(x), but I will tell you what they are. They are e^{ipx/ℏ} by square root of L. Then I said, “If you manage to write the ψ that’s given to you in this form, summing over all the allowed values of p, then the probability that you will get one of those allowed values is equal to the absolute square of that coefficient.” And then final question can be, “How do I know what A(p) is given some function ψ?,” and the answer is A(p) is the integral of the complex conjugate of this ψ(x) dx. In our case it was a ring of length L. Then it’s 0 to L, otherwise it’s whatever your allowed region is. Yep? Student: Can you explain ________ one more time what… Professor Ramamurti Shankar: What this guy means? Student: Yeah. Professor Ramamurti Shankar: So let me give an example. Suppose ψ of p was cosine 4Πx over L? Then that looks like e^{4Πix/L} + e^{4Πix/L} divided by 2. Let me put the 2 here, put the 2 here. Then my ψ is not quite in the standard form because I don’t have the square root of Ls. So let me put a square root of L here, a square root of L here and also put one outside. So my ψ looks like a square root of L. Let me see. Oh, I know what the problem is. This ψ is not normalized because the rule is if you don’t — first let’s say it’s not normalized, so don’t worry about any of these numbers. Can you see that it’s made up of two such functions with equal weight? That means if you measure the momentum you’ll only get the answer corresponding to what’s here and what’s here. If you compare it to e^{ipx/ℏ} you can see the p that goes here is really 4Πℏ/L or 4Πℏ/L. So a particle in this state can have only two values of momentum if you measure it, and the probability for the two you can see by symmetry must be 50/50 for each because they come with equal weight. But if you don’t want to do that you can, if you like, like this ψ, normalize it, and then we square it, integrate it and put a number in front so it comes out to have length 1. Then the coefficients here directly when squared will give you probabilities. Yes? Oh, a lot of questions. Yeah? Student: So A_{p} has to be a real number, right? Professor Ramamurti Shankar: No. A_{p} does not have to be a real number just like ψ(x) does not have to be a real number A_{p} — because I’m taking the absolute value of A^{2}. So even if it’s a complex number the absolute value of A will always come out to be real and positive. Is that your question? See A_{p} is not a probability. The absolute value squared of A_{p} is a probability which, of course, must be positive. But A does not have to be positive, does not have to be real. Yeah? Student: You said A_{p} like the integral if you’re multiplying ψ* times ψ would that be in real numbers? Professor Ramamurti Shankar: You should be very careful. Student: Really? Professor Ramamurti Shankar: A particular ψ times that same ψ* is a real positive number. This is not the same function. I think I remember now from previous years that students will get confused on this. This ψ(x) is the function that was given to me of which it was asked, “If I measure momentum what answer will I get?” This ψ_{p} is not the function that was given to me. This is another function that describes a particle whose momentum is p. So you’ve got to multiply that function with the given function to get these coefficients. So you can change your mind and say, “Oh, I’m not interested in that ψ. I want to know the answer to that ψ.” Well, you take that ψ and put it here. These guys will not change. Yeah? Student: I’m sort of confused. Right next to what you were explaining you have written ψ of p. Professor Ramamurti Shankar: Oh, I’m sorry. No, no, no. You have every right to be confused. I did not mean that. This is ψ of x. Student: And is that just ψ of x or that a ψ_{p} also? Professor Ramamurti Shankar: Here? Student: That’s just the ψ of x… Professor Ramamurti Shankar: Yes, this is ψ of x. Student: Okay. Professor Ramamurti Shankar: This ψ(x) — well, if I’m more disciplined, I can do that for you. Maybe I should do it one more time. It doesn’t hurt. So let’s so let’s say ψ(x) is cosine 4Πx/L. Now first job is to rescale the ψ so that if you square it and integrate it you get 1. I know the answer to that is square root of 2/L. You can check that. If you take this guy and you square it and integrate it it’ll work because the square of this will be 2/L, and the average value of cosign squared is half. And we integrate it from 0 to L it will cancel. You’ll get 1. If you do this particular function then all you have to do is write this in terms of — first write this one, 4Πix/L divided by 2 + e^{4Πix/L} divided by 2 because exponential is cosine θ is e + iθ + e to the iθ over 2. Now I’m going to write this as 1/√2 times e^{4Πix/L} divided by square root of L + e^{4ΠIx/L}. Divided by square root of L. I’m just manipulating the function that was given to me. Why do I like this form? Because this is of the form I wanted, A of p times ψ(p) where ψ(p) I this and that’s also ψ(p). It’s a normalized function associated with momentum p. By comparing this you can see that A is equal to 1 over square root of 2 for p = 4Πℏ/L, and A’s also equal to 1/√2 for p = 4Πℏ/L. How do I know what p is? States of definite momentum look like e^{ipx/ℏ}. So whatever multiplies ix/ℏ is the momentum that you can see it’s 4Πℏ. You see what I’m saying? If you want to put an ℏ down here and put it on the top and make the comparison. So these are the only two values for momentum. So this is a state which is simultaneously in the state of momentum 4Πℏ/L, and 4Πℏ/L. It has no definite momentum yet. If you measure it then you can get this answer with probability 1 over root 2 squared, which is 1 half, or you can get this answer 1 over root 2 squared which is 1 half and no probability for anything else. You cannot get any other value. See, normally when you take a ψ(x), a generic ψ(x) that I draw, it typically has some nonzero value in all of space, so you can find it anywhere. But this guy has only two values of p in the sum and only those will be possible. Yeah? Student: ________ you’re drawing ℏ over there ________. Professor Ramamurti Shankar: This guy here? Student: Yeah. Professor Ramamurti Shankar: Yes, I borrowed an ℏ, put it on top and bottom so you can compare the expression. I’m just saying compare, if you want without these ℏs, compare this expression to that and you can see p is equal to this. Now this problem was simple in the sense that you didn’t have to do that integral, but last time before end of class I took the following function, ψ(x) = e^{−αx}. On a ring the function looked like this, falling very rapidly and dying within a width roughly Δx = 1/α. Beyond that it is gone. Then if you square that and normalize it in the circle you find this is the correct normalization. In other words, if you took this ψ and you square and integrate it you’ll get 1. If you’re very, very careful it is not strictly 1 because this function the x goes from −L/2 to +L/2 whereas in the integral I took it to go from minus to plus infinity. That’s because this function is falling so rapidly. The function is falling so rapidly you don’t care if you cut if off at L/2 or go to infinity, so I did that to simplify the math, but the idea is the same. Then if you want to know what’s the probability to get some number p you’ve got to take A of p a square root of α times e^{ipx/ℏ} square root of L times e^{−αx} dx. You understand that? This is the ψ with the root α in it. That’s my ψ. That’s my ψ_{p}*(x). If you come to me with a new function tomorrow, I don’t know, maybe a function that looks like this, a constant in some region and 0 beyond. We’ll put that function here and do these integrals. You’ll get a different set of A of p’s. So for every function ψ(x) that’s a set of A_{p}’s that you find by doing the integral. Yeah? Student: So if you measure the momentum and you get one of those answers so it becomes that answer? Professor Ramamurti Shankar: Right. So I’m coming to that point. So the second thing is another postulate. In the end I’ll give you the list of postulates. And the postulate says that if you measure the momentum you’ll get only those values for which the corresponding A_{p} is not 0, and the ones for which it is not 0 the probability for getting the number is the absolute value squared of that A_{p}. And right after the measurement the wave function will change from this sum over all kinds of momenta to the one term corresponding to the one answer you got. You understand that? So in this simple example there are only two values for p you can get. No other values are possible. They’re all theoretically allowed. For example, 6Πℏ/L is a perfectly allowed momentum in that ring because it corresponds to a periodic function, but that’s not contained in this particular wave function. This wave function is built out of only two of them, namely 4Πℏ/L and 4Πℏ/L. Once you measure it suppose you got 4ℏ/L. The whole wave function, which I had here in the simple example, has got only two terms. This part will simply disappear, and the function after the measurement will be this. The rough logic is that if I measure momentum, and I got an answer, that’s got to mean something in any real sense that that is the momentum of the particle, therefore it has to be true if I immediately remeasure it. Immediately remeasure momentum, and I want to get exactly the same answer, that means the function after the first measurement should contain only one momentum in its expansion, so it’ll reduce to that one term. Yep? Student: Is there a difference between ψ(x) and negative ψ(x) because it seems like everything’s just… Professor Ramamurti Shankar: That is correct. I think I also explained the other day that in quantum theory ψ(x) and say 92i times ψ(x) are treated as physically equivalent because they contain the same relative probabilities. Of course if ψ(x) had been properly normalized to 1 this guy would not be normalized to 1. It’s as if you live in a world where all you care about is the direction of the vector, but not the length of the vector. We don’t live in such a world. If you say, “Where is Stop & Shop,” you say, “Well, go in that direction,” and you’re not told how far, well, you’re missing some information. Imagine a world where all you need to know is which way to go. In fact, maybe if you go to some town and ask people where is something they’ll say, “Go in that direction,” that’s useful information. It’s as if the direction is the only thing that matters, not the length of the vector. The analogous thing here is if you multiply the function by any number you don’t change the information. They’re all considered as the same state. And I told you from this huge family of functions all describing the same state I will pick that function, that member whose square integral happens to be 1. Yep? Student: So after the wave function breaks down how long do you have to wait for the wave function to rematerialize, or does it… Professor Ramamurti Shankar: Oh, it’s not break down. It’s another wave function because e^{ipx/ℏ}, which is ψ(p), this guy is as much a wave function as any other ψ(x) you write down. So these are not different creatures. It’s like saying the following. I think the analogy may be helpful to you. It may not be. ψ is like some vector in three dimensions, and you know there are these three unit vectors, i, j and k, and you can write the vector as some V_{x} times i + V_{y}times j + V_{z}times k where V_{y} = V⋅j etcetera. Think of ψ as a vector and each of these directions is corresponding to a possible momentum, and you want to expand that vector in terms of these vectors. This is ψ_{p} for p = p_{1}. This is ψ_{p} for p = p_{2} and so on. So it’s how much of the vector is in each direction that determines the likelihood you’ll get that answer, that answer or that answer. But right after the measurement, if you’ve caught it in this direction, the entire vector, only the component of the vector in the direction where you got your answer remains. The rest of it gets chopped out. It’s really like Polaroid glasses. If you’ve got Polaroid glasses light can come in polarized this way or polarized that way, but once it goes through the glass it’s polarized only in the one direction corresponding to the way the polaroid works. If it filters light with the e field going this way the light on the other side will have only up and up. So measurement is like a filtering process. It filters out of the sum over many terms the one term which corresponds to the one answer you got. So I don’t mind telling this to you any number of times, but that is the way that quantum mechanics works. Similarly, if you’re chosen to measure position starting with this ψ, and you found the guy here, once you found it — see, before you found it the odds may vary like this. Once you found it it’s there, and if you remeasure it infinitesimally later it’d better be there. So the only function which has the property is the big spike at wherever you found it. That’s called a collapsible of the wave function. The real difference is the following. I gave you an example of somebody tracking me to see where I am. I drew a probability graph, and the probability graph may look like this. And if you caught me here, if someone says, “Where will I find him next?” The answer is right there because the probability has collapsed from being anywhere to where we just saw him. The only difference between me and the quantum particle is if you got me here I really was here. You couldn’t have gotten me anywhere else at that right now. Can anybody find me anywhere else except right now here? You cannot. That’s because my location’s being constantly measured. Photons are bouncing off to see where I am, so my measurement is constantly measured. A quantum particle with a function like this, which is then found here, was not really here. It was in this state of limbo which has no analog in which it could have been anywhere. It’s the act of measuring it that nailed it to there, but then right after that if you measure it we expect it to be still there. Now the time over which you can do that is infinitesimal. If you wait long enough the wave function will change; I will tell you the laws for the dynamics that will tell you how it will change, but we all believe that if a measurement is immediately repeated for that same variable we should get the same answer. Need not be, it could have been even worse, but at least that much is true. So it’s not going to get any more familiar. It’s a strange thing, but hopefully you will know what the rules are. When I said no one understands quantum mechanics what I meant was, of course, by now you know the recipe. It doesn’t mean you like it, or it doesn’t mean it looks like anything in daily life. Things in daily life they have a location before you measure it, while you measure it and right after you measure it. They’re always in one place doing one thing. Only in the quantum world they can be in many places at the same time. But you should be very careful if this is a wave function for an electron. The charge of the electron is not spread out. I told you that. An electron is one little guy you’ll only catch in one place. It’s the wave function that’s spread out. Anyway, I think what’ll happen is you will have to do a lot of problems, and you will have to talk to a lot of people, and you have to read a lot of stuff. You can teach quantum mechanics for a whole semester, sometimes I taught it for a whole year. There’s more and more stuff. I want you at least to know how it works. In real life when you go forward I don’t think you need all of this. If you go into physics, or something, or chemistry, they’ll teach you quantum mechanics again, but I wanted people doing something else; they are the people I want to send a message to. Is here’s a part of the world. If you ever hear the word quantum, where does the word quantization come from? What’s the funny business in the microscopic world? I want you to have a felling for how that works. And I claim that the mathematics you need is not very much. You should know how to do integrals, and you should know what e^{iθ} is. So the postulates right now are the state has given me a function ψ of x. State of momentum, definite momentum looks like this, and if you want the odds for any particular momentum expand it in this fashion which is the same as A of p ψ_{p} of x, but A of p, again, is that integral. I don’t want to write it over. Then if you measure p you’ll get one of the answers with the probability given by the square of the corresponding number, and right after the measurement the function collapses. So it always collapses to whatever variable you measure. Yes, another question? Student: It looks like the opposite. If you collapse it and you measure x then what happens to p? Professor Ramamurti Shankar: Very good. His question was, “If you collapse it to x what happens to p?” I’m glad you asked. I mean, that is the real problem. You had the same question? Student: No. I was going to say well don’t you not know about p? Professor Ramamurti Shankar: That is correct. So I will repeat what she said, and I’ll repeat what you said. I want to have this discussion with you guys because it’s very important because everyone’s thinking the same thing. If I first take a generic ψ, so I want everyone to know what the answer’s going to be to any of these questions. So I take this ψ somebody prepared for me. Let’s not worry about how that person knew this is ψ, you’re given. The electron is in this stage ψ. If you measure x, and you got x equal to this point the function, of course, becomes a big spike. This spike in principle should be infinitesimally thin, but I don’t care. Let this be the width of a proton. If after that you say, “What momentum will I get,” well, you know what you have to do. You’ve got to write spike function equal to sum over these functions, right? You’ve got to take the integral of the spike with this exponential and do that integral, and you’ll get a bunch of numbers A of p for all values of p. And then if you measure momentum, and you got the one corresponding to p = 6Πℏ/L, the state which contains many, many things will reduce to the one term corresponding to p = 6Πℏ/L. If you plot that function it will be some oscillatory function. The real part and imaginary part will both oscillate with some wavelength given by p. Be very careful. This is not the absolute value of ψ. It’s the real or imaginary part. They’re both sines and cosines. Absolute value will be flat. So it’ll go from a particle of known location to a particle whose probability’s completely flat on the circle. You understand? The wave function can look like 6Πi/L square root of L right after the measurement. Let’s call is ψ_{6}. The absolute value of ψ is a constant, but the real and imaginary parts of ψ oscillate with some wavelength. So right after the measurement of momentum you don’t know where the guy is, and you say, “Let me find this fellow.” You catch it somewhere then it’s a spike at that point, but then you have no guarantee on the momentum. So you can never produce for me something of perfectly well defined position and momentum because once you squeeze it in x it gets broad it in p. Once you squeeze it in p it gets broad in x. This is really mathematical property of Fourier analysis, that functions which are very narrow in x when you do the Fourier expansion have many, many wavelengths in them. And likewise, a function with a very well defined wavelength, because it’s a complex exponential, has a magnitude which is flat. So now I’m going to ask the following question. So when I did ψ(x) the probability for x was very easy, squared the ψ. When I said, “Okay, I want to look at momentum,” the answer was long and complicated, namely, take these exponential functions, write the ψ in terms of those, find the coefficient, etcetera. Now I can say I want some other variable I’m interested in. I want to know what happens if I measure energy. Energy is a very, very important variable. It’s very, very important because it turns out that if a particle starts out in a state of definite energy, I will show that to you later, it remains in that state. That’s the only state that will remain the way it is. If you start in a state of definite momentum two seconds later it can have a different momentum or it can be a mixture of different momenta. But if it starts in the state of definite energy it will remain that way. That’s not obvious. I’m going to prove that to you later. That’s why it’s very important. So most atoms are in a state of definite energy and they can stay that way forever, but once in a while when they’re tickled by something they will either absorb light or they will emit light. So we draw a picture like this. We will see that the allowed energies of the systems are some special values. Not every value’s allowed. And this can be called n=1, n=2, n=3, etcetera. And an atom, for example, can sometimes jump from doing that to doing that, and in that process it will emit an energy which is E(n=3) − E(n=1). That difference of energy will come in the form of a photon and the energy of the photon is ℏω, or if you like, 2Πℏf where f is what you and I call frequency. And from the frequency you can find the wavelength. The wavelength is just the velocity of light divided by frequency. So an atom will have only certain allowed energies and when it jumps form one allowed energy to another allowed energy it will emit a photon whose frequency — in fact, you should probably call this frequency f_{31}meaning what I get when I jump from the level 3 to level 1. Similarly, if you shine light on this atom it won’t take any frequency. It’ll only take those frequencies that connect it from one allowed energy to another allowed energy. That’s the fingerprint of the atom. Both emission and absorption betray the atom. That’s how we know what atoms there are in this star, or that star, or what the composition is. No one’s gone to any of these stars, but we know because of the light they emit, and it’s all controlled by energy. So the question I’m going to ask is here’s the function ψ(x). Someone gave it to me in some context, and I say if I measure the energy of this particle what are the answers, and what are the odds? So how do you think that will play out? You have to make a guess. Suppose you’re inventing quantum mechanics and someone says, “What do you think is going to be the deal with energy?” You know what the scenario might look like? You can take a guess. I mean, as I told you many times I don’t expect you to invent quantum mechanics on the fly, but you should be able to guess. What form do you think the answer will take? You want to guess? Student: ________ they’ll be ________? Professor Ramamurti Shankar: Okay, but if I want to know what energies I can get and with what odds in analogy with momentum what do you think will happen? Yep? Student: Would you use the formula p^{2}/2m? Professor Ramamurti Shankar: He said use p^{2}/2m. That’s a good answer. His answer was we know that the energy is equal to 1 half m v squared which I can write as p^{2}/2m, right? So you’re saying if I measure the momentum and I got a certain answer, well, the energy’s that p^{2}/2m. That’s actually correct except the energy of a particle is not always just the kinetic energy. For a free particle this is the kinetic energy. For a particle moving in a potential you know that you have to add V(x). That is the total energy. Now that’s when we have a problem. In classical mechanics once I measure the x and p of the particle I don’t have to make another measurement of energy. Do you understand that? I just plug the values I got into this formula. For example, particle connected to a spring V is ½kx^{2} where k is the force constant of the spring, and the kinetic energy’s always p^{2}/2m. Say if I measured p and I got some number, and measured x I got some number, I can put that in the formula and get the energy. You don’t have to do another energy measurement. And you don’t have to do an angular momentum measurement either. In higher dimensions angular momentum is r x p, and if you already measured the position and you measured the momentum just take the cross product. So in classical mechanics you only need to measure x and p. In quantum mechanics he made a pretty good guess that if you measured p, the p^{2}/2m is the energy. That is true if the particle is not in a potential. But if the particle is in a potential can you tell me how to compute p^{2}/2m + V(x)? You realize you cannot really compute it because if you knew the p exactly you have no idea where it is, and if you knew the x exactly you don’t know what the momentum is. Maybe you know a little bit of both, but still, how are you going to find the energy? So the answer is you have to do a separate energy measurement. You cannot infer that from x and p, because first of all you cannot even get a pair of x and p at a given time. I hope I convinced you. You measure this guy you screw up that guy. Measure that one you mess up this one. You can never get a state of well defined x and p anyway. So the way to find energy is to do a whole other calculation. So I will tell you what the answer is, and hopefully you will realize it’s not completely different from the recipe we had before. I’m going to give you a rule for the functions which correspond to a state in which the particle has a definite energy E. Let’s not worry about how you get it, some function. You find all those functions, or you’re given all those functions, then can you imagine what will happen next? If I give you all those functions what do you think the rule is going to be? Yep? Student: You separate it into a sum of all the energies with coefficients. Professor Ramamurti Shankar: Very good. Let me repeat what she said. I hope at least some of you were thinking about the same answer. Her answer is take the function ψ, write it in terms of these functions ψ E of x with some coefficient A_{E} summing this over the allowed values of E, whatever they may be. And now that you said that what do you think A_{E} is going to be in a given case? Would you like to continue? Student: It’s going to be the integral or the wave function ________________. Professor Ramamurti Shankar: That’s right. That is correct. The recipe’s almost complete except you don’t know what these functions are, but if you knew these functions you have to write the given function, given wave function, as a sum of these functions with some suitable coefficients. Coefficients are found by the same rule, and then the probability that you’ll find an energy E is, again, A_{E}^{2}. And everything else will be also true. Once you measure energy you’ve got energy corresponding to E_{1}, or E_{2} or E_{3}. Let’s say you’ve got E_{3}, the third possible value. The whole wave function will collapse from being a sum over many things to just this one guy, E_{3}. The collapse is the same. The probability rule is the same. The only thing you don’t know is who are these functions ψ of E? You understand? So, again, the analogy’s the following. There is a vector that we call ψ. Sometimes you want to write it in terms of i, j, and k. They are like the A(p). Sometimes you may pick three other mutually perpendicular vectors, i’, j’ and k’. And if you know those coefficients you’ll get the probability for some other variable. So you’re expanding the same function over and over in many possible ways depending on what variable is of interest to you. If it’s momentum you expand it in terms of exponential ipx of ℏ. If it’s energy you expand it in terms of these functions. So the question is what is the recipe going to be for these functions ψ_{E} of x. By what means do I find them? Now you’re getting more and more and more recipes every day, but it’s going to stop pretty soon. This is about the last of the recipes. Even this recipe I’ll tell you how to get from a master recipe, so it’s not that many recipes, but I have to reveal that to you one at a time. Now you can say, “Okay, what function do you want me to use for every energy, what is this function? After all, when it was momentum you came right out and gave this answer. Why don’t you do the same thing here,” and that’s a problem here. The problem is the energy of a particle depends on what potential it is in because it’s got a kinetic and a potential part, so I cannot give you a universal answer for ψ_{E}(x). I will have to first ask you; “Tell me the potential the particle is in.” Once I know the potential I will give you the recipe. Imagine the potential has been given to me, for example, ½kx^{2}, or it could be a particle in what’s called a well. You make a hole in the ground or you build a little barrier. That’s a possible potential. You can have a potential that looks like harmonic isolator. That’s a possible potential. You can have an electron and a hydrogen atom 1/r. That’s a possible potential. There are manypotentials, and the answer’s going to vary on the problem. There’s no universal answer for the energy functions. You tell me what the electron’s doing, what field it is in, what field of force it is in. Then for each field of force, or for each potential, there’s a different answer. Chapter 2: The Schrödinger Equation [00:40:13]And here is the master formula. This is the great Schrödinger equation. So the answer looks like this. The function ψ_{E} obeys the following equation, ℏ^{2}/2m times the second derivative of ψ + V(x) times ψ(x) is equal to E times ψ(x). Do not worry. I will see you through this equation. Everything you need to know I will tell you, but you should not be afraid of what the equation says. It says those functions that are allowed corresponding to definite energy will have the property that if you took the second derivative, multiplied it by this number, add it to that V(x) times ψ_{E} you’ll get some function. That function should be some number times the very same function. If you can find me those functions then you will find out when I will show you mathematically that there are many solutions to the equation, but they don’t occur for every energy. Only some energies are allowed, and the energies are usually labeled by some integer n, and for every n, 1, 2, 3, 4, you’ll get a bunch of energies, ψ_{E1}, ψ_{E2}, ψ_{E3}, 3, and you write those functions down. Then you can do everything thing I said. But the only thing is here you have to do some hard work. Whereas for momentum I gave you that and for a state of definite positions x = x_{0}I told you spike at x_{0}. You didn’t have to do much work. For this you have to solve an equation before you an even start, but we’ll see how to do that. So the first problem I want to solve is the problem where there is no potential. That is called a free particle. A free particle is one for which V is 0. And let me imagine it’s living on this line, the circle of length 2ΠR = L. Oh, by the way, I should mention something else. In terms of all the postulates you notice I never mentioned the uncertainty principle today, ΔxΔp should be bigger than ℏ. I didn’t mention it as a postulate because once you tell me ψ is given by wave function and that states are definite momentum have definite wavelength it follows from mathematics that you cannot have a function of well defined periodicity and wavelength also localized in space. It’s a mathematical consequence. Similarly, who told me that I can expand every given ψ as a sum over these functions? There’s a very general mathematical theorem that tells you in what situations you can actually expand any given function in terms of a set of functions, namely are they like unit vector i, j and k? Suppose I had only i and j and I don’t have k. I cannot expand ever vector in 3D using i and j. So you’ve got to make sure you’ve got enough basis functions and the theory tells you that if you find all the solutions to that equation together they can expand any function. Similarly the rule for expansion is also not arbitrary. It all comes from that. It’s a very, very beautiful theory of the mathematics behind quantum mechanics. If you learn linear algebra one day, or if you’ve already learned it, it’s all linear algebra. So every great discovery in physics is accompanied by some mathematical stuff you need. Like all of Newtonian mechanics requires calculus. Without calculus you cannot do Newtonian mechanics. Maxwell’s theory for electromagnetism requires vector calculus. Einstein’s nonrelativistic theory doesn’t require anything; it’s algebra, but the general theory requires what’s called tensor calculus, and quantum mechanics requires linear algebra. And string theory we don’t know what it requires. People are still discovering new mathematics. But it’s very true that very often new mathematics is needed to express the new laws of physics. And if you don’t know the laws you may find out you’re not able to write it down. If you didn’t know what a gradient was, or if you did not know what a curl is, and so on, then you cannot write down the laws of electricity and magnetism. So we have to solve this equation, and I’m going to solve it for the easiest problem in the world, a particle moving on a ring of length L with no potential energy. So what does that equation look like? It says ℏ;/2m d^{2}ψ/dx^{2} + no potential = E times ψ_{E}(x). So let me rearrange the equation so it looks like this, d^{2}ψ/dx^{2} + k^{2}ψ = 0 where k^{2} is defined to be 2mE/ℏ^{2}= k^{2}. All I’ve done is just taken everything to one side and multiplied everything by 2m/ℏ^{2}and called that combination as k^{2}. So who is this number k? Let’s see. The energy is ℏk^{2}/2m, but we also know energy’s p^{2}/2m. So this number k will turn out to be just momentum divided by ℏ. Well, momentum has not entered the picture, but we will see. Let’s solve this equation now. I say the solution to this equation is ψ(x) = to any number A times e^{ikx} + any number B times e^{ikx}. Let’s see if that is true. Take two derivatives of ψ. What do you get? You understand every time I take a derivative you pull down an ik? If you pull it twice you’ll pull an ik squared which is k^{2}ψ, and the same thing will happen to this term that it’ll pull down a minus ik, but if you do it twice you’ll again get −k^{2}. So both of them will have the property that the second derivative of ψ will be equal to k^{2}ψ, and that is the equation you want to solve. And A and B is whatever you like. A and B are not fixed by the equation because for any choice of A and any choice of B this’ll work. So let me write it as follows. Ae to the i. Let me write k, k was a shorthand for square root of 2mE/ℏ^{2} x + B to the i square root of 2mE/ℏ^{2} x. I’m trying to show you that these are really functions of definite energy ψ_{E} and here is how the energy appears. So what does it look like to you? Have you seen these functions before? Yes? What does it look like? Student: One of the spring wave theories? Professor Ramamurti Shankar: Pardon me? Student: One of the spring wave theories? Professor Ramamurti Shankar: The equation is like the spring equation. That is absolutely correct, but what does this function look like? That function look like something to you? Yep? Student: Isn’t that a cosine function? Professor Ramamurti Shankar: It’s a cosine only if A is equal to B. Forget the sines and cosines. They are your old flames. What’s their new quantum flame? What’s the function that means a lot more in quantum mechanics than sines and cosines? No? I think I told you long back that e to the i times dog x over ℏ is a state where the momentum is equal to dog. In other words, you can put anything you want in the exponent. If a function looks like that, that fellow there is the momentum. So this is a state of momentum, of definite momentum. In fact, look at this, e to the i square root of 2mE/ℏ x + B times e to the minus that. Do you understand? This must be the momentum. It is the momentum because e^{ipx/ℏ}is a state of momentum p, but the momentum here can either have one value, square root of 2mE, or it can have another value minus square root of 2mE. So what do you think the particle is doing in these solutions? Yep? Student: It’s jumping back and forth _______________ one of those states. Professor Ramamurti Shankar: But in any one of them what is it doing here? What is the sign of the momentum here? Student: Oh, positive. Professor Ramamurti Shankar: Positive, and here it’s got negative momentum. And how much momentum does it have? The momentum it has, if you look at any of these things, is that p^{2}/2m = E is what is satisfied by the p that you have here. In other words, what I’m telling you is in quantum theory, in classical theory if I said, “I’ve got a particle of energy E, what is its momentum?” You would say, “Well, E is p^{2}/2m, therefore p is equal to plus or minus square root of 2mE,” because in one dimension when I give you the energy, the Kinetic energy if you like, the particle has to have a definite speed, but it can be to the left or it can be to the right. And the momentum is not arbitrary. If the energy is E the momentum has to satisfy the condition p^{2}/2m = E. That’s also exactly what’s happening in the quantum theory. The state of definite energy is a sum over two possible things, one where the momentum is the positive value for root of 2mE; other is a negative value, and these are the two allowed values even in classical mechanics for a particle of definite energy. But what’s novel in quantum mechanics, whereas in classical mechanics if it’s got energy E it can only be going clockwise or anticlockwise, but this fellow can be doing both because it’s not in a state of clockwise or anticlockwise. In fact, the probability for clockwise is proportional to A^{2}. Probability for anticlockwise is proportional to B^{2}. I’ve not normalized it, but the relative odds are simply proportional to A^{2}and B^{2}. So that’s what is bizarre about quantum mechanics that the particle has indefinite sign of momentum. Yes? Student: Do we have problem with m because m is too small to ________ properly? Professor Ramamurti Shankar: Which one? Student: With the mass is there going to be some kind of issue if that’s small? Professor Ramamurti Shankar: What about the m? I’m sorry. Student: m stands for the mass, right? Professor Ramamurti Shankar: Yeah, m is the mass of the particle, that’s right. Student: But is it too small in some way for our equations to work because we had that issue in previous equations? Professor Ramamurti Shankar: When did we have the issue? m is whatever the mass of the particle is. It can be small. It can be large. I didn’t understand. Student: [inaudible] Professor Ramamurti Shankar: No, no, go ahead. I want to know. Student: [inaudible] Professor Ramamurti Shankar: There are no restrictions on the correctness of this. If the particle weighs a kilogram then you will find that — well, we’re coming to this. So you find that these are the allowed values of p. There are two values, but p itself is not arbitrary, p itself is not continuous. Maybe that’s what you meant. There’s a restriction on the allowed values of p, and therefore a restriction on the allowed values of E. All I’m telling you now is that if you want to solve that equation it is obviously made up of sines and cosines as you recognize from the oscillator, or in the quantum world it’s more natural to write them in terms of exponentials, e^{ikx} and e^{ikx} where k is not independent of e, k satisfies this condition, and if you call ℏk as p, p satisfies this condition. This is just a classical relation between energy and momentum. The other difference is that not every value of momentum is allowed. Not every value of momentum is allowed for the same reason as when I did particles of definite momentum. In other words, if the particle is living on a circle and the state of energy E it’s given by Ae^{ipx/ℏ} + Be^{−ipx/ℏ} where p is related to E by E = p^{2}/2m. We have the requirement that when you go around a circle you’ve got to come back to where you start. Chapter 3: Quantization of Energy [00:54:21]And that condition, if you remember, says that p times L over ℏ has to be a multiple of some integer, or that the allowed values of p are labeled by some index n, which is 2Πℏ/L times n. I don’t want to use m because m stands for particle mass, n is the integer now. In other words, when we studied the state of definite momentum, namely the first function, we realized even then that p is quantized. Because of the single valued condition p is quantized. And if energy functions are made up of such functions they also have to be singlevalued. That means the p here or the p here both have to satisfy the condition given by this. Therefore, the allowed energies are also labeled by an integer n and they are really p n squared over 2m where p n is 2Πℏ n over L . You see that? This is the quantization of energy. So a particle in a ring has only these allowed values of energy. So in a way this problem is somewhat easy because it’s a free particle. Once you’ve understood the particles in terms of allowed momenta it turns out the allowed momentum states are also allowed energy states. The allowed momentum states, you remember, they either look this or they look like this. And I can pick A and B to be arbitrary, so one choice is to pick A equal to 1/√L and just take e^{ipx/ℏ}. Other is to pick B equal to 1/√L and pick e^{−ipx/ℏ}, but you can also mix them up. You don’t have to mix them. If you don’t mix them up you have a particle here which has a welldefined energy and a welldefined momentum. This guy also has a welldefined energy and a welldefined momentum. This guy only has a welldefined energy, but not a welldefined momentum because there’s a twofold ambiguity in momentum. You understand? Even in classical mechanics it’s true. If I give you the momentum you can find the energy. If I give you the energy you cannot find the momentum because there are two square roots you can take because p^{2}/2m is E, p is plus or minus square root of 2mE. It’s the same uncertainty even in classical mechanics. So what happens in quantum theory is if you pick any particle of definite momentum on the ring it’ll already have definite energy which is simply that momentum squared over 2m. You don’t need to find a new function. What is novel is that since the energy depends only on p^{2} you can take a function with one value of p, and you can take a function with the opposite value of p and add them, there will still be a state of definite energy because whether it’s doing this or whether it’s doing that the energy will always be p^{2}/2m and the minus signs drop out of that. So what is novel here is what’s called degeneracy. Degeneracy is the name, but there’s more than one solution for a given value of the variable you’re interested in. You saw the energy looks like this. Therefore, you’ll find there’s a state E_{0} which is 0 squared over 2m. Then there are two states E_{1}. One has got momentum p going clockwise and one has momentum going anticlockwise. So they look like (1/√L) e^{2Πix/L}. That’s this guy. Then you can have (1/√L) e^{2Πix/L} which is the other guy. So at everyenergy, allowed energy except 0, there’ll be two solutions. There are two quantum states with the same energy. When you studied hydrogen atom in high school maybe you remember there are these shells with 2, and 4, and 8, and 10 and so on. They are called degeneracies where the energy is not enough to tell you what it’s doing. There at everyenergy it can have a different angular momentum. Here at everyenergy it can have two different momenta, clockwise and anticlockwise. Yep? Student: Going back can you explain how you got the ________ of p L over ℏ ________? Professor Ramamurti Shankar: Here? Student: To the left. Professor Ramamurti Shankar: Here? You mean this one? Student: I don’t where that came from. Professor Ramamurti Shankar: It came just like in the momentum problem. If you’ve got a function like this you have a right to demand that if you go a distance L around the circle you come back to where you start. So if you take any x here and add to it an L it should not make a difference. And what you’re adding is pL/ℏ and that better be a multiple of 2Π. It’s the same single valuedness condition. So the momentum problem pretty much does this problem for you, all the singlevalued stuff we dealt with before. What is novel here is that demanding energy have one value fixes the momentum to be one of two values, and that double valuedness is the same as in classical theory that a particle of energy E can have two possible momenta plus or minus square root of 2mE, and the quantum theory, then, is the state of E is a sum of one value of p and the other value of p with any coefficient you like. So if this atom makes a jump from, or this system makes a jump from somewhere there to somewhere there you can find the frequency of the photons it will emit because here are my allowed energies. E_{n} is this. Suppose it jumps from n=4 to n=3. The energy that’s liberated that goes to the photon is ℏω will be 4Π^{2} ℏ^{2}over 2mL squared times 4 squared  3 squared. I’m just using this formula with n=4, and n=3, and take the difference. It’s 16  9 which 7. You plug all that in you can solve for the ω, or if you like frequency you can write it as 2Πℏf and you can find the frequencies. This is actually true. If you’ve got a charged particle moving in a ring and you want to excite it from one state to a higher state you will have to give it only one of these frequencies so that the frequency you give it, ℏω, must match the energy difference of the particle. You understand? So I’ve drawn these levels here. If you want the electrons, say, in a metallic ring going in the lowest possible energy state, if you want to jump, if you want to crank it up to the next level you’ve got to have photons at that energy, or that energy. Well, that happens to be the same as this, but these are the only frequencies it’ll absorb from you. And when it cools down it’ll emit back those frequencies, and that’s something you can test. By the way, do you know why there’s only one state at E_{0}and not two? Yep? Student: Because +0 equals 0. Professor Ramamurti Shankar: That’s right. The solution plus or minus 2m square root of 2mE has only oneanswer when E is 0 because 0 momentum and 0 momentum are the same. Otherwise, any finite positive momentum has a partner which is minus that momentum. There’s a very interesting piece of work being done experimentally at Yale which is the claim that if you took a metallic ring in a magnetic field it will have a current going one way or the other way, unbalanced current, and it’s not driven by a battery. It’s not driven by anything. Normally if you took an ordinary ring the lowest energy state will be a field of zero current while you go one way or the other, but if you put it in the magnetic field one can show it likes to go one way or the other way. And now measurements are being done at Yale where you can actually measure the tiny current due to one electron, or one net electron going one way or the other. So this L is either a mathematical convenience if you’re talking about free space and you want to be able to normalize your wave functions, or it really is the circumference of a real system. Now that we can probe nano systems very well we can vary the L and we can find out all the energy levels. Chapter 4: Particle in a Box [01:03:45]All right, so now I’m going to do the one problem which is really a very standard pedagogical exercise, and that’s called a particle in a box. I remember this example the first time I remember seeing quantization, which is more interesting than on a ring. A box is the following. If you dig a hole in the ground and you are standing somewhere here you realize you’re kind of trapped unless you can scale this wall. Now, you can call that as a ground level and think of it as a hole in the ground, or you can think of this as the ground level and that’s the height of your barrier. So imagine a particle living in a barrier that looks like this. This is the potential energy. This is like the height above the ground, if you like, and it has a height V_{0} t here, and is 0 here, and it goes from some −L/2 to +L/2. If the barrier V_{0} goes to infinity it’s called a box. So a barrier that goes to infinity, I’ll just show you the part that you can see here. Here V is infinity, and here V is infinity and inside V is 0. That’s a particle in a box. So if the particle goes hits against the wall it cannot, no matter how fast it’s moving, go over the top because it’s infinitely high. And that’s not realistic. Every barrier is finite, but just to teach you the principles we always pick the simple example. So now I want to solve this problem. What are the allowed wave functions, ψ, for a particle in this potential? So let’s go back to the same equation which says ℏ^{2}/2m d^{2}ψ/dx^{2} = E  V ψ. What I’ve done is I’ve taken V to the other side. Now the potential is a constant in the three regions. This is region I where the potential is 0. This is region II where the potential is infinity. This is region III where the potential is infinity. I’m going to first look at the region here, region III. What’s the solution going to look like in region III is what I’m asking. So in this region the energy’s something I don’t know what the allowed values are. Whatever the value is V is an extremely large number. Can you see that? In this region don’t let the barrier be infinity. Imagine it’s one zillion, a very high barrier. Then what’s the solution? The solution will look like d^{2}ψ/dx^{2} is = 2m/ℏ^{2} times V  E ψ. Then the solution to that is very easy, ψ = Ae^{κx} + Be^{κx} where κ is equal to all of this, square root of 2m/ℏ^{2} times V  E. I’m saying this is what I’m calling κ^{2}. Look, I’m asking you give me a function whose second derivative is some number times the function. Well, that’s obviously exponential and it’s the real exponential because it’s a positive number times the function. Now A and B are free parameters, whatever they are it’ll solve the equation, but we don’t want a function that’s growing exponentially when you go to infinity because that means that particle would rather be at infinity than in your box or near your box. So for that mathematical, for that physical reason we junk this function. You pick it on physical grounds as not having a part growing exponentially when you go to infinity, but you do admit a part that’s falling exponentially when you go to infinity. That’s okay. But how fast is it falling? It’s going like B e to the minus some blah, blah, blah times square root of V  E. That’s all I want you to look at. Forget all the ℏ’s and m’s. Make V larger, and larger, and larger, and just tell me what you think it will do. If V is larger and larger it’s like this e^{αx}I wrote for you where α’s very large. So this function will fall faster, and faster, and faster, and in the limit in which V goes to infinity it will vanish. That basically means the particle cannot be found outside the box. So your function ψ is 0 here and 0 here. Because if you made the barrier height finite you will find it’s falling exponentially on either side, but the exponential becomes narrower and narrower as the barrier becomes higher and higher, and in the limit in which the wall is infinitely tall there is nothing outside. The wave function is non0 only inside. So what’s the solution inside? Let me call this 0. Let me call this L. Inside the box there is no potential, so this equation is d^{2}ψ/dx^{2} + (2mE/ℏ^{2})ψ = 0. It’s like a free particle in the box, but it cannot leave the box. This is what I called k, remember? So now I’m going to purposely write the solution in terms of trigonometric functions. You’ll see in a minute why. So I’m going to write this Ae^{ikx} + Be^{ikx}. You can see if I take any of these solutions it’s going to satisfy the equation, but k better be related to E in this form, 2mE/ℏ^{2} is equal to k^{2}. Sorry that’s k^{2}. Because two derivatives of this will give me −k^{2}, and if you put that particular value for −k^{2} here these two will cancel. But I’m somehow going for — I’m sorry, I’m going to not write it this way, but write it as C cosine kx + D sine kx. Do you realize I can always go back and forth between exponentials and trigonometric functions because one is a linear combination of the other? If you want, write this as cos + i sine and cos  i sine and rearrange the coefficients. It’ll look like something, something cosine. I want to call that a C, and something, something sine which I want to call this D, D and C may be complex. I’m not saying anything, but you can write a solution either in terms of the sines and cosines or E to the plus or minus something. So here’s the function ψ. It looks like I have an answer for every energy I want, because pick any energy you like, find the corresponding k, you put it here and you’re done. But that’s not allowed because we have an extra condition which is the ψ was identically 0 here, ψ was identically 0 here. We’re going to demand that at the two ends, it can do whatever it wants in the middle, it must vanish at the two ends for the continuity of ψ because if ψ had two values you’re getting two different probabilities for the same point, so that’s not allowed. So ψ must have — must match at the two ends. But look at this function. It’s got to vanish at the left end at x = 0 and it’s got to vanish at the right end. At x = 0 you can see ψ of 0 is simply C because sine vanishes. Cosine is 1 and C has to be 0 then because this guy has no business being nonzero on the left end. That’s good. This guy vanishes at the left end so I allow it, but I have another condition. It should also vanish at the right end. If it should vanish at the right end I demand that sine kL should be 0 because sine kL vanishing is fine. Then it’ll vanish at both ends. But sine kL = 0 means kL is a multiple of Π. That means k is nΠ/L. This means the allowed values of k in the problem are very special so here is n = 1, ψ looks like sin(Πx/L), and equal to 2 looks like sin(2Πx/L), and so on with some numbers in front which I have not chosen yet. If you plot them they look like this. That’s one guy. At a higher energy I’ve got that guy. Then I’ve got that and so on because these are exactly like waves on a string, a violin string clamped at two ends. In fact, this wave equation’s identical to the wave equation on a string. The only requirement is that the string is clamped at two ends. Here the ψ is clamped at the two ends because it’s got to vanish on either side outside the box. So the allowed wavelengths are the same except here the wavelength is connected to the momentum that’s connected to energy and E, you remember, is ℏ^{2}k^{2}/2m then becomes ℏ^{2}/2m times k will be n^{2}Π^{2}/L^{2}. So this particle in a box can have only these particular energies. So let me write it for you nicely. The allowed energies are ℏ^{2}Π^{2}/2mL squared times an integer n squared, and the corresponding wave functions look like this. They’re waves in which you’ve got half an oscillation, or 2 half oscillations and 3 half oscillations, but you’ve got to start and finish at 0. So it’s the quantization. This is why when Schrödinger came up with this equation everybody embraced it right away because you suddenly understood why energy’s quantized. You’re trying to fit some number of waves into an interval and only some multiple of half wavelengths are allowed, but wavelength translates into momentum. That translates into energy. Suddenly you understand the quantization of energy. So it’s got 1 state n = 1, and only one state n = 2, one state n = 3. I’m going to come back to this next time, but you should think about this. [end of transcript] Back to Top 
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