Advanced Quantum Mechanics | Lecture 7
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Susskind, Leonard
(November 4, 2013) Leonard Susskind extends the presentation of quantum field theory to multi-particle systems, and derives the particle creation and annihilation operators.
 Stanford University. Well, I don't know, should we go over again the basic construction of a simple quantum field? Why not? It's not exactly what I have written here, but we can do it again. Not again, but maybe slightly different. Yeah, we're ready. Okay, so as I said, the trick of introducing fields from particles, see you can go two ways. You can start with particles and reconstruct why they're described by fields, or you can start with fields and then work down to why fields are described by particles, or why fields, waves, wave fields have quanta. You can go either way, and we'll go both ways. So we started by starting with particles, and we said, look, it's fine to be talking about one particle or two particles or three particles, a fixed number of particles. But gee, why not try to study all possible numbers of particles? And by particles now, I'm speaking about a single species of particle. We can come later tonight, what happens if you have more than one species, but for the moment, one kind of particle and only one kind of particle. Let's make a theory in which we can simultaneously, in one fell swoop, so to speak, talk about any number of particles and all the things that multiple particles can do, including situations where the number of particles change. Now, why would be ever be led to that? Well, in real life, numbers of particles change. The number of photons in this room is constantly changing. They're being emitted by the electric light bulbs. In other contexts, particles decay, turning into other kinds of particles. So the neutron can decay into a proton, an electron, and a neutrino. And in the process, the number of neutrons changes. The number of protons change. The number of electrons, the number of neutrinos. And in fact, the total number of particles change. So however we have to think about particles, we have to think not only about the quantum mechanics of a fixed number of them, we have to think of that number as changeable. So we need a flexible enough tool to be studying what happens in a single quantum mechanics which embraces essentially any number of particles and even changing number of particles. That's what the point is. And that's what the point of inventing fields is. Sorry, any questions about that? And I'll go over it again since the mathematics is not hard. But it's abstract. It's abstract. And face it, we're not as young as we used to be for one thing, none of us. And the mind is not necessarily attuned instantly to orthogonal functions and all that kind of thing. And so it's good to go slow. I'd rather you learn something rather than nothing. OK, so it's very easy to learn nothing. It's a little harder to learn something. OK, so we start with single particle quantum mechanics. Single particle, the description of one single particle, quantum mechanically, that means that we have state vectors but state vectors for one particle. We have wave functions for those particles. If we have a state of a single particle, let's call it psi, little psi, little psi will also stand for its wave function. If we have a state vector for a single particle, then the wave function is just the inner product of the state vector with a particle at position x. So a state labeled x means a particle at point x. The state psi means some general state but of a single particle. And that's called psi of x. What do you do with it? Here's a formula. Psi star of x times psi of x. And these are wave functions now. These are not quantum fields. When I go to quantum fields, I will use large capitalized letters even if they happen to be the same as this. What is that object? The object I respond to, that's the probability to find the particle at x. It's the probability density. The probability density to find the particle at x. Probability density. Prob-dense. You can fill in the rest of the letters. A fact about the probability density is that when you integrate it over position, what do you get? You get one. It happens in this case that one happens to be the number of particles, that isn't it? So you get one, but you could also say you get the number of particles. Well, that's not very interesting because we only had one particle to begin with and we were doing one particle quantum mechanics. Nevertheless, it's a true fact. Okay. Now, we next introduced a collection of one particle states that formed a basis, an orthonormal orthogonal basis. Now, we can do that in a number of ways. First of all, the x states themselves are orthogonal to each other and they form a basis. We'll come back to that in a minute. But there are many, many ways of choosing bases. Essentially, you take the eigenvectors of any Hermitian operator, we'll form a basis. And in particular, if we're interested, we may be especially interested in energy eigenstates. For example, of a particle in a box that represents a box, the horizontal axis is the position, the vertical axis is the potential energy. When you get to the edge of the box, the potential energy becomes huge and the particle just reflects off, it can't get through. In that case, the ground state wave function looks like that. The first, and we'll call that psi one of x. The next energy eigenstate looks like that. We'll call it psi two of x. And these are basically signs. Sign of two nodes and they're characterized by the number of nodes. Rather peculiarly, I started with psi one which has no nodes, psi two has one node, but so be it. Psi i of x. And since these are supposed to be energy eigenstates, they're orthogonal to each other. They form an orthonormal basis. So let's write the equation for that. The equation for that is that integral of psi star i of x with psi j of x, that's the inner product, the x. What is that? Anybody remember? Delta ij. Delta ij. If i is equal to j, they're the same state. The inner product is one, normalized state. If they're different, they're orthogonal, corresponding, let's say, to different energies. Sorry, delta ij, not zero, delta ij. That's another fact. Now I'm going to show you another little manipulation. This one, yeah? Why does the probability density have that particular form? That goes back to the postulates of quantum mechanics. The postulates of quantum mechanics said that probabilities are psi of x. This is the inner product of psi with a state localized at x. It's called the probability amplitude. The postulates of quantum mechanics, which I invite you to go back to. They're in the book. Say that the square of that is the probability. So OK, here's a little trick, a little mathematical trick that occurs over and over again. Take your orthonormal, your favorite orthonormal basis of functions. They could be sines. They could be cosines. They could be harmonic oscillator wave functions, whatever. Some collection of size which are orthonormal. And if they're orthonormal and they form a basis, then if you take this object, what is this object? This object is a operator. It operates on a state. Let's see what it does. Let's just explore what it does when it acts on a state. What does it do? If it acts on a state psi, then the beauty of the Dirac notation is you can just take the formula that you've written down here. When you act with this thing on a state, psi with i is a number. It's the inner product of psi with i times i. That's what it does. Now, here's a fact about any orthonormal basis. If you sum over all elements of the orthonormal basis, in other words, in this case, all the energy eigenvectors up to infinity, then what do you get? The unit operator, i, the identity operator, the operator that when it acts on any vector, it acts the same vector. That's a marvelous theorem. It's in the notes. You should go back to it. Don't take my word for it. You can even prove it. But it's a useful thing to know. Let's use that now in the following way. Let's take a project. Let's act with this thing to the right and to the left. What we're going to do is we're going to take, what are we going to act on? We're going to act on a state to the right, which is localized at position x. And on the left, we're going to act, well, actually what we're doing is I'm taking the inner product of a state with a particle at x with a state at y. Anybody know what the mathematical symbol for that is? It's a Dirac delta function. But let's come to that for a moment. Let's not write that down yet. Let's write that that is equal to the sum on i, y, i, i, x. All I've done is plug in the unit operator here. That does nothing. But now I have a formula here. The inner product of i with y is psi i of y. Let's see. Yeah, the inner product of any state with an eigenvector of position is the wave function of that state. Here, the state is i and it has wave function psi i of y. Here's over here, psi i of x. And what does that have to equal? It has to equal delta of x minus y. Now, sum over i. Yeah, there is. The second one. Yes, the second one. I assume that's what you were asking me. Yeah, the star. OK, now we have a marvelous theorem. If we take any orthonormal basis, they could be the eigenvectors of any Hermitian operator, but in particular the energy. And you multiply for each one of those states, psi star times psi, and then sum over them, you'll always get a delta function. That's a good thing to know, as we'll see. It's useful. It means there's many, many ways of writing the delta function. Just take any orthonormal basis and do this. These could be sines and cosines. Sines and cosines are another way of writing a delta function. But this is the basic mathematics, normal mathematics, that's it, I think. OK, now let's come back. Let's go to the problem of studying many particles all at the same time. OK, now how are we going to characterize a state? One set of states, which is particularly useful, is a set of states in which you say how many particles there are with each one of these energies, for example. How many particles are there with state with psi one, psi two, and psi three, and so forth. And so we can take the multi-particle system and characterize it by a basis of states, which are just n1, n2, dot, dot, dot, and i, dot, dot, dot. The ends are the occupation numbers. They're the occupation numbers which tell you how many particles there are with each energy level. And those are states. They're not single particle states. They don't even belong in the same vector space with the single particle wave functions. The single particle wave functions are themselves a whole vector space. But we're now talking about a bigger thing, a bigger system. And so we should not confuse the single particle states with the more general structures that we're talking about. We use the same notation for them because nobody ever invented a better notation in Dirac's, the Crosse and Kett's. OK, next, we would like to be able to change the number of particles. Maybe we do, or maybe we don't want to change the total number of particles. That's not clear yet. But we'd like to be able to change them. And so we'd like to be able to raise and lower the ends. So we introduce a oscillator for each i. For each i, we introduce an oscillator. And at the moment, this has nothing to do with oscillation. This is just a trick. This is just a bookkeeping trick. Although it will be more than that later. For each one of these i's, we introduce a creation operator or a raising operator. And we introduce a creation operator since it's now going to create a particle. A creation operator and an annihilation operator. And the creation and annihilation operators, it's convenient to make them just work exactly the same way as harmonic oscillators. It's just a convenient way of dealing with them. A plus i and A minus i. Now, another convention for the same things, I'll write this in red, is to call this A dagger i. That looks like a plus sign. Not quite, but nevertheless, same thing. And just calling this one ai. You'll often see it written that way. It's the same thing. A dagger creates a particle. A does not. But the correspondence is exactly this. Okay. This space of states of the many particle system is called Fox space. FOCK. Fox was a Russian physicist who somehow got his name attached to this. I don't really know how. Okay. Now we're going to find a very convenient, and you'll see how it works, to introduce the notion of a field operator. The field operator has the same symbol as the wave function psi of x. But it's not a wave function. It's not a wave function. It's not a state vector for one particle. It's an operator. And it's an operator that acts on the space of states of the many particle system. It itself contains creation and annihilation operators. All right. So let's write the notion of a field operator. And the best way to motivate it is to use it and see what it does. Now I use a uppercase psi. I hope. I'm a little bit afraid that you won't be able to tell my uppercases from my lower cases. All right. That's an uppercase psi, right? Its definition is the following. It's the sum of i of, let me get it straight, the annihilation operator for a particle in state i times the wave function of a particle if it were in state i. And that's it. Summed over all i. Now somebody asked me last week, does this have anything to do with Fourier transforms? And you bet it does. If the wave functions here were the wave functions of a free particle, the eigenstates of energy of a free particle, and not in a box just in an infinitely big box, what are those eigenstates, those eigenwave functions? They're plane waves. e to the ikx and e to the minus i, or e to the ipx and e to the minus ipx. So in that case, if we were using free particles in an infinitely big box, this would be an expansion in sines and cosines, and those a's would be the coefficients of sines and cosines. They would be the Fourier coefficients of a Fourier expansion, except that they're not numbers. They're operators. They're operators, the things that you measure, or things that play the role of operators in quantum mechanics. Okay, so that's one thing. The complex, or the Hermitian conjugate of it, so a dagger of x, this is of x. Yeah, what does it mean to say we have an operator which depends on x? Well, it just means we have a different operator at every x, that's all. We have a lot of creation and annihilation operators. Out of them, we can build a lot of independent operators. In fact, we can build an operator at each point x by combining the a's with size of x, at point x. So if we do this, we make a different operator at every position, and so we have an operator function of position, okay, or operator-valued function of position. It's complex, or it's Hermitian conjugate, is the same thing, except creation operators here. This is a, I get so used to, yeah, okay, a minus a plus times psi dagger, so I'm psi star i of x. Now, it doesn't matter, it's rather weird. Some people use a notation where this is called psi star, they use a one where this is called psi and they get all mixed up. These are definitions, and so you can do what you want, but here's the definition we use. As I said, there are two operator functions. Now, neither one of these is a Hermitian operator. They have the creation operators and annihilation operators are not Hermitian. Their conjugates are each other. If you add them, you make something that's Hermitian. If you take any operator and add it to a Hermitian conjugate, it's Hermitian. So if we add these two, they become Hermitian, and Hermitian means observable, means something you can measure in the laboratory. So, the sum of them is the thing you can measure. And the difference is the difference Hermitian. What happens if you take the difference of a number and it's complex conjugate? You get something real? You get something pure imaginary, but if you multiply it by i, then you get something Hermitian. So, each, the sum and difference of these defines some Hermitian observables. But right now, we're not observing them. We're not going to use them as bookkeeping devices. Okay, now, let me show you another interpretation of psi of x, what it does. In particular, let's ask what it does when it acts on a vacuum. The vacuum means a state with no particles in it at all. And so the vacuum, let's call it vacuum, which we usually denote just by O, is really the Fox-based state in which all the occupation numbers are zero. All the same thing here. Vacuum, Fox-based state with which all the occupation numbers are zero, sometimes just called the vacuum or just called the state zero. Okay, let's act, yeah, let's not act yet. Let's write down the following formula. This is a one-particle state now, one particle and only one particle. And let's write it in the following form. Let's use this, anybody know what the name for this kind of formula is? Projection. No, it's not, well, it is a project, it's the trivial projection operator. It's one. It's called a resolution of the identity. It's a way of writing the unit operator as a sum over a lot of orthogonal vectors. All right, so let's plug in the resolution of the identity, sum on I, I, I, psi. I, what is I? I is a state in which there's one particle and only one particle in the ith energy level here. Okay, what is this thing here? This is, sorry, X. X, sum over I, X, I. What is this object? Let's go up to the top formula and you see that when you project on to X, you get a wave function. This here is psi star I of X. It's the complex conjugate of the wave function of the ith energy level. The ith energy level has a wave function, this is its complex conjugate. The wave function itself would have the opposite order. What about this state here? That's a state with one particle in the ith energy level. So I'm going to write it this way. I'm going to take a creation operator for making a particle in the ith energy and multiplying it by the vacuum. I've created one particle in the state I by applying the creation operator on the vacuum and that's exactly the state here. On the other hand, this is just a complex conjugate of the wave function and my rule is sum. Did I erase the defi- no, there's the definition of the field operator. Okay, go to the second row. Psi star, the field operator, is sum over I, a plus of I, times the complex conjugate of the wave function. That's exactly what you see here, right? So this is exactly equal to the field operator at position X on the vacuum. That gives us a new interpretation of psi. It's the operator which creates out of the vacuum a particle at position I. I think that's a fairly intuitive thing that it's a quantum operator at position X. Sorry. Good. So the AIs create particles with certain energy levels. The psi I creates particles at certain locations. It's also a kind of creation operator. The field operator itself is a kind of creation operator, but instead of creating a particle with an energy, with a definite energy, it creates a particle with a definite position. You want to dagger there and the particle to start at the same time. Here? Yeah, it should be dagger. Good. You never use star for operators. Use dagger. Good. This assumes that the creation operator and the wave function can use. No. Yeah, wave functions are numbers. They're numbers. They're not trivial operators. They're numbers. Sometimes ordinary numbers are called C numbers. This is another invention of Dirac. C stands for classical. These are classical numbers. Operators are called Q numbers for quantum. Q numbers don't always commute. C numbers always commute with themselves, but they also commute with everything else because they're just numbers. OK. So now we have a lovely new interpretation of the field operator. The creation operator for a particle at a point of space. Let's look at some interesting quantities. I think we'll, well, we can put this up. Just to get a little more experience with this, let's play with it a little bit. Here's an interesting quantity, integral dx. Operator psi dagger of x times psi of x. Now, is this the same thing as up on the top of the blackboard up there? The x psi star of x psi of x. No, it's not the same thing. It looks very similar, but it's not the same thing. In the top formulas up on top, the size is just wave functions, one particle wave functions. They're not operators. The other side is just one by total probability. This object here is an operator. Psi is an operator, psi dagger is an operator. When it acts on a state, it does something. This one creates a particle, this one annihilates a particle. So, but nevertheless, it has some formal structural similarity with the thing up on top. One wonders, what is it? Does it have some meaning? Is it something that has a physical significance? Is it something that you can measure? Incidentally, it's Hermitian, because if you take any operator and multiply it by a Hermitian conjugate, the result is Hermitian. That's like multiplying a complex number by its complex conjugate. The result is always real. Okay, so let's see if we can figure out what this thing is. First step. Let's replace the psi's by their definitions. So, this is integral dx. And now, psi star, that's a sum over i. It's going to get messy, but then it'll get easy. Ai dagger, or ai plus, times psi star i of x. Without the what? Without the which? No, that's right. Now, I'm going to write the psi. That has another separate sum, not the sum over the same i. It's a sum over a different i, but I can just call it j. It's a double sum. There's a double sum here, and the double sum has a j sin. Psi j of x. It looks a bit complicated. An integral over x of a double sum over i and j, plus the fact that the integral is over all space, and i and j go from one to infinity. So, if you are actually to calculate it, to put it on your computer, your computer would run forever and ever and ever and ever and would never halt. However, we can do better than that. Let's do the integral over x first. Now, I think we have a formula here. Psi i of x. There it is. There it is right here. Psi i of x with psi j. That's just the orthogonality relation. So, when I integrate over x, psi dagger times psi, I just get delta ij. That's nice. Let's write it in a simpler way. This is equal to, we've done the integral, sum over i and j, a plus i, a j, times the chronicle delta. In other words, we get a term for each i and j, except only when i and j are equal. And so this is equal to the sum over i of a dagger i, a i. What is a dagger i times a i for a harmonic oscillator? It's just a number operator. The thing whose eigenvalues are the n's, the little n's. This is the big n, so this can be written as the sum over i of the number operator for each one of these oscillators. In other words, it's just the total number of particles. The number of particles in the first mode plus some first wave functions plus the number of particles in the second wave function plus the number of particles in the third wave function. So what is this thing? It's an operator that represents the total number of particles. For each one of these Fox states, it's just a sum of the little n's. Now, there's an infinite number of little n's, right? There's n1, there's n2, there's n3. Is this going to be a finite number? Well, remember that most of those i's correspond to very, very high energies. If the state that we're talking about is not dominated by insanely high energies, then these n's must fall off quickly. The number of particles with energies of a zillion, zillion, zillion must be essentially zero. So in order that the total energy be finite, you certainly have to have the total number of particles finite. And this thing here is typically finite. Notice its formal similarity with the integral psi star psi. The integral psi star psi was also the total number of particles, but only in the case with number of particles was known to be one. And that's just the accident. The total probability is one. It happens that the total number of particles is one. And so that's up there. But here, exactly the same formal looking structure is not telling us total probability, it's telling us how many particles we have. So if that's true, then what corresponds to the probability density at a point of space? The probability density at a point of space. Why not just the density of particles at a point of space? If the probability density, you want a function of all of the x's of all the particles, right? This would be the kind of function. Right. Right. Remember, this is not, okay, the answer to the question is this thing here is the density of particles. Okay. So we can write then that why, because its integral is the total number of particles. Now, I'm not going to write that it's the probability density. I'm going to write that it's the density. But the density is not necessarily definite. Remember, we're doing quantum mechanics. A thing can be unobservable, but there are states, generally, typical states, the thing is not going to have a definite value. You can measure it. You can measure the density of particles. Strictly speaking, you wouldn't measure the density of particles at a point. What would you do? You'd integrate over a little volume and ask how many particles are in that volume. So if I integrate this over a small volume, not over all space, that tells me how many particles are in that small volume, I can measure it. I won't always get the same answer. I measure it in one state. I measure it in another state. It's like measuring any other quantum mechanical variable. You can get a variety of answers. But this will tell you what the number of particles is in a small volume. That's its role. So it's a kind of density. It is a density. It's literally the density of particles. I have a question. Sit down on the operators, a sub i, dagger, and so on. This formalism is that they only operate on the state vectors, but not on the wave functions. Not on the single particle wave functions. Right. Now, let's come back for a moment. This is not the most general state that you can have. This is one of a basis of states. You can superpose these just like in any vector space. You can add vectors. You can add vectors with different numbers of particles. You can add, even if you have the same number of particles, you can add vectors in which there are five particles here, seven particles here. You can add that to another vector where the sevens and the fives are interchanged. Seven and five is 12. You could add it together with a vector with 12 particles, or the same i. There are many states and superpositions of states. Many, many possibilities. In any part, whatever the state is, though, the operator representing the number of particles in a small volume is psi star psi. It's a thing that you measure. It's a thing that you measure when you look in that small volume and you ask how many particles are there. So, although I haven't told you how to measure psi itself, psi star psi is a familiar thing. It's just a number of particles and a little volume. If you divide this operator at the top of there by n, by the number of particles, it would be fair to interpret it as a probability, a finding of particles at any particle that's voids x to the same. It's about the link on your face that's like, yes, but you don't find it. No, it's okay. Well, you could say this volume has a certain fraction of that volume. Yeah. That's not the point. Supposing you have a lot of particles, even just classically, you have a lot of particles and you want to know what's the probability that any one particle is in a little volume. Well, to a high approximation, the number of particles are right. All you have to do is one measurement. Measure how many particles are in that small volume and divide by the total number. That's not the definition of the probability. And you can get, you know, you could accidentally get zero. You could accidentally get too many. But one of the times within a margin of error, you will get the number of particles being the probability for a particle to be in there. The number of particles divided by the number. All right, so it's not an identity to say that this thing, appropriately divided by the total. It's not an identity to say that it is the probability for a particle to be in that region. The probability can be a half, two regions of space. Probability a half here and a half here and you have a million particles. You could accidentally discover that all the particles are on one side. Not very likely, but you could discover it. Okay, you make that measurement and the answer is a million on one side, none on the other side. Not equal to the probabilities. But most of the time within the margin of error, within about a square root of the number of particles, you'll get the probability. So that's the sense in which I was a little waffling on whether you want to call it the probability or not. This thing really stands for the observable, which really corresponds to the number of particles that you actually get when you look in there. However, yeah. Yeah, that's a bad thing to do. Right, but what you could do, no, that normally doesn't occur, but what you could do is have a superposition of states in which there's some number of electrons distributed throughout the different states and add that to some number which is distributed differently, same number of electrons. Photons, however, because they don't have charge, you have no problem superposing different numbers of photons. Now, even with electrons, you can, don't forget there are other species of particles in the world. In particular, there are positrons. So you can superpose states with different number of electrons and positrons as long as the total charge adds up to the same number. Okay, but right now we're just inventing for ourselves the concept of a quantum field. All right, let me do one more little manipulation to show you another example of things that you can build. I was going to start working out the expression, the operator, made out of a size which stands for the energy of the whole system. We probably won't get quite there tonight, but let's begin. Here's what we found. We found that the integral of x of psi, dagger psi, is the same as the sum of the number of particles in each, that was what these little manipulations did. Told us that this is the same as this. Now, let's do a slightly different thing. Let's start with a formula for the total energy. Now, I'm interested now only in systems with independent particles. Particles are not interacting with each other. They're not exactly in approximation. It tends to work when the particles are pretty dilute. Even photons, if you make them dense enough, will start to interact with each other. And then the energy will not just be the sum of the energies of the photons. Now, how dense do you have to make them plenty dense? Much denser than you can make in the laboratory. But still, in principle, energy is not equal to the sums of the energies of all the particles in the system. But typically, if the particles are not too dense, and there aren't too many of them per unit volume, it's generally a good approximation. So let's take that approximation and think of it now as a principle of free particles. Free particles, particles that don't interact with each other. Then what is the energy? The energy is the sum of all modes. Mode means one of these single particle states. It's the terminology, mode. Mode means a single particle state. Sum of i of the number of particles in that state times the energy associated with that state. We called it last time omega i. We could call it omega h bar i, but I'm going to leave the h bar out for now. I'll put it back tomorrow. Omega i and energy are the same thing. I just used omega because we're working with oscillators, but it is just the energy of the i-th mode. Okay, this is the energy of, we can also write that as the sum of i of a dagger i, a i times omega i. It looks a lot like this over here. The only difference is that each i is weighted with a factor of omega i. What I want to do and I won't do it tonight is to go from an expression like this back to an expression like this, except for this beast over here. What we're going to get is something involving the field operators, and that thing that involves a field operator is going to be the energy of the fields. It's going to be the energy, be the sum of all the energies of all the particles, but it's going to be something that's expressed in terms of the field operators, and we'll just call it the energy of the field. That's going to be something reasonably familiar, but we won't do it tonight. Tonight, yeah. The part of the section was last week. Oh, I thought it was a... It did go fast. It did go fast. That was... Oh, right, right, right. Time is not invariant. What's happened to the energy? Okay. Okay. I was wondering about that. Okay, we'll do a little more then. Yeah. The omega i is the function of position. No. They're just the energies of the i-state. The wave functions are functions of position. That's a function of position, but the energy that's associated with it is just the energy of the first of the ground state of the system, of the ground state of the one particle state. No, the omegas are not functions of position. They're functions of i. They're functions of which energy level we're talking about. Okay. Okay, so what are the omega i's? The omega i's are the energy levels which are the solutions of the Schrödinger equation, the time-independent Schrödinger equation. To find the energy levels of a system, you solve the Schrödinger equation, and you find out which eigenvalues on the right-hand side of the time-independent Schrödinger equation give you solutions. All right, so let's write out the time-independent Schrödinger equation. Now we're going to write an equation which determines the omegas. Okay. Are you saying that there are or not interactions? No, no interactions. You weren't listening. Three particles, no interaction. Okay. Yes, we are. No, we are not assuming there are interactions. Yes, we are assuming there are no interactions. Or at least that we're in a situation where interactions are very unimportant. Particles too far from each other. Okay. All right, so what do we know about omega? Omega is a solution of the time-independent Schrödinger equation. Let's write down the Schrödinger equation. The Schrödinger equation or the time-independent Schrödinger equation. Of course, it's just now we're talking about the one particle Schrödinger equation. The Schrödinger equation for only one particle. And what does it look like? Well, it's the one particle Hamiltonian acting on a wave function is equal. Some psi i and it's equal to the eigenvalue omega. That's the energy level times psi i. And what's the Hamiltonian? The Hamiltonian is, let's just take the simplest thing, p squared over 2m plus a potential energy. The particles may be interacting with a potential energy, but we're not going to let them interact with each other. We're not going to be electrons orbiting the sun. We're not treating the sun quantum mechanically. We are treating the electrons quantum mechanically. But we're taking the approximation that the electrons don't interact with each other. They only interact with the sun. So what do we write down for h? h is potential kinetic energy, p squared over 2m plus v of x, is potential energy. Let me write it a little clearer. Multiplying psi of x equals omega i, and this is the i-th psi of x, the i-th energy level is omega psi i of x. Now, this is an odd thing here. What do I mean p squared over 2m? We know what v of x times psi of x means. That just means multiply psi of x by v of x, where v is some potential energy. But what does this thing mean? p squared over 2m. We use p equals minus i d by dx. And you can, as homework, put in the h-bars. This is, of course, a component for each dimension of space. So this becomes px squared plus py squared plus pz squared. This thing here becomes, well, if I want to write it in three-dimensional language, it becomes del squared, which means second derivative with respect to x plus with respect to y with respect to z. A minus sign because of the i-squares, i times i gives us a minus sign, and we divide it by 2m. Now, this is the thing we know what to do with this on psi. It just differentiates psi twice. It's just the sum of the second derivatives in the various directions of space. Plus v of x, psi of x, plus v of x, let's put it all in one bracket, plus v of x, psi i equals omega psi i, omega i psi i. Omega depends on which energy level we're talking about. Well, we're not going to solve any Schrodinger equations tonight. We're just going to take omega i times psi i, which notice, let's see, it hasn't appeared yet. We're just going to use the formula without solving it. I think the easiest way to get there is to write down the answer and then prove that the answer is the same as this. It's going to be the easiest way to get there. So let's put this up on top. I'm going to write down the answer, and then we're going to prove it's the same as that. OK. Before I do, let me point out something about ordinary quantum mechanics, one particle quantum mechanics. Suppose that I wrote down the expression integral dx, psi star of x, any wave function, whatever, times minus del squared over 2m plus v of x psi of x. What would this thing be? Anybody know? It's the expectation value of the energy. This is h, and this is basically sandwiching h between bra and ket vectors. It's the expectation value of the energy. So here's a formula for the expectation value of the energy of one particle. What I'm going to tell you is that the actual operator representing the Hamiltonian of the many particle system, can anybody guess from this what the answer is? What's that? Well, no, we know that. But I'm looking for another expression for the same thing, and I want to guess at this point and then show it's the same as that. What's that? Yeah, we can do something like that, but I can make a guess. What did we do to go from the probability to find a particle to get to the number of particles in a small region? Indeed. We just replaced the little size by the big size. And that sort of gave us the sum over all the particles of sort of the same question. So let's try. Here's a guess. No, it's not a guess for me. It would be a guess for you except, of course, it's not a guess. I remember learning about this. When did I learn about it? 19, probably 1961. And it was kind of blown away. This, I'm reading Dirac's book, I think. Reading Dirac's book, this blew me away that this is the same as this. Let's prove it. Let's go through the steps. The first thing to do is to replace psi by the definitions up there. So let's replace psi by the definitions. This becomes integral dx. Now, this one here is sum over i ai dagger psi i star of x. That's this one over here. And this one over here is sum over j psi j of x times a j. Everybody happy with that? A minus j, I guess. That's all I did. I plugged in the formula. This can be written as a sum over i and j and replaced the second sum, summation sign there, just get rid of it. And now sandwich this minus del squared over 2m plus v of x. Now, what the devil does this mean for this to act on this object here? What does it mean? Well, it simply means that you do these operations on ordinary wave functions. This is an ordinary wave function. We know how to differentiate it. We know how to multiply it by v of x. And so we have an operation that we know how to do. We might not be able to do it because it might be too complicated, but we can at least define the operation. This differentiates psi twice. This multiplies by v. But we already know, at least I hope we know, somewhere on the blackboard here. Nope. Schrodinger equation. Here's the Schrodinger equation. The Schrodinger equation tells us exactly what this is. Minus del squared over 2m plus v of x acting on psi just gives us omega i times psi. So we can erase. I don't like erasing. I won't erase. We can write this integral dx sum over i and j, a plus i. Let's group it together with a minus j over there. Now there's psi i star of x. Now there's this beast multiplying psi. That's just omega j psi of j. What's the next step? Integrate over x. Integrate over x. When you integrate over x psi star of x i times psi j, that's just kernicah delta. That's just equal to sum over i and j. Plus i, a minus j, delta ij. But with one new factor that we didn't have when we did the probability, or when we did just the counting of particles, omega j. And that's exactly this. We have a lovely expression now. This is the energy of a field. Even in terms of something quadratic in the field with a certain differential operator, what can you compare that with? An electrodynamics. An electrodynamics, very, very similar kinds of things. Or in any field theory, you take the fields, you differentiate them a couple of times, make a quadratic expression, and integrate over space. And that's a characteristic formula for the energy of a field. All field theories have that property that the energy is equal typically to something quadratic in the field with a couple of derivatives acting on the field. The derivatives act on psi here. So this is an example of the energy of a field, but it's also the Hamiltonian. These things are operators. The left-hand side is an operator. It's not a number. It's an operator. What operator? Well, it's a thing which has to do with the energy. It's the Hamiltonian of the field theory. It's the Hamiltonian of the field theory. But it's also the energy. I went through every single step. I belabored the steps. It's a function of position. It's an observable, or at least psi plus psi dagger is a quantum mechanical observable. If you measure it at one place, you can measure it at the other place at the same time. So what more do you want? It's a function of position. It's an observable quantity that the... So we're going to define this thing because of these properties as we might call it a field. I mean, mathematically, it is a field. That's what we're saying. You can measure it. It depends on position. There is no other definition. Now, you could ask, what field? Is it the electromagnetic field? No, it's not the electromagnetic field. It stands in the same relationship to the electromagnetic field as photons would be in a certain relationship to the electromagnetic field. Whatever kind of particles we're talking about here, these are nonrelativistic particles. They have a mass. They could be protons or neutrons moving slowly. They could be electrons moving slowly. As long as they're not relativistic, as long as the energy is given by p squared over 2m, this field is the field for those particles. Those particles are the quanta of that field in the same precise sense that photons are the quanta of the electromagnetic field. The electromagnetic field would have an expression for the energy. It also has two derivatives, a field squared, and it's quite similar, although it's not identical. This one here? Right? OK, so we've invented a quantum field. We've invented two observables. One of them is the particle density. The other is now. We have the energy, but the energy is the integral of a space of something. What would you call a thing that's being integrated? The energy density. It's the energy density. So what's in here? Let's get another color. What's in here is the energy density, the local energy density. By measuring it or integrating it over a small volume, you find out how much energy is in that small volume. So it's not the integrated overall space? You could integrate overall space, and it's the total energy. That's this. But if we have something and the total energy is equal to an integral of it, it stands to reason, or we can call that thing, or we might, it's a definition of the energy density. It's not the only definition. There's a little bit of ambiguity, but... I have a question, please. In that formula, e equals sum n i to omega i, capital N, how is it different from the little n's? The little n's are the eigenvalues of a big n. Big n is an operator, the little n's are just numbers. I have a question in syntax. Can you express that field equation in terms of... Well, it's not a field equation yet. But in terms of Dirac notation? Because it's very analogous to what you have on the upper board, which you can express in the Dirac's Karat notation, whatever it's called. Brakin kets? Yeah, the brakin, yeah. No, no, no, no, but these are not states. These are operators. They're not brakin kets. The brakin kets, these are the brakin kets of the system. Robert, but you expanded the... to do the little n's that's really sort of an elaboration of it. It's not traditionally what brakin notation is, really. Brakin notation is a notation for a state vector of a system. That's a state, that's a state. It's a state with n-1 particles in the first state, then two... Yeah, you have to learn to separate the single particle quantum mechanics from this multi-particle quantum mechanics. In this multi-particle quantum mechanics, there are things which are operators which don't even occur in the single particle. The... Right. Now, this formalism or this technology does allow you to recover everything that we've ever said about single particle quantum mechanics. You just go to the states in which the total number of particles is one. And you work in that set of states. All of this formalism just reproduces the physics of the single particle quantum mechanics. But it allows you to deal with any number of particles. And in particular, there are states... there are states with large numbers of particles, typically. Remember, these are observables now. They're operators. When there are a large number of particles in the same state, then these size tend to behave very much like classical fields. And we'll talk about that a little more. How is this connected to classical field theory? Well, it's connected to classical theory like any quantum mechanics is connected to its classical counterpart. For example, expectation values of things follow the same kinds of equations of motion as the classical equations of motion. So you could ask, for example, what happens to the expectation values of psi? Psi is an operator. It has an expectation value at any time. We know the Hamiltonian of the system. We can use it to work out the time dependence of the state. And then we can use that to work out the time dependence of the expectation value of psi. Guess what? It essentially satisfies the classical equations of motion of wave fields with the same classical Hamiltonian. In fact, it satisfies the time-dependent Schrodinger equation. So, as I said, this is a very beautiful formalism. It is a bookkeeping device. But it's also the fundamental connection between particles and wave fields. And there is no other connection. This is it. That's the connection. So let's explore it in a great deal of exploration. Okay. I've been struggling with the manipulation of the creation and annihilation operators inside that integral. You seem to treat them almost like numbers. No, I never switched them. I never commuted them. I commuted them and switched that order. I know that they depend on the way some of them are. Yeah. They're numbers. They are numbers. Here they are. I'm sorry. The size of numbers, the a's are operators. But we need, okay. So if the a operates on the a's, say the last pic, when a j minus operates on side j, it doesn't operate on side j. No, no. It may operate on some n1, n2, n3, and so forth. It took advantage of the fact that the size are numbers. And because they're numbers, you can pull out the whole delta function. And then you're just left with the a's and a's. The only manipulation that you wouldn't want to do freely without thinking about it is to move an a minus past an a plus. They don't commute with each other. That's where the operator character of them would come in. But as far as moving them past ordinary numbers, that's fine. The derivatives here don't act on them. The derivatives act on the size. So there's no cause to worry about any of the operations I did. There would be a cause to worry if I tried to move a minus past a plus. But we did interchange the i minus i plus, but it introduced a plus 1 in certain things that you did. Oh, yeah. We just have to be careful about it. That's all. Yeah. I don't think you play some funny game with the derivative of psi j. But these are only derivatives acting on ordinary functions. I'm just saying we sort of pulled it all out into a delta. That's the integral over x. Right. Yeah. So I mean, I didn't say it wasn't valid. I was just saying that's a little tricky. That's a little tricky. Well, we did. We couldn't do the integral over x in this form. What we did was say that this operation here just multiplies psi j by omega. Once it became omega, then we could do the integral over x. Yeah. Yeah. All right. So I didn't cheat you. I didn't do anything that- Just a number. So you- in Schrodinger's formalism, say, cheat operators as things that can operate on functions, on the wave functions. What's that again? In Schrodinger's formalism, operators can operate on wave functions. Right. In this form, it was in these operators that just operate on state vectors and not on wave. Right. You can imagine how confused they were in the early days before Dirac mastered all of this and put it together. They were very confused about the meaning of the wave functions. Some people thought it was a field that you- you know, a real field, like the electromagnetic field. Some people thought it was just an auxiliary device for computing probabilities. And if that's all it was, then what does it have to do with the electromagnetic field? And what's the connect- there were several different ideas about what the connections between particles and waves were. On the one hand, the single particle wave functions satisfied a wave equation, the Schrodinger equation. Look, sort of suspiciously like Maxwell's equations. It does, roughly speaking, not quite. It looks like Maxwell's equations. Maxwell's equations were not equations as far as anybody knew for quantum amplitudes. They were equations for measurable values of the field. But still, there was some connection, some funny connection between these fields and particles. Namely, the field was made out of quanta called photons. So there was a lot of confusion. And this is the resolution of that confusion. That don't- I said for them, maybe. What did you help to- The question? The same letters. No, no, this is it. That's it. That's all there is. Did you actually have some background in over its faces, vector's faces? No. I don't think so. So you just had to- No, he made it all up himself. Which is fortunate, because had he used the notations that were there already, you know? I taught quantum mechanics last year for them in the mathematics department, which was a lot of fun. There was a crazy point at which I was explaining Dirac notation. And I told them that every vector, we call these a vector space, or whatever, psi. And the official term, sort of the ket, and the official mathematical term for the bras is the dual vector space. It means complex conjugate essentially, the dual vector space. And I learned to say dual, that made them happy. And then I said, there's an inner product between every vector and its dual. And they said, no, no, no, no, no, no, you can't do that. You cannot have an inner product between a vector and its dual. You can only have an inner product between a vector and itself. And I said, look, I'm telling you, there's a- just, trust me, I know what I'm talking about. You can have an inner product between a vector- No, no, no, no, illegal, not allowed, not allowed, not allowed. Okay? Well, of course, it turned out that they just have a notation for the inner product of two vectors. Take two vectors, and you put them next to each other this way, and then you put some funny symbol around them like this. And that's the inner product between psi and phi. What is it equal to? It's equal to psi-phi. When they finally got the hang of this, the mathematics professors all said, wow, this is powerful. It makes everything work simply. For example, they were again blown away by this formula that ii is equal to one. And all you have to do to multiply it by a state is just do what it says, psi. And this is the inner product. They've never seen that before. Of course, they saw it. Of course, they've seen it. They understood it all completely, but they had never seen it in this compact notation where you almost can't make mistakes. Anything you do turns out to be right. So Dirac was clever. He also, if you ask, was his mind an ordinary mind, it was not. He was probably autistic or something close to it. So he saw things in different patterns than most of us do. Going back to what you were saying a little while ago about this elegant formalism, it's sort of begging the question for me, are fields just formalism? No, you can measure them. But are you just measuring properties of collections of particles? Yes, but you're measuring properties that you might not have guessed. No, you're not measuring the properties of classical particles. You're measuring the properties of quantum particles. And quantum particles have properties like interference and that the classical particles don't. So yes, you're only dealing with a system of particles, that's all. But because they're quantum mechanical, they have kinds of properties, interference, entanglement and other things which ordinary particles don't have, classical particles. And this accounts for all of it. Is there some sort of identification between, for example, let's say the density of photons and the electrical field strength? Yeah, well, one has to be a little bit careful. When you're dealing with relativistic particles, and I've been careful not to deal with relativistic particles, only non-relativistic, when you're dealing with relativistic particles, the notion of the momentum of a particle is very well defined. The notion of a position of a particle is not. And if the notion of the position of a particle is not defined, then the notion of the density of particles is not defined. What is defined is the density of energy. And the density of energy has a similar form. But if we don't split too many hairs, then the square of the electric field is qualitatively related to the density of photons. So in relativity, you have a defined momentum, but not so much with position, is that what you're saying? Yeah, position is not a good concept in relativistic systems. Yeah, you can even make, if you define, this is not a special useful, but it's a fact, if you define a vector psi out of the electromagnetic field, you define it as the electric field plus i times the magnetic field. Take that quantity, psi star. Now there are three components to this, so it's not quite the same as this. Three components, electric field minus i times the magnetic field. Then psi star psi is not quite the density of photons, it's the energy density. So it's the density of photons weighted with the energy of each photon. That's a good concept for photons. The density of photons itself is not a good concept. So psi star psi is e squared plus b squared. If you know any electrodynamics, you know that e squared plus b squared is the energy density. And psi satisfies a Schrodinger-like equation. It's fun to work out from Maxwell's equations what psi, and this is psi dagger. If you know Maxwell's equations, you can work out the equations for psi. It's a simple equation, simple than Maxwell's equations themselves. I have a question. Energy equals the integral of capital Phi dagger phi. No. I'm just deviating that. Yeah, yeah, psi dagger confirms this differential operator here at times psi. If you switch the psi and the psi dagger, does that have any meaning? It's clearly not equal to this. You're not allowed to pass them through each other without taking care of the commutator. So it has a meaning, but it's what it is. It corresponds to an extra additive constant, the ground state energy. Let's see if we can make a guess. I'm not going to try to prove this. It's easier than what I did here, and you can try it. For the single particle quantum mechanics, momentum P is equal to minus i d by dx. There's some Planck's constant in there. What's the expectation value? Supposing I have a single particle wave function psi of x, what's the average or the expectation value of the momentum? Well, you take the ket vector psi, you take the momentum operator, and you sandwich it, the usual sandwich to create the expectation value. That means is you integrate over x, the x, psi star, times minus i d psi by dx. So let's put the minus i on the outside here. Psi star times d psi by dx, and you integrate it over all space. What about the momentum operator? Not the expectation value of the momentum, but the momentum operator for this field theory. Do fields have momentum? Fields have momentum. If you take the electromagnetic beam from the laser and you shine it on the wall, if the laser beam is powerful enough, it will blow a hole in the wall. Why? Because it carries a lot of momentum. It exerts a force on the wall. Fields have momentum. Classical fields, quantum fields, in the quantum case, the field momentum is just the momentum of all the photons. So here we have an expression for the average momentum of a particle in quantum mechanics. Can you guess what the expression for the total momentum of all the particles is? Yeah. Yeah? What? You changed operator. Just replace the size by size of bigger size. Sci. This is the operator that represents the momentum. If you measure the momentum, how do you measure the momentum? You see how big a hole you blow in the wall by, right? You measure the momentum, not the average of the momentum, the actual momentum in a wave going by you. How do you measure it? You see how hard it hits you, how much force it exerts on something. This is the operator, the observable that corresponds to that, this is the mathematical Hermitian operator that corresponds to that observable. You can also calculate its expectation value. This is an observable. It is not the expectation value. If you give me a state, I can calculate the expectation. This is something we'll do next week. These are operators. They have expectation values. They are not the expectation values. You need a state. You need an operator and a state to construct an expectation value. Where's a state? Give me a state. Well, you could take those in states up there, in one and two and three, and calculate the expectation value of the momentum. Okay. I think we'll finish for tonight unless somebody has some good questions because I've, I've I've, I have not run out of things to talk about, but it's a natural breaking point. Next time we'll talk about processes in particle physics and how they're represented by these fields in a very simple context. In a very simple context, we'll talk about how field operators implement processes that take place when particles decay, when particles scatter, and so forth. For more, please visit us at stanford.edu.
 Stanford University.  Well, I don't know, should we go over again the basic construction of a simple quantum  field?  Why not?  It's not exactly what I have written here, but we can do it again.  Not again, but maybe slightly different.  Yeah, we're ready.  Okay, so as I said, the trick of introducing fields from particles, see you can go two  ways.  You can start with particles and reconstruct why they're described by fields, or you can  start with fields and then work down to why fields are described by particles, or why  fields, waves, wave fields have quanta.  You can go either way, and we'll go both ways.  So we started by starting with particles, and we said, look, it's fine to be talking about  one particle or two particles or three particles, a fixed number of particles.  But gee, why not try to study all possible numbers of particles?  And by particles now, I'm speaking about a single species of particle.  We can come later tonight, what happens if you have more than one species, but for the  moment, one kind of particle and only one kind of particle.  Let's make a theory in which we can simultaneously, in one fell swoop, so to speak, talk about  any number of particles and all the things that multiple particles can do, including  situations where the number of particles change.  Now, why would be ever be led to that?  Well, in real life, numbers of particles change.  The number of photons in this room is constantly changing.  They're being emitted by the electric light bulbs.  In other contexts, particles decay, turning into other kinds of particles.  So the neutron can decay into a proton, an electron, and a neutrino.  And in the process, the number of neutrons changes.  The number of protons change.  The number of electrons, the number of neutrinos.  And in fact, the total number of particles change.  So however we have to think about particles, we have to think not only about the quantum  mechanics of a fixed number of them, we have to think of that number as changeable.  So we need a flexible enough tool to be studying what happens in a single quantum  mechanics which embraces essentially any number of particles and even changing number of particles.  That's what the point is.  And that's what the point of inventing fields is.  Sorry, any questions about that?  And I'll go over it again since the mathematics is not hard.  But it's abstract.  It's abstract.  And face it, we're not as young as we used to be for one thing, none of us.  And the mind is not necessarily attuned instantly to orthogonal functions and all that kind  of thing.  And so it's good to go slow.  I'd rather you learn something rather than nothing.  OK, so it's very easy to learn nothing.  It's a little harder to learn something.  OK, so we start with single particle quantum mechanics.  Single particle, the description of one single particle, quantum mechanically,  that means that we have state vectors but state vectors for one particle.  We have wave functions for those particles.  If we have a state of a single particle, let's call it psi, little psi,  little psi will also stand for its wave function.  If we have a state vector for a single particle, then the wave function is just the inner product  of the state vector with a particle at position x.  So a state labeled x means a particle at point x.  The state psi means some general state but of a single particle.  And that's called psi of x.  What do you do with it?  Here's a formula.  Psi star of x times psi of x.  And these are wave functions now.  These are not quantum fields.  When I go to quantum fields, I will use large capitalized letters  even if they happen to be the same as this.  What is that object?  The object I respond to, that's the probability to find the particle at x.  It's the probability density.  The probability density to find the particle at x.  Probability density.  Prob-dense.  You can fill in the rest of the letters.  A fact about the probability density is that when you integrate it over position,  what do you get?  You get one.  It happens in this case that one happens to be the number of particles,  that isn't it?  So you get one, but you could also say you get the number of particles.  Well, that's not very interesting because we only had one particle to begin with  and we were doing one particle quantum mechanics.  Nevertheless, it's a true fact.  Okay.  Now, we next introduced a collection of one particle states that formed a basis,  an orthonormal orthogonal basis.  Now, we can do that in a number of ways.  First of all, the x states themselves are orthogonal to each other  and they form a basis.  We'll come back to that in a minute.  But there are many, many ways of choosing bases.  Essentially, you take the eigenvectors of any Hermitian operator,  we'll form a basis.  And in particular, if we're interested, we may be especially interested in energy eigenstates.  For example, of a particle in a box that represents a box,  the horizontal axis is the position, the vertical axis is the potential energy.  When you get to the edge of the box, the potential energy becomes huge  and the particle just reflects off, it can't get through.  In that case, the ground state wave function looks like that.  The first, and we'll call that psi one of x.  The next energy eigenstate looks like that.  We'll call it psi two of x.  And these are basically signs.  Sign of two nodes and they're characterized by the number of nodes.  Rather peculiarly, I started with psi one which has no nodes,  psi two has one node, but so be it.  Psi i of x.  And since these are supposed to be energy eigenstates,  they're orthogonal to each other.  They form an orthonormal basis.  So let's write the equation for that.  The equation for that is that integral of psi star i of x with psi j of x,  that's the inner product, the x.  What is that?  Anybody remember?  Delta ij.  Delta ij.  If i is equal to j, they're the same state.  The inner product is one, normalized state.  If they're different, they're orthogonal, corresponding,  let's say, to different energies.  Sorry, delta ij, not zero, delta ij.  That's another fact.  Now I'm going to show you another little manipulation.  This one, yeah?  Why does the probability density have that particular form?  That goes back to the postulates of quantum mechanics.  The postulates of quantum mechanics said that probabilities  are psi of x.  This is the inner product of psi with a state localized at x.  It's called the probability amplitude.  The postulates of quantum mechanics, which I invite you to go back to.  They're in the book.  Say that the square of that is the probability.  So OK, here's a little trick, a little mathematical trick that occurs over and over again.  Take your orthonormal, your favorite orthonormal basis of functions.  They could be sines.  They could be cosines.  They could be harmonic oscillator wave functions, whatever.  Some collection of size which are orthonormal.  And if they're orthonormal and they form a basis, then if you take this object,  what is this object?  This object is a operator.  It operates on a state.  Let's see what it does.  Let's just explore what it does when it acts on a state.  What does it do?  If it acts on a state psi, then the beauty of the Dirac notation is you can just take  the formula that you've written down here.  When you act with this thing on a state, psi with i is a number.  It's the inner product of psi with i times i.  That's what it does.  Now, here's a fact about any orthonormal basis.  If you sum over all elements of the orthonormal basis, in other words,  in this case, all the energy eigenvectors up to infinity, then what do you get?  The unit operator, i, the identity operator, the operator that when it acts on any vector,  it acts the same vector.  That's a marvelous theorem.  It's in the notes.  You should go back to it.  Don't take my word for it.  You can even prove it.  But it's a useful thing to know.  Let's use that now in the following way.  Let's take a project.  Let's act with this thing to the right and to the left.  What we're going to do is we're going to take, what are we going to act on?  We're going to act on a state to the right, which is localized at position x.  And on the left, we're going to act, well, actually what we're doing is I'm taking  the inner product of a state with a particle at x with a state at y.  Anybody know what the mathematical symbol for that is?  It's a Dirac delta function.  But let's come to that for a moment.  Let's not write that down yet.  Let's write that that is equal to the sum on i, y, i, i, x.  All I've done is plug in the unit operator here.  That does nothing.  But now I have a formula here.  The inner product of i with y is psi i of y.  Let's see.  Yeah, the inner product of any state with an eigenvector of position is the wave  function of that state.  Here, the state is i and it has wave function psi i of y.  Here's over here, psi i of x.  And what does that have to equal?  It has to equal delta of x minus y.  Now, sum over i.  Yeah, there is.  The second one.  Yes, the second one.  I assume that's what you were asking me.  Yeah, the star.  OK, now we have a marvelous theorem.  If we take any orthonormal basis,  they could be the eigenvectors of any Hermitian operator,  but in particular the energy.  And you multiply for each one of those states,  psi star times psi, and then sum over them,  you'll always get a delta function.  That's a good thing to know, as we'll see.  It's useful.  It means there's many, many ways of writing the delta function.  Just take any orthonormal basis and do this.  These could be sines and cosines.  Sines and cosines are another way of writing a delta function.  But this is the basic mathematics, normal mathematics,  that's it, I think.  OK, now let's come back.  Let's go to the problem of studying many particles all at the same time.  OK, now how are we going to characterize a state?  One set of states, which is particularly useful,  is a set of states in which you say how many particles there are  with each one of these energies, for example.  How many particles are there with state with psi one,  psi two, and psi three, and so forth.  And so we can take the multi-particle system  and characterize it by a basis of states,  which are just n1, n2, dot, dot, dot, and i, dot, dot, dot.  The ends are the occupation numbers.  They're the occupation numbers which tell you  how many particles there are with each energy level.  And those are states.  They're not single particle states.  They don't even belong in the same vector space  with the single particle wave functions.  The single particle wave functions are themselves a whole vector space.  But we're now talking about a bigger thing, a bigger system.  And so we should not confuse the single particle states  with the more general structures that we're talking about.  We use the same notation for them  because nobody ever invented a better notation in Dirac's,  the Crosse and Kett's.  OK, next, we would like to be able to change the number of particles.  Maybe we do, or maybe we don't want to change the total number of particles.  That's not clear yet.  But we'd like to be able to change them.  And so we'd like to be able to raise and lower the ends.  So we introduce a oscillator for each i.  For each i, we introduce an oscillator.  And at the moment, this has nothing to do with oscillation.  This is just a trick.  This is just a bookkeeping trick.  Although it will be more than that later.  For each one of these i's, we introduce a creation operator  or a raising operator.  And we introduce a creation operator since it's now going to create a particle.  A creation operator and an annihilation operator.  And the creation and annihilation operators,  it's convenient to make them just work exactly the same way as harmonic oscillators.  It's just a convenient way of dealing with them.  A plus i and A minus i.  Now, another convention for the same things,  I'll write this in red, is to call this A dagger i.  That looks like a plus sign.  Not quite, but nevertheless, same thing.  And just calling this one ai.  You'll often see it written that way.  It's the same thing.  A dagger creates a particle.  A does not.  But the correspondence is exactly this.  Okay.  This space of states of the many particle system is called Fox space.  FOCK.  Fox was a Russian physicist who somehow got his name attached to this.  I don't really know how.  Okay.  Now we're going to find a very convenient, and you'll see how it works,  to introduce the notion of a field operator.  The field operator has the same symbol as the wave function psi of x.  But it's not a wave function.  It's not a wave function.  It's not a state vector for one particle.  It's an operator.  And it's an operator that acts on the space of states of the many particle system.  It itself contains creation and annihilation operators.  All right.  So let's write the notion of a field operator.  And the best way to motivate it is to use it and see what it does.  Now I use a uppercase psi.  I hope.  I'm a little bit afraid that you won't be able to tell my uppercases from my lower cases.  All right.  That's an uppercase psi, right?  Its definition is the following.  It's the sum of i of, let me get it straight,  the annihilation operator for a particle in state i times the wave function of a particle  if it were in state i.  And that's it.  Summed over all i.  Now somebody asked me last week, does this have anything to do with Fourier transforms?  And you bet it does.  If the wave functions here were the wave functions of a free particle,  the eigenstates of energy of a free particle, and not in a box just in an infinitely big box,  what are those eigenstates, those eigenwave functions?  They're plane waves.  e to the ikx and e to the minus i, or e to the ipx and e to the minus ipx.  So in that case, if we were using free particles in an infinitely big box,  this would be an expansion in sines and cosines, and those a's would be the coefficients of sines and cosines.  They would be the Fourier coefficients of a Fourier expansion,  except that they're not numbers.  They're operators.  They're operators, the things that you measure,  or things that play the role of operators in quantum mechanics.  Okay, so that's one thing.  The complex, or the Hermitian conjugate of it,  so a dagger of x, this is of x.  Yeah, what does it mean to say we have an operator which depends on x?  Well, it just means we have a different operator at every x, that's all.  We have a lot of creation and annihilation operators.  Out of them, we can build a lot of independent operators.  In fact, we can build an operator at each point x by combining the a's with size of x, at point x.  So if we do this, we make a different operator at every position,  and so we have an operator function of position, okay,  or operator-valued function of position.  It's complex, or it's Hermitian conjugate, is the same thing,  except creation operators here.  This is a, I get so used to, yeah, okay, a minus a plus times psi dagger,  so I'm psi star i of x.  Now, it doesn't matter, it's rather weird.  Some people use a notation where this is called psi star,  they use a one where this is called psi and they get all mixed up.  These are definitions, and so you can do what you want,  but here's the definition we use.  As I said, there are two operator functions.  Now, neither one of these is a Hermitian operator.  They have the creation operators and annihilation operators are not Hermitian.  Their conjugates are each other.  If you add them, you make something that's Hermitian.  If you take any operator and add it to a Hermitian conjugate, it's Hermitian.  So if we add these two, they become Hermitian,  and Hermitian means observable,  means something you can measure in the laboratory.  So, the sum of them is the thing you can measure.  And the difference is the difference Hermitian.  What happens if you take the difference of a number and it's complex conjugate?  You get something real?  You get something pure imaginary,  but if you multiply it by i, then you get something Hermitian.  So, each, the sum and difference of these defines some Hermitian observables.  But right now, we're not observing them.  We're not going to use them as bookkeeping devices.  Okay, now, let me show you another interpretation of psi of x, what it does.  In particular, let's ask what it does when it acts on a vacuum.  The vacuum means a state with no particles in it at all.  And so the vacuum, let's call it vacuum, which we usually denote just by O,  is really the Fox-based state in which all the occupation numbers are zero.  All the same thing here.  Vacuum, Fox-based state with which all the occupation numbers are zero,  sometimes just called the vacuum or just called the state zero.  Okay, let's act, yeah, let's not act yet.  Let's write down the following formula.  This is a one-particle state now, one particle and only one particle.  And let's write it in the following form.  Let's use this, anybody know what the name for this kind of formula is?  Projection.  No, it's not, well, it is a project, it's the trivial projection operator.  It's one.  It's called a resolution of the identity.  It's a way of writing the unit operator as a sum over a lot of orthogonal vectors.  All right, so let's plug in the resolution of the identity,  sum on I, I, I, psi.  I, what is I?  I is a state in which there's one particle and only one particle in the ith energy level here.  Okay, what is this thing here?  This is, sorry, X.  X, sum over I, X, I.  What is this object?  Let's go up to the top formula and you see that when you project on to X,  you get a wave function.  This here is psi star I of X.  It's the complex conjugate of the wave function of the ith energy level.  The ith energy level has a wave function, this is its complex conjugate.  The wave function itself would have the opposite order.  What about this state here?  That's a state with one particle in the ith energy level.  So I'm going to write it this way.  I'm going to take a creation operator for making a particle in the ith energy  and multiplying it by the vacuum.  I've created one particle in the state I by applying the creation operator on the vacuum  and that's exactly the state here.  On the other hand, this is just a complex conjugate of the wave function  and my rule is sum.  Did I erase the defi- no, there's the definition of the field operator.  Okay, go to the second row.  Psi star, the field operator, is sum over I, a plus of I,  times the complex conjugate of the wave function.  That's exactly what you see here, right?  So this is exactly equal to the field operator at position X on the vacuum.  That gives us a new interpretation of psi.  It's the operator which creates out of the vacuum a particle at position I.  I think that's a fairly intuitive thing that it's a quantum operator at position X.  Sorry.  Good.  So the AIs create particles with certain energy levels.  The psi I creates particles at certain locations.  It's also a kind of creation operator.  The field operator itself is a kind of creation operator,  but instead of creating a particle with an energy, with a definite energy,  it creates a particle with a definite position.  You want to dagger there and the particle to start at the same time.  Here?  Yeah, it should be dagger.  Good.  You never use star for operators.  Use dagger.  Good.  This assumes that the creation operator and the wave function can use.  No.  Yeah, wave functions are numbers.  They're numbers.  They're not trivial operators.  They're numbers.  Sometimes ordinary numbers are called C numbers.  This is another invention of Dirac.  C stands for classical.  These are classical numbers.  Operators are called Q numbers for quantum.  Q numbers don't always commute.  C numbers always commute with themselves,  but they also commute with everything else because they're just numbers.  OK.  So now we have a lovely new interpretation of the field operator.  The creation operator for a particle at a point of space.  Let's look at some interesting quantities.  I think we'll, well, we can put this up.  Just to get a little more experience with this,  let's play with it a little bit.  Here's an interesting quantity, integral dx.  Operator psi dagger of x times psi of x.  Now, is this the same thing as up on the top of the blackboard up there?  The x psi star of x psi of x.  No, it's not the same thing.  It looks very similar, but it's not the same thing.  In the top formulas up on top, the size is just wave functions,  one particle wave functions.  They're not operators.  The other side is just one by total probability.  This object here is an operator.  Psi is an operator, psi dagger is an operator.  When it acts on a state, it does something.  This one creates a particle, this one annihilates a particle.  So, but nevertheless, it has some formal structural similarity  with the thing up on top.  One wonders, what is it?  Does it have some meaning?  Is it something that has a physical significance?  Is it something that you can measure?  Incidentally, it's Hermitian, because if you take any operator  and multiply it by a Hermitian conjugate,  the result is Hermitian.  That's like multiplying a complex number by its complex conjugate.  The result is always real.  Okay, so let's see if we can figure out what this thing is.  First step.  Let's replace the psi's by their definitions.  So, this is integral dx.  And now, psi star, that's a sum over i.  It's going to get messy, but then it'll get easy.  Ai dagger, or ai plus, times psi star i of x.  Without the what?  Without the which?  No, that's right.  Now, I'm going to write the psi.  That has another separate sum, not the sum over the same i.  It's a sum over a different i, but I can just call it j.  It's a double sum.  There's a double sum here, and the double sum has a j sin.  Psi j of x.  It looks a bit complicated.  An integral over x of a double sum over i and j,  plus the fact that the integral is over all space,  and i and j go from one to infinity.  So, if you are actually to calculate it,  to put it on your computer, your computer would run forever  and ever and ever and ever and would never halt.  However, we can do better than that.  Let's do the integral over x first.  Now, I think we have a formula here.  Psi i of x.  There it is.  There it is right here.  Psi i of x with psi j.  That's just the orthogonality relation.  So, when I integrate over x,  psi dagger times psi, I just get delta ij.  That's nice.  Let's write it in a simpler way.  This is equal to, we've done the integral,  sum over i and j, a plus i, a j, times the chronicle delta.  In other words, we get a term for each i and j,  except only when i and j are equal.  And so this is equal to the sum over i of a dagger i, a i.  What is a dagger i times a i for a harmonic oscillator?  It's just a number operator.  The thing whose eigenvalues are the n's, the little n's.  This is the big n, so this can be written as the sum over i  of the number operator for each one of these oscillators.  In other words, it's just the total number of particles.  The number of particles in the first mode plus some first wave  functions plus the number of particles in the second wave  function plus the number of particles in the third wave  function.  So what is this thing?  It's an operator that represents the total number of particles.  For each one of these Fox states, it's just a sum of the little n's.  Now, there's an infinite number of little n's, right?  There's n1, there's n2, there's n3.  Is this going to be a finite number?  Well, remember that most of those i's correspond to very, very high energies.  If the state that we're talking about is not dominated by  insanely high energies, then these n's must fall off quickly.  The number of particles with energies of a zillion, zillion, zillion  must be essentially zero.  So in order that the total energy be finite, you certainly have to have  the total number of particles finite.  And this thing here is typically finite.  Notice its formal similarity with the integral psi star psi.  The integral psi star psi was also the total number of particles,  but only in the case with number of particles was known to be one.  And that's just the accident.  The total probability is one.  It happens that the total number of particles is one.  And so that's up there.  But here, exactly the same formal looking structure is not telling us  total probability, it's telling us how many particles we have.  So if that's true, then what corresponds to the probability density at a point of space?  The probability density at a point of space.  Why not just the density of particles at a point of space?  If the probability density, you want a function of all of the x's of all the particles, right?  This would be the kind of function.  Right.  Right.  Remember, this is not, okay, the answer to the question is this thing here is the density of particles.  Okay.  So we can write then that why, because its integral is the total number of particles.  Now, I'm not going to write that it's the probability density.  I'm going to write that it's the density.  But the density is not necessarily definite.  Remember, we're doing quantum mechanics.  A thing can be unobservable, but there are states, generally, typical states,  the thing is not going to have a definite value.  You can measure it.  You can measure the density of particles.  Strictly speaking, you wouldn't measure the density of particles at a point.  What would you do?  You'd integrate over a little volume and ask how many particles are in that volume.  So if I integrate this over a small volume, not over all space,  that tells me how many particles are in that small volume, I can measure it.  I won't always get the same answer.  I measure it in one state.  I measure it in another state.  It's like measuring any other quantum mechanical variable.  You can get a variety of answers.  But this will tell you what the number of particles is in a small volume.  That's its role.  So it's a kind of density.  It is a density.  It's literally the density of particles.  I have a question.  Sit down on the operators, a sub i, dagger, and so on.  This formalism is that they only operate on the state vectors,  but not on the wave functions.  Not on the single particle wave functions.  Right.  Now, let's come back for a moment.  This is not the most general state that you can have.  This is one of a basis of states.  You can superpose these just like in any vector space.  You can add vectors.  You can add vectors with different numbers of particles.  You can add, even if you have the same number of particles,  you can add vectors in which there are five particles here,  seven particles here.  You can add that to another vector where the sevens and the fives are interchanged.  Seven and five is 12.  You could add it together with a vector with 12 particles,  or the same i.  There are many states and superpositions of states.  Many, many possibilities.  In any part, whatever the state is, though,  the operator representing the number of particles in a small volume  is psi star psi.  It's a thing that you measure.  It's a thing that you measure when you look in that small volume  and you ask how many particles are there.  So, although I haven't told you how to measure psi itself,  psi star psi is a familiar thing.  It's just a number of particles and a little volume.  If you divide this operator at the top of there by n,  by the number of particles,  it would be fair to interpret it as a probability,  a finding of particles at any particle that's voids x to the same.  It's about the link on your face that's like,  yes, but you don't find it.  No, it's okay.  Well, you could say this volume has a certain fraction of that volume.  Yeah.  That's not the point.  Supposing you have a lot of particles, even just classically,  you have a lot of particles and you want to know  what's the probability that any one particle is in a little volume.  Well, to a high approximation, the number of particles are right.  All you have to do is one measurement.  Measure how many particles are in that small volume  and divide by the total number.  That's not the definition of the probability.  And you can get, you know, you could accidentally get zero.  You could accidentally get too many.  But one of the times within a margin of error,  you will get the number of particles being the probability for a particle to be in there.  The number of particles divided by the number.  All right, so it's not an identity to say that this thing,  appropriately divided by the total.  It's not an identity to say that it is the probability for a particle to be in that region.  The probability can be a half, two regions of space.  Probability a half here and a half here and you have a million particles.  You could accidentally discover that all the particles are on one side.  Not very likely, but you could discover it.  Okay, you make that measurement and the answer is a million on one side,  none on the other side.  Not equal to the probabilities.  But most of the time within the margin of error,  within about a square root of the number of particles,  you'll get the probability.  So that's the sense in which I was a little waffling  on whether you want to call it the probability or not.  This thing really stands for the observable,  which really corresponds to the number of particles that you actually get when you look in there.  However, yeah.  Yeah, that's a bad thing to do.  Right, but what you could do, no, that normally doesn't occur,  but what you could do is have a superposition of states  in which there's some number of electrons distributed throughout the different states  and add that to some number which is distributed differently, same number of electrons.  Photons, however, because they don't have charge,  you have no problem superposing different numbers of photons.  Now, even with electrons, you can, don't forget there are other species of particles in the world.  In particular, there are positrons.  So you can superpose states with different number of electrons and positrons  as long as the total charge adds up to the same number.  Okay, but right now we're just inventing for ourselves the concept of a quantum field.  All right, let me do one more little manipulation to show you another example  of things that you can build.  I was going to start working out the expression, the operator,  made out of a size which stands for the energy of the whole system.  We probably won't get quite there tonight, but let's begin.  Here's what we found.  We found that the integral of x of psi, dagger psi,  is the same as the sum of the number of particles in each,  that was what these little manipulations did.  Told us that this is the same as this.  Now, let's do a slightly different thing.  Let's start with a formula for the total energy.  Now, I'm interested now only in systems with independent particles.  Particles are not interacting with each other.  They're not exactly in approximation.  It tends to work when the particles are pretty dilute.  Even photons, if you make them dense enough,  will start to interact with each other.  And then the energy will not just be the sum of the energies of the photons.  Now, how dense do you have to make them plenty dense?  Much denser than you can make in the laboratory.  But still, in principle, energy is not equal to the sums of the energies of all the particles in the system.  But typically, if the particles are not too dense,  and there aren't too many of them per unit volume,  it's generally a good approximation.  So let's take that approximation and think of it now as a principle of free particles.  Free particles, particles that don't interact with each other.  Then what is the energy?  The energy is the sum of all modes.  Mode means one of these single particle states.  It's the terminology, mode.  Mode means a single particle state.  Sum of i of the number of particles in that state times the energy associated with that state.  We called it last time omega i.  We could call it omega h bar i,  but I'm going to leave the h bar out for now.  I'll put it back tomorrow.  Omega i and energy are the same thing.  I just used omega because we're working with oscillators,  but it is just the energy of the i-th mode.  Okay, this is the energy of, we can also write that as the sum of i of a dagger i, a i times omega i.  It looks a lot like this over here.  The only difference is that each i is weighted with a factor of omega i.  What I want to do and I won't do it tonight is to go from an expression like this  back to an expression like this, except for this beast over here.  What we're going to get is something involving the field operators,  and that thing that involves a field operator is going to be the energy of the fields.  It's going to be the energy, be the sum of all the energies of all the particles,  but it's going to be something that's expressed in terms of the field operators,  and we'll just call it the energy of the field.  That's going to be something reasonably familiar, but we won't do it tonight.  Tonight, yeah.  The part of the section was last week.  Oh, I thought it was a...  It did go fast.  It did go fast.  That was...  Oh, right, right, right.  Time is not invariant. What's happened to the energy?  Okay.  Okay.  I was wondering about that.  Okay, we'll do a little more then.  Yeah.  The omega i is the function of position.  No.  They're just the energies of the i-state.  The wave functions are functions of position.  That's a function of position, but the energy that's associated with it is just the energy of the first of the ground state of the system,  of the ground state of the one particle state.  No, the omegas are not functions of position.  They're functions of i.  They're functions of which energy level we're talking about.  Okay.  Okay, so what are the omega i's?  The omega i's are the energy levels which are the solutions of the Schrödinger equation, the time-independent Schrödinger equation.  To find the energy levels of a system, you solve the Schrödinger equation,  and you find out which eigenvalues on the right-hand side of the time-independent Schrödinger equation give you solutions.  All right, so let's write out the time-independent Schrödinger equation.  Now we're going to write an equation which determines the omegas.  Okay.  Are you saying that there are or not interactions?  No, no interactions.  You weren't listening.  Three particles, no interaction.  Okay.  Yes, we are.  No, we are not assuming there are interactions.  Yes, we are assuming there are no interactions.  Or at least that we're in a situation where interactions are very unimportant.  Particles too far from each other.  Okay.  All right, so what do we know about omega?  Omega is a solution of the time-independent Schrödinger equation.  Let's write down the Schrödinger equation.  The Schrödinger equation or the time-independent Schrödinger equation.  Of course, it's just now we're talking about the one particle Schrödinger equation.  The Schrödinger equation for only one particle.  And what does it look like?  Well, it's the one particle Hamiltonian acting on a wave function is equal.  Some psi i and it's equal to the eigenvalue omega.  That's the energy level times psi i.  And what's the Hamiltonian?  The Hamiltonian is, let's just take the simplest thing, p squared over 2m plus a potential energy.  The particles may be interacting with a potential energy,  but we're not going to let them interact with each other.  We're not going to be electrons orbiting the sun.  We're not treating the sun quantum mechanically.  We are treating the electrons quantum mechanically.  But we're taking the approximation that the electrons don't interact with each other.  They only interact with the sun.  So what do we write down for h?  h is potential kinetic energy, p squared over 2m plus v of x,  is potential energy.  Let me write it a little clearer.  Multiplying psi of x equals omega i,  and this is the i-th psi of x,  the i-th energy level is omega psi i of x.  Now, this is an odd thing here.  What do I mean p squared over 2m?  We know what v of x times psi of x means.  That just means multiply psi of x by v of x,  where v is some potential energy.  But what does this thing mean?  p squared over 2m.  We use p equals minus i d by dx.  And you can, as homework, put in the h-bars.  This is, of course, a component for each dimension of space.  So this becomes px squared plus py squared plus pz squared.  This thing here becomes, well,  if I want to write it in three-dimensional language,  it becomes del squared, which means second derivative  with respect to x plus with respect to y with respect to z.  A minus sign because of the i-squares,  i times i gives us a minus sign, and we divide it by 2m.  Now, this is the thing we know what to do with this on psi.  It just differentiates psi twice.  It's just the sum of the second derivatives in the various  directions of space.  Plus v of x, psi of x, plus v of x,  let's put it all in one bracket, plus v of x,  psi i equals omega psi i, omega i psi i.  Omega depends on which energy level we're talking about.  Well, we're not going to solve any Schrodinger equations  tonight.  We're just going to take omega i times psi i, which notice,  let's see, it hasn't appeared yet.  We're just going to use the formula without solving it.  I think the easiest way to get there is to write down the answer  and then prove that the answer is the same as this.  It's going to be the easiest way to get there.  So let's put this up on top.  I'm going to write down the answer,  and then we're going to prove it's the same as that.  OK.  Before I do, let me point out something about ordinary  quantum mechanics, one particle quantum mechanics.  Suppose that I wrote down the expression integral dx,  psi star of x, any wave function, whatever, times minus  del squared over 2m plus v of x psi of x.  What would this thing be?  Anybody know?  It's the expectation value of the energy.  This is h, and this is basically sandwiching h  between bra and ket vectors.  It's the expectation value of the energy.  So here's a formula for the expectation value of the energy  of one particle.  What I'm going to tell you is that the actual operator  representing the Hamiltonian of the many particle system,  can anybody guess from this what the answer is?  What's that?  Well, no, we know that.  But I'm looking for another expression for the same thing,  and I want to guess at this point and then show it's the same  as that.  What's that?  Yeah, we can do something like that, but I can make a guess.  What did we do to go from the probability to find a particle  to get to the number of particles in a small region?  Indeed.  We just replaced the little size by the big size.  And that sort of gave us the sum over all the particles of sort  of the same question.  So let's try.  Here's a guess.  No, it's not a guess for me.  It would be a guess for you except, of course, it's not a guess.  I remember learning about this.  When did I learn about it?  19, probably 1961.  And it was kind of blown away.  This, I'm reading Dirac's book, I think.  Reading Dirac's book, this blew me away that this is the same as this.  Let's prove it.  Let's go through the steps.  The first thing to do is to replace psi by the definitions up there.  So let's replace psi by the definitions.  This becomes integral dx.  Now, this one here is sum over i ai dagger psi i star of x.  That's this one over here.  And this one over here is sum over j psi j of x times a j.  Everybody happy with that?  A minus j, I guess.  That's all I did.  I plugged in the formula.  This can be written as a sum over i and j and replaced the second  sum, summation sign there, just get rid of it.  And now sandwich this minus del squared over 2m plus v of x.  Now, what the devil does this mean for this to act on this object here?  What does it mean?  Well, it simply means that you do these operations on ordinary wave functions.  This is an ordinary wave function.  We know how to differentiate it.  We know how to multiply it by v of x.  And so we have an operation that we know how to do.  We might not be able to do it because it might be too complicated,  but we can at least define the operation.  This differentiates psi twice.  This multiplies by v.  But we already know, at least I hope we know,  somewhere on the blackboard here.  Nope.  Schrodinger equation.  Here's the Schrodinger equation.  The Schrodinger equation tells us exactly what this is.  Minus del squared over 2m plus v of x acting on psi just gives us omega i times psi.  So we can erase.  I don't like erasing.  I won't erase.  We can write this integral dx sum over i and j, a plus i.  Let's group it together with a minus j over there.  Now there's psi i star of x.  Now there's this beast multiplying psi.  That's just omega j psi of j.  What's the next step?  Integrate over x.  Integrate over x.  When you integrate over x psi star of x i times psi j,  that's just kernicah delta.  That's just equal to sum over i and j.  Plus i, a minus j, delta ij.  But with one new factor that we didn't have when we did the probability,  or when we did just the counting of particles, omega j.  And that's exactly this.  We have a lovely expression now.  This is the energy of a field.  Even in terms of something quadratic in the field with a certain differential  operator, what can you compare that with?  An electrodynamics.  An electrodynamics, very, very similar kinds of things.  Or in any field theory, you take the fields, you differentiate them a couple of  times, make a quadratic expression, and integrate over space.  And that's a characteristic formula for the energy of a field.  All field theories have that property that the energy is equal typically to  something quadratic in the field with a couple of derivatives acting on the  field.  The derivatives act on psi here.  So this is an example of the energy of a field, but it's also the Hamiltonian.  These things are operators.  The left-hand side is an operator.  It's not a number.  It's an operator.  What operator?  Well, it's a thing which has to do with the energy.  It's the Hamiltonian of the field theory.  It's the Hamiltonian of the field theory.  But it's also the energy.  I went through every single step.  I belabored the steps.  It's a function of position.  It's an observable, or at least psi plus psi dagger is a quantum mechanical  observable.  If you measure it at one place, you can measure it at the other place at the  same time.  So what more do you want?  It's a function of position.  It's an observable quantity that the...  So we're going to define this thing because of these properties as we might  call it a field.  I mean, mathematically, it is a field.  That's what we're saying.  You can measure it.  It depends on position.  There is no other definition.  Now, you could ask, what field?  Is it the electromagnetic field?  No, it's not the electromagnetic field.  It stands in the same relationship to the electromagnetic field as photons would  be in a certain relationship to the electromagnetic field.  Whatever kind of particles we're talking about here, these are nonrelativistic  particles.  They have a mass.  They could be protons or neutrons moving slowly.  They could be electrons moving slowly.  As long as they're not relativistic, as long as the energy is given by p squared  over 2m, this field is the field for those particles.  Those particles are the quanta of that field in the same precise sense that  photons are the quanta of the electromagnetic field.  The electromagnetic field would have an expression for the energy.  It also has two derivatives, a field squared, and it's quite similar, although  it's not identical.  This one here?  Right?  OK, so we've invented a quantum field.  We've invented two observables.  One of them is the particle density.  The other is now.  We have the energy, but the energy is the integral of a space of something.  What would you call a thing that's being integrated?  The energy density.  It's the energy density.  So what's in here?  Let's get another color.  What's in here is the energy density, the local energy density.  By measuring it or integrating it over a small volume, you find out how much  energy is in that small volume.  So it's not the integrated overall space?  You could integrate overall space, and it's the total energy.  That's this.  But if we have something and the total energy is equal to an integral of it,  it stands to reason, or we can call that thing, or we might,  it's a definition of the energy density.  It's not the only definition. There's a little bit of ambiguity, but...  I have a question, please.  In that formula, e equals sum n i to omega i, capital N,  how is it different from the little n's?  The little n's are the eigenvalues of a big n.  Big n is an operator, the little n's are just numbers.  I have a question in syntax. Can you express that field equation in terms of...  Well, it's not a field equation yet.  But in terms of Dirac notation?  Because it's very analogous to what you have on the upper board,  which you can express in the Dirac's Karat notation, whatever it's called.  Brakin kets?  Yeah, the brakin, yeah.  No, no, no, no, but these are not states.  These are operators.  They're not brakin kets.  The brakin kets, these are the brakin kets of the system.  Robert, but you expanded the...  to do the little n's that's really sort of an elaboration of it.  It's not traditionally what brakin notation is, really.  Brakin notation is a notation for a state vector of a system.  That's a state, that's a state.  It's a state with n-1 particles in the first state, then two...  Yeah, you have to learn to separate the single particle quantum mechanics  from this multi-particle quantum mechanics.  In this multi-particle quantum mechanics, there are things which are operators  which don't even occur in the single particle.  The...  Right.  Now, this formalism or this technology does allow you to recover everything  that we've ever said about single particle quantum mechanics.  You just go to the states in which the total number of particles is one.  And you work in that set of states.  All of this formalism just reproduces the physics of the single particle quantum mechanics.  But it allows you to deal with any number of particles.  And in particular, there are states...  there are states with large numbers of particles, typically.  Remember, these are observables now.  They're operators.  When there are a large number of particles in the same state,  then these size tend to behave very much like classical fields.  And we'll talk about that a little more.  How is this connected to classical field theory?  Well, it's connected to classical theory like any quantum mechanics  is connected to its classical counterpart.  For example, expectation values of things  follow the same kinds of equations of motion as the classical equations of motion.  So you could ask, for example, what happens to the expectation values of psi?  Psi is an operator. It has an expectation value at any time.  We know the Hamiltonian of the system.  We can use it to work out the time dependence of the state.  And then we can use that to work out the time dependence of the expectation value of psi.  Guess what?  It essentially satisfies the classical equations of motion of wave fields  with the same classical Hamiltonian.  In fact, it satisfies the time-dependent Schrodinger equation.  So, as I said, this is a very beautiful formalism.  It is a bookkeeping device.  But it's also the fundamental connection between particles and wave fields.  And there is no other connection. This is it.  That's the connection.  So let's explore it in a great deal of exploration.  Okay.  I've been struggling with the manipulation of the creation and annihilation operators inside that integral.  You seem to treat them almost like numbers.  No, I never switched them.  I never commuted them.  I commuted them and switched that order.  I know that they depend on the way some of them are.  Yeah.  They're numbers. They are numbers. Here they are.  I'm sorry.  The size of numbers, the a's are operators.  But we need, okay.  So if the a operates on the a's, say the last pic,  when a j minus operates on side j, it doesn't operate on side j.  No, no. It may operate on some n1, n2, n3, and so forth.  It took advantage of the fact that the size are numbers.  And because they're numbers, you can pull out the whole delta function.  And then you're just left with the a's and a's.  The only manipulation that you wouldn't want to do freely without thinking about it  is to move an a minus past an a plus.  They don't commute with each other.  That's where the operator character of them would come in.  But as far as moving them past ordinary numbers, that's fine.  The derivatives here don't act on them.  The derivatives act on the size.  So there's no cause to worry about any of the operations I did.  There would be a cause to worry if I tried to move a minus past a plus.  But we did interchange the i minus i plus, but it introduced a plus 1 in certain things that you did.  Oh, yeah. We just have to be careful about it.  That's all. Yeah.  I don't think you play some funny game with the derivative of psi j.  But these are only derivatives acting on ordinary functions.  I'm just saying we sort of pulled it all out into a delta.  That's the integral over x.  Right.  Yeah.  So I mean, I didn't say it wasn't valid.  I was just saying that's a little tricky.  That's a little tricky.  Well, we did.  We couldn't do the integral over x in this form.  What we did was say that this operation here just multiplies psi j by omega.  Once it became omega, then we could do the integral over x.  Yeah.  Yeah.  All right.  So I didn't cheat you.  I didn't do anything that-  Just a number.  So you- in Schrodinger's formalism, say, cheat operators as things that can operate on functions,  on the wave functions.  What's that again?  In Schrodinger's formalism, operators can operate on wave functions.  Right.  In this form, it was in these operators that just operate on state vectors and not on wave.  Right.  You can imagine how confused they were in the early days before Dirac mastered all of this  and put it together.  They were very confused about the meaning of the wave functions.  Some people thought it was a field that you- you know, a real field, like the electromagnetic field.  Some people thought it was just an auxiliary device for computing probabilities.  And if that's all it was, then what does it have to do with the electromagnetic field?  And what's the connect- there were several different ideas about what the connections between particles and waves were.  On the one hand, the single particle wave functions satisfied a wave equation, the Schrodinger equation.  Look, sort of suspiciously like Maxwell's equations.  It does, roughly speaking, not quite.  It looks like Maxwell's equations.  Maxwell's equations were not equations as far as anybody knew for quantum amplitudes.  They were equations for measurable values of the field.  But still, there was some connection, some funny connection between these fields and particles.  Namely, the field was made out of quanta called photons.  So there was a lot of confusion.  And this is the resolution of that confusion.  That don't-  I said for them, maybe.  What did you help to-  The question?  The same letters.  No, no, this is it. That's it. That's all there is.  Did you actually have some background in over its faces, vector's faces?  No. I don't think so.  So you just had to-  No, he made it all up himself.  Which is fortunate, because had he used the notations that were there already, you know?  I taught quantum mechanics last year for them in the mathematics department, which was a lot of fun.  There was a crazy point at which I was explaining Dirac notation.  And I told them that every vector, we call these a vector space, or whatever, psi.  And the official term, sort of the ket, and the official mathematical term for the bras is the dual vector space.  It means complex conjugate essentially, the dual vector space.  And I learned to say dual, that made them happy.  And then I said, there's an inner product between every vector and its dual.  And they said, no, no, no, no, no, no, you can't do that.  You cannot have an inner product between a vector and its dual.  You can only have an inner product between a vector and itself.  And I said, look, I'm telling you, there's a- just, trust me, I know what I'm talking about.  You can have an inner product between a vector-  No, no, no, no, illegal, not allowed, not allowed, not allowed.  Okay? Well, of course, it turned out that they just have a notation for the inner product of two vectors.  Take two vectors, and you put them next to each other this way,  and then you put some funny symbol around them like this.  And that's the inner product between psi and phi.  What is it equal to? It's equal to psi-phi.  When they finally got the hang of this, the mathematics professors all said, wow, this is powerful.  It makes everything work simply.  For example, they were again blown away by this formula that ii is equal to one.  And all you have to do to multiply it by a state is just do what it says, psi.  And this is the inner product. They've never seen that before.  Of course, they saw it. Of course, they've seen it.  They understood it all completely, but they had never seen it in this compact notation where you almost can't make mistakes.  Anything you do turns out to be right.  So Dirac was clever.  He also, if you ask, was his mind an ordinary mind, it was not.  He was probably autistic or something close to it.  So he saw things in different patterns than most of us do.  Going back to what you were saying a little while ago about this elegant formalism,  it's sort of begging the question for me, are fields just formalism?  No, you can measure them.  But are you just measuring properties of collections of particles?  Yes, but you're measuring properties that you might not have guessed.  No, you're not measuring the properties of classical particles.  You're measuring the properties of quantum particles.  And quantum particles have properties like interference and that the classical particles don't.  So yes, you're only dealing with a system of particles, that's all.  But because they're quantum mechanical, they have kinds of properties, interference,  entanglement and other things which ordinary particles don't have, classical particles.  And this accounts for all of it.  Is there some sort of identification between, for example, let's say the density of photons and the electrical field strength?  Yeah, well, one has to be a little bit careful.  When you're dealing with relativistic particles, and I've been careful not to deal with relativistic particles,  only non-relativistic, when you're dealing with relativistic particles,  the notion of the momentum of a particle is very well defined.  The notion of a position of a particle is not.  And if the notion of the position of a particle is not defined,  then the notion of the density of particles is not defined.  What is defined is the density of energy.  And the density of energy has a similar form.  But if we don't split too many hairs, then the square of the electric field is qualitatively related to the density of photons.  So in relativity, you have a defined momentum, but not so much with position, is that what you're saying?  Yeah, position is not a good concept in relativistic systems.  Yeah, you can even make, if you define, this is not a special useful, but it's a fact,  if you define a vector psi out of the electromagnetic field,  you define it as the electric field plus i times the magnetic field.  Take that quantity, psi star.  Now there are three components to this, so it's not quite the same as this.  Three components, electric field minus i times the magnetic field.  Then psi star psi is not quite the density of photons, it's the energy density.  So it's the density of photons weighted with the energy of each photon.  That's a good concept for photons.  The density of photons itself is not a good concept.  So psi star psi is e squared plus b squared.  If you know any electrodynamics, you know that e squared plus b squared is the energy density.  And psi satisfies a Schrodinger-like equation.  It's fun to work out from Maxwell's equations what psi, and this is psi dagger.  If you know Maxwell's equations, you can work out the equations for psi.  It's a simple equation, simple than Maxwell's equations themselves.  I have a question.  Energy equals the integral of capital Phi dagger phi.  No.  I'm just deviating that.  Yeah, yeah, psi dagger confirms this differential operator here at times psi.  If you switch the psi and the psi dagger, does that have any meaning?  It's clearly not equal to this.  You're not allowed to pass them through each other without taking care of the commutator.  So it has a meaning, but it's what it is.  It corresponds to an extra additive constant, the ground state energy.  Let's see if we can make a guess.  I'm not going to try to prove this.  It's easier than what I did here, and you can try it.  For the single particle quantum mechanics, momentum P is equal to minus i d by dx.  There's some Planck's constant in there.  What's the expectation value?  Supposing I have a single particle wave function psi of x, what's the average or the expectation value of the momentum?  Well, you take the ket vector psi, you take the momentum operator, and you sandwich it,  the usual sandwich to create the expectation value.  That means is you integrate over x, the x, psi star, times minus i d psi by dx.  So let's put the minus i on the outside here.  Psi star times d psi by dx, and you integrate it over all space.  What about the momentum operator?  Not the expectation value of the momentum, but the momentum operator for this field theory.  Do fields have momentum?  Fields have momentum.  If you take the electromagnetic beam from the laser and you shine it on the wall,  if the laser beam is powerful enough, it will blow a hole in the wall.  Why?  Because it carries a lot of momentum.  It exerts a force on the wall.  Fields have momentum.  Classical fields, quantum fields, in the quantum case, the field momentum is just the momentum  of all the photons.  So here we have an expression for the average momentum of a particle in quantum mechanics.  Can you guess what the expression for the total momentum of all the particles is?  Yeah.  Yeah?  What?  You changed operator.  Just replace the size by size of bigger size.  Sci.  This is the operator that represents the momentum.  If you measure the momentum, how do you measure the momentum?  You see how big a hole you blow in the wall by, right?  You measure the momentum, not the average of the momentum, the actual momentum in a wave  going by you.  How do you measure it?  You see how hard it hits you, how much force it exerts on something.  This is the operator, the observable that corresponds to that, this is the mathematical  Hermitian operator that corresponds to that observable.  You can also calculate its expectation value.  This is an observable.  It is not the expectation value.  If you give me a state, I can calculate the expectation.  This is something we'll do next week.  These are operators.  They have expectation values.  They are not the expectation values.  You need a state.  You need an operator and a state to construct an expectation value.  Where's a state?  Give me a state.  Well, you could take those in states up there, in one and two and three, and calculate the  expectation value of the momentum.  Okay.  I think we'll finish for tonight unless somebody has some good questions because I've, I've  I've, I have not run out of things to talk about, but it's a natural breaking point.  Next time we'll talk about processes in particle physics and how they're represented by these  fields in a very simple context.  In a very simple context, we'll talk about how field operators implement processes that  take place when particles decay, when particles scatter, and so forth.  For more, please visit us at stanford.edu.
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