 So, the third possibility of quantum mechanics has told us that the allowed values of any observable property are the eigenvalues of the operator associated with that property. That doesn't quite answer the question, what value we're actually going to see when we make a measurement. And that question is a little bit difficult to answer because quantum mechanics is inherently probabilistic. It's not deterministic. So, what I mean by that is, for example, if we remember for the one-dimensional particle in a box where the wave function, let's say, let's say in particular wave function number one. So, our normalized wave function looks like sine of n pi x over a with n equal to one. So, the wave function itself looks like a sine wave. The probability of finding the particle at any particular position is psi squared. The wave function times its complex conjugate. So, the square of this wave function is sine squared pi x over a. And graphically, what that looks like. So, if I plot probability or psi squared, the graph of sine squared looks like this. It's zero. One x is equal to zero. It's also zero when x is equal to a. And it goes through a maximum at the middle when x is equal to a over two. So, this is the position of the particle. So, since we know probabilities related to wave function squared, all we know about the particle is the probability of finding it at any particular location. For a particle in the size of one wave function, in the ground state wave function for the particle in a box, it's most likely to be found at the middle of the box. The probability is highest that it's found in the middle of the box, but it's got some non-zero probability of being found here or being found here or being found here. The only place is we're not going to find it where the probability is zero is at the edges of the box or outside of the box where the probability is zero. So, all we know about the particle is its probability that it's found at different locations. I can't say definitively that it necessarily will be found at a over two or a over four at any particular value. It has some probability of being measured to be at any particular point within the box. I mean, when I say quantum mechanics is probabilistic, often all we know is the probability of finding a particle at certain locations or with certain properties. But that doesn't mean we don't know anything about the value that we'll measure when we measure a particular property. And that's the subject of the fourth postulate of quantum mechanics, which says, if I have a system with a particular wave function, if this is the wave function of the system, and I want to know the value of a particular property, turns out what I can predict is something called the expectation value, which is really just another word for the mean or the average. So if I have some property A that I'm interested in, maybe that property is energy, position, momentum, whatever it is. We know that property has some operator associated with it. And if I want to know the average value that I'll get after making multiple measurements, and this is how we write the expectation value of the average, you might be perhaps more familiar with thinking of an average as an A with a bar on top of it. I mean exactly the same thing when I say A with these angle brackets around the side of it. The average or the expectation value of that property can be calculated by taking the operator acting on the wave function with it, pre-multiplying by the wave function with the complex conjugate. So I've sandwiched the operator in between the wave function twice here. If I integrate that over everywhere, the answer to that question, doing that integral will give me a number. And that number will be the average value of the property I expect when I make multiple measurements of the particular property. So if I ask if the property is, where is the position of the particle? On average, it's going to be in the middle of the box. Sometimes it's on the left side of the box, sometimes it's on the right side of the box. On average, I expect to find it in the middle. Any one measurement is going to be unpredictable, but if I make a number of measurements one after the other, I can predict what the average of those measurements is going to be. So we can work some examples and make sure we understand how that works. So notice, in order to calculate an expectation value, I need to know what property I'm interested in, what the operator of that property is, and I need to know which wave function the system is occupying. So if I'm interested in the average value for the position, this exact question right here, average value of the position when the wave function is in the ground state wave function, size of one, then the fourth postulate tells us how to calculate that. We need to calculate the wave function times the operator associated with the position. Remember that operator is just the x times operator. So x times, so I've got wave function being acted on by the position operator, which is just multiplication by x, also another copy of the wave function here. So writing out what that is equal to, the wave functions each look like this wave function here, a sine of pi x over a with a normalization constant. The x operator is just multiply by x, and I've forgotten the normalization constant here, square root of 2 over a. So I want to do that integral over x. I want to do the integral over everywhere the particles allowed to be from 0 all the way up to a. So we'll actually not complete this integral. We can simplify it a little bit. That would be 2 over a integral from 0 to a of x sine squared of pi x over a. We can certainly do that integral. It's possible to do that integral. The trick would be using the double-angle formula and then using integration by parts. But for now, we'll leave that as something you can work on on your own or look up in an integral table. But if you complete that integral, you won't be surprised for now to see that the answer to that question is a over 2. The average value of where I find the particle, if I do this integral, works out to be a over 2 in the middle of the box. Again, by symmetry, we knew that was going to be the answer because it's found on the left as often as on the right. And this particular integral gives us a numerical answer. We can work another example where we'll actually compute the integral. So let's ask ourselves what would be the average for something that we can't guess the answer to quite as easily. And that would be what is the average moment if I have a particle in this box. The wave function itself doesn't look like it tells me very much about the momentum of the particle. It tells us directly about the probability of where I'm going to find the particle, but it doesn't tell me with that little more work what the momentum of the particle is going to be. But the fourth postulate tells us the momentum on average, the expectation value of the momentum, is again going to be the wave function sandwiched around the operator associated with that property. So the operator associated with the momentum, remember, is minus i h over 2 pi times the derivative. And that's the operator that acts on the wave function. All right, so now we have a slightly different integral to do instead of x times the wave function twice. I have some constants times the derivative of the wave function, times the original wave function. Derivative of sine is going to give me a cosine, and it's going to pull out some constants. So what this is going to look like if I collect the powers of 2 over a. Derivative of sine pi x over a gives me a pi over a times a cosine. So and then I've also got these constants minus i h over 2 pi. Those all multiply my integral, which now looks like sine of pi x over a cosine of pi x over a. So now we have that integral to perform, integral of sine times cosine. There's two ways we can think about solving that integral. One, if we could just think about solving it directly, the technique to use here would be u substitution. So in this next step, we're going to say if u is sine of pi x over a, then du, the differential of that would be cosine pi x over a dx. And again, that derivative pulls out a factor of pi over a. So that's what we can use to solve that integral. Or a simpler step would be to use symmetry. If you're comfortable looking at this and thinking of the symmetry of the problem, a sine function is symmetric within this integral interval, 0 to a. The cosine function, on the other hand, starts positive and goes negative, is asymmetric about the middle of that integral. So we can look ahead and predict that the answer to this integral is going to be 0 by symmetry, but we'll go ahead and work it out. So the result of the u substitution, so the pi over a from the du, I'm going to pull out of the integral. So dx is equal to a over pi du. So I'm left with the integral of sine pi x over a, which I'm now calling u, cosine pi x over a, which I'm calling du. So a over pi du is equal to cosine pi x over a dx. So I've replaced this with this. And I have the integral of u du to perform. But we have to also remember the limits of the integral x. And this integral goes from 0 to a. And this integral is u goes from different values. So u is equal to sine of pi x over a. So I have to substitute the limits into our u substitution. So when x is equal to 0, u is equal to the sine of 0, which is 0. When x is equal to a, u is equal to the sine of pi a over a. So the sine of pi, and that's also 0. So regardless of what this interior of this integral works out to, it's 1 half u squared. But it's evaluated between 0 and 0. And it's multiplied by all these constants out front. 2 over a pi over a a over pi. And these negative constants minus i h over 2 pi. So we could cancel lots of things in there and make it a little bit simpler. But the end result is when I evaluate u squared between 0 and 0, that all comes about equal to 0. And again, that's what we could have expected with a little less work by noticing the symmetry of the integral of the sine times cosine over this interval to begin with. So that's the details of how the math is performed. What we've discovered is the average momentum of the particle for a particle in a box is 0 when it's in the psi 1 wave function. Why is that true? And it's precisely because a particle in this one-dimensional particle in a box, psi sub 1 wave function, is going to be moving to the left as often as it's moving to the right and vice versa. So the average momentum, when it's moving to the positive x direction with a positive momentum, it equally often is found moving in the negative direction with a negative momentum. And those values cancel each other out. So on average, the expectation value of the momentum is going to be 0. So that's the physical explanation for why the momentum turns out to be 0. Mathematically, we can just calculate the result and it works out to be 0. So those are two examples of how to use the expectation value, how to use this postulate to calculate the expectation value for a property. And what we've seen is that we can predict the average value of a property even when we can't necessarily predict the specific value that a property is going to have. I can't tell you exactly where the particle is, but I can tell you if I make lots of measurements on average where I find it. On average, I find it in the middle of the box. However, there's one exception to my claim that quantum mechanics is probabilistic and not deterministic. There's one case where we can tell exactly what the result of a measurement is going to be. And that's in a case that we can illustrate with this example. Let's take one more property, the energy. So let's ask for the same particle in a one-dimensional particle in a box with wave function size of 1. In fact, let's say it's not required that it be in size of 1. Let's say that it's in wave function size of n. And I want to know what is the average energy? Again, using the postulate, average energy is going to be integral of the wave function sandwiched around the operator associated with this property. And remember, the operator associated with the property of the energy is the Hamiltonian operator. Hamiltonian is the property that gives us the energy. So when I do this integral, I don't even have to worry about whether this is one-dimensional particle in a box or which particular n value it is. I know that if this is the wave function, h psi is equal to e psi. So Hamiltonian acting on a wave function psi sub n gives me the energy of that wave function e sub n multiplying psi sub n. This e sub n is not an operator anymore. It's just a constant, so I can pull it out of the integral. And then what's left in the integral is just the square of the wave function with one of the two being the complex conjugate. So we know if the wave function is normalized, that integral is equal to 1. So the answer here is just e sub n. The average value of the energy for a particle that's in a wave function psi sub n, because psi sub n is an eigenvalue of that operator, we get just the single value, the only value the energy is allowed to have is the energy that it has in that wave function. And so the average value, of course, is also that same value. And that's true not just for the energy, but also for any property. If I have a wave function that is an eigenvalue of the operator that I'm asking about, then the only possible value of the property is the eigenvalue. I'm sorry, if the wave function is an eigenfunction of that operator, the only possible measured value is that particular eigenvalue. And because the one-dimensional particle in a box functions are not eigenfunctions of the position operator or of the momentum operator, we have lots of different values. And all we can predict is the average value. But for the energy, the wave functions are eigenvalues of the Hamiltonian. So they'll always have that particular value. So this tells us what to do with the fourth postulate. And we have one more postulate to go. And that's what we'll cover in the next video.