Shall we begin now? So as usual, and as promised, I will tell you what we have done so far. And that's all you really need to know to follow what's going to happen next. The first thing we learned is, if you're studying a

particle living in one dimension, that's all I'm going to do the whole time, because it's mathematically easier and there's not many new things you gain by going to higher dimensions. There's a particle somewhere in this one-dimensional universe. Everything you need to know about that particle is contained

in a function called a wave function, and it's denoted by the symbol ψ. By the way, everything I'm doing now is called kinematics. In other words, kinematics is a study of how to describe a system completely at a given time. For example, in classical mechanics,

for a single particle, the complete description of that particle is given by two things, where is it and what its momentum is. Dynamics is the question of how this changes with time. If you knew all you could know about the particle now,

can you predict the future? By predict the future, we mean, can you tell me what the future is, and x will be at a later time and p will be at a later time. That's Newtonian mechanics. So the kinematics is just how much do you need to know at a given time? Just x and p. Once you've got x and p, everything follows. As I mentioned to you, the kinetic energy, for example, which you write as ½ mv

squared is given just in terms of p. Or in higher dimensions, the angular momentum is some cross product of position and momentum. So you can get everything of interest just by giving the position and momentum. I claim now the equivalent of this pair of numbers in this

quantum world is one function, ψ of x. So it's a lot more information than you had in classical physics. In classical physics, two numbers tell you the whole story. Quantum theory says, give me a whole function, and we all know a function is really an infinite amount of information, because at every point x,

the function has a height, and you've got to give me all that. Only then you've told me everything. That's the definition. And the point is ψ can be real, ψ can be complex. And sometimes ψ is complex. So we can ask, you've got this function. You say it tells me everything I can know. Well, what can I find out from this function?

The first thing is that if you took the absolute square of this function, that is the probability density to find it at the point x. By that I mean, if you multiply both by some infinitesimal Δx, that is the probability that the particle will be between x and x dx.

That means you take this ψ and you square it. So you'll get something that will go to 0 here, go up, go to 0, do something like that. This is ψ squared. That's your probability density. And what we mean by density is if the function p of x has an area with the x axis, like px dx,

that's the actual probability that if you look for this guy, you will find him or her or it in this interval, okay? So we will make the requirement that the total probability to find it anywhere add up to 1.

That is a convention because, well, in some sense, it's up to you how you want to define probability. You say, what are the odds I will get through this course? 50-50? Okay, that doesn't add up to 1.

That adds up to 100. But it gives you the impression the relative odds are equal. So you can always give odds. Jimmy the Greek may tell you something. 7 is to 4, something's going to happen. They don't add up to 1 either. It means 7 divided by 11 is one thing, and 4 divided by 11 is the absolute probability. So in quatern theory, the wave function you're given

need not necessarily have the property that its square integral is 1, but you can rescale it by a suitable number. I mean, if it's not 1, but if it's 100, then you divide it by 10, and that function will have a square integral of 1. That's a convention, and it's a convenience.

And I will generally assume that we have done that. And I also pointed out to you that the function ψ and the function 3 times ψ stand for the same situation in quantum mechanics. So this ψ is not like any other ψ. If ψ is a water wave, 3 inches versus 30 inches are not the same situation. They describe completely different things.

But in quantum mechanics, ψ and a multiple of ψ have the same physics, because the same relative odds are contained in them. So given 1 ψ, you're free to multiply it by any number, in fact, real or complex. That doesn't change any prediction. So normally, you multiply it by that number, which makes the square integral in all of space equal to

1. Such a function is said to be normalized. If it's normalized, the advantage is the square directly gives you the absolute probability density, and integral of that will give you 1. That's one thing we learned. Do you understand now? What are the possible functions I can ascribe with a particle?

Whatever you like. Within reason, it's got to be a single value, and it cannot have discontinuous jumps. Beyond that, anything you write down is fine. That's like saying, what are the allowed positions or allowed momentum for a particle in classical mechanics? Anything. There are no restrictions, except x should be real and

p should be real. You can do what you want. Similarly, all possible functions describe possible quantum states. It's called a quantum state. It's this crazy situation where you don't know where it is, and you give the odds by squaring ψ. That's called a quantum state, and it's given by a function ψ. All right, now I also said there's one case where I know

what's going on. So let me give you one other case. Maybe I will ask you to give me one case. The particle is known to be very close to x equal to 5, because I just saw it there,

and there's an ε later. I know it's still got to be there because I just saw it. Now what function will describe that situation? You guys know this. Want to guess? What will ψ look like so that the particle is almost

certainly near x equal to 5? Centered where? At 5. Everybody agree with that? I mean, the exact shape we don't know. Maybe that's why you're hesitating. But here is the possible function that describes a particle that's location is known to within some accuracy.

So one look at it, it tells you, hey, this guy's close to 5. I agree that you can put a few wiggles on it, or you can make it taller or shorter, change the shape a bit. But roughly speaking, here is what functions describing particles of reasonably well-known location

look like. They're centered at the point which is a well-known location. On the other hand, I'm going to call it ψx equal to 5. That is a function, and the subscript you put on the function is a name you give the function. We don't go to a party and say, hi, I'm human, right?

You say, I'm so-and-so, because that tells you a little more than whatever species you belong to. Similarly, these are all normalizable wave functions, but x equal to 5 is one member of this family, which means I'm peaked at x equal to 5. Another function I mentioned is the function ψp

of x. That's a function that describes a particle of momentum p. We sort of inferred that by doing the double slit experiment. That function looks like this, some number a times e to the ipx over h-bar.

Now, you can no longer tell me you have no feeling for these exponentials, because it's going to be all about the exponential. I've been warning you the whole term. Get used to those complex exponentials. It's got a real part. It's got an imaginary part, but much more natural to think of a complex number as having a modulus in the phase, and I'm telling you it's a constant modulus. I don't know what it is, but the phase factor should

look like ipx over h-bar. So if I wrote a function e to the i times 96x over h-bar, and I say, what's going on? Well, that's a particle whose momentum is 96. So the momentum is hidden in the function right in the exponential. It's everything except the i, the x and the h-bar,

whatever is sitting there. That's the momentum. I'm going to study such states pretty much all of today. So let's say someone says, look, I produced a particle in a state of momentum p. Here it is. Let's normalize this guy. To normalize the guy, you've got to take ψ squared, and you've got to take the

dx and you've got to get it equal to 1. If you take the absolute square of this, A absolute squared is some fixed number. I hope you all know the absolute value of that is 1, because that times its conjugate, which is e to the ipx over h-bar, will just give you e to the 0, which is 1.

I want 1 times dx over all of space to be equal to 1. That's a hopeless task, because you cannot pick an That's not going to happen, because all of space, the integral of dx over all of space is the length of the

universe you're living in, and if that's infinite, no finite A will do it. So that poses a mathematical challenge, and people circumvent it in many ways. One is to say, let's pretend our universe is large and finite. It may even be the case, because we don't know. And I'm doing quantum mechanics, which I'm fooling

around in a tiny region like atoms and molecules, and it really doesn't matter if the universe even goes beyond this room, right? It goes beyond this room, goes beyond the planet, goes beyond solar system. I grant you all that, but I say, allow me to believe that if it goes sufficiently far enough, it's a closed universe. So a closed universe is like this, a closed one-dimensional

universe. It's a circle. In that universe, if you throw a rock, it'll come back and hit you from behind. In fact, you can see the back of your head in the universe, because everything goes around in a circle, okay? All right, so that's the world you take.

Now, that looks kind of artificial for the real world, which we all agree seems to be miles and miles long. But I don't care for this purpose, what L is, as long as it's finite. If L is finite, any number you like, then the ψ of p of x will look like 1 over square root of L e to the ipx over h-bar.

Do you agree? If you take the absolute square of this, you'll get a 1 over L. This will become 1. Integral of 1 over L over the length of space is just 1. So that's the normalized wave function. Now, this is also a very realistic thing if in practice your particle is restricted to live in a circle.

Again, there are a lot of experiments being done, including at Yale, where there's a tiny metallic ring, a nanoscale object, and the electrons are forced to live in that ring. So the ring has a radius, and this L is just 2ΠR, where R is the radius.

There, L is very real. It's not something we cooked up. It's the size of the ring. But sometimes, even if you're not doing particle in a ring, if you're doing particle on a line, you just pretend that the line closes in on itself. If you start at the origin and you go on the two sides, it closes in. If this is x equal to 0, you go right and you go left, they all meet at the back, and it's a closed ring.

So let's imagine what life looks like for a particle forced to live in a ring of circumference L. The normalized wave function looked like this of momentum p, this 1 over square root of L,

e to the ipx over h-bar. The probability to find this guy at some point x, which the absolute value of Ψ squared is just 1 over L. That means the probability is constant over the entire ring. We don't know where it is. You can find it anywhere with equal probability.

It's always true that if you know the momentum, you don't know where it is. Okay, now let's ask one other question. Here is the circle in which I'm living. If you took the real part and imaginary part,

which are sines and cosines, they kind of oscillate. Do you understand that? When x varies, they oscillate. And one requirement you make is that if you go all the way around and come back, it's got to close in on itself, because if you increase x from anywhere by an amount L, which is going all the way

around the circle, you've got to come back to where you start. The function has to come back to where you started, meaning it's a single valued function. If you say, what's Ψ here? You've got to get one number. That means if you start at some point, you go on moving, you follow this Ψ, you go all the way around and you come back, you shouldn't have a mismatch. You should agree. So the only allowed functions are those obeying the condition

Ψ of x L equal to Ψ of x. Do you understand that? Take any point you like. Follow the function for a distance L. That means you come back and the function better come back. So a function like e to the minus x squared is not a good

function. You go around a circle and come back. If you go in real space, if you go on all the way, it doesn't come back to where you are. So you've got to write functions which have the property that when you add L to them, you come back where

you are. That's a condition of single valuedness. So that means in this function, if I take it at some point e to the ipx, at some point x, then I add to it L.

Namely, I'm getting e to the ipx over h-bar times e to the ipl over h-bar. That has to agree with the starting value, which is e to the ipx over h-bar. In other words, I'm comparing the wave function at the point x and the point x L.

See, if your world is infinite, this is x. This is x L, this is x 2L. They're all different points. What the function does here has nothing to do with what it does here. But if you wrapped out the region back into itself, what it does when you go a distance L is no longer independent of what it does at the starting point.

It has to be the same thing. Therefore, I'm just applying the test. I'm saying take the function at x L, factorize it this way, demand that it be equal to the function at x. These cancel and you learn then that e to the ipl over h-bar should be 1.

Now how can that be 1? Well, one way is p equal to 0. But I hope you know that it's not the only way. Do you know how else it can be 1? Should p be 0? Pardon me?

I didn't hear that. Sorry? I didn't hear that again. You still have to say it louder. Any other answers? Yeah? Is that what you said, multiple of 2Π?

Good. I'm glad you didn't say it loud enough. Okay, good. It's a multiple of 2Π, because any trigonometric function, if you add 2Π or 4Π or 6Π to its argument, doesn't change. And you should never forget the fact that this exponential is a sum of two trigonometric functions, cosine i sine, and they all come back.

Therefore, it is too strong to say p should be 0. So p should be such that pL over h-bar is 2Π times any integer, where the integer can be 0, plus minus 1, plus minus 2, etc.

That means p has to be equal to 2Π h-bar over L times m, where m is an integer. Positive, negative, 0.

Now this is a very big moment in your life. Why is it a big moment? Yes? You don't know? Well, if you haven't done it before, this is the first time you're able to deduce the quantization of a dynamical variable.

This is the first time you realize this is the quantum of quantum theory. The allowed momenta for this particle, you might think it can zip around at any speed it likes. It cannot, especially in a ring of nano proportions. These values of p are all discrete.

p times 2Π h-bar over L times 0 is here. Times 1 is here. Times 2 is here. 3 is here. Minus 1, minus 2. These are the only allowed values of p. That's the case of quantization. And the quantization came from demanding that the wave function

had a certain behavior that's mathematically required. The behavior in question is single valuedness. Now I can say this in another way. Let's write the same relationship in another way. So let me go here. So I had the allowed values of p are 2Π h-bar over L times m.

I can write it as p times L over 2Π is equal to mh-bar. L over 2Π is the radius of the circle. So I find that p times r is equal to mh-bar.

You guys know what p times r is for a particle moving in a circle, momentum p? What is the momentum times the radial distance to the center? Angular momentum. That's usually denoted by the quantity L in quantum mechanics. So angular momentum is an integral multiple of h-bar.

That's something you will find even in high school. People tell you angular momentum is an integral multiple of h-bar. Where does it come from? How does it come from quantum mechanics? Here is one simple context in which you can see that angular momentum is quantized to these values.

Now, what I will do quite often is to write the state ψ as ψ of p this way, e to the i px over h-bar. But to remind myself that p is quantized to be integer multiples of some basic quantum,

that the multiple is just m. I may also write the very same function as ψm. It's 1 over square root of L e to the ix over h-bar, but for p you put the other value, which is 2Πh-bar over L times m, and what do you get?

You get 1 over square root of L times e to the 2Πi mx over L. They're all oscillating exponentials, but you realize that the label p and the label m are

equally good. If I tell you what m is, you know what the momentum is because you just multiply by this number. So quite often I'll refer to this wave function by this label m, which is as good as the label p. The nice thing about the label m is that m ranges over all integers.

p is a little more complicated. p is also quantized, but the allowed values are not integers, but integers times this funny number. In the limit in which L is very, very large compared to h-bar, the spacing between the allowed values becomes very, very close, and you may not even realize that p is taking only discrete

values. So when you do a macroscopic problem where L is 1 meter, the spacing between one allowed p and another allowed p will be so small you won't notice it. So even if you lived on a real ring of circumference 1 meter, the momenta that you'll find in the particle will look like any momentum is possible. That's because the allowed values of p are so densely close, packed, that you don't know whether

you got this one or that one or something in between, because you cannot measure it that well. So this will smoothly go into the classical world of all allowed momenta if L becomes macroscopic. And the notion of macroscopic is how big is big? Well, it should be comparable to h-bar. I mean, it should be much bigger than h-bar,

then it'll become continuous. Okay, so now I'm going to ask the following question. If I have a particle in this world, in this one-dimensional ring, and I plot the wave function,

some function, ψ of x. Suppose it's not one of these functions. It's not e to the ipx over h-bar. It's some random thing I wrote here. Of course, it meets itself when you go around a circle.

That's a periodic function. Be very careful. A periodic function doesn't mean it oscillates with a period. In this case, periodic means when I go around a loop, it comes back to a starting value. It doesn't do something like this where the two don't match. That's what I mean by periodic. It doesn't mean it's a nice oscillatory function. These guys are periodic and oscillatory with a period.

These are periodic only in the sense that if you go around the ring, you come back to the starting value of the function. So I give you some function like this, and I ask you, what's going on with this guy? What can you say about the particle? So can you tell me anything now, given this function ψ of x?

Does it tell you any information? You mean if I draw the function like that, you get no information from it? Is that what you're saying? You must know. Yes? Yes, look, if you knew that, you've got to say that,

because if I think that you didn't realize that, I know we are both in serious trouble. Okay, that's correct. I want to reinforce the notion over and over again. Wave functions should tell you something. Square it, you get the probability. If you don't square it, don't think it's a probability, because it can be negative. It can even be complex. Don't forget. If you square it, you get the probability.

To find a certain position, that means if you went around with this little Heisenberg microscope all over the ring, and you catch it and you say, good, I found it here, and you do that many, many times. By many, many times, I mean you take a million particles in a million rings, each in exactly this quantum state and make measurements, then your histogram will look

like the square of this function. But there is more to life than just saying, where is the particle? Because in classical mechanics, you also ask, what is its momentum? The question I'm asking you is, what is the momentum you will get when you measure the momentum of a particle in this quantum state?

You know the answer only on a few special cases. If by luck, your function happened to look like one of these functions. Right now, I've not told you what ψ is. It's whatever you like. But if it looked like one of these, you're in good shape, because then the momentum is whatever p you find up here. But it may not look like that. This guy doesn't look like that.

So then the question is, if you measure momentum, what answer will you get? Now that is something. Anybody know what the answer is? Okay, now I accept your silence, because you're not supposed to know this. This is a postulate in quantum mechanics.

It's not logic. It's not mathematics. No one could have told you 300 years ago, this is the right answer. So this is another postulate. Just like saying ψ squared is the probability to find it at x, there is a new postulate. It addresses the following question. If I shift my attention from position to momentum,

and I ask, what are the odds for getting different answers for momentum, that answer is actually contained in the same function ψ of x, in the following fashion. Take this function ψ of x and write it as a sum over p of this function ψp of x with some

coefficient Ap. I can also write it equally well as the same function labeled by integer m, and I want to call the coefficient m. They are exactly the same thing. p and m are synonymous. The same function, this function is called

ψm, ψm, because they contain the same information on the momentum. So either you can write it this way if you want to see the momentum highlighted, or you can write the function this way if you want to see the quantum number m highlighted, but they stand for the same

physics. If m is equal to 4, it means the cosine and the sine contained in the complex exponential finish four complete oscillations to go around the cycle. All right, so here is what you're told. Take the arbitrary periodic function, write it as a sum of these functions, each of definite momentum with

some coefficient. Then the probability that you will get a momentum p when you measure it, which is the same as the probability you will get the corresponding m, is nothing but the absolute square of this coefficient. In other words, anyway, this is a postulate,

okay? Let me repeat the postulate. Somebody gives you a function, you write that function as a sum of all these periodic functions, each with an index

m. Multiply each with a suitable number so that they add up to give you the function that's provided to you. Once you have done that, the coefficient square with the particular value m is the probability you will get that particular value for m, or the corresponding momentum.

So there are two questions one can ask at this point. The first question is, what makes you think that you can write any function I give you as a sum of these functions? Realize what this means. I'm saying I can write any function as e to the 2Π i m x over L square root of L times am.

I'm writing out this function explicitly for you. Now that is a mathematical result. I will not prove here. It's called Fourier series, and it tells you that every periodic function, namely that which comes back to itself when you go around a period length L, can be written as a sum of these periodic functions with

suitably chosen coefficients. It can always be done. That is analogous to the statement that if you are, say, living in three dimensions, and that is a vector vector v, and you pick for yourself three

orthonormal vectors i, j and k, then any vector v can be written as vx times i vy times j vz times k. In other words, I challenge you to write any arrow in three dimensions starting from the origin and

pointing in any direction of any finite length. I'll build it up for you using some multiple of i, some of j and some of k. We know that can be done. In fact, that's if you want the technical definition of three dimensions. Yes?

Good point. That's correct. That'll be a unique way. That is. You agree in three dimensions, there is no other mixture except this one. If somebody comes with a second way of writing it, you can show that the second way will coincide with the first way. So the expansion of a function in what are called these

trigonometric or exponential functions, and it's called the Fourier series of the function, has unique coefficients. And I'll tell you right now what the formula for the coefficient is. But first I'm telling you that just like it's natural to build a vector out of some building blocks, i, j and k, it's natural to build up periodic functions with these building blocks,

size of m. The only difference is there you need only three guys. Here you need an infinite number of them because the range of m goes from minus to infinity, all integers. But it's still remarkable that given all of them you can build any function you like, including this thing I just wrote down arbitrarily can be built.

The fact that you can prove it, that you can do it, I don't want to prove. It's kind of tricky, but I will prove the second part of it, which is given that such an expansion exists, how do I find these coefficients given a function? So let's ask a similar question. In the usual case of vectors, how do I find the coefficient?

Suppose I write the vector v as E1 times v1 E2 times v2 E3 times v3. Don't worry, E1, E2 and E3 are the usual guys. E1 is i, E2 is j and E3 is k.

People like to do that because in mathematics you may want to go to 96 dimensions, but you've only got 26 letters, so if you're stuck to i, j and k you're going to have trouble, but with numbers you never run out of numbers. So you label all the dimensions by some number,

which in this case happens to go from 1 to 3. You also know that these vectors i and j have some very interesting properties. That means i⋅i is 1. That's the same as j⋅j. That's the same as k⋅k. And that i⋅k and i⋅j are 0 and so on.

Normally the dot product of one guy with himself is 1, and anyone with anything else is 0. That just tells you they all have unit length and they're mutually perpendicular. I want to write this as Ei⋅ej, but this could be 1,2 or 3. That could be 1,2 or 3. I want to say this is equal to 1 if i is equal to

j. This is equal to 0 if i is not equal to j. This is usual vector analysis. I'm just saying the dot product of basis vectors has this property. 1 if they match, 2 if they are different. So that's a shortcut for this, and that's write the symbol Δij.

Δij is called Kronecker's delta. Kronecker's done a lot of things, but this is something, one place where his name has been immortalized. He just said, instead of saying this all the time, 1 if they're equal, 0 if they're different, why don't you call it my symbol, Kronecker's symbol?

Δij is understood that this whole thing you simply say is Δij is a shorthand. That means if on the left hand side there are two guys with indices i and j, if the indices are equal, right hand side is 1, indices are unequal, right hand side is 0. Do you understand that this gives you the fact that each vector has a length 1 and that each is perpendicular to the

other two? Now that is what we can use now to find out. So let me write the vector v in this notation as ei times vi, i from 1 to 3. You're all familiar with this way to write the sum?

So I come with a certain vector. The vector is not defined in any axis. It's just an arrow pointing in some direction. It's got a magnitude and direction. And I say, can you write the vector in terms of i, j and k? And the answer is, yes, of course I can. So here is your vector v. Here is, if you like, e1, e2 and e3.

The claim is some mixture of e1, e2, e3 will add up to this v. That's granted, but how much e1 do I need? How much e2 do I need? That's a very simple trick for that. Anybody know what that trick is? You might know that trick. You've seen it anywhere?

Okay, here is the trick. Suppose you want to find v2. You take the dot product of both these things with e2. Take e2 dot this and take dot product with e2. What happens is you will get the dot product can go inside.

You'll get ei dot e2 times vi, where i goes from 1 to 3. What is ei dot e2? a dot e2 is Δi2. That means if this index i is equal to 2 you get 1. If it's not equal to 0 you strike out, you get 0.

So of all the three terms in this, only 1 will survive. That's the one when i is equal to 2, in which case you will get 1. That's multiplying v2, so it will give you v2. So to find the component number 3, you take the dot product of the given vector with e3.

And that will give you, you can see, that will give you v3 or v2 or whatever you like. I'm going to use a similar trick now in our problem. The trick I'm going to use is the following. So the analogy is, just like you had v equal to

sum over i ei times vi, i have ψ of x is equal to sum over m of some am times the function ψ sub m of x. You understand that ψ sub m is a particular function, which I don't want to write every time,

but if you insist it is 2Πi mx over l divided by square root of l. So I'm trying to find this guy. How much is it, right? That's the question. Now here we had a nice rule. The rule says ei dot ej is Δij.

That was helpful in finding the coefficients. There's a similar rule on the right hand side, which I will show you and we can all verify it together. The claim is ψm star of x times ψn of

x dx from 0 to l is in fact Δmn. So the basis vectors ψm are like ei. The dot product of two basis vectors being Δij is the same

here, like saying the integral of one of them star times the other one is 1 if it's the same function, and 0 if it's not the same function. Let's see if this is true. If m is equal to n, can you do this in your head? If m is equal to n, what do we have?

This number times its conjugate is just 1. You just get 1 over l, an integral of 1 over l dx. So if you want I will write it here. It's 2Πi times n minus mx over l dx from 0 to l.

Do you understand that? You take the conjugate of the first function. That's why it's a minus m here. And take the second function, which is e to the 2Πi n over l times x. I'll wait till you have time to digest that.

The product of ψm star with ψn, you combine the two exponentials, but the thing that had m in it has a minus m here because you conjugated everything. So I'm saying, what is the integral going to be? If n is equal to m, you can see that in your head. This is e to the 0.

That is just 1. The integral of dx is l. That cancels the l. You get 1. That's certainly true if m is equal to n. If m is not equal to n, suppose it is 6. It doesn't matter. This will complete 6 full oscillations in a period,

the sine and the cosine. But whenever you integrate a sine or a cosine over some special number of full periods, you get 0. So this exponential, when integrated over a full cycle, if it's got a non-zero exponent, integer exponent will give you 0. So you see the remarkable similarity between the usual

vector analysis and these functions. This is an arbitrary vector. This is an arbitrary function. e1, e2, e3 are basis vectors. ψm or if you want, basis function. I can build any vector out of these unit vectors. I can build any function out of these basis functions.

And finally, if I want to find out a coefficient, what should I do? If I want to find the coefficient An, what did I do here to find V sub i or V sub j, you take V⋅ej, where j could be whatever number you picked, because if you take the dot product of the two sides,

when you take dot product of ej, the only guy who survives is i equal to j, that will give you the V⋅j. So similarly here, I claim the following is true. If you do the integral of ψn x times the given function dx from 0 to L,

you will get An. So once I show this, I'm done. Now if you don't have the stomach for this proof, you don't have to remember this proof. That's up to you. See, I don't know how much you guys want to know. I'm trying to keep the stuff I just tell you without proof to a minimum. I felt bad telling you that every function can be expanded

this way, but the coefficient that's being given by the formula is not too far away. So I want to show you how it's done. You may not know the details why this is working, but you should certainly know that if you want coefficient number 13, you've got to take ψ13 star and multiply with the given function and integrate.

That you're supposed to know. So why does this work? So let's see what this does. We are trying to take ψn star of x. The unknown function and the given function looks like a sum am ψm of x dx 0 to L summed over m.

So this summation, you can bring the ψn here if you like. It doesn't matter. Then do the integral over x. Then you will find this is giving me, maybe I'll write it here, that is going to be equal to

sum over m am of ψm of ψn star x ψm of x dx, and that is going to be Δmn. That means I will vanish unless m equals n, and when m equals n, I will give you 1.

That means the only term that survives from all these terms is the one where m matches n, so the thing that comes out is an. Once again, you will see this in my notes.

Do you have any idea what I did or where this is going? In quantum theory, if you want to know what will happen if I measure momentum for a particle living in a ring, you have to write the given function in terms of these special functions,

each identified as a definite momentum with suitable coefficients. The rule for finding coefficient a sub n is to do this integral of ψn star this one. This is the rule. a sub n is the integral of ψn star ψdx.

And once you've found the coefficients for all possible n, then the probability that you will have some momentum corresponding to m is just am squared. That is the recipe. What the recipe tells you is, if your function is made up, is a special function with a definite momentum,

of course you will get that momentum as the answer when you measure it. If your function is a sum over many different momentum functions, then you can get any of the answers in the sum, but if it had a big coefficient in the expansion, it's more likely to be that answer. If it had a small coefficient, it's less likely.

If it had no coefficient, you won't get that momentum at all. That's like saying if you had ψ of x, it's likely where it's big, unlikely where it's small, and impossible where it is 0. So that's your job. Any time someone gives you a function, you have to find these coefficients a n, then look at them. They'll tell you what the answer is. So that's what I'm going to do now.

I'm going to take some trial functions and go through this machinery of finding the coefficients and reading off the answers. So maybe if I do an example, you will see where this is going. So let's take an example where I'm going to pick,

first of all, a very benign function, then maybe a more difficult function. The function I want to pick is this, sum number n cosine 6Πx over L. Somebody gives you that function. That is not a state of definite momentum because it's not

e to the i something x. So we already know when you measure momentum, you won't get a unique answer. You'll get many answers. But what are the many answers? What are the many odds is what we're asking. I forgot to mention one thing in my postulate.

What I forgot to mention is that for all this to be true, it is important that the momentum functions are all normalized and the given function is also normalized. You should first normalize your function, then expand it in terms of these normalized functions of definite momentum. Only then the squares of the coefficient are the absolute

probabilities. By that I mean, if you do this calculation and you then went and added all these A of m squares, you will find amazingly it adds up to 1, because you can show mathematically that if ψ star ψ dx is 1, then the coefficient of

expansion squared will also be 1. So if you took a normalized ψ, then the probabilities you get are absolute probabilities, because they will add up to 1. That's also a mathematical result which I'm not showing.

All right, so let's go to this problem. So the first job is normalize your ψ. So how to normalize this function? I'm going to demand that if you square ψ and you integrate it, so I say n squared times integral of cosine squared 6Πx over L dx from 0 to L should be 1.

Now you don't have to look up the table of integrals, because when you take a cosine squared or a sine squared over any number of full cycles, the average value is 1 half. That means over a length L, this integral will be L over 2.

So you want n squared L over 2 to be 1, or you want n to be square root of 2 over L. So the normalized function looks like square root of 2 over L cosine 6Πx over L.

So that's the first job. Second is, you can ask now, what are the coefficients an? So an is going to be integral 1 over square root of L, e to the minus 2Πinx over L times this function,

square root of 2 over L cosine 6Πx over L dx. Are you with me? That's the rule. Take the function, multiply it by ψn*, which is this guy here. This is the ψn*.

This is the ψ that's given to you, and you're integrating the product in the interval. That will give you an. But I claim, yeah, if you want you can do those integrals, but I think there's a quicker way to do that. Have any idea what the quicker way is just by looking at it?

In other words, if you can guess the expansion, I will say I will take it. Can you guess what the answer looks like without doing the work? In other words, I want you to write this function as a sum over exponentials just by looking at

it. Can you tell me what it is? I want you to write it in the form sum over m A m e to the 2Πi mx over L. You can find the A m by doing all this nasty work, but I'm saying in this problem there's a much quicker way.

You see it anywhere? What's the relation between trigonometric functions and exponential functions? Yep, and what does it say?

Yes, but now I want to go the other way. It's certainly true that cosine θ, I'm sorry, e to the iθ is cosine θ i sine θ. I told you, you forget the formula at your own peril.

The conjugate of that is e to the minus iθ is cosine θ minus i sine θ. If you add these two, you will find cosine θ is e to the iθ e to the minus iθ over 2. And if you subtract them and divide by 2i,

you will find sine θ is e to the minus over 2i. Now I wanted you to be familiar with complex numbers enough so that when you see a cosine, the two exponentials jump out at you. Otherwise, you will be doing all these hard integrals that you don't have to do.

That's like saying sine square cosine square is 1, but you don't look it up, right? That's something you know. What you look up is your social security number or mother's maiden name. You are allowed to forget those things. You look in a book, okay? That's the name. But this you have to know, because you cannot go anywhere without this.

If you don't know it all the time, you must have done these trigonometry calculations in school, right? You've got to plug in the right stuff at the right time for things to all cancel out. You cannot say, I will look it up, because you don't know what to look up. So everything should be in your head. And the minimum you should know is this, the minimum.

Therefore, I can come to this function here, and I can write it in terms of what I'm telling you is cosine of x which was given to us is equal to the square root of 2 over L times cosine 6Πx over L.

But I'm going to write it as square root of 2 over L times 1 over 2 e to the 6Πix over L plus e to the minus 6Πix over L.

So I will write it very explicitly as 1 over root 2 times e to the 6Πix over L divided by square root of L, plus 1 over root 2 e to the minus 6Πix over L divided by square root of L.

In other words, what I've done is, rather than do any integrals, I just massage the given function and manage to write it as a sum over normalized functions associated with definite momentum with some

coefficients. In other words, here it is in the form that we want, A m e to the 2Πi mx over L. You agree that I have got it to the form I want. Do you see that?

You take the cosine, write it as sum of exponentials, then put factors of root L so that that's a normalized function and that's a normalized function. Everybody else is A. So what do you find here? What are the As for this problem? By comparing the two, what do you find are the As

in this problem? What is A14? Let me ask you this. If you compare this to this one, what m are you getting? Do that in your head. If you compare this to this one, what is m?

Pardon me? m is 3 for this guy. So this is really ψ sub 3, and this is ψ sub minus 3, because you can see there are two momenta in the problem,

and you can stare at them. You can see right away. Therefore, A sub 3 is 1 over root 2, and A sub minus 3 is also 1 over root 2. Therefore, the probability for m equal to 3 is

1 half, and the probability for m equal to minus 3 is 1 half, and nothing else. All other As are 0, because they don't have a role. So you might think every term must appear.

It need not. So this is a good example that tells you only two of the m's make it to the final summation. All the other m's are 0. They happen to come with equal coefficients, 1 over root 2. The square of that is 1 over 2, and you can see these probabilities nicely add up to 1. I told you, if you normalize the initial function,

the probability for everything will add up to 1. So this is a particle. If a particle is in a wave function cosine 6Πx over L, which is a real wave function, and you can plot that guy going around the circle, it describes a particle whose momenta, if you measure, will give you only one of two answers.

m equal to 3 is p equal to 2Πh-bar over L times 3, or minus 3, plus or minus 6Πh-bar over L. You can see that too. I mean, just look at this function. If you call it e to the ipx, p is 6Πh-bar over L.

So this particle has only two possible answers when you measure momentum. So it is not as good as the single exponential, which has only one momentum in it. This is made up of two possible momenta. But you won't get anything else. You will not get any other momentum if you measure this.

So if you got m equal to 14 as a momentum, there is something wrong. It's still probabilities, but it tells you that there is non-zero probabilities only for these two. Now, in a minute, I will take on a more difficult

problem where you cannot look at the answer, you cannot look at the function ψ, and just by fiddling with it, bring it to this form. Do you understand? It is very fortunate for you that I gave you a cosine, which is just the sum of two exponentials. So the two, what are called plane waves, exponentials are staring at you and you pick them up. I can write other functions, crazy functions for which you

will have to do the integral to find the coefficient. But I'm going to tell you one other postulate of quantum mechanics. In the end, I will assemble all the postulates for you. I'm going to tell you one more postulate. It's called the measurement postulate. The measurement postulate says that if you make a measurement

and you found the particle to be at a certain location x, then right after the measurement, the wave function of the particle will be a spike at the point x. Because you know that you found it at x, it means if that has any meaning at all,

if you repeat the measurement immediately afterwards, you've got to get the same answer. Therefore, your initial function could have looked like that, but after the measurement, it collapses to function at the point where you found it. This is called the collapse of the wave function. It goes from being able to be found anywhere to being able

to be found only where you found it just now. It won't stay that way for long, but right after the measurement, that's what it will be. But the state of the system changes following the measurement. And if you measured x, it will turn into a wave function with well-defined x, which we know is a spike at x equal to wherever you found it.

Similarly, if you measured momentum and you found it at the measure of momentum, first of all, you will get only one of two answers, m equal to 3 or m equal to minus 3. If you got m equal to 3, the state after the measurement will reduce to this. This guy will be gone. This will be the state.

So from being able to have two momenta which are equal or opposite, the act of measurement will force it to be one or the other. It can give you either one, but once you got it, that's the answer. It's like the double slit. It can be here and it can be there. It is not anywhere in particular, but if you shine

light and you catch it, right after the measurement, it is in front of one slit or the other. Think of this as double slit. There's some probability for 3 and some for minus 3, but if you catch it at 3, it will collapse to this one. So what happens is, in the sum over many terms, the answer will correspond to one of them, and whichever one you got, only that one term will

remain. Everything will be deleted from the wave function. The act of measurement filters out from the sum the one term corresponding to the one answer that you got. This is called the collapse of the wave function. So in classical mechanics, when you measure the position

of a particle, nothing much happens. It doesn't even know you measured it, because there are ways to measure it without affecting it in any way. So if it had a momentum p and a position x before the measurement, it's the answer right after the measurement, because you can do non-invasive measurements. In quantum theory, there are no such measurements in general. In general, the measurement will change the

wave function from being in one of many options to the one option you got. But the answer depends on what you measure. With the same wave function on a ring, let's say, if you measured position and you found it here, that will be the answer. If you measured momentum and you got 5, it will be something with 5 oscillations in it.

So it will collapse to that particular function. And what it collapses, it depends on what you measure. And for a long time, I'm going to focus only on x and p. Of course, there are other things you can measure, and I don't want to go there right now, but if you measured x, it collapses to a spike at that location. If you measure p, it collapses to the one term.

whatever that was, that 1 e to the ipx in the sum. So another interesting thing is the only answers you will get in the measurement are the allowed values of momentum. You will never get a momentum that's not allowed. And once you get one of the allowed answers,

there are two things I'm telling you. One is the probability you will get that answer is proportional to the square of that coefficient in the expansion of the given function. In our problem there are only two non-zero coefficients. Both happen to be 1 over root 2. It happened to be an equal mixture, but you can easily imagine some other problem where this is 1 over root 6,

and something else is 1 over root 6, and something else is 1 over root 3. They should all add up to 1 when you square and add them. Then you can get all those answers with those probabilities. Okay, so the last thing I'm going to do is just one more example of this where you actually have to do an integral and you cannot just read off the answer by looking at it.

Then we are done with this whole momentum thing. I will write the postulates later, so I don't want to write it in my handwriting, but one final time. Measurement of x collapses the function ψ

of x to a spike at x. Measurement of p collapses ψ to that particular plane wave with that particular p or m on the exponent. And after that, that's what's taken to be. People always ask, how do we know what state the quantum system is in? Who tells you what ψ of x is? It's the act of measurement.

If you measured the guy and you found him at x equal to 5, the answer is ψ is a big spike there. You measure momentum and you've got m equal to 3. The answer is that particular wave function with 3 on the exponent. So measurements are a way to prepare states. Okay, by the way, it's very important that if you

had a system like this, the probability for getting a certain x, say at this x, is proportional to the square of that number. Once you took the state and didn't measure x, but measured momentum, it'll become one of these oscillatory functions with a definite wavelength. It's a complex exponential, but I'm showing you the real

part. Now if you measure position, in fact, the square of this will be flat. Remember, e to the i px is flat. So in classical mechanics, if I measure position and I measure momentum, and I measure position again, I'll keep getting the same answers, right? I see where it is.

I see how fast it's moving. Again, I see where it is. If I do all of this in rapid succession, you'll get the same answer, xp, xp, xp. In quantum theory, once you measure x, it'll become a big spike at that point. You can get all kinds of answers p. If you measure p, you've got an answer. That answer is a new wave function which is completely flat.

That means if you measure x, you'll no longer get the old x. In fact, you can get any x. That's why you can never filter out a state with well-defined x and p, because states of well-defined x are very spiky, and states of well-defined p are very broad. So you cannot have it both ways. And that's what I want to show you in this final example.

I'm going to take the following wave function on this ring. The function I'm going to take is ψ is some n e to the minus α mod x. Mod x means this is x equal to 0. To the right it falls exponentially, and to the left it falls exponentially.

Is it clear what I'm saying here? This is my ring, and I'm trying to plot the function. It is highest at the origin and falls exponentially equally for positive and negative x. So that's the meaning of mod x. So how far can you go before ψ becomes negligible?

Well, that's when e to the minus x is a big number. So I'm going to assume that when you go all the way around half the circle, I'm going to assume α L is much bigger than 1. That means this function dies very quickly. It's pretty much negligible beyond some distance.

How far does the function live? Roughly speaking, you can go a distance Δx so that α times Δx is roughly 1. Because if you plot this exponential, you ask, when does it come to, say, half its value or one-fourth of its value? You'll find a sum number of order 1 over α.

So this is a particle whose position has an uncertainty of order 1 over α. So you can make it very narrow in space, so you know pretty much where it is, or you can make it broad. But I want to consider only those problems where even if it's broad, it's dead by the time you go to the other side,

to the back. That's for mathematical convenience. So this is the state. And I want to ask myself any question we can ask. The first question I can ask is, if I look for its position, what will I find? I think we have done it many, many times. You square the guy, you get n squared e to the

minus 2α mod x. So that's the shape. That shape looks the same. If you square an exponential, you get another exponential. But now let me demand that this integral dx be equal to 1. So what is that integral?

I hope you can see that this function is an even function of x, because it's mod x. So it's double the answer I get for positive x. Oh yeah, sorry, not minus infinity. That is minus l over 2 to plus l over 2, right? You can go this way l over 2, or you can go that way

l over 2. So it's n squared is 2 times e to the minus αx 2αx dx. Pardon me? Yes, thank you.

Right. So I'm going to now assume that in the upper limit of the integration, instead of going up to l over 2, I go to infinity. It doesn't matter, because this guy's dead long before that. That's why I made that choice, so that I'm going to write this 2 times n squared times 0 to infinity e to the minus

2αx dx. So that's a pretty trivial integral. It's just 1 over 2α. If you want, I will write it out. It's e to the minus 2αx divided by e to the minus 2α from 0 to infinity. At the upper limit when you put infinity you get 0.

At the lower limit when you put x equal to 0 you get 1. There's a minus sign from this, and you get n squared over α. It should be 1, or n is equal to the square root of α.

So my normalized wave function, this is the first order of business, is square root of α e to the minus α mod x. I just squared it and I integrated it. The only funny business I did was, instead of cutting up the

upper integral at l over 2, I cut it off at infinity, because e to the minus l over 2 is e to the minus 9 million, let's say. So I don't care if it's 9 million or infinity. To make the life simpler, I just did it that way. Okay, now I want to ask you, what is A of p?

A of p, the coefficient of the expansion, you remember is e to the minus ipx over h-bar over square root of l times ψ of x dx from minus l over 2 to

plus l over 2. Now I'm going to work with p rather than m. I will go back and forth. You should use the notion that the momentum can be either labeled by the actual momentum or the quantum number m, which tells you how much momentum you have in multiples of 2Πh-bar over l.

So you've got to do this integral. So let's write this integral. This looks like minus l over 2 to l over 2. In fact, I'm going to change this integral to minus infinity to plus infinity, because this function,

e to the ipx over h-bar times e to the minus α mod x dx. Do you understand that these limits can be made to be plus and minus infinity, because area under a graph that's falling so rapidly, whether it's between minus and

plus l over 2 or minus and plus infinity, is going to be the same. It's just that this integral is so much easier to do. Now, you cannot jump out and do this integral, because it's a mod x here. The mod x is not x. It is x when it's positive. It's minus x when it's negative.

So you've got to break this integral into two parts. One part, where x is positive, from 0 to infinity. I'm sorry, I also forgot a root α and a root l. Do you see that? So this is really square root of α over l times

e to the minus αx e to the minus ipx over h-bar dx, plus another integral from minus infinity to 0, e to the plus αx times e to the minus ipx over h-bar dx.

I split the integral into two parts. So I didn't make a mistake here. This really is e to the plus αx, because x is negative. So what one does in such situations, I'm going to do it quickly and you can go home and check it.

Just calculus. Change the variable from x to minus x everywhere. In terms of new variable, this will become minus x. That will become plus x. dx will become minus of the new variable. The limits will be plus infinity to 0, and you can flip that for another change of sign, 0 to infinity.

So that was a very rapid sleight of hand, but I don't want to delay that. This is just, you go home and if you want to check that if x goes to minus x you get that. So you notice this is the complex conjugate of this one. Whatever function I'm integrating here, this is the conjugate of it, because this is real,

αx, and minus ipx has become plus ipx. So if I find the first part of the integral, I just take that times its conjugate and I'm done. So what do I get for that? I get square root of α over L times, remember, integral e to the minus αx

dx from 0 to infinity is 1 over α. But what I have here is dx e to the minus α Ip over h bar x, 0 to infinity, plus the complex conjugate.

Now, you may be very nervous about doing this integral with a complex number in it. If it's real, we all know the integral is just 1 over α e to the minus α. It turns out it is true even if it's got an imaginary part, as long as you have a positive real part. In other words, the answer here doesn't depend

on this guy being real. So it's really α over L times 1 over α Ip over h bar, plus the complex conjugate, which is α minus Ip over h bar. Now, you should be able to combine these two denominators,

and you get square root of α over L divided by α squared plus p squared over h bar squared times 2α. Again, this is something you can go and check. I don't want to wait till everyone can do this thing,

because I want to tell you the punch line. So this is what it looks like. A of p looks like a whole bunch of numbers I'm not going to worry about, but look at the denominator. It's α squared p squared over h bar squared, or if you want, multiply by h bar squared. That's why there are some other numbers I'm not interested in.

The numbers are not important. How does it vary with p is all I'm asking you to think about. I'm sorry, A of p, this is just A of p, but I want the A of p squared. That looks like the square of this.

All I want you to notice is that this function is peaked at p equal to 0 and falls very rapidly as p increases. When p is 0, you've got the biggest height. When does it become half as big or one-fourth as big? Roughly when p squared is equal to h squared α squared, because that's when these numbers become

comparable. Therefore when p is of order h bar α, this function will have a denominator which is twice what it had here, or maybe one-fourth. I'm not worried about factors like 1 and 2. The point is, in momentum space,

in momentum, you can get all kinds of values of p, but the odds decrease very rapidly for p bigger than h bar α. So the most likely value is 0, but that's the spread, and Δp is h bar α, or α times Δp is h bar, and that is the uncertainty

principle, because α is just Δx. By the way, I will publish these notes too so you don't have to worry if you didn't write everything down. I suggest you, yes, complex conjugate. That's what I did. Whatever number this one is, the other guy is obtained by

changing i to −i. Look, all I want you to notice is this. I took a function whose width is roughly 1 over α. Then I looked at what kind of momenta I can get. Then I find that the narrower the function, the bigger the spread in the possible momenta you can get.

So squeezing it in x broadens it out in p, and that's the origin of the uncertainty principle. It's simply a mathematical result that functions which are narrow in x have a Fourier series which is very broad in p. The quantum mechanics relates p to momentum, and therefore the uncertainty principle. So anyway, I will give you some homework on this,

and you can also fill in the blanks of this derivation, which I think is very useful.