So today, before we get into the meat of today's lecture, Matt has very kindly, Professor Evans, has very kindly agreed to do an experiment. So for those of you all who are in recitations, both he and Barton talked about polarization in recitation

last week, and Matt will pick it up from there. So back to the ancient past. This was a week ago. We had our hyper intelligent monkeys that were sorting things that all seemed very theoretical. And in recitation, I said things about polarizers and said, look, if we use polarizers, we can do exactly the same thing as these monkeys.

We just need to set up a little polarization experiment, and the results are identical. You can use one to figure out the other. But I didn't have this or nice polarizers right behind me to give a demo, so here we go. What I'm going to show you is that if we start with something polarized here with all white,

and that's right now I have all vertical polarization here. And if I just put on this other box there, which is going to be another polarizer, if I put it the same way, this is all of our white electrons coming through all white, so it doesn't really do much. And if I look at the black output over here, the second color sorting box, that's

the same as turning my polarizer 90 degrees. So nothing comes out black. So this is our, if we remove this guy from the middle, you have just exactly what you'd expect. You sort here, you have white, and you get all white out. Great. Everyone thought that was easy. We all had that figured out. This box got thrown in the center here, and it became sort of confusing because you thought,

well, they were white, so I'm going to throw my box in the middle here. That's this guy at 45 degrees. And then if I throw this guy on the end again, the idea was, well, they were all white here, so maybe this guy identified the soft ones from the white ones. And now we have white and soft, and these would still all be white, right? So I put this guy on up here.

They should all come out, but sort of don't. And we say, well, are they black? Well, no, they're not really black either. They're some sort of strange combination of the two. So that's this experiment done in polarizers, but let me just play the polarizer trick a little bit because it's fun. So this is if I say vertical polarization,

and then how many of them come out horizontal? So here I'm saying white, and how many of them come out black? That's the analogy. And the answer is none of them. And strangely, if I take this thing, which seems to just attenuate, this is our middle box here, and I just stuff it in between them, I can get something to come out,

even though I still have cross polarizers on the side. So you can see the middle region is now brighter, and you can still see the dark corners there of the cross polarizers. And as I turn this guy around, I can make that better or worse. The maximum is somewhere right there. And it goes off again. So this is a way of understanding our electron sorting

hyper-intelligent monkeys in terms of polarizations. And here it's just a vector projected on another vector, projection on another vector, something everybody knows how to do. So here's the polarization analogy of this Dan Garlic experiment. All right, awesome.

So the polarization analogy for interference effects in quantum mechanics is a canonical one in the text of quantum mechanics. So you'll find lots of books talking about this.

It's a very useful analogy, and I encourage you to read more about it. We won't talk about it a whole lot more, but it's a useful one. All right, before I get going, any questions from last lecture? Last lecture was pretty much self-contained. It was experimental results. No, nothing? All right. OK.

The one thing that I want to add to the last lecture, the one last experimental observation, I glossed over something that's kind of important, which is the following. So we started off by saying, look, we know that if I have a ray of light, it's a wave. It's an electromagnetic wave, and it has some wavelength lambda.

And yet the photoelectric effect tells us that in addition to having a wavelength lambda, the energy, it has a frequency as well, a frequency in time. And the photoelectric effect suggested that the energy is proportional to the frequency.

And we wrote this as h nu. And h bar is equal to h upon 2 pi. And omega is equal to 2 pi nu.

So this is just the angular frequency rather than the number per time frequency. And h bar is the reduced Planck constant. So I'll typically write h bar omega rather than h nu because these two pis will just cause us endless pain if we don't use the bar. Anyway, so to an electromagnetic wave, we have a wavelength and a frequency.

And the photoelectric effect led us to predict that the energy is linearly proportional to the frequency with linear proportionality coefficient h bar, Planck's constant. And the momentum is equal to h upon lambda, also known as h bar k, which is equal to h upon lambda, where here

again h bar is h upon 2 pi. And so k is equal to 2 pi upon lambda. So k is called the wave number, and you should have seen this in 803.

So these are our basic relations for light. We know that light as an electromagnetic wave has a frequency and a wavelength, or a wave number, an inverse wavelength. And the claim of the photoelectric effect is that the energy and the momenta of that light are thus quantized. That light comes in chunks. So it has a wave-like aspect, and it also

has properties that are more familiar from particles. Now, early on, shortly after Einstein proposed this, a young French physicist named de Broglie said, well, look, OK, this is true of light. Light has both wave-like and particle-like properties. Why is it just light?

The world would be much more parsimonious if this relation were true not just of light, but also of all particles. I am thus conjecturing with no evidence whatsoever that, in fact, this relation holds not just for light, but for any object. Any object with momentum p has associated to a wavelength or a wave number, which is p upon h bar.

Every object that has energy has associated with it a wave with frequency omega. To those electrons that we send through the Davisson-Germer experiment apparatus, which are sent in with definite energy, there must

be a frequency associated with it, omega, and a wavelength, lambda, associated with it. And what we saw from the Davisson-Germer experiment was experimental confirmation of that prediction, that electrons have both particulate and wave-like features simultaneously. So these relations are called the de Broglie relations,

or de Broglie. I leave it up to you to decide how to pronounce that. And those relations are going to play an important role for us in the next few lectures. I just want to give them a name and a little context. This is a good example of parsimony and theoretical elegance leading you to an idea that turns out

to be true of the world. Now, that's a dangerous strategy for finding truth. Boy, wouldn't it be nice if we didn't have to pay taxes, but we also had Medicare. So that's not a terribly useful guide all the time,

but sometimes it really does lead you in the right direction. And this is a great example of physical intuition, wildly divorced from experiment, pushing you in the right direction. I'm making it sound a little more shocking than, well, it was shocking. It was just shocking. So with that said, let me introduce the moves

for the next few lectures. For the next several lectures, here's what we're going to do. I am not going to give you experimental motivation. I've given you experimental motivation. I'm going to give you a set of rules, a set of postulates. These are going to be the rules of quantum mechanics. And what quantum mechanics is for us is a set of rules to allow us to make predictions

about the world. And these rules will be awesome if their predictions are good. And if their predictions are bad, these rules will suck. We will avoid bad rules to the degree possible. I'm going to give you what we've learned over the past 100 years of developing quantum mechanics. That is amazing. Wow.

OK, yeah, over the past 100 years of developing quantum mechanics. And I'm going to give them to you as a series of postulates. And then we're going to work through the consequences. And then we're going to spend the rest of the semester studying examples to develop an understanding for what the rules of quantum mechanics are giving you. OK? So we're just going to scrap classical mechanics and start over from scratch.

So let me do that. And to begin, let me start with the definition of a system. And to understand that definition, I want to start with classical mechanics as a guide.

So in classical mechanics, let's think about the easiest classical system you can, just a single particle sitting somewhere. In classical mechanics of a single particle, how do you specify the configuration or the state? Just different word for the same thing. How do you specify the configuration or state of the system?

By a specified position of momentum. A specified position of momentum, exactly. So in classical mechanics, if you want to completely specify the configuration of the system, all you have to do is give me x and p for my particle. And if you tell me this, I know everything. If you know these numbers, you know everything.

In particular, what I mean by saying you know everything is that if there's anything else you want to measure, the energy, for example, the energy is just some function of the position and momentum. And if you know the position and momentum, you can unambiguously calculate the energy. Similarly, the angular momentum, which is a vector, you can calculate it if you know x and p, which is just r cross p. So this gives you complete knowledge of the system.

There's nothing more to know if you know that data. Now, there are certainly still questions that you can't answer given knowledge of x and p. For example, are there 14 invisible monkeys

standing behind me? I'm here. I'm not moving. Are there 14 invisible monkeys standing behind me? You can't answer that. It's a stupid question. OK, let me give you another example. The electron is x and p, some position. Is it happy? So there are still questions you can't answer.

The point is complete knowledge of the system to answer any physically observable question, any question that could be meaningfully turned into an experiment. The answer is contained in knowing the state of the system. OK. But this can't possibly be true in quantum mechanics

because as you saw in the problem set and as we've discussed previously, there's an uncertainty relation which says that your knowledge or your uncertainty, rather, in the position of a particle quantum mechanically, and I'm not even going to say quantum mechanically. I'm just going to say the real world. So in the real world, our uncertainty

in the position of our point-like object and our uncertainty in the momentum is always greater than or roughly equal to something that's proportional to Planck's constant. OK, you can't be arbitrarily confident of the position and of the momentum simultaneously. You worked through a good example of this on the problem set. We saw this in the two-slit experiment and the interference of electrons.

This is something we're going to have to deal with. So as a consequence, you can't possibly specify the position and the momentum with confidence of a system. You can't do it. OK, this was a myth. It was a good approximation, turned out to be false. So the first thing we need is to specify

the state, the configuration of a system. So what specifies the configuration of a system? And so this brings us to the first postulate, the configuration or state of a system. And here, again, just for simplicity, I'm going to talk about a single object.

Of a quantum object is completely specified, completely specified, by a single function, a wave

function, which I will denote generally psi of x, which is a complex function.

The state of a quantum object is completely specified once you know the wave function of the system, which is a function of position. Let me emphasize that this is a first pass at the postulates. What we're going to do is go through the basic postulates

of quantum mechanics. Then we'll go through them again and give them a little more generality. And then we'll go through them again and give them full generality. That last pass is 805. So let me give you some examples. Let me just draw some characteristic wave functions. And these are going to turn out to be useful for us.

So for example, consider the following function. So here is 0. And we're plotting as a function of x. And I'm plotting the real part of psi of x.

So first consider a very narrowly supported function. It's basically 0 everywhere except it has some particular spot at what I'll call x1. Here's another wave function, 0. It's basically 0 except for some special spot at x2.

And again, I'm plotting the real part of psi. And I'm plotting the real part of psi because A, psi is a complex function. At every point, it specifies a complex number. And B, I can't draw complex numbers.

So to keep my head from exploding, I'm just plotting the real part of the wave function. But you should never forget that the wave function is complex. So for the moment, I'm going to assume that the imaginary part is 0. I'm just going to draw the real parts. So let me draw a couple more examples. What else could be a good wave function?

Well, those are fine. What about, again, we want a function of x and I'm going to draw the real part and another one. So this is going to be a perfectly good wave function.

And let me draw two more. So what else could be a reasonable wave function? Well, this is harder than you'd think.

Oh, god.

OK, so that could be the wave function, I don't know. And that is actually my signature. My wife calls it a little gift.

OK, so here's the deal. Psi is a complex function. Psi also needs to not be a stupid function. So you have to ask me, look, could it be any function, any arbitrary function? So this is going to be a job for us. We're going to have to define what it means to be a not stupid function. Well, this is a completely reasonable function.

This is a reasonable function. Another reasonable function, reasonable, that's a little weird. But it's not horrible. That's stupid, right? So we're going to have to come up with a good definition of what not stupid means. So fine, these are all functions. One of them is multivalued. And that looks a little worrying.

But they're all functions. So here's the problem. What does it mean? So postulate two, the meaning of the wave function is that the probability that upon measurement the object is

found at the position x is equal to the norm squared of psi of x. If you know the system is described by the wave function psi and you want to look at point x and you want to know with what probability will I find the particle there, the answer is psi squared.

Notice that this is a complex number, but absolute value squared or norm squared of a complex number is always a real, non-negative number. And that's important because we want our probabilities to be real, non-negative numbers. Could be 0, right? Could be 0 chance of something. Could be negative 7 chance. Incidentally, there also can't be probability 2, right?

So that means that the total probability had better be normalized. So let me just say this in words though first. So p, which is the norm squared of psi, determines the probability and in particular the probability density that the object in state

psi in the state given by the wave function psi of x will be found.

So there's the second postulate. So in particular, when I say it's a probability density, what I mean is the probability that it is found between the position x and x plus dx

is equal to p of x dx, which is equal to psi of x. Squared dx. OK?

Does that make sense? So the probability that it's found in this infinitesimal interval is equal to this density times dx, or psi squared dx. Now again, it's crucial that the wave function is in fact properly normalized because if I say, look, something could either be here or it could be here, what's the sum of the probability that's here

plus the probability that it's here? It had better be 1, right? Or there's some other possibility. So probabilities have to sum to 1. Total probability that you find something somewhere must be 1. So what that tells you is that total probability, which is equal to the integral over all possible values of x. So if I sum over all possible values of x, p of x,

all values, should be equal to 1. And we can write this as integral dx over all values of x. And I write all here rather than putting minus infinity

to infinity because some systems will be defined from 1 to minus 1. Some systems will be defined from minus infinity to infinity. All just means integrate over all possible values. Hold on one sec. Of psi squared. Yep. I'm not going to.

Probability density is going to have just one argument. And total probability is going to have an interval as an argument. So they're distinct, and this is just the notation I like. Other questions? OK. OK. Just as a side note, what are the dimensions of the wave

function? Everyone think about this one for a second?

What are dimensions? 1 over the square root of length. Awesome. Yes. It's 1 over root length. Dimensions of psi are 1 over root length. And the way to see that is that this should be equal to 1. It's a total probability.

This is an infinitesimal length, so this has dimensions of length. This has no dimensions, so this must have dimensions of 1 over length. And so psi itself of x must have dimensions of 1 over root length. Now something I want to emphasize, I'm going to emphasize over and over in this class, is dimensional analysis. You need to become comfortable with dimensional analysis.

It's absolutely essential. It's essential for two reasons. First off, it's essential because I'm going to be merciless in taking off points if you write down a dimensionally false thing. If you write down something on a problem set or an exam that's like a length is equal to a velocity, ooh, not good. But the second thing is, forget the fact that I'm going to take off points.

Dimensional analysis is an incredibly powerful tool for you. You can check something that you've just calculated, and even better yet, sometimes you can just avoid a calculation entirely by doing a dimensional analysis and seeing that there's only one possible way to build something of dimensions length in your system. So we'll do that over and over again. But this is a question I want you guys to start asking yourselves at every step along the way of a calculation, what

are the dimensions of all the objects in my system? Something smells like smoke. So with that said, if that's the meaning of the wave function, what physically can we take away

from knowing these wave functions? Well, if this is the wave function, let's draw the probability distribution. What's the probability distribution? p of x. And the probability distribution here is really very simple. Again, 0 squared is still 0.

So it's still just a big spike at x1. And this one is a big spike at x2. Everyone cool with that?

So what do you know when I tell you that this is the wave function describing your system? You know that with great confidence, you will find the particle to be sitting at x1 if you look. So what this is telling you is you expect x is roughly x1 and our uncertainty in x is small.

Everyone cool with that? Similarly here, you see that the position is likely to be x2. And your uncertainty in your measurement, your confidence in your prediction is another way to say it, is quite good. So your uncertainty is small.

Now, what about these guys? Well, now it's norm squared. I need to tell you what the wave function is. Here, the wave function that I want, so here's 0. The wave function is e to the i k1 x. And here, the wave function is equal to e to the i k2 x.

And remember, I'm drawing the real part because of practical limitations. So the real part is just a sinusoid, or in fact, a cosine. And similarly, here the real part is a cosine. And I really should put 0 in the appropriate place, but that worked out well.

OK, good. Now the question is, what's the probability distribution, p of x, associated to these wave functions? So what's the norm squared of minus e to the i kx? If I have a complex number, a phase, e to the i alpha,

and I take its norm squared, what do I get? 1, right? So remember complex numbers. If we have a complex number alpha, or sorry, if we have a complex number of beta, then beta squared is, by definition, beta complex conjugate times beta.

So e to the i alpha, if the complex conjugate is e to the minus i alpha, e to the i alpha times e to the minus i alpha, they cancel out. That's 1. So if this is the wave function, what's the probability distribution? Well, it's 1. It's independent of x. So from this, we've learned two important things.

The first is this is not properly normalized. And that's not so key. But the most important thing is, if this is our wave function, and we subsequently measure the position of the particle, we look at it, we say, ah, there's the particle, where are we likely to find it?

Yeah, it could be anywhere. So what's the value of x you expect? Typical x. I have no idea. No information whatsoever. None. And correspondingly, what is our uncertainty in the position of x that we will measure? It's very large, exactly.

Now, in order to tell you it's actually infinite, I need to stretch this off and tell you that it's actually constant, off to infinity, and my arms aren't that long, so I'll just say large. Similarly here, if our wave function is e to the i k2 x, here k2 is larger, the wavelength is shorter, what's the probability distribution?

It's, again, constant. So this is 0, 0. So again, x, we have no idea, and our uncertainty in the x is large. And in fact, it's very large.

Everyone cool with that? Questions? What about these guys? OK, this is the real challenge. OK, so if this is our wave function, and let's just say that it's real, hard as it is to believe that, then what's our probability distribution? Well, it's something like, I don't know, it's something.

You get the point. So where is, so if this is our probability distribution,

where are we likely to find the particle? Well, now it's a little more difficult, right? Because we're unlikely to find it here. Well, it's reasonably likely to find it here. Unlikely here, reasonably likely, unlikely, like, you know, it's a mess. So where is this? I'm not really sure. What's our uncertainty? Well, our uncertainty is not infinite, right?

Because, OK, my name ends at some point, so this is going to go to 0. It's going to go to 0. So whatever else we know, we know it's in this region. So it's not infinite, it's not small, we'll say. But it's not arbitrarily small. It's not tiny, or sorry, it's not gigantic, is what I meant.

Our uncertainty is not gigantic. Not gigantic. But it's still pretty non-trivial, because I can say with some confidence that it's more likely to be here than here, but I really don't know which of those peaks it's going to be found. OK, now what about this guy? What's the probability distribution? Well, now you see why this is a stupid wave function.

Because it's multiply valued. It has multiple different values at every value of x. So what's the probability? Well, it might be it's root 2, maybe it's 1 over root 3. I'm really not sure, right? So this tells us an important lesson. This is stupid. And what I mean by stupid is it is multiply valued.

So the wave function, we've just learned a lesson, should be single valued. And we will explore some more on your problem set, which will be posted immediately after lecture. There are problems that walk you through a variety of other potential pathologies of the wave function and guide you to some more intuition.

For example, the wave function really needs to be continuous as well. You'll see why. All right, questions at this point. OK, so these look like pretty useful wave function,

because they correspond to the particle being at some definite spot. And I, for example, had a reasonably definite spot. These two wave functions, though, look pretty much useless, because they give us no information whatsoever about what the position is. Everyone agree with that? Except, remember, the de Broglie relations.

The de Broglie relations say that associated to a particle is also some wave. And the momentum of that particle is determined by the wavelength. It's inversely related to the wavelength. It's proportional to the wave number. And the energy is proportional to the frequency. Now, look at those wave functions.

Those wave functions give us no position information whatsoever, but they have very definite wavelengths. Those are periodic functions with definite wavelengths. In particular, this guy has a wavelength of from here to here. Yeah? It has a wave number k, k1.

So that tells us that if we measure the momentum of this particle, we can be pretty confident, because it has a reasonably well-defined wavelength corresponding to some wave number k, 2 pi upon the wavelength. It has some momentum, and if we measure it, we should be pretty confident that the momentum will be h bar k1.

Everyone agree with that? Looks like a sine wave. And de Broglie tells us that if you have a wave with wavelength lambda, that corresponds to a particle having momentum p. Now, how confident can we be in that estimation of the momentum?

Well, if I tell you it's e to the ikx, that's exactly a periodic function with wavelength lambda, 2 pi upon k. So how confident are we? Pretty confident. So our uncertainty in the momentum is tiny. Everyone agree? Similarly, for this wave, again, we

have a wavelength. It's a periodic function, but the wavelength is much shorter. If the wavelength is much shorter, then k is much larger. The momentum is much larger. So the momentum we expect to measure, which is roughly h bar k2, is going to be much larger. What about our uncertainty? Again, it's a perfect periodic function.

So our uncertainty in the momentum is small. Everyone cool with that? And that comes, again, from the de Broglie relations.

So good. So questions at this point? You guys are real quiet today. No questions? Yeah?

So delta p is 0, basically? It's pretty small. Now, again, I haven't drawn this off to infinity. But if it's exactly e to the ikx, then yeah, it turns out to be 0. Now, an important thing. So let me rephrase your question slightly. So the question was, is delta p 0? Is it really 0? So here's the problem for us right now. We don't have a definition of delta p.

So what is the definition of delta p? I haven't given you one. So here, when I've said delta p is small, what I mean is, intuitively, just by eyeball, our confidence in that momentum is pretty good using the de Broglie relations. I have not given you a definition, and that will be part of my job over the next couple lectures. Very good question. Yeah? So how would you encode noise in that function

if you just have different wavelengths as you go along? Awesome. So for example, this. Does it have a definite wavelength? Not so much. So hold that question, and wait until you see the next examples that I put on this board. And if that doesn't answer your question, ask it again, because it's a very important question. When you talk about a photon with a given,

well, you always say a photon has a certain frequency. Doesn't that mean that it must be a wave because you have to fix the wave number k? Awesome question. Does every wave packet of light that hits your eye, does it always have a single unique frequency?

No, you can take multiple frequency sources and superpose them. An interesting choice of words I use there. So the question is, since light has some wavelength, does every chunk of light have a definite? This is the question roughly. And the answer is, light doesn't always have a single wavelength. You can have light coming at you that has many different wavelengths

and put it in a prism and break it up into its various components. So you can have a superposition of different frequencies of light. We'll see the same effect happening for us. OK, so again, de Broglie made this conjecture

that e is h bar omega and p is h bar k. This was verified in the Davis and Germer experiment that we ran. But here, one of the things that's sort of latent in this is what he means is, look, associated to every particle with energy e and momentum p

is a plane wave of the form e to the i kx minus omega t. And this is properly in three dimensions. It should be k dot x. But at this point, this is an important simplification. For the rest of 804, we are going

to be doing one-dimensional, or for the rest of 804 until otherwise specified. We are going to be doing one-dimensional quantum mechanics. So I'm going to remove arrow marks and dot products. There's going to be one spatial dimension and one time dimension. We're always going to have just one time dimension. But sometimes we'll have more spatial dimension. But it's going to be a while until we get there. So for now, we're just going to have kx.

So this is a general plane wave. And what de Broglie really was saying is that somehow associated to the particle with energy e and momentum p should be some wave, a plane wave, with wave number k and frequency omega. And that's the wave function associated to it. The thing is, not every wave function is a plane wave.

Some wave functions are well localized. Some of them are just complicated morasses. Some of them are just a mess. So now is the most important postulate in quantum mechanics.

I remember vividly, vividly, when I took the analog of this class. It's called Physics 143A at Harvard. And the professor at this point said, I know him well now.

He's a friend. He said, this is what quantum mechanics is all about. And I was so psyched. And then he told me. And it was like, that's ridiculous. Seriously? That's what quantum mechanics is all about? So I always thought this is some weird thing where old physicists go crazy. But it turns out, I'm going to say exactly the same thing. This is the most important thing in all of quantum mechanics. It is all contained in the following proposition, everything.

The two slit experiments, the box experiments, all the cool stuff in quantum mechanics, all the strange and counterintuitive stuff comes directly from the next postulate. So here it is. I love this.

Three. Put a star on it. Given two possible wave functions or states, I'll say configurations, of a quantum system.

I wish someone was like, Ride of the Valkyries, playing in the background. So of a quantum system, corresponding to two distinct wave functions, f with an upper ns

is going to be my notation for functions, because I have to write it a lot, psi 1 and psi 2. And I'll say of x, of x.

The system is down. The system can also be in a superposition of psi 1 and psi

2, where alpha and beta are complex numbers.

Given any two possible configurations of the system, there is also an allowed configuration of the system corresponding to being in an arbitrary superposition of them. If an electron can be hard and it can be soft,

it can also be in an arbitrary superposition of being hard and soft. And what I mean by that is that hard corresponds to some particular wave function. Soft will correspond to some particular wave function. And a superposition corresponds to a different wave function, which is a linear combination of them.

Yeah. Yeah, OK, that's a very good question. So alpha and beta are some complex numbers subject to the normalization condition. So indeed, this wave function should be properly normalized

at the end. Now let me step back for a second. There's an alternate way to phrase the probability distribution here, which goes like this. And I'm going to put it here. The alternate statement of the probability distribution is that the probability density at x is equal to psi of x norm squared

divided by the integral over all x dx of psi squared. So notice that if we properly normalize the wave function, this denominator is equal to 1. But if we haven't properly normalized,

and so it's not there, then it's equivalent. But if we haven't properly normalized it, then this probability distribution is automatically properly normalized. Because this is a constant. When we integrate the top, that's equal to the bottom. It integrates to 1. Cool? So I prefer, personally, in thinking about this for the first pass, to just require that we always be careful to choose some normalization.

That won't always be easy. And so sometimes it's useful to forget about normalizing and just define the probability distribution that way. Is that cool? OK. This is the beating soul of quantum mechanics. Everything in quantum mechanics is in here. Everything in quantum mechanics is forced on us

from these few principles and a couple of requirements of matching to reality. Yes. Can you get interference? Yes. So the question is, when you have a sum of two wave functions, can you get some sort of interference effect?

And the answer is absolutely. And that's exactly what we're going to do next. So in particular, let me look at a particular pair of superpositions. So let's swap these boards around. The parallelism is a little more obvious. OK. So let's scrap these rather silly wave functions and come up with something that's

a little more interesting. So instead of using those as characteristic wave functions, I want to build superpositions. So in particular, I want to start by taking an arbitrary. Both these wave functions are simple interpretations. This corresponds to a particle being here. This corresponds to a particle being here. I want to take a superposition of them. So here's my superposition.

OK. Oops. Start that again. Good. And my superposition, so here's 0, and here's x1, and here's x2. My superposition is going to be some amount times the first one plus some amount times the second one.

There's a superposition. Similarly, I could have taken a superposition of the two functions on the second chalkboard. And again, I'm taking a superposition of the complex e to the ik1x and e to the ik2x and then taking the real part.

So that's a particular superposition, a particular linear combination. So now let's go back to this. This was a particle that was here. This was a particle that was there.

When we take the superposition, what is the probability distribution? Where is this particle? Well, there's some amplitude that it's here, and there's some amplitude that it's here. There's rather more amplitude than it's over here, but there's still some probability that it's over here. Where am I going to find the particle?

I'm not so sure anymore. It's either going to be here or here, but I'm not positive. And it's more likely to be here than it is to be here, but not a whole lot more. So where am I going to find the particle? Well, now we have to define this, where I'm going to find the particle. Look, if I did this experiment a whole bunch of times, it'd be over here more than it would be over here.

So the average will be somewhere around here. It'll be in between the two. So x is somewhere in between. That's where we expect to find it, on average. What's our uncertainty in the position?

Well, it's not that small anymore. It's now of order x1 minus x2. Everyone agree with that? Bless you.

Now, what about this guy? Well, does this thing have a period? Does it have a single wavelength, sorry? No. This is like light that comes at you from the sun. It has many wavelengths. In this case, it has just two. I've added those two together. So this is a plane wave, which is psi is e to the ik1x

plus e to the ik2x. So in fact, it has two wavelengths associated with it, lambda 1 and lambda 2. And so the probability distribution now, if we take the norm squared of this,

the probability distribution now, which is the norm squared of this guy, is no longer constant. But there's an interference term. And let's just see how that works out.

We need to be very explicit about this. Note that the probability in our superposition of psi 1 plus psi 2, which I'll call e to the ik1x plus e to the ik2x, the probability is equal to the norm squared of the wave function, which is the superposition psi 1 plus beta psi 2, which is equal to alpha squared psi 1 squared

plus beta squared psi 2 squared plus alpha star psi 1 star plus alpha star psi 1 star beta psi 2

star plus alpha psi 1 beta star psi 2 star, where star means complex conjugation. But notice that this is equal to that first term is alpha squared times the first probability,

or the probability of this thing, of alpha psi 1. This is equal to probability 1. This term, beta squared psi 2 squared, is the probability that the second thing happens. But these terms can't be understood in terms of the probabilities of psi 1 or the probability of psi 2 alone.

They're interference terms. So the superposition principle, together with the interpretation of the probability as the norm squared of the wave function, gives us a correction to the classical addition

of probabilities, which is these interference terms. Everyone happy with that? And here's something very important to keep in mind. These things are norm squareds of complex numbers. That means they're real, but in particular, they're non-negative. So these two are both real and non-negative. But what about this?

This is not the norm squared of anything. However, this is its complex conjugate. When you take something in its complex conjugate, you add them together, you get something that's necessarily real. But it's not necessarily positive. So this is a funny thing, the probability that something happens if we add together two configurations, we superpose two configurations, has positive probability terms, but it's also

got terms that don't have a definite sign. It could be negative. It's always real. And you can check that this quantity is always greater than or equal to 0. It's never negative, the total quantity. So remember Bell's inequality that we talked about? Bell's inequality said, look, the probability of one thing happening, if we have the probability of one thing happening being p1 and the probability

of the other thing happening being p2, the probability of both things happening is just p1 plus p2. And here we see that in quantum mechanics, probabilities don't add that way. The wave functions add, and the probability is the norm squared of the wave function. The wave functions add, not the probabilities.

And that is what underlies all of the interference effects we've seen. And it's going to be the heart of the rest of quantum mechanics. So you're probably all going in your head, more or less like I was when I took intro quantum, like, yeah, but I mean, you're adding complex numbers. But trust me on this one.

This is where it's all starting. OK, so let's go back to this. Similarly, let's look at this example. We've taken the norm squared, and now we have an interference effect. And now our probability distribution, instead of being totally trivial and containing no information, our probability distribution now contains some information about the position of the object.

It's likely to be here. It is unlikely to be here, likely and unlikely. We now have some position information. We don't have enough to say where it is, but x is still, you have some information. OK, now our uncertainty is still gigantic. Delta x is still huge. But OK, we just added together two plane waves.

Yeah? Big, small. Excellent. This was the real part of the wave function. And the wave function is a complex quantity. When you take e to the ik1, let me do this on the chalkboard.

When we take e to the ik1x plus e to the ik2x, let me write this out differently. e to the ia plus e to the ib and take its norm squared. So this is equal to, I'm going to write this in a slightly more suggestive way, the norm squared of e to the ia times 1 plus e

to the ib minus a, norm squared, parentheses, norm squared. So first off, the norm squared of a product of things is the product of the norm squareds. So I can do that. And this overall phase, the norm squared of a phase is just 1.

So that's just 1. So now we have the norm squared of 1 plus a complex number. And so the norm squared of 1 is going to give me 1. The norm squared of the complex number is going to give me 1. And the cross terms are going to give me the real part, twice the real part, of e to the ib minus a,

which is going to be equal to cosine of b minus a. And so what you see here is that you have a single frequency in the superposition. Cool? OK. So again, here, oh, so good.

So our uncertainty is large. So let's look at this second example in a little more detail. By superposing two states with wavelength lambda 1 and lambda 2, or k1 and k2, with wavelength lambda 1

and lambda 2, we get something that's, OK, it's still not well localized. We don't know where the particle is going to be. But it's better localized than it was before. What happens if we superpose with three wavelengths, or four or more? So for that, I want to pull out a Mathematica package.

You guys should all have seen Fourier analysis in 1803. But just in case, I'm putting on the web page, on the Stellar page, a notebook that walks you

through the basics of Fourier analysis in Mathematica. You should all be fluent in Mathematica. If you're not, you should probably come up to speed on it. That's not what we wanted. Let's try that again. There we go. Oh, that's awesome.

Or by awesome, I mean not. It's coming. OK, good. I'm not even going to mess with the screens after last time.

So I'm not going to go through the details of this package. But what this does is walk you through the superposition of wave packets. So here, I'm looking at the probability distribution coming from summing up a bunch of plane waves with some definite frequency.

So here, it's just one. That's one wave. So first we have, let me make this bigger. Yes, stupid Mathematica tricks. So here we have the wave function. And here we have the probability distribution,

the norm squared. And it's sort of badly normalized here. So that's for a single wave. And as you see, the probability distribution is constant. And that's not 0. That's 0.15. It's just I arbitrarily normalized this. So let's add two plane waves. And now what you see is the same effect as we had here.

You see a slightly more localized wave function. Now you have a little bit of structure in the probability distribution. So there's the structure in the probability distribution. We have a little more information about where the particle is. More likely to be here than it is to be here. Let's add one more. And as we keep adding more and more plane waves to our superposition, the wave function and the probability distribution associated with it become more

and more well localized. Until as we go to very high numbers of plane waves that we're superposing, we get an extremely narrow probability distribution, and wave function for that matter, corresponding to a particle that's very likely to be here and unlikely to be anywhere else. Everyone cool with that?

What's the expense? What have we lost in the process? Well, we know with great confidence now that the particle will be found here upon observation. But what will its momentum be? Yeah, now it's a superposition of a whole bunch of different momenta. So if it's a superposition of a whole bunch of different momenta, here, this

is like superposition of a whole bunch of different positions. Likely to be here, likely to be here, likely to be here, likely to be here. What's our knowledge of its position? It's not very good. Similarly, now that we have superposed many different momenta with comparable strength, in fact, here they were all with unit strength, we now have no information about what the momentum is anymore.

It could be anything in that superposition. So now we're seeing, quite sharply, the uncertainty relation. And here it is. So the uncertainty relation is now pretty clear from these guys.

Oops, that didn't work. Good. So the uncertainty relation is, oh, and I'm going to leave it alone. This is enough for the Fourier analysis. But that Fourier package is available with extensive commentary on the cellular page. Yeah? Sharp definition in the position caused by the interference between all those waves and all that.

That's exactly what it is. Precisely. It's precisely the interference between the different momentum modes that leads to certainty in the position. That's exactly right. Yeah? As we start to know the position, we will not start to know the position. Exactly. Here we are. So in this example, we have no idea what the position is.

We're quite confident of the momentum. Here we have no idea what the position is. But we have great confidence in the momentum. Similarly here, we have less perfect confidence of the position. And here we have less perfect confidence in the momentum. It would be nice to be able to estimate what our uncertainty is in the momentum here

and what our uncertainty is in the position here. So we're going to have to do that. That's going to be one of our next tasks. Other questions? Yeah? Yeah? You said, obviously, if we do it a bunch of times, we'll have more in the x2 than in the x1.

Yes. The average, it will never physically be at that point. Yeah, that's right. Because it's a probability distribution, it won't be exactly at that point, but it'll be nearby. So in order to be more precise, and so for example, for this, here's a quick question. How well do you know the position of this particle?

Pretty well, right? But how well do you know its momentum? Well, so we'd all like to say not very, but tell me why. Why is your uncertainty in the momentum of the particle large? Yeah, but that's a cheat, because we haven't actually proved Heisenberg's uncertainty principle. It's just something we're inheriting.

I believe it, too. But I want a better argument, because I believe all sorts of crazy stuff. So I really do. So black holes, fluids, I mean, don't get me started. Yeah, excellent. OK, we'll get to that in just one sec. So before taking a Fourier transform, so the answer was, just take a Fourier transform.

It's going to give you some information. We're going to do that in just a moment. But before we do a Fourier transform, just intuitively, why would de Broglie look at this and say, no, that doesn't have a definite momentum. Yeah, there's no wavelength. It's not periodic by any stretch of the imagination. It doesn't look like a thing with a definite wavelength. And de Broglie said, look, if you

you have a definite wavelength, then you have a definite momentum. And if you have a definite momentum, you have a definite wavelength. This is not a wave with a definite wavelength. So it is not corresponding to the wave function for a particle with a definite momentum. So our momentum is unknown. So this is large. And similarly here, our uncertainty in the momentum is large.

So to do better than this, we need to introduce the Fourier transform. And I want to do that now. So you should all have seen Fourier series in 8.03. Now we're going to do the Fourier transform. And I'm going to introduce this to you in 8.04 conventions

in the following way. And the theorem says the func, we're not going to prove it by any stretch of the imagination, but the theorem says any function f of x that is sufficiently well-behaved, it shouldn't be discontinuous, it shouldn't be singular, any reasonably well-behaved non-stupid function f of x can be built by enough plane waves

of the form e to the ikx.

Enough may be infinite. So f of x, any function f of x can be expressed as 1 over root 2 pi. And this root 2 pi is a choice of normalization. Everyone has their own conventions. And these are the ones we'll be using in 8.04 throughout. Minus infinity to infinity dk f tilde of k e to the ikx.

So here what we're doing is we're summing, or summing

over plane waves of the form e to the ikx. These are modes with a definite wavelength 2 pi upon k. f tilde of k is telling us the amplitude of the wave with wavelength lambda or wave number k.

And we sum over all possible values. And the claim is any function can be expressed as a superposition of plane waves in this form. Cool? And this is for functions which are non-periodic on the real line, rather than periodic functions on the interval, which is what you should have seen in 8.03.

Now conveniently, if you know f tilde of k, you can compute f of x by doing a sum. But suppose you know f of x and you want to determine what the coefficients are, the expansion coefficients. That's the inverse Fourier transform. And the statement for that is that f tilde of k is equal to 1 over root 2 pi integral from minus infinity

to infinity dx f of x e to the minus ikx. And that's sometimes referred to as the inverse Fourier transform. And here's something absolutely essential.

f tilde of k, the Fourier transform coefficients of f of x are completely equivalent. If you know f of x, you can determine f tilde of k. And if you know f tilde of k, you can determine f of x by just doing a sum, by just adding them up. So now here's the physical version of this.

Oh, I can't slide that up. The physical version of this I'm now going to put here. Oh, no, I'm not. I'm going to put that down here. So the physical version of this is that any wave function psi

of x can be expressed as the superposition in the form psi

of x is equal to 1 over root 2 pi integral from minus infinity to infinity dk psi tilde of k e to the ikx. Whoops.

Of states, or wave functions, with a definite momentum,

p is equal to h bar k. And so now it's useful to sketch the Fourier transforms

of each of these functions. So in fact, we want this up here. So there is erasers. Whoops, carnage down there.

So here we have the function and its probability distribution. Now I want to draw the Fourier transforms of these guys. So here is psi tilde of k, function of a different variable than of x.

But nonetheless, it's illuminating to draw them next to each other. And again, I'm drawing the real part. And here, x2, had I had my druthers about me,

I would have put x2 at a larger value. Good. So it's further off to the right there.

And similarly, I'm so loathe to erase the superposition principle. But fortunately, I'm not there yet. Let's look at the Fourier transform of these guys.

The Fourier transform of this guy is now, whoops, this is k psi tilde of k.

Well, that's something with a definite value of k. And its Fourier transform, this is 0, there's k1. And for this guy, there's 0, k2.

And now if we look at the Fourier transforms of these guys, this way I don't

have to erase the superposition principle, the Fourier transforms of these guys and the Fourier transform of this guy.

So note that there's a sort of pleasing symmetry here. If your wave function is well localized, corresponding to a reasonably well-defined position, then your Fourier transform is not well localized,

corresponding to not having a definite momentum. On the other hand, if you have definite momentum, your position is not well defined. But the Fourier transform has a single peak. The Fourier transform has a single peak at the value of k corresponding to the momentum of your wave,

of your wave function. Everyone cool with that? So here's a question. Sorry, there was a raised hand. Yeah? Are we going to learn, in this class, how to determine the Fourier transforms of these non-steward functions? Yes, that will be your homework.

On your homework is an extensive list of functions for you to compute Fourier transforms of. And that will be the job of problem sets and recitation. So Fourier series and computing, yeah, you know what's coming. Fourier series are assumed to have been covered for everyone in 803 and 1803 in some linear combination thereof.

And Fourier transforms, I couldn't help it. So Fourier transforms are slight in beginning of the space of Fourier series, because we're not looking at periodic functions. Yeah? Fourier transform of a wave function,

we're basically writing it as a continuous set of different waves. Can we write it as a discrete set, so use a Fourier series? Absolutely. So however, what is true of a Fourier series? When you use a discrete set of momentum, which are linear, it must be periodic function, exactly. So the question is, do you always have to, so here what we've done is we've said, look,

we're writing our wave function, our arbitrary wave function, as a continuous superposition of a continuous value of possible momenta. This is absolutely correct. It's exactly what we're doing. However, that's kind of annoying, because continuous, maybe you just want one momentum, and two momentum, and three momentum. What if you want a discrete series? So discrete is fine, but if you make that discrete series integer related to each other,

which is what you do with Fourier series, you force the function f of x to be periodic. And we don't want that in general, because life isn't periodic. Thank goodness, right? I mean, there's like one film in which it is, but so, it's a good movie. So no, so that's the central difference between Fourier series and Fourier transforms.

Fourier transforms are continuous in k, and do not assume periodicity of the function. Other questions? Yeah? Of associates an amplitude and a phase for each of the individual momenta? Precisely. Precisely correct.

So let me say that again. So the question was, so a Fourier transform effectively associates a magnitude and a phase for each possible wave vector. And that's exactly right. So here, there's some amplitude and phase. This is a complex number, because this is a complex function. There's some complex number which is an amplitude and a phase associated to every possible momentum going into the superposition.

That amplitude may be 0. There may be no contribution for a large number of momenta, or maybe insignificantly small. But it is indeed doing precisely that is associated an amplitude and a phase for every plane wave with every different value of momenta. And you can compute before panicking,

you can compute precisely what that amplitude and phase is by using the inverse Fourier transform. So there's no magic here. You just calculate it. You can use your calculator, literally. Hate that word. So now here's a natural question.

So if this is the Fourier transform of our wave function, this wave function, we already knew that this wave function corresponded to having a definite from de Broglie. We know that it has a definite momentum. We also see that its Fourier transform looks like this. So that leads to a reasonable guess. What do you think the probability distribution p of k

is, probability density, the probability density to find the momentum to have wave vector k, h bar k. Yeah, that's a pretty reasonable guess. So here's just, we're totally pulling this out of the dark, psi of k norm squared. OK, well, let's see if that works. So psi of k norm squared for this is going to give us. Nice, well localized function.

Similarly, psi of k for this. And so that makes a lot of sense. It's exactly what we expected. Definite value of p with very small uncertainty. Similarly here, definite value of p with very small uncertainty. Rock on. However, let's look at this guy. What is the expected value of p if this

is the Fourier transform? Well, remember, we have to take the norm squared and psi of k was e to the i kx1, the Fourier transform. You will do much practice on taking Fourier transforms on the problem set. Where did my eraser go? There it is.

Farewell, principle one. So what does norm squared of psi tilde look like? Well, just like before, the norm squared is constant, because the norm squared of a phase is constant. And again, the norm squared, so this is psi tilde of k norm squared. We believe we're conjecturing this is p of k.

You will prove this relation on your problem set. You'll prove that it follows from what we said before. And similarly, this is constant, e to the ikx2. So now we have no knowledge of the momenta.

So that also fits. The momenta is we have no idea, and uncertainty is large. And the momentum is we have no idea, and the uncertainty is large. So in all of these cases, we see that we satisfy quite nicely the uncertainty relation, small position momentum, large momentum uncertainty. Large position uncertainty, we're allowed

to have small momentum uncertainty. And here, it's a little more complicated. We have a little bit of knowledge of the position, and we have a little bit of knowledge of the momenta. We have a little bit of knowledge of position, and we have a little bit of knowledge of the momenta. So we'll walk through examples with superposition like this on the problem set. OK. OK.

Last questions before we get going. OK, so I have two things to do before we're done. The first is, after lecture ends, I have clickers. And anyone who wants to borrow clickers, you're welcome to come down and pick them up in a first come, first serve basis. I will start using the clickers in the next lecture.

So if you don't already have one, get one now. But the second thing is, don't get started yet. I have a demo to do. And last time I told you, this is awesome. It's like having an experimentalist for a day. Last time I told you that one of the experimental facts of lice, one of the experimental facts of life

is that there is uncertainty in the world and that there is probability, that there are unlikely events that happen with some probability, some finite probability. And a good example of the randomness of the real world involves radiation. So hopefully you can hear this.

Apparently, I'm not very radioactive. You'd be surprised at the things that are radioactive. Oh, got a little tick. This is a plate sold at an Amish county fair.

It's called Vaseline Ware, and it's made of local clays. It's got uranium in it. But even more exciting, I want to emphasize exactly when something goes click, it sounds pretty random.

And it's actually a better random number generator than anything you'll find in Mathematica or C. In fact, for some purposes, the decay of radioactive isotopes is used as the perfect random number generator, because it really is totally random as far as anyone can tell. But here's my favorite.

People used to eat off these. See you next time.