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  • 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

  • is 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 the wave function, and is 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 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

  • ½mv^(2), 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(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 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 have 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 that you take this ψ and you square it.

  • So you will get something that will go to 0 here,

  • go up, go to zero, do something like that.

  • This is ψ^(2), and that's your probability

  • density and what we mean by density is,

  • if the function p(x) has an area with the x-axis

  • like p(x)dx that's the actual probability that if you

  • look for this guy you'll 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?"

  • 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.

  • I mean, 7 divided by 11 is one thing and 4 divided by 11 is the

  • absolute probability.

  • So in quantum 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 the 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.

  • And 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 one ψ you're free to multiply it by

  • any number, in fact, real or complex and

  • 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.

  • You understand now?

  • What are the possible functions I can ascribe with the 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 is 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 = 5

  • because I just saw it there and 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 = 5?

  • Student: >

  • Prof: Yeah?

  • Centered where?

  • Student: Centered at 5.

  • Prof: 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 isn't

  • 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 if you change the shape

  • a bit, but roughly speaking here is

  • what functions, describing particles of

  • reasonably well know location, look like.

  • They're centered at the point which is the well-known

  • location.

  • On the other hand, I'm going to call is ψ

  • x = 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 am human."

  • 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 = 5 is one member of the family,

  • which means I'm peaked at x = 5.

  • Another function I mentioned is the function

  • ψ_p(x).

  • That's the 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^(ipx/ℏ).

  • 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 more natural to think of a

  • complex number as having a modulus and a 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/ℏ.

  • So if I wrote a function e to the i times

  • 96x/ℏ, and I said, "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 x of the i, the x and the

  • ℏ.

  • Whatever is sitting there that's the momentum.

  • I am 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^(2) 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

  • square is some fixed number.

  • I hope you all know the absolute value of that is 1

  • because that times its conjugate,

  • which is e^(−ipx/â„ ) will just give you

  • e^(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 A to

  • make that 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 in 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.

  • It goes beyond this room.

  • It goes beyond the planet.

  • It goes beyond the 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 is 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 this universe

  • because everything goes around in a circle.

  • 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/√L e^(ipx/â„ ).

  • Do you agree?

  • If you take the absolute square of this you'll get a 1/L.

  • This'll become 1.

  • The integral of 1 of 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 nano scale object,

  • and the electrons are forced to lived 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 you 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 in a

  • line you just pretend that the line closes in on itself.

  • That if you start at the origin and you go on the two sides it

  • closes in.

  • If this is x = 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 looks like this of momentum

  • p.

  • This is 1/√L e^(ipx/â„ ).

  • The probability to find this guy at some point x which

  • is the absolute value of ψ^(2) is just 1/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.

  • Now let's ask one other question.

  • Here is the circle in which I'm living.

  • If you look at the real part and imaginary part with their

  • sines and cosines they kind of oscillate.

  • Do you understand that?

  • When the x varies they oscillate.

  • And one requirement you make 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 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 the ψ,

  • you go all the way around and you come back you shouldn't have

  • a mismatch.

  • It should agree.

  • So the only allowed functions are those obeying the condition

  • ψ(x L) = ψ(x).

  • 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 −x^(2) is not a

  • good function to 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 the right 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^(ipx) at some point x then I add to it

  • L, namely I'm getting

  • e^(ipx/ℏ) times e^(ipL/ℏ).

  • That has to agree with the starting value which is

  • e^(ipx/ℏ).

  • 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 a function of x.

  • These cancel and you learn then that e^(ipL/ℏ) should

  • be 1.

  • Now how can that be 1?

  • Well, one way is p = 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?

  • Student: >

  • Prof: Pardon me?

  • Student: >

  • Prof: I didn't hear that, sorry.

  • Student: It's infinite.

  • Prof: I didn't hear that again.

  • Student: It's infinite >

  • Prof: You still have to say it louder.

  • Student: If it is infinite

  • >

  • Prof: Any other answers?

  • Yeah?

  • Student: If it's a multiple of 2Π.

  • Prof: Is that what you said multiple of 2Π?

  • Student: Sure.

  • Prof: Good.

  • I'm glad you didn't say it loud enough.

  • Okay good.

  • It's a multiple ofbecause any trigonometry

  • function if you add 2Π, or 4Π, or

  • whose argument doesn't change then you should never forget the

  • fact that this exponential is the sum of two trigonometric

  • functions cosine i sine, and they all come back.

  • Therefore, it is too strong to say p should be 0.

  • p should be such that pL/â„ is

  • times any integer when the integer can be 0,

  • plus minus 1, plus minus 2 etcetera.

  • That means p has to be equal to 2Πâ„ /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 living in there,

  • 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..

  • 2Πâ„ /L times 0 is here, times 1 is here,

  • times 2 is here, 3 is here, -1,

  • -2.

  • These are the only allowed values of p.

  • So 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 2Πâ„ /L

  • times m.

  • I can write it as p times L/2Π

  • is equal to mℏ.

  • L/2Π is the radius of the circle.

  • So I find that p times R is equal to

  • mℏ.

  • You guys know what p times R is for a particle

  • moving in a circle at momentum p?

  • What is the momentum times the radial distance to the center?

  • Student: Angular momentum.

  • Prof: Angular momentum.

  • That's usually denoted by a quantity L in quantum

  • mechanics.

  • So angular momentum is an integral multiple of ℏ.

  • That's something you will find even in high school people will

  • tell you angular momentum is an integral multiple of ℏ.

  • Where does it come from and how does it come from quantum

  • mechanics?

  • Here is one simple context in which you can see the 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^(2Π), I'm sorry,

  • e^(ipx/ℏ), but to remind myself that p

  • is quantized to be integer multiples of some basic quantum

  • where the multiple is just m,

  • I may also write the very same function the ψ's of m

  • is 1/√L e^(ix/â„ ),

  • but for p you put the allowed value which is

  • 2Πâ„ /L times m.

  • And what do you get?

  • You get 1/√L times e^(2Πimx/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 ℏ,

  • 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 microscopic 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 in 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 of so densely

  • close, packed, that you don't know

  • whether you've got this one, or that one or something in

  • between because you cannot measure it that well.

  • So it'll 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 ℏ.

  • I mean, it should be much bigger thanthen

  • it'll become continuous.

  • 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(x), suppose it's not one of these

  • functions?

  • It's not e^(ipx/ℏ).

  • 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 the 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 the 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(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?

  • Student: You square it you get the probability

  • Prof: 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.

  • 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 the probability

  • because it can be negative.

  • It can even be complex.

  • So don't forget if you square it you get the probability.

  • Define 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?

  • 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^(2) is probability defined 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's actually contained in the same function

  • ψ of x in the following fashion.

  • Take this function ψ(x) and write it as a sum over

  • p of these functions, ψ_p of x

  • with some coefficient A_p.

  • I can also write it equally well as the same function

  • labeled by integer m and I want to call the coefficient m.

  • They're exactly the same thing, p and m are

  • synonymous.

  • The same function.

  • This function is called ψ_m 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 4 complete oscillations as they go

  • around the cycle.

  • All right, so here's 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.

  • Let me repeat the postulate.

  • Somebody gives you a function.

  • You write the function as a sum of all these periodic functions,

  • each with the 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've done that the coefficient squared with the

  • particular value m is the probability you will get that

  • value for m or the corresponding momentum.

  • So there are two questions one can ask at this point.

  • 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^(2Πimx/L)√L times A_m.

  • 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 there is a vector

  • V, and you pick for yourself three

  • orthonormal vectors,

  • i, j and k, then any vector

  • V can be written as V_x times

  • i V_y times j

  • V_z 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-dimension.

  • Yes?

  • Student: With the Fourier case would there be one

  • unique way of writing ________?

  • Prof: Good point.

  • That's correct.

  • There will be a unique way.

  • Student: There is?

  • Prof: There is.

  • You agree in three dimensions there is no other mixture except

  • this one.

  • If somebody comes with the 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 the ψ_m.

  • The only difference is there you need only three guys.

  • Here you need an infinite number of them because a range

  • of m goes from minus to plus 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 because 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 coefficients?

  • Suppose I write the vector V as e_1

  • times V_1 e_2 times

  • V_2 e_3 times

  • V_3.

  • Don't worry e_1,

  • e_2 and e_3 are the

  • usual guys, e_1 is

  • i, e_2 is j and

  • e_3 is k.

  • People like to do that because in mathematics you may want to

  • go to 96 dimensions, but we've only got 26 letters,

  • so if you 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,

  • 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.

  • Namely the dot product of one guy with himself is 1,

  • and any one 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 e_i

  • ⋅ e_j,

  • 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 a usual vector analysis.

  • I'm just saying the dot product of basis vectors has this

  • property, 1 if they match,

  • 2 if they're different, so there's a shortcut for this

  • and that's write the symbol δ_ij.

  • δ_ij is called Kronecker's

  • delta.

  • Kronecker's in a lot of things, but this is 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, the Kronecker's symbol,

  • δ_ij.

  • It's understood that this whole thing simply says

  • δ_ij.

  • This is shorthand.

  • That means if on the left hand side there are two guys with

  • indices i and j if the indices are equal the 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 is of 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

  • e_i times V_ii 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's your vector V.

  • Here is, if you like, e_1,

  • e_2 and e_3.

  • The claim is some mixture of e_1,

  • e_2, e_3 will add

  • up to this V.

  • That's granted, but how much?

  • How much e_1 do I need?

  • How much e_2 do I need?

  • There's a very simple trick for that.

  • Anybody know what that trick is?

  • You might know that trick.

  • You've seen it anywhere?

  • Here is the trick.

  • Suppose you want to find V_2?

  • You take the dot product of both these things with

  • e_2.

  • Take e_2 dot this, and take dot product with

  • e_2.

  • What happens is you will get--the dot product can go

  • inside.

  • You'll get e_i ⋅

  • e_2 times V_i where

  • i goes from 1 to 3.

  • What is e_i ⋅

  • e_2?

  • e_i ⋅

  • e_2 is δ_i2.

  • That means if this index size is equal to 2 you get 1.

  • If not equal to 0 you strike out.

  • You get 0.

  • So of all the three terms in this only one will survive.

  • That's the one when i is equal to 2 in which case you

  • will get 1.

  • That's multiplying V_2 so it'll

  • give you V_2.

  • So to find the component number 3 you take the dot product of

  • the given vector with e_3,

  • and that will give you, you can see,

  • that will give you V_3 or

  • V_2 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 = sum over i

  • e_i times V_i,

  • I have ψ of x is equal to sum

  • over m of some A_m times the

  • function ψ_m of x.

  • You understand that?

  • ψ_m is a particular function which I

  • don't want to write every time, but if you insist it is

  • imx/L divided by √L.

  • So I'm trying to find this guy.

  • How much is it?

  • That's the question.

  • Now here we had a nice rule.

  • The rule says e_i ⋅

  • e_j is δ_ij.

  • 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 ψ_m*(x)

  • times ψ_n(x) dx

  • from 0 to L is in fact δ_mn.

  • So the basis vectors ψ_m are like

  • e_i.

  • The dot product of two basis vectors being

  • δ_ij is the same here,

  • like same integral, but 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 this conjugate is just 1.

  • You just get 1/L.

  • An integral of 1/L dx, so if you want I will write it

  • here, it's 2Πi times

  • n −mx/L dx from 0 to L.

  • Do you understand that?

  • You take the conjugate of the first function.

  • That where's it's a -m here, and take the second function

  • which is e^(2Πin)^(/L)

  • times x.

  • I'll wait until you have time to digest that.

  • The product of ψ_m* with

  • ψ_n you combine the two exponentials,

  • but the thing that had m in it has a -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^(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 the period,

  • the sine and the cosine, but whenever you integrate a

  • sine or a cosine over some 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 usual vector

  • analysis and these functions.

  • This is an arbitrary vector.

  • This is an arbitrary function.

  • e_1, e_2,

  • e_3 are basis vectors;

  • ψ_m are 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?

  • You want to find the coefficient

  • A_n.

  • What did I do here to find V_j?

  • You take V⋅ e_j 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 e_j the only

  • term who survives is i = j.

  • That'll give you the V_j.

  • So similarly here I claim the following is true.

  • If you do the integral of ψ_n* x

  • times the given function dx from 0 to L you

  • will get A_n.

  • 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 coefficients 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 ψ_13* 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 ψ_n*(x).

  • The given function looks like a sum A_m

  • ψ_m(x) dx 0 to L summed over

  • m.

  • So this summation you see you can bring the ψ

  • in 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

  • A_m of ψ_m of

  • ψ_n* x ψ_m(x)

  • dx, and that is going to be

  • δ_mn.

  • 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 A_n.

  • Once again, you will see this in my notes, but do you have any

  • idea of what I did or where this is going?

  • In quantum theory if you want to know what'll 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 item defined as a definite momentum with suitable

  • coefficients.

  • The rule for finding coefficient A_n

  • is to do this integral of ψ_n*.

  • This one, this is the rule.

  • A_n is the integral of

  • ψ_n*ψ dx.

  • And once you found the coefficients for all possible

  • n then the probability that you will have some momentum

  • corresponding to m is just A_m^(2).

  • That is a recipe.

  • What the recipe tells you is if your function is made up as 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 is 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(x) it's likely

  • where it's big, unlikely where it's small and

  • impossible where it is 0.

  • So that's your job.

  • Anytime 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'll 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.

  • Some number n cosine 6Πx/L,

  • somebody gives you that function.

  • That is not a state of definite momentum because it is 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 functions, 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_m_

  • squares, you will find amazingly it adds

  • up to 1.

  • Because you can show mathematically that if

  • ψ*ψ dx is 1 then the coefficients 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.

  • There's also a mathematical result which I am 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^(2) times integral of cosine square as

  • 6Πx/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 a half.

  • That means over the length L this integral will be

  • L/2.

  • So you want N^(2)L/2 to be 1 or you want n to be square

  • root of 2/L.

  • So the normalized function looks like square root of

  • 2/L cosine 6Πx/L.

  • So that's the first job.

  • Second is you can ask now what are the coefficients

  • A_n.

  • So A_n is going to be integral

  • 1/√L e^(-2Πinx/L)

  • times this function square root of 2/L cosine 6Πx/L

  • dx.

  • Are you with me?

  • That's the rule.

  • Take the function, multiply it by

  • ψ_n*, which is this guy here,

  • this is the ψ_n*, this is the ψ

  • that's given to you, and you're integrating the

  • product in the integral.

  • That'll give you A_n.

  • But I claim, yeah, if you want you can do

  • this integrals, but I think there's a quicker

  • way to do that.

  • Have any idea of what the quicker way just by looking at

  • it?

  • In other words, if you can guess the expansion

  • I'll 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^(2Πimx/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?

  • Student: Euler's formula?

  • Prof: And what does it say?

  • Student: e^(ix) = cosine x i sine

  • x.

  • Prof: Yes, but now I want to go the other

  • way.

  • It is certainly true that cosine θ,

  • I'm sorry, e^(iθ) is cosine θ

  • I sine θ.

  • I told you, you forget the formula at your own peril.

  • The conjugate of that is e^(-iθ) is cosine

  • θ -i sine θ.

  • If you add these two you'll find cosine θ

  • is (e^(iθ) e^(-iθ))/2,

  • and if you subtract them and divide by 2i you'll find

  • sine θ is e to the minus over 2i.

  • 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 squared cosine squared is 1,

  • but you don't look it up.

  • That's something you know.

  • What you look up is your Social Security number or mother's

  • maiden name.

  • You're allowed to forget those things.

  • You look in a book.

  • That's the name.

  • But this you have to know because you cannot go anywhere

  • without this because if you don't know it all the time you

  • must have done these trigonometry calculations in

  • school.

  • 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

  • ψ(x), which was given to us,

  • is equal to the square root of 2/L times cosine

  • 6Πx/L, but I'm going to write it as

  • square root of 2/L times ½e^(6Πix/L)

  • e^(-6Πix/L).

  • So that I will write it very explicitly as 1/√2 times

  • e^(6Πix/L)/√L (1/√2)e^(-6Πix/L)

  • divided by √L.

  • In other words, what I've done is rather than

  • doing integrals I just massaged the given function and managed

  • to write it as a sum over normalized functions,

  • as I said, with definite momentum with some coefficients.

  • In other words, here it is in the form that we

  • want.

  • A_m e^(2Πimx/L).

  • You agree that I have got it to the form I want?

  • You see that?

  • You take the cosine, write it as sum of

  • exponentials, then put factors of

  • √L so that that's a normalized function,

  • and that's a normalized function, and everybody else is

  • A.

  • So what do you find here?

  • What are the A's for this problem?

  • By comparing the two what do you find are the A's in

  • this problem?

  • What is A_14?

  • 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?

  • Student: 3.

  • Prof: Pardon me?

  • Student: 3.

  • Prof: m is 3 for this guy,

  • so this is really ψ_3 and this is

  • ψ_-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_3 is 1/√2,

  • and A_-3 is also 1/√2.

  • Therefore, the probability for m = 3 is ½,

  • and the probability for m = -3 is ½ and nothing else.

  • All other A's 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 and 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/L,

  • which is a real wave function, and you can plot that guy going

  • around the circle, it describes a particle whose

  • momentum, if you measure,

  • will give you only 1 of 2 answers,

  • m = 3 is p = 2Πâ„ /L times 3 or - 3

  • plus or minus 6Πâ„ /L.

  • You can see that too.

  • I mean, just look at this function.

  • If you call it e^(ipx) p is 6Πâ„ /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 = 14 as a momentum there's something

  • wrong.

  • It's still probabilities, but it tells you that there is

  • non-0 probability 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.

  • You understand?

  • It is very fortunate for you that I gave you cosine which is

  • just the sum of two exponentials so the two what are called plane

  • wave 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 a 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

  • measurement that's what it'll be.

  • But the state of the system changes following the

  • measurement.

  • And if you measured x it'll 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--

  • if you measure momentum, first of all you'll get only

  • one of two answers, m = 3 or m = -3.

  • If you got m = 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 and

  • opposite the act of measurement will force it to be one or the

  • other.

  • It can give you either one, but once you've 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 -3,

  • but if you catch it at 3 it'll 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.

  • So 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 had a momentum p at a position x before

  • the measurement it's the answer right after the measurement

  • because you can do noninvasive 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 measure position and you found it here that'll be the

  • answer.

  • If you measured momentum and you got 5 it'll be something

  • with 5 oscillations in it.

  • So it'll 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 measure x it collapses to a spike at that

  • location.

  • If you measure p it collapses to the one term

  • wherever that was, that one e^(ipx) in the

  • sum.

  • So another interesting thing is the only measurement,

  • only answers you will get in the measurement are the allowed

  • values of momentum.

  • You'll 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/√2.

  • It happened to be an equal mixture,

  • but you can easily imagine some other problem where this is

  • 1/√6, and something that is

  • 1/√6, and something that is

  • 1/√3, they should all add up to 1

  • when you square and add them.

  • Then you can get all those answers with those

  • probabilities.

  • 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're 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(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(x) is?

  • It's the act of measurement.

  • If you measure the guy and you found him at x = 5 the

  • answer is ψ is a big spike there.

  • If you measure momentum and you've got m = 3 the

  • answer is that particular wave function with 3 on the exponent.

  • So measurements are a way to prepare states.

  • 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^(ipx) 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.

  • 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 will 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

  • ne^(−α|x|), mod x [

  • |x| ]means this is x = 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^(-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 that

  • αL is much bigger than 1.

  • That means its function dies very quickly spreading

  • negligible beyond some distance.

  • How far does a 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 it's some number of

  • order 1/α.

  • So this is a particle whose position has an uncertainty of

  • order 1/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.

  • 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^(2)e^(-2α|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.

  • It's −L/2 to L/2.

  • You can go this way L/2 and we can go that way

  • L/2.

  • So it's n^(2) is 2 times e to the

  • x 2α x dx.

  • Student: Should it be L over 2?

  • Prof: Pardon me?

  • Student: Should it be L over 2.

  • Prof: Yes, thank you.

  • So I'm going to now assume that in the upper limit of the

  • integration instead of going up to L/2 I go to infinity.

  • It doesn't matter because this guy is dead long before that.

  • That's why I made that choice.

  • So that I'm going to write this 2 times N^(2) times 0 to

  • infinity e^(-2αx) dx.

  • That's a pretty trivial integral.

  • It's just 1/2α.

  • If you want I will write it out.

  • It's e^(-2αx)/-2α

  • from 0 to infinity.

  • At the upper limit when you put infinity you get 0.

  • At the lower limit when you put x = 0 you get 1.

  • There's a minus sign from this and you get

  • N^(2)/α.

  • It should be 1, or N is equal to the

  • √α.

  • So my normalized wave function, this is the first order of

  • business, is square root of α

  • e^(−α|x )^(|).

  • I just squared it and I integrated it.

  • The only funny business I did was instead of cutting off the

  • upper integral at L/2 I cut if off at infinity because

  • e^(−L/2) is e to the minus nine

  • million, let's say.

  • So I don't care if it's nine million or infinity.

  • To make the life simpler I just did it that way.

  • Now I want to ask you, what is A(p)?

  • A(p), the coefficient of the

  • expansion remember is e^(−ipx/â„ )/√L

  • times ψ(x) dx from −L/2 to

  • L/2.

  • Now I'm going to work with p rather than m.

  • I will go back and forth.

  • You should use to 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

  • 2Πâ„ /m.

  • So you've got to do this integral.

  • So let's write this integral.

  • This looks like −L/2 to L/2.

  • In fact I'm going to change this integral to minus infinity

  • to plus infinity because this function e^(ipx/ℏ)

  • times e^(−α|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/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.

  • So mod x is not x, it is x when

  • it's positive and it's -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√L.

  • Do you see that?

  • So this is really square root of α/L times

  • e^(-αx) e^(−ipx/â„ ) dx

  • plus another integral from minus infinity to 0 e^(

  • αx) times e^(−ipx/â„ ) dx.

  • I split the integral into two parts.

  • So I didn't make a mistake here.

  • This really is e^( 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 -x everywhere.

  • In the terms of new variable this will make a -x.

  • That'll become 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 slight-of-hand,

  • but I don't want to delay that.

  • This is just--you go home and if you want check that if

  • x goes to -x you get that.

  • So you notice this is the complex conjugate of this one.

  • Whatever function I'm integrating here is the

  • conjugate of it because this is real, αx,

  • and -ipx has become 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/L times--

  • remember integral e^(-αx)

  • dx from 0 to infinity is 1/α,

  • but what I have here is dx e^(-α) ip/â„

  • x_0 to infinity plus the complex conjugate.

  • Now you may be very nervous about doing this integral with a

  • complex number in it because real we all know the integral is

  • just 1/α e^(-α).

  • It turns out that it's 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/â„ plus a complex conjugate which is

  • α − ip/â„ .

  • Now you should be able to combine these two denominators

  • and you get square root of α/L divided by α^(2)

  • plus p^(2) over â„ ^(2) times 2α.

  • Again, this is something you can go and check,

  • but I don't want to wait until everyone can do this thing

  • because I want to tell you the punch line.

  • So this is what it looks like.

  • A(p) looks like a whole bunch of numbers I'm not going

  • to worry about, but look at the denominator.

  • It's α^(2) p^(2) over â„ ^(2),

  • or if you want multiply by ℏ^(2).

  • 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(p), this is just A(p),

  • but I want the A(p) squared that looks like the

  • square of this.

  • All I want you to notice is that this function is peaked at

  • p = 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^(2) is equal to h^(2)α^(2)

  • because that's when these numbers become comparable,

  • therefore when p is â„ 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 â„ α.

  • So the most likely value is 0, but that's the spread.

  • And Δp is â„ α,

  • or α times Δp is â„ ,

  • 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?

  • Student: Can you go over again what cc is?

  • Prof: 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/α.

  • Then I looked at what kind of momenta I can get.

  • Then I find that the narrower the function 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.

  • The 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'll give you some homework on this,

  • and you can also fill in the blanks of this derivation,

  • which I think is very useful.

Shall we begin now?

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