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 of 2Π
because any trigonometry function if you add 2Π,
or 4Π, or 6Π 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 2Π 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 than ℏ then 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 2Πimx/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 -α 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.
