Prof: So I got an
interesting email this weekend from two of your buddies,
Jerry Wang and Emma Alexander. The title of the email is this.
It says Casaplanka,
so I already know these guys are up to no good.
And the question is the
following. They said, "You've written
a wave function ψ(x) and A(p).
There seems to be no reference
of time, so where is time in this," and they couldn't
stop there. That would be a good place to
stop but they went on to say, "Are you saying the ψ
is just a ψ as time goes by?"
So that's the kind of stuff
that appeals to me, so I don't care if you don't
learn any quantum mechanics, but if you can do this kind of
stuff I'm not worried about you. But now I have to give a
serious answer to that serious question,
which I sort of mentioned before, which is that everything
I've done so far is for one instant in time.
I hope I made it very clear.
So here's the analogy.
If someone comes to you and
says, "Tell me all about Newtonian mechanics.
How does it work?
What's the scheme like?"
You say, "At any instant
someone has to give you the x and p of that particle at that
time," and that's all you need to know.
That's all you can ask and
that's all you need to know. Everything about that one
particle is completely given by this x and p.
That is called a state.
That's the complete description
of any system is called at state,
and a state in classical mechanics for one particle in
one dimension is just a pair of numbers.
It's a state because given that
I can predict the future for you.
I can predict the future
because if I knew the initial momentum, namely initial
velocity. I know how fast the guy is
moving, and I know where it is at the initial x.
A tiny amount of time later
it'll be at x vt, but t is very small.
Then since I know the
acceleration from Newton's Laws, which is rate of change of
velocity, if I knew the initial velocity
or initial momentum I can get the momentum a little later.
Then in that manner I can inch
forward in time. So you need Newton's Laws,
and Newton's Laws tell you, if you like,
rate of change of momentum is the force.
So if you like you can write
this mdx/dt squared is the force.
So this part is called dynamics
and this part is called kinematics.
So dynamics is how do things
change with time? Kinematics is what do you need
to know about the particle? In the quantum version of this
if you say, "What constitutes complete
knowledge of the particle?,"
the answer is this function ψ(x).
If you knew this function you
know all that there is to know about the particle.
We're not saying it's real.
It can be complex.
Then you can ask,
"If you knew the ψ at one time what's it going to
be later on?" That's the analog of this one.
So I have not come to that yet.
So this is a pretty long story.
It takes so long just to tell
you what to do at one time. Then I'm going to tell you how
to go from now to later. In Newtonian mechanics it's
just a one-word answer, x and p now is the whole story.
So anyway, since we've got four
lectures including today I will, of course, tell you before it's
all over what is the analog of this.
Namely I'll tell you a formula
for dψ/dt. How does ψ change with time?
So let's summarize what we know.
I want to do this every class
so that it gets into your system.
If it is true that ψ(x)
tells me everything I need to know let's ask questions of this
ψ. We can say, "Where is the
particle?" So there is some ψ.
You're told then the
probability that you will find it somewhere is the absolute
value. Probability density is
proportional to ψ^(2). So it doesn't tell you where
the particle is. It gives you the odds,
and the odds are this. Then you can say,
"Okay, x is just one of the variables of interest to me.
How about momentum?
If I measure momentum what
answers will I get, and what are the odds for the
different answers?" That question has a longer
answer and it goes as follows. Take the ψ
that's given to you, and write is as a sum with some
coefficients of functions that I call ψ_p(x),
but I will tell you what they are.
They are e^(ipx/ℏ) by square
root of L. Then I said,
"If you manage to write the ψ that's given to you in
this form, summing over all the allowed
values of p, then the probability that you
will get one of those allowed values is equal to the absolute
square of that coefficient."
And then final question can be,
"How do I know what A(p) is given some
function ψ?," and the answer is A(p)
is the integral of the complex conjugate of this ψ(x) dx.
In our case it was a ring of
length L. Then it's 0 to L,
otherwise it's whatever your allowed region is.
Yep?
Student: Can you
explain ________ one more time what...
Prof: What this guy
means? Student: Yeah.
Prof: So let me give an
example. Suppose ψ
of p was cosine 4Πx over L?
Then that looks like
e^(4Πix/L) e^(-4Πix/L) divided by 2.
Let me put the 2 here,
put the 2 here. Then my ψ
is not quite in the standard form because I don't have the
square root of Ls. So let me put a square root of
L here, a square root of L here and also put one outside.
So my ψ looks like a square
root of L. Let me see.
Oh, I know what the problem is.
This ψ is not normalized
because the rule is if you don't--
first let's say it's not normalized,
so don't worry about any of these numbers.
Can you see that it's made up
of two such functions with equal weight?
That means if you measure the
momentum you'll only get the answer corresponding to what's
here and what's here. If you compare it to
e^(ipx/ℏ) you can see the p that goes here is really
4Πâ„ /L or -4Πâ„ /L. So a particle in this state can
have only two values of momentum if you measure it,
and the probability for the two you can see by symmetry must be
50/50 for each because they come with equal weight.
But if you don't want to do
that you can, if you like, like this ψ,
normalize it, and then we square it,
integrate it and put a number in front so it comes out to have
length 1. Then the coefficients here
directly when squared will give you probabilities.
Yes?
Oh, a lot of questions.
Yeah?
Student: So
A_p has to be a real number, right?
Prof: No.
A_p does not have to
be a real number just like ψ(x) does not have to be a
real number A_p-- because I'm taking the absolute
value of A^(2). So even if it's a complex
number the absolute value of A will always come out to be real
and positive. Is that your question?
See A_p is not a
probability. The absolute value squared of
A_p is a probability which, of course,
must be positive. But A does not have to be
positive, does not have to be real.
Yeah?
Student: You said
A_p like the integral if you're multiplying ψ*
times ψ would that be in real numbers?
Prof: You should be very
careful. Student: Really?
Prof: A particular ψ
times that same ψ* is a real positive number.
This is not the same function.
I think I remember now from
previous years that students will get confused on this.
This ψ(x) is the function
that was given to me of which it was asked,
"If I measure momentum what answer will I get?"
This ψ_p is not
the function that was given to me.
This is another function that
describes a particle whose momentum is p.
So you've got to multiply that
function with the given function to get these coefficients.
So you can change your mind and
say, "Oh, I'm not interested in that
ψ. I want to know the answer to
that ψ." Well, you take that ψ
and put it here. These guys will not change.
Yeah?
Student: I'm sort of
confused. Right next to what you were
explaining you have written ψ of p.
Prof: Oh, I'm sorry.
No, no, no.
You have every right to be
confused. I did not mean that.
This is ψ of x.
Student: And is that
just ψ of x or that a ψ_p also?
Prof: Here?
Student: That's just
the ψ of x... Prof: Yes,
this is ψ of x.
Student: Okay.
Prof: This
ψ(x)--well, if I'm more disciplined,
I can do that for you. Maybe I should do it one more
time. It doesn't hurt.
So let's so let's say ψ(x)
is cosine 4Πx/L. Now first job is to rescale the
ψ so that if you square it and integrate it you get 1.
I know the answer to that is
square root of 2/L. You can check that.
If you take this guy and you
square it and integrate it it'll work because the square of this
will be 2/L, and the average value of cosign
squared is half. And we integrate it from 0 to L
it will cancel. You'll get 1.
If you do this particular
function then all you have to do is write this in terms of--
first write this one, 4Πix/L divided by 2
e^(-4Πix/L) divided by 2 because exponential is cosine
θ is e iθ e to the -iθ
over 2. Now I'm going to write this as
1/√2 times e^(4Πix/L) divided by square root of L
e^(-4ΠIx/L). Divided by square root of L.
I'm just manipulating the
function that was given to me. Why do I like this form?
Because this is of the form I
wanted, A of p times ψ(p) where ψ(p) I this and that's
also ψ(p). It's a normalized function
associated with momentum p. By comparing this you can see
that A is equal to 1 over square root of 2 for p = 4Πâ„ /L,
and A's also equal to 1/√2 for p = -4Πâ„ /L.
How do I know what p is?
States of definite momentum
look like e^(ipx/ℏ). So whatever multiplies ix/ℏ
is the momentum that you can see it's 4Πâ„ .
You see what I'm saying?
If you want to put an ℏ down
here and put it on the top and make the comparison.
So these are the only two
values for momentum. So this is a state which is
simultaneously in the state of momentum 4Πâ„ /L,
and -4Πâ„ /L. It has no definite momentum yet.
If you measure it then you can
get this answer with probability 1 over root 2 squared,
which is 1 half, or you can get this answer 1
over root 2 squared which is 1 half and no probability for
anything else. You cannot get any other value.
See, normally when you take a
ψ(x), a generic ψ(x) that I draw,
it typically has some non-zero value in all of space,
so you can find it anywhere. But this guy has only two
values of p in the sum and only those will be possible.
Yeah?
Student: ________
you're drawing ℏ over there ________.
Prof: This guy here?
Student: Yeah.
Prof: Yes,
I borrowed an ℏ, put it on top and bottom so you
can compare the expression. I'm just saying compare,
if you want without these ℏs, compare this expression to that
and you can see p is equal to this.
Now this problem was simple in
the sense that you didn't have to do that integral,
but last time before end of class I took the following
function, ψ(x) =
e^(−α|x|). On a ring the function looked
like this, falling very rapidly and dying within a width roughly
Δx = 1/α. Beyond that it is gone.
Then if you square that and
normalize it in the circle you find this is the correct
normalization. In other words,
if you took this ψ and you square and integrate it
you'll get 1. If you're very,
very careful it is not strictly 1 because this function the x
goes from −L/2 to L/2 whereas in the integral I took
it to go from minus to plus infinity.
That's because this function is
falling so rapidly. The function is falling so
rapidly you don't care if you cut if off at L/2 or go to
infinity, so I did that to simplify the
math, but the idea is the same.
Then if you want to know what's
the probability to get some number p you've got to take A of
p a square root of α times e^(ipx/â„ ) square root
of L times e^(−α|x|) dx.
You understand that?
This is the ψ
with the root α in it.
That's my ψ.
That's my ψ_p*(x).
If you come to me with a new
function tomorrow, I don't know,
maybe a function that looks like this, a constant in some
region and 0 beyond. We'll put that function here
and do these integrals. You'll get a different set of A
of p's. So for every function ψ(x)
that's a set of A_p's that you find by doing the
integral. Yeah?
Student: So if you
measure the momentum and you get one of those answers so it
becomes that answer? Prof: Right.
So I'm coming to that point.
So the second thing is another
postulate. In the end I'll give you the
list of postulates. And the postulate says that if
you measure the momentum you'll get only those values for which
the corresponding A_p is not 0,
and the ones for which it is not 0 the probability for
getting the number is the absolute value squared of that
A_p. And right after the measurement
the wave function will change from this sum over all kinds of
momenta to the one term corresponding to the one answer
you got. You understand that?
So in this simple example there
are only two values for p you can get.
No other values are possible.
They're all theoretically
allowed. For example,
6Πâ„ /L is a perfectly allowed momentum in that ring
because it corresponds to a periodic function,
but that's not contained in this particular wave function.
This wave function is built out
of only two of them, namely 4Πâ„ /L and
-4Πâ„ /L. Once you measure it suppose you
got -4ℏ/L. The whole wave function,
which I had here in the simple example, has got only two terms.
This part will simply
disappear, and the function after the measurement will be
this. The rough logic is that if I
measure momentum, and I got an answer,
that's got to mean something in any real sense that that is the
momentum of the particle, therefore it has to be true if
I immediately remeasure it. Immediately remeasure momentum,
and I want to get exactly the same answer,
that means the function after the first measurement should
contain only one momentum in its expansion,
so it'll reduce to that one term.
Yep?
Student: Is there a
difference between ψ(x) and negative ψ(x) because it
seems like everything's just... Prof: That is correct.
I think I also explained the
other day that in quantum theory ψ(x) and say 92i
times ψ(x) are treated as physically equivalent because
they contain the same relative probabilities.
Of course if ψ(x) had been
properly normalized to 1 this guy would not be normalized to
1. It's as if you live in a world
where all you care about is the direction of the vector,
but not the length of the vector.
We don't live in such a world.
If you say, "Where is Stop
& Shop," you say,
"Well, go in that direction,"
and you're not told how far, well, you're missing some
information. Imagine a world where all you
need to know is which way to go. In fact, maybe if you go to
some town and ask people where is something they'll say,
"Go in that direction,"
that's useful information. It's as if the direction is the
only thing that matters, not the length of the vector.
The analogous thing here is if
you multiply the function by any number you don't change the
information. They're all considered as the
same state. And I told you from this huge
family of functions all describing the same state I will
pick that function, that member whose square
integral happens to be 1. Yep?
Student: So after the
wave function breaks down how long do you have to wait for the
wave function to rematerialize, or does it...
Prof: Oh,
it's not break down. It's another wave function
because e^(ipx/ℏ), which is ψ(p),
this guy is as much a wave function as any other ψ(x)
you write down. So these are not different
creatures. It's like saying the following.
I think the analogy may be
helpful to you. It may not be.
ψ is like some vector in
three dimensions, and you know there are these
three unit vectors, i, j and
k, and you can write the vector as some
V_x times i V_y
times j V_z times
k where V_y =
V⋅j etcetera.
Think of ψ
as a vector and each of these directions is corresponding to a
possible momentum, and you want to expand that
vector in terms of these vectors.
This is ψ_p for
p = p_1. This is ψ_p for
p = p_2 and so on.
So it's how much of the vector
is in each direction that determines the likelihood you'll
get that answer, that answer or that answer.
But right after the
measurement, if you've caught it in this direction,
the entire vector, only the component of the
vector in the direction where you got your answer remains.
The rest of it gets chopped out.
It's really like Polaroid
glasses. If you've got Polaroid glasses
light can come in polarized this way or polarized that way,
but once it goes through the glass it's polarized only in the
one direction corresponding to the way the polaroid works.
If it filters light with the e
field going this way the light on the other side will have only
up and up. So measurement is like a
filtering process. It filters out of the sum over
many terms the one term which corresponds to the one answer
you got. So I don't mind telling this to
you any number of times, but that is the way that
quantum mechanics works. Similarly, if you're chosen to
measure position starting with this ψ,
and you found the guy here, once you found it--
see, before you found it the odds may vary like this.
Once you found it it's there,
and if you remeasure it infinitesimally later it'd
better be there. So the only function which has
the property is the big spike at wherever you found it.
That's called a collapsible of
the wave function. The real difference is the
following. I gave you an example of
somebody tracking me to see where I am.
I drew a probability graph,
and the probability graph may look like this.
And if you caught me here,
if someone says, "Where will I find him
next?" The answer is right there
because the probability has collapsed from being anywhere to
where we just saw him. The only difference between me
and the quantum particle is if you got me here I really was
here. You couldn't have gotten me
anywhere else at that right now. Can anybody find me anywhere
else except right now here? You cannot.
That's because my location's
being constantly measured. Photons are bouncing off to see
where I am, so my measurement is constantly measured.
A quantum particle with a
function like this, which is then found here,
was not really here. It was in this state of limbo
which has no analog in which it could have been anywhere.
It's the act of measuring it
that nailed it to there, but then right after that if
you measure it we expect it to be still there.
Now the time over which you can
do that is infinitesimal.
If you wait long enough the
wave function will change; I will tell you the laws for
the dynamics that will tell you how it will change,
but we all believe that if a measurement is immediately
repeated for that same variable we should get the same answer.
Need not be,
it could have been even worse, but at least that much is true.
So it's not going to get any
more familiar. It's a strange thing,
but hopefully you will know what the rules are.
When I said no one understands
quantum mechanics what I meant was, of course,
by now you know the recipe. It doesn't mean you like it,
or it doesn't mean it looks like anything in daily life.
Things in daily life they have
a location before you measure it, while you measure it and
right after you measure it. They're always in one place
doing one thing. Only in the quantum world they
can be in many places at the same time.
But you should be very careful
if this is a wave function for an electron.
The charge of the electron is
not spread out. I told you that.
An electron is one little guy
you'll only catch in one place. It's the wave function that's
spread out. Anyway, I think what'll happen
is you will have to do a lot of problems,
and you will have to talk to a lot of people,
and you have to read a lot of stuff.
You can teach quantum mechanics
for a whole semester, sometimes I taught it for a
whole year. There's more and more stuff.
I want you at least to know how
it works. In real life when you go
forward I don't think you need all of this.
If you go into physics,
or something, or chemistry,
they'll teach you quantum mechanics again,
but I wanted people doing something else;
they are the people I want to send a message to.
Is here's a part of the world.
If you ever hear the word
quantum, where does the word quantization come from?
What's the funny business in
the microscopic world? I want you to have a felling
for how that works. And I claim that the
mathematics you need is not very much.
You should know how to do
integrals, and you should know what e^(iθ) is.
So the postulates right now are
the state has given me a function ψ
of x. State of momentum,
definite momentum looks like this,
and if you want the odds for any particular momentum expand
it in this fashion which is the same as A of p
ψ_p of x, but A of p,
again, is that integral. I don't want to write it over.
Then if you measure p
you'll get one of the answers with the probability given by
the square of the corresponding number,
and right after the measurement the function collapses.
So it always collapses to
whatever variable you measure. Yes, another question?
Student: It looks like
the opposite. If you collapse it and you
measure x then what happens to p?
Prof: Very good.
His question was,
"If you collapse it to x what happens to
p?" I'm glad you asked.
I mean, that is the real
problem. You had the same question?
Student: No.
I was going to say well don't
you not know about p? Prof: That is correct.
So I will repeat what she said,
and I'll repeat what you said. I want to have this discussion
with you guys because it's very important because everyone's
thinking the same thing. If I first take a generic
ψ, so I want everyone to know what the answer's going to
be to any of these questions. So I take this ψ
somebody prepared for me. Let's not worry about how that
person knew this is ψ, you're given.
The electron is in this stage
ψ. If you measure x,
and you got x equal to this point the function,
of course, becomes a big spike. This spike in principle should
be infinitesimally thin, but I don't care.
Let this be the width of a
proton. If after that you say,
"What momentum will I get," well,
you know what you have to do. You've got to write spike
function equal to sum over these functions, right?
You've got to take the integral
of the spike with this exponential and do that
integral, and you'll get a bunch of
numbers A of p for all values of p.
And then if you measure
momentum, and you got the one corresponding to p =
6Πâ„ /L, the state which contains many,
many things will reduce to the one term corresponding to
p = 6Πâ„ /L. If you plot that function it
will be some oscillatory function.
The real part and imaginary
part will both oscillate with some wavelength given by
p. Be very careful.
This is not the absolute value
of ψ. It's the real or imaginary part.
They're both sines and cosines.
Absolute value will be flat.
So it'll go from a particle of
known location to a particle whose probability's completely
flat on the circle. You understand?
The wave function can look like
6Πi/L square root of L right after the
measurement. Let's call is ψ_6.
The absolute value of ψ
is a constant, but the real and imaginary
parts of ψ oscillate with some wavelength.
So right after the measurement
of momentum you don't know where the guy is, and you say,
"Let me find this fellow."
You catch it somewhere then
it's a spike at that point, but then you have no guarantee
on the momentum. So you can never produce for me
something of perfectly well defined position and momentum
because once you squeeze it in x it gets broad it in
p. Once you squeeze it in p
it gets broad in x. This is really mathematical
property of Fourier analysis, that functions which are very
narrow in x when you do the Fourier expansion have many,
many wavelengths in them. And likewise,
a function with a very well defined wavelength,
because it's a complex exponential, has a magnitude
which is flat. So now I'm going to ask the
following question. So when I did ψ(x)
the probability for x was very easy, squared the ψ.
When I said,
"Okay, I want to look at momentum,"
the answer was long and complicated,
namely, take these exponential functions,
write the ψ in terms of those,
find the coefficient, etcetera. Now I can say I want some other
variable I'm interested in. I want to know what happens if
I measure energy. Energy is a very,
very important variable. It's very, very important
because it turns out that if a particle starts out in a state
of definite energy, I will show that to you later,
it remains in that state. That's the only state that will
remain the way it is. If you start in a state of
definite momentum two seconds later it can have a different
momentum or it can be a mixture of different momenta.
But if it starts in the state
of definite energy it will remain that way.
That's not obvious.
I'm going to prove that to you
later. That's why it's very important.
So most atoms are in a state of
definite energy and they can stay that way forever,
but once in a while when they're tickled by something
they will either absorb light or they will emit light.
So we draw a picture like this.
We will see that the allowed
energies of the systems are some special values.
Not every value's allowed.
And this can be called
n=1, n=2, n=3, etcetera.
And an atom,
for example, can sometimes jump from doing
that to doing that, and in that process it will
emit an energy which is E(n=3) −
E(n=1). That difference of energy will
come in the form of a photon and the energy of the photon is
â„ ω, or if you like,
2Πâ„ f where f is what you and I call
frequency. And from the frequency you can
find the wavelength. The wavelength is just the
velocity of light divided by frequency.
So an atom will have only
certain allowed energies and when it jumps form one allowed
energy to another allowed energy it will emit a photon whose
frequency-- in fact, you should probably
call this frequency f_31 meaning
what I get when I jump from the level 3 to level 1.
Similarly, if you shine light
on this atom it won't take any frequency.
It'll only take those
frequencies that connect it from one allowed energy to another
allowed energy. That's the fingerprint of the
atom. Both emission and absorption
betray the atom. That's how we know what atoms
there are in this star, or that star,
or what the composition is. No one's gone to any of these
stars, but we know because of the light they emit,
and it's all controlled by energy.
So the question I'm going to
ask is here's the function ψ(x).
Someone gave it to me in some
context, and I say if I measure the energy of this particle what
are the answers, and what are the odds?
So how do you think that will
play out? You have to make a guess.
Suppose you're inventing
quantum mechanics and someone says, "What do you think is
going to be the deal with energy?"
You know what the scenario
might look like? You can take a guess.
I mean, as I told you many
times I don't expect you to invent quantum mechanics on the
fly, but you should be able to guess.
What form do you think the
answer will take? You want to guess?
Student: ________
they'll be ________? Prof: Okay,
but if I want to know what energies I can get and with what
odds in analogy with momentum what do you think will happen?
Yep?
Student: Would you use
the formula p^(2)/2m? Prof: He said use
p^(2)/2m. That's a good answer.
His answer was we know that the
energy is equal to 1 half m v squared which I can write as
p^(2)/2m, right?
So you're saying if I measure
the momentum and I got a certain answer, well,
the energy's that p^(2)/2m.
That's actually correct except
the energy of a particle is not always just the kinetic energy.
For a free particle this is the
kinetic energy. For a particle moving in a
potential you know that you have to add V(x).
That is the total energy.
Now that's when we have a
problem. In classical mechanics once I
measure the x and p of the particle I don't
have to make another measurement of energy.
Do you understand that?
I just plug the values I got
into this formula. For example,
particle connected to a spring V is
½kx^(2) where k is the force constant
of the spring, and the kinetic energy's always
p^(2)/2m. Say if I measured p and
I got some number, and measured x I got
some number, I can put that in the formula and get the energy.
You don't have to do another
energy measurement. And you don't have to do an
angular momentum measurement either.
In higher dimensions angular
momentum is r x p,
and if you already measured the position and you measured the
momentum just take the cross product.
So in classical mechanics you
only need to measure x and p.
In quantum mechanics he made a
pretty good guess that if you measured p,
the p^(2)/2m is the energy.
That is true if the particle is
not in a potential. But if the particle is in a
potential can you tell me how to compute p^(2)/2m
V(x)? You realize you cannot really
compute it because if you knew the p exactly you have no
idea where it is, and if you knew the x exactly
you don't know what the momentum is.
Maybe you know a little bit of
both, but still, how are you going to find the
energy? So the answer is you have to do
a separate energy measurement. You cannot infer that from x
and p, because first of all you cannot
even get a pair of x and p at a given time.
I hope I convinced you.
You measure this guy you screw
up that guy. Measure that one you mess up
this one. You can never get a state of
well defined x and p anyway.
So the way to find energy is to
do a whole other calculation. So I will tell you what the
answer is, and hopefully you will realize it's not completely
different from the recipe we had before.
I'm going to give you a rule
for the functions which correspond to a state in which
the particle has a definite energy E.
Let's not worry about how you
get it, some function. You find all those functions,
or you're given all those functions, then can you imagine
what will happen next? If I give you all those
functions what do you think the rule is going to be?
Yep?
Student: You separate
it into a sum of all the energies with coefficients.
Prof: Very good.
Let me repeat what she said.
I hope at least some of you
were thinking about the same answer.
Her answer is take the function
ψ, write it in terms of these
functions ψ E of x with some
coefficient A_E summing this over the allowed
values of E, whatever they may be.
And now that you said that what
do you think A_E is going to
be in a given case? Would you like to continue?
Student: It's going to
be the integral or the wave function ________________.
Prof: That's right.
That is correct.
The recipe's almost complete
except you don't know what these functions are,
but if you knew these functions you have to write the given
function, given wave function,
as a sum of these functions with some suitable coefficients.
Coefficients are found by the
same rule, and then the probability that
you'll find an energy E is,
again, A_E^(2). And
everything else will be also true.
Once you measure energy you've
got energy corresponding to E_1,
or E_2 or E_3.
Let's say you've got
E_3, the third possible value.
The whole wave function will
collapse from being a sum over many things to just this one
guy, E_3. The collapse is the same.
The probability rule is the
same. The only thing you don't know
is who are these functions ψ of E?
You understand?
So, again, the analogy's the
following. There is a vector that we call
ψ. Sometimes you want to write it
in terms of i, j,
and k. They are like the A(p).
Sometimes you may pick three
other mutually perpendicular vectors, i',
j' and k'. And if you know those
coefficients you'll get the probability for some other
variable. So you're expanding the same
function over and over in many possible ways depending on what
variable is of interest to you. If it's momentum you expand it
in terms of exponential ipx of ℏ.
If it's energy you expand it in
terms of these functions. So the question is what is the
recipe going to be for these functions ψ_E of
x. By what means do I find them?
Now you're getting more and
more and more recipes every day, but it's going to stop pretty
soon. This is about the last of the
recipes. Even this recipe I'll tell you
how to get from a master recipe, so it's not that many recipes,
but I have to reveal that to you one at a time.
Now you can say,
"Okay, what function do you want me to use for every
energy, what is this function? After all, when it was momentum
you came right out and gave this answer.
Why don't you do the same thing
here," and that's a problem here.
The problem is the energy of a
particle depends on what potential it is in because it's
got a kinetic and a potential part,
so I cannot give you a universal answer for
ψ_E(x). I will have to first ask you;
"Tell me the potential the particle is in."
Once I know the potential I
will give you the recipe. Imagine the potential has been
given to me, for example, ½kx^(2),
or it could be a particle in what's called a well.
You make a hole in the ground
or you build a little barrier. That's a possible potential.
You can have a potential that
looks like harmonic isolator. That's a possible potential.
You can have an electron and a
hydrogen atom -1/r. That's a possible potential.
There are many-potentials,
and the answer's going to vary on the problem.
There's no universal answer for
the energy functions. You tell me what the electron's
doing, what field it is in, what field of force it is in.
Then for each field of force,
or for each potential, there's a different answer.
And here is the master formula.
This is the great
Schr�dinger equation. So the answer looks like this.
The function ψ_E
obeys the following equation, -â„ ^(2)/2m times
the second derivative of ψ V(x) times
ψ(x) is equal to E times ψ(x).
Do not worry.
I will see you through this
equation. Everything you need to know I
will tell you, but you should not be afraid of
what the equation says. It says those functions that
are allowed corresponding to definite energy will have the
property that if you took the second derivative,
multiplied it by this number, add it to that V(x)
times ψ_E you'll get some function.
That function should be some
number times the very same function.
If you can find me those
functions then you will find out when I will show you
mathematically that there are many solutions to the equation,
but they don't occur for every energy.
Only some energies are allowed,
and the energies are usually labeled by some integer
n, and for every n,
1,2, 3,4, you'll get a bunch of energies,
ψ_E1, ψ _E2,
ψ_E3, 3, and you write those
functions down. Then you can do everything
thing I said. But the only thing is here you
have to do some hard work. Whereas for momentum I gave you
that and for a state of definite positions x = x_0
I told you spike at x_0.
You didn't have to do much work.
For this you have to solve an
equation before you an even start, but we'll see how to do
that. So the first problem I want to
solve is the problem where there is no potential.
That is called a free particle.
A free particle is one for
which V is 0. And let me imagine it's living
on this line, the circle of length
2ΠR = L. Oh, by the way,
I should mention something else.
In terms of all the postulates
you notice I never mentioned the uncertainty principle today,
ΔxΔp should be bigger than ℏ.
I didn't mention it as a
postulate because once you tell me ψ is given by wave
function and that states are definite momentum have definite
wavelength it follows from mathematics that you cannot have
a function of well defined periodicity and wavelength also
localized in space. It's a mathematical consequence.
Similarly, who told me that I
can expand every given ψ as a sum over these functions?
There's a very general
mathematical theorem that tells you in what situations you can
actually expand any given function in terms of a set of
functions, namely are they like unit
vector i, j and k.?
Suppose I had only i and
j and I don't have k.
I cannot expand ever vector in
3D using i and j. So you've got to make sure
you've got enough basis functions and the theory tells
you that if you find all the solutions to that equation
together they can expand any function.
Similarly the rule for
expansion is also not arbitrary. It all comes from that.
It's a very,
very beautiful theory of the mathematics behind quantum
mechanics. If you learn linear algebra one
day, or if you've already learned it, it's all linear
algebra. So every great discovery in
physics is accompanied by some mathematical stuff you need.
Like all of Newtonian mechanics
requires calculus. Without calculus you cannot do
Newtonian mechanics. Maxwell's theory for
electromagnetism requires vector calculus.
Einstein's non-relativistic
theory doesn't require anything; it's algebra,
but the general theory requires what's called tensor calculus,
and quantum mechanics requires linear algebra.
And string theory we don't know
what it requires. People are still discovering
new mathematics. But it's very true that very
often new mathematics is needed to express the new laws of
physics. And if you don't know the laws
you may find out you're not able to write it down.
If you didn't know what a
gradient was, or if you did not know what a
curl is, and so on, then you cannot
write down the laws of electricity and magnetism.
So we have to solve this
equation, and I'm going to solve it for
the easiest problem in the world,
a particle moving on a ring of length L with no
potential energy. So what does that equation look
like? It says -â„ ;/2m
d^(2)ψ/dx^(2) no potential = E times
ψ_E(x). So let me rearrange the
equation so it looks like this, d^(2)ψ/dx^(2)
k^(2)ψ = 0 where k^(2) is
defined to be 2mE/ℏ^(2) = k^(2).
All I've done is just taken
everything to one side and multiplied everything by
2m/ℏ^(2) and called that combination as
k^(2). So who is this number k?
Let's see.
The energy is
ℏk^(2)/2m, but we also know energy's
p^(2)/2m. So this number k will
turn out to be just momentum divided by ℏ.
Well, momentum has not entered
the picture, but we will see. Let's solve this equation now.
I say the solution to this
equation is ψ(x) = to any number A times
e^(ikx) any number B times e^(-ikx).
Let's see if that is true.
Take two derivatives of ψ.
What do you get?
You understand every time I
take a derivative you pull down an ik?
If you pull it twice you'll
pull an ik squared which is -k^(2)ψ,
and the same thing will happen to this term that it'll pull
down a minus ik, but if you do it twice you'll
again get −k^(2). So both of them will have the
property that the second derivative of ψ
will be equal to -k^(2)ψ,
and that is the equation you want to solve.
And A and B is
whatever you like. A and B are not
fixed by the equation because for any choice of A and
any choice of B this'll work.
So let me write it as follows.
Ae to the i.
Let me write k,
k was a shorthand for square root of 2mE/ℏ^(2)
x B to the -i square root of 2mE/ℏ^(2)
x. I'm trying to show you that
these are really functions of definite energy
ψ_E and here is how the energy appears.
So what does it look like to
you? Have you seen these functions
before? Yes?
What does it look like?
Student: One of the
spring wave theories? Prof: Pardon me?
Student: One of the
spring wave theories? Prof: The equation is
like the spring equation. That is absolutely correct,
but what does this function look like?
That function look like
something to you? Yep?
Student: Isn't that a
cosine function? Prof: It's a cosine only
if A is equal to B. Forget the sines and cosines.
They are your old flames.
What's their new quantum flame?
What's the function that means
a lot more in quantum mechanics than sines and cosines?
No?
I think I told you long back
that e to the i times dog x over ℏ
is a state where the momentum is equal to dog.
In other words,
you can put anything you want in the exponent.
If a function looks like that,
that fellow there is the momentum.
So this is a state of momentum,
of definite momentum. In fact, look at this,
e to the i square root of 2mE/ℏ x
B times e to the minus that.
Do you understand?
This must be the momentum.
It is the momentum because
e^(ipx/ℏ) is a state of momentum p,
but the momentum here can either have one value,
square root of 2mE, or it can have another value
minus square root of 2mE. So what do you think the
particle is doing in these solutions?
Yep?
Student: It's jumping
back and forth _______________ one of those states.
Prof: But in any one of
them what is it doing here? What is the sign of the
momentum here? Student: Oh, positive.
Prof: Positive,
and here it's got negative momentum.
And how much momentum does it
have? The momentum it has,
if you look at any of these things,
is that p^(2)/2m = E is what is satisfied
by the p that you have here.
In other words,
what I'm telling you is in quantum theory,
in classical theory if I said, "I've got a particle of
energy E, what is its momentum?"
You would say, "Well,
E is p^(2)/2m,
therefore p is equal to plus or minus square root of
2mE," because in one dimension when I
give you the energy, the Kinetic energy if you like,
the particle has to have a definite speed,
but it can be to the left or it can be to the right.
And the momentum is not
arbitrary. If the energy is E the
momentum has to satisfy the condition p^(2)/2m
= E. That's also exactly what's
happening in the quantum theory. The state of definite energy is
a sum over two possible things, one where the momentum is the
positive value for root of 2mE;
other is a negative value, and these are the two allowed
values even in classical mechanics for a particle of
definite energy. But what's novel in quantum
mechanics, whereas in classical mechanics
if it's got energy E it can only be going clockwise or
anti-clockwise, but this fellow can be doing
both because it's not in a state of clockwise or anti-clockwise.
In fact, the probability for
clockwise is proportional to A^(2).
Probability for anti-clockwise
is proportional to B^(2). I've not normalized it,
but the relative odds are simply proportional to A^(2)
and B^(2). So that's what is bizarre about quantum
mechanics that the particle has indefinite sign of momentum.
Yes?
Student: Do we have
problem with m because m is too small to
________ properly? Prof: Which one?
Student: With the mass
is there going to be some kind of issue if that's small?
Prof: What about the
m? I'm sorry.
Student: m
stands for the mass, right?
Prof: Yeah,
m is the mass of the particle, that's right.
Student: But is it too
small in some way for our equations to work because we had
that issue in previous equations?
Prof: When did we have
the issue? m is whatever the mass
of the particle is. It can be small.
It can be large.
I didn't understand.
Student:
> Prof: No,
no, go ahead. I want to know.
Student:
> Prof: There are no
restrictions on the correctness of this.
If the particle weighs a
kilogram then you will find that--well, we're coming to
this. So you find that these are the
allowed values of p. There are two values,
but p itself is not arbitrary, p itself is
not continuous. Maybe that's what you meant.
There's a restriction on the
allowed values of p, and therefore a restriction on
the allowed values of E. All I'm telling you now is that
if you want to solve that equation it is obviously made up
of sines and cosines as you recognize from the oscillator,
or in the quantum world it's more natural to write them in
terms of exponentials, e^(ikx) and
e^(-ikx) where k is not independent of e,
k satisfies this condition, and if you call
ℏk as p, p satisfies this
condition. This is just a classical
relation between energy and momentum.
The other difference is that
not every value of momentum is allowed.
Not every value of momentum is
allowed for the same reason as when I did particles of definite
momentum. In other words,
if the particle is living on a circle and the state of energy
E it's given by Ae^(ipx/â„ )
Be^(−ipx/â„ ) where p is related to E
by E = p^(2)/2m. We have the requirement that
when you go around a circle you've got to come back to where
you start. And that condition,
if you remember, says that p times
L over ℏ has to be a multiple of some integer,
or that the allowed values of p are labeled by some
index n, which is 2Πâ„ /L
times n. I don't want to use m
because m stands for particle mass,
n is the integer now. In other words,
when we studied the state of definite momentum,
namely the first function, we realized even then that
p is quantized. Because of the single valued
condition p is quantized. And if energy functions are
made up of such functions they also have to be singlevalued.
That means the p here or
the -p here both have to satisfy the condition given by
this. Therefore, the allowed energies
are also labeled by an integer n and they are really
p n squared over 2m where p
n is 2Πâ„ n over L .
You see that?
This is the quantization of
energy. So a particle in a ring has
only these allowed values of energy.
So in a way this problem is
somewhat easy because it's a free particle.
Once you've understood the
particles in terms of allowed momenta it turns out the allowed
momentum states are also allowed energy states.
The allowed momentum states,
you remember, they either look this or they
look like this. And I can pick A and
B to be arbitrary, so one choice is to pick
A equal to 1/√L and just take
e^(ipx/â„ ). Other is to pick B equal
to 1/√L and pick e^(−ipx/â„ ),
but you can also mix them up. You don't have to mix them.
If you don't mix them up you
have a particle here which has a well-defined energy and a
well-defined momentum. This guy also has a
well-defined energy and a well-defined momentum.
This guy only has a
well-defined energy, but not a well-defined momentum
because there's a two-fold ambiguity in momentum.
You understand?
Even in classical mechanics
it's true. If I give you the momentum you
can find the energy. If I give you the energy you
cannot find the momentum because there are two square roots you
can take because p^(2)/2m is E,
p is plus or minus square root of 2mE.
It's the same uncertainty even
in classical mechanics. So what happens in quantum
theory is if you pick any particle of definite momentum on
the ring it'll already have definite energy which is simply
that momentum squared over 2m.
You don't need to find a new
function. What is novel is that since the
energy depends only on p^(2) you can take a
function with one value of p,
and you can take a function with the opposite value of
p and add them, there will still be a state of
definite energy because whether it's doing this or whether it's
doing that the energy will always be p^(2)/2m
and the minus signs drop out of that.
So what is novel here is what's
called degeneracy. Degeneracy is the name,
but there's more than one solution for a given value of
the variable you're interested in.
You saw the energy looks like
this. Therefore, you'll find there's
a state E_0 which is 0 squared over
2m. Then there are two states
E_1. One has got momentum p
going clockwise and one has momentum going anti-clockwise.
So they look like
(1/√L) e^(2Πix/L).
That's this guy.
Then you can have
(1/√L) e^(-2Πix/L) which is
the other guy. So at every-energy,
allowed energy except 0, there'll be two solutions.
There are two quantum states
with the same energy. When you studied hydrogen atom
in high school maybe you remember there are these shells
with 2, and 4, and 8, and 10 and so on.
They are called degeneracies
where the energy is not enough to tell you what it's doing.
There at every-energy it can
have a different angular momentum.
Here at every-energy it can
have two different momenta, clockwise and anti-clockwise.
Yep?
Student: Going back can
you explain how you got the ________ of p L over ℏ
________? Prof: Here?
Student: To the left.
Prof: Here?
You mean this one?
Student: I don't where
that came from. Prof: It came just like
in the momentum problem. If you've got a function like
this you have a right to demand that if you go a distance
L around the circle you come back to where you start.
So if you take any x here and
add to it an L it should not make a difference.
And what you're adding is
pL/ℏ and that better be a multiple of 2Π.
It's the same single valuedness
condition. So the momentum problem pretty
much does this problem for you, all the singlevalued stuff we
dealt with before. What is novel here is that
demanding energy have one value fixes the momentum to be one of
two values, and that double valuedness is
the same as in classical theory that a particle of energy
E can have two possible momenta plus or minus square
root of 2mE, and the quantum theory,
then, is the state of E is a sum of one value of
p and the other value of p with any coefficient
you like. So if this atom makes a jump
from, or this system makes a jump
from somewhere there to somewhere there you can find the
frequency of the photons it will emit because here are my allowed
energies. E_n is this.
Suppose it jumps from
n=4 to n=3. The energy that's liberated
that goes to the photon is â„ ω will be
4Π^(2) â„ ^(2) over 2mL squared times 4
squared - 3 squared. I'm just using this formula
with n=4, and n=3,
and take the difference. It's 16 - 9 which 7.
You plug all that in you can
solve for the ω, or if you like frequency you
can write it as 2Πâ„ f and you can find the
frequencies. This is actually true.
If you've got a charged
particle moving in a ring and you want to excite it from one
state to a higher state you will have to give it only one of
these frequencies so that the frequency you give it,
â„ ω, must match the energy
difference of the particle. You understand?
So I've drawn these levels here.
If you want the electrons,
say, in a metallic ring going in the lowest possible energy
state, if you want to jump,
if you want to crank it up to the next level you've got to
have photons at that energy, or that energy.
Well, that happens to be the
same as this, but these are the only
frequencies it'll absorb from you.
And when it cools down it'll
emit back those frequencies, and that's something you can
test. By the way, do you know why
there's only one state at E_0 and not
two? Yep?
Student: Because 0
equals -0. Prof: That's right.
The solution plus or minus
2m square root of 2mE has only one-answer
when E is 0 because 0 momentum and -0 momentum are the
same. Otherwise, any finite positive
momentum has a partner which is minus that momentum.
There's a very interesting
piece of work being done experimentally at Yale which is
the claim that if you took a metallic ring in a magnetic
field it will have a current going one way or the other way,
unbalanced current, and it's not driven by a
battery. It's not driven by anything.
Normally if you took an
ordinary ring the lowest energy state will be a field of zero
current while you go one way or the other,
but if you put it in the magnetic field one can show it
likes to go one way or the other way.
And now measurements are being
done at Yale where you can actually measure the tiny
current due to one electron, or one net electron going one
way or the other. So this L is either a
mathematical convenience if you're talking about free space
and you want to be able to normalize your wave functions,
or it really is the circumference of a real system.
Now that we can probe nano
systems very well we can vary the L and we can find out
all the energy levels. All right, so now I'm going to
do the one problem which is really a very standard
pedagogical exercise, and that's called a particle in
a box. I remember this example the
first time I remember seeing quantization,
which is more interesting than on a ring.
A box is the following.
If you dig a hole in the ground
and you are standing somewhere here you realize you're kind of
trapped unless you can scale this wall.
Now, you can call that as a
ground level and think of it as a hole in the ground,
or you can think of this as the ground level and that's the
height of your barrier. So imagine a particle living in
a barrier that looks like this. This is the potential energy.
This is like the height above
the ground, if you like,
and it has a height V_0 t here,
and is 0 here, and it goes from some
−L/2 to L/2. If the barrier
V_0 goes to infinity it's called a box.
So a barrier that goes to
infinity, I'll just show you the part that you can see here.
Here V is infinity,
and here V is infinity and inside V is 0.
That's a particle in a box.
So if the particle goes hits
against the wall it cannot, no matter how fast it's moving,
go over the top because it's infinitely high.
And that's not realistic.
Every barrier is finite,
but just to teach you the principles we always pick the
simple example. So now I want to solve this
problem. What are the allowed wave
functions, ψ, for a particle in this
potential? So let's go back to the same
equation which says -â„ ^(2)/2m
d^(2)ψ/dx^(2) = E - V ψ.
What I've done is I've taken
V to the other side. Now the potential is a constant
in the three regions. This is region I where the
potential is 0. This is region II where the
potential is infinity. This is region III where the
potential is infinity. I'm going to first look at the
region here, region III. What's the solution going to
look like in region III is what I'm asking.
So in this region the energy's
something I don't know what the allowed values are.
Whatever the value is V is an
extremely large number. Can you see that?
In this region don't let the
barrier be infinity. Imagine it's one zillion,
a very high barrier. Then what's the solution?
The solution will look like
d^(2)ψ/dx^(2) is = 2m/â„ ^(2) times
V - E ψ. Then the solution to that is
very easy, ψ = Ae^(κx)
Be^(-κx) where κ is equal to all of this,
square root of 2m/ℏ^(2) times V -
E. I'm saying this is what I'm
calling κ^(2). Look, I'm asking you give me a
function whose second derivative is some number times the
function. Well, that's obviously
exponential and it's the real exponential because it's a
positive number times the function.
Now A and B are
free parameters, whatever they are it'll solve
the equation, but we don't want a function
that's growing exponentially when you go to infinity because
that means that particle would rather be at infinity than in
your box or near your box. So for that mathematical,
for that physical reason we junk this function.
You pick it on physical grounds
as not having a part growing exponentially when you go to
infinity, but you do admit a part that's
falling exponentially when you go to infinity.
That's okay.
But how fast is it falling?
It's going like B e to
the minus some blah, blah, blah times square root of
V - E. That's all I want you to look
at. Forget all the ℏ's and
m's. Make V larger,
and larger, and larger, and just tell me what you think
it will do. If V is larger and
larger it's like this e^(-αx) I wrote for
you where α's very large. So this function will fall
faster, and faster, and faster, and in the limit in
which V goes to infinity it will vanish.
That basically means the
particle cannot be found outside the box.
So your function ψ
is 0 here and 0 here. Because if you made the barrier
height finite you will find it's falling exponentially on either
side, but the exponential becomes
narrower and narrower as the barrier becomes higher and
higher, and in the limit in which the
wall is infinitely tall there is nothing outside.
The wave function is non-0 only
inside. So what's the solution inside?
Let me call this 0.
Let me call this L.
Inside the box there is no
potential, so this equation is
d^(2)ψ/dx^(2) (2mE/â„ ^(2))ψ
= 0. It's like a free particle in
the box, but it cannot leave the box.
This is what I called k,
remember? So now I'm going to purposely
write the solution in terms of trigonometric functions.
You'll see in a minute why.
So I'm going to write this
Ae^(ikx) Be^(-ikx).
You can see if I take any of
these solutions it's going to satisfy the equation,
but k better be related to E in this form,
2mE/ℏ^(2) is equal to k^(2).
Sorry that's k^(2).
Because two derivatives of this
will give me −k^(2), and if you
put that particular value for −k^(2) here these
two will cancel. But I'm somehow going for--I'm
sorry, I'm going to not write it this way, but write it as
C cosine kx D sine kx.
Do you realize I can always go
back and forth between exponentials and trigonometric
functions because one is a linear combination of the other?
If you want,
write this as cos i sine and cos - i sine and
rearrange the coefficients. It'll look like something,
something cosine. I want to call that a C,
and something, something sine which I want to
call this D, D and C may be
complex. I'm not saying anything,
but you can write a solution either in terms of the sines and
cosines or E to the plus or minus something.
So here's the function ψ.
It looks like I have an answer
for every energy I want, because pick any energy you
like, find the corresponding k,
you put it here and you're done. But that's not allowed because
we have an extra condition which is the ψ was identically 0
here, ψ was identically 0 here.
We're going to demand that at
the two ends, it can do whatever it wants in
the middle, it must vanish at the two ends
for the continuity of ψ because if ψ
had two values you're getting two different probabilities for
the same point, so that's not allowed.
So ψ must have--must match
at the two ends. But look at this function.
It's got to vanish at the left
end at x = 0 and it's got to vanish at the right end.
At x = 0 you can see
ψ of 0 is simply C because sine vanishes.
Cosine is 1 and C has to
be 0 then because this guy has no business being non-zero on
the left end. That's good.
This guy vanishes at the left
end so I allow it, but I have another condition.
It should also vanish at the
right end. If it should vanish at the
right end I demand that sine kL should be 0 because
sine kL vanishing is fine.
Then it'll vanish at both ends.
But sine kL = 0 means kL
is a multiple of Π. That means k is
nΠ/L. This means the allowed values
of k in the problem are very special so here is n = 1,
ψ looks like sin(Πx/L),
and equal to 2 looks like sin(2Πx/L),
and so on with some numbers in front which I have not chosen
yet. If you plot them they look like
this. That's one guy.
At a higher energy I've got
that guy. Then I've got that and so on
because these are exactly like waves on a string,
a violin string clamped at two ends.
In fact, this wave equation's
identical to the wave equation on a string.
The only requirement is that
the string is clamped at two ends.
Here the ψ
is clamped at the two ends because it's got to vanish on
either side outside the box. So the allowed wavelengths are
the same except here the wavelength is connected to the
momentum that's connected to energy and E,
you remember, is
ℏ^(2)k^(2)/2m then becomes
ℏ^(2)/2m times k will be
n^(2)Π ^(2)/L^(2).
So this particle in a box can
have only these particular energies.
So let me write it for you
nicely. The allowed energies are
â„ ^(2) Π^(2)/2 mL squared times an
integer n squared, and the corresponding wave
functions look like this. They're waves in which you've
got half an oscillation, or 2 half oscillations and 3
half oscillations, but you've got to start and
finish at 0. So it's the quantization.
This is why when Schr�dinger
came up with this equation everybody embraced it right away
because you suddenly understood why energy's quantized.
You're trying to fit some
number of waves into an interval and only some multiple of half
wavelengths are allowed, but wavelength translates into
momentum. That translates into energy.
Suddenly you understand the
quantization of energy. So it's got 1 state
n = 1, and only one state
n = 2, one state n =
3. I'm going to come back to this
next time, but you should think about this.
