Prof: Okay,
so I'm going to begin with something I did last time,
because it was not completed, then we'll take it from there.
And we had a lot more questions
last time, which is good, very helpful,
so I know what may be bothering you.
So don't hesitate to stop me.
The new thing we did last time
was, if I give you some wave
function, ψ(x), and I ask you what happens if I
measured energy, the recipe is the same as in
momentum. You've got to find functions
called ψ_E(x),
which are supposed to be states in which,
if the electron was there, it'll have a definite energy.
Now, not worrying for a moment
on how you get them, the recipe for what happens
when you measure energy is that if you're given some arbitrary
function ψ, that's a state the particle is
in. And you're asking,
"What happens if I measure energy?
What will I get?"
The instruction is to write
that function as a sum over these functions,
each associated with a definite energy.
And once you have done that,
the probability that you will get a certain energy E =
the absolute value squared of that particular coefficient.
Not different from
ψ_p , same rule.
And the rule for
A_E is also the same, or similar.
A_E is the
integral of complex conjugate of this function associated with
definite energy, times the function you're
thinking about, integrated over space.
It's the same rule.
So it can happen,
the problem can take many forms.
Sometimes one can give you a
function ψ(x). Then you take all these
functions, do all these integrals, find all the
coefficients and you square them.
Other times the coefficients
may be given to you; you're just asked,
"What's the odd for this energy or that energy?"
But now the question is,
what are these functions, ψ_E(x).
Whereas functions of definite
momentum are given once and for all by this formula,
e^(ipx/ℏ), I wish I could give you a
function of definite energy and say,
"These are functions of definite energy.
Maybe instead of p,
there'll be some E on the top."
No, it's not like that.
You've got to do a lot more
work, because the reason is that the functions of definite energy
depend on what the energy is in classical mechanics.
And in classical mechanics,
the energy depends on the momentum and on the potential
energy. That means on the position.
Every problem has a different
formula for E. If it's a harmonic oscillator,
this will be ½kx^(2).
If it's a particle moving up
and down in gravity, it could be mgy.
If it's two particles connected
by a spring, it will be something else.
So in each problem,
our electron in Hydrogen atom, is the coulomb force,
coulomb potential, e_1e
_2/r^(2) times some 4Π's.
So in each problem,
there's a different V and for each V,
you've got to find the E--I mean,
you've got to find the functions.
And what is the rule?
The rule is the following.
You have to solve the following
equation, called the Schrödinger
equation, d^(2)ψ/dx^(2)
V(x)ψ(x) = E times--
you just have to solve this equation.
So let's not worry about how
you solve it. That's a problem in
mathematics, but the point is that,
given a potential V, you come to this equation,
and you go to the math department and say,
"Give me all the solutions of this equation."
And they'll give you all the
solutions of the equation. They will find,
in fact, only for certain special values of E that
the equation has acceptable solutions.
And for each one,
they'll give you the solution and you start with that solution
and do all of this. You follow that?
That's the program.
So I did a very simple example
last time, which was two examples.
First I said,
take the problem with no potential, no force,
particle is free of all forces. Then the equation is very
simple, -ℏ^(2)/2m d^(2)ψ--
I'm sorry, d^(2)ψ/dx^(2).
There is no Vψ.
That is just Eψ.
So let's rearrange this to the
following form: d^(2)ψ/dx^(2)
k^(2)ψ = 0, where k^(2) is just
2mE/ℏ^(2). You know if you multiply by
2m/ℏ^(2), you'll get this equation.
Since this combination appears
all the time, I want to give it a name,
k^(2). But you must understand,
what's the relation of k to anything else you know will
be clear in a minute. So what's the solution to this
one? You can write them in terms of
sines and cosines, but I want to write them in
terms of exponentials and the solution was--there are two
solutions. One is e^(ikx),
where k is defined that way.
Other is e^(-ikx).
You can check.
If you took this function and
take two derivatives of either this term of that term,
both will give you −k^(2) times the
function itself. Every time you take a
derivative, you bring down an ik or a -ik.
If you do it twice,
it's −k^(2). And A and B are
arbitrary right now. There is no restriction coming
from that equation. And what is k?
You know that functions of
definite momentum look like e^(ipx/ℏ),
and I have something that says e^(ikx).
It's very clear that p
is just ℏk, or k is p/ℏ.
So if you like,
this equation I got, we can write as
ℏ^(2)k^(2) is 2mE.
That means p^(2) is
2mE. That means E =
p^(2)/2m. So let's understand what that
means. Yes.
Student: Just jumping
back for a second. At the bottom of the top board,
you wrote the E for classical mechanics.
Did the equation to the right
of that come from that? Prof: In a way, it does.
That's correct.
I will tell you how it does.
But right now,
I'm just giving you a recipe. If in classical mechanics,
you know what the kinetic and potential energy of the particle
are. Kinetic is always
p^(2)/2m. Potential is V(x) where
V depends on the context. Then in the quantum theory,
the functions of definite energy obey an equation which
looks a lot like that in many ways, right?
So I've given you some clues or
some more discretion in my notes on what is the universal recipe
for guessing the equation in every context.
I'll talk about it a little bit.
But right now,
you can take the view that this guy just tells me that if you
want states of definite energy, solve that equation.
Look, so what is this saying?
Think about this.
It's saying,
if you want a state of definite energy,
it is really made up of states of definite momentum,
and the momenta that appear here are not arbitrary.
They are simply the momenta
that you will have, even in classical mechanics,
for that definite energy, because if E is
p^(2)/2m, p would be
�√2mE. That's exactly what we have.
If k = here,…
this says k = √2mE/â„ ,
and â„ k is just √2mE.
So that looks like classical
mechanics, right? Even before you took the
course, if I said there's a particle on a ring.
Its energy is E,
what is its momentum? You will say since energy is
p^(2)/2m, p must be
�√2mE and they mean going one way,
and going the other way. These are the only two options.
That part is very much like
classical mechanics. The k is related to
E exactly the way it would be.
What is novel here is that this
state, simultaneously can yield either the clockwise momentum or
the counterclockwise momentum. That's the sense in which it's
a different thing, you understand?
A classical particle of energy
E will have one of these two momenta,
but on any given day, on any given trial,
it will have only one of these two.
It cannot have both the signs.
Either it's going like this or
going like that. In quantum theory,
it's in the state of limbo. Not this alone,
not that alone, both.
The probability of going
clockwise is proportional to A^(2),
and the probability of going counterclockwise is proportional
to B^(2). If you want,
you can put the square root of L's and so on,
but they're all common to everything,
so the relative probability is just in the ratio of
A^(2) is to B^(2). It's an admixture.
And there is no restriction on
how much of A and how much of B you should put
into your answer. You can pick the extreme case
where only A is non-zero. That's fine.
That's a particle who has a
definite energy and a definite momentum.
You can pick this one,
a definite energy in the opposite momentum,
but in general, both can be non-zero.
It's a state of definite energy
whose momentum is known in magnitude but not in sign,
because both allowed signs appear here.
The other result from quantum
mechanics, which is surprising,
is whereas a particle in a ring can have any momentum,
in quantum theory, the demand that when you go
around the ring, if you add an L to the
x, you should come back to the
same function, makes the requirement that
kL is some multiple of 2Π.That means k =
2Πn/L, or p is
2Πâ„ /L times n.
The allowed values of k
or p are restricted in this way,
and the allowed values of energy then,
which is p^(2)/2m, are restricted to be
4Π^(2)â„ ^(2) over 2m L^(2) times an
integer n^(2). And this is a problem where
energy is quantized. It's really quantized because
momentum is quantized and the energy in this simple example is
simply p^(2)/2m. That means if you shine light
on the system, it will absorb light only at
certain frequencies, and it will emit light only at
certain frequencies. And what are those frequencies?
Well, all of this there's a
bunch of numbers. ℏ, m, L they're not
changing. n can be 0 or - 1,
or - 2 etc. For every n you put in,
you'll get this basic number times 0 squared or 1 squared or
2 squared. So you will find that will be
the state E_0, which is n=0.
Then above that,
the energy of that will be 2Π^(2)â„ ^(2) over
mL squared. Then there'll be n=1,
and there'll be n=-1. Both will have the same energy,
because energy depends only on n^(2).
The energy of that will be--I'm
sorry. I made a mistake.
E_0 is simply
0, because that corresponds to n=0.
E_1 will be
what I wrote here, with n=1,
which will be 2Π^(2)â„ ^(2) over
mL squared. So what I'm saying is,
at a given energy, you can have a state of right
moving clockwise or anticlockwise momentum,
or a crazy mixture of the two. But normally,
if you wish, you may select from these
mixtures, two special cases, one which is all A and
one which is all B. We like those because now we
can visualize them as states having a definite momentum
clockwise, a definite momentum counterclockwise.
That's what one really means by
drawing two lines at that energy.
So if you absorb a photon,
you've got to give an energy equal to the difference of these
two numbers, which is that. And that will be the
â„ ω of your photon.
And if you emit a photon and
come from there down here, that photon frequency will be
given by that. So this is how a quantum system
is usually probed. We cannot actually look in and
see anything. We deal with it by shining
light and we see what light it absorbs, what light it emits.
The frequency of light is a
clue as to energy level differences, you understand?
Not to the energy themselves,
but to the energy level difference that turns into
photon energy. But you can put together the
levels by putting together all the differences,
so you measure all the differences.
So one famous line is the
hydrogen 21 centimeter line. Hydrogen atom has two levels
which are close, and the wavelength of that is
21 centimeters. There's a standard fingerprint
of hydrogen anywhere in the universe.
Sometimes, if the wavelength is
not 21 but 22, you can say,
"Why is it 22? The wavelength is longer.
Maybe it's not hydrogen.
Maybe it's somebody else."
But the answer is that it is
hydrogen, but that galaxy is moving away from you,
so that its light is Doppler shifted into the red.
If something was coming towards
you, it would be blue shifted. So you can tell,
if you believe that hydrogen atoms all over the universe are
the same hydrogen atoms, and the frequency shift is only
due to the motion of the galaxy, you find two things.
First, what is it made of and
secondly, what's the speed of the galaxy?
So the red shift is what one
uses to find the whole story about galaxies.
Yes?
Student: In the context
of a photon, or an atom being ___________ by a photon,
what does the L correspond to? Prof: What is L
for an atom, right? That's your question.
Now, this is a fake atom.
This fake atom is an electron
moving on a ring, correct?
That L is actually the
size of that ring. You can go to a lab and you can
make a ring of certain micron, then the mass of the electron
is known. That L would be the
length of the ring. For an atom,
what you have to do is that you must solve it,
solve this equation, where V(x) = what?
In three dimensions,
it will be the charge of the nucleus,
which is Z times e, charge of the electron,
and it's attractive 4Πε
_0r, which is the separation between
them. So you'll have to solve a
similar equation, but that equation will have
d/dx and d/dy and d/dz and so on,
right? It's in three dimensions.
But this V(r) will
determine the allowed functions in that problem.
That won't be an L,
but a certain size will emerge even from that problem.
That size will be the size of
the orbit. For example,
if you look at the function, the lowest energy state will
look like e to the -r/a_0 times
some number, where a_0 is
called the Bohr radius. That will be roughly 10^(-8)
centimeters. That a_0 will
be a length--you can say where is the length and the height of
an atom, right? It's no box.
The length is the length made
out of eℏm. You can make a length out of
those things. That's what will come out.
So if you want,
you can take any textbook that gives you the formula for
hydrogen's Bohr radius, that's supposedly the radius of
the orbit, and you will find its length
a_0 made out of the parameters in the problem,
which is the charge of the nucleus,
the charge of the electron, the mass of the electron.
Okay, so this is,
if you want, a simple atom that we have
manufactured, because we can see quantization
of energy. Yes?
Student: When you go to
the potential of 0, is it possible to measure both
momentum and position, in that if you measure the
energy and a position, don't you know something about
the momentum? Prof: But you cannot
measure energy and position at the same time either,
because energy--what you can really measure at the same time
is energy and momentum. If you can measure p,
you can measure p^(2), or p^(3) or any function
of p. Because knowing a state of well
defined momentum, in this problem,
doesn't do anything to the energy,
because once you know that momentum is p,
p^(2)/2m is the energy. So it turns out,
this is a fortunate case where the two variables,
energy and momentum, are simply related to each
other in the fact that one is essentially the square of the
other. But if you have two variables,
one involving x and p,
at the same time, then you cannot measure them
simultaneously with either x or with p.
Yes?
Student: What is
a_0 again? Prof:
a_0. Gee, I don't know.
The Bohr radius.
I think it's ℏ squared
over me squared, but you have to check.
I don't guarantee that answer,
but I think it's something like that.
You can see if it's got at
least units of length. I'm too lazy to try that.
I think it's something like
this. Anyway, suppose it is something
like this. My main point is,
it is a length made out of Planck's constant,
the mass of the electron and the electric charge.
Together they make a length,
and that's the length that controls the size of the orbit.
All right, so this was the
simpler problem. Now I want to take a more
complicated problem of what I call a particle in a box.
So that is a potential.
See, normally,
if you make a hole in the ground,
it may take this shape, and this is the location and
this is the height, but the height is just the
potential energy directly, because it's mg times the
height. So imagine a particle somewhere
here. In classical mechanics--by the
way, the rule for doing quantum mechanics is,
in classical theory, what's the potential
experienced by the particle? That's what you ask.
And it is that potential that
enters this crazy Schrödinger equation.
So you need to know classical
mechanics before you can do quantum mechanics.
You should know what potential
the particle experiences in classical theory,
and stick that into this equation to find the allowed
energies in quantum theory. So if it's a vibrating system,
there is some omega for vibrations.
That ω
is √k/m, and you can get k from
that one. Or if it's a system like an
atom, where there's an electric charge between them,
electric force between them, you should know the electric
potential. So you should know the
potential in classical mechanics, and then you do the
quantum theory from there. So imagine a particle in this
potential in classical theory, and you ask yourself,
what does it do in quantum theory?
Well, classical theory says,
let's plot the energy of the particle, right?
It's some number that's not
supposed to change as the particle moves.
That means here,
if the energy is kinetic potential, here the potential is
0, it's all kinetic. But as you start going near the
edge, there are two parts here. That's the potential and that's
the kinetic. And right here,
all the energy is potential, and E = potential.
There is no K. What does
that mean? That means the particle comes
here, slows down, comes to a rest,
then goes back. It can never be found here if
it had this energy, because it's got more--let's
see. It's got more potential energy
than total energy. That means it's got negative
kinetic energy. That's not possible.
So every classical particle
will execute bounded oscillations at any energy,
between what are called the turning points.
The left turning point and the
right turning point is as far as you can go at that energy.
If you've got more energy,
of course you can go further out,
and if you've got even more energy,
you can just roll up to the top of the hill and escape.
That's classical mechanics.
Now we want to ask ourselves,
in quantum mechanics, what does the particle do?
That's our goal.
So I'm going to take a
following simpler problem, because it's mathematically
easier. I'm going to take a potential
that looks like this. It is V_0 to
the right of L. It's V_0 to
the left of the origin, and it is 0 here.
It's got very sharp walls,
because that makes it easy to solve the equation.
In real life,
potentials are rounded off. My fake potential has got sharp
corners. I want to take a particle of
some energy and ask, what are the allowed energies
in this problem? What do the corresponding
functions look like? That's what we're asking.
So the rule is to go back to
that equation, which now takes the following
form: d^(2)ψ/dx^(2). I'm going to stop writing the
subscript e. From now on,
you should understand, I'm trying to find states of
definite energy E. 2m times E - V ψ
= 0. I've just rearranged the
Schrödinger equation, so that it looks like second
derivative all kinds of stuff. So how do we solve this problem?
Well, V varies with
position, so there are three regions in the problem.
There's region I inside the
well. Inside this well, V is 0.
Outside this well in region II,
it = V_0, some constant.
Likewise in region III to the
right of the well, it's also V_0.
So we've got to solve this
equation in 3 different regions, middle, left and right.
First let's solve it in the
middle because it's very easy. What is the middle defined as?
Middle, region I, V = 0.
So the equation is
d^(2)ψ/dx^(2) 2mE/â„ ^(2) is 0.
So let me write it as
d^(2)ψ/dx^(2) k^(2)ψ = 0.
Just like before.
This k is the same as
before, k = √2mE/â„ ,
in region I. So what's the answer to that
one? Again, it's exponentials,
but I'm purposely going to write it in region 1 as follows:
as Acoskx Bsinkx.
I hope you all know that this
has exponentials, our old friends the
trigonometric functions, are also the solution to this
equation, because the second derivative
of cosine is roughly - the cosine.
The second derivative of sine
will also give you −k^(2) times the
sine. So rather than using
exponentials, I'm going to use sine and
cosine, because I have an ultimate motive.
You realize that exponentials
and sines and cosines are all connected by Euler's great
formula, so if I want,
I can take e^(ikx) and write it as cos i sine.
e^(-ikx),
write that as cos - i sine.
Combine the cosines,
and call the coefficient A.
Combine the coefficients for
sine and call it B, and get this form.
There are two alternative ways.
You just write this,
you will see why I picked that. That's the answer in region I.
We're all set.
Right now, A is
arbitrary, B is arbitrary and k is arbitrary.
There is no restriction at all
on any of them. Do you understand that?
Because if you take this
function, it satisfies that equation absolutely.
So the free parameters are
A, B and k, are all arbitrary.
That means it can have any
k. But the energy is
ℏ^(2)k^(2)/2m, so it can have any energy.
But we're not done yet,
because we only looked at region I.
Have to look at region II and
region III, and then you will find out what happens.
So let's go to region II So in
region II, the equation obeyed by the
ψ, is ψ (let me call it region
II) over dx squared = -2m/â„ ^(2) times
V_0 - E ψ. I just put it on the other side
now. Now this is all a constant.
That's why we picked the
potential, which is piecewise constant, so it's some constant.
You're asking yourself,
what ψ has the property that when I take 2 derivatives,
it looks like this number times ψ?
So let me call this whole
combination kappa squared. Then the answer is ψ
of 2 = C times e^(κx) D
times e^(-κx). Why?
Because two derivatives of this
guy will give me kappa squared times itself;
likewise here. And kappa has been cooked up so
that kappa squared is what I want.
Yes, now so κ
= 2m/â„ ^(2) times V_0 - E under
root. So these are now real
exponentials and not imaginary exponentials.
The reason that used to be
sines and cosines, or imaginary exponentials,
is that inside the well, E - V is a positive
number. Outside the well,
I'm assuming that E - V is a negative number.
In other words,
I'm going to consider the well to be, in the end,
infinitely deep. And the energy of the particle,
whatever it is, is some finite number.
I want to send the walls to
infinity. That's called particle in a box.
So I'm taking
V_0 finite but very, very large.
Then it's very clear that
V_0 - E is a positive number.
This kappa is what I have here.
Now look at these two functions.
This function grows
exponentially as you go to the right, to infinity.
That means the wave function
describes a particle that would rather be at infinity than
anywhere near this well. So it's not the problem we're
describing where the particle we expect is somewhere near the
well. Furthermore,
the square of ψ can never be normalized to 1,
because an exponentially growing function,
the area under that is exponentially large.
There's no way to fix that.
So on physical grounds,
we drop the solution, because we don't have to pick
C to be non-zero. We have the freedom,
and the physical condition tells you, no one wants a
function growing at infinity. We want functions falling at
infinity. This is the only function that
falls at infinity. And how does it fall at
infinity, how does it fall off? It looks like e to the
-√2m/â„ ^(2) times V_0(x).
I'm ignoring e compared to this
very large V_0. I'm taking V_0
much, much bigger than any energy I'm looking at.
So it's falling exponentially,
and the number on top is like square root of
V_0. So what do you think happens if
V_0 becomes very, very large?
Well, if you plot the function
that's exponentially falling, it falls like that.
But as you increase
V_0, it falls more and more quickly.
You follow that?
And eventually,
even V_0 goes to infinity;
it just doesn't have a life outside.
It just vanishes immediately.
Whatever you start with goes to
0 immediately. That basically means
ψ_II(x) is just 0 in region III,
if V_0 goes to infinity.
So let me emphasize what we are
doing here. We have a particle in some kind
of well, which is very, very tall, not yet infinite,
and you give it some energy E that cannot change.
In this region,
in classical mechanics, that would be the entire
kinetic energy, because potential energy is 0.
What about here?
This is the total energy
available to you. Potential energy is all of
that, so kinetic energy has to be minus some number.
That can never be.
That means that particle will
never be found in the region outside this well.
Do you understand?
As long as the energy is less
than it takes to go to the top of the barrier and climb out,
a classical particle--imagine rounding it off.
You can come rolling with any
energy. If you don't have enough energy
to go over the top, you cannot be found in this
region. But in quantum theory,
as long as the well is not infinitely large,
there's an exponentially small decaying wave function.
So a classical particle will
turn around at the classical turning points.
A quantum particle will go
slightly outside the forbidden region, into the forbidden
region in both directions of the well.
That is a purely quantum
phenomenon. It is forbidden,
meaning it's discouraged from going there, but it's not
absolutely forbidden. That's called going over
the--well, it's going into the classical
forbidden region, but as the barrier gets higher
and higher, the excursion into the
forbidden region becomes less and less,
because the function falls very rapidly.
And I'm considering,
for simplicity, a barrier which is infinitely
tall, in which case it doesn't go there at all,
even in quantum theory. Therefore, by a similar
argument, ψ is 0 here and ψ
is 0 here, and in region I,
I've shown you, ψ = Acoskx
Bsinkx, where the energy is related to
k by ℏ^(2)k^(2)/2m.
So first when we did only
region I, you get the impression that A and B and
k can be whatever you like.
But now we have an extra
restriction. The restriction is that if you
draw the function in this well, it can do whatever it wants,
but it must vanish at the two ends,
because it's 0 here and it's 0 here.
Just by continuity of ψ,
every ψ that's allowed must vanish at the two ends.
That's a new restriction,
because the function that I have here in general will not
vanish at the two ends, but I'm going to demand that it
vanish. That will tell me something
about the allowed values of A, B and k.
You will see what it is.
First I demand that ψ
should vanish on the left end of the wall, which is the wall
at x = 0. This is the wall at x =
L. ψ of 0 should be 0.
I take the function I have.
When this is 0,
x is 0, that guy is gone.
I get A times cosine 0
which is just A. But that has to vanish.
That means the coefficient
A has to vanish. It is not allowed.
So even though it looked like,
before we went to the other regions, you can have any
A and any B. If you're only in the middle
region, that's correct. But if you want to connect your
answer in the middle region to a 0 on either side,
then the coefficient has to be such that the ψ
you have in the middle vanishes at the left end.
So A is gone.
So I'm now left with only the
following thing: ψ(x) =
Bsinkx. We did not have to kill
B because that x = 0 sine vanished,
so it is allowed. But now I demand furthermore it
should vanish at x = L. Well, at x = L,
what do I find? I want 0 to =
BsinkL. That should be 0.
Now one simple way to get there
is to kill B, say B = 0, but then you've got
no solution. You killed A and you
killed B. It's certainly true that ψ
= 0 is the solution to this equation, but it doesn't
describe anything. So you don't want to kill
B, so you want to blame the 0 on the sine,
and the sine, as you know,
vanishes for any multiple of Π.
You can have 0 times Π,
2Π, 3Π, 4Π, -6Π.
They all give you 0 for sine.
Therefore k is quantized to be
one of these numbers. Therefore k has got only values
nΠ/L. I'm going to call it
k_n. And the energy,
which is really ℏ^(2)k^(2)/2m L
squared-- sorry, k^(2)/2m, becomes
now â„ ^(2)Π^(2)/2m L^(2) times n^(2).
And ψ looks like B-- ψ's of n looks like
Bsin(nΠx/L). You guys see that?
These are the allowed function.
The sine function is very nice.
It vanishes on the left end,
because when x is 0, it's gone.
When x=L,
the argument of the sine is a multiple of Π,
it vanishes. So it looks like this when
n = 1. It looks like that when
n=2, and so on. So it's got more and more
wiggles in the region in between, but they are designed
so that they vanish precisely at the edges.
So it's like it's exactly the
case of a violin string that's clamped to two ends,
which has to vanish at the two ends,
because you don't let it move there.
In between, it can do this and
it can do that. It can have more and more
wiggles. So what you find is,
of the three things that we thought were arbitrary,
A is gone, and k is not arbitrary.
k is again quantized to
be nΠ/L, so that the function does
n half oscillations between one end and the other
end. And therefore there's a
corresponding quantization of energy.
Yes?
Student: What does that
tell you? If you square
ψ_n, does that give the probability
of finding it in each one, the energy levels?
Prof: No.
Let's talk about that now.
So let's write down the
function. The ψ sub energy that we
were looking for, we have finally found out,
so let's have a picture in our mind.
Here is the well.
It's going all the way to
infinity. There is the state n=1,
and the function looks like this.
Then there is the state--let me
draw it right there, it looks like that,
n=2 and so on. In the ψ_n (let
me label the energy by n) equals some constant B
times sin(nΠx/L). That means if you want a
particle in the well to have a definite energy,
the wave function must look like one of these sine
functions. It must look like this or it
must look like that, or the higher harmonics.
They are the only allowed
functions for states of definite energy.
Now before I answer any further
questions, the allowed values of n
here you might think are 0, or - 1, or - 2 etc.,
but I claim they're just 1, 2,3 and so on.
You should think about why.
What's wrong with n=0?
Is that a possible value?
Can anybody tells me what
happens to this function if--yes?
Student: It identically
vanishes Prof: If n=0,
the function identically vanishes, right?
Because sine of 0 times
Πx/L is identical. That's not what you want.
You want a nontrivial solution.
So we drop this guy.
But how about n=-1?
Why don't I keep that?
Yes?
Oh, I thought I imagined a hand
going up. Yes?
Student: It's the same
thing as -B sine of the positive value.
Prof: That's correct.
If you put n=-1,
the function ψ of -1 is B
sin(-Πx/L), but that = -1 times B
sin(Πx/L). This guy is just
ψ_1. So it's just a multiple of the
ψ_1 function, because -1 times the state is
the same state. You don't get a new function;
you just get a multiple of the old function,
and in quantum mechanics, they're all the same.
But be very careful.
When I did a particle on a
ring, I did e^(ipx/ℏ), where I said p =
2Πâ„ /L times any integer n.
There I let n = 0, - 1, - 2 etc.
Why did I allow that?
First of all,
in that case, e to the i times 0 is not 0.
It's in fact constant.
So for a particle going in a
ring, the 0 momentum state is actually allowed.
And 1 and -1 are not the same,
because e^(ipx) and e^(-ipx),
forget the ℏ, they are not proportional to
each other. You understand that?
You can never write
e^(ipx) as a multiple of e^(-ipx).
You see that?
If there were such a multiple,
let me call the multiple as N, then N is
really e^(2ipx) by bringing it the other side.
That's certainly not a constant.
So e^(ipx) can never be
converted to e^(-ipx) by multiplying by any number,
so it's an independent solution.
On the other hand,
sinx and sin(-x) are related to each other by a
single factor of -1. So that's the reason why in
that case of a particle going in a ring,
the solutions are labeled by all integers,
positive, negative and 0, but a particle in a box,
we label only by positive integers.
Negative integers give you the
same function and 0 kills the function, so that's not allowed.
Okay, so what about B?
What shall I do with B?
What equation's going to
determine B? Yes, any idea?
Yes?
Student: Normalize right?
Prof: Right.
So first of all,
the equation will never tell you what B is.
That's what I want you to
understand. The equation that I've written
here for the energy functions on the top,
if size of B is a solution, you can multiply both
sides by any number, like 15.
Then 15 times ψ
will also be a solution. That's a property of a linear
equation that the answer is not given in overall scale.
Give me one answer;
I rescale it, it's still an answer.
So linear equation will never
tell me the overall size of the function.
That is picked by the extra
convenient restriction that ψ^(2)dx = 1.
That will determine B
and I'm claiming the answer is the √(2/L) is what
you want B to be. The answer is again simple.
The average value of sine
squared over 1 half period is 1 half.
If you square or integrate or
whatever, you'll get L/2,
so B^(2) times L/2 should be 1,
and B should therefore be the √(2/L).
So let me write down now the
final result of this problem, which I think is very,
very instructive. If you can follow this much,
you've really got 80 percent of the quantum mechanics I want to
teach you. If you're comfortable with this
example, because we're trying to find
out, in quantum theory, you've got a particle in a box,
which is like a hole in the floor,
which is so deep that it's infinitely deep.
But even if it's not infinite
but very deep, the answer's good enough.
You want to know what it will
do. In classical mechanics,
what will it do? If you're a particle in a box,
what's your lowest energy state?
Yes?
Student: Floor of the
box. Prof: If you're in
prison with infinitely tall walls, you just sit on the floor
and mope, right? That's the lowest energy state.
But that's not allowed in
quantum mechanics, because you have a definite
x, which is wherever you're
sitting, a definite p, you're not moving.
Such a state is not allowed.
So in quantum mechanics,
a particle cannot simply sit still.
Therefore, it will necessarily
have an uncertainty in position, and it will try to spread
itself over the box. And the typical uncertainty is
of order. If I say, "Where is this
particle?" I don't know anything.
It's got to be somewhere in the
box. That's the estimate for
Δx. So Δp has got to
be bigger than â„ /L. Forget all the 2Π's.
So it cannot have 0 momentum.
It's momentum has to be
fluctuating with that range. And the kinetic energy,
which is in fact the total energy inside the box,
which is p^(2)/2m, the size of that is
ℏ^(2)/2m L^(2). So that is your estimate for
what should be the energy of a particle in a box of size
L. So what happens is, the box of size L forces
it to have a spread in momentum, because you've squeezed it to a
position known to within L.
The spread in momentum
translates into spread in kinetic energy,
and if I take the spread in kinetic energy to be a rough
estimate of the lowest kinetic energy I could have,
you get a number which except for factors of Π
is in fact the right answer. So people always estimate
energies of particles by doing this calculation.
For example,
take a nucleus, which is 10^(-13) centimeters
across. I put a proton there.
What's the minimum kinetic
energy of that proton? You find by the following
method: Δx is 10^(-13), because it's somewhere
in the nucleus. Δp is that,
and you do that, you will get some energy.
That will be the typical energy
of a particle inside the nucleus.
So particles inside the nucleus
cannot sit still. More you cram them,
it's like children. You put them in a tighter room,
they're jiggling more and more. That's called zero point motion.
You cannot just nail them in
x without letting them spread in p.
That means you cannot get the
lowest energy So lowest energy is some number you can estimate
by the uncertainty principle. So the next question was,
let's take one of these functions, say ψ's of 3,
the square root of (2/L)sin(
3Πx/L). If you plot that guy,
that function will look like this.
Now what was your question?
If I measure a particle's
position, what's the probability?
You've just got to square this
function. It will look like this.
That's the probability.
If you measure the energy,
what will you get? You will get exactly one energy
corresponding to n=3. It's in a state of definite
energy, but not definite position.
Position has got some ups and
downs. The probability for x looks
like that. So everyone with me?
This is like knowing the
spectrum of this artificial atom.
In a real atom,
the electron is forced to stay near the nucleus,
because the 1 over r potential,
if you plot it as a function of r,
looks like a well. And the particle is somewhere
here. In this problem,
the particle is confined by the box.
Whenever you have a confining
thing that keeps the particle in a certain region,
the energy gets quantized. And we found out in our first
example the quantized energy levels of a particle in a box.
And you can probe the particle
by shining light and the energy levels are plotted like this.
This is n=1,
this is n=2, which is n=3.
Zero is here.
The zero state is not allowed.
You've got to have n=1
and 4 times that will be n=2, and 9 times that
will be n=3. Energy grows like n^(2).
And then you can come from
there to there emitting a photon,
or you can go from there to there emitting a photon,
or you can go from here to here absorbing a photon.
You can do all kinds of things.
That's how if you want you can
test quantum mechanics. Put a known particle inside the
well and shine light and see what light it absorbs.
Yes.
Student: So the ψ
of 3, is that the ψ of the energy function
_________? Prof: Yes.
Exactly right.
So what I should really write,
he's absolutely right, I should write E =
â„ ^(2)Π^(2)/2m L squared 3 squared,
right? The third energy level.
So I'm short circuiting all of
that and just calling it ψ_3.
Normally, if you put a label 3,
it's not clear what we're talking about.
Maybe it's the position
x=3. Maybe it's the momentum
p=3. But in this lecture,
since we're talking only about energy,
I thought I will just call it ψ_3,
but at least I should call it ψ_E3.
Student:
> Prof: Yeah,
I just did that. Student:
> Prof: You can take the
complex conjugate, but for a real function,
you can just square it, right?
You understand?
If ψ is real,
ψ*is the same as ψ. So ψ*is also
(2/L)sin(3 Πx/L).
So ψ*ψ is just a square.
Student: On the first
board, you have A_E = _________
ψ*E ψ _______.
I'm wondering why there's no
_______________. Prof: No.
I think I know what the problem
is you're having now. That formula is not relevant
now. I'll tell you when that formula
will come into play. It will come into play in the
following situation. I have a particle in a box
whose wave function looks like this.
Somebody put it in that state,
okay? It's an allowed wave function
because it vanishes at the ends like it should.
Beyond that,
it does anything it wants. That is my ψ(x),
okay? If I measure the energy of that
guy, now you tell me, if I measure the energy in that
state, what answer will I get? Student:
> Prof: Pardon me?
Yes.
I'll get one of those quantized
values, do you understand? A particle in a box under
energy measurement can give only one of those numbers.
Now this guy is not one of
those functions, you understand.
This is a randomly chosen
function. But when I measure energy,
so what will I do? Maybe you can guess now.
Student: Write it as a
sum of energy functions Prof: That's correct.
You will have to write this guy
as a sum of A--let me call it n for energy now.
A √(2/L)
sin(nΠx/L). I mean, this is just
A_E ψ's of E of x.
Except E is labeled by an
integer, so I'm writing it this way.
Then I have to find
A_n. And if I want to find
A_n, that = integral of (2/L)
sin(nΠx/L). If you want,
you can put a conjugate, but it doesn't matter,
because sine is real. Multiply it by Mr. Crazy
here, whatever it is, and do their integral.
It will give you some number
for every n. A_1,
A_2, A_3,
A_4, and the square of every one of
them will give you the odds that you will find this energy or
that energy or some other energy.
So what quantum mechanics tells
you is, an energy measurement of a particle in a box cannot give
a random answer. It can only give one of these
answers. That's the quantization of
energy. It does not mean every particle
in a box is in a state of definite energy.
A particle in a box can be any
wave function, provided it vanishes at the
ends and doesn't spill out of the infinitely tall wall.
So for every one of them,
you can ask the usual questions: what happens when I
measure position? Square the function.
What happens when I measure
momentum? Write it in terms of e
to the ipx functions. What happens when I measure
energy? Write it in terms of these
functions. Okay?
All right, so I want to do one
last topic here, which is called scattering,
but I don't want to proceed till you guys have understood
this part. This is covered in all the
books and maybe you can read the books or you can read my notes.
You can talk to everybody.
But we have solved--from this,
all quantum problems that we all do are exactly isomorphic to
this problem. It's the same thing we do over
and over again, except the equations you solve
are more difficult. They can be in higher
dimensions. They can involve more than one
particle. The potential could be
complicated, but the philosophy is always the same.
There'll be some
Schrödinger equation, and you want to find all the
functions that satisfy it, find the allowed energies and
the allowed functions. We will see that states of
definite energy have a very privileged role in quantum
mechanics. That will be clear in the next
two lectures. But right now,
here's what I want to do. This is a very small variation
of what we have done, so most of the hard work has
been done. I'm just going to cash in on
that. Take the following problem in
classical mechanics: here is the level ground.
Then I make a little change in
the landscape so that this has got certain height and therefore
certain potential, V_0. And from
this end, I roll a marble and see what
happens. The marble is given some energy
E, which because there is no potential here,
it's just p^(2)/2m. Question is, what will happen?
Can you tell me what will
happen in this problem, to this marble?
Student: Don't we have
to use the Schrödinger equations?
Prof: No.
Classical mechanics.
Student: Oh,
the marble will only go up as far as it has the amount of
energy that that would equal the potential energy at that point.
Prof: That's correct.
So he said it will go as far as
energy equals potential, so let p^(2)/2m have
this value. That's my energy.
And it does not change.
That's the law of conservation
of energy. So here, since energy is
kinetic potential, it's all kinetic.
Somewhere here,
that part of it is potential; the rest of it is kinetic.
Here kinetic vanishes.
Here you're not allowed.
So you will go,
you will climb that far and you'll come back.
But if your initial energy was
that, then you will come here, going very fast.
You will slow down,
but you still make it on the other side, but with lower
velocity. That's very simple.
There's only one answer you can
get. You throw something.
If it comes back,
it means the barrier is taller than what you have.
Suppose you cannot see this
barrier. All you have is a marble,
and you want to know how tall the barrier is.
Very easy.
Throw the marble at a known
speed. If it comes back,
then you haven't reached the top of the barrier.
Keep on increasing the kinetic
energy till one day it doesn't come back.
That's when you're sitting
right here. That's one way to tell how tall
this hill is without actually going there.
But now what I want to do is do
a quantum mechanical experiment in exactly the same situation.
I want to send a particle from
the left and I want to see what happens.
So now, to simplify the
problem, I'm going to take the potential to again have only 2
values. At x = 0,
it jumps to a value V_0.
So this problem is just the
Schrödinger equation in the left region, Schrödinger
equation in the right region. So I'm going to spare you some
of the calculations, because it's so familiar.
One will look like Ae^(ikx)
Be^(-ikx), where k^(2) is just
2mE/ℏ^(2), here.
First, let's take a case where
E is bigger than V_0,
okay? Then I go to region II.
Region II I've got to solve the
Schrödinger equation with a V_0 in it.
Then you will find in region
II, ψ II looks like Ce^(ik'x)
De^(−ik'x), where k'^(2) will be
2m times E - V_0 over
ℏ^(2). Because when you have a
potential, you must always have--
look at the Schrödinger equation here--
2m/ℏ^(2) at E - V.
So E is not 0.
The k' now,
so what does this mean? You can see very simply,
it's got some momentum here. It's got a smaller momentum
here, because k'^(2) is going to be smaller than
k^(2), because V_0 is non-zero
here. That is just the statement that
when the particle's moving here, it's slower than when it's
here, because it has to climb up the hill.
So the wave functions in this
region has got an incoming, right moving wave of momentum,
ℏk. It's got to reflect that wave
of momentum -ℏk. And in this region again,
it's got a right moving and a left moving, because this
k can be of either sign. So you can pick the
coefficients, A,
B, C, D arbitrarily, but I'm going to choose
D = 0, because what I have in mind is
a problem where the particle actually came in from the left.
If it came in from the left,
I'm prepared for some of it coming back reflected,
and some of it getting transmitted.
There's nobody coming in from
the right at infinity, because I'm not sending
anything from there. I'm sending my particle from
the left. So I pick a solution in which
D, which is anything, I pick that to be 0.
So here are my two solutions.
I hope you understand that
k and k' are not the same number.
k' is smaller because
the momentum on the right hand side is going to be less,
because some of it's eaten up by potential energy.
So what people do is they draw
a picture like this. They draw a wave that's going
at some speed, and it comes in this region and
it slows down. That's what the function will
look like. The real function's made of
exponentials, but the real and imaginary
parts will have rapid oscillations on the left and
slow oscillations on the right, because k is bigger than
k'. Very good.
So now I have the following
condition: at the interface where the two regions meet,
the ψ must match from the left and from the right.
So x = 0,
coming from the left, I want ψ_1 at 0
should be the same at ψ_2 at 0.
That will tell me that A B =
C. Because if you put x =
0, all the exponentials become one, and A B must =
C. There's the condition that if
you come from the two sides, the wave function has to match.
But in the Schrödinger
theory, you make a second requirement that the derivatives
also match. The two functions coming from
the left and right cannot join like this.
They must join with the same
slope. The reason the slope must be
the same is that if the slope changed abruptly at one point,
the rate of change of slope will be infinite there,
because d^(2)ψ/dx^(2) is the rate of change of the
slope. But in the Schrödinger
equation, the second derivative ψ,
if it's infinite, there is no compensating terms
in the equation, because E and V
are all finite. So that infinity will not be
balanced, so you can never have that.
So second derivative must be
finite; first derivative must be
continuous. So I make the requirement that
dψ/dx in region I at x = 0 must match
dψ II dx at x = 0.
What does that give me?
Well, take the derivative and
equate them at x = 0, you'll find ik times A - B
= ik' times C. Take the derivative and then
equate the derivatives at x = 0.
Derivatives brings down
ik for this, -ik for that.
Blah, blah, you put them
together. So these are the two conditions.
Now what causes mild panic is
that I have three unknowns and two equations.
I've got A,
B and C, and I've got only two
equations. So what do you think is going
on? Anybody have an idea?
It's that even in the particle
in a box, you'll remember the overall
height of the wave function is never determined by the
Schrödinger equation. It's whatever you like.
So what people do is they say,
"Let's pick A = 1."
That sets the overall height
arbitrarily. Then I get 1 B = C and
ik times 1 - B = ik' [C}.
Now of course you can solve
for B and you can solve for C.
I'm not going to do that here;
it's very easy. You're going to find B = k
− k' over k k'. You're going to find C =
2k over k k'. That's simple algebra of
simultaneous equations. I want you to look at the
answer. You remember now what A,
B and C are. A comes in,
B goes back and C goes to the right.
What do you find strange about
the answer? It's that if you came in from
the left with an energy bigger than the barrier,
there's a certain probability that you will go backwards.
In classical mechanics,
if your energy is more than the barrier, you will never come
back. The probability is 0.
In quantum theory,
even though your energy is more than the barrier,
there's a certain probability you'll get reflected and certain
probability to get transmitted. The square of this is like the
odds of going to the right. The square of this is like the
odds of getting reflected. The point is the reflection
probability does not vanish even if the energy is bigger than the
barrier height. That's because this is quantum
mechanics and you're really dealing with waves and not
particles. And waves have the property
that when the medium changes, there is some reflection.
And to the quantum mechanical
wave ψ, a barrier looks like a change
in refractive index, so it will get reflected and
some will get transmitted. But the interesting thing is,
in the quantum problem, you can send your billiard ball
to find how tall the mountain is.
And even if the energy was more
than the mountain, sometimes it'll come back.
So that's not a reliable way to
measure that. The last example is,
what if your energy is less than this?
What if your energy is like
here, lower than the barrier? If you're lower than the
barrier, you will find that k' will equal i
times some κ, because it will become
imaginary, and you will have then a function that looks like
this. The function will oscillate,
then it will die exponentially in this region.
It's exactly what I did
earlier, like particle in a box. In the classically forbidden
region, it goes a little distance but not very far.
So it does get reflected if the
energy is less, but you can actually find it in
a region where it's not supposed to be.
And you can say that's going to
be very embarrassing for the particle to be found here,
because its kinetic energy should be negative here.
So what's the particle say to
you when you catch it there? It will say,
"In order to locate me here,
you shine some light on me, and the energy I got from the
photon made me legitimate in that region."
It will come back to positive
energy. So the act of measurement will
prevent it from ever being caught in that region,
but it's got an amplitude, a probability to be caught
there. Final thing I want to mention
is, if this barrier, instead of going on forever,
terminated there and became 0 again,
this will leak out here and start going out with the same
wavelength as that one. That's called barrier
penetration. That means if you send a
particle here with an energy not enough to overcome the barrier
in classical mechanics, in quantum mechanics,
it's got a small chance of being found on the other side.
On the other side,
it can go scot free. It's not allowed in this
region, because there the kinetic energy is negative.
Here kinetic energy is the same
as before, except the height may be very small.
So this means no barrier is
completely safe in quantum theory.
So I'm going to give you my
final survival tip. I've told you so many
situations you can be in where this course will help you.
So here's another one.
You are in a prison and it
doesn't have infinite walls, but it's got some walls.
What's your strategy?
I say go ram yourself into that
wall as often as you can, because there's a small
probability that you can suddenly find yourself on the
other side. This is what happens in alpha
decay. In alpha decay,
there's helium nuclei sitting inside a nucleus.
There's a barrier that keeps it
from coming out of the nucleus, but the barrier goes up and
comes down. So if you go a certain distance
from the nucleus, you're again free.
So the alpha particle does
exactly what I told you. It goes rattling back and forth
inside the nucleus and once in a while,
it penetrates and comes to the other side,
and that's the alpha decay of the nucleus.
So this really happens.
It may not happen to you in the
situation I described, but the probability is not
zero.
