WEBVTT

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Prof: Okay,
so I'm going to begin with

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something I did last time,
because it was not completed,

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then we'll take it from there.

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And we had a lot more questions
last time, which is good,

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very helpful,
so I know what may be bothering

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you.

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So don't hesitate to stop me.

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The new thing we did last time
was,

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if I give you some wave
function, ψ(x),

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and I ask you what happens if I
measured energy,

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the recipe is the same as in
momentum.

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You've got to find functions
called

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ψ_E(x),
which are supposed to be states

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in which,
if the electron was there,

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it'll have a definite energy.

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Now, not worrying for a moment
on how you get them,

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the recipe for what happens
when you measure energy is that

00:57.268 --> 01:00.408
if you're given some arbitrary
function ψ,

01:00.408 --> 01:02.598
that's a state the particle is
in.

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And you're asking,
"What happens if I measure

01:04.102 --> 01:04.382
energy?

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What will I get?"

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The instruction is to write
that function as a sum over

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these functions,
each associated with a definite

01:14.899 --> 01:15.899
energy.

01:15.900 --> 01:20.270
And once you have done that,
the probability that you will

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get a certain energy E =
the absolute value squared of

01:24.875 --> 01:27.325
that particular coefficient.

01:27.330 --> 01:29.320
Not different from
ψ_p ,

01:29.322 --> 01:29.972
same rule.

01:29.970 --> 01:33.150
And the rule for
A_E is also the

01:33.150 --> 01:34.410
same, or similar.

01:34.410 --> 01:38.200
A_E is the
integral of complex conjugate of

01:38.200 --> 01:41.350
this function associated with
definite energy,

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times the function you're
thinking about,

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integrated over space.

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It's the same rule.

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So it can happen,
the problem can take many

01:53.617 --> 01:54.147
forms.

01:54.150 --> 01:58.470
Sometimes one can give you a
function ψ(x).

01:58.470 --> 02:01.660
Then you take all these
functions, do all these

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integrals, find all the
coefficients and you square

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them.

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Other times the coefficients
may be given to you;

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you're just asked,
"What's the odd for this

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energy or that energy?"

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But now the question is,
what are these functions,

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ψ_E(x).

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Whereas functions of definite
momentum are given once and for

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all by this formula,
e^(ipx/ℏ),

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I wish I could give you a
function of definite energy and

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say,
"These are functions of

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definite energy.

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Maybe instead of p,
there'll be some E on

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the top."

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No, it's not like that.

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You've got to do a lot more
work, because the reason is that

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the functions of definite energy
depend on what the energy is in

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classical mechanics.

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And in classical mechanics,
the energy depends on the

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momentum and on the potential
energy.

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That means on the position.

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Every problem has a different
formula for E.

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If it's a harmonic oscillator,
this will be

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If it's a particle moving up
and down in gravity,

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it could be mgy.

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If it's two particles connected
by a spring, it will be

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something else.

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So in each problem,
our electron in Hydrogen atom,

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is the coulomb force,
coulomb potential,

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e_1e
_2/r^(2)

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times some 4Π's.

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So in each problem,
there's a different V

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and for each V,
you've got to find the

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E--I mean,
you've got to find the

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functions.

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And what is the rule?

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The rule is the following.

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You have to solve the following
equation,

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d^(2)ψ/dx^(2)
V(x)ψ(x) = E

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times--
you just have to solve this

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equation.

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So let's not worry about how
you solve it.

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That's a problem in
mathematics, but the point is

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that,
given a potential V,

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you come to this equation,
and you go to the math

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department and say,
"Give me all the solutions

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of this equation."

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And they'll give you all the
solutions of the equation.

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They will find,
in fact, only for certain

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special values of E that
the equation has acceptable

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solutions.

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And for each one,
they'll give you the solution

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and you start with that solution
and do all of this.

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You follow that?

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That's the program.

04:33.519 --> 04:37.709
So I did a very simple example
last time, which was two

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examples.

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First I said,
take the problem with no

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potential, no force,
particle is free of all forces.

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Then the equation is very
simple, -ℏ^(2)/2m

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d^(2)ψ--
I'm sorry,

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d^(2)ψ/dx^(2).

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There is no Vψ.

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That is just Eψ.

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So let's rearrange this to the
following form:

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d^(2)ψ/dx^(2)
k^(2)ψ = 0,

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where k^(2) is just
2mE/ℏ^(2).

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You know if you multiply by
2m/ℏ^(2),

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you'll get this equation.

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Since this combination appears
all the time,

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I want to give it a name,
k^(2).

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But you must understand,
what's the relation of k

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to anything else you know will
be clear in a minute.

05:42.759 --> 05:45.099
So what's the solution to this
one?

05:45.100 --> 05:48.210
You can write them in terms of
sines and cosines,

05:48.209 --> 05:51.839
but I want to write them in
terms of exponentials and the

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solution was--there are two
solutions.

05:54.560 --> 05:58.690
One is e^(ikx),
where k is defined that

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way.

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Other is e^(-ikx).

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You can check.

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If you took this function and
take two derivatives of either

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this term of that term,
both will give you

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−k^(2) times the
function itself.

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Every time you take a
derivative, you bring down an

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ik or a -ik.

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If you do it twice,
it's −k^(2).

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And A and B are
arbitrary right now.

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There is no restriction coming
from that equation.

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And what is k?

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You know that functions of
definite momentum look like

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e^(ipx/ℏ),
and I have something that says

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e^(ikx).

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It's very clear that p
is just ℏk,

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or k is p/ℏ.

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So if you like,
this equation I got,

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we can write as
ℏ^(2)k^(2) is

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2mE.

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That means p^(2) is
2mE.

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That means E =
p^(2)/2m.

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So let's understand what that
means.

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Yes.

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Student: Just jumping
back for a second.

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At the bottom of the top board,
you wrote the E for

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classical mechanics.

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Did the equation to the right
of that come from that?

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Prof: In a way, it does.

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That's correct.

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I will tell you how it does.

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But right now,
I'm just giving you a recipe.

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If in classical mechanics,
you know what the kinetic and

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potential energy of the particle
are.

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Kinetic is always
p^(2)/2m.

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Potential is V(x) where
V depends on the context.

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Then in the quantum theory,
the functions of definite

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energy obey an equation which
looks a lot like that in many

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ways, right?

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So I've given you some clues or
some more discretion in my notes

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on what is the universal recipe
for guessing the equation in

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every context.

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I'll talk about it a little bit.

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But right now,
you can take the view that this

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guy just tells me that if you
want states of definite energy,

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solve that equation.

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Look, so what is this saying?

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Think about this.

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It's saying,
if you want a state of definite

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energy,
it is really made up of states

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of definite momentum,
and the momenta that appear

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here are not arbitrary.

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They are simply the momenta
that you will have,

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even in classical mechanics,
for that definite energy,

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because if E is
p^(2)/2m,

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p would be
�√2mE.

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That's exactly what we have.

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If k = here,…
this says k =

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√2mE/ℏ,
and ℏk is just

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√2mE.

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So that looks like classical
mechanics, right?

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Even before you took the
course, if I said there's a

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particle on a ring.

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Its energy is E,
what is its momentum?

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You will say since energy is
p^(2)/2m,

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p must be
�√2mE and they

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mean going one way,
and going the other way.

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These are the only two options.

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That part is very much like
classical mechanics.

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The k is related to
E exactly the way it

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would be.

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What is novel here is that this
state, simultaneously can yield

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either the clockwise momentum or
the counterclockwise momentum.

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That's the sense in which it's
a different thing,

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you understand?

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A classical particle of energy
E will have one of these

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two momenta,
but on any given day,

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on any given trial,
it will have only one of these

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two.

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It cannot have both the signs.

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Either it's going like this or
going like that.

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In quantum theory,
it's in the state of limbo.

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Not this alone,
not that alone,

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both.

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The probability of going
clockwise is proportional to

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A^(2),
and the probability of going

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counterclockwise is proportional
to B^(2).

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If you want,
you can put the square root of

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L's and so on,
but they're all common to

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everything,
so the relative probability is

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just in the ratio of
A^(2) is to B^(2).

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It's an admixture.

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And there is no restriction on
how much of A and how

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much of B you should put
into your answer.

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You can pick the extreme case
where only A is non-zero.

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That's fine.

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That's a particle who has a
definite energy and a definite

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momentum.

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You can pick this one,
a definite energy in the

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opposite momentum,
but in general,

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both can be non-zero.

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It's a state of definite energy
whose momentum is known in

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magnitude but not in sign,
because both allowed signs

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appear here.

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The other result from quantum
mechanics,

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which is surprising,
is whereas a particle in a ring

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can have any momentum,
in quantum theory,

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the demand that when you go
around the ring,

10:48.250 --> 10:50.680
if you add an L to the
x,

10:50.678 --> 10:52.738
you should come back to the
same function,

10:52.740 --> 10:59.100
makes the requirement that
kL is some multiple of

10:59.096 --> 11:04.526
2Π.That means k =
2Πn/L,

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or p is
2Πℏ/L times

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n.

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The allowed values of k
or p are restricted in

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this way,
and the allowed values of

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energy then,
which is p^(2)/2m,

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are restricted to be
4Π^(2)ℏ^(2) over

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2m L^(2) times an
integer n^(2).

11:31.440 --> 11:35.440
And this is a problem where
energy is quantized.

11:35.440 --> 11:39.830
It's really quantized because
momentum is quantized and the

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energy in this simple example is
simply p^(2)/2m.

11:44.210 --> 11:46.540
That means if you shine light
on the system,

11:46.535 --> 11:49.125
it will absorb light only at
certain frequencies,

11:49.133 --> 11:52.113
and it will emit light only at
certain frequencies.

11:52.110 --> 11:53.930
And what are those frequencies?

11:53.928 --> 11:57.108
Well, all of this there's a
bunch of numbers.

11:57.110 --> 11:58.590
ℏ, m, L they're not
changing.

11:58.590 --> 12:06.280
n can be 0 or - 1,
or - 2 etc.

12:06.278 --> 12:09.408
For every n you put in,
you'll get this basic number

12:09.407 --> 12:11.777
times 0 squared or 1 squared or
2 squared.

12:11.778 --> 12:15.998
So you will find that will be
the state E_0,

12:16.000 --> 12:17.550
which is n=0.

12:17.548 --> 12:22.118
Then above that,
the energy of that will be

12:22.119 --> 12:27.449
2Π^(2)ℏ^(2) over
mL squared.

12:27.450 --> 12:32.400
Then there'll be n=1,
and there'll be n=-1.

12:32.399 --> 12:35.529
Both will have the same energy,
because energy depends only on

12:35.529 --> 12:36.299
n^(2).

12:36.298 --> 12:39.708
The energy of that will be--I'm
sorry.

12:39.710 --> 12:41.040
I made a mistake.

12:41.038 --> 12:46.638
E_0 is simply
0, because that corresponds to

12:46.636 --> 12:47.846
n=0.

12:47.850 --> 12:49.960
E_1 will be
what I wrote here,

12:49.960 --> 12:54.850
with n=1,
which will be

12:54.845 --> 13:03.095
2Π^(2)ℏ^(2) over
mL squared.

13:03.100 --> 13:06.200
So what I'm saying is,
at a given energy,

13:06.197 --> 13:09.917
you can have a state of right
moving clockwise or

13:09.916 --> 13:14.406
anticlockwise momentum,
or a crazy mixture of the two.

13:14.408 --> 13:15.978
But normally,
if you wish,

13:15.976 --> 13:19.296
you may select from these
mixtures, two special cases,

13:19.298 --> 13:23.058
one which is all A and
one which is all B.

13:23.058 --> 13:25.768
We like those because now we
can visualize them as states

13:25.770 --> 13:28.480
having a definite momentum
clockwise, a definite momentum

13:28.482 --> 13:29.502
counterclockwise.

13:29.500 --> 13:34.440
That's what one really means by
drawing two lines at that

13:34.436 --> 13:35.226
energy.

13:35.230 --> 13:38.310
So if you absorb a photon,
you've got to give an energy

13:38.307 --> 13:41.837
equal to the difference of these
two numbers, which is that.

13:41.840 --> 13:44.840
And that will be the
ℏω of your

13:44.840 --> 13:45.440
photon.

13:45.440 --> 13:48.110
And if you emit a photon and
come from there down here,

13:48.106 --> 13:50.376
that photon frequency will be
given by that.

13:50.379 --> 13:52.739
So this is how a quantum system
is usually probed.

13:52.740 --> 13:54.990
We cannot actually look in and
see anything.

13:54.990 --> 13:58.210
We deal with it by shining
light and we see what light it

13:58.207 --> 13:59.987
absorbs, what light it emits.

13:59.990 --> 14:04.240
The frequency of light is a
clue as to energy level

14:04.240 --> 14:06.790
differences, you understand?

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Not to the energy themselves,
but to the energy level

14:08.990 --> 14:10.810
difference that turns into
photon energy.

14:10.808 --> 14:14.268
But you can put together the
levels by putting together all

14:14.269 --> 14:16.529
the differences,
so you measure all the

14:16.534 --> 14:17.494
differences.

14:17.490 --> 14:20.850
So one famous line is the
hydrogen 21 centimeter line.

14:20.850 --> 14:23.370
Hydrogen atom has two levels
which are close,

14:23.373 --> 14:26.073
and the wavelength of that is
21 centimeters.

14:26.070 --> 14:29.080
There's a standard fingerprint
of hydrogen anywhere in the

14:29.078 --> 14:29.658
universe.

14:29.658 --> 14:33.498
Sometimes, if the wavelength is
not 21 but 22,

14:33.504 --> 14:36.414
you can say,
"Why is it 22?

14:36.409 --> 14:38.119
The wavelength is longer.

14:38.120 --> 14:38.920
Maybe it's not hydrogen.

14:38.919 --> 14:40.979
Maybe it's somebody else."

14:40.980 --> 14:44.410
But the answer is that it is
hydrogen, but that galaxy is

14:44.408 --> 14:47.408
moving away from you,
so that its light is Doppler

14:47.408 --> 14:48.938
shifted into the red.

14:48.940 --> 14:51.960
If something was coming towards
you, it would be blue shifted.

14:51.960 --> 14:54.400
So you can tell,
if you believe that hydrogen

14:54.399 --> 14:57.669
atoms all over the universe are
the same hydrogen atoms,

14:57.668 --> 15:00.788
and the frequency shift is only
due to the motion of the galaxy,

15:00.789 --> 15:01.959
you find two things.

15:01.960 --> 15:07.000
First, what is it made of and
secondly, what's the speed of

15:07.000 --> 15:08.130
the galaxy?

15:08.129 --> 15:11.139
So the red shift is what one
uses to find the whole story

15:11.144 --> 15:12.064
about galaxies.

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Yes?

15:12.710 --> 15:17.490
Student: In the context
of a photon, or an atom being

15:17.485 --> 15:22.095
___________ by a photon,
what does the L correspond to?

15:22.100 --> 15:24.000
Prof: What is L
for an atom, right?

15:24.000 --> 15:25.600
That's your question.

15:25.600 --> 15:29.590
Now, this is a fake atom.

15:29.590 --> 15:32.180
This fake atom is an electron
moving on a ring,

15:32.184 --> 15:32.754
correct?

15:32.750 --> 15:35.610
That L is actually the
size of that ring.

15:35.610 --> 15:38.570
You can go to a lab and you can
make a ring of certain micron,

15:38.570 --> 15:40.560
then the mass of the electron
is known.

15:40.558 --> 15:42.158
That L would be the
length of the ring.

15:42.158 --> 15:45.938
For an atom,
what you have to do is that you

15:45.936 --> 15:48.916
must solve it,
solve this equation,

15:48.922 --> 15:51.472
where V(x) = what?

15:51.470 --> 15:54.960
In three dimensions,
it will be the charge of the

15:54.960 --> 15:57.180
nucleus,
which is Z times e,

15:57.182 --> 15:59.902
charge of the electron,
and it's attractive

15:59.895 --> 16:02.585
4Πε
_0r,

16:02.590 --> 16:05.570
which is the separation between
them.

16:05.570 --> 16:07.540
So you'll have to solve a
similar equation,

16:07.543 --> 16:10.273
but that equation will have
d/dx and d/dy and

16:10.269 --> 16:11.819
d/dz and so on,
right?

16:11.820 --> 16:13.270
It's in three dimensions.

16:13.269 --> 16:17.749
But this V(r) will
determine the allowed functions

16:17.750 --> 16:19.190
in that problem.

16:19.190 --> 16:22.340
That won't be an L,
but a certain size will emerge

16:22.335 --> 16:23.735
even from that problem.

16:23.740 --> 16:26.040
That size will be the size of
the orbit.

16:26.038 --> 16:28.828
For example,
if you look at the function,

16:28.830 --> 16:32.730
the lowest energy state will
look like e to the

16:32.729 --> 16:36.039
-r/a_0 times
some number,

16:36.038 --> 16:38.148
where a_0 is
called the Bohr radius.

16:38.149 --> 16:41.659
That will be roughly 10^(-8)
centimeters.

16:41.658 --> 16:44.308
That a_0 will
be a length--you can say where

16:44.307 --> 16:46.467
is the length and the height of
an atom, right?

16:46.470 --> 16:47.760
It's no box.

16:47.759 --> 16:53.979
The length is the length made
out of eℏm.

16:53.980 --> 16:55.530
You can make a length out of
those things.

16:55.529 --> 16:57.849
That's what will come out.

16:57.850 --> 17:00.100
So if you want,
you can take any textbook that

17:00.104 --> 17:02.714
gives you the formula for
hydrogen's Bohr radius,

17:02.710 --> 17:04.740
that's supposedly the radius of
the orbit,

17:04.740 --> 17:07.820
and you will find its length
a_0 made out of

17:07.817 --> 17:10.657
the parameters in the problem,
which is the charge of the

17:10.657 --> 17:13.637
nucleus,
the charge of the electron,

17:13.635 --> 17:16.075
the mass of the electron.

17:16.078 --> 17:18.158
Okay, so this is,
if you want,

17:18.163 --> 17:20.963
a simple atom that we have
manufactured,

17:20.964 --> 17:24.274
because we can see quantization
of energy.

17:24.269 --> 17:26.189
Yes?

17:26.190 --> 17:28.930
Student: When you go to
the potential of 0,

17:28.930 --> 17:32.940
is it possible to measure both
momentum and position,

17:32.940 --> 17:36.540
in that if you measure the
energy and a position,

17:36.538 --> 17:40.758
don't you know something about
the momentum?

17:40.759 --> 17:43.919
Prof: But you cannot
measure energy and position at

17:43.919 --> 17:47.329
the same time either,
because energy--what you can

17:47.333 --> 17:51.503
really measure at the same time
is energy and momentum.

17:51.500 --> 17:55.030
If you can measure p,
you can measure p^(2),

17:55.029 --> 17:57.879
or p^(3) or any function
of p.

17:57.880 --> 18:00.370
Because knowing a state of well
defined momentum,

18:00.368 --> 18:04.418
in this problem,
doesn't do anything to the

18:04.423 --> 18:06.443
energy,
because once you know that

18:06.444 --> 18:09.474
momentum is p,
p^(2)/2m is the energy.

18:09.470 --> 18:12.010
So it turns out,
this is a fortunate case where

18:12.009 --> 18:14.179
the two variables,
energy and momentum,

18:14.184 --> 18:16.984
are simply related to each
other in the fact that one is

18:16.979 --> 18:18.959
essentially the square of the
other.

18:18.960 --> 18:21.330
But if you have two variables,
one involving x and

18:21.330 --> 18:22.870
p,
at the same time,

18:22.867 --> 18:26.127
then you cannot measure them
simultaneously with either

18:26.134 --> 18:27.894
x or with p.

18:27.890 --> 18:28.650
Yes?

18:28.650 --> 18:32.270
Student: What is
a_0 again?

18:32.269 --> 18:34.129
Prof:
a_0.

18:34.130 --> 18:34.700
Gee, I don't know.

18:34.700 --> 18:36.180
The Bohr radius.

18:36.180 --> 18:39.260
I think it's ℏ squared
over me squared,

18:39.258 --> 18:40.828
but you have to check.

18:40.828 --> 18:45.248
I don't guarantee that answer,
but I think it's something like

18:45.251 --> 18:45.761
that.

18:45.759 --> 18:51.719
You can see if it's got at
least units of length.

18:51.720 --> 18:52.910
I'm too lazy to try that.

18:52.910 --> 18:56.910
I think it's something like
this.

18:56.910 --> 18:58.840
Anyway, suppose it is something
like this.

18:58.838 --> 19:01.848
My main point is,
it is a length made out of

19:01.853 --> 19:05.083
Planck's constant,
the mass of the electron and

19:05.076 --> 19:06.756
the electric charge.

19:06.759 --> 19:10.599
Together they make a length,
and that's the length that

19:10.602 --> 19:12.952
controls the size of the orbit.

19:12.950 --> 19:16.180
All right, so this was the
simpler problem.

19:16.180 --> 19:20.370
Now I want to take a more
complicated problem of what I

19:20.365 --> 19:22.455
call a particle in a box.

19:22.460 --> 19:23.730
So that is a potential.

19:23.730 --> 19:27.890
See, normally,
if you make a hole in the

19:27.892 --> 19:30.532
ground,
it may take this shape,

19:30.531 --> 19:33.951
and this is the location and
this is the height,

19:33.950 --> 19:37.550
but the height is just the
potential energy directly,

19:37.548 --> 19:39.978
because it's mg times the
height.

19:39.980 --> 19:42.830
So imagine a particle somewhere
here.

19:42.828 --> 19:46.558
In classical mechanics--by the
way, the rule for doing quantum

19:46.559 --> 19:48.579
mechanics is,
in classical theory,

19:48.577 --> 19:51.817
what's the potential
experienced by the particle?

19:51.819 --> 19:52.969
That's what you ask.

19:52.970 --> 19:55.720
And it is that potential that
enters this crazy

19:57.390 --> 19:59.840
So you need to know classical
mechanics before you can do

19:59.836 --> 20:00.706
quantum mechanics.

20:00.710 --> 20:03.210
You should know what potential
the particle experiences in

20:03.211 --> 20:05.451
classical theory,
and stick that into this

20:05.450 --> 20:08.680
equation to find the allowed
energies in quantum theory.

20:08.680 --> 20:12.000
So if it's a vibrating system,
there is some omega for

20:11.996 --> 20:12.806
vibrations.

20:12.808 --> 20:14.648
That ω
is √k/m,

20:14.646 --> 20:16.926
and you can get k from
that one.

20:16.930 --> 20:19.850
Or if it's a system like an
atom, where there's an electric

20:19.845 --> 20:22.255
charge between them,
electric force between them,

20:22.258 --> 20:24.418
you should know the electric
potential.

20:24.420 --> 20:26.430
So you should know the
potential in classical

20:26.426 --> 20:29.116
mechanics, and then you do the
quantum theory from there.

20:29.118 --> 20:32.098
So imagine a particle in this
potential in classical theory,

20:32.099 --> 20:34.469
and you ask yourself,
what does it do in quantum

20:34.473 --> 20:35.033
theory?

20:35.029 --> 20:39.879
Well, classical theory says,
let's plot the energy of the

20:39.880 --> 20:41.440
particle, right?

20:41.440 --> 20:46.010
It's some number that's not
supposed to change as the

20:46.013 --> 20:47.513
particle moves.

20:47.509 --> 20:50.999
That means here,
if the energy is kinetic

20:51.003 --> 20:55.813
potential, here the potential is
0, it's all kinetic.

20:55.808 --> 21:00.148
But as you start going near the
edge, there are two parts here.

21:00.150 --> 21:04.510
That's the potential and that's
the kinetic.

21:04.509 --> 21:09.129
And right here,
all the energy is potential,

21:09.128 --> 21:12.028
and E = potential.

21:12.028 --> 21:15.098
There is no K. What does
that mean?

21:15.098 --> 21:19.348
That means the particle comes
here, slows down,

21:19.348 --> 21:22.488
comes to a rest,
then goes back.

21:22.490 --> 21:26.580
It can never be found here if
it had this energy,

21:26.584 --> 21:29.574
because it's got more--let's
see.

21:29.568 --> 21:32.788
It's got more potential energy
than total energy.

21:32.788 --> 21:34.428
That means it's got negative
kinetic energy.

21:34.430 --> 21:36.070
That's not possible.

21:36.068 --> 21:39.328
So every classical particle
will execute bounded

21:39.327 --> 21:43.067
oscillations at any energy,
between what are called the

21:43.071 --> 21:44.391
turning points.

21:44.390 --> 21:48.670
The left turning point and the
right turning point is as far as

21:48.673 --> 21:50.613
you can go at that energy.

21:50.608 --> 21:52.778
If you've got more energy,
of course you can go further

21:52.778 --> 21:54.608
out,
and if you've got even more

21:54.605 --> 21:57.575
energy,
you can just roll up to the top

21:57.583 --> 21:59.493
of the hill and escape.

21:59.490 --> 22:01.480
That's classical mechanics.

22:01.480 --> 22:05.010
Now we want to ask ourselves,
in quantum mechanics,

22:05.011 --> 22:06.991
what does the particle do?

22:06.990 --> 22:10.790
That's our goal.

22:10.788 --> 22:14.078
So I'm going to take a
following simpler problem,

22:14.076 --> 22:16.606
because it's mathematically
easier.

22:16.608 --> 22:20.618
I'm going to take a potential
that looks like this.

22:20.618 --> 22:26.318
It is V_0 to
the right of L.

22:26.318 --> 22:29.738
It's V_0 to
the left of the origin,

22:29.737 --> 22:31.007
and it is 0 here.

22:31.009 --> 22:34.439
It's got very sharp walls,
because that makes it easy to

22:34.440 --> 22:35.750
solve the equation.

22:35.750 --> 22:38.640
In real life,
potentials are rounded off.

22:38.640 --> 22:43.570
My fake potential has got sharp
corners.

22:43.568 --> 22:46.668
I want to take a particle of
some energy and ask,

22:46.673 --> 22:49.783
what are the allowed energies
in this problem?

22:49.779 --> 22:51.919
What do the corresponding
functions look like?

22:51.920 --> 22:55.740
That's what we're asking.

22:55.740 --> 22:59.000
So the rule is to go back to
that equation,

22:58.996 --> 23:03.876
which now takes the following
form: d^(2)ψ/dx^(2).

23:03.880 --> 23:05.940
I'm going to stop writing the
subscript e.

23:05.940 --> 23:07.810
From now on,
you should understand,

23:07.813 --> 23:10.903
I'm trying to find states of
definite energy E.

23:10.900 --> 23:21.560
2m times E - V ψ
= 0.

23:25.635 --> 23:30.605
so that it looks like second
derivative all kinds of stuff.

23:30.609 --> 23:33.909
So how do we solve this problem?

23:33.910 --> 23:37.240
Well, V varies with
position, so there are three

23:37.244 --> 23:38.794
regions in the problem.

23:38.788 --> 23:42.318
There's region I inside the
well.

23:42.319 --> 23:45.149
Inside this well, V is 0.

23:45.150 --> 23:49.740
Outside this well in region II,
it = V_0,

23:49.740 --> 23:51.030
some constant.

23:51.029 --> 23:54.139
Likewise in region III to the
right of the well,

23:54.143 --> 23:56.333
it's also V_0.

23:56.328 --> 24:02.498
So we've got to solve this
equation in 3 different regions,

24:02.500 --> 24:05.160
middle, left and right.

24:05.160 --> 24:09.460
First let's solve it in the
middle because it's very easy.

24:09.460 --> 24:11.050
What is the middle defined as?

24:11.049 --> 24:18.519
Middle, region I, V = 0.

24:18.519 --> 24:27.899
So the equation is
d^(2)ψ/dx^(2) 2mE/ℏ^(2)

24:27.904 --> 24:29.974
is 0.

24:29.970 --> 24:36.860
So let me write it as
d^(2)ψ/dx^(2) k^(2)ψ

24:36.864 --> 24:38.194
= 0.

24:38.190 --> 24:43.450
Just like before.

24:43.450 --> 24:51.780
This k is the same as
before, k =

24:51.782 --> 24:58.532
√2mE/ℏ,
in region I.

24:58.529 --> 25:00.219
So what's the answer to that
one?

25:00.220 --> 25:05.040
Again, it's exponentials,
but I'm purposely going to

25:05.040 --> 25:10.340
write it in region 1 as follows:
as Acoskx

25:10.336 --> 25:12.696
Bsinkx.

25:12.700 --> 25:16.210
I hope you all know that this
has exponentials,

25:16.210 --> 25:18.310
our old friends the
trigonometric functions,

25:18.308 --> 25:21.228
are also the solution to this
equation,

25:21.230 --> 25:25.520
because the second derivative
of cosine is roughly - the

25:25.518 --> 25:26.218
cosine.

25:26.220 --> 25:28.140
The second derivative of sine
will also give you

25:28.144 --> 25:29.664
−k^(2) times the
sine.

25:29.660 --> 25:32.160
So rather than using
exponentials,

25:32.164 --> 25:35.964
I'm going to use sine and
cosine, because I have an

25:35.961 --> 25:37.481
ultimate motive.

25:37.480 --> 25:41.590
You realize that exponentials
and sines and cosines are all

25:41.594 --> 25:44.294
connected by Euler's great
formula,

25:44.288 --> 25:46.678
so if I want,
I can take e^(ikx) and

25:46.679 --> 25:48.499
write it as cos i sine.

25:48.500 --> 25:51.060
e^(-ikx),
write that as cos - i

25:51.061 --> 25:51.471
sine.

25:51.470 --> 25:53.670
Combine the cosines,
and call the coefficient

25:53.670 --> 25:54.220
A.

25:54.220 --> 25:56.440
Combine the coefficients for
sine and call it B,

25:56.438 --> 25:57.258
and get this form.

25:57.259 --> 25:59.119
There are two alternative ways.

25:59.118 --> 26:02.738
You just write this,
you will see why I picked that.

26:02.740 --> 26:04.320
That's the answer in region I.

26:04.319 --> 26:05.919
We're all set.

26:05.920 --> 26:09.680
Right now, A is
arbitrary, B is arbitrary

26:09.680 --> 26:11.630
and k is arbitrary.

26:11.630 --> 26:13.420
There is no restriction at all
on any of them.

26:13.420 --> 26:14.740
Do you understand that?

26:14.740 --> 26:18.360
Because if you take this
function, it satisfies that

26:18.357 --> 26:19.917
equation absolutely.

26:19.920 --> 26:27.790
So the free parameters are
A, B and k,

26:27.794 --> 26:30.514
are all arbitrary.

26:30.509 --> 26:32.979
That means it can have any
k.

26:32.980 --> 26:37.150
But the energy is
ℏ^(2)k^(2)/2m,

26:37.150 --> 26:40.070
so it can have any energy.

26:40.068 --> 26:42.598
But we're not done yet,
because we only looked at

26:42.599 --> 26:43.179
region I.

26:43.180 --> 26:46.290
Have to look at region II and
region III, and then you will

26:46.291 --> 26:47.581
find out what happens.

26:47.578 --> 26:51.748
So let's go to region II So in
region II,

26:51.750 --> 26:54.780
the equation obeyed by the
ψ,

26:54.779 --> 27:08.159
is ψ (let me call it region
II) over dx squared =

27:08.163 --> 27:21.323
-2m/ℏ^(2) times
V_0 - E ψ.

27:21.318 --> 27:24.428
I just put it on the other side
now.

27:24.430 --> 27:25.710
Now this is all a constant.

27:25.710 --> 27:28.490
That's why we picked the
potential, which is piecewise

27:28.487 --> 27:30.267
constant, so it's some constant.

27:30.269 --> 27:32.569
You're asking yourself,
what ψ has the property

27:32.573 --> 27:35.433
that when I take 2 derivatives,
it looks like this number times

27:35.429 --> 27:35.889
ψ?

27:35.890 --> 27:42.960
So let me call this whole
combination kappa squared.

27:42.960 --> 27:48.650
Then the answer is ψ
of 2 = C times

27:48.652 --> 27:55.992
e^(κx) D
times e^(-κx).

27:55.990 --> 27:56.470
Why?

27:56.470 --> 27:59.900
Because two derivatives of this
guy will give me kappa squared

27:59.896 --> 28:02.296
times itself;
likewise here.

28:02.298 --> 28:14.708
And kappa has been cooked up so
that kappa squared is what I

28:14.707 --> 28:16.177
want.

28:16.180 --> 28:23.180
Yes, now so κ
= 2m/ℏ^(2) times

28:23.176 --> 28:29.216
V_0 - E under
root.

28:29.220 --> 28:33.550
So these are now real
exponentials and not imaginary

28:33.547 --> 28:34.817
exponentials.

28:34.818 --> 28:38.038
The reason that used to be
sines and cosines,

28:38.042 --> 28:41.712
or imaginary exponentials,
is that inside the well,

28:41.705 --> 28:44.485
E - V is a positive
number.

28:44.490 --> 28:47.420
Outside the well,
I'm assuming that E - V

28:47.420 --> 28:49.260
is a negative number.

28:49.259 --> 28:52.419
In other words,
I'm going to consider the well

28:52.420 --> 28:55.020
to be, in the end,
infinitely deep.

28:55.019 --> 28:58.579
And the energy of the particle,
whatever it is,

28:58.575 --> 29:00.425
is some finite number.

29:00.430 --> 29:02.730
I want to send the walls to
infinity.

29:02.730 --> 29:04.860
That's called particle in a box.

29:04.858 --> 29:07.378
So I'm taking
V_0 finite but

29:07.381 --> 29:08.471
very, very large.

29:08.470 --> 29:10.850
Then it's very clear that
V_0 - E is a

29:10.846 --> 29:11.636
positive number.

29:11.640 --> 29:14.270
This kappa is what I have here.

29:14.269 --> 29:16.789
Now look at these two functions.

29:16.788 --> 29:21.228
This function grows
exponentially as you go to the

29:21.228 --> 29:23.128
right, to infinity.

29:23.130 --> 29:26.600
That means the wave function
describes a particle that would

29:26.596 --> 29:29.706
rather be at infinity than
anywhere near this well.

29:29.710 --> 29:32.200
So it's not the problem we're
describing where the particle we

29:32.200 --> 29:33.670
expect is somewhere near the
well.

29:33.670 --> 29:36.010
Furthermore,
the square of ψ

29:36.007 --> 29:39.237
can never be normalized to 1,
because an exponentially

29:39.240 --> 29:40.870
growing function,
the area under that is

29:40.872 --> 29:41.812
exponentially large.

29:41.809 --> 29:44.129
There's no way to fix that.

29:44.130 --> 29:47.520
So on physical grounds,
we drop the solution,

29:47.518 --> 29:51.828
because we don't have to pick
C to be non-zero.

29:51.828 --> 29:53.968
We have the freedom,
and the physical condition

29:53.969 --> 29:56.619
tells you, no one wants a
function growing at infinity.

29:56.618 --> 29:58.378
We want functions falling at
infinity.

29:58.380 --> 30:01.840
This is the only function that
falls at infinity.

30:01.838 --> 30:04.768
And how does it fall at
infinity, how does it fall off?

30:04.769 --> 30:10.599
It looks like e to the
-√2m/ℏ^(2) times

30:10.602 --> 30:13.132
V_0(x).

30:13.130 --> 30:17.860
I'm ignoring e compared to this
very large V_0.

30:17.858 --> 30:20.728
I'm taking V_0
much, much bigger than any

30:20.730 --> 30:21.940
energy I'm looking at.

30:21.940 --> 30:25.220
So it's falling exponentially,
and the number on top is like

30:25.218 --> 30:27.328
square root of
V_0.

30:27.328 --> 30:29.528
So what do you think happens if
V_0 becomes

30:29.530 --> 30:30.240
very, very large?

30:30.240 --> 30:34.490
Well, if you plot the function
that's exponentially falling,

30:34.488 --> 30:35.998
it falls like that.

30:36.000 --> 30:39.700
But as you increase
V_0,

30:39.703 --> 30:42.763
it falls more and more quickly.

30:42.759 --> 30:46.109
You follow that?

30:46.108 --> 30:48.338
And eventually,
even V_0 goes

30:48.336 --> 30:50.146
to infinity;
it just doesn't have a life

30:50.154 --> 30:50.534
outside.

30:50.529 --> 30:53.109
It just vanishes immediately.

30:53.108 --> 30:55.738
Whatever you start with goes to
0 immediately.

30:55.740 --> 31:03.240
That basically means
ψ_II(x) is

31:03.236 --> 31:10.876
just 0 in region III,
if V_0 goes to

31:10.878 --> 31:12.788
infinity.

31:12.789 --> 31:25.279


31:25.278 --> 31:28.308
So let me emphasize what we are
doing here.

31:28.308 --> 31:32.898
We have a particle in some kind
of well, which is very,

31:32.901 --> 31:37.581
very tall, not yet infinite,
and you give it some energy

31:37.577 --> 31:40.297
E that cannot change.

31:40.298 --> 31:43.398
In this region,
in classical mechanics,

31:43.397 --> 31:46.577
that would be the entire
kinetic energy,

31:46.577 --> 31:49.347
because potential energy is 0.

31:49.349 --> 31:52.029
What about here?

31:52.029 --> 31:54.839
This is the total energy
available to you.

31:54.838 --> 31:57.578
Potential energy is all of
that, so kinetic energy has to

31:57.583 --> 31:58.713
be minus some number.

31:58.710 --> 32:01.500
That can never be.

32:01.500 --> 32:07.260
That means that particle will
never be found in the region

32:07.259 --> 32:09.279
outside this well.

32:09.279 --> 32:10.679
Do you understand?

32:10.680 --> 32:14.530
As long as the energy is less
than it takes to go to the top

32:14.525 --> 32:18.305
of the barrier and climb out,
a classical particle--imagine

32:18.306 --> 32:19.606
rounding it off.

32:19.608 --> 32:21.068
You can come rolling with any
energy.

32:21.068 --> 32:24.168
If you don't have enough energy
to go over the top,

32:24.173 --> 32:26.473
you cannot be found in this
region.

32:26.470 --> 32:30.260
But in quantum theory,
as long as the well is not

32:30.258 --> 32:33.968
infinitely large,
there's an exponentially small

32:33.968 --> 32:36.098
decaying wave function.

32:36.098 --> 32:39.638
So a classical particle will
turn around at the classical

32:39.635 --> 32:40.705
turning points.

32:40.710 --> 32:44.500
A quantum particle will go
slightly outside the forbidden

32:44.501 --> 32:48.431
region, into the forbidden
region in both directions of the

32:48.430 --> 32:49.040
well.

32:49.038 --> 32:51.888
That is a purely quantum
phenomenon.

32:51.890 --> 32:55.020
It is forbidden,
meaning it's discouraged from

32:55.019 --> 32:58.429
going there, but it's not
absolutely forbidden.

32:58.430 --> 33:01.020
That's called going over
the--well,

33:01.019 --> 33:02.959
it's going into the classical
forbidden region,

33:02.960 --> 33:05.640
but as the barrier gets higher
and higher,

33:05.640 --> 33:07.870
the excursion into the
forbidden region becomes less

33:07.874 --> 33:10.824
and less,
because the function falls very

33:10.817 --> 33:11.517
rapidly.

33:11.519 --> 33:13.839
And I'm considering,
for simplicity,

33:13.839 --> 33:17.749
a barrier which is infinitely
tall, in which case it doesn't

33:17.751 --> 33:20.671
go there at all,
even in quantum theory.

33:20.670 --> 33:24.830
Therefore, by a similar
argument, ψ

33:24.833 --> 33:28.453
is 0 here and ψ
is 0 here,

33:28.450 --> 33:31.890
and in region I,
I've shown you,

33:31.888 --> 33:37.768
ψ = Acoskx
Bsinkx,

33:37.769 --> 33:42.829
where the energy is related to
k by

33:42.826 --> 33:45.906
ℏ^(2)k^(2)/2m.

33:45.910 --> 33:49.300
So first when we did only
region I, you get the impression

33:49.297 --> 33:52.737
that A and B and
k can be whatever you

33:52.744 --> 33:53.284
like.

33:53.279 --> 33:55.869
But now we have an extra
restriction.

33:55.868 --> 34:01.178
The restriction is that if you
draw the function in this well,

34:01.180 --> 34:04.520
it can do whatever it wants,
but it must vanish at the two

34:04.520 --> 34:07.120
ends,
because it's 0 here and it's 0

34:07.123 --> 34:07.553
here.

34:07.548 --> 34:11.928
Just by continuity of ψ,
every ψ that's allowed must

34:11.925 --> 34:13.775
vanish at the two ends.

34:13.780 --> 34:17.650
That's a new restriction,
because the function that I

34:17.648 --> 34:21.518
have here in general will not
vanish at the two ends,

34:21.518 --> 34:24.718
but I'm going to demand that it
vanish.

34:24.719 --> 34:27.359
That will tell me something
about the allowed values of

34:27.358 --> 34:29.018
A, B and k.

34:29.019 --> 34:30.509
You will see what it is.

34:30.510 --> 34:34.690
First I demand that ψ
should vanish on the left end

34:34.688 --> 34:38.478
of the wall, which is the wall
at x = 0.

34:38.480 --> 34:40.430
This is the wall at x =
L.

34:40.429 --> 34:43.389
ψ of 0 should be 0.

34:43.389 --> 34:45.569
I take the function I have.

34:45.570 --> 34:47.560
When this is 0,
x is 0,

34:47.556 --> 34:48.856
that guy is gone.

34:48.860 --> 34:54.660
I get A times cosine 0
which is just A.

34:54.659 --> 34:56.429
But that has to vanish.

34:56.429 --> 34:58.829
That means the coefficient
A has to vanish.

34:58.829 --> 35:01.429
It is not allowed.

35:01.429 --> 35:04.199
So even though it looked like,
before we went to the other

35:04.197 --> 35:06.817
regions, you can have any
A and any B.

35:06.820 --> 35:08.450
If you're only in the middle
region, that's correct.

35:08.449 --> 35:12.119
But if you want to connect your
answer in the middle region to a

35:12.115 --> 35:15.055
0 on either side,
then the coefficient has to be

35:15.057 --> 35:18.017
such that the ψ
you have in the middle vanishes

35:18.023 --> 35:19.153
at the left end.

35:19.150 --> 35:21.230
So A is gone.

35:21.230 --> 35:24.820
So I'm now left with only the
following thing:

35:24.818 --> 35:28.088
ψ(x) =
Bsinkx.

35:28.090 --> 35:32.620
We did not have to kill
B because that x =

35:32.621 --> 35:35.591
0 sine vanished,
so it is allowed.

35:35.590 --> 35:40.680
But now I demand furthermore it
should vanish at x = L.

35:40.679 --> 35:44.089
Well, at x = L,
what do I find?

35:44.090 --> 35:50.070
I want 0 to =
BsinkL.

35:50.070 --> 35:52.350
That should be 0.

35:52.349 --> 35:55.279
Now one simple way to get there
is to kill B, say

35:55.277 --> 35:57.547
B = 0, but then you've got
no solution.

35:57.550 --> 35:59.150
You killed A and you
killed B.

35:59.150 --> 36:02.300
It's certainly true that ψ
= 0 is the solution to this

36:02.295 --> 36:04.775
equation, but it doesn't
describe anything.

36:04.780 --> 36:09.220
So you don't want to kill
B, so you want to blame

36:09.224 --> 36:11.734
the 0 on the sine,
and the sine,

36:11.728 --> 36:14.958
as you know,
vanishes for any multiple of

36:14.960 --> 36:15.850
Π.

36:15.849 --> 36:19.239
You can have 0 times Π,
2Π, 3Π,

36:19.239 --> 36:20.629
4Π, -6Π.

36:20.630 --> 36:23.780
They all give you 0 for sine.

36:23.780 --> 36:28.720
Therefore k is quantized to be
one of these numbers.

36:28.719 --> 36:32.759
Therefore k has got only values
nΠ/L.

36:32.760 --> 36:35.200
I'm going to call it
k_n.

36:35.199 --> 36:38.809
And the energy,
which is really

36:38.809 --> 36:43.139
ℏ^(2)k^(2)/2m L
squared--

36:43.139 --> 36:52.929
sorry, k^(2)/2m, becomes
now ℏ^(2)Π^(2)/2m

36:52.931 --> 37:03.951
L^(2) times n^(2).
And ψ looks like B--

37:03.949 --> 37:13.139
ψ's of n looks like
Bsin(nΠx/L).

37:13.143 --> 37:16.263
You guys see that?

37:16.260 --> 37:18.620
These are the allowed function.

37:18.619 --> 37:21.049
The sine function is very nice.

37:21.050 --> 37:24.550
It vanishes on the left end,
because when x is 0,

37:24.554 --> 37:25.324
it's gone.

37:25.320 --> 37:28.110
When x=L,
the argument of the sine is a

37:28.106 --> 37:30.086
multiple of Π,
it vanishes.

37:30.090 --> 37:33.550
So it looks like this when
n = 1.

37:33.550 --> 37:37.020
It looks like that when
n=2, and so on.

37:37.018 --> 37:41.428
So it's got more and more
wiggles in the region in

37:41.429 --> 37:47.009
between, but they are designed
so that they vanish precisely at

37:47.010 --> 37:48.270
the edges.

37:48.268 --> 37:51.808
So it's like it's exactly the
case of a violin string that's

37:51.809 --> 37:54.459
clamped to two ends,
which has to vanish at the two

37:54.461 --> 37:56.181
ends,
because you don't let it move

37:56.184 --> 37:56.534
there.

37:56.530 --> 37:58.850
In between, it can do this and
it can do that.

37:58.849 --> 38:04.219
It can have more and more
wiggles.

38:04.219 --> 38:07.539
So what you find is,
of the three things that we

38:07.543 --> 38:10.373
thought were arbitrary,
A is gone,

38:10.373 --> 38:12.783
and k is not arbitrary.

38:12.780 --> 38:17.630
k is again quantized to
be nΠ/L,

38:17.625 --> 38:22.765
so that the function does
n half oscillations

38:22.773 --> 38:26.613
between one end and the other
end.

38:26.610 --> 38:28.780
And therefore there's a
corresponding quantization of

38:28.775 --> 38:29.145
energy.

38:29.150 --> 38:29.960
Yes?

38:29.960 --> 38:30.960
Student: What does that
tell you?

38:30.960 --> 38:32.490
If you square
ψ_n,

38:32.489 --> 38:35.249
does that give the probability
of finding it in each one,

38:35.253 --> 38:36.343
the energy levels?

38:36.340 --> 38:36.760
Prof: No.

38:36.760 --> 38:38.310
Let's talk about that now.

38:38.309 --> 38:40.979
So let's write down the
function.

38:40.980 --> 38:43.660
The ψ sub energy that we
were looking for,

38:43.659 --> 38:46.989
we have finally found out,
so let's have a picture in our

38:46.994 --> 38:47.534
mind.

38:47.530 --> 38:48.510
Here is the well.

38:48.510 --> 38:50.900
It's going all the way to
infinity.

38:50.900 --> 38:56.720
There is the state n=1,
and the function looks like

38:56.724 --> 38:57.444
this.

38:57.440 --> 39:02.110
Then there is the state--let me
draw it right there,

39:02.110 --> 39:06.050
it looks like that,
n=2 and so on.

39:06.050 --> 39:12.400
In the ψ_n (let
me label the energy by n)

39:12.403 --> 39:18.553
equals some constant B
times sin(nΠx/L).

39:18.550 --> 39:22.510
That means if you want a
particle in the well to have a

39:22.510 --> 39:25.660
definite energy,
the wave function must look

39:25.664 --> 39:28.384
like one of these sine
functions.

39:28.380 --> 39:30.920
It must look like this or it
must look like that,

39:30.922 --> 39:32.302
or the higher harmonics.

39:32.300 --> 39:36.170
They are the only allowed
functions for states of definite

39:36.168 --> 39:36.778
energy.

39:36.780 --> 39:40.040
Now before I answer any further
questions,

39:40.039 --> 39:45.619
the allowed values of n
here you might think are 0,

39:45.619 --> 39:50.069
or - 1, or - 2 etc.,
but I claim they're just 1,

39:50.070 --> 39:51.550
2,3 and so on.

39:51.550 --> 39:55.680
You should think about why.

39:55.679 --> 40:01.319
What's wrong with n=0?

40:01.320 --> 40:07.510
Is that a possible value?

40:07.510 --> 40:09.850
Can anybody tells me what
happens to this function

40:09.851 --> 40:10.331
if--yes?

40:10.329 --> 40:11.869
Student: It identically
vanishes

40:11.869 --> 40:15.079
Prof: If n=0,
the function identically

40:15.077 --> 40:16.207
vanishes, right?

40:16.210 --> 40:19.020
Because sine of 0 times
Πx/L is identical.

40:19.019 --> 40:19.909
That's not what you want.

40:19.909 --> 40:21.709
You want a nontrivial solution.

40:21.710 --> 40:24.720
So we drop this guy.

40:24.719 --> 40:28.499
But how about n=-1?

40:28.500 --> 40:35.720
Why don't I keep that?

40:35.719 --> 40:37.009
Yes?

40:37.010 --> 40:38.710
Oh, I thought I imagined a hand
going up.

40:38.710 --> 40:39.500
Yes?

40:39.500 --> 40:42.830
Student: It's the same
thing as -B sine of the

40:42.829 --> 40:43.789
positive value.

40:43.789 --> 40:45.429
Prof: That's correct.

40:45.429 --> 40:50.389
If you put n=-1,
the function ψ

40:50.389 --> 40:55.469
of -1 is B
sin(-Πx/L),

40:55.469 --> 41:01.799
but that = -1 times B
sin(Πx/L).

41:01.800 --> 41:04.920
This guy is just
ψ_1.

41:04.920 --> 41:09.900
So it's just a multiple of the
ψ_1 function,

41:09.902 --> 41:14.012
because -1 times the state is
the same state.

41:14.010 --> 41:17.190
You don't get a new function;
you just get a multiple of the

41:17.188 --> 41:18.948
old function,
and in quantum mechanics,

41:18.947 --> 41:20.057
they're all the same.

41:20.059 --> 41:20.919
But be very careful.

41:20.920 --> 41:25.730
When I did a particle on a
ring, I did e^(ipx/ℏ),

41:25.728 --> 41:29.948
where I said p =
2Πℏ/L times any

41:29.947 --> 41:31.717
integer n.

41:31.719 --> 41:38.929
There I let n = 0, - 1, - 2 etc.

41:38.929 --> 41:41.549
Why did I allow that?

41:41.550 --> 41:43.910
First of all,
in that case,

41:43.913 --> 41:46.643
e to the i times 0 is not 0.

41:46.639 --> 41:49.429
It's in fact constant.

41:49.429 --> 41:53.009
So for a particle going in a
ring, the 0 momentum state is

41:53.014 --> 41:54.214
actually allowed.

41:54.210 --> 41:58.680
And 1 and -1 are not the same,
because e^(ipx) and

41:58.677 --> 42:01.147
e^(-ipx),
forget the ℏ,

42:01.150 --> 42:04.660
they are not proportional to
each other.

42:04.659 --> 42:05.959
You understand that?

42:05.960 --> 42:09.650
You can never write
e^(ipx) as a multiple of

42:09.650 --> 42:10.980
e^(-ipx).

42:10.980 --> 42:12.360
You see that?

42:12.360 --> 42:16.620
If there were such a multiple,
let me call the multiple as

42:16.615 --> 42:20.425
N, then N is
really e^(2ipx) by

42:20.425 --> 42:22.735
bringing it the other side.

42:22.739 --> 42:25.989
That's certainly not a constant.

42:25.989 --> 42:29.319
So e^(ipx) can never be
converted to e^(-ipx) by

42:29.315 --> 42:31.925
multiplying by any number,
so it's an independent

42:31.931 --> 42:32.641
solution.

42:32.639 --> 42:36.439
On the other hand,
sinx and sin(-x)

42:36.440 --> 42:40.640
are related to each other by a
single factor of -1.

42:40.639 --> 42:44.749
So that's the reason why in
that case of a particle going in

42:44.751 --> 42:46.711
a ring,
the solutions are labeled by

42:46.706 --> 42:48.976
all integers,
positive, negative and 0,

42:48.981 --> 42:53.301
but a particle in a box,
we label only by positive

42:53.300 --> 42:54.360
integers.

42:54.360 --> 42:58.050
Negative integers give you the
same function and 0 kills the

42:58.045 --> 43:00.165
function, so that's not allowed.

43:00.170 --> 43:02.950
Okay, so what about B?

43:02.949 --> 43:05.049
What shall I do with B?

43:05.050 --> 43:11.910
What equation's going to
determine B?

43:11.909 --> 43:13.359
Yes, any idea?

43:13.360 --> 43:14.030
Yes?

43:14.030 --> 43:15.870
Student: Normalize right?

43:15.869 --> 43:16.369
Prof: Right.

43:16.369 --> 43:18.519
So first of all,
the equation will never tell

43:18.516 --> 43:19.636
you what B is.

43:19.639 --> 43:21.709
That's what I want you to
understand.

43:21.710 --> 43:25.800
The equation that I've written
here for the energy functions on

43:25.802 --> 43:27.712
the top,
if size of B is a

43:27.710 --> 43:30.370
solution, you can multiply both
sides by any number,

43:30.369 --> 43:32.099
like 15.

43:32.099 --> 43:34.759
Then 15 times ψ
will also be a solution.

43:34.760 --> 43:38.140
That's a property of a linear
equation that the answer is not

43:38.141 --> 43:39.551
given in overall scale.

43:39.550 --> 43:41.260
Give me one answer;
I rescale it,

43:41.257 --> 43:42.657
it's still an answer.

43:42.659 --> 43:45.999
So linear equation will never
tell me the overall size of the

43:45.998 --> 43:46.608
function.

43:46.610 --> 43:55.010
That is picked by the extra
convenient restriction that

43:55.010 --> 43:58.900
ψ^(2)dx = 1.

43:58.900 --> 44:04.550
That will determine B
and I'm claiming the answer is

44:04.550 --> 44:10.200
the √(2/L) is what
you want B to be.

44:10.199 --> 44:11.319
The answer is again simple.

44:11.320 --> 44:14.950
The average value of sine
squared over 1 half period is 1

44:14.945 --> 44:15.395
half.

44:15.400 --> 44:17.100
If you square or integrate or
whatever,

44:17.099 --> 44:19.479
you'll get L/2,
so B^(2) times

44:19.480 --> 44:23.300
L/2 should be 1,
and B should therefore

44:23.300 --> 44:25.600
be the √(2/L).

44:25.599 --> 44:29.079
So let me write down now the
final result of this problem,

44:29.083 --> 44:31.653
which I think is very,
very instructive.

44:31.650 --> 44:34.950
If you can follow this much,
you've really got 80 percent of

44:34.954 --> 44:37.424
the quantum mechanics I want to
teach you.

44:37.420 --> 44:40.330
If you're comfortable with this
example,

44:40.329 --> 44:43.429
because we're trying to find
out, in quantum theory,

44:43.429 --> 44:48.259
you've got a particle in a box,
which is like a hole in the

44:48.260 --> 44:50.550
floor,
which is so deep that it's

44:50.545 --> 44:51.545
infinitely deep.

44:51.550 --> 44:53.230
But even if it's not infinite
but very deep,

44:53.233 --> 44:54.293
the answer's good enough.

44:54.289 --> 44:56.739
You want to know what it will
do.

44:56.739 --> 45:00.429
In classical mechanics,
what will it do?

45:00.429 --> 45:04.979
If you're a particle in a box,
what's your lowest energy

45:04.978 --> 45:05.638
state?

45:05.639 --> 45:06.429
Yes?

45:06.429 --> 45:07.669
Student: Floor of the
box.

45:07.670 --> 45:10.620
Prof: If you're in
prison with infinitely tall

45:10.617 --> 45:13.507
walls, you just sit on the floor
and mope, right?

45:13.510 --> 45:15.640
That's the lowest energy state.

45:15.639 --> 45:17.669
But that's not allowed in
quantum mechanics,

45:17.670 --> 45:19.510
because you have a definite
x,

45:19.510 --> 45:21.610
which is wherever you're
sitting, a definite p,

45:21.610 --> 45:22.650
you're not moving.

45:22.650 --> 45:25.250
Such a state is not allowed.

45:25.250 --> 45:29.320
So in quantum mechanics,
a particle cannot simply sit

45:29.322 --> 45:29.952
still.

45:29.949 --> 45:34.459
Therefore, it will necessarily
have an uncertainty in position,

45:34.458 --> 45:37.948
and it will try to spread
itself over the box.

45:37.949 --> 45:42.209
And the typical uncertainty is
of order.

45:42.210 --> 45:43.560
If I say, "Where is this
particle?"

45:43.559 --> 45:44.569
I don't know anything.

45:44.570 --> 45:46.560
It's got to be somewhere in the
box.

45:46.559 --> 45:48.609
That's the estimate for
Δx.

45:48.610 --> 45:54.250
So Δp has got to
be bigger than ℏ/L.

45:54.250 --> 45:56.020
Forget all the 2Π's.

45:56.019 --> 45:57.399
So it cannot have 0 momentum.

45:57.400 --> 46:00.450
It's momentum has to be
fluctuating with that range.

46:00.449 --> 46:03.229
And the kinetic energy,
which is in fact the total

46:03.231 --> 46:08.581
energy inside the box,
which is p^(2)/2m,

46:08.577 --> 46:15.647
the size of that is
ℏ^(2)/2m L^(2).

46:15.650 --> 46:18.540
So that is your estimate for
what should be the energy of a

46:18.539 --> 46:21.429
particle in a box of size
L. So what happens is,

46:21.429 --> 46:25.169
the box of size L forces
it to have a spread in momentum,

46:25.170 --> 46:28.190
because you've squeezed it to a
position known to within

46:28.186 --> 46:28.786
L.

46:28.789 --> 46:31.179
The spread in momentum
translates into spread in

46:31.182 --> 46:33.342
kinetic energy,
and if I take the spread in

46:33.344 --> 46:36.024
kinetic energy to be a rough
estimate of the lowest kinetic

46:36.021 --> 46:39.091
energy I could have,
you get a number which except

46:39.085 --> 46:42.385
for factors of Π
is in fact the right answer.

46:42.389 --> 46:46.079
So people always estimate
energies of particles by doing

46:46.076 --> 46:47.346
this calculation.

46:47.349 --> 46:49.089
For example,
take a nucleus,

46:49.092 --> 46:51.612
which is 10^(-13) centimeters
across.

46:51.610 --> 46:52.920
I put a proton there.

46:52.920 --> 46:55.850
What's the minimum kinetic
energy of that proton?

46:55.849 --> 46:58.839
You find by the following
method: Δx is

46:58.838 --> 47:01.768
10^(-13), because it's somewhere
in the nucleus.

47:01.768 --> 47:03.858
Δp is that,
and you do that,

47:03.858 --> 47:05.268
you will get some energy.

47:05.268 --> 47:09.398
That will be the typical energy
of a particle inside the

47:09.398 --> 47:10.148
nucleus.

47:10.150 --> 47:13.130
So particles inside the nucleus
cannot sit still.

47:13.130 --> 47:15.320
More you cram them,
it's like children.

47:15.320 --> 47:18.350
You put them in a tighter room,
they're jiggling more and more.

47:18.349 --> 47:20.759
That's called zero point motion.

47:20.760 --> 47:23.300
You cannot just nail them in
x without letting them

47:23.302 --> 47:24.242
spread in p.

47:24.239 --> 47:27.869
That means you cannot get the
lowest energy So lowest energy

47:27.867 --> 47:31.737
is some number you can estimate
by the uncertainty principle.

47:31.739 --> 47:36.249
So the next question was,
let's take one of these

47:36.251 --> 47:40.581
functions, say ψ's of 3,
the square root of

47:40.577 --> 47:44.147
(2/L)sin(
3Πx/L).

47:44.150 --> 47:49.300
If you plot that guy,
that function will look like

47:49.295 --> 47:50.025
this.

47:50.030 --> 47:51.790
Now what was your question?

47:51.789 --> 47:55.209
If I measure a particle's
position, what's the

47:55.206 --> 47:56.266
probability?

47:56.268 --> 47:57.728
You've just got to square this
function.

47:57.730 --> 47:59.430
It will look like this.

47:59.429 --> 48:03.009
That's the probability.

48:03.010 --> 48:04.890
If you measure the energy,
what will you get?

48:04.889 --> 48:11.229
You will get exactly one energy
corresponding to n=3.

48:11.230 --> 48:14.350
It's in a state of definite
energy, but not definite

48:14.347 --> 48:15.017
position.

48:15.018 --> 48:18.128
Position has got some ups and
downs.

48:18.130 --> 48:25.700
The probability for x looks
like that.

48:25.699 --> 48:26.749
So everyone with me?

48:26.750 --> 48:30.620
This is like knowing the
spectrum of this artificial

48:30.617 --> 48:31.147
atom.

48:31.150 --> 48:34.210
In a real atom,
the electron is forced to stay

48:34.210 --> 48:36.730
near the nucleus,
because the 1 over r

48:36.733 --> 48:39.103
potential,
if you plot it as a function of

48:39.101 --> 48:41.411
r,
looks like a well.

48:41.409 --> 48:44.749
And the particle is somewhere
here.

48:44.750 --> 48:50.770
In this problem,
the particle is confined by the

48:50.771 --> 48:51.541
box.

48:51.539 --> 48:54.499
Whenever you have a confining
thing that keeps the particle in

48:54.503 --> 48:56.743
a certain region,
the energy gets quantized.

48:56.739 --> 49:00.399
And we found out in our first
example the quantized energy

49:00.396 --> 49:02.446
levels of a particle in a box.

49:02.449 --> 49:05.869
And you can probe the particle
by shining light and the energy

49:05.871 --> 49:07.611
levels are plotted like this.

49:07.610 --> 49:10.770
This is n=1,
this is n=2,

49:10.769 --> 49:12.599
which is n=3.

49:12.599 --> 49:15.609
Zero is here.

49:15.610 --> 49:18.110
The zero state is not allowed.

49:18.110 --> 49:20.680
You've got to have n=1
and 4 times that will be

49:20.677 --> 49:23.097
n=2, and 9 times that
will be n=3.

49:23.099 --> 49:25.219
Energy grows like n^(2).

49:25.219 --> 49:28.259
And then you can come from
there to there emitting a

49:28.262 --> 49:29.792
photon,
or you can go from there to

49:29.786 --> 49:32.196
there emitting a photon,
or you can go from here to here

49:32.199 --> 49:33.179
absorbing a photon.

49:33.179 --> 49:35.349
You can do all kinds of things.

49:35.349 --> 49:38.119
That's how if you want you can
test quantum mechanics.

49:38.119 --> 49:41.689
Put a known particle inside the
well and shine light and see

49:41.688 --> 49:43.138
what light it absorbs.

49:43.139 --> 49:43.859
Yes.

49:43.860 --> 49:47.180
Student: So the ψ
of 3, is that the ψ

49:47.175 --> 49:49.445
of the energy function
_________?

49:49.449 --> 49:50.949
Prof: Yes.

49:50.949 --> 49:51.829
Exactly right.

49:51.829 --> 49:55.259
So what I should really write,
he's absolutely right,

49:55.260 --> 50:01.140
I should write E =
ℏ^(2)Π^(2)/2m L

50:01.135 --> 50:04.925
squared 3 squared,
right?

50:04.929 --> 50:06.399
The third energy level.

50:06.400 --> 50:09.650
So I'm short circuiting all of
that and just calling it

50:09.653 --> 50:10.863
ψ_3.

50:10.860 --> 50:13.950
Normally, if you put a label 3,
it's not clear what we're

50:13.954 --> 50:14.844
talking about.

50:14.840 --> 50:16.940
Maybe it's the position
x=3.

50:16.940 --> 50:18.880
Maybe it's the momentum
p=3.

50:18.880 --> 50:21.130
But in this lecture,
since we're talking only about

50:21.130 --> 50:22.940
energy,
I thought I will just call it

50:22.940 --> 50:25.590
ψ_3,
but at least I should call it

50:25.590 --> 50:26.770
ψ_E3.

50:26.768 --> 50:33.918
Student:
>

50:33.920 --> 50:35.350
Prof: Yeah,
I just did that.

50:35.349 --> 50:41.019
Student:
>

50:41.018 --> 50:43.988
Prof: You can take the
complex conjugate,

50:43.989 --> 50:46.959
but for a real function,
you can just square it,

50:46.958 --> 50:47.588
right?

50:47.590 --> 50:49.030
You understand?

50:49.030 --> 50:54.260
If ψ is real,
ψ*is the same as ψ.

50:54.260 --> 50:58.370
So ψ*is also
(2/L)sin(3

50:58.367 --> 51:00.607
Πx/L).

51:00.610 --> 51:02.710
So ψ*ψ is just a square.

51:02.710 --> 51:05.970
Student: On the first
board, you have

51:05.969 --> 51:09.909
A_E = _________
ψ*E ψ

51:09.909 --> 51:10.819
_______.

51:10.820 --> 51:14.800
I'm wondering why there's no
_______________.

51:14.800 --> 51:15.320
Prof: No.

51:15.320 --> 51:18.150
I think I know what the problem
is you're having now.

51:18.150 --> 51:20.900
That formula is not relevant
now.

51:20.900 --> 51:25.720
I'll tell you when that formula
will come into play.

51:25.719 --> 51:29.109
It will come into play in the
following situation.

51:29.110 --> 51:34.640
I have a particle in a box
whose wave function looks like

51:34.637 --> 51:35.327
this.

51:35.329 --> 51:40.229
Somebody put it in that state,
okay?

51:40.230 --> 51:42.860
It's an allowed wave function
because it vanishes at the ends

51:42.862 --> 51:43.612
like it should.

51:43.610 --> 51:46.220
Beyond that,
it does anything it wants.

51:46.219 --> 51:50.139
That is my ψ(x),
okay?

51:50.139 --> 51:52.809
If I measure the energy of that
guy, now you tell me,

51:52.811 --> 51:56.101
if I measure the energy in that
state, what answer will I get?

51:56.099 --> 52:00.229
Student:
>

52:00.230 --> 52:00.960
Prof: Pardon me?

52:00.960 --> 52:01.440
Yes.

52:01.440 --> 52:04.780
I'll get one of those quantized
values, do you understand?

52:04.780 --> 52:08.350
A particle in a box under
energy measurement can give only

52:08.349 --> 52:09.789
one of those numbers.

52:09.789 --> 52:11.659
Now this guy is not one of
those functions,

52:11.655 --> 52:12.405
you understand.

52:12.409 --> 52:15.009
This is a randomly chosen
function.

52:15.010 --> 52:16.870
But when I measure energy,
so what will I do?

52:16.869 --> 52:18.739
Maybe you can guess now.

52:18.739 --> 52:21.409
Student: Write it as a
sum of energy functions

52:21.409 --> 52:22.239
Prof: That's correct.

52:22.239 --> 52:28.019
You will have to write this guy
as a sum of A--let me call it n

52:28.016 --> 52:29.596
for energy now.

52:29.599 --> 52:35.069
A √(2/L)
sin(nΠx/L).

52:35.070 --> 52:38.610
I mean, this is just
A_E ψ's of

52:38.612 --> 52:40.212
E of x.

52:40.210 --> 52:44.310
Except E is labeled by an
integer, so I'm writing it this

52:44.309 --> 52:44.749
way.

52:44.750 --> 52:47.420
Then I have to find
A_n.

52:47.420 --> 52:51.150
And if I want to find
A_n,

52:51.148 --> 52:56.118
that = integral of (2/L)
sin(nΠx/L).

52:56.119 --> 52:57.929
If you want,
you can put a conjugate,

52:57.927 --> 53:00.237
but it doesn't matter,
because sine is real.

53:00.239 --> 53:03.959
Multiply it by Mr. Crazy
here, whatever it is,

53:03.960 --> 53:06.110
and do their integral.

53:06.110 --> 53:09.500
It will give you some number
for every n.

53:09.500 --> 53:10.870
A_1,
A_2,

53:10.867 --> 53:12.277
A_3,
A_4,

53:12.280 --> 53:15.020
and the square of every one of
them will give you the odds that

53:15.018 --> 53:17.358
you will find this energy or
that energy or some other

53:17.356 --> 53:18.016
energy.

53:18.018 --> 53:21.498
So what quantum mechanics tells
you is, an energy measurement of

53:21.498 --> 53:24.258
a particle in a box cannot give
a random answer.

53:24.260 --> 53:26.390
It can only give one of these
answers.

53:26.389 --> 53:28.939
That's the quantization of
energy.

53:28.940 --> 53:32.910
It does not mean every particle
in a box is in a state of

53:32.911 --> 53:34.191
definite energy.

53:34.190 --> 53:37.010
A particle in a box can be any
wave function,

53:37.014 --> 53:40.614
provided it vanishes at the
ends and doesn't spill out of

53:40.608 --> 53:42.468
the infinitely tall wall.

53:42.469 --> 53:44.409
So for every one of them,
you can ask the usual

53:44.405 --> 53:46.505
questions: what happens when I
measure position?

53:46.510 --> 53:47.750
Square the function.

53:47.750 --> 53:49.270
What happens when I measure
momentum?

53:49.268 --> 53:51.798
Write it in terms of e
to the ipx functions.

53:51.800 --> 53:53.310
What happens when I measure
energy?

53:53.309 --> 53:56.809
Write it in terms of these
functions.

53:56.809 --> 54:03.579
Okay?

54:03.579 --> 54:07.119
All right, so I want to do one
last topic here,

54:07.121 --> 54:11.281
which is called scattering,
but I don't want to proceed

54:11.280 --> 54:14.670
till you guys have understood
this part.

54:14.670 --> 54:17.410
This is covered in all the
books and maybe you can read the

54:17.409 --> 54:18.969
books or you can read my notes.

54:18.969 --> 54:21.039
You can talk to everybody.

54:21.039 --> 54:24.529
But we have solved--from this,
all quantum problems that we

54:24.525 --> 54:27.405
all do are exactly isomorphic to
this problem.

54:27.409 --> 54:30.009
It's the same thing we do over
and over again,

54:30.007 --> 54:33.007
except the equations you solve
are more difficult.

54:33.010 --> 54:33.930
They can be in higher
dimensions.

54:33.929 --> 54:37.259
They can involve more than one
particle.

54:37.260 --> 54:40.110
The potential could be
complicated, but the philosophy

54:40.110 --> 54:41.240
is always the same.

54:43.039 --> 54:45.969
and you want to find all the
functions that satisfy it,

54:45.969 --> 54:49.759
find the allowed energies and
the allowed functions.

54:49.760 --> 54:52.070
We will see that states of
definite energy have a very

54:52.070 --> 54:53.770
privileged role in quantum
mechanics.

54:53.768 --> 54:57.638
That will be clear in the next
two lectures.

54:57.639 --> 55:01.939
But right now,
here's what I want to do.

55:01.940 --> 55:05.100
This is a very small variation
of what we have done,

55:05.099 --> 55:07.639
so most of the hard work has
been done.

55:07.639 --> 55:10.319
I'm just going to cash in on
that.

55:10.320 --> 55:15.570
Take the following problem in
classical mechanics:

55:15.574 --> 55:18.474
here is the level ground.

55:18.469 --> 55:23.979
Then I make a little change in
the landscape so that this has

55:23.980 --> 55:28.940
got certain height and therefore
certain potential,

55:28.940 --> 55:31.810
V_0. And from
this end,

55:31.809 --> 55:35.619
I roll a marble and see what
happens.

55:35.619 --> 55:40.729
The marble is given some energy
E, which because there is

55:40.733 --> 55:44.553
no potential here,
it's just p^(2)/2m.

55:44.550 --> 55:48.600
Question is, what will happen?

55:48.599 --> 55:52.309
Can you tell me what will
happen in this problem,

55:52.313 --> 55:53.633
to this marble?

55:55.885 --> 55:56.365
equations?

55:56.369 --> 55:57.039
Prof: No.

55:57.039 --> 55:57.919
Classical mechanics.

55:57.920 --> 56:00.330
Student: Oh,
the marble will only go up as

56:00.326 --> 56:03.276
far as it has the amount of
energy that that would equal the

56:03.284 --> 56:05.044
potential energy at that point.

56:05.039 --> 56:05.549
Prof: That's correct.

56:05.550 --> 56:09.170
So he said it will go as far as
energy equals potential,

56:09.170 --> 56:11.870
so let p^(2)/2m have
this value.

56:11.869 --> 56:12.809
That's my energy.

56:12.809 --> 56:13.889
And it does not change.

56:13.889 --> 56:15.959
That's the law of conservation
of energy.

56:15.960 --> 56:20.080
So here, since energy is
kinetic potential,

56:20.083 --> 56:21.953
it's all kinetic.

56:21.949 --> 56:26.879
Somewhere here,
that part of it is potential;

56:26.880 --> 56:28.740
the rest of it is kinetic.

56:28.739 --> 56:31.279
Here kinetic vanishes.

56:31.280 --> 56:34.020
Here you're not allowed.

56:34.018 --> 56:37.308
So you will go,
you will climb that far and

56:37.311 --> 56:38.801
you'll come back.

56:38.800 --> 56:45.020
But if your initial energy was
that, then you will come here,

56:45.023 --> 56:46.893
going very fast.

56:46.889 --> 56:50.799
You will slow down,
but you still make it on the

56:50.798 --> 56:53.958
other side, but with lower
velocity.

56:53.960 --> 56:55.010
That's very simple.

56:55.010 --> 56:56.320
There's only one answer you can
get.

56:56.320 --> 56:57.470
You throw something.

56:57.469 --> 56:59.899
If it comes back,
it means the barrier is taller

56:59.902 --> 57:00.992
than what you have.

57:00.989 --> 57:03.579
Suppose you cannot see this
barrier.

57:03.579 --> 57:05.909
All you have is a marble,
and you want to know how tall

57:05.914 --> 57:06.654
the barrier is.

57:06.650 --> 57:07.480
Very easy.

57:07.480 --> 57:09.130
Throw the marble at a known
speed.

57:09.130 --> 57:12.370
If it comes back,
then you haven't reached the

57:12.367 --> 57:13.877
top of the barrier.

57:13.880 --> 57:17.080
Keep on increasing the kinetic
energy till one day it doesn't

57:17.079 --> 57:17.719
come back.

57:17.719 --> 57:19.699
That's when you're sitting
right here.

57:19.699 --> 57:23.069
That's one way to tell how tall
this hill is without actually

57:23.072 --> 57:23.862
going there.

57:23.860 --> 57:27.680
But now what I want to do is do
a quantum mechanical experiment

57:27.684 --> 57:29.664
in exactly the same situation.

57:29.659 --> 57:38.359
I want to send a particle from
the left and I want to see what

57:38.362 --> 57:39.792
happens.

57:39.789 --> 57:43.289
So now, to simplify the
problem, I'm going to take the

57:43.293 --> 57:45.943
potential to again have only 2
values.

57:45.940 --> 57:49.920
At x = 0,
it jumps to a value

57:49.922 --> 57:52.542
V_0.

57:59.250 --> 58:03.260
So I'm going to spare you some
of the calculations,

58:03.255 --> 58:05.415
because it's so familiar.

58:05.420 --> 58:13.240
One will look like Ae^(ikx)
Be^(-ikx),

58:13.240 --> 58:21.240
where k^(2) is just
2mE/ℏ^(2),

58:21.239 --> 58:22.839
here.

58:22.840 --> 58:27.340
First, let's take a case where
E is bigger than

58:27.344 --> 58:29.814
V_0,
okay?

58:29.809 --> 58:32.449
Then I go to region II.

58:35.148 --> 58:36.388
V_0 in it.

58:36.389 --> 58:41.719
Then you will find in region
II, ψ II looks like

58:41.715 --> 58:46.305
Ce^(ik'x)
De^(−ik'x),

58:46.309 --> 58:52.199
where k'^(2) will be
2m times E -

58:52.201 --> 58:57.011
V_0 over
ℏ^(2).

58:57.010 --> 58:59.220
Because when you have a
potential, you must always

59:00.617 --> 59:04.377
equation here--
2m/ℏ^(2) at E -

59:04.378 --> 59:05.198
V.

59:05.199 --> 59:06.549
So E is not 0.

59:06.550 --> 59:09.560
The k' now,
so what does this mean?

59:09.559 --> 59:13.579
You can see very simply,
it's got some momentum here.

59:13.579 --> 59:16.499
It's got a smaller momentum
here, because k'^(2) is

59:16.496 --> 59:18.796
going to be smaller than
k^(2), because

59:18.800 --> 59:20.950
V_0 is non-zero
here.

59:20.949 --> 59:24.809
That is just the statement that
when the particle's moving here,

59:24.813 --> 59:28.373
it's slower than when it's
here, because it has to climb up

59:28.371 --> 59:29.171
the hill.

59:29.170 --> 59:34.140
So the wave functions in this
region has got an incoming,

59:34.137 --> 59:38.127
right moving wave of momentum,
ℏk.

59:38.130 --> 59:41.730
It's got to reflect that wave
of momentum -ℏk.

59:41.730 --> 59:45.800
And in this region again,
it's got a right moving and a

59:45.795 --> 59:50.235
left moving, because this
k can be of either sign.

59:50.239 --> 59:51.379
So you can pick the
coefficients,

59:51.384 --> 59:53.244
A,
B, C, D arbitrarily,

59:53.242 --> 59:55.802
but I'm going to choose
D = 0,

59:55.800 --> 59:59.930
because what I have in mind is
a problem where the particle

59:59.929 --> 1:00:02.279
actually came in from the left.

1:00:02.280 --> 1:00:05.580
If it came in from the left,
I'm prepared for some of it

1:00:05.579 --> 1:00:08.219
coming back reflected,
and some of it getting

1:00:08.219 --> 1:00:09.179
transmitted.

1:00:09.179 --> 1:00:11.369
There's nobody coming in from
the right at infinity,

1:00:11.371 --> 1:00:13.351
because I'm not sending
anything from there.

1:00:13.349 --> 1:00:15.169
I'm sending my particle from
the left.

1:00:15.170 --> 1:00:19.200
So I pick a solution in which
D, which is anything,

1:00:19.202 --> 1:00:20.762
I pick that to be 0.

1:00:20.760 --> 1:00:24.890
So here are my two solutions.

1:00:24.889 --> 1:00:28.769
I hope you understand that
k and k' are not

1:00:28.771 --> 1:00:30.021
the same number.

1:00:30.018 --> 1:00:32.428
k' is smaller because
the momentum on the right hand

1:00:32.425 --> 1:00:34.655
side is going to be less,
because some of it's eaten up

1:00:34.664 --> 1:00:35.664
by potential energy.

1:00:35.659 --> 1:00:38.469
So what people do is they draw
a picture like this.

1:00:38.469 --> 1:00:41.609
They draw a wave that's going
at some speed,

1:00:41.606 --> 1:00:45.106
and it comes in this region and
it slows down.

1:00:45.110 --> 1:00:47.220
That's what the function will
look like.

1:00:47.219 --> 1:00:48.829
The real function's made of
exponentials,

1:00:48.829 --> 1:00:51.139
but the real and imaginary
parts will have rapid

1:00:51.141 --> 1:00:54.241
oscillations on the left and
slow oscillations on the right,

1:00:54.239 --> 1:00:58.109
because k is bigger than
k'.

1:00:58.110 --> 1:00:58.540
Very good.

1:00:58.539 --> 1:01:02.799
So now I have the following
condition: at the interface

1:01:02.804 --> 1:01:07.234
where the two regions meet,
the ψ must match from the

1:01:07.228 --> 1:01:09.438
left and from the right.

1:01:09.440 --> 1:01:12.340
So x = 0,
coming from the left,

1:01:12.335 --> 1:01:16.245
I want ψ_1 at 0
should be the same at

1:01:16.246 --> 1:01:18.356
ψ_2 at 0.

1:01:18.360 --> 1:01:24.650
That will tell me that A B =
C.

1:01:24.650 --> 1:01:30.170
Because if you put x =
0, all the exponentials become

1:01:30.166 --> 1:01:33.716
one, and A B must =
C.

1:01:33.719 --> 1:01:36.699
There's the condition that if
you come from the two sides,

1:01:36.695 --> 1:01:38.415
the wave function has to match.

1:01:41.894 --> 1:01:44.974
requirement that the derivatives
also match.

1:01:44.969 --> 1:01:49.689
The two functions coming from
the left and right cannot join

1:01:49.690 --> 1:01:50.650
like this.

1:01:50.650 --> 1:01:53.890
They must join with the same
slope.

1:01:53.889 --> 1:01:57.459
The reason the slope must be
the same is that if the slope

1:01:57.460 --> 1:02:01.160
changed abruptly at one point,
the rate of change of slope

1:02:01.159 --> 1:02:03.939
will be infinite there,
because d^(2)ψ/dx^(2)

1:02:03.942 --> 1:02:05.762
is the rate of change of the
slope.

1:02:07.320 --> 1:02:09.910
the second derivative ψ,
if it's infinite,

1:02:09.909 --> 1:02:12.359
there is no compensating terms
in the equation,

1:02:12.360 --> 1:02:14.630
because E and V
are all finite.

1:02:14.630 --> 1:02:17.260
So that infinity will not be
balanced, so you can never have

1:02:17.257 --> 1:02:17.567
that.

1:02:17.570 --> 1:02:19.010
So second derivative must be
finite;

1:02:19.010 --> 1:02:21.000
first derivative must be
continuous.

1:02:21.000 --> 1:02:26.990
So I make the requirement that
dψ/dx in region I at

1:02:26.987 --> 1:02:32.087
x = 0 must match
dψ II dx at

1:02:32.090 --> 1:02:33.760
x = 0.

1:02:33.760 --> 1:02:35.680
What does that give me?

1:02:35.679 --> 1:02:42.929
Well, take the derivative and
equate them at x = 0,

1:02:42.929 --> 1:02:51.069
you'll find ik times A - B
= ik' times C.

1:02:51.070 --> 1:02:55.440
Take the derivative and then
equate the derivatives at

1:02:55.442 --> 1:02:56.682
x = 0.

1:02:56.679 --> 1:02:58.449
Derivatives brings down
ik for this,

1:02:58.445 --> 1:02:59.365
-ik for that.

1:02:59.369 --> 1:03:02.569
Blah, blah, you put them
together.

1:03:02.570 --> 1:03:05.980
So these are the two conditions.

1:03:05.980 --> 1:03:10.490
Now what causes mild panic is
that I have three unknowns and

1:03:10.487 --> 1:03:11.707
two equations.

1:03:11.710 --> 1:03:13.810
I've got A,
B and C,

1:03:13.807 --> 1:03:15.967
and I've got only two
equations.

1:03:15.969 --> 1:03:22.079
So what do you think is going
on?

1:03:22.079 --> 1:03:29.319
Anybody have an idea?

1:03:29.320 --> 1:03:31.660
It's that even in the particle
in a box,

1:03:31.659 --> 1:03:34.479
you'll remember the overall
height of the wave function is

1:03:37.059 --> 1:03:38.219
It's whatever you like.

1:03:38.219 --> 1:03:43.119
So what people do is they say,
"Let's pick A =

1:03:43.119 --> 1:03:43.979
1."

1:03:43.980 --> 1:03:46.370
That sets the overall height
arbitrarily.

1:03:46.369 --> 1:03:54.789
Then I get 1 B = C and
ik times 1 - B =

1:03:54.793 --> 1:03:56.653
ik' [C}.

1:03:56.650 --> 1:03:59.930
Now of course you can solve
for B and you can solve

1:03:59.931 --> 1:04:00.741
for C.

1:04:00.739 --> 1:04:03.519
I'm not going to do that here;
it's very easy.

1:04:03.518 --> 1:04:11.728
You're going to find B = k
− k' over k k'.

1:04:11.730 --> 1:04:19.350
You're going to find C =
2k over k k'.

1:04:19.349 --> 1:04:22.579
That's simple algebra of
simultaneous equations.

1:04:22.579 --> 1:04:27.119
I want you to look at the
answer.

1:04:27.119 --> 1:04:30.809
You remember now what A,
B and C are.

1:04:30.809 --> 1:04:34.899
A comes in,
B goes back and C

1:04:34.900 --> 1:04:36.570
goes to the right.

1:04:36.570 --> 1:04:41.940
What do you find strange about
the answer?

1:04:41.940 --> 1:04:47.460
It's that if you came in from
the left with an energy bigger

1:04:47.458 --> 1:04:51.758
than the barrier,
there's a certain probability

1:04:51.760 --> 1:04:54.660
that you will go backwards.

1:04:54.659 --> 1:04:56.529
In classical mechanics,
if your energy is more than the

1:04:56.525 --> 1:04:57.765
barrier, you will never come
back.

1:04:57.769 --> 1:04:58.639
The probability is 0.

1:04:58.639 --> 1:05:00.949
In quantum theory,
even though your energy is more

1:05:00.947 --> 1:05:03.877
than the barrier,
there's a certain probability

1:05:03.876 --> 1:05:08.326
you'll get reflected and certain
probability to get transmitted.

1:05:08.329 --> 1:05:10.879
The square of this is like the
odds of going to the right.

1:05:10.880 --> 1:05:13.530
The square of this is like the
odds of getting reflected.

1:05:13.530 --> 1:05:17.410
The point is the reflection
probability does not vanish even

1:05:17.411 --> 1:05:20.701
if the energy is bigger than the
barrier height.

1:05:20.699 --> 1:05:23.709
That's because this is quantum
mechanics and you're really

1:05:23.710 --> 1:05:25.770
dealing with waves and not
particles.

1:05:25.768 --> 1:05:28.938
And waves have the property
that when the medium changes,

1:05:28.940 --> 1:05:30.470
there is some reflection.

1:05:30.469 --> 1:05:33.849
And to the quantum mechanical
wave ψ,

1:05:33.849 --> 1:05:36.539
a barrier looks like a change
in refractive index,

1:05:36.539 --> 1:05:39.809
so it will get reflected and
some will get transmitted.

1:05:39.809 --> 1:05:42.469
But the interesting thing is,
in the quantum problem,

1:05:42.474 --> 1:05:45.554
you can send your billiard ball
to find how tall the mountain

1:05:45.550 --> 1:05:45.910
is.

1:05:45.909 --> 1:05:48.229
And even if the energy was more
than the mountain,

1:05:48.231 --> 1:05:49.561
sometimes it'll come back.

1:05:49.559 --> 1:05:53.449
So that's not a reliable way to
measure that.

1:05:53.449 --> 1:05:57.149
The last example is,
what if your energy is less

1:05:57.154 --> 1:05:58.104
than this?

1:05:58.099 --> 1:06:03.479
What if your energy is like
here, lower than the barrier?

1:06:03.480 --> 1:06:07.960
If you're lower than the
barrier, you will find that

1:06:07.956 --> 1:06:12.516
k' will equal i
times some κ,

1:06:12.518 --> 1:06:17.578
because it will become
imaginary, and you will have

1:06:17.576 --> 1:06:21.516
then a function that looks like
this.

1:06:21.518 --> 1:06:24.938
The function will oscillate,
then it will die exponentially

1:06:24.938 --> 1:06:25.938
in this region.

1:06:25.940 --> 1:06:29.410
It's exactly what I did
earlier, like particle in a box.

1:06:29.409 --> 1:06:34.559
In the classically forbidden
region, it goes a little

1:06:34.556 --> 1:06:37.326
distance but not very far.

1:06:37.329 --> 1:06:40.189
So it does get reflected if the
energy is less,

1:06:40.190 --> 1:06:44.110
but you can actually find it in
a region where it's not supposed

1:06:44.108 --> 1:06:44.728
to be.

1:06:44.730 --> 1:06:47.950
And you can say that's going to
be very embarrassing for the

1:06:47.951 --> 1:06:50.791
particle to be found here,
because its kinetic energy

1:06:50.791 --> 1:06:52.321
should be negative here.

1:06:52.320 --> 1:06:55.180
So what's the particle say to
you when you catch it there?

1:06:55.179 --> 1:06:57.509
It will say,
"In order to locate me

1:06:57.510 --> 1:06:59.430
here,
you shine some light on me,

1:06:59.434 --> 1:07:02.494
and the energy I got from the
photon made me legitimate in

1:07:02.485 --> 1:07:03.605
that region."

1:07:03.610 --> 1:07:05.510
It will come back to positive
energy.

1:07:05.510 --> 1:07:08.490
So the act of measurement will
prevent it from ever being

1:07:08.487 --> 1:07:11.037
caught in that region,
but it's got an amplitude,

1:07:11.041 --> 1:07:13.011
a probability to be caught
there.

1:07:13.010 --> 1:07:16.010
Final thing I want to mention
is, if this barrier,

1:07:16.010 --> 1:07:20.830
instead of going on forever,
terminated there and became 0

1:07:20.829 --> 1:07:23.569
again,
this will leak out here and

1:07:23.570 --> 1:07:27.530
start going out with the same
wavelength as that one.

1:07:27.530 --> 1:07:30.010
That's called barrier
penetration.

1:07:30.010 --> 1:07:32.640
That means if you send a
particle here with an energy not

1:07:32.635 --> 1:07:35.305
enough to overcome the barrier
in classical mechanics,

1:07:35.309 --> 1:07:38.059
in quantum mechanics,
it's got a small chance of

1:07:38.057 --> 1:07:39.927
being found on the other side.

1:07:39.929 --> 1:07:42.069
On the other side,
it can go scot free.

1:07:42.070 --> 1:07:44.980
It's not allowed in this
region, because there the

1:07:44.983 --> 1:07:46.713
kinetic energy is negative.

1:07:46.710 --> 1:07:51.390
Here kinetic energy is the same
as before, except the height may

1:07:51.391 --> 1:07:52.581
be very small.

1:07:52.579 --> 1:07:56.509
So this means no barrier is
completely safe in quantum

1:07:56.512 --> 1:07:57.182
theory.

1:07:57.179 --> 1:08:00.659
So I'm going to give you my
final survival tip.

1:08:00.659 --> 1:08:02.939
I've told you so many
situations you can be in where

1:08:02.938 --> 1:08:04.188
this course will help you.

1:08:04.190 --> 1:08:05.730
So here's another one.

1:08:05.730 --> 1:08:09.290
You are in a prison and it
doesn't have infinite walls,

1:08:09.293 --> 1:08:11.013
but it's got some walls.

1:08:11.010 --> 1:08:12.620
What's your strategy?

1:08:12.619 --> 1:08:16.359
I say go ram yourself into that
wall as often as you can,

1:08:16.358 --> 1:08:19.488
because there's a small
probability that you can

1:08:19.494 --> 1:08:22.504
suddenly find yourself on the
other side.

1:08:22.500 --> 1:08:25.580
This is what happens in alpha
decay.

1:08:25.578 --> 1:08:29.108
In alpha decay,
there's helium nuclei sitting

1:08:29.105 --> 1:08:30.625
inside a nucleus.

1:08:30.630 --> 1:08:34.170
There's a barrier that keeps it
from coming out of the nucleus,

1:08:34.168 --> 1:08:36.508
but the barrier goes up and
comes down.

1:08:36.510 --> 1:08:38.830
So if you go a certain distance
from the nucleus,

1:08:38.832 --> 1:08:39.802
you're again free.

1:08:39.800 --> 1:08:41.970
So the alpha particle does
exactly what I told you.

1:08:41.970 --> 1:08:44.460
It goes rattling back and forth
inside the nucleus and once in a

1:08:44.462 --> 1:08:46.442
while,
it penetrates and comes to the

1:08:46.438 --> 1:08:48.658
other side,
and that's the alpha decay of

1:08:48.658 --> 1:08:49.338
the nucleus.

1:08:49.340 --> 1:08:51.240
So this really happens.

1:08:51.238 --> 1:08:54.628
It may not happen to you in the
situation I described,

1:08:54.627 --> 1:08:56.797
but the probability is not
zero.

1:08:56.800 --> 1:09:02.000