WEBVTT
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PROFESSOR: Delta
function potential.
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So it's still a one-dimensional
potential-- potential
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is a function of x.
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We'll write it this
way-- minus alpha delta
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of x, where alpha is positive.
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So this is a delta function
in a negative direction.
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So if you want to
draw the potential--
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there's no way to draw really
nicely a delta function.
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So you just do a thick
arrow with it pointing down.
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It's a representation of
a potential that, somehow,
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is rather infinite
at x equals zero--
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but infinite and negative.
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It can be thought of as the
limit of a square well that
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is becoming deeper and deeper.
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And, in fact, that
could be a way
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to analytically
calculate the energy
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levels-- by taking carefully
the limit of a potential.
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It is becoming thinner and
thinner, but deeper and deeper,
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which is the way you define or
regulate the delta function.
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You can imagine
the delta function
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as a sequence of
functions, in which it's
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becoming more and more narrow--
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but deeper at the same time.
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So that the area under the
curve is still the same.
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So, at any rate, the delta
function potential is
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a potential that should be
understood as 0 everywhere ,
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else except at the delta
function where it becomes
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infinite.
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And there are all kinds
of questions we can ask.
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OK.
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Are there bound states?
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What are bound
states in this case?
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They are energy eigenstates
with energy less than zero.
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So bound states, which
means e less than zero.
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Do they exist?
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Does this potential
have bound states?
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And, if it does, how
many bound states?
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1, 2, 3?
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Does It depend on the intensity
of the delta function?
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When you get more bound states,
the deeper the potential is.
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Well, we'll try to figure out.
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In fact, there's a lot
that can be figured out
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without calculating, too much.
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And it's a good habit to try
to do those things before you--
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not to be so impatient
that you begin,
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and within a second
start writing
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the differential equation
trying to solve it.
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Get a little intuition about
how any state could look like,
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and how could the answer
for the energy eigenstates--
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the energies--
what could they be?
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Could you just reason
your way and conclude
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there's no bound states?
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Or one bound state?
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Or two?
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All these things
are pretty useful.
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So one way, as you can
imagine, is to think of units.
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And what are the
constants in this problem?
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In this problem we'll
have three constants.
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Alpha, the mass of the
particle, and h-bar.
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So with alpha, the
mass and the particle,
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and h-bar you can ask, how
do I construct the quantity
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with units of energy?
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If there there's only one
way to construct the quantity
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with units of energy, then
the energy of a bound state
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will be proportional
to that quantity--
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because that's the only quantity
that can carry the units.
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And here, indeed,
there's only one way
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to construct that quantity
with units of energy--
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from these three.
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That's to be expected.
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With three constants that
are not linearly dependent--
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whatever that is
supposed to mean--
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you can build anything that has
units of length, mass, or time.
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And from that you
can build something
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that has units of energy.
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So you can now decide, well,
what are the units of alpha?
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The units of alpha
have give you energy,
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but the delta function has
units of one over length.
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This has one over length.
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, Remember if you integrate over
x the delta function gives you
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1.
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So this has units
of 1 over length.
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And, therefore,
alpha has to have
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units of energy times length.
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So this is not quite enough
to solve the problem,
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because I want to write e--
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think of finding how do
you get units of energy
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from these quantities?
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But l-- we still don't
have a length scale either.
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So we have to do a
little more work.
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So from here we say that units
of energy is alpha over l.
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There should be a
way to say that this
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is an equality between units.
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I could put units or
leave it just like that.
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So in terms of units, it's this.
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But in terms of units, energy--
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you should always remember--
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is p squared over m.
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And p is h-bar over a length.
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So that's p squared
and that's m.
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So that's also units of energy
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From these two you can get
what has units of length.
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Length.
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You pass the l to
this side-- the l
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squared to this left-hand side.
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Divide.
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So you get l is h
squared over m alpha.
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And if I substitute
back into this l here,
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e would be alpha over l, which
is h squared, alpha squared, m.
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So that's the quantity
that has units of energy.
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M alpha squared over h
squared has units of energy.
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If this has units of energy--
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the bound state energy.
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Now, if you have a
bounce state here,
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it has to decay in order
to be normalizable.
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In order to be normalizable
it has to decay,
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so it has to be in the
forbidden region throughout x.
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So the energy as we
said is negative,
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energy of a bound
state-- if it exists.
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And this bound
state energy would
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have to be negative some number
m alpha squared over h squared.
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And that's very
useful information.
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The whole problem
has been reduced
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to calculating a number.
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It better be and the answer
cannot be any other way.
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There's no other way to
get the units of energy.
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So if a bound state
exists it has to be that.
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And that number could be
pi, it could be 1/3, 1/4,
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it could be anything.
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There's a naturalness
to that problem
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in that you don't expect
that number to be a trillion.
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Nor do you expect that number
to be 10 to the minus 6.
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Because there's no way-- where
would those numbers appear?
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So this number should be
a number of order one,
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and we're going to wait
and see what it is.
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So that's one thing we know
already about this problem.
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The other thing we can do is
to think of the regulated delta
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function.
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So we think of this as a
potential that has this form.
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So here is v of
x, and here is x.
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And for this potential--
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if you have a bound state--
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how would the wave
function look?
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Well, it would have to--
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suppose you have
a ground state--
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it's an even potential.
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The delta function is even, too.
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It's in the middle.
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It's symmetric.
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There's nothing asymmetric
about the delta function.
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So if it's an even
potential the ground states
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should be even, because
the ground state
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is supposed to have no nodes.
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And it's supposed to be even
if the potential is even.
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So how will it look?
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Well, it shouldn't be
decaying in this region.
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So, presumably, it decays here.
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It decays there-- symmetrically.
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And in the middle it curves
in the other direction.
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It is in an allowed region--
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and you remember that's
kind of allowed this way.
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So that's probably
the way it looks.
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Now, if that bound
state exists, somehow,
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as I narrow this and go down--
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as it becomes even more narrow,
very narrow now, but very deep.
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This region becomes smaller.
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And I would pretty much
expect the wave function
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to have a discontinuity.
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You basically don't
have enough power
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to see the curving
that is happening here.
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Especially because the
curving is going down.
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The distance is going down.
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So if this bound state
exists, as you approach
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the limit in which this
becomes a delta function
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the energy moves a little, but
stays finite at some number.
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And the curvature that is
created by the delta function
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is not visible, and the thing
looks just discontinuous
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in its derivative.
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So this is an intuitive
way to understand
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that the wave
function we're looking
00:10:57.480 --> 00:11:02.330
for is going to be
discontinuous on its derivative.
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Let's write the
differential equation,
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even though we're still
not going to solve it.
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So what is the
differential equation?
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Minus h squared over
m, psi double prime,
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is equal to E psi.
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And, therefore-- and I
write this, and you say,
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oh, what are you writing?
00:11:25.730 --> 00:11:28.220
I'm writing the
differential equation
00:11:28.220 --> 00:11:31.500
when x is different from 0.
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No potential when x
is different from 0.
00:11:38.380 --> 00:11:41.710
So this applies for
positive x and negative x.
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It doesn't apply at x equals 0.
00:11:44.140 --> 00:11:46.280
We'll have to deal
with that later.
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So then, no potential
for x different from 0.
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And this differential
equation becomes
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psi double prime equals minus
2m e over h squared psi.
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And this is equal
to kappa squared
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psi, where kappa squared is
minus 2me over h squared.
00:12:09.880 --> 00:12:13.350
And it's positive.
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Let's make that positive.
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It's positive because
the energy is negative
00:12:19.180 --> 00:12:22.450
and we're looking
for bound states.
00:12:22.450 --> 00:12:26.690
So we're looking for
bound states only.
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Kappa squared is positive.
00:12:28.560 --> 00:12:32.470
And this differential
equation is just this.
00:12:32.470 --> 00:12:35.690
I'll copy it again here.
00:12:35.690 --> 00:12:38.990
Kappa squared psi.
00:12:38.990 --> 00:12:42.920
And the solutions of this
equation are-- solutions--
00:12:46.130 --> 00:12:52.640
are e to the minus kappa
x and e to the kappa x.
00:12:52.640 --> 00:12:59.840
Or, if you wish, cosh
kappa x and sinh kappa x--
00:12:59.840 --> 00:13:00.830
whichever you prefer.
00:13:13.920 --> 00:13:16.560
This is something we
now have to use in order
00:13:16.560 --> 00:13:17.550
to produce a solution.
00:13:20.200 --> 00:13:24.900
But now, let's see
if I can figure out
00:13:24.900 --> 00:13:26.940
how many bound states there are.
00:13:33.240 --> 00:13:37.840
If there is one bound state,
it's going to be even.
00:13:37.840 --> 00:13:39.180
It's the ground state.
00:13:39.180 --> 00:13:40.975
It has no nodes.
00:13:40.975 --> 00:13:45.800
It has to be even, because
the potential is even.
00:13:45.800 --> 00:13:50.990
If I have the first excited
state after the ground state,
00:13:50.990 --> 00:13:53.330
it will have to be odd.
00:13:53.330 --> 00:13:59.320
It would have to vanish at x
equals 0, because it's odd.
00:13:59.320 --> 00:14:01.010
There is it's node--
00:14:01.010 --> 00:14:02.390
it has to have one node.
00:14:04.910 --> 00:14:10.800
For an odd bound state--
00:14:14.200 --> 00:14:17.310
or first excited state--
00:14:21.250 --> 00:14:26.590
you'd have to have psi
equals 0 at x equals 0.
00:14:32.670 --> 00:14:37.920
And the way to do that would
be to have a sinh, because this
00:14:37.920 --> 00:14:39.530
doesn't vanish at zero.
00:14:39.530 --> 00:14:40.870
This doesn't vanish at zero.
00:14:40.870 --> 00:14:43.050
And cosh doesn't vanish at zero.
00:14:43.050 --> 00:14:51.210
So you would need psi of
x equals sinh of kappa x.
00:14:56.730 --> 00:14:59.230
But that's not good.
00:14:59.230 --> 00:15:06.420
sinh of kappa x is
like this and blows up.
00:15:06.420 --> 00:15:08.030
Blows down.
00:15:08.030 --> 00:15:10.590
It has to go like this.
00:15:10.590 --> 00:15:13.230
It is in a forbidden
region, so it has
00:15:13.230 --> 00:15:17.230
to be convex towards the axis.
00:15:17.230 --> 00:15:18.300
And convex here.
00:15:18.300 --> 00:15:19.930
But it blows up.
00:15:19.930 --> 00:15:23.370
So there's no such solution.
00:15:23.370 --> 00:15:24.240
No such solution.
00:15:26.750 --> 00:15:29.630
You cannot have an
odd bound state.
00:15:29.630 --> 00:15:33.870
So since the bound states
alternate-- even, odd, even,
00:15:33.870 --> 00:15:36.270
odd, even, odd--
00:15:36.270 --> 00:15:37.560
you're stuck.
00:15:37.560 --> 00:15:41.720
You only will have a ground
state-- if we're lucky--
00:15:41.720 --> 00:15:47.970
but no excited state that is
bound, while a finite square
00:15:47.970 --> 00:15:48.470
will.
00:15:48.470 --> 00:15:51.590
You remember this
quantity z0 that
00:15:51.590 --> 00:15:56.420
tells you how many bound
states you can have.
00:15:56.420 --> 00:16:00.080
Probably you're anticipating
that in the case
00:16:00.080 --> 00:16:02.630
of the delta function
potential, you can only
00:16:02.630 --> 00:16:05.630
have one bound state, if any.
00:16:05.630 --> 00:16:07.960
The first excited
state would not exist.
00:16:07.960 --> 00:16:11.630
So, enough preliminaries.
00:16:11.630 --> 00:16:14.410
Let's just solve that now.