Lecture 13: Delta function potential. Justifying the node theorem. Simple harmonic oscillator.

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L13.1 Delta function potential I: Preliminaries (16:14)

L13.2 Delta function potential I: Solving for the bound state (15:21)

L13.3 Node Theorem (13:01)

L13.4 Harmonic oscillator: Differential equation (16:45)

L13.5 Behavior of the differential equation (10:31)

L13.1 Delta function potential I: Preliminaries (16:14)

MITOCW | watch?v=vcuY46RwoV0
PROFESSOR: Delta function potential. So it’s still a one-dimensional potential-- potential is a function of x.
We’ll write it this way-- minus alpha delta of x, where alpha is positive. So this is a delta
function in a negative direction.
So if you want to draw the potential-- there’s no way to draw really nicely a delta function. So
you just do a thick arrow with it pointing down. It’s a representation of a potential that,
somehow, is rather infinite at x equals zero-- but infinite and negative.
It can be thought of as the limit of a square well that is becoming deeper and deeper. And, in
fact, that could be a way to analytically calculate the energy levels-- by taking carefully the limit
of a potential. It is becoming thinner and thinner, but deeper and deeper, which is the way you
define or regulate the delta function.
You can imagine the delta function as a sequence of functions, in which it’s becoming more
and more narrow-- but deeper at the same time. So that the area under the curve is still the
same.
So, at any rate, the delta function potential is a potential that should be understood as 0
everywhere , else except at the delta function where it becomes infinite. And there are all kinds
of questions we can ask.
OK. Are there bound states? What are bound states in this case? They are energy eigenstates
with energy less than zero. So bound states, which means e less than zero. Do they exist?
Does this potential have bound states?
And, if it does, how many bound states? 1, 2, 3? Does It depend on the intensity of the delta
function? When you get more bound states, the deeper the potential is. Well, we’ll try to figure
out.
In fact, there’s a lot that can be figured out without calculating, too much. And it’s a good habit
to try to do those things before you-- not to be so impatient that you begin, and within a
second start writing the differential equation trying to solve it. Get a little intuition about how
any state could look like, and how could the answer for the energy eigenstates-- the energies–
what could they be?
Could you just reason your way and conclude there’s no bound states? Or one bound state?
Could you just reason your way and conclude there’s no bound states? Or one bound state?
Or two? All these things are pretty useful. So one way, as you can imagine, is to think of units.
And what are the constants in this problem? In this problem we’ll have three constants. Alpha,
the mass of the particle, and h-bar. So with alpha, the mass and the particle, and h-bar you
can ask, how do I construct the quantity with units of energy?
If there there’s only one way to construct the quantity with units of energy, then the energy of a
bound state will be proportional to that quantity-- because that’s the only quantity that can
carry the units.
And here, indeed, there’s only one way to construct that quantity with units of energy-- from
these three. That’s to be expected. With three constants that are not linearly dependent–
whatever that is supposed to mean-- you can build anything that has units of length, mass, or
time. And from that you can build something that has units of energy.
So you can now decide, well, what are the units of alpha? The units of alpha have give you
energy, but the delta function has units of one over length. This has one over length. ,
Remember if you integrate over x the delta function gives you 1. So this has units of 1 over
length. And, therefore, alpha has to have units of energy times length.
So this is not quite enough to solve the problem, because I want to write e-- think of finding
how do you get units of energy from these quantities? But l-- we still don’t have a length scale
either. So we have to do a little more work.
So from here we say that units of energy is alpha over l. There should be a way to say that this
is an equality between units. I could put units or leave it just like that. So in terms of units, it’s
this. But in terms of units, energy-- you should always remember-- is p squared over m. And p
is h-bar over a length. So that’s p squared and that’s m. So that’s also units of energy
From these two you can get what has units of length. Length. You pass the l to this side-- the l
squared to this left-hand side. Divide. So you get l is h squared over m alpha. And if I
substitute back into this l here, e would be alpha over l, which is h squared, alpha squared, m.
So that’s the quantity that has units of energy. M alpha squared over h squared has units of
energy. If this has units of energy-- the bound state energy. Now, if you have a bounce state
here, it has to decay in order to be normalizable. In order to be normalizable it has to decay,
so it has to be in the forbidden region throughout x.
So the energy as we said is negative, energy of a bound state-- if it exists. And this bound
state energy would have to be negative some number m alpha squared over h squared. And
that’s very useful information.
The whole problem has been reduced to calculating a number. It better be and the answer
cannot be any other way. There’s no other way to get the units of energy. So if a bound state
exists it has to be that. And that number could be pi, it could be 1/3, 1/4, it could be anything.
There’s a naturalness to that problem in that you don’t expect that number to be a trillion. Nor
do you expect that number to be 10 to the minus 6. Because there’s no way-- where would
those numbers appear? So this number should be a number of order one, and we’re going to
wait and see what it is.
So that’s one thing we know already about this problem. The other thing we can do is to think
of the regulated delta function. So we think of this as a potential that has this form. So here is v
of x, and here is x. And for this potential-- if you have a bound state-- how would the wave
function look?
Well, it would have to-- suppose you have a ground state-- it’s an even potential. The delta
function is even, too. It’s in the middle. It’s symmetric. There’s nothing asymmetric about the
delta function. So if it’s an even potential the ground states should be even, because the
ground state is supposed to have no nodes. And it’s supposed to be even if the potential is
even.
So how will it look? Well, it shouldn’t be decaying in this region. So, presumably, it decays
here. It decays there-- symmetrically. And in the middle it curves in the other direction. It is in
an allowed region-- and you remember that’s kind of allowed this way. So that’s probably the
way it looks.
Now, if that bound state exists, somehow, as I narrow this and go down-- as it becomes even
more narrow, very narrow now, but very deep. This region becomes smaller. And I would
pretty much expect the wave function to have a discontinuity.
You basically don’t have enough power to see the curving that is happening here. Especially
because the curving is going down. The distance is going down. So if this bound state exists,
as you approach the limit in which this becomes a delta function the energy moves a little, but
stays finite at some number. And the curvature that is created by the delta function is not
visible, and the thing looks just discontinuous in its derivative.
So this is an intuitive way to understand that the wave function we’re looking for is going to be
discontinuous on its derivative. Let’s write the differential equation, even though we’re still not
going to solve it. So what is the differential equation? Minus h squared over m, psi double
prime, is equal to E psi.
And, therefore-- and I write this, and you say, oh, what are you writing? I’m writing the
differential equation when x is different from 0. No potential when x is different from 0. So this
applies for positive x and negative x. It doesn’t apply at x equals 0. We’ll have to deal with that
later.
So then, no potential for x different from 0. And this differential equation becomes psi double
prime equals minus 2m e over h squared psi. And this is equal to kappa squ

L13.1 Delta function potential I: Preliminaries (16:14)

L13.2 Delta function potential I: Solving for the bound state (15:21)

L13.3 Node Theorem (13:01)

L13.4 Harmonic oscillator: Differential equation (16:45)

L13.5 Behavior of the differential equation (10:31)

L13.1 Delta function potential I: Preliminaries (16:14)

MITOCW | watch?v=vcuY46RwoV0
PROFESSOR: Delta function potential. So it’s still a one-dimensional potential-- potential is a function of x.
We’ll write it this way-- minus alpha delta of x, where alpha is positive. So this is a delta
function in a negative direction.
So if you want to draw the potential-- there’s no way to draw really nicely a delta function. So
you just do a thick arrow with it pointing down. It’s a representation of a potential that,
somehow, is rather infinite at x equals zero-- but infinite and negative.
It can be thought of as the limit of a square well that is becoming deeper and deeper. And, in
fact, that could be a way to analytically calculate the energy levels-- by taking carefully the limit
of a potential. It is becoming thinner and thinner, but deeper and deeper, which is the way you
define or regulate the delta function.
You can imagine the delta function as a sequence of functions, in which it’s becoming more
and more narrow-- but deeper at the same time. So that the area under the curve is still the
same.
So, at any rate, the delta function potential is a potential that should be understood as 0
everywhere , else except at the delta function where it becomes infinite. And there are all kinds
of questions we can ask.
OK. Are there bound states? What are bound states in this case? They are energy eigenstates
with energy less than zero. So bound states, which means e less than zero. Do they exist?
Does this potential have bound states?
And, if it does, how many bound states? 1, 2, 3? Does It depend on the intensity of the delta
function? When you get more bound states, the deeper the potential is. Well, we’ll try to figure
out.
In fact, there’s a lot that can be figured out without calculating, too much. And it’s a good habit
to try to do those things before you-- not to be so impatient that you begin, and within a
second start writing the differential equation trying to solve it. Get a little intuition about how
any state could look like, and how could the answer for the energy eigenstates-- the energies–
what could they be?
Could you just reason your way and conclude there’s no bound states? Or one bound state?
Could you just reason your way and conclude there’s no bound states? Or one bound state?
Or two? All these things are pretty useful. So one way, as you can imagine, is to think of units.
And what are the constants in this problem? In this problem we’ll have three constants. Alpha,
the mass of the particle, and h-bar. So with alpha, the mass and the particle, and h-bar you
can ask, how do I construct the quantity with units of energy?
If there there’s only one way to construct the quantity with units of energy, then the energy of a
bound state will be proportional to that quantity-- because that’s the only quantity that can
carry the units.
And here, indeed, there’s only one way to construct that quantity with units of energy-- from
these three. That’s to be expected. With three constants that are not linearly dependent–
whatever that is supposed to mean-- you can build anything that has units of length, mass, or
time. And from that you can build something that has units of energy.
So you can now decide, well, what are the units of alpha? The units of alpha have give you
energy, but the delta function has units of one over length. This has one over length. ,
Remember if you integrate over x the delta function gives you 1. So this has units of 1 over
length. And, therefore, alpha has to have units of energy times length.
So this is not quite enough to solve the problem, because I want to write e-- think of finding
how do you get units of energy from these quantities? But l-- we still don’t have a length scale
either. So we have to do a little more work.
So from here we say that units of energy is alpha over l. There should be a way to say that this
is an equality between units. I could put units or leave it just like that. So in terms of units, it’s
this. But in terms of units, energy-- you should always remember-- is p squared over m. And p
is h-bar over a length. So that’s p squared and that’s m. So that’s also units of energy
From these two you can get what has units of length. Length. You pass the l to this side-- the l
squared to this left-hand side. Divide. So you get l is h squared over m alpha. And if I
substitute back into this l here, e would be alpha over l, which is h squared, alpha squared, m.
So that’s the quantity that has units of energy. M alpha squared over h squared has units of
energy. If this has units of energy-- the bound state energy. Now, if you have a bounce state
here, it has to decay in order to be normalizable. In order to be normalizable it has to decay,
so it has to be in the forbidden region throughout x.
So the energy as we said is negative, energy of a bound state-- if it exists. And this bound
state energy would have to be negative some number m alpha squared over h squared. And
that’s very useful information.
The whole problem has been reduced to calculating a number. It better be and the answer
cannot be any other way. There’s no other way to get the units of energy. So if a bound state
exists it has to be that. And that number could be pi, it could be 1/3, 1/4, it could be anything.
There’s a naturalness to that problem in that you don’t expect that number to be a trillion. Nor
do you expect that number to be 10 to the minus 6. Because there’s no way-- where would
those numbers appear? So this number should be a number of order one, and we’re going to
wait and see what it is.
So that’s one thing we know already about this problem. The other thing we can do is to think
of the regulated delta function. So we think of this as a potential that has this form. So here is v
of x, and here is x. And for this potential-- if you have a bound state-- how would the wave
function look?
Well, it would have to-- suppose you have a ground state-- it’s an even potential. The delta
function is even, too. It’s in the middle. It’s symmetric. There’s nothing asymmetric about the
delta function. So if it’s an even potential the ground states should be even, because the
ground state is supposed to have no nodes. And it’s supposed to be even if the potential is
even.
So how will it look? Well, it shouldn’t be decaying in this region. So, presumably, it decays
here. It decays there-- symmetrically. And in the middle it curves in the other direction. It is in
an allowed region-- and you remember that’s kind of allowed this way. So that’s probably the
way it looks.
Now, if that bound state exists, somehow, as I narrow this and go down-- as it becomes even
more narrow, very narrow now, but very deep. This region becomes smaller. And I would
pretty much expect the wave function to have a discontinuity.
You basically don’t have enough power to see the curving that is happening here. Especially
because the curving is going down. The distance is going down. So if this bound state exists,
as you approach the limit in which this becomes a delta function the energy moves a little, but
stays finite at some number. And the curvature that is created by the delta function is not
visible, and the thing looks just discontinuous in its derivative.
So this is an intuitive way to understand that the wave function we’re looking for is going to be
discontinuous on its derivative. Let’s write the differential equation, even though we’re still not
going to solve it. So what is the differential equation? Minus h squared over m, psi double
prime, is equal to E psi.
And, therefore-- and I write this, and you say, oh, what are you writing? I’m writing the
differential equation when x is different from 0. No potential when x is different from 0. So this
applies for positive x and negative x. It doesn’t apply at x equals 0. We’ll have to deal with that
later.
So then, no potential for x different from 0. And this differential equation becomes psi double
prime equals minus 2m e over h squared psi. And this is equal to kappa squ

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