In the previous video we saw that with 
the shallow quantum well, can be 
faithfully represented, as an attractive 
derived delta potential. 
and, in this video we are going to solve 
the Schrodinger equation, with a delta 
potential in an arbitrary dimension 1, 2, 
or 3. 
And we're going to use a very powerful 
method, mathematical method, that, as a 
matter of fact, was used by Leon Cooper 
to solve the key problem in the theory of 
superconductivity, and the problem of 
electron pairing, which brought him 
eventually a Nobel prize and now this 
phenomenon of pairing is now called 
Cooper pairing. 
But well, we'll talk a little bit about 
this later but at this stage we're going 
to follow on this single particle 
Schrodinger equation with this sort of 
short range attraction. 
And, so before going through the 
technicalities I want to give you the 
bottom line, the main result if you sort 
of get lost in these technical details. 
So please I would like you to know. 
the very important outcome of this 
compilation we’re going to see, and so 
the outcome would be the following. 
So we will see that in one and two 
dimensions, any, it doesn't matter how 
weak an attractive potential gives rise 
to bounce the, of the quantum particle. 
So it doesn’t matter how shallow a well 
we have, it always will bind the 
particle. 
But in three dimensions, it turns out 
that if you have a very weak potential, 
very shallow potential, it's not enough 
to give the particle, to localize the 
particle. 
And this difference qualitative 
difference actually have a very important 
consequences in various fields of 
physics, so you may want to remember 
that. 
But now, let me go into the technical 
part. 
Okay? 
So, and I will start with a certain math 
reminder, which basically will remind you 
what a Fourier transform is. 
So, we can solve the problem, we will 
solve the problem using only the Fourier 
transform. 
There will be nothing else involved 
essentially in this solution. 
Fourier transform, and inverse Fourier 
transform. 
So what is a Fourier transform? 
So if you have an arbitrary function, 
well reasonable function g of x, in our 
case it will be way function and various 
derivatives of it. 
So the Fourier transform of this function 
g of x, which we'll call f of g of x, so 
g of ke is an integral of g of x with 
this exponential. 
So physically this k in our case, will 
represent the wave vector, and h times k 
is going to be the particles momentum. 
But in principle you can view this wave 
transform as a mathematical procedure. 
An important fact here is that if you, if 
you know the free transformer of a 
function, you can restore the function 
itself by doing the inverse free 
transform. 
so g of x is equal to an integral of 
there should be a g tilde of k, either 
the power minus ikx dk over 2 pi. 
So I should remind you or emphasize that 
where to put this 2 pi is actually a 
matter of convention or convenience so we 
could have put the 2 pi in the Fourier 
transform or we could have put, we could 
have separated them sort of like this. 
So we could have put a 1 over square root 
of 2 pi here, 1 over square root of 2 pi 
here or any other combination of powers 
so that the total power of 2 pi in the 
denominator is equal to 1. 
And this is a matter of convention in 
quantum mechanics the convention is that 
you know two pi appears in this inverse 
reaction which, which features the wave 
transfer or the momentum so I'll erase 
this off here. 
So anyway, so this what we know now, the 
simplest perhaps example of Fourier 
transform will be exactly Fourier 
transform of the delta function. 
So if we want the Fourier transform, the 
delta function, we know that the delta 
function when integrated in infinity is 
just peaks the value of the function is 
being integrated with at x equals 0. 
So, and if we integrate delta of x with 
the e to the power of ikx from minus 
infinity to plus infinity, we simply will 
get 1, because e exponent of 0 is equal 
to 1. 
So therefore, the Fourier image of the 
delta function is 1. 
Just see the identity in this case base. 
So and we can also do the inverse Fourier 
transform, which basically gives us the 
representation of the delta function. 
It's an integral who minus infinity to 
plus infinity is exponential dk over 2 
pi. 
We have seen it already, actually, 
before. 
So in this identities, actually will be 
enough again for us to solve the problem. 
So, now the problem that we actually want 
to solve mathematically is the 
Schrodinger equation which is presented 
here. 
So the left hand side of this equation 
here represents the ah,[INAUDIBLE] so the 
[INAUDIBLE] in this part is the kinetic 
energy, the usual kinetic energy written 
explicitly in the quardenential 
presentation, so that's just p square 
over 2m. 
And this guy is our potential, delta 
potential. 
So later on we're going to generalize it 
to arbitrary dimension, but at this stage 
I'm just doing one dimension for 
simplicity. 
So of course we're dealing now with time 
and temperature in your equation. 
And our, the goal of this calculation 
basically or the goal of our evaluation 
would be to determine the value of the 
energy, and in particular we will, we 
will, we would want to figure out whether 
or not there are bound states, with 
negative energy. 
So essentially if we want to sort of draw 
symbolically the potential that we're 
dealing with so it's zero everywhere but 
in the vicinity of x equals zero there is 
a sharp sort of well and the question is 
whether there is a bound state or bound 
states in this well with some negative 
energy So this is basically the question. 
And so now, let me do the actual 
calculation using this reminder in the 
Schrodinger equation. 
So what we're going to do, we're going to 
represent this f of x as a Fourier 
transform of the wave function in 
momentum representation. 
So t's going to be an integral of gk over 
2 pi of sound side tilde sub k. 
I'm going to write it, it's from minus 
ikx from minus infinity to plus infinity, 
and just for the sake of gravity, I'm 
going to call this integral I saw so that 
I don't have to write the standard 
factors and limits again. 
So it's identical equal in my notations 
to this. 
So there are three types of terms that 
appear in this equation, so the way 
function itself appears in the right hand 
side, multiply by the energy, in the left 
and side, what we have for instance this 
guy which is the second derivative of the 
delta Psi double front. 
So if we differentiate this free 
transform twice over the coordinate, 
we're going to pull out the, so the 
beginning of the exponential is the 
exponential itself, but they're going to 
pull out this minus i t squared. 
So therefore the we can represent this as 
a derivative as psi tilde sub k times 
minus ik squared, e to the power minus 
ikx. 
So this guy is the Fria transform of psi 
double prime. 
And finally we also have this term, so 
delta of x, of psi of x. 
But since delta of x is a sharply peaked 
function of x equals 0, so it essentially 
pulls out only the value of the wave 
function of x equals 0. 
So all other values of this wave function 
is In this product it basically don't 
matter. 
So in this theorem is going to be 
identical and equal to psi of 0, delta of 
x. 
Okay? 
And delta of x, we know that the delta of 
x, the Fourier image of delta of x as we 
discussed in the previous the Slight this 
is simply equal to 1. 
So we can represent this as of size 0 and 
integral over k again is implied that is 
for minus infinity to plus infinity e to 
the minus ikx that's it. 
So this is, this is the basically the way 
we can write this 
In this situation. 
And so, if you put everything together, 
so we're going to have, in the left hand 
side, we're going to have minus h squared 
over 2 m, psi tilde sub k minus i k 
squared, e to the bar minus i k x. 
So this term, the second term is going to 
be minus alpha Size of 0 which here is 
simply a constant. 
So we have pulled out the value of the 
wave function x equal 0 is just a 
constant and here we have this integral e 
to the power minus ikx. 
In the right hand side we're going to 
have simply a energy times the wave 
function. 
Coming from here, k psi tilde is to the 
power minus ks. 
So we see that in all three terms we have 
this integrals and we have this 
exponential so that we can simply focus 
on the overall coefficients that 
multiplies all this exponential and 
essentially get what we can call it. 
Schrodinger equation in a momentum. 
So in some sense, we can just get rid of 
these integrals and let me have some 
space here. 
Let me erase most of this stuff. 
So as a result, if we basically look at 
this equation without the integrals for a 
particular key. 
So this term, the first term on the 
left-hand side, is going to give us now 
plus h squared over 2m. 
k squared psi tilde of k. 
the second term is going to give us minus 
alpha times side of 0. 
So this psi of 0 is 
[UNKNOWN]. 
And the right hand side is going to be a 
the energy that we're looking for so this 
is our unknown, this is what we're 
looking for, psi tilde of k . 
So we see that instead of solving the 
complicated differential equation as it 
looked in the beginning, we've now 
reduced the problem to solve essentially. 
A linear equation that, we know how to do 
from high school math. 
So we simply can put this psi tilde in 
the left hand side, this guy in the right 
hand side and so we're going to have h 
squared k squared over 2m minus energy 
times psi tilde. 
So these two guys. 
And in the right hand side we're going to 
have plus alpha psi of 0. 
So, the way function, the other unknown, 
the way function therefore is simply, 
equal to alpha side of zero way function 
in real space divided by h squared t 
square over two m minus energy. 
And, well in some sense this is already, 
partially, solution to our problem. 
It's a solution in the sense that it 
determines the wave function in momentum 
space and if we want for instance to find 
the wave function in real space, all we 
have to do is just inverse Fourier 
transform it back to x. 
But our main interest again is to answer 
the question of whether or not there are 
bounce states in this potential. 
And, so this, well I just copied here, 
the same equation we derived previously. 
So emphasize that this is really what 
we're after here. 
So, the question is, are there any 
available energy levels, leveled in this 
shallow quantum well, which responds to 
particle beam localized in the vicinity 
of the So it turns out that this question 
can be answered from this equation in 
just two simple steps. 
And those of you like mathematical 
thought puzzles, sort of, sort of 
aficionados of, of you know, these 
puzzles, may just want to stop the 
presentation and stare at this equation. 
And try to come up with a trick that will 
allow you to figure out this enumeration. 
But if you just want to know the trick, 
let me show you how it works. 
And it's actually very neat, it doesn't 
require any super complicated 
calculations. 
So this stage, let me recall that psi of 
x basically the can be represented as a 
the inverse Fourier transform of psi 
tilde of k, e to the power minus ikx. 
So therefore psi of 0, which appears in 
the right hand side of this equation. 
Psi of 0 corresponds simply to setting x 
to 0 in this equation. 
So it's going to be an integral of all, 
over all k over 2 pi of this psi tilde of 
k. 
So let me apply this integral in both 
sides of this equation. 
So if I have an equation, I can take 
integral of both sides of this equation. 
And if I do so in there, left hand side, 
I'm going to have psi tilde of k which 
per this relation, gives me simply psi of 
0. 
While in the right-hand side I will have 
this coefficient, which is simply a 
constant alpha psi of 0, an integral, 
over k 1 over h squared, k squared over 
2m minus epsilon. 
So the neat thing that happens here is 
that this, we see that there is a psi of 
zero that we don't really know at this 
stage. 
But this sign of zero appears in both 
sides of this equation so we can simply 
cancel it. 
And so we are where we end up with an 
equation that doesn't have the wave 
function anymore. 
So it has only one unknown and this 
unknown is exactly the energy we are 
after. 
So, if we can solve our equation and find 
the value of the energy, this would imply 
that there exists a bound state in our 
potential. 
And remember that the bound state has 
negative energy so it must be below the 
continuum. 
And this would correspond sort of to 
localize particles. 
So we can simply let us say models here 
and the plus sign. 
So this would correspond to our image 
we're looking for and also if we want 
sort of restore normal form of the 
integral, so we can 
Recall that my notation was simply to 
save some space. 
So this is really what, what we have. 
So let me skip the remaining technical 
step which essentially boils down just to 
copulating this integral which is an 
elementary integral. 
And just present the result. 
And so here I sort of rewritten this 
equation in 
Sort of slightly different form and this 
is the result of the calculations. 
If you calculate the integral, you will 
see that there is, exists one and only 
one solution for the energy. 
And the energy is equal to minus m alpha 
squared over 2h squared. 
So alpha here, remember, alpha here is 
the parameter. 
Which multiplies the delta potential. 
So, in some sense, alpha controls the 
strength of the potential. 
The smaller the alpha, the weaker the 
potential that acts on the particle. 
And the result here is that, in the 
shallow quantum well, in one dimension, 
or in the corresponding delta potential 
well, there exists one, and only one, 
bound state with this energy, which 
scales so/g, with the strength of the 
potential. 
So in the second part of this video we're 
going to discuss how this result is 
modified in a very essential way if we go 
to more physical dimensions to 2 and 3 
dimensions.