So now, we're going to address the same 
question we addressed in the previous 
video. 
namely the question about the existence 
of a bound state in an arbitrarily weak 
shallow potential wells but now will do 
so in more than one dimension. 
I should mention that there are actually 
very few problems in quantum mechanics in 
2D and 3D. 
We should allow relatively simple 
analytical solution. 
Most often in these cases we have to 
resolve to either numerical analysis or 
rather complicated, technically 
complicated mathematical calculations. 
Now, the solution I am going to present 
is not exactly trivial but will require a 
lot of thinking. 
But otherwise, it will follow almost one 
to one the steps that we used in the 
solution of the one dimensional 
delta-potential. 
however very importantly, the result of 
this steps is going to be quite different 
in 2D and 3D. 
So namely for those of you who want to 
skip the technical part and just to get 
through the bottom lines, I will just 
tell you the results right away. 
So unlike in 1 dimension, where the 
energy of a bound state, first of all the 
bound state always exists and the energy 
of the bound state in the weak potential 
scales sort of as a power law. 
The second power law of the strength of 
the potential, in 2 dimensions, it turns 
out that the level also always exists. 
But the energy of the level is extremely 
extremely small. 
The level is very shallow and the energy 
of the level scales exponentially in a 
very unusual form with the strength of 
the potential. 
And this actually has very important 
consequences in many cases and in 
particular and it release to certain 
calculation we're going to discuss in the 
theory of super conductivity a little 
later. 
And also in three dimensions it turns out 
that there is no level at all so if you 
have a very weak attraction between 
particles. 
For example, a very weak potential well, 
there is, so these attractions, these 
weak attractions are not going to induce 
a bond state and this fact is also very 
important. 
Now, let me prove the statements I just 
made, and the proof, the solution is 
going to follow very closely as I already 
mentioned, the solution that we had in 
one dimension. 
We started from this Schrodinger equation 
for the delta potential, so now of course 
we have to deal with this Schrodinger 
equation, in higher dimension, where wave 
function now is the wave function, is a 
function of a vector r. 
And the kinetic energy now involves this 
low placement , and, and counters to this 
derivative over x, and finally the delta 
function is the delta function is sort of 
a sharply big function in thee d 
dimensional space. 
So let's say in 3D, so, this delta so far 
is simply the product of the delta 
function facts, delta function of y, and 
delta function of z. 
So the first step in the solution, just 
like in 1 z, involves retransforming our 
equation. 
So essentially we go from the way 
function and the real space to the image 
of the way function momental space or 
more precisely here is a function of the 
way key. 
So the only difference in entire 
dimension is that the Fourier transform 
is slightly more complicated so it 
involves Fourier transforms along. 
Each of the directions, x, y, z, 
etcetera, so if you're in higher 
dimensions. 
And so if we perform this slightly more 
complicated Fourier transform so the 
kinetic energy which is basically this 
derivative of[INAUDIBLE] becomes minus k 
squared. 
The delta function, the Fourier image of 
the delta function, just like the one I 
mentioned is simply 1. 
And so we get to the following equation, 
which is just the linear equation before 
the image of the wave function. 
And as you see there's essentially no 
difference between one dimensional keys 
and D dimensional keys apart from from 
the fact that this guy now is the reactor 
and while this guy was just one 
dimensional component of the wave effort. 
And so again we can of course solve this 
equation for psi tilde of k. 
So we're going to have the same psi of 0 
divided by h squared. 
K squared over 2m and I will write it as 
plus absolute value of the energy. 
Again, we are looking for a bound state, 
and we know that the bound state energy 
must be negative, if it exists at all. 
And using the exactly same trick as 
before, we're going to integrate this 
equation, both sides of this equation, 
now, over the entire D dimensional space 
of this wave action. 
So if we do so, we, come up with the self 
consistency equation which again in one d 
and in higher dimensions look almost the 
same. 
So the only difference is that the 
integral here is going to over a higher 
dimensional space. 
So it turns out that this difference the 
fact that we now integrating over high 
dimension space lead a very important 
consequences for the actual solution for 
this equation. 
Here, we knew that doesn't matter how 
small off way is as long as it's non 
zero, we can always find an energy that 
will satisfy this equation. 
So now, it' turns out that higher 
dimension is not going to happen, 
necessary for d greater than 2. 
There will be no solution. 
So to see this, we actually have to 
calculate this integral that appears in 
the cell consistency equation that we 
just derived. 
So to do so we should first know is that 
the function that we're actually 
integrating, does not depend on the 
directions of the wave action It only 
depends on this absolute value. 
So let's see in two dimensions we can 
write d 2 k as a d k times k which is 
essentially corresponds to an integrating 
over the absolute value of k. 
Times d 5 sub k, which is the angle that 
the vector k forms with the x direction. 
So let's see, this is our two dimensional 
k space, this would by k sub y, this 
would be k sub x. 
And so this is our vector k, so this 
would be the angle 5 sub k. 
So, and what I'm saying here is that the 
equation, this equation that we are 
looking at, so can be written as alpha 
d2k, in this case, over two pi squared. 
so this function, and we can write it as 
simply, well, alpha over 2 pi squared. 
Oops. 
Squared is here. 
So dk times k over this function are from 
zero to infinity. 
So we integrate all the possible reactors 
and then the integral over the angles but 
nothing dependent then , so this integral 
will simply give us the y. 
And so in 3D, we can do the same thing so 
in the new dimension, we can do this 
procedure. 
We can sort of integrate the end of those 
and just separate the integral over, the, 
magnitude of k. 
So let me rather for my, self consistency 
equation in this form. 
I will write it as a new dimension. 
It's alpha s sub d minus one, which I 
will, 
Explain what I mean by that. 
two pi, to the power of d. 
So in here I'm going to write it as d k 
to the power of g minus 1 h square k 
square over 2m plus absolute value of the 
energy. 
So this s sub d minus 1 is the integral 
that comes out of the angles. 
So in two dimensions, it will simply be 2 
pi. 
Okay? 
So, and I wrote, in this forum because it 
actually corresponds sort of to a D minus 
1 dimensional theories. 
In this case, 2 dimensions would be 
basically integrating over this circle, 
in 3 dimensions, it will be integrating 
over this solid angles that cover the 
entire sphere. 
So let's say as as 1 is equal to 2pi, 
which just found out as, s2 is going to 
be equal to 4pi. 
Well we can formally introduce as 0, 
which is just 2 et cetera. 
So many dimension in principle can just 
calculated. 
So for practical purposes, we don't 
really need to know anything about these 
three numbers because in real life we 
don't deal with anything about 3 
dimensions, 2 dimensions or 1 dimension. 
Now, the problem we have that appears 
here in, already in 2 dimensions is that 
well when we integrate let's see. 
From zero to infinity, so the integral is 
actually diversion at large k. 
So let's see if these equal to 2 and 
especially if these equal to 3, so the 
higher up we go, the more diversion it 
becomes as we go to larger k. 
So this integral is actually equal to 
infinity. 
So, if we look at it sort of without 
thinking. 
And well, when there is infinity so we 
have to think about why it appears. 
So remember that actually this constant 
alpha came out, well we are solving 
essentially The problem for delta 
potential. 
So our v of r is alpha times delta of r 
and delta function there are the delta 
function is a very artificial function. 
It was basically introduced to model a 
shallow potential, a physical potential 
that has so has a certain with, with 
equality a and it has a certain depth. 
So which means which will be called, you 
know, u naught. 
So, and alpha is of the order of u naught 
e to the power of d, where d is the 
dimentionality of our space. 
But in any case, so this delta function 
potential is sort of introduce, represent 
real physical potential, that exists in 
reality, which has, actually, a finite 
width and finite depth. 
So when we go to a large key, large wave 
vectors, actually the large wave vectors 
respond to small distances. 
And if we, consider and if we, if we 
encounter a divergence at large K which 
we do here. 
So it means that something bad happens in 
small smaller, small distances. 
And so this something bad means well I 
mean there is no really, there are no 
real delta potentials in nature. 
Okay so all potentials that have a finite 
radiance. 
And so we can copy all of this in some 
sense and physical diversions by 
introducing a parameter 1 over e, where e 
is the radius of our real physical 
potential that we model with the delta 
potential. 
So this is actually a sort of a solution 
to this otherwise puzzle because 
otherwise, it would get infinity and it 
shouldn't get infinity. 
So basically what I'm saying here is that 
if you were to put a real potential in 
the very beginning, and do everything 
properly then this cut would appear 
actually naturally in our calculations 
but here we would just introduce it by 
hand. 
So now we're entering a, a final part of 
our derivation where we will rely on the 
integral we just derived. 
So here, there are no more angle angular 
components of the wave vector. 
There's just one integral involved. 
And we cut it off at the 1 over a. 
So and also, we modified it a little bit, 
so there's 1 over alpha. 
It's not a. 
It's alpha is the strength of the 
potential in the left hand side. 
So again, the main question we've been 
actually asking is whether there is a 
bound state in a weak potential. 
That is, in a potential where, where 
alpha goes to zero, and therefore, if 
alpha goes to zero, 1 over alpha goes to 
infinity. 
And so the question that we are now 
asking is whether we can find a solution 
to this equation when the left-hand side 
is very large. 
So you may ask me, you know, you just got 
rid of infinity , so what are you doing? 
You know? 
This is this is really would be 
reasonable questions let's say from a 
mathematician. 
But you know, this calculation actually 
illustrates in a very nice way the 
difference between doing mathematically 
calculation and physics and math. 
So well, the infinity I just got in rid 
off was an unphysical infinity, it's not 
real. 
It was sort of an artifact of our model. 
And also this infinity was not dependent 
on the energy. 
So whether or not we put energy to 0 or 
to for something finite, you know, these 
so called ultraviolet diversions would 
have existed. 
And so we cut it off, we sort of 
remembered where these things came from 
and got rid of these errors. 
Now, the question is whether we can have, 
and on arbitrary, large integral by 
adjusting the value of the energy, okay. 
And so here is basically this integral, 
so I don't write here at this overall 
coefficient because it's just a number. 
It can't be made large or small, it is 
just a number assigned. 
2D is 1 over 2 pi and 3D is 1 over 2 pi 
squared, it's just, because it's 
unimportant. 
So, how do we make how do we answer this 
question. 
So, if we look at it we see that the 
energy, absolute value of energy appears 
in the denominator, okay? 
So the larger the energy, the smaller 
this integral. 
It's very clear cut tendency. 
So these integral is the largest when 
energy's equal to zero. 
So therefore, the questions we're really 
asking is whether we can, you know, if we 
set this energy to zero, and the 
remaining integral will be k to the power 
of d minus 1 divided by k squared, or 
just simply this. 
So, the question we're asking whether the 
integral below can diverge. 
So we, we got rid of divergence in the 
previous slide. 
In this slide, we're asking whether we 
can get it back. 
But the important thing here is that we 
want to get it back not from a large 
momentum but from small momentum. 
So the latter is called infrared 
divergence. 
Okay. 
And in physics, for those of you who will 
be thinking about becoming theoretical 
physicists. 
You may remember that infrared diversions 
is if you find something like that well, 
either you made the mistake or you should 
if it's new, you should think about 
number of parameters. 
Okay, so it's very important. 
Infrared diversion is real unlike 
ultraviolet diversions which are sort of 
artificial. 
So and, this boils down basically this 
question. 
Now, clearly if if let's say d is equal 
to 3, so the dimension of space is equal 
to 3, the space in which we live. 
S then we simply have a integral from 0 
to 1 or a, or where d k is finite, there 
is no diversions. 
And therefore the answer to the question 
is the question of existence of the bound 
state is that in 3D there is no bound 
state. 
Okay, so if alpha is sufficiently small 
the right hand side can't possibly 
compete with the large value in the left 
hand side. 
But if we go to the lower dimension and 
among the reasonable dimensions 2 comes 
to mind. 
So unless we consider some fractals or 
whatever, so that's the only thing we can 
have 1 and 2. 
So in 2 dimensions, this integral becomes 
an integral from 0 to 1 over a, dk over 
k. 
And we get a very weak simple logarithmic 
divergence so at the lower limit we're 
going to get 1 over 0. 
So it's equal to infinity. 
Okay? 
So the Integral divergence, which is 
great. 
It means that we have a bound state. 
So this really the summary of what we 
learned from just simply considering this 
integral to which we simplified our 
problem. 
So, if our dimensionality is smaller or 
equal to 2, so the integral diversions 
are physical. 
It means that there exists always exists 
a bond state, doesn't matter how weak the 
attraction is. 
So otherwise the integral is finite and 
they're no bond states. 
So even though there's a well there's 
nothing in there. 
There's nothing in this well. 
So the last thing I'm going to do is that 
I'll estimate the value of the energy of 
the bond. 
State in two dimensions. 
So we just found there is one, so let's 
estimate its energy. 
And of course, for this I have to solve 
this equation, specifically in 2D. 
So, let me do so, let me just calculate 
this integral, actually estimate this 
integral. 
I can, of course, calculate it exactly. 
But let me estimate it, it's easier. 
So in the first thing I'm going to do, 
I'm going to factor out this coefficient 
here in front of the integral. 
I'm going to have m over pi a squared. 
And so I will get an integral of 0 to 1 
over a, and let me write it in this form, 
d k times k. 
And so here I'm going to have k squared 
plus square root of 2 m e over h squared. 
So I just wrote in this form, you will 
understand why. 
So if I set energy to 0 I have the 
logarithmic diversions that well, as we 
know allows me allows the existence of 
the of the bounce state. 
Now but, it is finite so these diversions 
disappear. 
So since the smaller the e, the larger 
the diversion. 
So now to estimate the value of the 
integral, I can simply move this sort of 
cutoff of these diversions from the 
denominator to the lower limit of 
integration. 
This method is called well this is 
asymptotic method is called logarithmic 
accuracy. 
So if we have integrals of this type with 
logarithmic diversions, if we don't want 
to perform exact calculations. 
Sometimes it's not possible so one can 
just do this trick and this provides a 
reasonable estimate. 
So if we do that. 
So what we're going to get is 
approximately, let me write it this way, 
approximately m over pi h squared in 
integral, now from this 2 m e, h squared, 
1 over e. 
And so here is I will call simply dk over 
k and so this of course is a logarithm so 
the result is then m over pi h squared 
log of 1 over a the upper limit times h 
over 2me, the lower limit. 
And so now I am in the position to find 
the energy so I can simply solve this 
equation for e by exponentiating both 
sides of this equation which gets results 
of the logarithm. 
And so the result is going to be so the 
energy or the absolute value of the 
energy is going to be, let me write out 
the order, a squared over m a squared 
exponential of, two pi h squared divided 
by alpha with a minus. 
So this is really the final result. 
And the result is very interesting. 
So for a theoretical physicist at least, 
because the function that we got is a 
very strange function. 
So remember that this parameter alpha 
controls the strength of the potential. 
And, we, consider the case when it's very 
small. 
So, e to the power minus 1 over alpha is 
really exponentially small, very shallow 
level and already shallow potential. 
But the function is strange because well 
if you write it, either to the power 
minus Something over alpha. 
So this function does not have a Taylor 
Expansion. 
So remember that, you know? 
If we have a, function, f of alpha. 
Normally, at small alpha, we can 
approximate it by f of 0. 
Plus, you know, f prime of zero, times 
alpha plus f2 prime subzero over 2. 
alpha square root, et cetera. 
So this is a regular Taylor expansion of 
a function. 
So in this case, all the derivatives, all 
the terms in this expansion are equal to 
0. 
And this actually would have pose the 
problem if we had try to solve the 
problem using so called petrubation 
theory which is[UNKNOWN]. 
Because oftentimes, it gives us access to 
solutions that are otherwise impossible 
to get. 
So we start with a normal solution and 
then do a Taylor expansion in the 
vicinity of the solution. 
So here, it wouldn't have been 
impossible. 
Exactly because of these weird properties 
of the function exists, we can plot it. 
We can look at it, but there is no Taylor 
expansion at all. 
So -- and it turns out that in the second 
lecture this week, we're going to study 
some basics about the theory of 
superconductivity. 
And there is a classic formula in this 
Nobel Prize-winning 
Barneen-Cooper-Schrieffer theory of 
superconductivity, which features this 
function. 
This sort of function and actually the 
reason, it took people, as, as I will 
talk about in the next lecture, the 
reason it took people almost 50 years to 
come up with a solution to the problem of 
superconductivity is partially because of 
this circumstance. 
That the result actually is something 
that can only be obtained by an exact 
non-interpretive calculation but it 
couldn't be obtained using the 
Perturbation theory.