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.