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.