In the end of the previous video, I advertised to some degree, the main
result that we are going to derive. Namely, that the electrons near the
[INAUDIBLE] surface at low temperatures are going to form these balance states of
2 electrons which are very condensed, and form this superconductor.
And now, we're entering the technical part of the lecture, where we actually
would want to get this calculation done and to see how it actually happens.
And it connects somewhat to the calculations we were doing in the
previous lecture, where we were studying particles in various potential So there,
we were dealing with single particle quantum mechanics, so it was an
extraordinarily imposed potential in which a single particle was moving, or
which was localized there. Now, what we want to do, we want to study
actually a two-particle problem with two particles here describing effectively the
two electrons near the [UNKNOWN] service for the counting, in the counting of
super conductivity. And it turns out that actually two
particles problems often times, in quantum mechanics reviews the single
particle problems. And in this segment I'm going to
illustrate how exactly it happens. So here we're interested in analyzing
this two-particle Schrodinger equation and to make things general for the sake
of lectures later in the course, I'm considering here actually two different
particles with masses m1 and m2. But in the next segment we'll work with a
two-electron problem where one is actually equal to two.
So so what we have here is essentially kinetic energy of two particles so this
corresponds to b1 squared over 2m1 kinetic energy of the first particle plus
b two square over 2m2 kinetic energy of the second particle.
And the interaction between them which we expect to depend only on the distance
between the particles. And now we have a wave function that
depends on both coordinates of the particle r1 and r2 and well, the right
hand side is the usual energy times of psi.
So we're going to solve this equation. Now, the goal of this, video is to show
that actually instead of solving this complicated, differential equation which
depends two vectors r one and r two we can actually reuse it.
To the good old single particle Schrodinger equations we really know how
to solve. And to see how it happens lets do the
following change of variables to go, we will go from the coordinates r1 and r2,
the actual coordinates of each individual particles, to this capital R and
uh,[UNKNOWN] are. So the former represents the center of
mass of our particles, so it's defined as so and the second one is the distance
between the particles. So the relative distance between
particles are one and are two. Now what we're going to do, we're
going to rewrite this equation in terms of this capital R and letter case R.
So in order to do this, we well, first we notice that the, of course the, potential
energy depends only on the distance, basically about the instruction, we've
chosen that, this way. And the only non trivial part, we have to
analyze is this kinetic energy. The derivatives.
This Laplacian which appear in this brackets.
And so we're going to work with this guys now.
So to calculate this part of kinitec energy, we can focus for the sake of
simplicity on just one component of this Laplacian.
So Laplacian is a sum of second derivatives to the respect of, with
respect to each other directions in our problem x, y, z so let's just focus this
one x. And in this case the capital is X is
going to be the center of mass x component and this is going to be a
relative coordinate in the X direction. Now in order to change the variables in
the Laplacian and the derivatives we can. go from the derivative with respect to X
one to a derivative with respect to the capital X and the lower case x.
And this transformation can be done by rewriting it soon.
So we basically add in some sense derivative in the numerator and
denominator and so this derivative[UNKNOWN] capital X over one is
calculated from here And derivative of d lowercase x over x1 is calculated from
here. And the first one is, as you can see,
uses m1 over the total mass and the second one is give us basically minus
sine. So this is the result.
of this derivative when expressed through capital s and small s.
We can do the same thing of course for x two and this is the result with the only
difference here being m two and here being the plus sign.
And that's, pretty much it. So we actually calculate the quantity
that we're after, written here. by, essentially squaring this result.
So this would give us the x particles[UNKNOWN] for the first particle
and this will, this guy squared will give us the, x component that applies into the
second particle. So therefore this guy goes here and you
know, this result goes here. And so you can see if we square this
operation differ it in all derivatives we're going to get now the second
derivative for this vector center of mass, the second derivative with respect
to the center of coordinate the, to the relative coordinate but also we're
going to get the cross terms and actually the coordinates were chosen in such a way
that the cross terms in these 2 brackets cancel each other out because of the
minus sign here and the plus sign here. So well you can verify this, of course,
just by writing it down, or just by staring it, staring at it for a minute or
so, but it, it's going to be pretty clear.
So in the, if you put everything together from these two brackets for a
straightforward calculation, this is what we're going to get.
So this is essentially a result for this, for the X component of this quantity we
are calculating. And so a Laplacian in terms of X1 and X2
can now be written essentially as a combination of 2 Laplacians with a Yeah,
there should be, of course, a plus sign here.
There's no minus sign so it was a typo. And so the first Laplacian with respect
to the center of mass involves the total mass in the denominator.
And the second applies within respect to the relative coordinate this guy, which
actually is called the reduced mass. So now the way we can write it as m1 m2
divided by m1 plus m2, and this is called reduced mass.
So, now we're in the position to actually put everything together.
So what we have proven, as a matter of fact, is that minus h squared over 2 m 1
Leplacian with respect to 1, the kinetic energy, the first particle minus h
squared over 2 m 2, leplacian of 2 the kinetic energy of the second particle.
Can be written as minus h squared over 2 m 1 plus m 2.
the glassian with respect to the sensor of mass minus h squared over 2 mew the
reduced mass that I introduced in the previous.
Slide, La Plassian with respect to the relative coordinate.
If I stick it into the Schongerd equation, I will get therefore, the full
length, I will get minus h squared over 2, this La Plassian of r and 1 plus m2
plus Laplacian lower case r. Over mu, plus the, potential energy and
the interaction energy between the two particles, which depends only on this
relative coordinate, acting on my wave function.
And the right-hand side is just the energy as usual.
So what I see from this equation and that, is that there is just one part
Which depends on the relative coordinate and which involves interaction.
And this part represents essentially a free Shrodinger equation for a combined
particle with the mass M1 plus M2. It essentially describes a free motion.
Of 2 particles as a whole. So for instance, if they were to form a
bounce state this exactly what we're going to be discussing in the last video
in this lecture. So this term essentially describes the
motion of this bounce state this sort of molecule like state as a whole.
And so to sort of extract this overall motion.
We can, just look for a solution. This psi tilda of capital R and, lower
case r can be written as a solution to this, sort of free Schrodinger equation,
which is simply a plane wave as we know. It's e to the power i.
Some momentum of this bound state or whatever it is.
That's capital r divided by h bar, times, well, the wave function, which depends on
the, relative coordinate that we actually want to find.
And, the energy of this, motion from here is going to be just this capital p.
squared divided by m one plus m two, twice and one plus m two.
So, which is just a parameter. It doesn't appear anywhere else in the
problem and we can just move it, to the, right hand side subtract the total energy
from the total energy of this. Energy.
That's it. So the resulting, the final result that
we're going to get, which doesn't lot include this theorem, is going to look,
if we just write it here, is going to look as so, so it's going to be minus
eight squared over two[UNKNOWN], so this is a reduced mass.
So this plus which I can write as D two over D R Squared plus v of r side of r.
And the left hand side is going to be this e minus p squared over 2 and 1 plus
m2. So let me just write it as e prime psi of
r. Okay and this as you can see looks like
Single particle Schrodinger equation. So, it depends only on, the wave function
depends only on one coordinate. Everything else is exactly the same.
The only difference is here I'm dealing with some reduced mass.
So, let's see in the context of the problem of two electrons if they're
actually going to start here. So m1 and so e, m1 is equal to m2.
So if m1 is equal to m2. So then fo course, the total mass is
equal to well, 2m. 2, 2m.
And the reduced mass of this new Is equal to m squared over 2m, or just m over 2.
So therefore, in our electron problem, this guy is going to be just the electron
mass. So and just to summarize what we have
proven By this sort of brute force straight forward calculation is that in
this simple two particle problem, in two particle quantum mechanics, we can always
get treat of one sort of unnecessary quantum.
It describes a motion of our two particles as a whole And we can reduce
our problem to the standard single-particle quantum mechanics which
makes things much easier because we can now use the result and some of the
conclusions of single-particle quantum mechanics that we saw in the previous
lecture.