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.