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In the end of the previous video, I 
advertised to some degree, the main 

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result that we are going to derive. 
Namely, that the electrons near the 

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[INAUDIBLE] surface at low temperatures 
are going to form these balance states of 

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2 electrons which are very condensed, and 
form this superconductor. 

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And now, we're entering the technical 
part of the lecture, where we actually 

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would want to get this calculation done 
and to see how it actually happens. 

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And it connects somewhat to the 
calculations we were doing in the 

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previous lecture, where we were studying 
particles in various potential So there, 

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we were dealing with single particle 
quantum mechanics, so it was an 

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extraordinarily imposed potential in 
which a single particle was moving, or 

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which was localized there. 
Now, what we want to do, we want to study 

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actually a two-particle problem with two 
particles here describing effectively the 

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two electrons near the [UNKNOWN] service 
for the counting, in the counting of 

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super conductivity. 
And it turns out that actually two 

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particles problems often times, in 
quantum mechanics reviews the single 

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particle problems. 
And in this segment I'm going to 

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illustrate how exactly it happens. 
So here we're interested in analyzing 

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this two-particle Schrodinger equation 
and to make things general for the sake 

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of lectures later in the course, I'm 
considering here actually two different 

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particles with masses m1 and m2. 
But in the next segment we'll work with a 

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two-electron problem where one is 
actually equal to two. 

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So so what we have here is essentially 
kinetic energy of two particles so this 

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corresponds to b1 squared over 2m1 
kinetic energy of the first particle plus 

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b two square over 2m2 kinetic energy of 
the second particle. 

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And the interaction between them which we 
expect to depend only on the distance 

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between the particles. 
And now we have a wave function that 

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depends on both coordinates of the 
particle r1 and r2 and well, the right 

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hand side is the usual energy times of 
psi. 

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So we're going to solve this equation. 
Now, the goal of this, video is to show 

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that actually instead of solving this 
complicated, differential equation which 

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depends two vectors r one and r two we 
can actually reuse it. 

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To the good old single particle 
Schrodinger equations we really know how 

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to solve. 
And to see how it happens lets do the 

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following change of variables to go, we 
will go from the coordinates r1 and r2, 

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the actual coordinates of each individual 
particles, to this capital R and 

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uh,[UNKNOWN] are. 
So the former represents the center of 

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mass of our particles, so it's defined as 
so and the second one is the distance 

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between the particles. 
So the relative distance between 

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particles are one and are two. 
Now what we're going to do, we're 

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going to rewrite this equation in terms 
of this capital R and letter case R. 

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So in order to do this, we well, first we 
notice that the, of course the, potential 

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energy depends only on the distance, 
basically about the instruction, we've 

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chosen that, this way. 
And the only non trivial part, we have to 

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analyze is this kinetic energy. 
The derivatives. 

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This Laplacian which appear in this 
brackets. 

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And so we're going to work with this guys 
now. 

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So to calculate this part of kinitec 
energy, we can focus for the sake of 

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simplicity on just one component of this 
Laplacian. 

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So Laplacian is a sum of second 
derivatives to the respect of, with 

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respect to each other directions in our 
problem x, y, z so let's just focus this 

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one x. 
And in this case the capital is X is 

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going to be the center of mass x 
component and this is going to be a 

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relative coordinate in the X direction. 
Now in order to change the variables in 

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the Laplacian and the derivatives we can. 
go from the derivative with respect to X 

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one to a derivative with respect to the 
capital X and the lower case x. 

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And this transformation can be done by 
rewriting it soon. 

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So we basically add in some sense 
derivative in the numerator and 

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denominator and so this 
derivative[UNKNOWN] capital X over one is 

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calculated from here And derivative of d 
lowercase x over x1 is calculated from 

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here. 
And the first one is, as you can see, 

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uses m1 over the total mass and the 
second one is give us basically minus 

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sine. 
So this is the result. 

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of this derivative when expressed through 
capital s and small s. 

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We can do the same thing of course for x 
two and this is the result with the only 

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difference here being m two and here 
being the plus sign. 

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And that's, pretty much it. 
So we actually calculate the quantity 

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that we're after, written here. 
by, essentially squaring this result. 

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So this would give us the x 
particles[UNKNOWN] for the first particle 

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and this will, this guy squared will give 
us the, x component that applies into the 

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second particle. 
So therefore this guy goes here and you 

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know, this result goes here. 
And so you can see if we square this 

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operation differ it in all derivatives 
we're going to get now the second 

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derivative for this vector center of 
mass, the second derivative with respect 

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to the center of coordinate the, to the 
relative coordinate but also we're 

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going to get the cross terms and actually 
the coordinates were chosen in such a way 

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that the cross terms in these 2 brackets 
cancel each other out because of the 

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minus sign here and the plus sign here. 
So well you can verify this, of course, 

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just by writing it down, or just by 
staring it, staring at it for a minute or 

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so, but it, it's going to be pretty 
clear. 

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So in the, if you put everything together 
from these two brackets for a 

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straightforward calculation, this is what 
we're going to get. 

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So this is essentially a result for this, 
for the X component of this quantity we 

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are calculating. 
And so a Laplacian in terms of X1 and X2 

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can now be written essentially as a 
combination of 2 Laplacians with a Yeah, 

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there should be, of course, a plus sign 
here. 

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There's no minus sign so it was a typo. 
And so the first Laplacian with respect 

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to the center of mass involves the total 
mass in the denominator. 

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And the second applies within respect to 
the relative coordinate this guy, which 

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actually is called the reduced mass. 
So now the way we can write it as m1 m2 

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divided by m1 plus m2, and this is called 
reduced mass. 

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So, now we're in the position to actually 
put everything together. 

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So what we have proven, as a matter of 
fact, is that minus h squared over 2 m 1 

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Leplacian with respect to 1, the kinetic 
energy, the first particle minus h 

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squared over 2 m 2, leplacian of 2 the 
kinetic energy of the second particle. 

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Can be written as minus h squared over 2 
m 1 plus m 2. 

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the glassian with respect to the sensor 
of mass minus h squared over 2 mew the 

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reduced mass that I introduced in the 
previous. 

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Slide, La Plassian with respect to the 
relative coordinate. 

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If I stick it into the Schongerd 
equation, I will get therefore, the full 

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length, I will get minus h squared over 
2, this La Plassian of r and 1 plus m2 

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plus Laplacian lower case r. 
Over mu, plus the, potential energy and 

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the interaction energy between the two 
particles, which depends only on this 

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relative coordinate, acting on my wave 
function. 

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And the right-hand side is just the 
energy as usual. 

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So what I see from this equation and 
that, is that there is just one part 

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Which depends on the relative coordinate 
and which involves interaction. 

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And this part represents essentially a 
free Shrodinger equation for a combined 

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particle with the mass M1 plus M2. 
It essentially describes a free motion. 

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Of 2 particles as a whole. 
So for instance, if they were to form a 

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bounce state this exactly what we're 
going to be discussing in the last video 

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in this lecture. 
So this term essentially describes the 

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motion of this bounce state this sort of 
molecule like state as a whole. 

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And so to sort of extract this overall 
motion. 

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We can, just look for a solution. 
This psi tilda of capital R and, lower 

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case r can be written as a solution to 
this, sort of free Schrodinger equation, 

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which is simply a plane wave as we know. 
It's e to the power i. 

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Some momentum of this bound state or 
whatever it is. 

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That's capital r divided by h bar, times, 
well, the wave function, which depends on 

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the, relative coordinate that we actually 
want to find. 

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And, the energy of this, motion from here 
is going to be just this capital p. 

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squared divided by m one plus m two, 
twice and one plus m two. 

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So, which is just a parameter. 
It doesn't appear anywhere else in the 

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problem and we can just move it, to the, 
right hand side subtract the total energy 

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from the total energy of this. 
Energy. 

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That's it. 
So the resulting, the final result that 

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we're going to get, which doesn't lot 
include this theorem, is going to look, 

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if we just write it here, is going to 
look as so, so it's going to be minus 

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eight squared over two[UNKNOWN], so this 
is a reduced mass. 

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So this plus which I can write as D two 
over D R Squared plus v of r side of r. 

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And the left hand side is going to be 
this e minus p squared over 2 and 1 plus 

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m2. 
So let me just write it as e prime psi of 

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r. 
Okay and this as you can see looks like 

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Single particle Schrodinger equation. 
So, it depends only on, the wave function 

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depends only on one coordinate. 
Everything else is exactly the same. 

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The only difference is here I'm dealing 
with some reduced mass. 

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So, let's see in the context of the 
problem of two electrons if they're 

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actually going to start here. 
So m1 and so e, m1 is equal to m2. 

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So if m1 is equal to m2. 
So then fo course, the total mass is 

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equal to well, 2m. 
2, 2m. 

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And the reduced mass of this new Is equal 
to m squared over 2m, or just m over 2. 

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So therefore, in our electron problem, 
this guy is going to be just the electron 

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mass. 
So and just to summarize what we have 

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proven By this sort of brute force 
straight forward calculation is that in 

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this simple two particle problem, in two 
particle quantum mechanics, we can always 

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get treat of one sort of unnecessary 
quantum. 

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It describes a motion of our two 
particles as a whole And we can reduce 

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our problem to the standard 
single-particle quantum mechanics which 

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makes things much easier because we can 
now use the result and some of the 

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conclusions of single-particle quantum 
mechanics that we saw in the previous 

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lecture.