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BARTON ZWIEBACH:
Hydrogen atom is

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the beginning of our analysis.

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It still won't solve
differential equations,

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but we will now two particles,
a proton, whose coordinates are

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going to be coordinate
of the proton,

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subbing Xp for the proton,
and momentum of the proton.

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And there's an electron.

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And there's the coordinates,
the three coordinates

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of the electron, and
the three components

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of the momenta of the electron.

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And these are your
canonical variables.

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This means that the
components of this object

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satisfy the standard
commutation relations.

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That is-- I have to write
it like the following.

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They considered the
coordinates of the proton,

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the i-th component.

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And the momentum of the proton,
the j-th component, that's

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equal to ih bar delta ij.

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You see, we used to code for
its x, y, and z, and momenta

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Px, Py, Pz.

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You could've called it X1,
X2, X3, momenta P1, P2, P3.

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And in that way, you can use
a parameter delta over here.

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So, these are the
commutation relations

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of X's and P's, but
X and P for a proton.

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So, the proton has
its X, has its P,

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and it behaves like an X
and P that we've studied.

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The electron has its X, its P,
and also behaves the same way

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as coordinates P of electron
j equal ih bar delta ij.

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You see, this is
because x operator, y

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operator, z operator--
you can call them

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xi operators with 1, 2, 3.

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The Px, Py, Pz operators
are better called P sub i,

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with i running from 1, 2,
and 3, and x being the first,

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y the second, z being the last.

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Then the fact that the x
just fails to commute with Px

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and y fails to commute with
Py, and z fails to commute

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with Pz is xi pj
equal ih bar delta ij.

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And this is what I'm saying
here with this notation

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because there are
too many subscripts.

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There's a P for
proton, but there's

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an i for the first, second
and third component.

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And there's the
momentum of proton,

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which has 1, 2, 3 components.

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So, those are our
dynamical variables.

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And then when you
try to solve this,

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you have the issue
of wave functions.

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What happens to
the wave functions?

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Should I invent a wave function
for the electron and a wave

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function for the proton?

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No.

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That's not good.

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You shouldn't a wave
function for everybody,

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which is a wave function that
depends on the coordinates

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X of the electron.

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And if you had just
an electron, you

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would have a wave function that
depends on X of the electron

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and a wave function that
depends on the coordinates

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of the proton.

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And how would you normalize it?

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Well, this thing times
d cubed Xe times d

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cubed Xp is the
probability, dP to find

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the electron around the little
cube about the point, Xe,

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and the proton around the
little cube around Xp.

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And then if you
want to see what is

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it probability to find
the electron at some point

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regardless of the proton,
you integrate over

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the whole proton.

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And then you should just
have a wave function.

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So, if you integrate this
psi squared-- oh, I'm sorry,

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psi is squared.

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If you this psi squared
over the proton,

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you're left with
some probability

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density for the electron.

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So, this is the good thing about
having more and more particles.

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You still have just one wave
function and one Schrodinger

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equation.

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You have more particles, and
it's just one wave function.

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That's why people eventually
talk about the wave

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function of the universe.

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I say, well, the whole
universe has one wave function

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for every, for every molecule,
for every elementary particle.

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There's just one wave function.

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That's nice about the
Schrodinger equation.

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And, OK, so we have this,
and what is the Hamiltonian?

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The Hamiltonian is
the kinetic energy

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associated to the proton plus
the kinetic energy associated

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to the electron plus the
potential energy, which

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just the depends on the
distance between the electron

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and the proton.

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And then and write the
differential equation,

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the Schrodinger equation.

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And the P of the proton would
be thought as ddx of the proton.

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And the P of the electron would
be like ddx of the electron.

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And that differential
equation would make sense.

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So what is our problem
showing that we have--

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can think in some way as a
potential or a Schrodinger

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equation that the most--
just some distance that

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separates the particles, and
some equation for the rest.

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So, we have to change variables.

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So what is the
change of variables?

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I'm going to motivate
some of that.

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Well, we have two pairs
of canonical variables.

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We have the proton canonical
variables, X and P,

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of the proton, and the electron
canonical variables, X and P,

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of the electron.

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So how can we have
different kind

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of variables that will lead
to a more interesting thing?

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We can imagine that maybe we
should define a new X that

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is the difference for which
the magnitude of that X

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will be what we call r.

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And that's true, but it's
not so easy to do that

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from the beginning, so let's
try to do what most people would

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do from experience
and say, I know

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there's something simple about
a system of doing the wrapping

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values, center of mass.

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If the center of mass moves
with constant velocity,

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that's what it does always.

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So, if classical notions of
how you treat two-body systems

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can be used in a
quantum setting,

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we should be able to
define a new quantum

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coordinate associated
to the center of mass,

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and a new quantum
momentum associated

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to the center of mass.

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And the center of mass
momentum is always

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the sum of the momenta, so we
will define at P, a capital P--

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I hope that my P's are
going to be always clear.

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They're not supposed to have
before this one, the little bar

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below.

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This is the total,
P, and it's going

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to be defined as the momentum
of the proton plus the momentum

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of the electrons.

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And now, I want to define
an X that goes with this P.

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And by that, I mean that this
X must have a commutation

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relation with this P
that tells, yes, you

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are an X because X with
P should be ih bar.

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So, whatever I define
here, that should happen.

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Now, you could imagine
that we usually

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do something like this--

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we multiply each particle
according to its mass.

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Pe.

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So we're weighting the
momenta associated to--

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oh, I'm sorry.

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The positions, Xp, Xe,
associated to the mass.

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And this, of course, would
not have the right units.

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It doesn't have
units of coordinate,

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so this is the center of mass.

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So, this will be center
of mass quantum variables.

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And in order to have units,
you must divide by a mass,

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so it's not too unreasonable
to put the sum of masses here.

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That's how you define it
anyway in classical mechanics.

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And now, we must ask is X
with P the i-th component

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of the j-th component.

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Is it really equal
to ih bar delta ij?

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And the answer, I
would say, is yes

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because, here,
let's do this one.

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You will have mp Xpi plus
me Xei over mp plus m e.

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You see, once you
do one, all the rest

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you should be able to do
just without thinking.

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P, on the other hand, is
P of the momentums of j

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plus the momentum of
the electron, so j.

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What is it equal to?

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And then you say, look,
from Xp with momentums of P,

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yes, they give me ih bar delta
ij, but there will be ih bar.

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The first factor there will be
an extra mp over mp plus m e.

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So this contribution comes from
the first term that only talks

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to the momentum of the proton.

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Doesn't talk to the
momentum of the electron.

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And the second term for the
coordinate of the electron

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only talks to the
momentum of the electron.

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And it does give you
an ih bar delta ij,

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but this time will be with
an m e over mp plus m e.

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And this factor is equal to 1.

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So, yes, we can box this.

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This is a pair of quantum
mechanical canonical variables.

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They have the right commutation.

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They have the right units,
the right commutator,

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the right everything.

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So next time when you
see this, you will say,

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X with P-- you say,
Xp with P is 1.

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Xe with P is 1.

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mp plus-- 1.

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Yes, it works.

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You forget ih bar delta ij.

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This is more clear
that it will work.

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Try it.

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Next, we need the second pair
of canonical variables, so--

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well, actually I should
use this one for this.

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So for the second pair,
it's reasonable to use the X

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that we anticipated, so we'll
have a relative coordinate, X.

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And I don't know how I call it--

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X relative, or--
no, just little x.

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Little x will be defined
as X of the electron.

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I think that's coordinates.

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X of the electron
minus X of the proton.

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This is natural.

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We want a little x like that.

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Already at this point, you
could be paralyzed with fear.

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Something could have gone
wrong at this moment.

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Imagine if this x doesn't
commute with these guys.

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Then it's all a
disaster because this

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should be electron and proton
commute with each other.

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I should have written
there, not only

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those are the
commutation relations,

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everything else is 0.

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Any X of electron with a
momentum of a proton is 0.

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That's why they're
two independent pairs

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of canonical variables
that we can treat.

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It better be that this
is an independent pair

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of canonical
variables, so it better

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be that this x and whatever
P I'm going to define here

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commute with these guys.

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And as far as this x,
happily it works out

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because X's commute
with any X's here,

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00:16:00,850 --> 00:16:05,530
so this little x definitely
commutes with the little x.

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00:16:05,530 --> 00:16:09,320
But the fact that this commutes
with P could have killed it,

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00:16:09,320 --> 00:16:12,680
but it doesn't because
of the minus sign.

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One of the commutators of the
little x with the capital P

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would give for the electron
an ih bar, and for the proton,

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a minus ih bar.

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And they cancel.

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So, all is good so far.

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So, this is so far so good--
commutes with capital X

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and capital P. So now,
we have here something,

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and we could put that
number, alpha Pe minus beta

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P of the proton.

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And we all know what
alpha and beta are.

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But now you can more
or less be confident.

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I need that this momentum
with this x give me 1, ih bar.

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So, that would put the
condition on alpha and beta.

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In fact, the condition
that xi with P bar j

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give you ih bar delta ij,
you can imagine what it is.

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It's that alpha plus
beta is equal to 1.

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And the condition
that this momentum

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commute with the center
of mass position, which

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it can fail to do so--

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what can it give?

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Well, W alpha of the P of
the electron goes with m e,

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00:18:23,930 --> 00:18:34,505
so it's something like alpha m
e minus beta mp is equal to 0.

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00:18:37,070 --> 00:18:43,080
And please-- oh, this
commutator should be 0.

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Please make sure you
know how to do this.

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You can get those conditions.

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I'm going maybe a little fast
for you to just follow it up.

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And so, at this moment, we
can solve for alpha and beta--

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00:18:59,570 --> 00:19:00,520
two equations.

245
00:19:00,520 --> 00:19:07,992
So alpha is equal to
mp over me plus mp.

246
00:19:07,992 --> 00:19:14,945
And beta is equal to
m e over m e plus mp.

247
00:19:22,865 --> 00:19:26,520
And we have a pair of
canonical variables

248
00:19:26,520 --> 00:19:31,065
as well here, therefore,
the relative coordinate

249
00:19:31,065 --> 00:19:35,220
and the relative
momentum, in some sense.

250
00:19:35,220 --> 00:19:38,220
It's useful to
define two symbols.

251
00:19:38,220 --> 00:19:46,294
So, we use mass m e
mp over mp plus mp.

252
00:19:46,294 --> 00:19:53,400
And the total mass,
which is m e plus mp.

253
00:19:53,400 --> 00:19:59,540
In the case of a heavy proton,
the reduced mass-- the proton

254
00:19:59,540 --> 00:20:01,343
is heavy compared
with the electron.

255
00:20:01,343 --> 00:20:05,045
You can ignore the electron
here, cancel them, e.

256
00:20:05,045 --> 00:20:10,050
And the reduced
mass is called this.

257
00:20:10,050 --> 00:20:12,570
So, in terms of
the reduced mass,

258
00:20:12,570 --> 00:20:22,140
alpha is equal to mu over m
e, and beta is mu over mp.

259
00:20:22,140 --> 00:20:24,030
Those are unit-free constants.

260
00:20:24,030 --> 00:20:28,184
So summarizing, the second
pair of canonical variables

261
00:20:28,184 --> 00:20:31,480
are Xe minus--

262
00:20:31,480 --> 00:20:33,470
Xe minus Xp.

263
00:20:36,090 --> 00:20:48,090
And P equals mu Pe over m
e minus P sub p over mp.

264
00:20:55,080 --> 00:21:03,280
All right, so--
so, the last step

265
00:21:03,280 --> 00:21:06,600
that we're going to follow,
and after that, we'll

266
00:21:06,600 --> 00:21:08,765
get the Hamiltonian
and stop there.

267
00:21:08,765 --> 00:21:13,446
And we'll discuss
solutions next time.

268
00:21:13,446 --> 00:21:19,530
But finally, we have
enough equations

269
00:21:19,530 --> 00:21:27,980
to solve for the momenta
we have in the Hamiltonian

270
00:21:27,980 --> 00:21:32,290
in terms of the
center of mass momenta

271
00:21:32,290 --> 00:21:35,220
and the relative momenta.

272
00:21:35,220 --> 00:21:42,590
So, from equation
star and double star

273
00:21:42,590 --> 00:21:49,290
here, you can solve
for these momenta.

274
00:21:49,290 --> 00:21:59,320
So, P of the proton will be
equal to mp over M, big momenta

275
00:21:59,320 --> 00:22:01,356
minus little momenta.

276
00:22:01,356 --> 00:22:10,028
And P of the electron is m e
over capital M, big momenta

277
00:22:10,028 --> 00:22:11,492
plus little momenta.

278
00:22:18,330 --> 00:22:24,440
Well, one more equation
that we have to write,

279
00:22:24,440 --> 00:22:26,810
and that's the key
to the simplicity

280
00:22:26,810 --> 00:22:28,702
of all what we've done.

281
00:22:28,702 --> 00:22:32,930
At this moment, we have a
very good physical insight

282
00:22:32,930 --> 00:22:35,180
into what we've
done, but we want

283
00:22:35,180 --> 00:22:38,210
to see if the math collaborates.

284
00:22:38,210 --> 00:22:40,400
And the good
physical insight has

285
00:22:40,400 --> 00:22:42,740
been that center of
mass motion should

286
00:22:42,740 --> 00:22:46,250
be independent of
the relative motion,

287
00:22:46,250 --> 00:22:50,180
and of we have the right X.

288
00:22:50,180 --> 00:22:52,790
So, the thing that
we have to compute

289
00:22:52,790 --> 00:22:57,530
is the first two terms
in the Hamiltonian,

290
00:22:57,530 --> 00:23:03,470
P of the proton squared
over 2m of the proton

291
00:23:03,470 --> 00:23:10,970
plus p of the electron squared
over 2 mass of the electron.

292
00:23:10,970 --> 00:23:15,600
So, I have 1 over
2m of the proton,

293
00:23:15,600 --> 00:23:26,250
and I have m of the proton over
capital M, P minus little p,

294
00:23:26,250 --> 00:23:31,024
and plus 1 over 2m
of the electron--

295
00:23:31,024 --> 00:23:33,900
this is squared--
m of the electron

296
00:23:33,900 --> 00:23:39,350
over capital M, big
P plus little p.

297
00:23:39,350 --> 00:23:44,150
And the plus and minus
are extremely reassuring

298
00:23:44,150 --> 00:23:49,370
because the cross terms
that couple the two

299
00:23:49,370 --> 00:23:51,040
are going to vanish.

300
00:23:51,040 --> 00:23:54,553
You can see cross terms will
have a minus sign. mp will

301
00:23:54,553 --> 00:23:55,618
be canceled.

302
00:23:55,618 --> 00:23:56,900
m e will cancel.

303
00:23:56,900 --> 00:23:58,780
They will cancel.

304
00:23:58,780 --> 00:24:02,902
So at the end of this
little calculation that

305
00:24:02,902 --> 00:24:12,980
takes couple of lines, you get
1 over 2 capital M total center

306
00:24:12,980 --> 00:24:25,990
of mass of this squared plus
1 over 2 mu little momentum

307
00:24:25,990 --> 00:24:26,660
squared.

308
00:24:26,660 --> 00:24:30,850
This is a kinetic operator.

309
00:24:30,850 --> 00:24:36,330
And then you say, great
success, the kinetic energy

310
00:24:36,330 --> 00:24:41,275
now has separated into a
center of mass contribution

311
00:24:41,275 --> 00:24:43,100
and a relative contribution.

312
00:24:43,100 --> 00:24:47,540
This will allow us to
now with a little step,

313
00:24:47,540 --> 00:24:52,060
separate the total Schrodinger
equation into center

314
00:24:52,060 --> 00:24:54,410
of mass motion and
relative motion,

315
00:24:54,410 --> 00:24:58,480
and the relative motion will
have [INAUDIBLE] potential.

316
00:24:58,480 --> 00:25:02,390
So we'll do the
punchline next time.