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PROFESSOR: Write w
equals sum from k

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equals 0 to infinity a k rho k.

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And plug in--

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I've suggested that usually the
thing that you should know when

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you plug in those equations
is to look for the power rho

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to the k in this equation.

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And just, since
everything is equal to 0,

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the coefficient of rho to the
k in this equation should be 0.

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And that's the easiest
way to select the powers.

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That's good practice.

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You should do it.

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I won't do it here.

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We've done it in a few cases.

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So this will relate a k plus 1.

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To a k.

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So that's algebra.

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It's a good skill
to be able to do it.

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But it would be
not very good use

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of our time to do it right now.

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So here is the answer.

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OK.

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This is more important.

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2 to the k plus l plus 1 minus
1 over kappa over k plus 1

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plus k plus 2l plus 2.

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OK.

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We've got our
recursion relation.

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And the issue is, again, what
happens with this coefficient

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as k goes to infinity?

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So as k goes large, a k plus
1 over a k goes like what?

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Well, we have a k that
is becoming large,

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and everything else
doesn't matter.

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There's a k and a k, so there's
going to be some cancellation.

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And this looks like
roughly 2 to the k.

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Now, you could change
these numbers a little bit.

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I'm going to do a tiny
trick to simplify it,

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but it's just a trick.

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Don't worry about it too much.

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I'll put 2 to the k plus 1 here.

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And I will say, look, if the
series diverges in this case,

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this coefficient is
bigger than that one.

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So it will certainly diverge
for this case, as well.

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So the coefficients
here are smaller

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than those ones, this ratio.

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So if the ratio
between coefficients

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here is such that
the series diverges,

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then it will even diverge
a little more in this case.

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And the reason I put it
here is because then this

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is kind of solvable, a k
plus 1 nicely solvable, very

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nicely solvable, 2k plus 1 a k.

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And the solution
of this is to say

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you can try with a 0,
what a 1 is, what a 2.

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a k is 2 to the k a
0 over k factorial.

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OK.

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With that we can reconstruct
what kind of function

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this series would be building
if the series doesn't terminate

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and will not be too surprising.

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So in this case, the sum
over k of a k rho k, which

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is the function we're building,
is roughly equal to this

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a k here, which is 2 to the k.

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a 0 can go out.

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k factorial rho k.

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So this is a 0 e to the 2 rho.

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It's kind of fair
of it to do that.

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It's kind of saying that if the
w solution doesn't truncate,

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it's going to go like e to
the 2 rho, which precisely

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with an e to the
minus rho is going

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to give you the other possible
behavior of the solutions.

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It happened for the
harmonic oscillator.

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So what this is saying is
that then w, which is this,

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would go roughly like that.

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And that's bad.

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So the series must truncate.

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So we must truncate the series.

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OK.

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So here comes the
interesting part

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because there's
lots of quantities,

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and that has to be done a little
slowly so that nobody gets

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confused of what's
going to happen.

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We have to terminate
this series.

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So how are we going to do it?

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I'm going to state
it the following way.

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I'm going to say
that let's assume

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that we want a polynomial
of degree capital N.

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There will be lots of little
constants, capital N, little n.

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I want a polynomial
of degree capital N.

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That means that a sub capital
N is different from 0.

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And a capital N plus
1 is equal to 0.

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That's what should happen.

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If you have constants
up to a capital N,

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you'll have rho
to the capital N,

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and you'll have a
polynomial of degree N.

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But that must happen that
the next one must be 0.

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I don't have to state
that all of the rest

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are 0 because it's a
one-step recursion relation.

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Once a 5 is 0, a 6, a 7,
a 8, all of them are 0.

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That's it.

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And we will have like
even or odd solutions

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that we had for the
harmonic oscillator

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because these are
functions of r.

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And r and minus r you should
not quite expect anything.

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Minus r doesn't exist.

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So this is what should happen.

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But if that happens,
think of this.

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You have a n plus 1 should be 0.

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So the numerator should have
become 0 for k equals 2N.

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So you have one over kappa
is equal to 2N plus l plus 1.

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And in a sense that's it.

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Whatever had to
happen, happened.

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Why?

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The energy got
quantized already.

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Somehow it did because
the energy is kappa.

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Remember, kappa
squared actually was

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the ratio of the energy divided
by the dimension [INAUDIBLE]..

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So here it's saying the
energy is some number that

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has to do with an
integer, which is

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the degree of the
polynomial you're going

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to use, an l integer, and 1.

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So this is, of course,
pretty important.

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So what values happen here?

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What are the
possible values of l?

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Well, l here, l can go
0, 1, 2, 3, all of them.

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All of them are possible.

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And why is that?

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It's because of the
physics of the problem.

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We assume we'll have a particle
in a central potential.

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All values of angular
momentum can exist.

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So we should be looking for
l's that take all these values.

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Moreover, N is the polynomial
that you can choose.

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And N can also be
all of those values.

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We can begin with 0.

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So a 0 would be a number.

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But then it dies.

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A polynomial of degree 0
would be just a constant.

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It's possible.

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1, 2, 3, all of
those are possible.

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And for each combination
will have some energy.

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But here you start
to see degeneracies,

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multiple degeneracies, because
if you have the number 100,000

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here, it can be built in
many, many ways, 100,000 and 1

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ways or something like
that, with two integers that

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have to add up to it.

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And all of them will
have the same energy.

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So the hydrogen atom is going
to have lots of degeneracy.

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So here is a little bit of
a definition that we follow.

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So all these are
allowed, all allowed,

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all combinations allowed.

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So l can be anything, and
capital N can be anything.

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And let's define a slightly
better version of this thing.

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So let's move the 2
down, 1 over 2 kappa.

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That's N plus l plus 1.

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And let's call all this n, or
the principal quantum number.

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So n is the principal
quantum number.

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And in some sense,
well, you know

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that n has to be
greater or equal than 1.

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It's an integer.

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And has to be greater or equal
than 1 because of this 1 here

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and because the other ones
cannot be negative either.

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So n is a principal
quantum number,

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and it's a fundamental number
because it immediately gives

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you the value of the
energy, which we will

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write more physically shortly.

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But it hides within
it a degeneracy

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that this allowed because
of these differing numbers.

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So these different
numbers have to do

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with the degree of the
polynomial and the value of l

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that you are using.

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So let's classify this and
understand it a little better.

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So what do we have
for the energy?

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Remember, the energy divided
by the dimensionless factor--

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well, to make it
dimensionless, z squared e

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squared over a 0 kappa squared.

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We wrote actually that e
divided by this quantity, which

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has units of energy,
was kappa squared.

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So kappa squared now,
kappa is 1 over 2n.

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So when we substitute here,
we get e is equal to minus

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z squared e squared over
2a 0 1 over n squared.

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It's probably the
most famous formula

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that you certainly have studied
in high school, the 1 over n

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squared of the energy
levels of the hydrogen atom.

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The units are nice.

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There's for z equals to 1,
there's the e squared over 2a 0

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that we mentioned a little
while ago as giving you

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the characteristic energy.

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And e squared over
a 0 was 27.2 EV.

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And therefore, e squared over
2 a 0 is the famous 13.6 EV.