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PROFESSOR: Solutions
can be organized.

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One way to do it, it's
not the standard way,

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is with plotting
n here and l here,

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and you have 0,
1, 2, 3, 1, 2, 3.

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You have all this points.

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Remember, every
point this allowed.

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Every integer
combination is allowed.

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All those are energy levels
of the hydrogen atom,

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but now you can see that
this point, when both are 0,

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corresponds to n equal to 1.

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Because if n and l are
0, n is equal to 1,

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and there's just one
solution with n equal to 1.

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These two points here,
when l is equal to 1

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and n equals 0, with n is
equal 1 and l equals 0,

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represent the two possibilities
that realize n equals to 2.

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n equals to 2 is realized by
having 1 and 0 or 1 and 0, two

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values.

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Similarly, there
are three things

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with n equal to 3, four things--

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my graph is not that
great-- with n equal to 4,

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and more and more states.

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So for each n you have that n
plus l is equal to n minus 1.

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And l can never
exceed n minus 1,

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and that's physically quite
something that people remember.

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But also n cannot exceed n
minus 1 or 0, both are there,

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limited by these quantities.

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So if you have some n, you
will have l and n, for example,

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for some quantum number n.

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You will have l equals 0
and n would be n minus 1.

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That would work out.

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l plus n would be n minus
1, or 1 and n minus 2,

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or all the things up
to n minus 1 and 0.

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So they take turns.

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They have to add
up to n minus 1.

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So let's plot.

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Let's actually, we can go here.

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We don't have too much
more to say at this moment.

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So let's have this.

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We can do a little counting
that is interesting,

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and I'll count the number
of states for a given n.

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So for example, for n equals
to 1, what can we have?

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We said, l equals 0, and
capital N is equal to 0 as well,

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and that's one state
for n equals 2.

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What can you have?

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n equals to 2, you could have
l equals 1, or l equals 0.

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So l equals 1, or l equals 0.

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And how many states
do we have here?

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Well, l equals 1 can have
m equals 1, 0, and minus 1.

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Remember, the m values is
another label for states.

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Those are different states.

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So here, there are 3 states,
plus 1 state, 3 plus 1 states,

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that's 4.

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n equals 3 will have l
equals 2, l equals 1,

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and l equals 0, which
is 5 states plus 3

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states plus 1 state, which is
9 states, and that's 3 squared.

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And 4 was actually 2 squared,
and if you go on to n

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you will have n square states.

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Something that perhaps
you could try to count,

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and show that that's true.

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So what are our quantum numbers
for the states of hydrogen?

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Well, our quantum numbers
are, it's our choice,

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but physically we
want to understand

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them each intuitively.

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So here we go.

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One most important quantum
number, and its name says so,

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is principle quantum number.

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So the quantum
numbers of hydrogen,

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and the first
important thing is n.

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We definitely cannot
do away with n.

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It fixes our energies.

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And now we have a possibility.

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We look at this and we
say, yeah well, I actually

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I either need to determine
what is l or what is n.

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So it doesn't even come close.

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Physicists will not say, oh,
I want to describe the quantum

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number by the degree of the
polynomial inside the solution.

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No, physicists will say, I want
to use the angular momentum.

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And certainly, if you
know l, and you know n,

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you know capital N.

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So capital N is a funny number.

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It has to do with the degree
of the polynomial that shows up

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in between this leading behavior
and that exponential behavior,

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very interesting, but
not directly physical.

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The l, however, is
directly associated

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to an observable
angular momentum.

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So to describe the
state that I have here,

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if I give you n and l,
you can see that you

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determine which state you are.

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So the second quantum
number is going to be l,

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and the third quantum
number is unavoidable.

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It's the z component of
angular momentum, should be m.

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That's also physical.

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We should not skip it.

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So these are our
quantum numbers,

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and they fix capital N,
in case you're interested,

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as n minus l plus 1, and
that's interesting information.

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So let's recall our variables.

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OK, a rho is here.

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That's very nice.

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So rho is 2 kappa
z over a naught r,

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but now we know what kappa is.

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Kappa is 1 over 2n.

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So actually, the rho
variable is tailored

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to the quantum numbers.

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It's just Zr over n a naught,
where n is the principle

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quantum number.

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So back to the solution.

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You see, we have
to recap quickly.

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Psi nlm is equal to U of the
energy of the radial equation--

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so n and l is sufficient
for that-- over r Ylm.

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Or the U is the thing
that we had here, Ul,

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and now it has an
energy into it.

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So it's a rho to the l plus 1,
still r and rho up to numbers.

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So this is like.

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A rho, a Wnl if we wish,
of rho e to the minus rho,

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and Ylm theta phi.

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Well, let's write one more
equation, and then finish.

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So just to give the
feeling of this solution,

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what does that give you?

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Rho to the l, a
polynomial of rho,

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which is a polynomial
of degree n.

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n, which is little n minus l
plus 1 times e to the minus rho

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Ylm.

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It's important for you to
see the whole solution.

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This is the whole solution
of the hydrogen atom.

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I'll write it in one more way.

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A, a constant, because
this is similar.

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Rho, well, rho is, in
terms of units, at least

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has r over a naught to the l.

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Here it is a polynomial in
r over a naught of degree.

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Little n minus l plus
1, and this polynomial,

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we could make a
whole study of it.

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These are Laguerre polynomials.

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We will not look into
them in this course.

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You may do it in a
more advanced course.

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It's interesting, but it's
better to just get an intuition

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as to what's happening here.

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There is an e to
the minus rho, which

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is interesting to have fully.

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So this is e to the
minus Zr over n a naught,

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and there's a Ylm
of theta and phi.

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So this is your whole solution
for the hydrogen atom.

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We should write the simplest
one psi 1, 0, 0, n equals 1, l

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equals 0, m equals 0,
spherically symmetric.

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Here it is, 1 over pi a cube,
e to the minus r over a naught

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For the KZ is equal to 1.

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Ground state of hydrogen.