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PROFESSOR: How do we state
the issue of normalization?

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See, the spherical harmonics
are functions of theta and phi.

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So it makes sense that you would
integrate over theta and phi--

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solid angle.

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The solid angle is the
natural integration.

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And it's a helpful integration,
because if you have solid angle

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integrals and then
radial integrals,

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you will have integrated
over all volume.

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So for this spherical
harmonic, solid angle

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is the right variable.

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And you may remember,
if you have solid angle,

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you have to integrate
over theta and phi.

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Solid angle, you think
of it as a radius of one.

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Here is sine theta.

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So what is solid angle?

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It's really the area on
a sphere of radius one.

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The definition of
solid angle is area

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over radius squared, the part of
the solid angle that you have.

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If you're working with
a sphere of radius one,

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it's the area element
is the solid angle.

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So the area element in here
would be, or the integral

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over solid angle--

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this is solid angle, d omega--

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you would integrate
from theta equals 0

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to theta equals pi of
sine theta, d theta.

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And then you would integrate
from 0 to 2 pi of d phi.

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Now, you've seen that
this equation that

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began as a differential equation
for functions of theta ending

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up being the
differential equation

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for a function of cosine theta.

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Cosine theta was
the right variable.

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Well, here it is as well,
and you should always

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recognize that.

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This is minus d of cosine theta.

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And here you would be degrading
from cosine theta equals 1

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to minus 1.

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But the minus and the order of
integration can be reversed,

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so you have the integral from
minus 1 to 1 of d cos theta

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and then the integral
from 0 to 2 pi of alpha.

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So this is the solid
angle integral.

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You integrate d of cosine
theta from minus 1 to 1 and 0

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to 2 pi of d phi.

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And we will many times use
this notation, the omega,

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to represent that integral so
that we don't have to write it.

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But when we have to
write it, we technically

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prefer to write it this way,
so that the integrals should

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be doable in terms
of cosine theta.

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So what does this all mean
for our spherical harmonics?

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Well, our spherical
harmonics turned out

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to be eigenfunctions
or Hermitian operators.

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And if they have different
l's and m's, they

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are having different
eigenvalues.

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So eigenfunctions of
Hermitian operators

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with different eigenvalues
have to be orthogonal.

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So we'll write the
main property, which

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is the integral of
2l, say l prime, m

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prime, of theta and phi.

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And he put the star here.

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I'll put the star just
at the Y, and he's

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complex conjugated
the whole thing.

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Remember that in our problem,
one wave function was complex

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conjugated.

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The other wave function
is not complex conjugated.

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They're different
ones because l and m

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and l prime and m prime
could be different.

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So orthogonality is warranted.

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Two different ones with
different l's and different m's

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should give you different
values of this integral.

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So at this moment, you
should get a delta l prime l

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delta m prime m.

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And if l and m are the same
as l prime and m prime,

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you have the same
spherical harmonic.

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And all this tremendous
formula over there,

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with 2l plus 1 and
all these figures,

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you're guaranteed that in
that case you get 1 here.

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So this formula is
correct as written.

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That is the orthonormality
of this solution.

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Probably, this stage might
be a little vague for you.

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We saw this a long time ago.

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We may want to review
why eigenfunctions

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of Hermitian operators
with different eigenvalues

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are orthogonal and see
if you could prove.

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And you do it, or is it kind
of a little fuzzy already?

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We saw it over a month ago.

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So time to go back to
the Schrodinger equation.

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So for that, we
remember what we have.

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We have minus h squared over
2m Laplacian of psi plus V of r

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psi equals E psi.

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And the Laplacian has
this form, so that we can

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write it the following way--

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minus h squared over 2m 1 over
r d second dr squared r psi--

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I won't close the
brackets here--

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minus this term.

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So I'll write it minus one
over h squared r squared l

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squared psi plus V of r
psi is equal to E psi.

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Now, you could be a little
concerned doing operators

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and say, well, am I
sure this l squared

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is to the right
of the r squared?

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r and l-- l has momentum.

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Momentum [INAUDIBLE] with r.

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Maybe there's a problem there.

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But rest assured, there
is no problem whatsoever.

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You realize that l squared was
all these things with dd phetas

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and dd phis.

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There was no r in there.

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It commutes with it.

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There is no ambiguity.

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We can prove directly that
l squared commutes with r,

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and it takes a little more work.

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But you've seen
what l squared is.

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It's the dd thetas and dd phis.

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It just doesn't have
anything to do with it.

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So now for the great
simplification.

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You don't want any of your
variables in this equation.

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You want to elicit
to a radial equation.

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So we try a factorized solution.

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Psi is going to be--

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of all the correlates--
is going to be a product

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a purely radial wave
function of some energy E

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times a Ylm of theta and phi.

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And we can declare
success if we can

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get from this
differential equation now

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a radial differential
equation, just for r.

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Forget thetas and phi.

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All that must have been taken
care of by the angular momentum

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operators.

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And we have hoped for that.

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In fact, if you look at it, you
realize that we've succeeded.

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Why?

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The right-hand side will have
a factor of y and m untouched.

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V of r times psi will have
a factor of Ylm untouched.

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This term, having
just r derivatives

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will have some things acting
on this capital R and Ylm

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untouched.

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The only problem is this one.

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But l squared on Ylm
is a number times Ylm.

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It's one of our eigenstates.

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Therefore, the Y's and
m's drop out completely

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from this equation.

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And what do we get?

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Well, you get minus h squared
over 2m 1 over r, d second,

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dr squared, r capital RE
minus l squared on psi lm--

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or Ylm now--

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is h squared times
l times l plus 1.

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So the h squared cancels.

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You get l times l
plus 1 r squared,

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and then we get the RE
of r times the psi lm

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that has already--

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I started to cancel it
from the whole equation.

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So I use here that L squared
from the top blackboard

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over there, has that eigenvalue,
and the psi lm has dropped out.

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Then I have the V of r
RE equals E time RE of r.

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So this is great.

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We have a simplified equation,
all the angular dependencies

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gone.

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I now have to solve this
equation for the radial wave

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function and then multiply
it by a spherical harmonic.

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And I got a solution
that represents

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a state of the system
with angular momentum l

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and with z component
of angular momentum m.

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The only thing I
have to do, however,

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is to clean up this
equation a little bit.

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And the way to clean it up is
to admit that, probably, it's

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better as an equation
for this product.

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So let's clean it up by
multiplying everything by r.

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dr squared of little r RE.

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If I multiplied
by r here, I will

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have plus h squared
over 2m l times

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l plus 1 over r squared
rRE plus V of r times

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r times RE equals E
times r times RE of r.

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So we'll call u of r r times RE
of r, and look what we've got.

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We've got something
that has been adjusted,

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but things worked out
to look just right.

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Minus h squared over 2m
d second, dr squared u

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of r plus--

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let me open a parentheses V
of r plus h squared over 2m r

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squared, l times l plus one u
of r is equal to E times u of r.

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Here it is.

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It's just a nice form of a
one-dimensional Schrodinger

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equation.

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The radial equation for the wave
function dependents, a long r,

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has become a radial
one-dimensional particle

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in that potential, in which
you should remember two things.

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That this u is not quite
the full radial dependent.

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The radial dependent is
RE, which is u over r.

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But this equation
is just very nice.

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And what you see is
another important thing.

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If you look at the given
particle in a potential,

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you have many options.

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You can look first for the
states that have 0 angular

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momentum-- l equals zero--

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and you must solve
this equation.

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Then you must look
at l equals 1.

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There can be states
with l equals one.

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And then you must
solve it again.

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And then you must solve for
l equals 2 and for l equals 3

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and for all values of l.

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So actually, yes, the
three-dimensional problem

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is more complicated than
the one-dimensional problem,

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but only because, in
fact, solving a problem

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means learning how to solve
it for all values of l.

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Now, you will
imagine that if you

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learn how to solve for one
value of l, solving for another

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is not that different.

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And that's roughly true, but
there's still differences.

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l equals 0 is the easiest thing.

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So if the particle is
in three dimensions

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but has no angular
momentum-- and remember,

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l equals 0 means no angular
momentum-- it's this case.

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l equals 0 means m equals 0.

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l squared is 0.

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lz is 0.

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This is 0 angular momentum.