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PROFESSOR: Last time we talked
about particle on a circle.

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Today the whole lecture is
going to be developed to solving

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Schrodinger's equation.

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This is very important,
has lots of applications,

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and begins to give you
the insight that you

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need to the solutions.

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So we're going to be
solving this equation

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all through this lecture.

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And let me remind
you what with had

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with a particle on a circle.

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The circle is segment 0 to
L, with L and 0 identified.

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More properly we actually
think of the whole x-axis

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with the identification
that two points related

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in this way that differ by
L, or therefore for 2L, 3L,

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are the same point.

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As a result, we
want wave functions

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that have this periodicity.

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And that implies the same
periodicity for the derivatives

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as well.

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We looked at the
Schrodinger equation

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and we proved that the
energy of any solution

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has to be positive or 0.

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And therefore the differential
equation, the Schrodinger

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differential equation,
can be read then

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as minus k squared psi, where
k squared is this quantity

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and it's positive, so
k is a real number.

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That makes sense.

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And finally, once you
have this equation,

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you know that if the second
derivative of a function

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is proportional to
minus the function,

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the solution are trigonometric
functions or exponentials.

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And we decided to
go for exponentials,

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that they are perhaps a
little more understandable,

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though we will go back to them.

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Now that's where we
stopped last time.

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And now we apply the
periodicity condition.

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So we must have e to the ik x
plus L equal to e to the ikx.

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If you cancel the e to
the ikx on both sides,

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you get to e to the ikL must
be equal to 1, which forces kL

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to be a multiple of pi.

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That is, kL equal 2 pi n--

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of 2 pi, I'm sorry--

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2 pi n, where n is an integer.

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So those are the values of k.

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We'll write them
slightly differently.

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We'll write kn with
the subscript n

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to represent the k
determined by the integer n.

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So it will be 2 pi n over
L. Now from that equation,

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for k squared equal
2mE over h squared,

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you get e is equal to h
squared k squared over 2m.

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And k in these solutions
represents therefore

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the momentum.

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That is, the momentum
Pn is h bar kn,

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and it's 2 pi h bar n over L.
And the energies associated

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with solutions
with kn value is En

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would be h squared
kn squared, so

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4 pi squared n squared
over L squared over 2m.

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So this is equal to 2
pi squared h squared

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n squared over m L squared.

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Those are numbers.

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It's good to have them.

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Our solution is psi n of
x is equal to e to the i,

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or is proportional
to, e to the i knx.

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So far so good.

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But we can now
normalize this thing.

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This is the beauty
of the problem

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of a particle on a circle.

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If you have a particle
in free space,

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psi squared is equal to 1
and the integral is infinite.

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On the other hand, these
ones are normalizable.

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That is, we can demand that the
infinite over the circle of psi

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n squared be equal to 1.

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So how do we do that?

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Well I'll write it a little
more explicitly here.

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Psi n of x will be some
constant times e to the iknx.

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And therefore this thing
is the integral from 0

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to L dx of the constant squared.

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The constant can be
chosen to be real.

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N squared times psi
n squared, which is--

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this exponential
squared is just 1.

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This is 1.

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So it just gives you L times
N squared is equal to 1.

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So N is equal to 1
over square root of L.

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And finally, our psi n's of
x are 1 over square root of L

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e to the iknx or 1 over square
root of L e to the 2 pi inx

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over L. Oops.

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All right.

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So these are our wave functions.

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These are our
energy eigenstates.

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Our full stationary
states, where we're

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finding stationary states--

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stationary states have a psi
of x times a time dependence.

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The time dependence is
e to the minus i e--

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so you could say that
psi n of x and t,

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the full stationary
state, is psi n of x times

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e to the minus i
e n t over h bar.

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And that solves the full
Schrodinger equation.

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That's our stationary state.

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So one thing that should
be emphasized here

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is the range of the integers.

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n is an integer and
we better realize

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if there are some exceptions.

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Maybe just the positive?

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Is 0 included?

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Is 0 not included?

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And here, it's really
as stated here.

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It's all the integers,
n from minus infinity

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to plus infinity.

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All of them must be included.

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The reason we can
understand that

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is that the momentum of each
of these states, the momentum

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is 2 pi h n over
L. And therefore

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these are all states
of different momentum.

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There's no question that
these are different states.

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It cannot be that one is
just the same as another one.

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They have different momentum.

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They represent
the particle going

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with some momentum
around the circle,

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and that momentum
is quantified by n

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and it could be in
the positive direction

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or negative direction.

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Now you could be suspicious
about n equals 0.

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But there's actually nothing
to be suspicious about it.

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It's surprising.

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But psi 0 is 1 over square
root of L, has no x dependence.

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And therefore it has 0 energy.

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And that's-- I'm sorry, here.

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There's some psi missing.

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The second derivative
of a constant is 0.

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And if e is equal to 0,
that's a consistent solution.

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The constant is important.

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And now you also
realize that psi,

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you have a nice phenomenon,
that psi minus 1 and psi 1,

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for example, they correspond
to n equals 1 and minus 1,

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have the same energy.

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Because energy
depends on n squared,

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so these are degenerate
states with energy E1

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equal to E minus 1.

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And so are psi 2
and psi minus 2.

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And of course just
psi minus k and psi k.

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They are degenerate states.

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And now this hits
into a property

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that is going to be
important in the future about

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degenerate states.

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Whenever somebody gives you a
couple of degenerate states,

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you know they have
the same energy.

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But you must not stop there.

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If they are degenerate states
and there are two states,

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it means that they
are not the same.

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So there must be something
physical about them

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that distinguishes them.

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Whenever you have
degenerate states,

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you have to work
until you figure out

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what is different about
one state and the other.

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And here we got the answer.

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The answer is simply that
they are degenerate states

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with a different momentum.

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So the momentum is an
observable that distinguishes

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those degenerate states.

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In fact, as we've
written here, p on psi n

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of x is equal to Pn psi n.

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And P n given by this quantity.

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OK.

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Our eigenstates are orthonormal.

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They're eigenstates.

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So why are they orthonormal?

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They are eigenstates
of a Hermitian operator

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with different eigenvalues.

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They're eigenstates of p
with different eigenvalues.

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So they're orthonormal.

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The argument with the
energy would have not

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worked out so well because there
you have degenerate states.

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So these two states are
degenerate with respect

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to energy.

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So you could wonder, how do
you know they are orthogonal?

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But in this case it's simple.

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They have different momentum.

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Momentum is a
Hermitian operator,

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and it should be orthogonal.

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So the states are orthogonal.

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They are complete.

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You could write any wave
function of the circle

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as a superposition
of those psi n's.

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So any psi of x periodic can be
written as psi of x the sum a n

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psi n's over all the integers .

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And one last remark.

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We could have worked
with sines and cosines.

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And therefore we
could have worked

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with psi k plus psi minus k.

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This psi k and psi minus
k have the same energy.

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Therefore this sum is an energy
eigenstate of that same energy.

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The Hamiltonian acting
on psi k gives you

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the energy times psi k.

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Here, the same energy
times psi minus k,

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so this is an energy eigenstate.

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And this is proportional
to cosine of kx.

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And this is an energy
eigenstate you know,

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because two
derivatives of a cosine

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will give you back that cosine.

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Similarly, psi k
minus psi of minus k

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is proportional to sine of kx.

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And that's also an
energy eigenstate.

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Both are energy eigenstates.

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So this is kind of the
way you can reformulate

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Fourier's theorem here.

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You could say anything
can be written

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as a superposition of
all the exponentials,

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including the
exponential with n equals

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0, which is just a constant.

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Or alternatively,
everything could

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be written in terms
of sines and cosines,

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which is another way of
doing the Fourier theorem.

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These are energy eigenstates,
but they're not P eigenstates

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anymore.

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This, when you take a
derivative, becomes a sine.

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When this, you take a
derivative, becomes a cosine.

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They're not energy.

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They're not momentum
eigenstates.

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So you can work with
momentum eigenstates,

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you can work with
energy eigenstates.

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It's your choice.

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It's probably easier to work
just with momentum, I can say.

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So that's it for the
particle in a circle.

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We have three problems
to solve today.

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Particle in a circle, particle
in a box, and particle

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in a finite well.