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PROFESSOR: Would solving this
equation for some potential,

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and since h is Hermitian,
we found the results

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that we mentioned last time.

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That is the
eigenfunctions of h are

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going to form an
orthonormal set of functions

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that span the space.

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You can expand
anything on there.

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This is what we proved for
a general condition operator

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to some degree.

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So the eigenfunctions
form an orthonormal set

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that spans the space.

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So you're going to define
that psi 1 with an E1

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and psi 2 with an E2,
and then this continues.

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And this is called the
spectrum of the theory

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because energy eigenstates are
considered the gold standard.

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If you want to find
solving a theory

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means finding the
energy eigenstates.

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Because if you find the energy
eigenstates, you can solve,

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you can write any wave
function of superposition

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of energetic states and
then just let them evolve.

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And the energetic
states involve easily

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because they are just
stationary states.

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So the spectrum of the
theory is the collection

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of numbers that are the
allowed energies and of course,

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the associated eigenfunctions.

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So the energies may be many,
maybe discrete, maybe it

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has a little bit of
continuous partners,

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all kind of varieties.

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But your task is to find
those for any problem.

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So the equation that
we're trying to solve

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is now re-written.

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We're going to try to solve it.

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So let's look at it.

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It's a second order
differential equation

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with a potential in general.

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So we had an example there.

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It's there.

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It's boxed.

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So we'll write it
slightly different,

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remove the potential
to the right-hand side

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and get rid of the
constants here.

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So d x squared is equal
to 2m over h squared.

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So this is the equation
we have to solve.

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So whenever you
have a problem, you

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may encounter a
potential, v of x.

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And the question is how
bad this potential can be.

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Well, the potential may be nice
and simple, or it may be nice

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but then has some jumps.

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It may have infinite
jumps, like a potential is

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a complete barrier, or it
may have delta functions.

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all these are v of
x equal possibles.

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All of them.

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Many things can happen
with a potential.

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In fact, the potential
can be as strange

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as you're one, depending on
what problems you want to solve.

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So it's your choice.

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Now, we're going to accept, in
fact, all of those potentials

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for our analysis.

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May be nice and smooth.

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There may have discontinuities.

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It may have infinite
discontinuities,

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and worse things
like delta function.

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But worse things than
that we will ignore,

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and there are worse
things than that.

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Maybe a potential discontinues
at every point, or maybe

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a potential has delta
functions and derivatives

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of delta functions.

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Or potentials that blow up
and do all kinds of things.

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And I'm not saying you
should never consider that.

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I'm saying that we don't know
of any very useful case where

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you get anything
interesting with that.

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But a conceivable a particular
time a singular potential one

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day could be used.

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So we'll look at
these potentials

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and try to understand how to
set up boundary conditions.

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And we're going to worry
about basically psi

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and how does it behave.

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And my first claim is that
psi of x has to be continuous.

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So psi of x cannot jump.

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The wave function move
along but cannot jump.

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And the reason is a
differential equation.

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Look, if psi of x
was not continuous,

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if psi of x was
like this, and just

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had a discontinuity,
psi of x equal to x,

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psi prime of x would
contain a delta function

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and this is continuity.

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The derivative is infinite.

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And psi double prime of
x, the second derivative,

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would have a derivative of a
delta function which is worse

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because a delta
function, we think of it

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as a spike that is becoming
thinner and higher,

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but the derivative of the delta
function first goes to infinity

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and then goes to minus infinity
and then comes back up.

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It's much worse in many ways.

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And look, if you have
this differential equation

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and psi is not continuous,
well, the right-hand side

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is not continuous.

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Or if you have a
delta function, then

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something not continuous,
but left-hand side,

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we've had a derivative of a
delta function that is nowhere

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on the right-hand side.

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On the right-hand side,
the worst that could exist

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is a delta function in v of x.

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But the derivative of a
delta function doesn't exist.

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So you cannot afford to have
a psi that is discontinuous.

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Psi has to be continuous.

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There's other ways
to argue this.

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You might put them
in your notes,

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but I'll leave it like that.

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Now how about the next case?

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I will say the
following happens too.

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Sine prime of x is
continuous unless v of x

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has a delta function.

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You see, potentials of
delta functions are nice,

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they are interesting.

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We will consider that.

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Delta functions potentials
can be attractive

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potentials, repulsive
potentials of [INAUDIBLE].

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So I claim now that psi prime
of x has to also be continuous.

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Why are we worrying
about psi and psi

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prime is because you need two
conditions whenever you're

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going to solve this differential
equation at an interface,

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you will need to know
psi is continuous

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and psi prime is
continuous because

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of second-order
differential equations.

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So suppose psi
prime is continuous.

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Then there is no problem.

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If psi prime is continuous,
the worse that can happen

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is that the second
derivative is discontinuous.

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And the second derivative
is discontinuous

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could happen with a potential
of this discontinuous,

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so one problem if psi
prime is continuous.

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But psi prime can fail to be
continuous if the potential has

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a delta function.

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And let's see that.

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If psi prime is
discontinuous, then

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psi double prime is proportional
to a delta function.

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If psi prime is
discontinuous, double prime

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is proportional to
a delta function.

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But here psi just
takes some value--

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there's nothing
strange about it--

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in order to have delta function,
which is psi double prime.

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To be equal to the
right-hand side, v of x

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must have a delta function.

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And v will have
a delta function.

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So it will be a somewhat
similar potential,

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but we're going to look at
them in about a week from now.

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But this will be our
guidance to solve problems.

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The continuity of
the wave function

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and the continuity of the
derivative of the wave

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function.

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And for this slightly
more complicated problems

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in which the potential
has a delta function,

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then you will have a
discontinuity in psi prime,

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and it will be calculable,
and it's manageable,

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and it's all very nice.

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Now, we do it a
little complicated,

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and everything is
mixed up, but you will

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see that it's quite doable.