# L10.4 Comments on the spectrum and continuity conditions.

PROFESSOR: Would solving this
equation for some potential, and since h is Hermitian,
we found the results that we mentioned last time. That is the
eigenfunctions of h are going to form an
orthonormal set of functions that span the space. You can expand
anything on there. This is what we proved for
a general condition operator to some degree. So the eigenfunctions
form an orthonormal set that spans the space. So you’re going to define
that psi 1 with an E1 and psi 2 with an E2,
and then this continues. And this is called the
spectrum of the theory because energy eigenstates are
considered the gold standard. If you want to find
solving a theory means finding the
energy eigenstates. Because if you find the energy
eigenstates, you can solve, you can write any wave
function of superposition of energetic states and
then just let them evolve. And the energetic
states involve easily because they are just
stationary states. So the spectrum of the
theory is the collection of numbers that are the
allowed energies and of course, the associated eigenfunctions. So the energies may be many,
maybe discrete, maybe it has a little bit of
continuous partners, all kind of varieties. But your task is to find
those for any problem. So the equation that
we’re trying to solve is now re-written. We’re going to try to solve it. So let’s look at it. It’s a second order
differential equation with a potential in general. So we had an example there. It’s there. It’s boxed. So we’ll write it
slightly different, remove the potential
to the right-hand side and get rid of the
constants here. So d x squared is equal
to 2m over h squared. So this is the equation
we have to solve. So whenever you
have a problem, you may encounter a
potential, v of x. And the question is how
bad this potential can be. Well, the potential may be nice
and simple, or it may be nice but then has some jumps. It may have infinite
jumps, like a potential is a complete barrier, or it
may have delta functions. all these are v of
x equal possibles. All of them. Many things can happen
with a potential. In fact, the potential
can be as strange as you’re one, depending on
what problems you want to solve. So it’s your choice. Now, we’re going to accept, in
fact, all of those potentials for our analysis. May be nice and smooth. There may have discontinuities. It may have infinite
discontinuities, and worse things
like delta function. But worse things than
that we will ignore, and there are worse
things than that. Maybe a potential discontinues
at every point, or maybe a potential has delta
functions and derivatives of delta functions. Or potentials that blow up
and do all kinds of things. And I’m not saying you
should never consider that. I’m saying that we don’t know
of any very useful case where you get anything
interesting with that. But a conceivable a particular
time a singular potential one day could be used. So we’ll look at
these potentials and try to understand how to
set up boundary conditions. And we’re going to worry
about basically psi and how does it behave. And my first claim is that
psi of x has to be continuous. So psi of x cannot jump. The wave function move
along but cannot jump. And the reason is a
differential equation. Look, if psi of x
was not continuous, if psi of x was
like this, and just had a discontinuity,
psi of x equal to x, psi prime of x would
contain a delta function and this is continuity. The derivative is infinite. And psi double prime of
x, the second derivative, would have a derivative of a
delta function which is worse because a delta
function, we think of it as a spike that is becoming
thinner and higher, but the derivative of the delta
function first goes to infinity and then goes to minus infinity
and then comes back up. It’s much worse in many ways. And look, if you have
this differential equation and psi is not continuous,
well, the right-hand side is not continuous. Or if you have a
delta function, then something not continuous,
but left-hand side, we’ve had a derivative of a
delta function that is nowhere on the right-hand side. On the right-hand side,
the worst that could exist is a delta function in v of x. But the derivative of a
delta function doesn’t exist. So you cannot afford to have
a psi that is discontinuous. Psi has to be continuous. There’s other ways
to argue this. You might put them
in your notes, but I’ll leave it like that. Now how about the next case? I will say the
following happens too. Sine prime of x is
continuous unless v of x has a delta function. You see, potentials of
delta functions are nice, they are interesting. We will consider that. Delta functions potentials
can be attractive potentials, repulsive
potentials of [INAUDIBLE]. So I claim now that psi prime
of x has to also be continuous. Why are we worrying
about psi and psi prime is because you need two
conditions whenever you’re going to solve this differential
equation at an interface, you will need to know
psi is continuous and psi prime is
continuous because of second-order
differential equations. So suppose psi
prime is continuous. Then there is no problem. If psi prime is continuous,
the worse that can happen is that the second
derivative is discontinuous. And the second derivative
is discontinuous could happen with a potential
of this discontinuous, so one problem if psi
prime is continuous. But psi prime can fail to be
continuous if the potential has a delta function. And let’s see that. If psi prime is
discontinuous, then psi double prime is proportional
to a delta function. If psi prime is
discontinuous, double prime is proportional to
a delta function. But here psi just
takes some value– there’s nothing
strange about it– in order to have delta function,
which is psi double prime. To be equal to the
right-hand side, v of x must have a delta function. And v will have
a delta function. So it will be a somewhat
similar potential, but we’re going to look at
them in about a week from now. But this will be our
guidance to solve problems. The continuity of
the wave function and the continuity of the
derivative of the wave function. And for this slightly
more complicated problems in which the potential
has a delta function, then you will have a
discontinuity in psi prime, and it will be calculable,
and it’s manageable, and it’s all very nice. Now, we do it a
little complicated, and everything is
mixed up, but you will see that it’s quite doable.

#### 3 thoughts on “L10.4 Comments on the spectrum and continuity conditions.”

• nickt says:

9:20 if you make V such that it looks like d'' (the spike up spike down), then won't psi be a step?

• Robin Nikolić says:

doesn't he actually mean continuously differentiable when he says continuous?

• Zachary Thatcher says:

He claims right that the begining that the eigenstates of the operator span the space, but thus far he has only made this claim for general wave equations and not stationary state equations. Can anyone clarify why this is not a big jump? Going back, I don't think it is ever proven for general wave equations.