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.

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

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

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.