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PROFESSOR: OK, so
I want to start out
by reviewing a few
things and putting
some machinery together.
Unfortunately, this
thing is sort of stuck.
We're going to need a
later, so I don't know.
I'll put it up for now.
So first just a bit of notation.
This symbol, you
should think of it
like the dot product,
or the inner product.
It's just saying that
bracket f g is the integral.
It's a number that you get.
So this is a number that
you get from the function
f and the function g
by taking f, taking
its complex conjugate,
multiplying it by g,
and then integrating
overall positions.
All right?
So it's a way to get a number.
And you should think about it
as the analog for functions
of the dot product for vectors.
It's a way to get a
number out of two vectors.
And so, for example, with
vectors we could do v dot w,
and this is some number.
And it has a nice
property that v dot v,
we can think it as v squared,
it's something like a length.
It's strictly positive,
and it's something
like the length of a vector.
Similarly, if I take f and
take its bracket with f,
this is equal to the
integral dx of f squared,
and in particular, f could be
complex, so f norm squared.
This is strictly non-negative.
It could vanish, but it's
not negative at a point,
hence the norm squared.
So this will be zero
if and only if what?
f is 0, f is the
0 function, right.
So the same way that
if you take a vector,
and you take its dot product
with itself, take it the norm,
it's 0 if an only
if the vector is 0.
So this beast satisfies a lot
of the properties of a dot
product.
You should think about
it as morally equivalent.
We'll talk about that
in more detail later.
Second, basic postulate
of quantum mechanics,
to every observable is
associated an operator,
and it's an operator acting
on the space of functions
or on the space
of wave functions.
And to every operator
corresponding
to an observable in
quantum mechanics
are associated a special
set of functions called
the eigenfunctions, such
that when the operator acts
on that function, it gives
you the same function
back times a constant.
What these functions
mean, physically,
is they are the wave functions
describing configurations
with a definite value of the
corresponding observable.
If I'm in an eigenfunction
of position with
eigenvalue x naught, awesome.
Thank you, AV person, thank you.
So if your system is described
by a wave function which
is an eigenfunction
of the position
operator with
eigenvalue x naught,
that means you can be
confident that the system is
in the configuration
corresponding
to having a definite
position x naught.
Right?
It's not a superposition
of different positions.
It is at x naught.
Similarly, momentum,
momentum has eigenfunctions,
and we know what these guys are.
These are the exponentials,
e to the iKX's.
They're the eigenfunctions, and
those are the wave functions
describing states with
definite value of the momentum,
of the associated observable.
Energy as an operator, energy is
described by an operator, which
has eigenfunctions which I'll
call phi sub n, with energy
as E sub n, those
are the eigenvalues.
And if I tell you that your
wave function is the state phi
sub 2, what that tells
you is that the system has
a definite energy, E
sub 2, corresponding
to that eigenvalue.
Cool?
And this is true for
any physical observable.
But these are sort
of the basic ones
that we'll keep focusing on,
position, momentum, and energy,
for the next while.
Now a nice property about
these eigenfunctions
is that for different
eigenvalues,
the associated wave functions
are different functions.
And what I mean by
saying they're different
functions is that
they're actually
orthogonal functions in the
sense of this dot product.
If I have a state
corresponding to be at x 0,
definite position
x 0, that means
they're in eigenfunction of
position with eigenvalue x 0,
and I have another
that corresponds
to being at x1, an eigenfunction
of the position operator
or the eigenvalue x1, then
these wave functions are
orthogonal to each other.
And we get 0 if x 0
is not equal to x1.
Everyone cool with that?
Now, meanwhile not only
are they orthogonal
but they're normalized
in a particular way.
The inner product gives
me a delta function,
which goes beep once, so that
if I integrate against it
I get a 1.
Same thing with momentum.
And you do this, this you're
checking on the problem set.
I don't remember if it
was last one or this one.
And for the energies,
energy 1, if I
know the system is in state
energy 1, and let's say e sub n
and e sub m, those
are different states
if n and m are not
equal to each other.
And this inner product
is 0 if n and m are not
equal to each other
and 1 if they are.
Their properly normalized.
Everyone cool with that?
Yeah.
AUDIENCE: Is it possible
that two eigenfunctions
have the same eigenvalue?
PROFESSOR: Absolutely.
It is absolutely possible
for two eigenfunctions
to have the same eigenvalue.
That is certainly possible.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, good.
Thank you, this is
a good technicality
that I didn't want to get into,
but I'll go and get into it.
It's a very good question.
So the question is,
is it possible for two
different eigenfunctions to
have the same eigenvalue.
Could there be two states
with the same energy ,
different states, same energy?
Yeah, that's
absolutely possible.
And we'll run into that.
And there's nice
physics encoded in it.
But let's think about
what that means.
The subsequent question is
well, if that's the case,
are they really
still orthogonal?
And here's the crucial thing.
The crucial thing is, let's
say I take one function,
I'll call the function phi 1,
consider the function phi 1.
And let it have energy E1,
so that E acting on phi 1
is equal to E1 phi 1.
And let there be
another function, phi 2,
such that the energy
operator acting on phi 2
is also equal to E1 phi 2.
These are said to be degenerate.
Degenerate doesn't mean you
go out and trash your car,
degenerate that the
energies are the same.
So what does this tell me?
This tells me a cool fact.
If I take a wave function phi,
and I will call this phi star,
in honor of Shri
Kulkarni, so I've
got this phi star, which is
a linear combination alpha
phi 1 plus beta phi 2,
a linear combination
of them, a superposition
of those two states.
Is this also an
energy eigenfunction?
Yeah, because if I act on phi
star with E, then it's linear,
so E acting on phi star is
E acting on alpha phi 1,
alpha's a constant,
doesn't care.
Phi 1 gives me an E1.
Similarly, E acting on
phi 2 gives me an E1.
So if I act with E
on this guy, this
is equal to, from both of these
I get an overall factor of E1.
So notice that we get the
same vector back, times
a constant, a common constant.
So when we have
degenerate eigenfunctions,
we can take arbitrary
linear combinations to them,
get another degenerate
eigenfunction.
Cool?
So this is like,
imagine I have a vector,
and I have another vector.
And they share the property
that they're both eigenfunctions
of some operator.
That means any linear
combination of them
is also, right?
So there's a whole
vector space, there's
a whole space of
possible functions
that all have the
same eigenvalue.
So now you say, well, look,
are these two orthogonal
to each other?
No.
These two?
No.
But here's the thing.
If you have a vector space,
if you have a the space,
you can always find
orthogonal guys and a basis
for that space, yes?
So while it's not true that
the eigenfunctions are always
orthogonal, it is true--
we will not prove this, but we
will discuss the proof of it
later by pulling the
mathematician out
of the closet--
the proof will say that
it is possible to find
a set of eigenfunctions which
are orthogonal in precisely
this fashion, even if
there are degeneracies.
OK?
That theorem is called
the spectral theorem.
And we'll discuss it later.
So it is always
possible to do so.
But you must be alert that
there may be degeneracies.
There aren't always
degeneracies.
In fact, degeneracies
are very special.
Why should two numbers
happen to be the same?
Something has to be forcing
them to be the same.
That's going to be an
important theme for us.
But it certainly is possible.
Good question.
Other questions?
Yeah.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, so using
the triangular brackets--
so there's another notation for
the same thing, which is f g,
but this carries some
slightly different weight.
It mean something slightly--
you'll see this in books,
and this means something
very similar to this.
But I'm not going to
use this notation.
It's called Dirac notation.
We'll talk about it
later in the semester,
but we're not going to
talk about it just yet.
But when you see
this, effectively it
means the same thing as this.
This is sort of like dialect.
You know, it's like
French and Quebecois.
Other questions?
My wife's Canadian.
Other questions?
OK.
So given this fact,
given the fact
that we can
associate observables
to operators, operators
come with special functions,
the eigenfunctions, those
eigenfunctions corresponding
to have a definite
value of the observable,
and they're orthonormal.
This tells us,
and this is really
the statement of the
spectral theorem,
that any function can be
expanded in a basis of states
with definite values
of some observable.
So for example,
consider position.
I claim that any
wave function can
be expanded as a
superposition of states
with definite position.
So here's an arbitrary
function, here's
this set of states with definite
position, the delta functions.
And I can write any
function as a superposition
with some coefficients of
states with definite position,
integrating over all
possible positions, x0.
And this is also sort
of trivially true,
because what's this integral?
Well, it's an integral, dx0
over all possible positions
of this delta function.
But we're evaluating
at x, so this is 0
unless x is equal to x0.
So I can just put
in x instead of x0,
and that gives me psi of x.
Sort of tautological We
can do the same thing
for momentum eigenfunctions.
I claim that any
function can be expanded
in a superposition of
momentum eigenfunctions, where
I sum over all possible
values in the momentum
with some weight.
This psi tilde of
K is just telling
me how much amplitude there
is at that wave number.
Cool?
But this is the Fourier theorem,
it's a Fourier expansion.
So purely mathematically,
we know that this is true.
But there's also the
physical statement.
Any state can be expressed
as a superposition of states
with definite momentum.
There's a math in here, but
there's also physics in it.
Finally, this is less obvious
from a mathematical point
of view, because I
haven't even told you
what energy is, any wave
function can be expanded
in states with definite energy.
So this is a state, my state
En, with definite energy,
with some coefficient summed
over all possible values
of the energy.
Given any physical observable,
any physical observable,
momentum, position,
angular momentum,
whatever, given any
physical observable,
a given wave function
can be expanded
as some superposition of
having definite values of that.
Will it in general have definite
values of the observable?
Well a general state be
an energy eigenfunction?
No.
But any state is a superposition
of energy eigenfunctions.
Will a random state
have definite position?
Certainly not.
You could have
this wave function.
Superposition.
Yeah.
AUDIENCE: Why is the
energy special such
that you can make an arbitrary
state with a countable number
of energy eigenfunctions
rather than having
to do a continuous spectrum?
PROFESSOR: Excellent question.
So I'm going to phrase
that slightly differently.
It's an excellent
question, and we'll
come to that at the
end of today's lecture.
So the question is,
those are integrals, that
is a sum over discrete things.
Why?
Why is the possible values
of the position continuous,
possible values of momentum
continuous, and possible values
of energy discrete?
The answer to this
will become apparent
over the course of your
next few problem sets.
You have to do some
problems to get
your fingers dirty to
really understand this.
But here's the
statement, and we'll
see the first version of this
at the end of today's lecture.
Sometimes the allowed energies
of a system, the energy
eigenvalues, are discrete.
Sometimes they are continuous.
They will be discrete when you
have bound states, states that
are trapped in some
region and aren't allowed
to get arbitrarily far away.
They'll be continuous
when you have states that
can get arbitrarily far away.
Sometimes the momentum will be
allowed to be discrete values,
sometimes it will be allowed
to be continuous values.
And we'll see exactly
why subsequently.
But the thing I
want to emphasize
is that I'm writing this to
emphasize that it's possible
that each of these can be
discrete or continuous.
The important thing is that once
you pick your physical system,
you ask what are the allowed
values of position, what
are the allowed
values of momentum,
and what are the allowed
values of energy.
And then you sum over
all possible values.
Now, in the examples we looked
at yesterday, or last lecture,
the energy could
have been discrete,
as in the case of
the infinite well,
or continuous, as in the
case of the free particle.
In the case of a
continuous particle
this would have
been an integral.
In the case of the system
such as a free particle, where
the energy could take any
of a continuous number
of possible values, this would
be a continuous integral.
To deal with that,
I'm often going
to use the notation, just
shorthand, integral sum.
Which I know is a
horrible bastardization
of all that's good and
just, but on the other hand,
emphasizes the fact
that in some systems
you will get continuous,
in some systems discrete,
and sometimes you'll have
both continuous and discrete.
For example, in
hydrogen, in hydrogen
we'll find that there
are bound states
where the electron is stuck
to the hydrogen nucleus,
to the proton.
And there are discrete
allowed energy levels
for that configuration.
However, once you
ionize the hydrogen,
the electron can add
any energy you want.
It's no longer bound.
It can just get
arbitrarily far away.
And there are an uncountable
infinity, a continuous set
of possible states.
So in that situation,
we'll find that we
have both the discrete
and continuous series
of possible states.
Yeah.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, sure,
if you work on a lattice.
So for example, consider the
following quantum system.
I have an undergraduate.
And that undergraduate has
been placed in 1 of 12 boxes.
OK?
Now, what's the state
of the undergraduate?
I don't know.
Is it a definite position state?
It might be.
But probably it's a
superposition, an arbitrary
superposition, right?
Very impressive
undergraduates at MIT.
OK, other questions.
Yeah.
AUDIENCE: Do these
three [INAUDIBLE]
hold even if the probability
changes over time?
PROFESSOR: Excellent question.
We'll come back to that.
Very good question,
leading question.
OK, so we have this.
The next thing is that
energy eigenfunctions satisfy
some very special properties.
And in particular,
energy eigenfunctions
have the property from the
Schrodinger equation i h
bar d t on psi of x and t is
equal to the energy operator
acting on psi of x and t.
This tells us that
if we have psi
x 0 time t 0 is equal to phi
n of x, as we saw last time,
then the wave function,
psi at x at time t
is equal to phi n of x.
And it only changes by an
overall phase, e to the minus i
En t over h bar.
And this ratio En
upon h bar will often
be written omega n is
equal to En over h bar.
This is just the
Dupre relations.
Everyone cool with that?
So are energy eigenfunctions--
how to say.
No wave function is more
morally good than another.
But some are
particularly convenient.
Energy eigenfunctions
have the nice property
that while they're not
in a definite position
and they don't necessarily
have a definite momentum,
they do evolve over time in
a particularly simple way.
And that and the
superposition principle
allow me to write the following.
If I know that this is my wave
function at psi at x at time 0,
so let's say in all these cases,
this is psi of x at time 0,
how does this state
evolve forward in time?
It's kind of complicated.
How does this description,
how does psi tilde of k
evolve forward in time?
Again, kind of complicated.
But when expressed in terms
of the energy eigenstates,
the answer to how it
evolves forward in time
is very simple, because
I know that this
is a superposition, a
linear combination of states
with definite energy.
States with definite
energy evolve with a phase.
And the Schrodinger
equation is linear,
so solutions of the
Schrodinger equation
evolve to become solutions
of the Schrodinger equation.
So how does this state
evolve forward in time?
It evolves forward with a phase,
e to the minus i omega n t.
One for every different
terms in this sum.
Cool?
So we are going to harp
on energy functions,
not because they're more moral,
or more just, or more good,
but because they're more
convenient for solving the time
evolution problem in
quantum mechanics.
So most of today is going
to be about this expansion
and qualitative features
of energy eigenfunctions.
Cool?
OK.
And just to close
that out, I just
want to remind you of
a couple of examples
that we did last time,
just get them on board.
So the first is a free particle.
So for free particle, we have
that our wave functions--
well, actually let me
not write that down.
Actually, let me skip
over the free particle,
because it's so trivial.
Let me just talk about
the infinite well.
So the potential is
infinite out here,
and it's 0 inside
the well, and it
goes from 0 to L. This is
just my choice of notation.
And the energy
operator, as usual,
is p squared upon
2m plus u of x.
You might say, where
did I derive this,
and the answer is I
didn't derive this.
I just wrote it down.
It's like force in
Newton's equations.
You just declare some
force and you ask,
what system does is model.
So here's my system.
It has what looks like a
classical kind of energy,
except these are all operators.
And the potential here is this
guy, it's 0 between 0 and L,
and it's infinite elsewhere.
And as we saw last
time, the solutions
to the energy
eigenvalue equation
are particularly simple.
Phi sub n of x is
equal to root properly
normalized 2 upon
L sine of Kn x,
where kn is equal to
n plus 1 pi, where
n is an integer upon L.
And these were chosen to
satisfy our boundary conditions,
that the wave function must
vanish here, hence the sine,
and K was chosen so that it
turned over and just hit 0
as we got to L. And that gave us
that the allowed energies were
discrete, because the En, which
you can get by just plugging
into the energy
eigenvalue equation,
was equal to h bar squared
Kn squared upon 2m.
So this tells us a nice thing.
First off, in this system,
if I take a particle
and I throw it in here
in some arbitrary state
so that at time t
equals zero the wave
function x 0 is equal to
sum over n phi n of x Cn.
OK?
Can I do this?
Can I just pick some
arbitrary function
which is a superposition
of energy eigenstates?
Sure, because any function is.
Any function can be
described as a superposition
of energy eigenfunctions.
And if I use the
energy eigenfunctions,
it will automatically satisfy
the boundary conditions.
All good things will happen.
So this is perfectly
fine initial condition.
What is the system at time t?
Yeah, we just pick
up the phases.
And what phase is this guy?
It's this, e to the
minus i omega n t.
And when I write omega n, let
me be more explicit about that,
that's En over h bar.
So that's h bar Kn
squared upon 2m t.
Cool?
So there is our solution for
arbitrary initial conditions
to the infinite square well
problem in quantum mechanics.
And you're going to study this
in some detail on your problem
set.
But just to start with a
little bit of intuition,
let's look at the wave
functions and the probability
distributions for the
lowest lying states.
So for example, let's
look at the wave
function for the ground state,
what I will call psi sub 0.
And this is from 0 to L.
And I put these bars here
not because we're
looking at the potential.
I'm going to be plotting the
real part of the wave function.
But I put these walls
here just to emphasize
that that's where the walls are,
at x equals 0 and x equals L.
So what does it look like?
Well, the first one is
going to sine of Kn x.
n is 0.
Kn is going to be pi upon L.
So that's again just this guy.
Now, what's the
probability distribution
associated with psi 0?
Where do you find the particle?
So we know that it's just
the norm squared of this wave
function and the norm
squared is here at 0, it's 0
and it rises
linearly, because sine
is linear for small values.
That makes this
quadratic, and a maximum,
and then quadratic again.
So there's our
probability distribution.
Now, here's a funny thing.
Imagine I take a particle,
classical particle,
and I put it in a box.
And you put it in a box,
and you tell it, OK, it's
got some energy.
So classically it's
got some momentum.
So it's sort of
bouncing back and forth
and just bounces off the
arbitrarily hard walls
and moves around.
Where are you most likely
to find that particle?
Where does it spend
most of its time?
It spends the same amount
of time at any point.
It's moving at
constant velocity.
It goes boo, boo,
boo, boo, right?
So what's the
probability distribution
for finding it at any
point inside, classically?
Constant.
Classically, the probability
distribution is constant.
You're just as likely
to find it near the wall
as not near the wall.
However, quantum mechanically,
for the lowest lying
state that is clearly not true.
You're really likely to
find it near the wall.
What's up with that?
So that's a question that
I want to put in your head
and have you think about.
You're going to see a similar
effect arising over and over.
And we're going to
see at the very end
that that is directly related,
the fact that this goes to 0,
is directly related,
and I'm not kidding,
to the transparency of diamond.
OK, I think it was pretty cool.
They're expensive.
It's also related
to the transparency
of cubic zirconium, which
I guess is less impressive.
So the first state,
again, let's look
at the real part of psi 1,
the first excited state.
Well, this is now a
sine with one extra--
with a 2 here, 2 pi,
so it goes through 0.
So the probability distribution
associated with psi 1,
and I should say write this
as a function of x, looks
like, well, again,
it's quadratic.
But it has a 0
again in the middle.
So it's going to look like--
oops, my bad art defeats me.
OK, there we go.
So now it's even worse.
Not only is unlikely
to be out here,
it's also very unlikely
to be found in the middle.
In fact, there is 0 probability
you'll find it in the middle.
That's sort of surprising.
But you can quickly
guess what happens as you
go to very high energies.
The real part of psi n let's
say 10,000, 10 to the 4,
what is that going to look like?
Well, this had no
0s, this had one 0,
and every time you
increase n by 1,
you're just going to add
one more 0 to the sign.
That's an interesting
and suggestive fact.
So if it's size of
10,000, how many nodes
are there going to be in
the middle of the domain?
10,000.
And the amplitude is
going to be the same.
I'm not to be able to do
this, but you get the idea.
And now if I construct the
probability distribution,
what's the probability
distribution going to be?
Probability of the 10,000th
psi sub 10 to the 4 of x.
Well, it's again going
to be strictly positive.
And if you are not able to make
measurements on the scale of L
upon 10,000, but just say like
L over 3, because you have
a thumb and you don't have an
infinitely accurate meter, what
do you see?
You see effectively a constant
probability distribution.
And actually, I
shouldn't draw it there.
I should draw it
through the half,
because sine squared over
2 averages to one half,
or, sorry, sine squared averages
to one half over many periods.
So what we see is that
the classical probability
distribution constant
does arise when we look
at very high energy states.
Cool?
But it is manifestly
not a good description.
The classical description
is not a good description.
Your intuition is
crappy at low energies,
near the ground state, where
quantum effects are dominating,
because indeed, classically
there was no minimum energy.
Quantum effects have
to be dominating there.
And here we see that even the
probability distribution's
radically different
than our intuition.
Yeah.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Keep working on it.
So I want you all to
think about what--
you're not, I promise
you, unless you've already
seen some quantum
mechanics, you're
not going to be able to
answer this question now.
But I want you to have it as
an uncomfortable little piece
of sand in the back
of your oyster mind--
no offense-- what
is causing that 0?
Why are we getting 0?
And I'll give you a hint.
In quantum mechanics,
anytime something interesting
happens it's because of
superposition and interference.
All right.
So with all that said, so any
questions now over this story
about energy
eigenfunctions expanding
in a basis, et cetera,
before we get moving?
No, OK.
In that case, get
out your clickers.
We're going to test
your knowledge.
Channel 41, for those of
you who have to adjust it.
[CHATTER]
Wow.
That's kind of worrying.
Aha.
OK, ready?
OK, channel 41, and here we go.
So go ahead and start now.
Sorry, there was a little
technical glitch there.
So psi 1 and psi
2 are eigenstates.
They're non-degenerate, meaning
the energies are different.
Is a superposition psi 1 plus
psi 2 also an eigenstate?
All right, four more seconds.
All right.
I want everyone to turn
to the person next to you
and discuss this.
You've got about 30 seconds
to discuss, or a minute.
[CHATTER]
All right.
I want everyone, now that you've
got an answer, click again,
put in your current best guess.
Oh, wait, sorry.
For some reason I have
to start over again.
OK, now click.
This is the best.
I'm such a convert to clickers,
this is just fantastic.
So you guys went from,
so roughly you all
went from about 30, 60,
10, to what are we now?
8, 82, and 10.
So it sounds like you guys
are predicting answer b.
And the answer is--
I like the suspense.
There we go.
B, good.
So here's a quick question.
So why?
And the reason why is that if
we have E on psi 1 plus psi 2,
this is equal to E on psi 1 plus
E on psi 2, operator, operator,
operator, but this
is equal to E 1
psi 1 E 2 psi 2, which if
E1 and E2 are not equal,
which is not equal to E
times psi 1 plus psi 2.
Right?
Not equal to E anything
times psi 1 plus psi 2.
And it needs to
be, in order to be
an eigenfunction, an
eigenfunction of the energy
operator.
Yeah.
AUDIENCE: So I was
thinking about this,
if this was kind of a
silly random case where
one of the energies is 0.
Does this only happen if you
have something that's infinite?
PROFESSOR: Yeah, that's
a really good question.
So first off, how do
you measure an energy?
Do you ever measure an energy?
Do you ever measure a
voltage, the actual value
of the scalar potential,
the electromagnetic scalar
potential?
No.
You measure a difference.
Do you ever measure the energy?
No, you measure a
difference in energy.
So the absolute value of energy
is sort of a silly thing.
But we always talk
about it as if it's not.
We say, that's got energy 14.
It's a little bit suspicious.
So to answer your
question, there's
nothing hallowed
about the number 0,
although we will often
refer to zero energy
with a very specific meaning.
What we really mean
in that case is
the value of the potential
energy at infinity.
So when I say energy,
usually what I mean
is relative to the
value at infinity.
So then let me ask
your question again.
Your question is it
possible to have energy 0?
Absolutely, and we'll see that.
And it's actually going to
be really interesting what's
true of states with
energy 0 in that sense.
Second part of your
question, though,
is how does energy
being 0 fit into this?
Well, does that save us?
Suppose one of
the energies is 0.
Then that says E on psi
1 plus psi 2 is equal to,
let's say E2 is 0.
Well, that term is gone.
So there's just the one E1.
Are we in energy eigenstate?
No, because it's
still not of the form
E times psi 1 plus psi 2.
So it doesn't save us, but
it's an interesting question
for the future.
All right.
Next question, four parts.
So the question says x
and p commute to i h bar.
We've shown this.
Is p x equal to i h
bar, and is ip plus cx
the same as cx plus ip?
If you're really unsure you
can ask the person next to you,
but you don't have to.
OK, so this is looking good.
Everyone have an answer in?
No?
Five, four, three,
two, one, OK, good.
So the answer is C, which most
of you got, but not everyone.
A bunch of you put D. So
let's talk through it.
So remember what the definition
of the commutator is.
x with p by definition
is equal to xp minus px.
If we change the
order here, px is
equal to minus
this, px minus xp.
It's just the definition
of the commutator.
So on the other hand, if you
add things, does 7 plus 6
equal 6 plus 7?
Yeah.
Well, of course 6
times 7 is 7 times 6.
So that's not a
terribly good analogy.
Does the order of addition
of operators matter?
No.
Yeah.
Yeah, exactly.
Exactly.
So it's slightly sneaky.
OK, next question.
OK, this one has five.
f and g are both wave functions.
c is a constant.
Then if we take the inner
product c times f with g,
this is equal to what?
Three, two, one, OK.
So the answer is--
so this one definitely discuss.
Discuss with the
person next to you.
[CHATTER]
All right.
OK, go ahead and enter
your guess again,
or your answer again,
let it not be a guess.
OK, 10 seconds.
Wow.
OK, fantastic.
That works like a champ.
So what's the answer?
Yes, complex conjugation.
Don't screw that one up.
It's very easy to forget,
but it matters a lot.
Cursor keeps disappearing.
OK, next one.
A wave function
has been expressed
as a sum of energy
eigenfunctions.
Here I'm calling them mu rather
than phi, but same thing.
Compared to the
original wave function,
the set of coefficients, given
that we're using the energy
basis, the set of coefficients
contains more or less
the same information, or
it can't be determined.
OK, five seconds.
All right.
And the answer is C, great.
OK, next one.
So right now we're normalizing.
OK.
All stationary states, or
all energy eigenstates,
have the form that spatial
and time dependence
is the spatial dependence,
the energy eigenfunction,
times a phase, so that the norm
squared is time independent.
Consider the sum of two
non-degenerate energy
eigenstates psi 1 and psi 2.
Non-degenerate means they
have different energy.
Is the wave function stationary?
Is the probability
distribution time
independent or is
it time dependent?
This one's not trivial.
Oh, shoot.
I forgot to get it started.
Sorry.
It's particularly non-trivial
if you can't enter your answer.
Right.
So go ahead and
enter your answer.
Whoo, yeah.
This one always kills people.
No chatting just yet.
Test yourself,
not your neighbor.
It's fine to look
deep into your soul,
but don't look deep into the
soul of the person sitting next
to you.
All right.
So at this point, chat
with your neighbor.
Let me just give
you some presage.
The parallel strategy's probably
not so good, because about half
of you got it right, and about
half of you got it wrong.
[CHATTER]
All right.
Let's vote again.
And hold on, starting now.
OK, vote again.
You've got 10 seconds
to enter a vote.
Wow.
OK, two seconds.
Good.
So the distribution on
this one went from 30, 50,
20 initially, to now
it is 10, 80, and 10.
Amazingly, you guys got worse.
The answer is C. And I want
you to discuss with each other
why it's C.
[CHATTER]
All right.
OK.
So let me talk you through it.
So the wave function,
we've said psi of x and t
is equal to phi 1 at x, e to
the minus i omega 1 t plus phi
2 of x e to the
minus i omega 2 t.
So great, we take
the norm squared.
What's the probability to
find it at x at time t.
The probability
density is the norm
squared of this guy,
psi squared, which
is equal to phi 1 complex
conjugate e to the plus i omega
1 t plus phi 2 complex conjugate
e to the plus i omega 2t times
the thing itself phi 1
of x e to the minus i
omega 1 t plus phi 2 of x e to
the minus i omega 2t, right?
So this has four terms.
The first term is
psi 1 norm squared.
The phases cancel, right?
You're going to see this
happen a billion times in 804.
The first term is going
to be phi 1 norm squared.
There's another term, which
is phi 2 norm squared.
Again the phases exactly
cancel, even the minus i omega 2
t to the plus i omega 2 t.
Plus phi 2 squared.
But then there are two cross
terms, the interference terms.
Plus phi 1 complex conjugate
phi 2 e to the i omega 1 t
e to the plus i omega 1 t, i
omega 1 t, and e to the minus
i omega 2t, minus omega 2.
So we have a cross-term which
depends on the difference
in frequencies.
Frequencies are like
energies modulo on h-bar,
so it's a difference
in energies.
And then there's
another term, which
is the complex
conjugate of this guy,
phi 2 star times phi 1 phi
2 complex conjugate phi 1
and the phases are also
the complex conjugate
e to the minus i omega 1 minus
omega 2 t of x of x of x of x.
So is there time dependence
in this, in principle?
Absolutely, from the
interference terms.
Were we not in
the superposition,
we would not have
interference terms.
Time dependence comes from
interference, when we expand
in energy eigenfunctions.
Cool?
However, can these vanish?
When?
Sorry, say again?
Great, so when omega 1
equals omega 2, what happens?
Time dependence goes away.
But omega 1 is e 1 over h bar,
omega 2 is e 2 over h bar,
and we started out by saying
these are non-degenerate.
So if they're non-degenerate,
the energies are different,
the frequencies are different,
so that doesn't help us.
How do we kill this
time dependence?
Yes.
If the two functions
aren't just orthogonal
in a functional sense, but
if we have the following.
Suppose phi 1 is like this.
It's 0 everywhere except for
in some lump that's phi 1,
and phi 2 is 0
everywhere except here.
Then anywhere that phi 1
is non-zero, phi 2 is zero.
And anywhere where phi 2
is non-zero, phi 1 is zero.
So this can point-wise vanish.
Do you expect this to
happen generically?
Does it happen for the
energy eigenfunctions
in the infinite square well?
Sine waves?
No.
They have zero at
isolated points,
but they're non-zero
generically.
Yeah, so it doesn't work there.
What about for
the free particle?
Well, those are
just plain waves.
Does that ever happen?
No.
OK, so this is an
incredibly special case.
We'll actually see
it in one problem
on a problem set later on.
It's a very special case.
So technically, the
answer is C. And I
want you guys to
keep your minds open
on these sorts of questions,
when does a spatial dependence
matter and when are
there interference terms.
Those are two
different questions,
and I want you to
tease them apart.
OK?
Cool?
Yeah?
AUDIENCE: Is a valid
way to think about this
to think that you're fixing
the initial frequency
but then you have that
group velocity is still
time dependent.
PROFESSOR: That's a very
good way to think about it.
That's exactly right.
That's a very,
very good question.
Let me say that subtly
differently, and tell me
if this agrees with what
you were just saying.
So I can look at
this wave function,
and I already know that
the overall phase of a wave
function doesn't matter.
That's what it is to say a
stationary state is stationary.
It's got an overall phase
that's the only thing,
norm squared it goes away.
So I can write this as e to
the minus i omega 1 t times phi
1 of x plus phi 2 of x
e to the minus i omega
2 minus omega 1 t.
Is that what you mean?
So that's one way to do this.
We could also do something else.
We could do e to the minus i
omega 1 plus omega 2 upon 2 t.
And this is more,
I think, what you
were thinking of, a sort
of average frequency
and then a relative
frequency, and then
the change in the frequencies
on these two terms.
Absolutely.
So you can organize
this in many, many ways.
But your question gets at
a very important point,
which is that the overall
phase doesn't matter.
But relative phases in a
superposition do matter.
So when does a phase
matter in a wave function?
It does not matter if
it's an overall phase.
But it does matter if it's a
relative phase between terms
in a superposition.
Cool?
Very good question.
Other questions?
If not, then I have some.
So, consider a system
which is in the state--
so I should give you five--
system is in a state which is a
linear combination of n equals
1 and n equals 2 eigenstates.
What's the probability
that measurement
will give us energy E1?
And it's in this superposition.
OK, five seconds.
OK, fantastic.
What's the answer?
Yes, C, great.
OK, everyone got that one.
So one's a slightly more
interesting question.
Suppose I have an infinite
well with width L.
How does the energy,
the ground state energy,
compare to that of a
system with a wider well?
So L versus a larger
L. OK, four seconds.
OK, quickly discuss amongst
yourselves, like 10 seconds.
[CHATTER]
All right.
Now click again.
Yeah.
All right.
Five seconds.
One, two, three,
four, five, great.
OK, the answer is A. OK,
great, because the energy
of the infinite well
goes like K squared.
K goes like 1 over L. So the
energy is, if we make it wider,
the energy if we make it
wider is going to be lower.
And last couple of questions.
OK, so t equals 0.
Could the wave function for an
electron in an infinite square
well of width a, rather than
L, be A sine squared of pi x
upon a, where A is suitably
chosen to be normalized?
All right, you've got
about five seconds left.
And OK, we are at chance.
We are at even
odds, and the answer
is not a superposition
of A and B,
so I encourage you to discuss
with the people around you.
[CHATTER]
Great.
What properties had it
better satisfy in order
to be a viable wave function?
What properties should
the wave function
have so that it's reasonable?
Yeah.
Is it zero at the ends?
Yeah.
Good.
Is it smooth?
Yeah.
Exactly.
And so you can write
it as a superposition.
Excellent.
So the answer is?
Yeah.
All right.
Vote again.
OK, I might have
missed a few people.
So go ahead and start.
OK, five more seconds.
All right.
So we went from 50-50 to 77-23.
That's pretty good.
What's the answer?
A. Why?
Is this an energy eigenstate?
No.
Does that matter?
No.
What properties had this
wave function better
satisfy to be a reasonable wave
function in this potential?
Say again?
It's got to vanish at the walls.
It's got to satisfy the
boundary conditions.
What else must be true
of this wave function?
Normalizable.
Is it normalizable?
Yeah.
What else?
Continuous.
It better not have
any discontinuities.
Is it continuous?
Great.
OK.
Is there any reason that this
is a stupid wave function?
No.
It's perfectly reasonable.
It's not an energy
eigenfunction, but--
Yeah, cool?
Yeah.
AUDIENCE: This is sort
of like a math question.
So to write that
at a superposition,
you have to write it like
basically a Fourier sign
series?
Isn't the [INAUDIBLE]
function even, though?
PROFESSOR: On this domain,
that and the sines are even.
So this is actually odd, but
we're only looking at it from 0
to L. So, I mean
that half of it.
The sines are odd, but we're
only looking at the first peak.
So you could just
as well have written
that as cosine of the
midpoint plus the distance
from the midpoint.
Actually, let me say
that again, because it's
a much better question I
just give it shrift for.
So here's the question.
The question is, look, so
sine is an odd function,
but sine squared is
an even function.
So how can you expand sine
squared, an even function,
in terms of sines,
an odd function?
But think about this physically.
Here's sine squared in our
domain, and here's sine.
Now what do you mean by even?
Usually by even we mean
reflection around zero.
But I could just as well
have said reflection
around the origin.
This potential is symmetric.
And the energy eigenfunctions
are symmetric about the origin.
They're not symmetric about
reflection around this point.
But they are symmetric about
reflection around this point.
That's a particularly
natural place to call it 0.
So I was calling them sine
because I was calling this 0,
but I could have
called it cosine
if I called this
0, for the same Kx.
And indeed, can we
expand this sine
squared function in terms
of a basis of these sines
on the domain 0 to L?
Absolutely.
Very good question.
And lastly, last
clicker question.
Oops.
Whatever.
OK.
At t equals 0, a particle is
described by the wave function
we just saw.
Which of the following is
true about the wave function
at subsequent times?
5 seconds.
Whew.
Oh, OK.
In the last few seconds we had
an explosive burst for A, B,
and C. So our
current distribution
is 8, 16, 10, and 67,
sounds like 67 is popular.
Discuss quickly, very quickly,
with the person next to you.
[CHATTER]
OK, and vote again.
OK, five seconds.
Get your last vote in.
All right.
And the answer is D. Yay.
So let's think about the logic.
Let's go through the logic here.
So as was pointed out by
a student up here earlier,
the wave function
sine squared of pi x
can be expanded in terms of
the energy eigenfunction.
Any reasonable function
can be expanded
in terms of a superposition
of definite energy states
of energy eigenfunctions.
So that means we can
write psi at some time
as a superposition Cn sine of n
pi x upon a e to the minus i e
n t upon h bar, since those are,
in fact, the eigenfunctions.
So we can do that.
Now, when we look at
the time evolution,
we know that each term
in that superposition
evolves with a phase.
The overall wave function
does not evolve with a phase.
It is not an energy eigenstate.
There are going to be
interference terms due
to the fact that
it's a superposition.
So its probability distribution
is not time-independent.
It is a superposition.
And so the wave function doesn't
rotate by an overall phase.
However, we can solve
the Schrodinger equation,
as we did before.
The wave function is
expanded at time 0
as the energy eigenfunctions
times some set of coefficients.
And the time
evolution corresponds
to adding two each independent
term in the superposition
the appropriate phase for
that energy eigenstate.
Cool?
All right.
So the answer is D. And that's
it for the clicker questions.
OK, so any questions on the
clicker questions so far?
OK, those are going to
be posted on the web site
so you can go over them.
And now back to
energy eigenfunctions.
So what I want to talk about
now is the qualitative behavior
of energy eigenfunctions.
Suppose I know I have
an energy eigenfunction.
What can I say generally
about its structure?
So let me ask the question,
qualitative behavior.
So suppose someone hands
you a potential U of x.
Someone hands you some
potential, U of x,
and says, look, I've
got this potential.
Maybe I'll draw it for you.
It's got some wiggles,
and then a big wiggle,
and then it's got a
big wiggle, and then--
do I want to do that?
Yeah, let's do that.
Then a big wiggle, and
something like this.
And someone shows
you this potential.
And they say, look, what are
the energy eigenfunctions?
Well, OK, free
particle was easy.
The infinite square
well was easy.
We could solve
that analytically.
The next involved solving
a differential equation.
So what differential equation
is this going to lead us to?
Well, we know that the
energy eigenvalue equation
is minus h bar
squared upon 2 m phi
prime prime of x plus U of x
phi x, so that's the energy
operator acting on
phi, is equal to,
saying that it's an energy
eigenfunction, phi sub E,
says that it's equal
to the energy operator
acting on this eigenfunction
is just a constant E phi sub
E of x.
And I'm going to work
at moment in time,
so we're going to drop all the
t dependence for the moment.
So this is the
differential equation
we need to solve where U of
x is this god-awful function.
Do you think it's very
likely that you're
going to be able to
solve this analytically?
Probably not.
However, some basic
ideas will help
you get an intuition for
what the wave function should
look like.
And I cannot overstate the
importance of being able
to eyeball a system and guess
the qualitative features of its
wave functions,
because that intuition,
that ability to estimate, is
going to contain an awful lot
of physics.
So let's try to extract it.
So in order to do
so, I want to start
by massaging this
equation into a form which
is particularly convenient.
So in particular, I'm going to
write this equation as phi sub
E prime prime.
So what I'm going to do is
I'm going to take this term,
I'm going to notice this
has two derivatives,
this has no derivatives,
this has no derivatives.
And I'm going to move this
term over here and combine
these terms into E
minus U of x, and I'm
going to divide each side by 2m
upon h bar squared with a minus
sign, giving me that
phi prime prime of E
of x upon phi E of x dividing
through by this phi E
is equal to minus 2m
over h bar squared.
And let's just get
our signs right.
We've got the minus
from here, so this
is going to be E minus U of x.
So you might look at
that and think, well,
why is that any better than
what I've just written down.
But what is the second
derivative of function?
It's telling you not its
slope, but it's telling you
how the slope changes.
It's telling about the
curvature of the function.
And what this is telling me is
something very, very useful.
So for example, let's
look at the function.
Let's assume that
the function is real,
although we know in
general it's not.
Let's assume that the function
is real for simplicity.
So we're going to plot
the real part of phi
in the vertical axis.
And this is x.
Suppose the real part of phi
is positive at some point.
Phi prime prime,
if it's positive,
tells us that not only
is the slope positive,
but it's increasing.
Or it doesn't tell us
anything about the slope,
but it tells us that whatever
the slope, it's increasing.
If it's negative, the slope is
increasing as we increase x.
If it's positive, it's
increasing as we increase x.
So it's telling us that the
wave function looks like this,
locally, something like that.
If phi is negative,
if phi is negative,
then if this
quantity is positive,
then phi prime prime
has to be negative.
But negative is curving down.
So if this quantity, which
I will call the curvature,
if this quantity is positive,
it curves away from the axis.
So this is phi prime prime
over phi greater than 0.
If this quantity is
positive, the function
curves away from the axis.
Cool?
If this quantity is negative,
phi prime prime upon
phi less than 0,
exactly the opposite.
This has to be negative.
If phi is positive, then phi
prime prime has to be negative.
It has to be curving down.
And similarly, if
phi is negative,
then phi prime prime
has to be positive,
and it has to curve up.
So if this quantity is positive,
if the curvature is positive,
it curves away from the axis.
If the curvature is negative,
if this quantity is negative,
it curves towards the axis.
So what does that tell you about
solutions when the curvature is
positive or negative?
It tells you the following.
It tells you that,
imagine we have
a function where phi prime
prime over phi is constant.
And in particular, let's
let phi prime prime over phi
be a constant,
which is positive.
And I'll call that positive
constant kappa squared.
And to emphasize
that it's positive,
I'm going to call
it kappa squared.
It's a positive thing.
It's a real number squared.
What does the
solution look like?
Well, this quantity is positive.
It's always going
to be curving away.
So we have solutions that
look like this or solutions
that look like this.
Can it ever be 0?
Yeah, sure, it could
be an inflection point.
So for example, here the
curvature is positive,
but at this point the curvature
has to switch to be like this.
What functions are of this form?
Let me give you another hint.
Here's one.
Is this curvature positive?
Yes.
What about this one?
Yup.
Those are all
positive curvature.
And these are exponentials.
And the solution to this
differential equation
is e to the plus kappa x
or e to the minus kappa x.
And an arbitrary
solution of this equation
is a superposition A e
to the kappa x plus B
e to the minus kappa x.
Everyone cool with that?
When this quantity
is positive, we
get growing and
collapsing exponentials.
Yeah?
On the other hand, if
phi prime prime over phi
is a negative number, i.e.
minus what I'll call k
squared, then the curvature
has to be negative.
And what functions have
everywhere negative curvature?
Sinusoidals.
Cool?
And the general solution
is A e to the i K x plus B
e to the minus i K x.
So that differential equation,
also known as sine and cosine.
Cool?
So putting that together
with our original function,
let's bring this up.
So we want to think about
the wave functions here.
But in order to think about
the energy eigenstates,
we need to decide on an energy.
We need to pick an
energy, because you
can't find the solution
without figuring the energy.
But notice something nice here.
So suppose the energy is e.
And let me just draw
E. This is a constant.
The energy is this.
So this is the value of E.
Here we're drawing potential.
But this is the value of the
energy, which is a constant.
It's just a number.
If you had a classical particle
moving in this potential,
what would happen?
It would roll around.
So for example, let's say
you gave it this energy
by putting it here.
And think of this as a
gravitational potential.
You put it here, you let
go, and it falls down.
And it'll keep rolling
until it gets up here
to the classical turning point.
And at that point,
its kinetic energy
must be 0, because
its potential energy
is its total energy,
at which point
it will turn around
and fall back.
Yes?
If you take your ball,
and you put it here,
and you let it
roll, does it ever
get here, to this position?
No, because it doesn't
have enough energy.
Classically, this is
a forbidden position.
So given an energy
and given a potential,
we can break the system up
into classically allowed zones
and classically forbidden zones.
Cool?
Now, in a classically
allowed zone,
the energy is greater
than the potential.
And in a classically
forbidden zone,
the energy is less
than the potential.
Everyone cool with that?
But this tells us
something really nice.
If the energy is greater
than the potential,
what do you know
about the curvature?
Yeah.
If we're in a
classically allowed zone,
so the energy is greater
than the potential,
then this quantity is positive,
there's a minus sign here,
so this is negative.
So the curvature is negative.
Remember, curvature
is negative means
that we curve towards the axis.
So in a classically
allowed region,
the wave function
should be sinusoidal.
What about in the classically
forbidden regions?
In the classically
forbidden regions,
the energy is less
than the potential.
That means in magnitude
this is less than this,
this is a negative
number, minus sign,
the curvature is
going to be minus
times a minus is a positive,
so the curvature's positive.
So the solutions are
either growing exponentials
or shrinking exponentials
or superpositions of them.
Everyone cool with that?
So let's think about
a simple example.
Let's work through this
in a simple example.
And let me give you a
little bit more board space.
Simple example would be a
potential that looks like this.
And let's just
suppose that we want
to find an energy eigenfunction
with energy that's E. Well,
this is a classically
allowed zone,
and these are the classically
forbidden regions.
Now I want to ask, what does
the wave function look like?
And I don't want to draw it
on top of the energy diagram,
because wave function
is not an energy.
Wave function is a
different quantity,
because it's got
different axes and I want
it drawn on a different plot.
So but as a function of x--
so just to get the
positions straight,
these are the bounds
of the classically
allowed and forbidden regions.
What do we expect?
Well, we expect that it's
going to be sinusoidal in here.
We expect that it's going
to be exponential growing
or converging out here, exp.
But one last important thing is
that not only is the curvature
negative in here in these
classically allowed regions,
but the magnitude
of the curvature,
how rapidly it's turning over,
how big that second derivative
is, depends on the
difference between the energy
and the potential.
The greater the difference,
the more rapid the curvature,
the more rapid the turning
over and fluctuation.
If the differences between the
potential and the true energy,
the total energy, is small, then
the curvature is very small.
So the derivative
changes very gradually.
What does that tell us?
That tells us that in here the
wave function is oscillating
rapidly, because the
curvature, the difference
between the energy and
the potential is large,
and so the wave function
is oscillating rapidly.
As we get out towards the
classical turning points,
the wave function will be
oscillating less rapidly.
The slope will be
changing more gradually.
And as a consequence,
two things happen.
Let me actually draw this
slightly differently.
So as a consequence
two things happen.
One is the wavelength
gets longer,
because the
curvature is smaller.
And the second is the
amplitude gets larger,
because it keeps on having
a positive slope for longer
and longer, and it takes
longer to curve back down.
So here we have
rapid oscillations.
And then the oscillations get
longer and longer wavelength,
until we get out to the
classical turning point.
And at this point, what happens?
Yeah, it's got to
be [INAUDIBLE]..
Now, here we have some sine,
and some superpositions
of sine and cosines,
exponentials.
And in particular, it
arrives here with some slope
and with some value.
We know this side we've
got to get exponentials.
And so this sum of sines
and cosines at this point
must match the sum
of exponentials.
How must it do so?
What must be true of the
wave function at this point?
Can it be discontinuous?
Can its derivative
be discontinuous?
No.
So the value and the
derivative must be continuous.
So that tells us precisely
which linear combination
of positive growing and
shrinking exponentials we get.
So we'll get some
linear combination,
which may do this for awhile.
But since it's got
some contribution
of positive exponential, it'll
just grow exponentially off
to infinity.
And as the energy gets
further and further away
from the potential, now in
their negative sine, what
happens to the rate of growth?
It gets more and more rapid.
So this just diverges
more and more rapidly.
Similarly, out here we
have to match the slope.
And we know that
the curvature has
to be now positive,
so it has to do this.
So two questions.
First off, is this sketch
of the wave function
a reasonable sketch,
given what we
know about curvature and this
potential of a wave function
with that energy?
Are there ways in which
it's a bad estimate?
AUDIENCE: [INAUDIBLE]
PROFESSOR: OK, excellent.
AUDIENCE: On the right side,
could it have crossed zero?
PROFESSOR: Absolutely, it
could have crossed zero.
So I may have drawn this badly.
It turned out it
was a little subtle.
It's not obvious.
Maybe it actually punched
all the way through zero,
and then it diverged
down negative.
That's absolutely positive.
So that was one of the
quibbles you could have.
Another quibble
you could have is
that it looks like I have
constant wavelength in here.
But the potential's
actually changing.
And what you should
chalk this up to,
if you'll pardon the pun, is
my artistic skills are limited.
So this is always going
to be sort of inescapable
when you qualitatively
draw something.
On a test, I'm not going to bag
you points on things like that.
That's what I want to emphasize.
But the second thing,
is there something
bad about this wave function?
Yes, you've already named it.
What's bad about
this wave function?
It's badly non-normalizable.
It diverges off to infinity
out here and out here.
What does that tell you?
It's not physical.
Good.
What else does it tell
you about this system?
Sorry?
Excellent.
Is this an allowable energy?
No.
If the wave function
has this energy,
it is impossible to
make it continuous,
assuming that I
drew it correctly,
and have it converge.
Is this wave function allowable?
No, because it does not satisfy
our boundary conditions.
Our boundary conditions are that
the wave function must vanish
out here and it must
vanish out here at infinity
in order to be normalizable.
Here we failed.
Now, you can imagine
that-- so let's decrease
the energy a little bit.
If we decrease the energy,
our trial energy just a little
tiny bit, what happens?
Well, that's going to decrease
the curvature in here.
We decrease, we bring the energy
in just a little tiny bit.
That means this is a
little bit smaller.
The potential stays the same.
So the curvature in
the allowed region
is just a little
tiny bit smaller.
And meanwhile,
the allowed region
has got just a
little bit thinner.
And what that will do is the
curvature's a little less,
the region's a little
less, so now we have--
Sorry, I get excited.
And if we tweak the energy,
what's going to happen?
Well, it's going to arrive
here a little bit sooner.
And let's imagine
something like this.
And if we chose the
energy just right,
we would get it to match
to a linear combination
of collapsing and
growing exponentials,
where the contribution from
the growing exponential
in this direction vanishes.
There's precisely one
value of the energy
that lets me do that with
this number of wiggles.
And so then it goes
through and does its thing.
And we need it to
happen on both sides.
Now if I take that solution,
so that it achieves convergence
out here, and it achieves
convergence out here,
and I take that energy and
I increase it by epsilon,
by just the tiniest little bit,
what will happen to this wave
function?
It'll diverge.
It will no longer
be normalizable.
When you have classically
forbidden regions,
are the allowed energies
continuous or discrete?
And that answers a question
from earlier in the class.
And it also is going
to be the beginning
of the answer to the question,
why is the spectrum of hydrogen
discrete.
See you next time.
