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PROFESSOR: So today
we begin our study
of solids and in
particular of conductivity
in solids and
periodic potentials,
and that diagram will mean
something in a few weeks.
So before I get
started, questions
on everything previous?
Yeah.
AUDIENCE: So when you're talking
about fermions and bosons
[INAUDIBLE].
Statistics.
So we said That you had to write
the statement particular way
for fermions with a
minus sign particular.
PROFESSOR: Yes.
AUDIENCE: Well, we
were assuming that you
had to write the statement a
product statement of the two.
Will it be that you
can't do that sometimes
because they can't be separated.
PROFESSOR: Absolutely.
That's a wonderful question.
We'll come to that in the second
next to the last lecture where
we'll talk about something
called entanglement, which
is really what
you're talking about.
But here's the crucial
thing, for fermions,
so what I wrote
last time, I said
suppose we have two particles
that are identical one is
in the state we know one
is in the state chi of x
and the other is in the
state described by phi of x.
I'm going to call the position
of the first vertical x
and the position of
the second particle y.
And I can write the two states
without property, well, three.
I could fill many, but two in
particular of the following
from, chi of x phi of y plus
or minus chi of y phi of x.
And the point of this
linear combination
is that when they exchange
positions, the x and y
So here this is the amplitude
for the first particle
will be an x, and the
second particle to be a y.
If I switch the sign or the
positions to psi plus minus
of yx or if the first
particle is at y,
the second particle is at x,
this is equal to just swap
these two terms, plus or
minus psi plus minus of x,y.
So this wave function
has been contracted
to ensure that under the
exchange of the position
of the two particles, the wave
function stays the same up
to a sign.
This plus sign is associated
with systems we call bosons,
and the minus sign is associated
with systems we call fermions.
Now as I've
discussed previously,
this is a persistent
property of a system.
If a system is bosonic,
at some moment in time,
it will always be bosonic.
If it's fermionic in
some moment in time,
it will always be
fermionic, and that
is a consequence of the fact
that the exchange operator that
swaps the two particles commutes
with the energy eigenfunction
if they're identical particles.
And so if it commutes with
the energy operator, then
its value is preserved
under time evolution
with that energy operator.
So now the question
that was asked is,
if this is our definition
of bosons and fermions,
is it true that you can
always write the wave function
in this form of differences
of products of two states?
Well, the answer
is both yes and no.
The answer is yes for
sufficiently-simple systems.
The answer is no when you
have many, many particles,
and we can't always write it as
a simple product of two states.
We can't always write
as the sum of two terms.
What you will discover if you
studied multi-particle systems
in 805 and 806 is that there's
a nice way of organizing
this anti
symmetrization property,
and it doesn't necessarily
have two terms,
but it may have
many, many terms.
With that said, this is
not the defining property.
This is really the
defining property,
whether your a
boson or a fermion.
OK?
Cool.
Yeah.
AUDIENCE: So I had a question
about fermions. [INAUDIBLE]
the explanation being
makes perfect sense
about why two things can't
be in the same state.
PROFESSOR: Yeah.
AUDIENCE: But Used in
804-4, during recitation
one of the instructors
said, well actually, this
takes like really hard
quantum field theory
to actually prove this.
What does that actually mean?
PROFESSOR: OK.
Good.
So the question
is, look, there's
this legendary
statement that this fact
that the Pauli exclusion
principle is incredibly hard
and requires the magical
details of quantum field theory
to explain, and that
some surprising.
So your recitation
lecturer said this to you.
AUDIENCE: Yes.
PROFESSOR: I'm sure I
love this person dearly,
but I disagree with him
in an important way.
So there are two
things are confusing.
Two things that require
explanation I should say.
One, we can explain
at the level of 804,
and the other does
require machinery,
but it's not crazy
complicated machinery.
So let me tell you what
we can explain with 804,
and then let me
sketch for you how
the more sophisticated
argument goes.
So this fact, the basic
fact that wave functions can
be symmetric or
anti-symmetric, and when
you have identical
particles, they
must be either symmetric
or anti-symmetric.
This follows from
the statement that we
have identical particles.
If you have identical
particles, it
must be true that
under the exchange,
they are either even or odd.
They're either symmetric or
anti-symmetric under exchange.
So if we have an exchange
of identical particles,
we need to have identical
symmetric or identical
anti-symmetric.
And now it's just an
empirical question.
What particles in the
world are identical bosons
and what particles are
identical fermions.
Identical fermions?
Well, how could you
tell the difference?
You can tell that
identical fermions do not
want to be the same state,
and identical bosons
do as we saw last time,
as an extra factor of two
for two particles that are
actually n for n particles.
So bosons want to be
in the same state.
Fermions want to not
be in the same state,
so if we just look
at the world and say,
which particles are
well modeled by assuming
we have bosons, which
particles are well
modeled by assuming
we have fermions.
And answer turns out to be the
following, any particle that
has intrinsic angular momentum,
which is a half integer,
any particle which
spin like the electron,
which has a limit
to 1/2 and we'll
study that in more detail
later, any particle that
has half-integer intrinsic
angular momentum empirically
turns out to be a fermion.
Take two electrons, take
their wave function,
and swap the positions
of the two electrons,
the way function picks
up the minus sign.
Take two neutrons,
also spin 1/2.
Swap them and you
pick up a minus sign.
It's a fermion.
Take two particles of
light, two photons,
swap them and you
get a plus sign,
and these guys
have integer spin.
Angular momentum of light
is 1, and similarly,
if we could, for
the Higgs boson,
if you take two Higgs
bosons and swap them,
you'll discover
that they're bosons
and as a consequence there's
a plus sign under exchange.
So that is not
surprising, it's just
an observable
property of the world.
The thing that's shocking
is why that's true, right?
Why is it that things that
have half-integer angular
momentum, which sounds like
an independent property
from whether they're bosonic
or whether identical bosons are
identified, why is
it that all things
with half-integer angular
momentum or spin are fermions,
and all things with integer
angular momentum like light
or the Higgs boson or
me we're for bosonic,
we're identical
with an even sign,
and that requires
quantum field theory.
But it's not insanely difficult.
It requires the bearest minima
of quantum field theory.
So anyone who wants
to understand,
come to my office hours
and I happily explain that.
It takes about 15
minutes just to give you
the basics of quantum field
theory, just the bare basics.
It's pretty straightforward.
But it's a beautiful thing about
relativistic quantum mechanics,
so I'm going to turn
this all around.
So let's look at the history.
The history of this was people
looked at atomic spectra
and said, this is bizarre.
It almost fits what we
get from central potential
except it does buy this
factor of two integer indices.
Ah-ha.
There must be an
extra quantum number
and on top of that, so the
spin, the two possible states
of the electron.
And secondly, it must
be that electrons
can't be in the same state.
So it sounds like
that's two hypotheses.
Now, if you take
those two hypotheses,
everything else follows.
Fine.
What Dirac discovered
when he studied
the relativistic version of
the Schrodinger equation, which
really is quantum
field theory, but when
he studied the
relativistic version
of electrons, quantum
mechanical relativistic version,
he discovered that these
two things are linked.
It is cool.
I grant you.
So you he discovered that
these two things are linked,
and he identified what's
referred to as the spin
statistics theorem.
If you know the spin, you
determine the statistics.
So exactly how that
works, that does
require a little bit
of effort-- that's
the 15 minutes-- but you can
dispense with that entirely
and not be at all shocked by
anything if you just accept
two principles instead of
one, first, that electrons are
anti-fermionic, so they don't
want to be the same state,
and secondly, they
have half-integer spin.
AUDIENCE: Given that the only
thing that's really measurable
is [? the square ?] or
non-square of the wave function
and the way the function itself
is never completely actually,
measurable.
Is there a
mathematical reason why
exchange can't transform
a wave function just
by arbitrary phase?
It has to be plus or minus 1?
PROFESSOR: Well,
the reason it has
to be plus or minus
1 for two particles,
so you said a couple of things.
Let me answer the first part.
So what I will call the first
part of your question is,
why did it have to
be plus or minus 1.
Why couldn't it have
been an arbitrary phase?
And the reason is that if you
do this exchange operation twice
for two particles, you
get back the same thing.
And that's what I mean by saying
I have identical particles.
Because I could
quibble with this.
You could say, well
look, you could
define some different
notion of exchange,
or under exchange, I
pick up an extra phase.
I swap them and then I swap
them and I swap them again,
I get a phase.
But here's the
problem with that.
The problem with that is,
and this can be dealt with,
but an interesting
question is if it can only
be dealt with in
two dimensions or it
can be dealt with in general
number of dimensions.
This is a story called antions,
which Frank Wilczek here
has really pounded hard on.
But here's the basic question.
So does the wave function
have a single value?
I'd like to think it does.
The wave function really
should have a single value.
If it has multiple
values, then it's
ambiguous what the value of
the wave function is there.
That's not good.
That mean you have to specify
more than just the wave
function, you have to
specify which of the values
the wave function takes
in a particular point
in the value of
the wave function.
So if under double
exchange we've
returned to the
original configuration,
you pick up a phase,
something is screwed.
You must be able to tell whether
or not you've exchanged twice,
and here's why you can tell.
Because while it's true that you
can't tell the overall phase,
imagine if I take my system
and I put in a splitter
and I have a beam
splitter I could think
of it as a two-split experiment,
one component of that wave
function I will double swap and
the other component I will not.
And then I will interfere them.
I will combine
them back together,
do an interference experiment.
So you measure that
relative phase.
So that relative phase
definitely matters.
Do you just see that point?
So it's true that
they overall phase
they can't measure, but
by doing a superposition
and only exchanging one of
the superposition pairs,
by physically
separating those, then
you can see that in
interference experiments again.
So you can still deal
with this, but it
requires being able
to know whether or not
you can exchange,
so there has to be
some way of telling that you
sort of entwined these guys,
and that's something you
could do in two dimensions.
It turns out to
be difficult to do
in general number of dimensions.
Well, it's an act
of research topic.
If you want to a more
detailed answer, ask me later.
OK.
One more question.
Yeah.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah. it has
precisely one observable
property, which is this sign,
this it's eigenvalue, value,
and you can tell because the
wave function either vanishes
when you take the two points
together or it doesn't.
So that's it.
AUDIENCE: So it's not like
the momentum [INAUDIBLE].
So it's only basically
just [INAUDIBLE].
PROFESSOR: Yeah.
It's less interesting
than position or momentum.
It has two eigenvalues
rather than many.
But it's no less interesting
than, for example, spin.
It contains information that
you can learn about system.
OK.
One more.
AUDIENCE: I was wondering why
you can write this fermion
as two electrons I guess when
it seems to me [INAUDIBLE].
PROFESSOR: That's a
fantastic question.
OK.
Good.
I was going to gloss over this.
That's a fantastic question.
So here's the question.
Look, I have in my box here,
I have a hydrogen atom,
and I've quantized
the cooling potential,
and I know what the energy
eigenvalues are for the cooling
potential, and I
put the electron
in one of those states.
And I say the wave function
describing my system as sign is
equal to phi [? nlm ?] for
some particular [? nlm. ?]
That's what we've been doing.
But there are a lot of
electrons in the world.
In fact, there are
electrons in the walls.
There are electrons
in your nose.
There are electrons everywhere,
and they are fermion.
So the wave function
describing the entire system
must be invariant up to a sign
under the exchange of any two
electrons.
If I take this
electron and I swap it
with an electron
in Matt's ear-- Hi,
Matt-- then the wave
function had better
pick up a minus sign.
So what business do I ever have
writing a single wave function
for a single electron.
Is that your question?
AUDIENCE: Yes.
PROFESSOR: That's an
excellent question.
So let me answer that.
So suppose I have
two electrons, and I
have the wave function
for two electrons x and y.
I mean, I write this as
if it's in one dimension.
This is some position,
some position,
but this generally is just an
arbitrary number of dimensions,
and one is in the
state chi of x,
and the other is in
the state phi of y.
But I know that I can't
have this be the state
because it must be
invariant under swapping
x and y of two minus signs.
So I need minus
chi at y phi at x,
and let's normalize
this, so 1 over 2.
So that's our electron.
Now, at this point,
you might really worry,
because say chi is the electron
is the wave function bound
to my atom, and phi is the wave
function for the electron bound
to an atom in Matt's ear.
So we've just in writing this.
Why can we do that?
Well, let's think about this.
What does the combination
of terms give us?
What do they tell us is that
they're interference term.
If I ask them what's
the probability
to find an electron at
x and an electron at y,
this is equal to, well,
it's this whole quantity
norm squared sine norm
squared, which is equal to 1/2,
and then there's a term
where this gets term squared,
this term squared, and
then two cross terms.
So the first term
gives us the chi
at x norm squared phi
at y norm squared.
The second term gives us plus
chi at y norm squared phi
at x norm squared plus
twice the real part
of I'm going to write
this subscript chi
sub x phi y, complex conjugate,
complex conjugate chi y phi x.
Everybody agree?
That's just the norm
squared of everything.
AUDIENCE: Shouldn't
that be minus?
PROFESSOR: Oh, shoot.
Sorry.
Thank you.
Minus.
Yes, minus.
Importantly minus.
Now, let's be a little
more explicit about this.
Saying that chi is the state
localized at the hydrogen,
then that's as if
the hydrogen is here.
If the proton is here, then
the wave function for chi
has a norm squared
that looks like this.
And if I say that phi is the
state corresponding to being
localized, at Matt's ear,
then here is the ear,
and here is the wave function
for it, so that's phi.
So let's then ask, what's
the probability that I find
the first electron here and
the second electron here?
Well, if this is
x, and this is y,
what's the probability
that I find
a particle there and
a particle there?
Well, what's this term?
Well, it's chi of x, which
is not so small-- that's just
some number-- times phi of
y, some value norm squared,
so this is some number.
What about this one?
What's chi of y?
Well, chi of y, this is all
sharply-localized wave function
over here chi, and
it's 0 out here.
The probability that an
electron bound to the hydrogen
is found far away is
exponentially small
as we've seen from the
wave functions of hydrogen.
It's bound.
So this is the negligibility
small as is phi of x.
Phi of x is
negligibility do small.
The wave function is
localized around the ear,
but x is way over
there by the hydrogen.
It's exponentially small.
This term is also
exponentially small.
Now in here, what about this?
Chi of x phi of y, that's good.
Those are both fine, but chi
of y phi of x phi of x, these
are both negligibility small,
so these terms go away.
So this is just equal to chi
of x norm squared phi of y norm
squared.
And this is what you
would get approximately,
and that approximately
is fantastically good
as they get far apart.
This chi of x is
just the probability
that the particle that x
given that it was in the wave
function bound to the
hydrogen, and ditto given
that it was bound to the ear.
Cool?
So the important thing is
when fermions are separated
and the wave functions
are localized,
you can treat them
independently.
When the fermions
are not separated
or when their wave
functions are not localized,
you can't treat
them independently
and you have to
include all the terms.
That cool?
OK.
Yeah.
AUDIENCE: That's
times 1/2, right?
PROFESSOR: Yeah,
that is times 1/2.
What did I do?
I'm screwing something up
with the normalization.
Ask me that after lecture.
It's a good question.
I'm screwing something up.
I'm not going to get the
factor of 2 straightened out
right now.
OK.
Does that answer your question?
OK.
Good.
This is a really deep-- it's
important that this is true.
I'm going to stop
questions because I
need to pick up with
what we need to do.
So this is a really
important question
because if it weren't true
that you could independently
deal with these electron,
then everything we
did up until now in
quantum mechanics
would have been totally useless
for any identical particles.
And since as far as we can
tell, all fundamental particles
are identical, this
would have been very bad.
So with that, let's move
on to the study of solids.
So I want to pick up where
your problem set left off.
The last part of your problem
set, which I hope everyone
did, the left part
of your problem set
involved looking
at this simulation
from the PhET
people of a particle
in a series of n wells.
And I want to look at the
result, at least the results
that I got for that simulation.
So let's see, here
are the data points
that I got-- shall
I make that larger?
Is that impossible to see?
Those are the data
points that I pulled off
of PhET just the
same way you did.
You look, you point,
you move the cursor,
and you pull off
the data points.
So those are approximately then.
And this is for 1, 2, 3, 4, 5,
6, 7, 8, 10 wells, and here's
the plot of them as a function
of the index which state they
are.
So as a function, this
is the first state.
This is the second state,
third, fourth, et cetera.
Here are the
energies vertically.
And hopefully, you all got
to plot some more of this
and they seem to become
bunched together,
and the energy of the ground
state without the potential
was this.
So I'm not going to add in
a parabola corresponding
to a free particle with
the associated wavelengths.
Remember I asked you to
say take that state that
looks most like a momentum.
eigenstate with a
definite wavelength
and calculate the energy
of a free particle
with the corresponding
wavelength.
So this parabola-- I
should make this smaller--
is a parabola describing
a free particle which
agrees with those free particle
energies at the three points
where you're supposed
to compute them.
So there's the parabola.
And so what I've done here
is I've grouped them together
into the bands of states.
So vertically we have energy.
On the horizontal
side, we have n
but remember that n is
also corresponding kind
to a momentum because each
state wiggles and especially
the states at the
top of each band
have a reasonably
well-defined momenta,
and so they look a lot
like momentum eigenstates.
So I'm going to interpret
this horizontal direction,
it's something like a momentum.
And at the top of the states,
it certainly makes sense,
but then what it actually is
is just the number the level,
and what we see is
that the energy is
a function of the level
is close-- near the top
anyway-- to the energy
of a free particle,
but the actual allowed
energies are bound.
So which energies correspond
to allowed states,
and which energies
have no allowed states
associate with them is
encoded in this diagram.
Are we cool with that?
In this shaded region,
there are states
with an energy in that region,
at least approximately,
and in this unshaded
region, there aren't any.
And in each band, there
are n energy eigenstates,
and each band corresponds
to one of the bound
states of the potential.
OK.
Everyone cool with that?
So does this look more or
less like what y'all got?
AUDIENCE: Yes.
PROFESSOR: Good.
OK.
Questions about this
before I move on?
It's an important one.
And let me just remind you of a
couple of facts about this PhET
simulation so here's what we
see if we have the single well.
We have three states 1, 0, two
nodes, two zeroes, and they
satisfy the node theorem.
If we have many states, then
if we look at the bottom
state in each well.
So there's the bottom
state and the bottom band.
It has no zeroes.
It satisfies the node theorem,
and the top one by comparison
looks extremely similar.
It looks extremely similar
to the ground state, which
is the orange one, the
lowest energy state.
It is a little higher energy.
You can see that because
the curvature is greater.
If you look at the top and
bottom of each wave function,
you see that the
curvature is greater
for the yellow guy, which is
the guy at the top of the band,
and the yellow guy,
meanwhile, looks
like a wave with a
definite wavelength.
But a general state
in here doesn't
have any simple
symmetry properties.
It's not periodic.
It's not approximately periodic.
Well, that one's kind of
approximately periodic.
But the top guy
and the bottom guy
look approximately periodic.
They have some nice
symmetry properties.
But in general, they're just
not translationally invariant.
They're not symmetric.
They don't have any
simple, nice structure.
They are just some
horrible, ugly things.
The bottom though
and the top guy
always have some nice
symmetry property.
Anyway to keep that in mind.
And the last thing I want
you to note from all of this
is that as we make the
potential barriers stronger,
two things happen.
First off, bands become very
narrow, the gaps between them
become very large, and
secondly, the wave functions
take now an even more sort
of complicated and messy,
which is more obvious that
the wave functions are not
periodic.
This thing it looks
almost periodic,
but is periodic with
some funny period,
and it's not a single period by
any stretch of the imagination.
OK.
So any questions about the
wave functions in here?
Yeah.
AUDIENCE: In this one, there is
an overall [INAUDIBLE] period.
[INAUDIBLE]
PROFESSOR: Excellent.
OK.
So let me give you a little
bit of intuition for that.
So suppose I have a big box.
What's the ground state?
What does the ground
state look like?
Yeah.
Good.
Everyone's doing this.
That's good.
So the ground state
looks like this,
and the first excited
state is going
to do something like
this, and so on.
Cool.
And what's the third excited
state going to look like?
I guess I shouldn't do that.
The third excited state is
going to do something like this.
Cool?
Now, meanwhile, imagine
I take my periodic well
and I add to it a bunch of
delta function scatterers
with some strength.
When the strength is zero, I
just recover my original well.
When the strength is
non-zero though, we
know that what a delta
function is going to do is,
it's going to make
a little kink.
It's going to induce
a little condition
on the first derivative
at the delta function.
So what's going to
this ground state
when we turn a little tiny
bit of that delta function?
Yeah.
What we're going
to get is, we're
going to get
something that kinks
at each of the delta functions.
But it still has to have
those boundary conditions.
Everybody cool with that?
And as we make the delta
function stronger and stronger,
the effect of this is
going to eventually
be to give us something
that looks like-- yeah?
Now that should look
a lot like that.
Everyone cool with that?
Meanwhile, what's going to
happen to the second guy?
Well the second guy, same thing.
The effect on the second
state is going to be-- cool?
I'm not sure of I need
the sound effects.
So let's magnify and let's
look at the second state here.
Ooh, it's going to be hard
to deal with that tiny energy
splitting.
Come on finger, you can do this.
There we go.
There is two.
Ah-ha.
That looks a lot like this.
Yes, that second state looks
an awful lot like this.
And let's go back
to that funny state
that we were looking at before.
Let' see if I can get it.
Should be the third guy here.
So there's a second
guy in the third band.
Again, you see an envelope
and then the fluctuation
from being in the third
state and in the second state
in this band as well.
So on the other hand, what's
the scale of the energy
splitting due to a
well of it this width
compared to the energy
splitting between states
due to a well of this width?
A wide well means that its
states are close together.
An ENDOR well means the
states are far apart.
So what state do we have?
We have the state
that was the ground
state of the whole
box with a correction
due to the delta function
versus states which
are, for example, excited
states in the box.
The splitting between
this state and this state
is going to be much
larger than the splitting
between this state
and this state,
just because this is
a tiny little well
and so the link scale
is much shorter,
the energy is much greater.
That's the intuition you
should have from these guys.
So when you see that there's
some approximate structure,
let's see let's make this
nice and separated again.
And Oh, yeah.
these are kind of
difficult to control
when they're so close together.
The trouble with touch pads.
There we go.
OK.
There we go.
That was the guy we
were looking at before.
And so you see that there
is an overall envelope which
has three maxima, and then
there's the n inside each well.
It's got two zeroes
inside each well.
And it's that envelope which
is coming from the fact
that you're in a box
and the two zeroes
inside each well, which
is coming from this.
All the states here in
this band are of the form
two zeroes inside the well.
All the states here are the
form two zeroes is inside
each well in this
band except they're
being modulated by an overall
sine wave due to the fact
that they're in a box.
And so they're
very closely split
states with a different
overall modulation
but with each state inside
each band corresponding
to either the ground state
for the individual well
or the first excited state
in the individual well
or the second excited state.
So each band corresponds
to which excited state
you are inside the well,
and which state you
are within the band
is your modulation
of the overall wave
from being inside a box.
That make sense?
AUDIENCE: I guess so, but why
does the overall modulation
have a smaller effect than the--
PROFESSOR: Great.
If I have a well
that's this wide,
what are the ground
state energies?
And let's say the
width is L. What
are the energies in this square
well, an infinite well with L?
Yeah, but what are
the eigenvalues?
AUDIENCE: [INAUDIBLE].
PROFESSOR: E sub n.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah.
Exactly.
So they go like, I'm just
going to write proportional to,
n squared over L
squared, because it's
a sine wave with
period n pi upon L.
So the k is n upon
L with constants,
and that means that the energy
again, constant h squared k
upon 2n is k squared, which
is n square over L squared.
But the important thing
is that the splitting
between subsequent energy levels
is controlled by the width.
It goes like 1 over L squared.
So if you have a very
wide well, the splittings
are very small,
the ground state.
On the other hand, if you
have a thin, narrow well,
then the splittings
also go like L squared,
but L is very small here so
the splittings are gigantic.
When we have a superposition
of a big box inside of which we
have a bunch of barriers,
then the splitting
due to being excited
in the individual wells
goes like 1 over the
little distance squared.
Let's call this a squared.
So these go like 1 over a
squared, delta a delta E,
goes like 1 over a
squared, and from here it
goes like if this is L, delta
E goes like 1 over L squared.
So the combined effect is that
you get splittings due to both.
These splittings are
tiny, and these splittings
are really large because this
is a much smaller distance,
and this is a much
larger distance.
Cool?
So the only question
is how big in amplitude
are these-- how effective
are these barriers.
If they were strict
delta functions, then
this is all we would get.
And if there were
no delta functions,
we would just get
precisely these.
And when we have
some barrier, we
get a combination of
the two, and that's
what we're seeing here
with the splitting.
So what I'm doing when I'm
tuning the separation here
I'm controlling effectively
how strong is that barrier.
And so as we make the
barriers stronger--
there we go-- then
the splittings
are controlled by
the individual wells,
an as we make the
barrier very inefficient,
then the splittings are
controlled by the overall box.
Cool?
Excellent.
Other questions
about the simulation?
Yeah.
AUDIENCE: How close are
the actual [? wavelength ?]
potentials or the combinations
of the original wavelength?
PROFESSOR: Excellent question.
So you can answer that
immediately from this.
So the question is this,
look, if we had arbitrarily
separated wells,
if we had something
that looked like-- so they're
very, very far separated.
Then the ground state would
be effectively degenerate,
because the wave function
or the ground state
would be the completely
symmetric combination
of these guys--
you guys actually
studied this in our problem
set-- then there's also
a combination where you
have this, this, this.
You could also take
this constant this.
There are many
things you could do.
I'm sorry, this,
many combinations.
The point is since
each wave function
for each well effectively
drops off to zero inside,
we can just linear
combinations of these,
and there's no penalty for using
the true ground seats in here
because the potential is so
high, that the true solution
has an exponential tail,
but that exponential tail is
ridiculously small if
the barrier is big.
So in the limit that the
barriers are gigantic,
the true energy eigenstates
are just arbitrary
linear combinations
of these guys,
of the individual eigenstates
of each individual well.
Agreed?
However, when the barriers
a large but finite,
then the true energetic states
have little tiny exponential
tails, and the curvature of that
little tiny exponential tail
will determine exactly
what the energy is.
So a state that curves a
little bit versus a state that
curves more will have
slightly different energies.
So when the barriers
are very, very large,
the linear combinations of the
individual well eigenstates
should be very good
approximations, but not exact.
And as the barriers become
less and less effective,
they should become
less and less exact,
and we can see that right here.
So let's take the ground state.
The bottom of the band is
the completely symmetric
combination of the
ground seat of each well.
Everyone see that?
It's just well,
well, well, well.
It's a completely
symmetric combination.
At the top of the band
is the completely anti
symmetric combination, and
let's put them for comparison.
The top of the band
is the orange one,
and the bottom of the
band is the yellow one.
So the bottom of the band
is the completely symmetric,
and the top of the band
is alternating combination
of the ground
state in each well.
And all the other states
inside here-- well,
which are extremely
difficult to-- Let's see.
There we go.
These states are also linear
combinations of the ground
state in each well with
different coefficients
in front of them.
There are the
different coefficients,
and they are almost degenerate.
But if we go to
higher energy states
where the barrier is less
effective-- because they have
high energy and the
ratio between the barrier
height and their
energy is small--
then you see that these
states are not particularly
well approximated by
linear combinations.
And the energies correspondingly
are not degenerating.
AUDIENCE: So you said
linear combinations
of the [INAUDIBLE].
PROFESSOR: That gives you
a better approximation,
but for the same reason.
It's a good approximation,
but it's not exact.
But it becomes excellent as
the barriers become infinite,
even when they also go over
to the infinites very well.
Very good question.
Yeah.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Very good question.
We'll come to that
in little bit.
We'll come to that shortly.
So I'm done at the moment
with the PhET simulations.
I'm motivated by all this, and
effectively by that question,
I'm going to ask
the following thing.
Look, real materials like metal
in my laptop, the real material
is actually a
periodic potential.
It's built out of a
crystal of metal atoms
that are bound
together, and each atom
is some
positively-charged beast,
to which some
positively-charged nucleus,
to which is bound an electron.
And if I want to understand
properties of solids,
like the fact that metals
conduct electricity,
a basic fact I'd
like to explain,
if I want to explain the
fact that metals conduct
electricity, but
plastic doesn't, which
we will be able to explain, and
diamond doesn't, which is cool,
if we want to explain
that property,
we probably ought to study
the physics of electrons
in periodic systems where
the potential is, atom,
I don't want to be stuck to you.
Atom, I don't want
to be stuck to you.
Atom I don't want
to be stuck to you.
So it's a periodic
well a potential.
So in order to study
the physics of solids,
I need to understand first
the physics of electrons
in periodic potentials.
And these PhET simulations
were a first start at that.
We did it n's wells.
I want to now think about
what happens if I take not
n, but an infinite
number of wells.
What if I really strictly
periodic lattice on the line?
What do we expect to happen?
Well, looking at the results
of these PhET simulations,
as you did this for different
numbers, n, what you found
was the same sort
of band structure.
You just find for
more and more wells,
you find more and more
states inside each band.
In fact, you find n
states within each band.
Both the top and
bottom of the band
quickly asymptote
to fixed values.
As we take at large, what
do you expect to happen?
Well, this was a problem
on your problem set.
What did you predict?
What should happen
when you take n large?
AUDIENCE: [? It ?]
depends on the experiment.
PROFESSOR: Yeah.
How many states should
there be in each bend?
AUDIENCE: N.
PROFESSOR: N. So there
are arbitrarily many.
Exactly.
So there should be arbitrarily
many states in each band,
but they all have to fit within
this energy band, the width,
which is asymptoting
to a constant.
So there has to be a
continuum of states,
and they're not going to
be free particle states
because the system
has a potential.
What is the shape going to be?
Well, a reasonable
guess is that the shape
is going to be something
that fills out this curve.
It's going to be a
little bit different
from the free particle state.
So let's find out.
Let's just solve the
problem of an electron
in a periodic
potential, and look,
the periodic
potential is extremely
similar to things
we've already solved.
So I want to walk you through
just a basic argument.
So periodic potentials.
Instead of n wells,
consider a system
of an infinite number
of wells, each one
identical and just
for symmetry purposes.
And let the width here be L, so
the period of the lattice is L.
And what I mean
by that is saying,
so this is V of x
seeing that it's
periodic is the statement
as that V of x plus L
is equal to V of x.
Potential doesn't change if I
shift it by a lattice vector,
Now there's a nice
way to say this
which is that, if I take
the potential, v of x, and I
translate it by L, I get the
same potential back V of x.
And more to the point,
if I take the translation
if I think of these as
operators expressions,
if I take TL and
act on Vx and then
the potential of Vx f of x,
then this is equal to V of x TL
f of x.
So if I think of these as
operators instead of just
little functions
operator, operator,
then if I take the
translator operator
and I translate V of x f of x,
I'm going to get V of x plus L
f of x plus L. And if I
do the right-hand side
I get translator of L that's
f of x plus L but just times V
of x.
But if V of x is equal to V of x
plus L-- this is V of x plus L.
So if I translate by L says take
this thing and translate it,
plus L f of x plus L. And
on the right-hand side,
we have V of x, not V of
x plus L translate by L f
of x, f of x plus L And these
are equal because V of x plus L
is equal to V of x.
Yeah.
So what that tells
you is as operators
translate by L and
V of x commute.
Equals 0.
Everyone happy with that?
So you should just be able
to immediately see this,
the potential is periodic with
period L if by translated by L,
nothing should happen.
So the potential respects
T of L it commutes with it.
So this tells you a neat thing.
Remember that translate by L
is equal to E to the i PL upon
H bar, so in particular,
TL commutes with P
because this is just a
polynomial in P dL P,
and in particular with
P squared is equal to 0,
and that just follows
from the definition.
So what that tells
us is that if we
take the system with
our periodic potential,
periodic V of x plus L
is equal to V of x, then
the energy, which is P squared
upon 2 m of V of x commutes
with TL.
So in this system, with
a periodic potential,
is momentum conserved?
What must be true for
momentum to be conserved?
So momentum conservation come
from a symmetry principle.
AUDIENCE: Translation
invariance.
PROFESSOR: Translation
invariance.
In order for momentum
be conserved,
the system must be
translationally invariant.
Is the system
translationally invariant?
No.
If I shift it by a little
bit, it's not invariant.
It is, however, invariant
under a certain subset
of translations, which
is finite shifts by L.
Yeah.
So it's invariant
under shifts by L,
which is a subset
of translations,
and the legacy of that is
the fact that the energy
commutes with translations by
L. The energy does not commute
with P, Because P acting on V
of x, it takes the derivative
and gives you the prime of x.
It is not the same thing.
But we have a subset
of translations
under which the system is
invariant, and that's good.
That's less powerful than
being a free particle,
but it's more powerful
than not knowing anything.
In particular, it
tells us that we
can find energy eigenfunctions,
which are simultaneously
eigenfunctions of the
energy, and, I'll call it
Q for a moment, they have
an eigenvalue under TL.
So we can find eigenfunctions
which are simultaneously
eigenfunctions the
energy operator you might
E of phi E is equal
to E phi Eq and which
are eigenfunctions under
translate by L on phi Eq
is equal to something
times phi Eq.
So we're going to find
simultaneous eigenfunctions.
Everyone cool with that?
Yeah.
AUDIENCE: Do you know if the
momentum operator is still
the momentum operator?
PROFESSOR: Sorry.
Say it again.
AUDIENCE: The momentum operator
is still in remission, right?
PROFESSOR: Yeah.
The momentum operator is
still the momentum operator.
[INAUDIBLE].
AUDIENCE: Do you
know if [INAUDIBLE]?
PROFESSOR: Excellent.
Excellent.
Good Yes.
OK.
So if T of L is unitary,
note, T of L is unitary.
And you actually showed this
on a problem set before,
and let me remind you a
couple of facts about it.
The first is that
T of L is unitary,
and that says that TL dagger is
equal to the identity, right?
But what is TL dagger?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah.
TL dagger, well, P is
remission, so we just
pick up a minus
sign, so TL dagger
is equal to e to the
minus i PL over H bar,
and we can do think
of this two ways.
First thing you can
notice that it's clearly
the inverse of that guy, by
definition of the exponential,
but the other is, TL is the
thing that translates you by L,
and this operator,
e to the minus i PL,
well, I could put my H wit h
the and that's just translation
by minus L.
So this is translated by L and
then translated by minus L.
And of course, if
you translate by L
and then you
untranslate by L, you
haven't done anything
with the identity.
What are the eigenvalues?
What are the form
of the eigenvalues
of a unitary operator?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Well, that's true.
The eigenfunctions
are orthonormal,
but the eigenvalues,
are they real numbers?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah.
So the eigenvalues of a unitary
operator are of the form,
well, imagine we're in
a moment eigenstate,
for a momentum eigenstate,
then P is a real number.
And then the eigenvalue of TL
is e to the i, the P over H bar,
that's a real number.
So is e to the i a real number?
That's a pure phase.
It's e to i a real number.
So the eigenvalues
are pure phases,
and this let's just
do a nice thing.
We can note that here the
translate by L of the phi,
if these are common
eigenfunctions of E
and of TL, well, TL is
unitary, its eigenvalue
must be a pure phase, e
to the i a real number.
So let's call that real number
alpha, e to the i alpha.
So maybe just label this
by alpha for the moment,
and you'll see why I want
to change this to q later,
but for the moment let's
just call it alpha.
Everyone cool with that?
So you actually show the
following thing on a problem
set, but I'm going to
re-drive it for you now.
We can say something
more about the form
of an eigenfunction of
the translation operator.
This is quite nice.
So suppose we have
a function phi
sub alpha such that TL and i is
equal to e to the i alpha phi
alpha.
I want to know what is the form.
What can I say about
the function phi f of x?
What can I say about the
form of this function?
We can actually say something
really useful for this.
This is going to turn out
to be necessary for us.
The first thing
I'm going to do is,
I'm going to just make
the following observation.
Define the function
u of x, which
is equal to e to the
minus i qx phi sub alpha.
So if phi sub alpha is
an eigenfunction of TL
with an eigenvalue
e to the alpha,
and I'm just defining
this function
u to be e to minus i
qx times phi sub alpha.
Cool?
It's just a definition.
I'm going to construct
some stupid u.
Note the following thing.
T sub L on u, if we
translate u of x, this
is u of x plus L.
Translating this,
I get e to the i qx goes to
e to the minus i qx plus L,
which is just the e to the
minus i qx times e to the minus
i qL e to the minus i qx
plus L phi alpha of x plus L.
But this is equal to from
here e to the minus i qL
e to the minus i qx.
And from phi of x plus a, I know
that if I translate by a to get
phi of x plus L, I just pick
up a phase, e to the phi alpha,
so e to the phi
alpha phi alpha of x.
But this is equal to e
to the i alpha minus qL,
putting these two
terms together,
and this is nothing but u of x.
Yeah.
So another following cool thing.
So here I define some function
u with some stupid value
q, which I just pulled
out of thin air,
and I'm just defining
this function.
What I've observed is that if
I translate this function by L,
it picks up an overall phase
times its original value
where the phase depends
on alpha and on q.
So then I'm going to
do the following thing.
This becomes really simple if
I pick a particular value of q,
for a special value of q, q is
equal to alpha divided by L,
then this is equal to e
to the i alpha minus qL.
That's going to be equal to e
to the i 0, which is just 1,
so I have the phase,
so this u of x.
So that means if I
translate by L on u
and I pick q is equal to
alpha over L, I get back u,
so u is periodic.
Everyone see that?
Question?
So as a result, I can always
write my wave function phi sub
e and we'll now call
this q or q is related
to alpha as q is
equal to alpha over L,
every energy eigenstate can find
a basis of energy eigenstates
with a definite eigenvalue
under the energy,
a definite eigenvalue
under T sub L,
and I can write
them in the form,
since u is equal to e
to the minus i qx phi,
then phi is equal
to e to the i qx u e
to the i qx u of
x where q, where
the eigenvalue, e
to the i phi alpha,
is equal to e to the i qL
and u of x is periodic.
Everyone cool with that?
It's not totally obvious how
much this is helping us here,
but what we've done
is, we've extracted,
we've observe that
there is some lingering
symmetry in the
system, and I've used
that symmetry to deduce
the form of the energy
eigenfunctions as
best as possible.
So I haven't completely
determined the energy
eigenfunctions,
I've just determined
that the energy eigenfunctions
are of the form, a phase,
e to the i qx, so
it varies as we
change x, times a
periodic function.
So the wave function
is periodic up
to an overall phase,
not a constant phase,
a position-dependent phase.
Does everybody agree with that?
Questions about that?
So a couple things
to note about this q,
the first is, the eigenvalue
under TL e to the i alpha, this
is the eigenvalue under L
also equal to e to the i qL,
but if I take q to
q plus 2 pi over L,
then nothing changes
to the eigenvalue,
because the 2 pi over
L times L is just 2 pi,
e to the 2 pi is 1, so we
don't change the eigenvalue.
So different ways
of q only correspond
to different
eigenvalues of translate
by L of a translation
by one lattice vector,
one lattice basing.
They only correspond to the
different eigenvalues of TL
if they don't differ
by 2 pi over L.
If different values of
q differ by 2 pi over L,
then they really
mean the same thing
because we're just talking
about translate by one period.
So q is equivalent to
q plus 2 pi over L.
So that's just the
first thing to know.
So what have we done so far?
What we've done so far
is nothing whatsoever
except extract, take advantage
of the translational symmetry
that's left, the
remaining lingering
little bit of
translational symmetry
to constrain the for of
the energetic functions.
What I want to do now is
observe some consequences
that follows
immediately from this.
So there are some
very nice things
to follow just from this.
We can learn something
great about the system
without knowing anything
else, without knowing anything
about the detailed
structure of the potential.
At this point, I've
made no assumptions
about the potential other than
the fact that it's periodic.
So the thing I'm
about to tell you
were to be true for
any periodic potential.
They follow only
from the structure.
So let's see what they are.
So the first one is
that the wave function
sine of x itself with
a definite value of e
and a definite value of
q is not periodic by L.
Because under a shift
by L, u is periodic,
but this part picks up
a phase e to the i qL.
That's what it is to say that an
eigenfunction, the translation
operator.
It's not periodic by L
unless q is equal to 0.
So for q equals 0,
there is precisely
one wave function
which is periodic by L.
Because if q is equal to
0, then this phase is 0.
So only if is equal to 0 is
the wave function periodic.
Cool?
So this should look familiar
because back in the band
structure, these guys
are not periodic.
So these states in the middle
of the band, they're horrible.
They're not periodic.
They're some horrible things,
but that top guy in the band
is periodic, and it
turns out that if we
had made the system
infinitely large,
it would become
exactly periodic.
The fact that it has an
envelope on it like this
is just the fact that we have
an finite number of wells,
we're in a box.
If we got rid of that box
and we made many, many wells,
we would find it
perfectly periodic.
Yeah.
AUDIENCE: Could q not be
something like 2 pi over L?
PROFESSOR: Yeah, but 2 pi
over L would be equal to,
it's equivalent to q
equals 0, because it
corresponds to the
same eigenvalue.
So q's that differ
by 2 pi over L,
I'm going to call
the same thing.
AUDIENCE: OK.
Got it.
PROFESSOR: Good.
Other questions?
So the wave function is
not periodic except for one
special case, and we already
have a guess as to which
special case.
It looks like it's the state
on the top of the energy
band, which is
approximately periodic.
Two, while the wave
function is not periodic,
the probability that you
find the particle at x plus L
is equal to the
probability that you
find the particle at x, and this
immediately falls in this form,
because when you take
the norm squared,
the phis cancels
out at every point
and we're left with
the periodic u of x.
So the probability
is equal to norm
squared of u of x, which is
equal to norm squared of u
of x plus L because
u is periodic.
So probability
distribution is periodic.
On the one hand,
this is reassuring,
because if you look
at the potential,
the potential is
perfectly periodic,
and it would be
really weird if you
could tell by some
probability, by some measure,
like how likely you would
find an electron here
as supposed to over here,
potential is the same,
the probability should
be the same find
the electron in each spot.
The problem with that logic
though, is that these are wells
and usually we think
that when we have wells,
we have bound states, and those
bound states are localized.
But the probability has to be
invariant under translations,
so it cannot be that the
electron wave functions are
delocalized.
The electrons wave
functions must
have equal overall amplitude
to be in any given well.
It must be invariant
under translations
by a lattice spacing.
So that's really weird.
We have a whole bunch
of quantum wells,
and yet there are
no localized states.
All the states are extended
just like free particle
states are extended.
It had to be this
way by periodicity,
but it should surprise
you a little bit.
Anyway it surprises me.
I shouldn't tell you
what to be surprised by.
Yeah.
AUDIENCE: To normalize
probability, what do you do?
PROFESSOR: Yeah.
Excellent.
OK.
So this is for the same
reasons that free particle wave
functions aren't
normalizable, these
aren't going to be
normalized either.
So what are we
going to have to do?
AUDIENCE: Build wave packets.
PROFESSOR: Yeah.
We've going to have
to build wave packets.
In order to really meaningfully
talk about that stuff,
we're going to have
to built wave packets.
We need wave packets to
make normalizable states.
But the localized wave
packets are necessarily
not going to be
energy eigenstates
just like for a free particle.
And so there's going
to be dispersion,
and the whole story
that you saw three
particles we're
going to see again
for the particles in
the periodic potential.
Cool?
So next thing.
Therefore, all phi
sub Eq extended.
This is going to have some
really cool consequences,
and understanding
this fact in detail
is going to explain to us the
difference between connectivity
in the middle and not
conductivity in plastic.
Although there's a
surprising hook in there.
Naively, it would have
gone the other way around.
So the last thing
to say three, is
that phi sub to Eq, so this is
the state with definite energy
and definite eigenvalue under
translation E to the i qL Yeah.
And you showed on a problem
set that if you take a wave
function, you multiply
it by E to the i qx, what
have you done to the expectation
value of the momentum?
AUDIENCE: You've
increased it by q.
PROFESSOR: You've
increased it by q, right?
This is the boost
it by q operation.
So q incidentally has units
of a wave number, so times
h bar has units of a momentum.
q is like a momentum.
Is q a momentum?
Well, we can answer
that question by asking,
is this an eigenfunction of P?
Is it an eigenfunction
a momentum operator?
And the answer is
definitely not.
That's an eigenfunction
of a momentum operator.
I take a derivative, I get q
times i and I multiply by h
bar, I get h bar q,
momentum, but when
I have this extra
periodic function,
this thing together is no
longer an eigenfunction of q,
because u is some periodic
function, which is necessarily
not a momentum eigenfunction.
So this guy is not a
momentum eigenfunction.
You can see that by just
taking the derivative.
I get one term that
picks up a q from this,
but I get another term that gets
a derivative with respect to u,
and that's not proportional with
the constant to u e to the i qx
unless u is in the
form e to the i kx.
But if u is in the
form e to the i kx,
this is just a free
particle wave function,
which it can't be if we
have a periodic potential.
So finally q is not a
momentum eigenstate.
And correspondingly,
q and more important
h bar q is not a momentum, I'll
say the momentum in the sense
that it's not the eigenvalue
of the momentum operator.
However, it has-- semi colon--
it has units of momentum.
And not only does it
have units of momentum,
it's the thing that tells
you how the state transforms
under translations.
It translates by an e to the
i qL under translations by L.
And there is usually
what momentum is.
it's the constant
that tells you how
the state transfers
under momentum.
So it's kind of like a
momentum, but it's not
the eigenvalue of the
momentum operator.
We have a name for it.
We call it the crystal
momentum, and we'll
understand what it does for us.
Part of the goal for
the next two lectures
is going to be to understand
exactly what this crystal
moment is.
So for the moment,
I'm just giving it
a name sort of
like the beginning
we give super position
a name, and we'll
exploit its properties
basically now.
So these are the things that
follow just from periodicity.
I didn't know anything
about the potential.
I just used periodicity.
So before we move on to talking
about a specific potential,
you all have questions
about the general structure
of periodic systems like this.
Yeah.
AUDIENCE: What's the
difference between the psi
and the [INAUDIBLE]?
PROFESSOR: Oh, excellent.
I shouldn't have used psi,
I should have used phi.
Thanks.
I'll use psi only when we talk
about general super positions
and wave packets.
Other questions?
All right.
So with all that, let's talk
about a specific potential.
So far we've extracted
about as much
as of the physics
out of the system
that we can just
from periodicity.
In order to make more
progress, in order
to talk about, for example,
what are the allowed energy
eigenvalues and for that matter,
how does that allowed energy
eigenvalue depend on q,
the crystal momentum,
in order to answer
that, we have to talk
about a specific system.
So let's go ahead and talk
about a specific system.
I'm going to pick my favorite.
Now the problem
set, you're going
to do something really
cool about it, which
I'll explain in a moment.
So you'll do a general
case in the problem set,
but for the moment, let's
work with a simple example.
And the simplest
example is going
to be a periodic potential
with the simplest
possible barriers
in the potential.
What's the simplest barrier?
Delta function.
Yes.
So the potential is going
to be for this example
V of x is equal to sum from n
is equal to minus infinity of h
bar squared over 2mL g
naught delta of x minus nL.
So what is this?
So this is just some
overall constant out front.
Normally we call this
V naught, but I've
made g naught dimensionless
by pulling out
an h bar squared over 2nL.
L is the spacing
between delta barriers.
So the potential
looks like this.
Here is 0.
We have a delta
function barrier.
We have a delta function barrier
L. We have a delta function
barrier at 2L, dot, dot,
dot, minus L, dot, dot, dot.
So there's our potential.
And I want to know what are
the energy eigenfunctions
for a single particle
in this potential.
And again, g naught is
the dimensionless strength
of the potential.
So how do we solve this problem?
Well, we've done
this so many times.
What we see is that in
between each barrier,
the particle is just free, so
we know the form of the wave
function between each
barrier, it's just
e to the i kx plus 2
minus e to the minus i kx,
where k is defined
from the energy h bar
squared k squared upon 2n is e.
At the barrier, we have to
satisfy appropriate matching
conditions.
So I could put a and b here
and c and d and e and f.
And under each one of these
imposed boundary conditions,
and this is going to be an
infinite number of coefficients
with an infinite number
of boundary conditions.
And that sounds like it's
going to take some time.
But we can use
something really nice.
We already know that the
wave function is periodic.
So suppose between 0 and
L, let's say less than x,
less than L the wave
function phi takes the form,
and we know that phi of eq
takes the form e to the i qx,
and now some periodic
function, which
would write as a e to
the i kx-- actually,
I'm not going to write that.
So in between 0 and L,
it's a free particle,
and it's equal to A e to the
i kx plus B e to the minus
akx, where k is defined
purely from the energy h bar
squared k squared
upon 2m is the energy.
So this is what we mean by k.
And in between 2L it will
take the similar form,
but what we can do
is, we can define,
so this is sine of x,
between L less than x less
than 2L phi eq is
equal to-- well,
I could define it
with new constants,
but I know what it has to be
because whatever the value is
here it's the same as the
value here up to a shift by L
the translation to
phase e to the i qL.
So this is equal
between L and 2L,
it's equal to e to the
i qL and periodicity
condition, the same
thing, A plus B,
with the same coefficients.
Everyone cool with that?
So I can now translate,
so let's think
about what the
boundary conditions are
going to have to be.
At each of the delta functions,
the delta function boundary
condition is going to say that
the slope here minus the slope
here is something proportional
to the amplitude here.
And the slope here is going to
be minus something proportional
to the amplitude.
But the slope here is the
same as the slope here up
to an overall phase
from the translation.
And the slope here is the same
as the slope here up to a phase
coming from the translation.
So we can now turn this
into instead of this,
the condition between
the slopes here
and here, I can turn that into
a condition between the slopes
here and here.
Wow.
This is horrible.
Here are my two delta functions.
The slope here and here, the
boundary condition I put at 0,
I can translate this
into the slope here up
to an overall phase
e to the i qL.
So I can turn the
boundary condition here
into a boundary condition
between these two
guys inside one domain.
And so I can turn this
into a problem that
just involved A and
B and that's it.
Everyone see that?
So let's do it.
So questions on the basic
strategy at that point?
So let's do it.
So what we need is we got
the general form of the wave
function, and I'm going to erase
this because we don't need it.
Got the general form
of the wave function,
and then we need to
impose the boundary
conditions of the
delta function.
And in particular, let's
just impose them at 0
because imposing
them at 0 is going
to be equivalent to
imposing them everywhere
else by periodicity if I use
the periodicity of the wave
function.
So the boundary conditions
at the delta functions.
So the first is
that phi at 0 plus
is equal to phi at
0 minus, so this
is a statement that the
wave function is continuous.
But phi at 0 minus is equal
to-- well, that's this point.
But this point is the same
as this point except the wave
function picks up a
phase e to the i qL.
This is equal to phi
at L minus, right here,
times e to the i qL.
And we have to be careful
because the statement is
at the wave function here,
is the wave function here
time e to the i qL.
So the wave function here times
the wave function here times e
to the i qL minus.
And this leads to phi zero
plus-- the value right here--
is A plus B because it's just
this wave function evaluated
at x equal 0, and
this guy at L is
going to be this
evaluated at x is
equal to L times e
to the minus i qL.
So this is equal to A e to
the i k minus qL plus B e
to the minus i k plus L, So
there's my first condition.
Second boundary condition
is the derivative condition,
and the derivative
condition is that phi prime
at 0 plus minus phi
prime at 0 minus
is equal to g naught
upon L phi at 0.
And this gives me that ik A
minus B. So that's this guy,
and now the derivative
of the second guy
picked up our extra
phase minus ik A
e to the ik minus q L
minus B e to the minus ik
plus qL-- this is
all in the notes
so I'm not going to
worry too much about it--
is equal to, on the right-hand
side is g naught upon L A
plus B.
So now note the following
properties of these equations.
These equations determine an
equation, so from the first,
I'll call this 1,
and I'll call this 2.
So 1 implies that A is
equal to B times something.
It's just a linear equation.
So if we pull the side
over, put this over here,
A is equal to B times something.
Everybody agree with that?
Some horrible expression
of k and q times.
And 2 also gives an expression
form A is equal to B times
a horrible expression in terms
of k and q but a different one.
Call this expression 1,
I'll call this expression 2.
So as usual, so this is
two different expressions
for A given B and
this only makes
sense it's only a solution if
these two horrible expressions
are equal to each other.
And so this is what we've
done many times before,
and if you've set these two
expressions equal to each other
and do a little bit
of trigonometry,
you get the following relation.
Cosine of qL is equal to
cosine of kL plus g naught
upon 2kL sine kL.
Where what k means is nothing
other than the square root
of 2ne upon h bar.
k is defined from that
relation, and what q means is q
is the eigenfunction or
e to the i qL anyway,
is the eigenfunction
of the wave function
under translation by L.
So k is really playing
the role of the energy.
Everyone cool with that?
I just skipped the algebra, but
if you do the horrible algebra
from setting these
two expressions
equal to each other,
then you get this.
Yeah.
AUDIENCE: Is q some like
undetermined thing as of right
now, or do we know--
PROFESSOR: Excellent.
What values of q are
allowed by this expression?
Is exactly the right
question to ask here.
What values of q's are
allowed for our wave function?
Is there any condition on
what the value q is so far?
No.
It could have been
absolutely any number.
For any number q, we
can find a eigenfunction
to translate by L, e to the i
qL as the eigenvalue, any value
of q.
However, q is equivalent
to q plus 2 pi upon L.
If you skipped by
2 pi over L, then
it's not really a
different eigenvalue.
It's the same eigenvalue.
So q is defined.
It's any continuous number
between let's say minus
pi over L and pi over
L for simplicity,
make it nice and small. i
could have said 0 and 2 pi.
It doesn't make any difference,
minus pi and pi over L.
So q is a free parameter.
So there exists a state.
Another way to say this
is that there exist states
and in particular eigenstates
for any value of q,
for any q, and any e, so
for any value of q and e
which satisfy this equation
corresponds to a state.
There exists states for any
q E satisfying this equation.
Cool?
This is just like
the finite well.
In the case of the
finite well, we
go through exactly
the same analysis
the only difference is
that we don't have the qL.
What happens in the
finite well case
we impose a boundary
condition off in infinity.
So the thing has normalized,
and we put another boundary
condition in infinity,
and what we ended up
getting is something of the form
roughly 1 is equal to something
like this.
It's actually not exactly
that, but here we have kL,
and if you combine
these two together,
you get something kL plus
a phase shift over kL
and then a 1.
If you multiply, you get
kL is equal to cosine
of kL plus a phase
shift, which is exactly
the form for the
delta potential.
Good.
So this is similar.
However, here we have
an extra parameter q,
which is free to vary.
q could be anything between
minus pi over L and pi over L
So what does cosine
of qL turn into?
Well, it's anything qL can
take q is 2 pi upon L or pi
upon L to minus pi over L. So qL
can go between pi and minus pi,
so cosine of qL
varies between cosine
of pi, which is 1 and cosine
of minus pi, which is minus 1.
So any value of cosine
of qL between 1 minus 1
is a valid value of
qL On the other hand,
if this is equal to
7, is there a q such
that this is equal to 7?
No.
So what we need to do now is,
we need to solve this equation,
but it's obviously a god
awful transcendental equation.
So how do we solve horrible
transcendental equations?
AUDIENCE: Graphically.
PROFESSOR: Graphically.
So let's solve it graphically.
So to solve it graphically, that
lets plot the left-hand side
and the right-hand side
as a function of qL
So in this direction, I'm
going to plot cosine of qL.
That's the left side.
And cosine of qL we know is
going to vary between plus 1
and minus 1.
Because q between q
is equal to pi over L,
and q is equal to
minus pi over L.
And in the horizontal
direction, I'm going to plot kL.
And when kL is 0, remember that
the definition of kL is that E,
the energy eigenvalue
is equal to h
bar squared k squared upon 2m.
And what we're looking
for are points,
are any common solution to
this equation where there's
a value of q and a
value of E of k such
that these two expressions
are equal to each other.
Yeah.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Sorry?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Oh, sorry.
Thank you.
This is qL is equal to
0 and qL is equal to pi.
Thank you.
Yes.
Are there types of
other questions?
So kL equals 0 is going to
be equal to 0, k equals 0,
and this is going to
be a function of kL.
I'm going to plot it
as a function of kL
because that's what shows up.
But remember that k is
nothing but code for E.
And so what we're looking
for a common solutions
of that equation.
Now let's first just plot.
So if this is 0, let's
just plot cosine of kL,
the right-hand side cosine of kL
plus g naught over 2kL sine kL.
And let's do this first
for the free particle.
Oh, I wish I had colored chalk.
For the free particle, free is
going to be g naught equals 0.
There is a potential
and the potential
is 0 times the delta
function, which
is not so much delta function.
So for free particle
when g is equal to 0,
that second term
vanishes, and we
have cosine of qL is
equal to cosine of kL.
Kale, if you fry
it, it's crispy.
So what is that going to be?
Well, this has an
obvious solution
which is that q is equal
to k, but the problem
is, q is only defined up to 2
pi over L, so mod 2 pi over L.
Well, let's see what
this looks like in here.
So for the free particle
what does this tell us?
So for the free particle,
let me just write this
as therefore, q is equal
to L or q is equal to k,
and therefore E is equal
to h bar squared is
equal to q squared.
So any q or E are allowed
such that q squared
is equal E times 2
upon h bar squared.
So that's just the
usual condition.
It's the free particle has
energy h bar squared k squared
upon 2n, and we just chose
to call it k equals q.
So what we're doing is
instead of organizing things
as momentum eigenstates
of the free particle
or organizing the free particle
energy eigenstates in terms
of transient by L eigenstates.
Perfectly valid.
It commutes with
the energy operator.
What does this look like?
Well, cosine of kL is cosine
of qL, when k is equal to 0,
q is equal to 0.
So we get this.
And the points, when the
points where qL is equal to pi
or minus 1, these
are the points where
we get the qL is equal to n pi
or equivalently plus or minus
kL is equal to n pi
plus or minus n pi.
So this is n is
1, n is 2, n is 3.
Everyone cool with that?
And these guys are 2 pi
times n or 2n plus 1 2n pi.
And I really want this
to be odd, so n odd.
And this is n pi kL
so n pi for n even.
Everyone cool with that?
So that's what it would look
like for a free particle.
What happens if we don't
have a free particle?
What happens if we have
the interacting potential?
So now let's have g
naught equal to 1 or 0.
And again, here is plus 1.
Here is minus 1.
And we're plotting
as a function of kL.
And now it changes,
so let g naught
be a small, positive number.
If g naught is a small, positive
number, then near k goes to 0.
Sine goes like it's argument,
and so that means kL over kL
we get cosine of 0, which
is 1, plus g naught over 2.
So at 0, the value of
the right-hand side
is greater than 1.
Is there an energy eigenvalue
with energy equal to 0?
No.
We have to wait until
cosine gets sufficiently
small and sine is increasing,
so what does this curve do?
It looks like this.
It's again periodic with
the appropriate period,
but the amplitude
of the deviation
from the free particle
is falling off
as we go to higher
and higher values.
So we've taken this
point and pushed it out.
So what does this tell us?
Well, remember that there
is an energy eigenfunction
with a given q and remember
this is cosine of qL,
there is an energy eigenvalue.
There exists an
energy eigenstate
with an e and TL eigenstate
for any E and q such
that there exists a solution.
And so what does that mean?
Well, is there an energy
eigenvalue with this value of E
or this value of k?
No.
Right?
Because here's the right-hand
side and the left-hand side
is any value in here.
There's a q for which
any left-hand side
is any value between
1 and minus 1.
So there's no value of q you can
pick so the left-hand side is
equal to the right-hand
side for the value of k.
And similarly here,
here, here, here, here,
specifically for this
value of k or this value
of the energy, E minimum,
there's one value of q.
This one. q is equal to 0 where
there is an energy eigenstate.
So there's a minimum
energy in the system.
What about for this
value of energy?
Yeah.
There's exactly
one solution there,
and actually if
you're careful, it's
2 because it's the
cosine of qL. q
could have been
positive or negative.
So we have a solution here
with this value of energy
and with this
value of cosine qL.
Any q that gives us this value
of cosine qL is going to work,
and that's one of two values.
And similarly for
each of these guys,
there's a state for every
value of energy in this region
until we get here, and when we
get here, this is E maximum.
If we go to any higher
energy, any higher k,
then there's no solution for any
value of q and any value of E.
So what we get are these
bands, continuous bands
for any energy separated
by gaps and then again
continuous bands for any energy
and bands for any energy.
So all of these
shaded in regions,
any energy in this
region corresponds
to a state and similarly here.
So here's the
upshot of all this.
We're going to study the details
of this in detail next time.
When we have a
periodic potential,
every energy
eigenfunction is extended
through the entire material.
Every energy
eigenfunction is extended
across the entire lattice.
None of them are localized.
And that's like a free particle.
However, not every energy
is an allowed energy.
Only some energies are allowed.
Some energies do not
correspond to allowed energy
eigenfunctions eigenvalues,
and some energies do.
And they come in continuous
bands of allowed energies
and continuous gaps of
disallowed energies.
And it's going to turn out to be
exactly this structure of bands
and gaps, or band gaps,
that is going to give us
the structure of
conductivity in metals
and explain to us why we don't
have conductivity in plastic.
We'll pick up on that next time.
