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ROBERT FIELD: Let's get started.
William Klemperer was my thesis
advisor, and he died yesterday.
It also happens that the
subject of this lecture
is really the core of
what I got from him.
He showed me how to
evaluate matrix elements
of many-electron operators,
which is the key to being able
to interpret--
not just tabulate--
electronic properties
of atoms and molecules.
Our goal is to be able
to reduce the complexity
of electronic structure,
which is really complicated.
The electrons interact with
each other really strongly,
and there are a lot of them.
And it's very hard to
separate the complexity
of the many-body
interactions into things
that we can put in our
head and interpret.
And the whole goal
of this course
is to give you the
tools to interpret
complicated phenomena.
We have the vibrational problem
as a way of understanding
internuclear interactions--
nuclear motions.
We have electronic structure
and the hydrogen atom
as a way of understanding
what electronic structure is,
and to reduce it to,
basically, the things
we learn about hydrogen.
When we go to molecular
orbital theory,
we take what we
know about atoms,
and build a minimally-complex
interpretive picture,
which is sort of a
framework for understanding
complicated molecular
interactions.
So one of the most important
things about understanding
electronic structure
is, how do we
deal with many-electron
wave functions?
And one of the terrible
problems is that the electrons
are indistinguishable.
And so we have to ensure
that the wave functions are
anti-symmetric with
respect to permutation
of every pair of
electrons, not just two.
In helium we just dealt with
two, and that wasn't so bad.
But when we deal with
n electrons, what
we are going to discover
is that in order
to anti-symmetrize
the wave function,
we have to write a determinantal
wave function, a determinant
of orbitals.
And when you expand
an n by n determinant,
you get n factorial times.
And when you calculate
matrix elements,
you have n factorial
squared integrals.
So you're not going to be
handling these one at a time,
and looking at them lovingly.
You're going to want to be
able to take these things
and extract what is
the important thing
about the electronic
structure that you're
going to need to know.
And as a graduate student,
I was collecting numbers.
I was collecting numbers about
spectroscopic perturbations,
where non-degenerate
perturbation 3 breaks down,
and interesting things happen.
But this was something
that nobody in the world
was interested in because
it was the breaking
of the usual patterns.
And I was convinced that I
had collected some stuff that
told an interesting story.
And I told Klemperer about
this, and he said, well,
have you thought about how
to evaluate these integrals--
these numbers that you are
extracting from the spectrum,
by doing some tricks with the
many-electron wave functions?
And then he showed me, on a
scrap of paper, how to do it.
And I was launched.
That was it.
That has been the
foundation of my career
for the last 50 years.
And I didn't think that
Klemperer knew that.
I didn't think anybody
knew it, because I
didn't think it was knowable.
But he just gave it to
me on a silver platter.
And so I'm going
to try to give you
at least the rudiments of
what it is you're up against,
and how you reduce
them to things
that you care about,
that you can think about.
And you can understand
the hydrogen atom
in rather complete detail.
Or at least you can understand
how one observable relates
to another.
And so the relationship
between the effective quantum
number and the ionization
energy of a state
then provides a
hydrogen-atom-based structural
model for everything
you can observe.
Now spectroscopists have
the unfortunate habit
of saying we're
interested in structure.
Structure is static.
Dynamics is magical,
and special, and hard.
But if you understand
structure in a way which is not
the exact eigenstates, not
the exact wave functions,
but something that the molecule
was trying to do and sort of
missed.
And the dynamics is
just what happens
when this preparation
isn't in eigenstate,
which would be boring.
And you get dynamics,
which you can understand,
as opposed to just
saying, I'm going
to tabulate the dynamics too.
You don't know anything unless
you have a reductionist picture
of what's going on.
And since the hardest part
of dealing with molecules
is the fact that they
have a lot of electrons,
this is really the core of being
able to do important stuff.
Now it's a horrendously
complicated problem,
and notationally awkward, too.
And let me just
try to explain it.
And I'm going to try to do
this without too much reliance
on my notes, because
they're terrible.
We talked about helium.
And helium has two electrons.
And there's this 1 over r12
interaction between electrons,
which looks innocent enough.
You can write it down.
You know, it's
just a few symbols.
And we can call it the
first order perturbation.
But that's really a
lie, because it's as
big as almost everything else.
And so, yeah, we can, in fact,
do a series of approximations.
One is, ignore it, the
non-interacting electron
approximation.
And that's basically
repackaging hydrogen,
and it's not quite enough.
And then we can say, OK, let's
calculate the first order
energy by calculating
expectation values of h12.
So that's E1.
And that's almost enough to give
us a sense of what is going on.
1 over r12, commutate with
any electron, that's not 0.
1 over r12, commutated with
any orbital angular momentum--
any momentum is not 0.
So that means that l and n
are not good quantum numbers.
What's a good quantum number?
What's the definition of
a good quantum number?
Come on, this is an
important question.
Yes.
AUDIENCE: [INAUDIBLE]
count, and then you
put them into some
formula, and then
you can read off eigenvalues.
ROBERT FIELD: That's
maybe 70% of what I want.
You can put it into a formula.
That means it's a
rigorously good thing.
It means it commutes
with the Hamiltonian.
A rigorously good quantum
number corresponds
to a eigenvalue of
an operator that
commutes with the Hamiltonian.
So hydrogen, we rely on n and
l to get almost everything.
But here we find that,
in addition to this
being that small, it destroys
the foundation of our picture.
And so how do we
think that we can
make any sense of many-electron
atoms and molecules?
Well it turns out we can
hide most of the complexity.
And most of the complexity
is just working out the rules
for calculating these
matrix elements.
The matrix elements of
operators that we care about,
like transition moments, spin
orbit, Zeeman effect, things
that correspond
to how we observe
atomic and molecular structure.
And so the main
obstacle to being
able to evaluate
these matrix elements
is the permutation requirement.
And it turns out that there is
a really simple way of dealing
with the requirements
for electron permutation,
and that is to write the wave
function as a determinant
of one-electron orbitals.
Because a determinant has three
really important properties.
One, it changes its sign when
you permute any two columns.
Two, it changes its sign when
you permute any two rows.
And three, if you have two
identical columns or rows,
it's 0.
That's really fantastic.
And that is the Pauli
exclusion principle-- not what
you learned in high school.
What you learned is a
small consequence of that.
So if we can build
anti-symmetric wave functions,
we have aufbau, we can only
put one electron in an orbital.
We have all sorts of stuff, but
it's too complicated to tell
a student in high
school that you can't--
just the question of
indistinguishable electrons
is such a subtle thing
that you can't say,
well, they have to
be anti-symmetric.
But it's easy to say, you can't
put more than one electron
in a spin orbital.
But we don't talk
about spin orbitals.
We say, we can't put more than
two electrons in an orbital,
because we're protecting you
from unnecessary knowledge.
OK, well, I'm not
going to protect you.
[LAUGHTER]
OK, so we know that
the Hamiltonian
has to commute with--
these capital letters mean,
many electrons' angular momenta.
And this is the spin, this is
the projection of the spin.
We know this is true because
the Hamiltonian doesn't
operate on spin.
It's a trivial result, but
it's a very important result.
OK, so we have to worry about
spin, and spin eigenstates,
and other things like that.
OK, so Slater determinants.
J.C. Slater was an MIT
professor in physics.
He invented these
things in 1929.
I have a reference.
I don't know if I've
ever read this paper,
but it's probably beautiful.
So basically what
Slater did is showed,
yeah, you can do the necessary
algebra to deal with any atom,
and to be able to reduce an atom
to a small number of integrals
that you really care about.
And there are two
ways of doing this.
One is the truth, and
one is the fit model.
Now the truth is really
boring, because you
lose all the insights,
and the fit model
gives you the things you have
to think about and understand.
And a fit model also tells you
what are the import the actors.
And maybe they're in
costume, maybe they're not,
but we can deal with them.
But the truth is really
very complicated.
And as I said many times, when
you go from hydrogen to helium,
you can't solve the
Schrodinger equation exactly.
This was perhaps a
little bit of a surprise,
but I think it was only
a surprise in newspapers.
I think physicists
knew immediately,
when you go for a
two-bodied problem
to a three-bodied
problem, there is no way
you can have an exact solution.
And that's the truth.
You can't solve helium or any
more-than-one-electron problem
exactly.
But you can do it
really well, and it just
costs computer time.
And if the computer is doing
the work, you don't really care.
Because once you've told
the computer the rules,
then it's off to the races.
You can go have lives or you can
go have a life, and come back,
and the computer will tell
you whether you made a mistake
and you're getting
a nonsense result,
or that you have
the correct result.
So what we know is this
permutation operator,
operating on any two-electron
function, has to make--
OK, I'm skipping steps, and my
notes are really kind of stupid
sometimes.
P 2, 1, which has to
be equal to minus.
And now if you apply the
permutation operator twice,
you get back to the same thing.
So there's only two
possible eigenvalues.
You can have minus 1 or plus 1.
And the minus 1
corresponds to fermions,
things that have half-integer
spin, like electrons.
And the plus 1 corresponds to
things that have integer spin,
like photons, and
vibrons, and other things.
And actually, it's harder to
construct a symmetric function
than an anti-symmetric function.
But the thing is, you've
got lots of electrons,
and you have very few quanta
of vibrations in a single mode,
and you have very few
photons interacting
with a molecule at once.
And so the boson
symmetry is less
important in most applications.
And so we just have
to kill this one.
OK, so suppose we want to talk
about something like this,
the 1s, 2s configuration.
A configuration is a list
of the occupied orbitals--
not the occupied
spin orbitals, which
is a spin associated
with an orbital.
The world of spin
orbitals is where I live,
but we do that for a reason.
And so this two-electron thing
can be expressed as a space
part--
there are various
conventions that--
times the spin part.
And alpha 1, beta 2, and then
we have minus or plus beta 1,
alpha 2.
I'm looking at my notes
because some people always
keep the electron in
the first position,
and some people keep always the
orbital in the first position.
And it doesn't
matter, because you
can permute rows or columns.
But I just want to write
what is in my notes.
OK, so this thing, this
two-electron function,
has two anti-symmetrized
possibilities.
And one is a singlet,
and one is a triplet.
So s equals zero, s equals 1.
We recognize this alpha
beta minus beta alpha
as the singlet spin state,
and alpha beta plus beta alpha
as the triplet spin state.
So we have alpha beta plus
beta alpha, and alpha alpha
and beta beta, and we have
alpha beta minus beta alpha.
So we call s = 0 a singlet,
and this a triplet,
because of the number of states.
And this wave function has
the necessary spin symmetry
and the necessary
permutation symmetry.
OK, so if, instead of
two electrons, we have 1,
2 dot dot dot N, then Mr.
Slater says we do this--
whoops.
OK, N, 1, and then K1,
and KN N. So that's
a determinant-- an
N by N determinant.
And the rows correspond
to the electrons,
and the columns corresponding
to the orbitals.
Now this, because of the
properties of the determinant,
is anti-symmetric with respect
to permutation of any two
electrons or any two orbitals.
But we don't really care
about the permutation
of the orbitals, because
it's really the same thing
is permuting the electrons.
And so this N factorial is a
consequence of normalization,
because when you expand
an N by N determinant,
you get N factorial, additive
products of N functions.
It looks horrible.
And because we're
normalizing, we
need this 1 over the
square root of N factorial
in order to have
this thing come out
to be 1 when you calculate
the normalization integral.
Now this notation is
horrible because you've
got too many symbols.
And so depending on what
you're trying to convey,
you reduce the symbols, and
you can reduce it simply
by, instead of writing psi every
time, just writing the state.
Or you can-- since
you don't need psi,
you don't need the state
letter, you can just
have the state number.
But the best way to do
this is simply to say--
this is just the main
diagonal of the determinant.
It conveys everything you need.
Again, if you permute any two of
these guys, any adjacent pair,
the sign changes.
And it contains
everything you need,
and it doesn't require
you to look at stuff
you're not going to use.
And your goal is going to
be to take these things,
and calculate matrix of them.
And so you'll be dealing
with the orbitals one or two
at a time.
And this is very convenient.
And soon, you start to
take this for granted.
And it's a very
simple thing, but it
isn't, because you're doing
a huge number of tricks.
OK, I'm going to skip
over what's in my notes.
Demonstrating that for a two
by two, that what I asserted
is correct, you can
do that very easily.
OK, so we can count,
and we have an atom,
and we know how many
electrons it has.
And so we immediately know
what our job is going to be.
We're going to be having to
write some Slater determinant
of those number of electrons.
And the goal is to be able
to do the algebra in a way
that maybe you can't
describe to your friends
because it's too complicated.
I'm faced with the problem
of trying to explain
how to do this algebra.
But it is something
that you can learn,
and you can ask a
computer to do it,
and there are all sorts
of intuitive shortcuts
where you can look at a
problem, and you could say,
I understand.
OK, so you're used to orbitals.
And that's perfectly reasonable,
because for hydrogen, we
have orbitals, and
there's only one orbital,
and it could have either spin.
We don't mess with that.
But now we're going to
talk about spin orbitals.
And that's just the
combination of the name
of the orbital with whether
the spin is up or down.
And the reason for this is
it's easier to do the algebra.
And the reason the algebra is--
it's initially harder
to do the algebra,
because there are certain
selection rules, and stuff
like that.
But once you know how to
do it you do the algebra.
And then all of a
sudden, everything
pops out in a very useful form.
So the stick diagrams
are very important.
But now I'm specifying
the stick diagrams
as spin orbitals
rather than orbitals.
Now another point, there
are rules for how--
the number of spin
orbitals is different
between the left-hand side and
the right-hand side of a matrix
element.
There are rules that are
easily described-- and so
for every kind of
orbital, an orbital that
is a scalar, that doesn't
depend on quantum numbers, that
has a selection
rule delta SO of 0;
and for something like a
one-electron operator that
has a selection rule delta
spin orbital of 1 and 0;
and then we have our
friend 1 over rij,
that has a selection rule delta
spin orbital of 2, 1, and 0.
And the algebra for each
is something you work out,
and then you know how to do it.
And I'm going to try to
give you just a little bit
of a taste of this.
So we already looked
at something like this.
But we use a slightly
different notation.
So I'm going to go back.
And we have 1s alpha, 1s
beta, 2s alpha, 2s beta.
And so for the ground
state of helium 1s
squared, and we would do this.
And the stick
diagrams are great,
because it's easier to see on
a picture, who are the actors,
and have I included all of them,
or have I left something out?
And so now we're interested
in the stick diagram
for the 1s 2s configuration.
And there are several kinds
of 1s/2s configurations,
depending on what the
alpha and beta are.
So we have 1s alpha, 2s alpha.
And we have 1s alpha, 2s beta.
1s beta, 2s alpha, 2s
alpha, 1s beta, 2s beta.
So there's four guys, and we can
put our arrows on these things,
and we know everything we
need to know about these guys.
It tells us what to do.
Well, when we do this, the
diagonal matrix elements
of the 1 over rij
Hamiltonian can be expressed.
And we use this notation,
J tilde minus K tilde.
So for every two-electron thing,
we're going to get this kind
of--
now these are simple integrals,
and some of them are 0.
Because this doesn't
operate on spins.
And so if you had a 1s alpha,
2s beta, 1 over rij, 1s beta,
2s alpha, then the 1s alpha
with the 1s beta is 0.
The 1s alpha with the 2s
alpha is not 0, et cetera.
There are all sorts of stuff.
But this tilde notation
says, well, this
is what we start with,
and we have to convert it
into things that really matter.
So the operation of
removing the tilde
requires a little bit of
work, a little bit of thought.
And that's why my
notes are crap,
because I can't explain it well
enough to really teach this.
So when we do the
1s squared, the J 1s
squared tilde, is equal
to the J 1s squared,
because the spins take
care of themselves.
But k tilde 1s
squared is equal to 0.
Because when we
do 1s squared, we
have an alpha with
an alpha for the J
term, and an alpha with
the beta for the K term.
And alpha with beta is 0,
because the operator cannot
change the alpha into beta.
So this tilde notation
is a convenient thing,
because you can use
any Slater determinant,
and you can express it
in terms of J's and K's.
And the sign comes
from switching
the order of the orbitals.
That's how the
determinants work.
And so you're going to see
a whole bunch of stuff.
But removing the tildes
is the tricky business.
OK, now when you
get a problem where
you have a configuration
where a single Slater is not
sufficient--
in other words, in order to
make an eigenstate of s squared
or sz, you sometimes
need two or more Slaters,
and you have to use a particular
linear combination of them
to get the right
value of s and sz.
And then what happens
is you're looking
at matrix elements of the 1 over
rij operator, between Slaters.
Now this is a headache.
And I could talk until
I'm blue in the face,
and I cannot make it
clear how to do this.
Because it's just awful.
But some things in life
are worth suffering for.
And so anyway, in
the 1s, 2s situation,
when you do everything
right, you get--
this is just a general notation
for a two-electron wave
function--
1 over r12, psi 2.
So these guys are eigenfunctions
of s squared and sz.
And when you do that,
you get 1/2 times 2 J,
1s, 2s, minus or
plus 2K, 1s, 2s.
Remember, when you have
mismatched alpha and beta,
the K's are 0.
But when you have K, the 1 over
rij matrix element between two
Slaters, you can fix that.
And so this is why it's so
hard to explain, because--
yes, I'm not even going
to apologize anymore.
OK, so this is what you do.
And the notes are pretty
clear about how to do them,
and what the problems are.
But lecturing on it would
be a little bit hard.
OK, so now what are we
going to do with it?
Well, we'd like qualitative
stuff and interpretive stuff.
Qualitative is Hund's rules.
Now if you looked at 100
textbooks, I think 95% of them
will have Hund's rules wrong.
You're never going
to make that mistake.
And interpretive-- well, we want
to know the trends of things,
and we want to be
able to do something
like what you did in freshman
chemistry on shielding.
Now you probably memorized some
rules about what shields what.
But I'm going to give
you a little bit more
insight into that.
So we're going to
talk about this
for the rest of the lecture.
OK, so you specify
a configuration.
And this configuration might
be two electrons, two spin
orbitals, two orbitals
times e, or three, or 10.
And often, when you specify
the occupied orbitals,
you neglect the
field ones, which
is nice, because you have
fewer things to worry about.
Because field
orbitals have spin 0.
And you don't have
to do anything.
They're automatically
asymmetrized,
and they basically act
as a charged distribution
in the core that is sampled
by the electrons outside.
And so you need some
sort of a set of rules
for how does that work.
And that's shielding.
So first, we specify
a configuration.
And you also learned-- in
high school, probably--
how to determine the
L, S, J terms that
result from this configuration
by some magical crossing-out
of boxes.
And if you didn't, I'm glad.
Because it would have
just clouded your mind,
and caused earlier
insanity than MIT causes.
So anyway, so we have
orbital angular momentum.
And we can add the
orbital angular
momenta of the electrons
following certain rules.
And we have spin
angular momentum.
And J is equal to the
vector sum of L and S.
And we say we have an LS term--
like triplet P.
And it can have J
is equal to L plus S, L plus
S minus 1, down to L minus S
absolute value.
These are the possible J's.
And so Hund's
rules is all about,
of all of the states that belong
to a particular configuration,
which one is the lowest?
One-- which one, not
the second lowest.
Which one is the lowest?
And why do we care?
Because in statistical
mechanics everything
is dominated by the
lowest energy state.
And so if you can figure out
what is the lowest energy
state, you've basically got
as much as most people are
going to want.
So you want to know what are L,
S, and J for the lowest energy
state of a configuration.
Configurations are typically
far apart in energy.
So if you know what the lowest
energy configuration is,
and the lowest energy state of
it, as far as your friends--
the statistical
machinations, you
can tell them how to write
their partition functions.
And the rest is details.
And mostly, you
don't want details.
If your friends tell
you they want details,
well, you tell them, this
is what you have to do,
but it's no simple
three Hund's rules.
OK, so Hund's rules--
you look at all of the
L, S, J states that
are possible for a
particular configuration.
And you can use the crossing
out of ML/MS boxes if you want.
And I could tell you
why you would do that.
But I don't want to cause
insanity at this stage, either.
But I'm an expert at that.
And you can also use
lowering operators
to generate all the states,
once you know stuff.
OK, so once you
know all the states,
Hund says, which one of
these has the largest S?
which one?
And that's easy to know.
And for example, if
you had 2p squared,
you're going to get singlet
D, triplet P, and singlet S.
And well, here's the triplet.
That has the largest
S. So the triplet P
is the lowest energy state.
Now if there were multiple
triplets, as there would be,
say, for 2p3d, then
you'd have to decide
which of those
triplets is the lowest.
And all you care about,
all you're allowed to say
is which one is the lowest.
And it's the one
with the maximum L.
And then the last step
is, what is the lowest
J for that LS state?
And that's kind of cute.
Because you have the
P shell, there's--
for a P shell, you can have six
P orbitals to fill the shells.
1 alpha, 1 beta, 0
alpha, 0 beta, et cetera.
So the degeneracy
of a P orbital is 6.
If you have p to N,
where N is less than 3,
you have a
less-than-half-filled shell.
And then lowest is J equal
L minus S, absolute value.
And if N is greater than 3,
you have the lowest being--
the highest possible value
of J is equal to L plus S--
so for L, N greater than 3.
And now when you have
a half-filled shell,
the lowest state is usually
an S state with maximum spin.
But it doesn't matter.
When you have less-than or
more-than-half-filled shell,
you have generally a state with
orbital angular momentum not
equal to 0.
And you have spin orbit
splitting of that.
And so you do want to
care what is the lowest J.
But when N is equal
to 3, the lowest state
is usually an S state.
It doesn't have a
spin orbit splitting,
and it just has one value of J,
which is whatever the spin is.
So Hund's rules tell
you how to identify,
without knowing beans, what
is the lowest energy state,
and it's never wrong.
Well, maybe sometimes
wrong, but that's
because of one of
my things where
you have a perturbation
between states belonging
to two configurations.
But people get really
excited when they discover
a violation of Hund's rules.
And it's just trivial.
So there is this.
What time is it?
I have a few minutes to talk
about shielding, and I will.
OK, so we have a nucleus.
And it has a charge of z.
Bare nucleus--
there's no electrons,
so the atomic number
is the charge.
And then we have a
filled shell around it.
It's spherically symmetric.
And so if we penetrate
inside of it, what we see--
suppose we penetrate inside,
to this point, what we see
is only the plus z, minus
the number of electrons
inside this sphere.
Now if you took Electricity and
Magnetism, you can prove this.
If you didn't,
you can accept it.
And so outside the
nucleus, the charge
is plus one, because
you have a neutral atom.
And then when you penetrate
inside this region
of dense charge, and all of
the spins are generally paired,
this is spherical.
So what you end up
seeing is z effective, as
opposed to z true.
And let's say here here's r0--
or this is r0.
And so beyond r0, the charge
that you see is plus 1.
Add 0, you see a charge of z.
And so what ends up happening
is you get z effective,
which is dependent on
distance from the nucleus.
And it goes from
the integer value
that you know, from the
atomic number, down to 1,
because you've taken
one electron away
from a neutral atom,
and taken it outside.
And now we have
this wonderful thing
called the centrifugal barrier.
So if we have a state
that has a non-zero of l--
well, if we have a state
with a zero value of l,
it can penetrate all the way
into the core, to the nucleus.
And so that means that
the shielding is less
for s orbitals.
And now if we have a non-zero
l, it can't get in so far.
And the larger
the l is, the less
it can see this extra charge.
So high l's are
very well shielded.
Low l's are not
so well shielded.
And the shielding goes s least
shielded, p, less, so on.
Now there's some other
interesting things.
Which, you know, I
hate to say this,
but comparing 5.111
or 5.112 to 3.091,
there is this business of what
happens when you start to--
you start with potassium, and so
you put an electron in the 4s,
not in the 3d orbital.
Right?
Why is that?
Well, the 4s sees the larger
charge is less shielded.
So it goes in.
Then when you go from
potassium to calcium,
you put another
electron in this.
And that's true.
So for calcium, you
have a 4s squared.
And 4s 3d is a
higher-lying state.
Now you take an electron--
I'm cooking my own goose.
If you take one of
these electrons away--
this is not the way I
wanted it to come out--
you find yourself in a 3d state.
Because 3D penetrates
a little bit under 4s.
I can't explain it in a way
that's going to make sense.
I really wanted to,
because I care so much
about these simple arguments.
But I will just be
wasting your time.
So the order in which
you feel orbitals
comes out, naturally, different
from the order in which you
remove electrons from orbitals.
And the shielding arguments
are capable of explaining that.
OK, so this is the end of atoms.
And I've asked you to observe
some complicated algebra which
you're never going
to do, or at least
never going to do much of.
Everything you need
to know about atoms,
you can tell a computer,
and it can do it.
Now molecules are
much more complicated.
And that's we're going
to start on next time.
We're going to start with
molecular orbital theory.
And I'm not going to be
presenting the normal textbook
approach.
I'm going to present an
interpretive approach, where
you understand
why things happen,
as opposed to memorize just
symmetries, and filling orders,
and so on.
OK, I'll see you on Wednesday.
