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PROFESSOR: Good afternoon.
So last week, we started
our discussion of atoms.
So these are, of
course, the key players
in a course of atomic physics.
And we will reveal the structure
of atoms going from the larger
energies to the
smaller energies.
That means we start with
electronic energies mainly
determined by Coulomb energy.
And then we go to
finer levels which
are fine structure, hyperfine
structure, Lamb shift, and all
that.
So we have started last week
to discuss the Schrodinger
equation, discuss
hydrogenic energy levels,
and I pointed out some important
results on typical length
scales and the scaling
of the wave function.
We will meet that later today.
But before I continue
with atomic structure,
I want to start today
to discuss the units.
The atomic units we are
using to describe the atom.
So the nat-- and for
every problem in physics,
you have what one may
call natural units.
And for us these are also
called the atomic units.
So what are atomic units?
Well, atomic units are the
units for length, for energy,
for velocity, for
electric field.
And all of these
units, you should
be able to construct them
out of fundamental constants.
The fundamental constants
which appear in the Schrodinger
equation for the
electron within the atom
are the charge of the electron,
the mass of the electron,
and h-bar.
Now, is there any other
fundamental constant
we should include here to
construct our natural units?
c, good question.
What about c?
The speed of light.
Should the speed
of light be part
of our system of atomic units?
Let's not go there.
[LAUGHTER]
Because if you set it to
one, you have made a choice.
You have constrained
your system.
And you're almost obscuring the
fact whether it should appear
or not.
Well, the strong message
I want to give you
is at the level of
electronic structure,
at the level of the
Schrodinger equation,
it should not be there.
Because we are
talking at this point
about solutions
of the Schrodinger
equation, the hydrogenic levels.
And there is no c, no speed
of light in the Schrodinger
equation.
So c is not part of the
fundamental constants
we have to consider now.
It will later come in the
fine-structure constant,
but this is a different story.
So we have three
units, e, m, and c.
And you can just play
the game of combinatorics
and see can you
find a length which
consists of those three units.
And, well, it's h
squared, m over e squared.
This is how you get
the unit of length.
And this length is
called the Bohr radius.
And indeed, this is
sort of the RMS size
of the electron in the 1s state.
You can play the
game again and ask,
can we construct an energy?
Well, you find that if
you take e to the four,
if you take the mass and
finally divide by h square,
then you have a unit of energy.
And this unit of
energy turns out
to be one Hartree or 27.2
electron volts or two Rydbergs,
twice the binding energy of
the electron in the 1s state.
And we had some discussion last
week that the factor of two
reflects the virial theorem.
This is actually that one
Hartree is the Coulomb energy
of the electron in the 1s state,
which is the binding energy.
But then half of it
is kinetic energy.
And, therefore, the total
energy of the 1s electron
is half the Coulomb energy.
And it's that.
But no.
OK, so, so far no c.
The energy, the energy
levels, the wave function.
If there is no c, no speed
of light in the Schrodinger
equation, there is no c in the
solution of the Schrodinger
equation.
And if you set it 1, sure,
it wouldn't-- I mean,
if your relativity
equation said c equals 1,
you obscure the fact.
But here, it's
definitely not there.
But now we can also
see, well, there
are other important energies.
One energy, and now
I bring in the c
just because I want to
compare two energies which
include the speed of light, the
rest energy of the electron.
Or a very fundamental unit
of length is h-bar over mc.
Which is the
Compton wavelengths.
So that's lambda Compton,
the Compton wavelengths
of the electron.
And, well, if we
try to figure out
what is the ratio of the
atomic unit of lengths.
And the Compton radius,
we have to multiply
with a dimensionless unit
which is hc over e squared.
Or if I take the reverse,
e squared over hc, h-bar c.
And similarly, here the
dimensionless quantity
to multiply is that.
So what do we find now here
is we find that what we get
is a quantity which
I want to call alpha,
the fine-structure constant.
And what I find here is the same
constant, alpha to the minus 1,
times the Compton wavelengths.
Let me discuss the
fine-structure constant
in a second, but there's
still two more atomic units
we want to discuss.
There is the velocity and
there is the electric field.
I can simply get the
velocity by saying, well,
the velocity enters the
kinetic energy in mv square.
And if I said mv square, I want
to skip all factors of unity.
So that's not one half,
it's just mv square.
And if I said mv square
equal to one Hartree,
then I find that the
atomic unit of velocity
is e squared over h-bar.
But this turns out
to be alpha times c.
So again, what we find is alpha,
the fine-structure constant.
Alpha is, of course,
dimensionless.
It is 1/137.
So therefore, we see that if the
velocity-- and this is actually
the orbital velocity
of the 1s electron.
If this is alpha
times c that confirms
that the electron
non-relativistic.
We have solved the
non-relativistic Schrodinger
equation for it.
And consistently, we find that
the velocity of the electron
is 1/137 of the speed of light.
It's actually physics trivia.
If somebody asks you
how fast is the electron
in the hydrogen atom, about
1% of the speed of light.
Of course, if you had solved
the non-relativistic Schrodinger
equation.
And you solve it for, let's
say, a naked uranium nucleus
where z, the charge
of the nucleus, is 92,
then you find that this
fine-structure constant times
z is on the order of unity.
You would find that the electron
moves at the speed of light.
And then you realize, gosh,
I've solved the wrong equation.
Because a non-relativistic
Schrodinger equation
when the solution
is that something
moves at the speed
of light, I'd better
start with a different equation.
But here, we find
we are consistent.
An electron in the hydrogen
atom for low nuclear charges
of a few hydrogen, helium,
and so on is non-relativistic.
So let's just finish that.
The electric field
is the electric field
felt by the electron
which orbits
the nucleus on the 1s shell.
And this is 5.1 times 10 to
the 9 volts per centimeter.
So everything I've constructed
here out of the three
fundamental
constants, e, m, and h
are typical for the 1s
electron for the ground
state of hydrogen.
We got the typical
lengths, the Bohr radius,
the typical energy, the
Hartree, the typical velocity,
which doesn't have a name,
and the electric field
experienced by this electron.
So let's now talk about alpha.
So alpha is dimensionless.
It is, you know, if a constant
has dimension like h-bar,
like c, actually the
value of it reflects
our system of metrology.
If you define the second
in a different way,
h-bar will change.
c will change.
So a lot of constants are
not fundamental constants,
being fundamental to
the physics at hand.
They are more kind
of translating
our metrological system
into the equations
we use to describe our system.
But if something
has no dimension,
it is not related to a unit
like the kilogram or the second,
it has really
fundamental importance.
So, therefore, alpha is
the fundamental constant
in atomic physics.
And if you have a
fundamental constant,
ultimately, there should
be a theory of everything
which should ultimately
predict the value of alpha.
Which, ultimately, we
predicted by a complete theory.
So alpha, the fine-structure
constant, is smaller than 1.
It's 1/137.
And the fact that
alpha is smaller like 1
is often phrased in these words.
That since alpha is
much smaller than 1
that this implies the
electromagnetic interactions
are weak.
OK, I want to explain that.
I mean, I've heard
this many, many times.
But what does it mean
interactions are weak?
So let me give you sort
of my, in 90 seconds,
sort of my spiel on it.
Why does alpha mean that the
electromagnetic interactions
are weak?
Well, we have to compare the
electromagnetic interaction
to something else.
And the result will be for
this situation I create,
the electromagnetic
interactions are weak.
Electromagnetic interactions
of the Coulomb field
gets, of course,
stronger, and stronger,
and stronger the closer you
move two charges together.
So what does it mean that the
Coulomb interaction is weak?
Weaker compared to
strong interactions
or other interactions.
Well, let me try to
justify it as follows.
We cannot go to epitary
small distances.
We can do it in
classical physics,
but not in quantum physics.
So if you go to very
small distances,
if you localize
particles very tightly,
they have a lot of
momentum uncertainty.
The momentum uncertainty
means energy uncertainty.
And the energy
uncertainty may mean
that we can create electrons
and positron pairs.
So the moment we have
an energy uncertainty,
by our definition of bringing
two particles close together,
and this energy uncertainty
is larger than the rest mass,
we have to be very careful.
We can no longer use a
single particle description,
or our concept of
single particle physics
breaks down if you have
seen the particles prepared
with an energy uncertainty,
which would spontaneously
create more particles.
So therefore, let me
postulate that our picture
how we think about
those interactions,
arranging two charges
and writing down
what the Coulomb energy is.
That if energy uncertainties
become on the order of the rest
energy, then the concept of
single particles breaks down.
Of course, that's not
the end of physics.
You need now a field theory
for particles where particles
are just excitations
in your field.
But here in atomic
physics, you want
to describe an electron
bound to a nucleus.
And we want to use
those concepts.
So let's just sort of say,
what does it mean with delta e
is mc squared?
Well, that means the
momentum uncertainty
is on the order of mc.
And with this
momentum uncertainty,
I can localize particles
to within h-bar over mc.
And this just turns
out to be the Compton
radius of the electron.
So, therefore, I
should be careful
when I talk about the Coulomb
energy between particles
if I would go closer
than the Compton radius.
So, therefore, let me compare
now the Coulomb interaction
at the Compton radius
to something else.
So these Coulomb energy is
e squared over the radius.
The Compton radius.
And this turns out to be
e squared, mc over h-bar.
So unless I want to
get into quantum field
theory of particles, before I
need a different description,
the strongest
Coulomb interaction
I can create by putting two
particles at the Compton radius
is that.
And I can now compare
this Coulomb interaction,
the Coulomb energy, to the
rest energy, mc squared.
And, well, if I take what I had
above, e squared mc over h-bar,
I divide by mc squared.
I find that e square, that
the result, the ratio,
is e squared over h-bar c.
And this is just alpha.
So in other words, what
I've shown to you, if you
try to bring two
charges as close as
possible before spontaneous
pair production sets in, then
you find that the
Coulomb energy is not
in the dominant
energy in the system.
The dominant energy
is the rest energy,
is the mass of the
electron itself.
And the ratio of those
two energies at this point
is, of course, completely
independent of what
metrological system you use
for energy lengths and such.
It's really something which
says something fundamental
about the nature of interaction.
And what I just presented to
you leads to the statement
that the Coulomb interaction,
the electromagnetic
interaction, is weak.
Because the
fine-structure constant
is much smaller than unity.
Of course, if you use a nucleus
of uranium, naked uranium,
and people have ion traps where
they create uranium 92 plus
and then they add
an electron, you're
really studying very
interesting physics.
You're studying the physics
of an electron for which
the effective
fine-structure constant is
on the order of unity.
And that's why people are
very interested in it.
And that's one area
of current research.
Any questions?
OK, so that's our little
excursion about units.
Let's talk now about
some general properties
of one electron
atoms with cores.
A lot of research in our field
is done with alkali atoms.
Alkali atoms are not
the hydrogen atom.
But they have one
outer electron.
So they are hydrogen-like.
And now we want to
sort of figure out
what is the main
difference because
for one outer electron
in rubidium and sodium.
And the electron in hydrogen.
Well, if we have an electron.
And this electron orbits around.
In the alkali atom,
there is an ionic core
which has a charge of a z plus.
But then in this
sort of compact core,
there are also c
minus 1 electron.
So the electron, the
outer electron pretty much
feels the electric field
of a single charge,
but there is a the of the core.
So what I want to
discuss with you is now
what is the leading correction
to the properties of this atom?
What is the leading correction
to the hydrogen-like wave
function due to the fact that
we have an ionic core and not
a proton.
So for hydrogen, we would
have the Rydberg formula.
that for principal quantum
number n, the energy is-- oh,
just one second.
Yes.
OK, the hydrogenic
energy would be z squared
times the Rydberg constant
divided by n squared.
I'm confused about the factor
of z squared, which I clearly
have in my notes.
I have to read up
something about it.
Let me take it out here.
You know, we have a charge of z.
But what I'm talking
about is one electron
which is very far
away from the nucleus.
And this one electron fields,
in effect, only a single charge.
And so, ultimately,
you would expect
that the Rydberg
spectrum, if you
go to higher and higher
end, would actually
converge to the Rydberg spectrum
of hydrogen which would not
have the factor of z squared.
I hope I'm not
overlooking something,
but I'm just correcting
my notes on the fly.
So this would simply be
what an electron would
do in the Coulomb field
of a single charge.
And the question is, what
is the leading correction
to this formula?
And I want to ask it
as a clicker question.
So that's the effect, that
we have an ionic core.
Does it make a constant
offset to the binding energy?
Or does it make the
correction which,
of course, is 1 over n squared.
This would just scale
the overall spectrum.
It would actually lead to
a modified rescaled Rydberg
constant.
Or is the correction
higher order in n?
Or finally, is it
a correction which
changes the effective
principle quantum number n?
Maybe you know the
answer or maybe you
want to try to guess the answer.
So these are the four choices.
Of course delta has
different units.
I just used the same symbol
for the correction term.
But depending whether it
appears it has units of energy
or if it appears in
the denominator with n,
it is dimensionless.
AUDIENCE: Aren't these questions
also C and D [INAUDIBLE]?
PROFESSOR: Yes, this
is the first thing
I wanted to tell you, that
answers C and D are the same.
Because if we assume
delta is small
and you do a Taylor
expansion of the denominator,
you just get this.
So c and d are
actually equivalent.
And so if I add
up 20 people voted
for c and d, which
is equivalent.
So this is equivalent
by Taylor expansion.
And, indeed, this
is also-- this time
there's two correct answers.
So let me quickly derive it.
The derivation is short.
And it adds some insight.
We want to do
perturbation theory.
And in perturbation
theory, we simply take
the wave function of the
simplified Hamiltonian
and ask what is the
energy correction
due to the fact that we
have a finite core size?
Since the finite core is near
r equals 0 at the origin,
we only need the scaling
of the wave function close
to the origin.
And this is r to the l.
And we have n to the power
3/2, as we discussed last week.
So our Hamiltonian is
now the Hamiltonian
of the hydrogen atom
plus a perturbation term.
And the perturbation
term is the derivation
of the potentially experienced
by the outer electron,
the deviation from a
pure Coulomb potential.
So the energy correction,
en, is the expectation value
of the hydrogenic wave
function with the perturbation
Hamiltonian.
And the only thing we have to
know about the perturbation
Hamiltonian is that
this Hamiltonian is
localized around the origin.
And then we immediately
find because of the scaling
of the wave function with
n that this is 1/n cubed.
And it's proportional.
And by factoring out
the Rydberg constant,
I can parameterize
this matrix element
with a quantity, delta l,
which is dimensionless.
So that means that
the binding energy
is the hydrogenic binding
energy plus this correction.
And then to leading
order in delta
l, it's identical
to this result.
And this parameter,
delta l, which
is characterizing a whole
Rydberg series for all n
values for given l, this is
called the quantum defect.
So people realized early
on in the early days
of quantum mechanics
before they understood it
that the spectrum of
many atoms followed
a formula which was not
1/n squared is hydrogen.
It was 1n minus delta l squared.
They didn't understand it.
But this is, of course,
now the derivation.
There are many other
derivations which
you may enjoy reading about
it in our atomic physics wiki.
There's a derivation using
the semi-classical approach
and using the very
nice physical picture
that an electron, when it
comes close to the core,
experiences scattering
phase shift.
And this scattering phase
shift is directly related
to the quantum defect.
Or you can make a
model Hamiltonian
which is exactly solvable where
the perturbation Hamiltonian is
not completely
localized at the core.
But it's proportional
to 1/r squared.
And then you can
exactly solve it
because you have
already a term which
is 1/r squared in your
Schrodinger equation.
Which is the centrifugal term.
So therefore, the
perturbation is only
redefining the centrifugal term.
It's redefining what l is.
And eventually, you
can solve it directly.
Questions?
OK.
Then let me spend
five to 10 minutes
on spectroscopic notation.
So the next five
or 10 minutes, how
we describe the
configuration of an atom,
well, I don't particularly
like to teach it.
Because it's more nomenclature
about old-fashioned symbols.
On the other hand, if
you're working with atoms,
you have to learn the language
how to describe atoms.
And I also know that
an appreciable fraction
of oral exams there will be
one person in the community
and say, what is
your favorite atom?
And what is the
configuration of your atom?
So it's something if
you're an atomic physicist
you're supposed to know.
So the spectroscopic
notation, the term designation
focuses on the fact that if
you have an isolated atom,
we have angular
momentum conservation.
And so we have at least
two quantum numbers.
Which are sometimes also
two good quantum numbers.
We have some approximate
quantum numbers
where we have additional terms
which break certain symmetries.
But an isolated atom
lives in isotropic space.
The total angular momentum
of this atom is conserved.
It's absolutely conserved.
It's absolutely
good quantum number.
And the good quantum numbers is
the total angular momentum, j,
and its protection energy.
So in the language of atomic
physics, we call j a level.
It's different from states.
So one level has now 2j
plus 1 sub-levels or states.
So usually when we
talk about a level,
we assume the level
has degeneracies
because there is the
Mj quantum number.
So j, you're talking about
electronic structure.
So j can have, when we
have an isolated atom,
can have contributions
from several electrons.
It can have contributions
and these electrons
can contribute through
spin, s, and orbital angular
momentum, l.
In many situations,
especially with alkali atoms,
the inner core is
completely field shell.
There is no s, no l
from the inner electrons
which contribute.
And all the contribution
with the angular momentum
comes from the outer electron.
Especially for
the lighter atoms.
The non-relativistic atoms.
The different electrons
undergo ls copying.
In other words, if you
have multiple electrons,
their orbital angular
momentum couple up
to the total orbital
angular momentum, l.
And all the spins couple
up to the total spin, s.
So therefore, before we
introduce spin orbit copying,
l is a good quantum number.
s is a good quantum number.
And then they couple to
a good quantum number, j.
Of course, once l and s
couple to j, orbital and spin
angular momentum precess around
the total angular momentum.
Anyway, I just
want to say when I
talk about j, what are
possible ingredients?
So let's assume we have an
atom which has total angular
momentum, j.
And which is the sum of orbital
angular momentum and spin
angular momentum.
And then in this case, we
use a term designation.
A level is designated
by a term which
is written as l, the value
of orbital angular momentum.
The spin multiplicity, 2s
plus 1 is an upper left index.
And a lower right index is j.
And of course if l is 0,
we use the letter s, p, d.
This is sort of the
historic letter designation
for l equals 0, 1 and 2.
So in other words, if you have
an atom where the total angular
momentum is composed of orbital
angular momentum and spin,
you can always
write this symbol.
And this symbol is
the term designation
which characterizes the state,
the ground state or an excited
state of your atom.
If you have the hydrogenic
atom, often you precede the term
by the principal
quantum number, n.
So let me give you an example.
If you have the sodium atom, the
outer electron has n equals 3.
It has 0 orbital
angular momentum.
It has spin 1/2.
And 2s plus 1 is 2.
And the total angular
momentum is 1/2.
If you go to the first excited
state, you're still in n
equals 3.
But you have promoted the
electron from an s state
to p state.
So, therefore, the
orbital angular momentum
is now 1 designated by p.
The spin is still spin 1/2.
But now orbital angular
momentum of 1 and spin 1/2
can form a total
angular momentum
which can either be 1/2 or 3/2.
So if you're asked what is the
state you prepare your atom in,
you would give it a
symbol 3 doublet p 1/2.
And I've explained
to you what it means.
There is one addition.
And sometimes you want
to not just mention
what is the principle quantum
number of the outer electrons.
Sometimes you want to specify
the whole configuration.
So this would mean you want to
sort of build up the electron
shell and say that I have
two electrons in 1s, two
electrons in 2s, one
electron in 2p, and so on.
So you use, I think this
would now be beryllium atom?
1s.
So 1s is hydrogen,
1s2 is helium.
Then we go to lithium.
AUDIENCE: [INAUDIBLE] is boron.
PROFESSOR: It's boron, no?
OK, so this would be boron.
So what we use is here
we use the products.
We use products of
symbols n, l, m.
So to come back to
the example of sodium,
so sodium is filled up.
As in the first shells.
So it's 2p6.
And then we have one electron,
the outer electron in 3s.
However, let me point
out that this way
to specify the
configuration strongly
depends on a hydrogenic model.
It assumes that the electrons
are non-interacting.
And is, therefore,
an approximation.
In contrast, the
term designation
with a total angular
momentum is always exact.
Well, at least the
total angular momentum
is an exact quantum number.
Whereas the
configuration is based
on the independent
electron approximation
in hydrogenic orbits.
And usually when
you have a real atom
and you calculate
with high precision
what the electronic
wave function is,
you find actually that total
many-body wave function
is a superposition of
many such configurations.
But as long as one
configuration is dominant,
this configuration
designation makes sense.
Any questions?
OK, so we come back
to the hydrogen atom
when we discuss smaller
features of the energy levels.
Fine structure, hyperfine
structure, and so on.
But in our discussion of
electronic energies, that's all
I want to say about
one electron atoms.
So let's now proceed and
discuss the helium atom.
So we want to understand now
what are the new effects when
we have not only one
electron, but two electrons.
And don't worry.
We are not proceeding to
three, four, five electrons.
I think to go from one
to two, we actually
capture the most
important state.
Namely, the interaction between
the two electrons and what
the results of that is.
For the helium atom, there
are some excellent treatments
in standard textbooks
of quantum mechanics.
One is the famous
quantum mechanic
text by Cohen-Tannoudji et al.
But also the text
of Gasiorowicz.
The reason why I have added the
helium atom to the curriculum
is because it is a
simple system where
we can discuss singlet
and triplet states.
And singlet and
triplet configuration
is important for population
of inoptical lattices
for quantum magnetism and such.
Actually, you can say if
you have two electrons
and they align in a triplet
state or a singlet state,
one you can say
is ferromagnetic.
The other one is paired
and antiferromagnetic.
It's the simplest example
where we can discuss magnetism.
So that's my motivation why
I want you to know something
about the helium atom.
So, therefore, let's now
discuss energy levels of helium.
And let's just start with
the most basic model.
The helium atom has two charges.
So if you regard it
as a hydrogen problem
and we put two electrons
into the 1s state,
we would expect that based
on the hydrogenic model
that the binding energy
of that is, per electron,
is the Rydberg energy
as in hydrogen.
But now we have to scale it with
z squared, the nuclear charge.
And this gives us a factor of 4.
So we would expect that per
electron, the binding energy
in the more simple hydrogenic
model is 54 electron volt.
So that would mean that the
binding energy of the ground
state is minus
108 electron volt.
However, the
experimental result is
that it's only 79 electron volt.
So we find that there is a
big discrepancy of 29 electron
volt.
Which is really huge.
So what is responsible
for this big discrepancy?
Well, what we have
neglected, of course,
is the interaction
between the two electrons.
So we can fix that
in the simplest way
by keeping the wave function
from the hydrogenic model.
But now calculating the
electronic energies,
the electron-electron
energy, by using
the electron-electron
interaction as a perturbation
operator.
So we still use as the wave
function for the ground
state, electron 1
in the 1s state.
So 1-0-0 is a designation for n,
l, m for the hydrogenic quantum
numbers.
And we assume that the
ground state is simply
the product of two
electrons in the 1s state.
So if I calculate for
this perturbation operator
the expectation value
with this ground state,
we find that there is an
energy correction which
is 34 electron volt.
So this removes most
of the discrepancy.
You can improve on it by
a variation wave function
if you use hydrogenic
eigenfunctions as your tile
wave function.
But you are now calculating
those hydrogenic wave functions
not for nuclear charge 2, but
in nuclear charge c star, which
you keep as a
variation of parameter.
You find that you find
even better wave functions.
And you can remove
2/3 of these remaining
discrepancy of five
electron volts.
This variational wave
function is left to you
as a homework assignment.
So where z is z
equals 2 is replaced
by a variation of parameter.
Anyway, that's all
I want to tell you
about the ground
state of helium.
It's pretty much
finding a wave function
which correctly captures
the Coulomb energies.
The interaction between
the two electrons.
But what is much
more interesting,
and this is what
I want to focus on
for the rest of this
lecture, is what
happens in the excited state.
Before I do that,
let me just tell you
what we just discussed.
So we have to this
hydrogenic estimate
and eventually the Coulomb
energy raises the energy level
to what we have just discussed.
But, you know, as the ground
state has no degeneracy.
And so we all talk about
quantitative shifts.
However, when we go
to the excited state,
we will find the degeneracies.
And degeneracies are
much more interesting
because something which
was degenerate can split.
You have two different terms.
So suddenly there
is richer physics.
So, therefore, we want to
discuss now the excited state.
So starting again with
the hydrogenic model.
In hydrogen, the
2s and 2p state are
degenerate so we have two
configurations contributing
to the same energy.
1s 2s and 1s 2p.
The binding energy in
the hydrogenic model
is a quarter of a Rydberg.
Rydberg over n square and ns2.
We have to scale it by z square.
And we find 13.6 electron volt.
But now what happens
is that we have
to introduce the Coulomb
energy between the electrons.
And if you do that, it shifts
up the levels in different ways.
So this is 1s, 2s.
This is 1s, 2p.
So why is this different?
Well, you can say the following.
You have two electrons.
And you have the helium nucleus.
And you first put
in the 1s electron.
And now the second electron
when it is in a p state,
in a 2p state, it's further out.
And it pretty much
experiences out
there the charge of
the helium nucleus
shielded by the 1s electron.
And, therefore, it sees
in effect a smaller
nuclear charge.
Whereas, the two s
electron penetrates deeper.
Gets closer to the
nucleus and will still
realize that the nucleus
has a charge of 2.
And not a shielded charge.
So, therefore, you would expect
that the shielding effect
due to the innermost
electron is more important,
has a bigger effect
for the 2p electron
then for the 2s electron.
So let me just write that down.
So the 2p electron it
sees a shielded nucleus.
In other words, what
it experiences more
as an inner core, the Coulomb
potential of helium plus.
And not so much
of helium 2 plus.
This is due to the 1s electron.
And, therefore, the 2p electron
has a smaller binding energy.
It's actually comparable
to the binding energy
of the 2p state in hydrogen.
Which is on the order
of 4 electron volt.
And this effect is
smaller for the 2s state.
So let's go back to
the energy diagram.
So we have now the situation
which I just described.
We have two degenerate
configurations
for in the hydrogen model.
And when we add in Coulomb
energy between the electrons,
there is quite a big splitting
of several electron volts.
But now, each label
undergoes further splitting.
And this is what
I want to discuss.
So we are still sorting
out just the preliminaries.
What I really want
to discuss with you
is the singlet
and triplet thing.
But now we are there
where we can do it.
So if you have a
configuration with 1s, 2s,
there are two possibilities
for the total angular momentum.
Two electrons in
an s state, there
is no orbital angular momentum.
But there are two spins, 1/2.
And they can add up to
1 or can add up to 0.
So, therefore, we will
have two different terms.
One is singlet s0.
And one is triplet s1.
And this splitting
is 0.8 electron volt.
And this is what we want to
discuss in the following.
For completeness, but
the physics is similar,
let me mention that the
1s, 2p state also gives
rise to two terms.
We have now the total
orbital angular momentum, p.
Orbital angular momentum of 1.
The spin angular
momentum is 0 or 1.
So we have singlet and triplet.
And the total angular
momentum is 1 in this case.
Or in this case, 2, 1, or 0.
And the splitting
in this situation
is also on the order of a
fraction of an electron volt.
So what we want
to understand now
is why do we observe a splitting
between those two levels
which seems to
depend on the spin?
How can the spin
cause a splitting?
Because the spin so far has not
appeared in our Hamiltonian.
We rarely have a
Hamiltonian which
has only the Coulomb energy.
And the spin is not part of it.
So we don't have
a magnetic field
to which the magnetic momentum
of the spin would couple.
And also we have not yet
introduced spin orbit coupling.
But if this is on your
mind, take it off your mind.
Spin orbit coupling is a
much, much smaller effect.
Energies on the order
of 1 electron volt,
you just cannot get from
spin orbit coupling.
Spin orbit coupling is smaller
than electronic energies,
as I will explain
to you on Friday,
is smaller by the fine
structure constant.
So the typical scale
for spin orbit coupling
is maybe 10 or 100
million electron volt.
It's much smaller.
So, therefore, we
want to understand now
why do we have a spin
dependent energy.
Also we haven't coupled at this
point the spin to any field.
OK, so we are focusing
now on the splitting.
So we have the 1s,
2s configuration.
We get two terms,
as I just discussed.
One is singlet and
one is triplet.
And still using the
hydrogenic model
of non-interacting
electrons, we want
to write down the wave function.
So the wave function
of the twin electrons
is we have one electron
in the 1s state.
We have one electron
in the 2s state.
But now since we
have fermionic atoms,
we have to correctly
symmetrize it.
So whether we want the symmetric
or antisymmetric combination,
we exchange the two electrons.
So now we have r2 and
r2 in reverse order.
Of course, the total wave
function for two fermions
has to be antisymmetric.
But the total wave function is
the product of the spatial wave
function which I just wrote down
times the spin wave function.
The spin wave function can be
antisymmetric and symmetric.
And the antisymmetric
spin wave function
has to combine with the
symmetric spatial wave
function, and vice
versa, to make sure
that the total wave
function is antisymmetric.
And the correct
description for fermions.
The designation here,
symmetric and antisymmetric
for the total wave function
reflects the spatial part.
The total wave function, of
course, including the spin wave
function is always
antisymmetric.
OK, so we have two
wave functions.
One has a symmetric
spatial wave function.
The other one, an antisymmetric.
The symmetric
spatial wave function
has an antisymmetric
spin wave function.
And that means we
have s equals 0.
That's up, down, minus
down, up is antisymmetric.
Where as the
antisymmetric spatial
wave function goes together
with the symmetric spin wave
function s equals 1.
So this is the situation
which gives rise
to that triplet s1 term.
And this here is
the singlet s0 term.
OK.
As long as we have
non-interacting electron,
the two wave functions
are degenerate.
But now we want to bring in the
Coulomb energy between the two
electrons which we had
already discussed before.
And if you calculate the energy
using this as a perturbation
operator, well, remember the
wave function had two parts.
It was 100, r1, 200, r2.
And the part where our 1
and our 2 were flipped.
And now if we have the wave
function, the perturbation
operator, we get a
total of 4 terms.
2 times 2.
We have to sort of x it out.
And we will then have the
sort of diagonal parts.
And we have the parts
which are off diagonal.
And for the off
diagonal parts, it
matters whether we had
the plus or minus sign.
You know, if you have plus,
plus and minus, minus,
you give a positive
contribution.
But if you connect
the wave function
before and after the
operator, plus with minus,
we get minus signs.
So in other words, we
have one contribution
where it doesn't
matter whether we
have the symmetric or
antisymmetric spatial wave
function.
But then we have
another term where
it matters whether we have the
symmetric and antisymmetric
wave function.
That's where the plus or minus
sign from the symmetrized wave
function appears.
So we have two contributions
now to the energy correction.
One is independent of
the spin wave function
whether we have the symmetric
or antisymmetric configuration.
The other one is not.
The first term is called
the Coulomb energy.
The second term is called
the exchange energy.
So what we find is-- we
have to trace back the sign,
but you find that the triplet
state symmetric in spin.
And antisymmetric in the
spatial wave function.
Has the lower energy,
is more strongly bound
because the antisymmetric
spatial wave function reduces
the repulsive interaction
between the two electrons.
So in other words,
we do not have
any spin term in
the Hamiltonian.
It's just that whether
the spin wave function is
symmetric or antisymmetric,
it requires the spatial wave
function to be the opposite.
And now when we calculate
the Coulomb energy
for the spatial or antisymmetric
spatial wave function,
we find a big difference.
And the big difference
is the exchange energy.
So it is a spin dependent
term for the total energy.
But what is behind this energy
is simply the Coulomb energy.
So the spin through the
symmetry of the wave function
leads to a difference
in the Coulomb energy.
And actually what
I'm telling you
is the explanation why we have
magnetism at room temperature.
It was Heisenberg's
idea when he realized
for the first time what
can cause ferromagnetism.
It was pretty much the model
of the helium atom expanded
to many, many
electrons in a lattice.
So the result is that the
Curie temperature of thousands
of degrees, we have magnetism
below 1,000 degrees.
This energy scale is
an electronic energy,
is in Coulomb energy scale
and not an energy scale
where spin interactions
comes into play.
If it were not for
the exchange energy,
we would not have magnetic
materials above 1 calorie.
So this is what you see
here in the helium atom.
How speed leads to an
energy splitting which
is an electronic Coulombic
energy splitting.
OK, let's make it maybe
even more obvious.
The above equation
can be rewritten.
So this energy
splitting can be written
as a constant, alpha,
plus a constant beta,
times the product of s1 and s2.
Well, what happens is-- let
me just give you as a sidebar,
the product of s1
and s2 can be written
as minus s1 square, minus
s2 square, plus s square.
s1 square is one half times 3/2,
it's 3/4. s2 squared is 3/4.
And s square is either
0 or 2 depending
whether you're in the
singlet or triplet state.
In other words, I just
want to with the sidebar,
which you have seen many,
many times, just remind you
that the product of s1 and
s2 has only two values.
One for the singlet state,
one for the triplet state.
So, therefore, if I
have a singlet level
and a triplet
level, I can always
parametrize it like this.
And I can make it more
obvious by showing
what this formula on the right
hand side has two values.
One for s equals 0.
And one for s equals 1.
And this can be
rewritten as alpha.
You find a more explicit
calculation in the [? video, ?]
but it's also just one
more line of algebra.
You can write it in
the following way.
So, therefore, the
alpha and beta parameter
are just a way of, you know,
you have two energy levels.
And with two constants,
you can always
describe two energy levels.
Here, I've done it
with alpha and beta.
But before, I did it
with the Coulomb energy.
And the exchange energy which
we obtained in this equation
when we perturbatively
calculated the integral.
So by writing it as
s1 dot s2, I even
suggest that the
two spins interact
like a dipole-dipole
interaction.
But they don't.
It comes from the
Coulomb interaction,
but it is equivalent to
a gigantic dipole-dipole
interaction.
So, therefore, the
conclusion of this
is that what I have derived
for you for the helium atom,
it looks like a ferromagnetic
spin-spin interaction.
And, well, it looks
like it, it is actually
an effective ferromagnetic
spin-spin interaction.
However, the coupling
is purely electrostatic
here and not magnetic.
Questions?
OK, yes?
AUDIENCE: So since we're still
using the hydrogenic wave
function as a basis,
how closely does this
get to the actual
measured values?
PROFESSOR: The question
is, how close do we
get with hydrogenic
wave functions
to the actual measured energy?
I don't know.
I don't know the exact
numbers for the excited state.
I assume that it's similar
to the ground state.
You saw that for
the ground state,
we had a big discrepancy.
Most of that was closed by
using hydrogenic wave functions.
And just calculating
the perturbative terms.
So you pretty much
get qualitatively or
semi-quantitatively, you
get the picture out of it.
And unless you're
really interested
in the absolute values,
you can stop there.
But one way to go further and
reduce the discrepancy by 75%
is to use this variational
wave function where
you use hydrogenic
wave function,
but you use z, the
nuclear charge,
as a variation of parameter.
And I mean, this is,
I mean, it's amazing.
I mean, you have a
two electron atom
and you use a
hydrogenic wave function
with one free parameter.
And you get binding
energies which
are on the order of 70
electron volt accurate to
within better than 3%, 4%, 5%.
But by adding other terms
or using a little bit
more fancy wave
functions, I'm sure you
can get further and further.
OK, so the last thing
I wanted to discuss
is the new feature
of two electrons
is that we have singlet
and triplet levels.
So we have a letter of states
which are singlet states.
And then, and this is
what we just discussed,
let me just make dashed lines.
Because of this ferromagnetic
spin-spin interaction,
we have triplets states
which have lower energy.
So these are n equals
2 triplet states.
n equals 2 triplet states.
j equals 1.
j equals 0, 1, 2.
So of course, there
are transitions
between those levels.
And form a p state, you can have
transitions down to an s state.
The question I have is what
about possible transitions
between triplet and singlet?
So what I want to ask you know
with a clicker question is
what kind?
So those transitions
here are transitions
between singlet and triplet.
And the technical term for
transitions between singlet
and triplet are
intercombination lines.
So the question I have for you
is what fields or couplings
drive singlet
triplet transitions?
So I want you to think about the
model we have discussed so far.
All we have is Coulomb energy.
Coulomb energy between the
nucleus and the electrons.
And between the electrons.
That's it.
We do not put any other
terms into the Hamiltonian.
And now we have obtained
those wave functions.
We have obtained those
energies splittings.
And now we want to ask are
there transitions possible?
So one possibility is that
we can drive the transition
with optical fields.
Let's say our dipole operator.
Which many of you have
encountered in life.
We have already discussed
rotating magnetic fields
in the first part of the course.
Is it possible to use
both magnetic fields
and optical fields?
Or the last answer
is none of the above.
It's not possible with
any of those fields
to create any kind of transition
between singlet and triplet.
OK, all right.
We have to spend the first 10
minutes on the class on Friday
to discuss that.
But the answer is none.
There is no way how you can get
a transition between singlet
and triplet using
the approximations
for the description
of the helium atom
we have done so far.
Since I don't want to end
with such a cliffhanger,
let me just say, we have
actually a selection rule
which says, we are not
changing the total spin.
And you would say, hey, come on.
Can't I just take a magnetic
field and flip one of the spin,
go from a triplet
to a singlet state?
Well, the answer is no.
And let me just give
it to you qualitatively
and deformalize it on Friday.
First, as far as transverse
magnetic fields, rotating
magnetic fields, are
concerned. transverse B fields.
Remember, transverse B fields
create some f of the spin.
But both spins precess equally.
So you can never
even classically,
you cannot change the angle
between the two spins.
When the two spins
are interparallel,
they stay interparallel.
When they stay parallel
when they are parallel,
they stay parallel.
Or to say the
transverse B field means
we have a coupling, sx, sy.
And sx and sy can be
written as letter operators,
s plus and s minus.
And as you know, s
plus and s minus only
change the magnetic
quantum number.
But they do not change
the value of the total s.
In other words, if
you have a spin which
is s equals 1 triplet
state, you can
change the angle, how
the spin equals 1 points.
But it still will
be spin 1 where
both electrons are
aligned parallel.
There is no magnetic field
which can selectively
talk to one spin and rotate
spin 1 with respect to spin 2.
And coming to the
other questions
about optical fields
with a dipole operator,
the answer is also
a resounding no.
Because the dipole operator
of the electromagnetic field,
the dipole operator acts only
on the spatial wave function.
Not on the spin part.
So a laser beam through
the dipole operator
can never ever flip the spin.
It only acts on the
spatial wave function.
So anyway, I wanted to just give
you the answer why you cannot
go from singlet to triplet
state with sort of our standard
operators, with the electric
dipole operator or with
the rotating field.
And I hope it was worthwhile
to explain in detail why each
of them cannot do it.
But on Friday, I
explained to you
that there is a general
symmetry behind it
that even more fancy
combinations will not
be able to do that.
