>> Welcome back to
Chemistry 131A.
Today, we're going to talk about
spin defector model and begin
to talk about hydrogen atoms,
a subject that will continue
on for a couple of lectures.
Recall that when Stern
and Gerlach did their
measurement there were 2 bands
that were observed
rather than a continuum
of silver atoms distributed.
And that was interpreted to mean
that there were 2 possible
states for the angler momentum,
the magnetic moment,
the intrinsic magnetic
moment of the electron.
And that would correspond
to m sub s equals
to plus 1/2 or minus 1/2.
And the interesting thing
is that it was a half rather
than integer which made it
seem mysterious at the time.
In the vector model of angular
momentum we visualize the
angular momentum vector as
lying somewhere on a cone
and the z component we
imagine that we've measured
and that has a definite value
and we've measured the
total angular momentum.
That also has a definite value.
But the x and y values are
indeterminate for reasons
that we'll see in a minute.
For an electron then, there
are just 2 such orientations
of this particular
cone and we tend
to just call them up and down.
And recall that we don't
actually physically think
anything is spinning or rotating
or there's a little planetary
model inside the electron.
We have no evidence at all
for anything like that.
This is just an intrinsic
property
that there is a magnetic
moment and it behaves
as if there were
an angular momentum
and it had the 2 projections
as m sub s equals
plus or minus 1/2.
We can do spectroscopy.
We can do magnetic resonance.
We can make transitions
between these 2 magnetic states.
And a lot of information
about molecules
and photosynthetic systems
and carious other kinds
of things have been
gleaned by actually seeing
if there is an unpaired electron
what the energy difference is
between the 2 magnetic states.
Then actually getting a spectrum
of the possible difference
and seeing how that
spectrum may change
if we irradiate the sample
with light or do other things
to influence the
molecular structure.
So for a spin 1/2 we've
got a picture like this,
of these 2 cones, 1
cone up, 1 cone down.
These are the 2 possibilities.
And there's a definite
value of the z component,
of the angular momentum,
h-bar over 2.
And the other possibility
is minus h-bar over 2.
For the x and y components,
though, they're indeterminate.
And that's why we just draw
this cone because that's meant
to show that the x
and y components could
be anywhere on this cone.
We'll see why in a second.
For a higher angular momentum,
we would have more cones
and we'll see that in a second.
What about the usual conditions
that the wave function,
whatever it is, be
single valued and have
at least a second derivative?
If we put m instead of
either the [inaudible] ,
if we imagine some kind of
thing there with m over 2,
then the problem is
that it changes sign.
But keep in mind that
spin is different.
Because when we derive that e
to the [inaudible] we actually
have a physical particle
on a physical ring.
And we've got those
wave functions
from the boundary
conditions there.
But here, we have an observation
of a magnetic moment.
But we don't know
necessarily that it corresponds
to anything like that.
There is no spatial
component to spin.
It exists just as the magnetic
moment of the electron.
But there's nothing
moving around in space
that we can ascertain.
And the spin part can actually
make some calculations quite
difficult because for
example if you have a molecule
with an unpaired
electron then it may be
that whether the spin is up
or down changes the
spatial electron density.
Why? Because there are a lot
of other magnetic particles
around the molecule, nuclei,
other electrons, and so on.
And it may be that if you --
the spin has 1 orientation,
the magnetic moment, that
it hangs out more over here.
And if the spin has it opposite
orientation, that it ends
up in a different
part of the molecule.
And that means that if
you're trying to keep track
of chemical reactions where
electrons can become unpaired
or various things can happen.
Sometimes you're calculation
gets much more complicated
than you would like because
you have to know something
about the spin or
average over it
or figure out what's going on.
The spin 1/2, if you
want to think about it,
is rather like a Mobius strip in
that if I put a twist in a piece
of paper, then rather than
coming around like a ring
and pointing the same way,
if I put just half a twist
and tape it together,
when I come around,
I'm actually pointing
inside rather that outside.
And when I come around
again, I'm point outside.
So the wave function repeats
when you rotate by 4pi.
That's so called
"spinner behavior".
And usually it's
extremely hard to observe,
but if you have a reference
state and then you have a state
that you're rotating, with
respect to the reference
and you can do this in
magnetic resonance experiments.
You can actually see that
when you apply a pulse
that you actually see
the thing change sign
if you go around by 2pi.
And if you go around
by 4pi it comes back
to the same sign again.
Here's a picture for
the spatial part.
Here are the 5 components of m
sub l for and l equals 2 state,
which we'll see is
called a d orbital.
And again the cones have to
match and the length of l has
to also match because we
imagine we've measured l.
So we know l times l plus 1.
And then we've measured the
z component, so we know that.
And you get this series
of cones and again the lx
and ly components are
completely indeterminate.
But you get these
pretty pictures.
And this is the so called
"vector model triangular
momentum".
We'll see in atomic spectroscopy
in a couple of lectures
that keeping track of
the angular momentum
of the electron, depending
what value of l it has,
and keeping track of the spin
depending on whether it's up
or down, is very
important for keeping track
of where the spectral
lines will appear
and to assign the spectrum.
To assign the spectrum means
that we know the energy level
where the electron
started and then it falls
down to a lower energy
level and emits light
of a certain frequency
that comes out someplace.
And we know then the
energy levels and we know
as many quantum numbers
about each energy level
as is possible to know.
If we know that, then we say
that we've assigned
the spectrum.
If we can match up everything
and then we can infer a lot
about the structure of the atom
by where these lines
actually happen to be.
Like the sodium d
lines for example
that give the characteristic
yellow color
of a sodium vapor lamp.
Why do we have this uncertainty?
Well, it turns out that this
is another manifestation
of the uncertainty principal
and at this point rather
than just talking
about delta p delta x,
we can make a connection with
a much more general idea.
Namely if we're going to
make a measurement of 1 thing
and another thing then if
we make the measurements
in reverse order, the question
is whether we get the same
results or not.
If we get the same result
by making the measurements
the other way around,
then we imagine that these
measurements are compatible
and it's okay.
If on the other hand, we
get a different result,
like we saw with the coin
where if we wanted to see
if it was heads or tails, or we
wanted to see how thick it was
but we ended up getting
different results at random
for whether it was
heads or tails
after we measured the thickness,
then they're incompatible.
And this incompatibility
is actually coded
in the operators themselves so
we don't have to guess about it.
We have a mathematical way
to analyze what the operators
are doing and we can figure out.
It turns out that the square
of the angular momentum,
l times l plus 1, or s times
s plus 1, and the z component,
or any other component, but when
we're picking a component we
pick z by convention,
they're compatible.
>> They can both be
determined, no problem at all.
No problem with the
coin on the edge.
But if we try to, in addition,
determine in addition to lz
and l squared, if we go
further and say, "Well,
I'd like to also know
lx and ly as well,"
then things get fouled up.
Then what happens is if
we measure 1 of those,
then lz gets changed at
random and if we go back
and measure lz, the other
1 gets changed at random.
And no matter how many
times we go back and forth,
we can't get a consistent result
because the result
we get depends
on the order that
we measured in.
And furthermore, it's random.
It doesn't depend like a flip
flop or anything like that,
something orderly
that we can predict.
It's just a random value.
And the key is whether
the operators commute.
And the commutator [assumed
spelling] really is you put 1
operator on 1 side and
the other 1 behind it
and then you do it
the reverse way.
You put this 1 on this
side and this 1 behind it.
And you put them both
on the wave function.
And then you subtract them.
And if you subtract them
and it comes to zero,
then they are compatible.
And you can measure them both
as accurately as you want.
And if they do not, then
they are incompatible.
And that means that making 1
measurement will destroy some
of the information
that you gleaned
from the previous measurement.
Let's just take our operators
for position then and momentum.
Position, remember the
x hat operator was just
to multiply the wave function by
x. And the p hat x operator was
to take the derivative minus
i h-bar times the derivative
with respect to x.
Suppose we want
to figure out the commutator.
Then we write the square bracket
and we write x comma p. And what
that means is that means take x
px minus px x. So let's do that.
The first term is x and then
on the right hand side
minus i h-bar d by dx.
That's p hat x. Minus
the opposite order
which is minus i
h-bar d by dx of x.
And what we do then is
we put our wave function
on the far right of this.
We don't insert it where
it might seem to go,
next to the derivative
past the x. We put it
on the far right hand side
of these operator equations.
Always remember that.
Don't start inserting
thing in the middle.
Keep them to the right because
the way operators work is left
to right by convention.
Let's apply this commutator
to some wave function.
Si of x, that's completely
arbitrary.
We don't care what it is as
long as it follows our rules
for the being continuous and
having derivatives and so on.
The first term then
is quite easy
because it's x times minus i
h-bar times the derivative of si
with respect to x.
Whatever that may be.
The second term is now plus
i h-bar because I have minus,
minus i h-bar times the
derivative of the product x si.
And so that means I have
to take the derivative of x
and then times si plus the
derivative of si times x.
And if I do that then I find
that 2 of the terms vanish.
The minus i h-bar x
deep side the x cancels
with the plus si h-bar x
deep side the x. But I end
up with a third term
because of the rules
for taking the derivative
which is i h-bar si.
And so the commutator
x hat p hat x applied
to the wave function si returns
a number which isn't real.
But it has i h --
it has h-bar in it.
I h-bar si.
Therefore since si is
arbitrary, what we say is
that the commutator -- this
relation has nothing to do
with si so at the
end, we throw si out
and we have an operator
relation.
The operator x hat comma p
hat x is equal to i h-bar.
And the key is it's not zero.
Since it's not zero, we conclude
that we cannot measure
x position
and x momentum simultaneously
to arbitrary precision.
And I've written this
relationship here on slide 312.
This is just an operator
relationship.
You can think of it as i h-bar
times the 1 operator, 1 hat,
where the 1 operator just
multiplies si by the number 1.
And any time in fact
the commutator
of 2 Hermitian operators is
nonzero, it doesn't matter
so much what it is, it gives
you a hint about how it's going
to behave, but the key
thing is, is it zero?
Or is it not zero?
If it's not zero, they
are not compatible.
If it is zero, they are.
And in fact, we can
take the 3 components
of angular momentum l
hat z, that's remember
that these components
are just r cross p,
and that where we take
the vector cross product.
So we can take l hat z, that's
the angular momentum operator
for the z component, that's
x hat py minus y hat px.
And lx is y hat pz
minus z hat py.
And ly is z hat px
minus x hat pz.
If we take these,
because we now know that x
and px don't commute,
do y and px commute?
Yes, they do.
Because there's no -- when I
take the derivative with respect
to y of x times the wave
function, it doesn't --
I don't need to use
the product rule
because these are
partial derivatives and so
if there's no y there,
then I don't bother
taking the product rule.
X is treated as a constant.
So I can measure
the momentum 1 way
and the position another
way, in principle,
as accurately as I like.
I just can't measure them both
and I certainly can't
measure all 3
to localize the particle
plus know its trajectory.
If you work this out, you
can verify on your own
that in fact the
commutation relations
between angular momentum are
a little bit more interesting.
The commutation between position
and momentum is just a number,
an imaginary number,
but still just a number.
In fact, if you take the
commutator of lx with ly,
what you get is i
h-bar times lz.
So you not only get
a number back
but you get another operator.
And if you take the
commutator of lz
with lx, you get i h-bar ly.
And if you take the
commutator with ly
and lz, you get i h-bar lx.
And when you look at
them, these things are
in a cycling permutation.
So if you start with the
3 operators like this,
you're allowed to throw
this 1 here, put this 1 here
and move that 1 there.
And you can do it again and
you always get the same result.
That's called a cyclic
permutation.
And it's very interesting
that these cyclic
permutations look very much
like what a rotation would
look like if we're going
around like this and every time
we take the commutator we go
around to the other component
and in fact there's a very deep
connection between the structure
of these commutators and
the fact that these are
in fact referring to a
rotation of something,
that they are angular momentum.
Therefore for the spin angular
momentum we don't have to -
we don't have r cross p.
We just have this behavior.
But now all we have to say
is, "We purpose that there is
such an operator as the
spin angular momentum.
We know that because we
have this magnetic moment.
We know it has the
value 1/2 for the spin
and that there are
2 orientations.
And we just purpose that
there are 3 operators.
We call them sx, sy and sc.
And that they follow
the same exact rules.
They follow the same exact rules
as the orbital angular momentum.
And that in fact that it's
not really the derivatives
and so forth that are important,
but what's more important
is whether you follow this
particular rule or not.
>> If you do you're going to
behave like an angular momentum.
And that's kind
of an interesting way
to formulate things.
And in fact, as I remark
here, in more advanced courses
that you take on quantum
mechanics we just simply purpose
that they exist.
There are operators
and they follow certain
commutation relations.
And we don't need to specify
what they are in concrete terms.
We don't have to say, "Well
this operator is a derivative
with respect to x,"
or anything like that.
We just say x hat p hat
x is equal to i h-bar.
And anything that follows
that is going to behave
like position momentum.
We don't have to
write derivatives
and various things like that.
Why would we want to do that?
Well, there's a kind
of a minimalist approach
to the theory.
The less things you
have to actually assume
that are detailed then the
stronger your theory is
because it doesn't depend
on whether those assumptions are
correct if you didn't make them,
for example, in place of them
taking the derivative si sort
of the idea that space
is a continuous thing.
That there is such a thing as dx
and there's an infinitesimal
unit of space.
We're taking the
derivative with respect to it
and using these nice
stylized [inaudible] reality.
But what if it isn't?
What if space is digital or sort
of like a billboard and you get
up close and you see
little dots and only
when you get back
away does it look
like a continuous
picture or a photograph.
Well then you'd have to
reformulate your whole formula
if it depended on all
that kind of thing.
Whereas if you just say, "Well,
I have got the commutators
and that's it."
You don't have to do anything
different because you come
up with something else that
has the same commutator.
And then everything else
in the theory still works.
Whereas if everything in the
theory depends on this lynchpin
of well there is a
derivative and all this stuff
that isn't quite true
then you are in trouble.
You have to start over.
Okay. We're going to talk
in a couple of lectures
about atomic spectroscopy.
And in all kinds
of spectroscopy,
in rotational spectroscopy
and vibrational spectroscopy
and atomic spectroscopy,
you run into this unit
of energy that seems a bit odd.
It's called the wave number.
And it has a very strange unit
because the unit is centimeters
to the minus 1 and
that doesn't seem
like a unit of energy at all.
And then why is it centimeters?
And so I want to take a little
bit of an aside here to go
into why we use this unit.
The main reason is that a wave
number is about the right size.
There's a reason why we have
measurements like the foot
and stuff like that because
it's about the right size
to measure things that we
encounter in real life.
And while we're writing down
energies and we're writing
down transitions and so forth,
if we express the energy
in wave numbers, we get numbers
like 1, 10, and 100 maybe 1,000.
But we don't get numbers
like 10 to the minus 34 or 10
to the plus 20, and chemists
don't like big numbers like that
because it makes it harder
to understand things and talk
about it and relate them.
It's always a little bit more
daunting when we've got huge
and tiny numbers and
we're dealing with them.
For atomic and molecular
spectroscopy then the
relationship between the energy
and the wave number is
e is equal to h new.
That we knew from the photon.
And we can write that in
terms of the speed of light
because the wavelength
times the frequency.
You can see that because if
you've got a certain wavelength
and the frequency changes,
that's how far it moved.
So that's wavelength
times frequency is
in fact the velocity of the wave
which [inaudible] is
always c in a vacuum.
So for the frequency new
we substitute c upon lambda
and then we said new over
lambda is new bar and we say
that the energy is,
is hc times new bar.
And new bar is the wave number.
That's what we're quoting
when we, when we do that.
And 1 wave number,
or 1 centimeter
to the minus 1 corresponds
to about 30 gigahertz
and the reason why
is that 3 times 10
to the 10 is how fast light
goes in centimeters per second.
So that's the conversion then
between wave numbers and hertz.
And it's much easier to
visualize 1 wave number,
1 unit as a wave number,
than 30 gigahertz,
3 times 10 to the 10 per
second, as a frequency.
They, they are the same thing.
And we can measure the
thermal energy and we can see
that the wave number is --
makes sense in that regard.
As you may have learned
in freshman chemistry the random
thermal energy that's around
and can be accessed by a system
at some temperature t in Kelvin,
of course is always
in Kelvin, is kt.
That's about the rant --
that's about the range.
K is Boltzmann's constant.
Let's figure out then how
many wave numbers this is.
So we take the kt, let's take
25 Celsius, 298.15 Kelvin,
put in all the constants,
k 1.38 times 10 minus
23 joules per Kelvin,
298.15 Kelvin, h 10 to the
[ inaudible ]
because 34 joules times seconds,
and then for c all we have
to do is put it into centimeters
per second because we see
that the joules go away,
the per second goes away,
the Kelvins go away,
and so if we just put c
in centimeters per second
rather than meters per second,
we get the answer in
inverse centimeters.
And what we find then is
random thermal motion is
about 207 wave numbers.
Therefore, if we've got energy
levels that are spaced less
than 207 wave numbers,
then just things knocking
around can knock stuff
into these higher levels
because there's plenty of
energy around to do that.
You can think of the 207 as
there's lots of 100 dollar bills
on the ground just lying
around, randomly, everywhere.
Then it's pretty easy
to buy a cup of coffee,
if that's the way it is.
If it's very cold,
there's no energy,
there's no money anywhere.
There's a couple of cents.
There's no way you're going
to buy a cup of coffee.
So you can't make any kind
of molecular transition
in that case.
Electronic excitation of
atoms is usually much higher
than this number
207.2 wave numbers.
And that makes sense because
materials only glow let's say
red hot.
So we know if they're glowing
they're emitting light,
that electrons are making
transitions in the material.
But that only happens
if it's red hot.
If it's actually very, very
hot, it's not going to happen
for any kind of metal
or material like that
at 200 -- at 25 Celsius.
They don't emit light and so
that explains why the
numbers we're going to get
for electronic transitions
in atoms are very,
very much higher
numbers than that.
And that also explains for
example, why infrared light,
just heat, as long as you
don't get too much of it
because that photons are low,
doesn't do anything
to chemical bonds.
And neither does the
radio waves or microwaves
from your cell phone,
regardless of what people say
to the contrary who
don't know anything
about molecular structure
or radiation.
Whereas UV light, if we figure
out how many wave numbers
that has, that's different.
That has the potential
to break some bonds.
And that's why we can
use it to sterilize water
when we go camping
because if we stick it
in an intense LED UV source
because it breaks bonds
and all the bacteria that are
swimming around in the water,
of course they're very minute
amounts, but the key is
when you drink it,
are they alive?
And then they start tripling
and quadrupling and so forth
and they make you very sick, or
are they there but they're dead
because their bonds have
been broken by UV light.
Same thing with your skin, you
get too much, you get sunburned.
It's very bad.
Now in atomic spectroscopy,
it was experiment first
and then theory next.
Most of the time
it's experiment first
and then theory comes
in afterwards.
Once you know the answer,
then you can appear
to be quite smart.
>>It's only very rarely,
let's say with the observation
of mercury by prediction
by special relativity
and a few cases like that
where the prediction was made
and then the experiment was
done and it actually agreed
with the prediction because
the theory was so deep.
And in this case a dedicated
spectroscopist, Ruberg,
a Swedish man, found that
in 1890, that the lines --
emission lines from a hydrogen
arc, so what's a hydrogen arc?
Well, I can take hydrogen
gas, that's a diatomic,
and I can like put a
lightning strike through it,
boom, this high voltage.
And 2 things will happen.
The bond will break.
I've got so much energy
there I've got a cascade
of electrons slamming
in, knocking things up.
And then the isolated hydrogen
atoms then will have electrons
in very high orbitals,
a random number
and depending how
big I make the arc.
And then they start
emitting light.
And then I record all
the light they emit
and I have a look at it.
And quite how he divined
this is a mystery,
but he must have been very,
very good with numbers
and some people are very
good with things like that
and others are not so strong.
And he looked at where these
lines were and he worked
out that they followed
this relationship
that the wave number
of the line appeared
to be a constant times
the difference in 1
over n 2 squared,
where n2 is an integer,
minus n over n 1 squared.
Fe it could be a 2 squared is
4, 1 squared is 1 so you get
that one and then there's
3 and 2 and so forth.
And he saw all these things
and this number, r sub h,
the Rydberg constant
referred to hydrogen,
was this number 109,670
so, grave numbers.
Why it followed that was a
mystery, but it was interesting
that it followed that.
And what was even
more interesting,
especially now, quantization.
Quantum mechanics, not
continuum mechanics,
is that these numbers
were integers.
That was a crucial observation
because now this
is a simple thing.
Hydrogen, 1 proton, 1 electron
and it's giving you this clue
that there are numbers
and they're integers
that are dictating where
the light comes out.
If you just take differences
between them you can work
all, all these states out.
It would make sense then, if
you get all these combinations,
1 over n squared minus
1 over m squared,
that the energy levels
themselves go
like 1 over m squared.
And m could be 1, that would be
down at the bottom,
2, 3 and so forth.
And maybe up to infinity.
And the zero of reference here
for this kind of atomic system
and all these systems
in general,
is that you take the
electron and the proton
and you move them apart so
their infinitely far apart
and you leave them at rest.
If they're infinitely far apart
their electrostatic energy
is zero.
And if they're at rest their
kinetic energy is zero,
that imaginary reference
state you call zero.
And then you compare what
the energy is as the atom.
And of course the energy
is going to be negative
in that case because the way to
think about it is if I'm stuck
down here where they've
attracted to each other,
and I want to pry
them back out to here,
I've got to put in
positive energy.
So if I go the other
way from the reference,
I had to go down in energy.
I couldn't go up in energy
or I'd be getting energy back
when I went off to infinity.
Therefore, we can
write the Hamiltonian
and the wave function for
the hydrogen atom, we've,
we've got the Schrodinger
equation --
time independent
Schrodinger equation.
We've got the clue from the
experiment as to what happened,
and so we'll write
down a Hamiltonian.
And it has the kinetic
energy of the nucleus,
which in this case
is a single proton.
It's got the kinetic
energy of the electron
and it's got the potential
energy between them.
So I've written this here.
H hat is minus h-bar
squared over 2 times the mass
of the electron times the, the
second derivative with respect
to where the electron
is minus h-bar squared
over 2 times the mass
of the nucleus times the
second derivative with respect
to where the nucleus is
minus e squared over 4 pi f,
so on that r where r
is a spatial thing,
it's the distance between them.
Now suppose we solve this
just written like this.
What would the wave
function depend on?
Well, it would depend at
the minimum on 6 coordinates
because it would depend on where
the electron is in x, y and z.
And then it would depend on
where the nucleus is in x, y,
and z. That's 6 things that I
have to put in to get a value
for the wave function.
Now how do I plot that?
There's no way I can plot
that because I can't
see what I'm doing.
It's as if I make a
value I have 6 axes.
All I can do is make
like contour plots
and make fixed certain values
and then cut through just
like we do with a mountain range
when we make a contour plot,
we turn a 3 dimensional
thing into a map that's flat
and we draw contours on it.
But here, I'd have to fix
a bunch of these guys.
And then let 2 of them vary and
then plot the wave function.
And it's really --
you're really blinded.
It's like being stuck
between 2 tall buildings
and you just can't see where
on earth you are in the city
because you're just going
down this narrow alley.
And so this is completely
unworkable.
And as you get more
particles involved,
the real wave function
itself is just impossible
to understand what it
actually looks like
and it's extremely hard to
calculate what it is as well.
Since the save function gives us
the most information possible,
most of the time we don't need
to know everything about it.
We just need to know enough
to calculate what we
want to calculate.
And there's a trick
I'd like to go
through to show you how we get
rid of 3 of these coordinates
so that we just have 3 things
and then we can make these plots
in 3D space where we use
transparency and shapes
to indicate a certain
percentage chance
that the electron is
in a certain area.
So the first thing we do
is our potential depends
on just the difference in
position between the 2 things.
So that's a clue that trying
to specify the exact position
of the nucleus in x, y, z
and the electron in x, y,
z is kind of a waste of time.
And I need a trick so that what
I'm looking at is the nucleus
and then I can pretend
the nucleus is at zero
and then it's out
of the picture.
And the way to do that is
to change our coordinates
so that 1 system is
the center of mass,
1 system is wherever the
proton is and the electron is,
I take their center of
mass, which is very close
to the proton, and
I call that big M.
And the other part is called
the reduced mass, which is --
should be familiar from
vibrational problems.
And that's given the symbol
usually u. And that's the mass
of the nucleus times the mass
of the electron over the mass
of the nucleus plus the
mass of the electron.
And when the nucleus is heavy,
u is pretty much the
mass of the electron.
The mass big M, the
center of mass just drifts
around as a free particle
because there is no potential
that refers to the coordinates
of the center of mass.
The potential only
refers to the difference
between the particles.
Not to where they are in the
universe as the center of mass.
And we know the solution
for that.
That's just our plane waves,
e to the i px upon h-bar,
or I've written it here
as si for this big M,
the center of mass is equal
to some constant times e
to the minus ik dot
r. And so that's done.
Now all we have to do is set a
derivative with respect to that.
That's all kinetic energy.
All the center of mass can
have is kinetic energy.
>> It can't ever have
any potential energy
in this kind of problem.
The other part which separates
only depends on the difference
between the particles, not
their absolute coordinates.
And we can write that then
as minus h-bar squared
over 2 times mu times
del squared,
where now del is referring
to the difference --
just the difference.
What's the difference
between x between the nucleus
and the electron,
y and so forth.
And then this is a dodge.
Usually we aren't interested
in the center if mass motion.
We know that things
kind of drift around.
In fact, we try to
design experiments
where things are either
very cold or very quiet
so that things aren't
drifting around too much
because if things drift
around we get slight
frequency shifts just
like a train whistle coming
towards you or going away.
And usually if we're trying
to do an accurate measurement
we don't want that stuff.
We'd like to know what
the train whistle is
when the train is stationary.
Mathematically though,
this lets us just forget
about the center of mass.
We just take the
coordinate of the nucleus
and we just artificially --
totally artificially
assume it's fixed.
We know it can't be fixed
because of the uncertainty
principal.
But we just assume that it is
for the purpose of
the calculation.
And then if we're really
doing something detailed,
we add the center of mass
calculation in later.
Now then this is great because
we're down to 1derivitive
and it's a 3 dimensional problem
but the potential has
spherical symmetry.
And therefore we just imagine
the nucleus at zero, zero, zero.
The electron then is
the spatial variable.
And we just go forward.
And as I remarked, of course,
in real fact the nucleus
can't literally be fixed.
We go ahead then -- we
know how to transform the d
by dx squared, d by dy
squared and so forth
into spherical polar
coordinates.
We get the partial derivative
-- the second partial derivative
with respect to r plus 2
over r times the derivative
with respect to r plus 1
over r squared times the
same thing we had before
for the particle on a sphere.
Recall that before
we got to this point
and then we just
said well r is fixed,
let's look at theta and phi.
Now we've got r in there,
but we've got the other thing
where we already know
what that other part does.
And so all we have
to really figure
out is what does this part r do?
Where instead of being fixed it
has this electrostatic potential
between the 2 particles
that only depends on r
but not on theta and phi.
So as I remarked, we've
solved that part --
that problem on a sphere.
No big deal.
And all we do then is we just
assume like we always did,
that the wave function,
whatever it is, is a product.
And it's a part in r that
I'm going to call big R of r
and as a part in theta and y
that I'm just going to call y
because that's the
conventional term
for the spherical
harmonics, ylm.
And you can verify on
your own as an exercise
that if you make
this substitution
into the Schrodinger equation
that it will separate.
You'll get 2 parts, 1 part that
depends only on theta and phi,
the other part depends
only on r,
therefore you can
do them separately.
And you get the following
2 equations.
You get minus h-bar squared
over 2 mu y times what
we call lambda squared
which was all the sign --
sign theta and d by d phi
times y as the constant.
And the other part we get
is minus h-bar squared
over 2 mu big R times
r squared d,
d squared big R dr squared plus
2r d big R dr plus dr squared
minus dr squared and that should
be a minus whatever the other 1
constant was so that the 2 of
them together add up to zero.
But they're separately
constants.
Once you pick 1 as a number,
the other is the same number
with the opposite sign.
We already know the top part is
just the particle on a sphere.
So good thing we did that.
We know that h-bar --
minus h-bar squared over 2 mu
times lambda squared y is equal
to h-bar squared over 2 mu
times l times l plus 1 times y.
And so the constant
is h-bar squared
over 2 mu times l
times l plus 1.
That's the constant we need.
And the second equation
we can simplify.
Instead of dealing with big R,
we can make a substitution
little --
u of r is equal to r times big
R. And let's see if we can --
let's do this together as a
practice problem now and see
if we can come to
some conclusion.
So here's practice
problem number 15.
Show that the differential
equation simplifies
substantially if we express
it in terms of u rather
than in terms of big R.
What is the connection
with the classical case?
Okay. Well, we're going
to write u as equal
to little r times big R
and we're going to have
to use the product rule
to take derivatives.
So that being the case, d u d
r is r d big R d r plus r times
the derivative of r with
respect to r which is 1.
And we can take the
second derivative.
We take the derivative
of the derivative
and we just follow
the same thing.
The first thing is r d squared
big R d r squared plus d r
over little d r plus
d r over little d r
and then we get terms
out of the end.
R d squared big R d r squared
plus times the derivative
of bir R with respect
to r. Then we just solve
for the second derivative
of big R. And what we get is
over r d squared u d r squared
minus 2 d r d big R d r.
And if we substitute that
into our original differential
equation, substitute the second
derivative of r, then it turns
out that minus 2 d big R d
little r cancels the other,
the other term that
I had perfectly
and we find the following
then that we've got the sum
of these terms plus v r squared
minus e r squared is equal
to the constant which
is minus h-bar squared
over 2 mu times l
times l plus 1.
After we cancel the
terms, here's what we get.
We get minus h-bar squared
over 2 mu times little r
over u times little r times
the second derivative of u
with respect to r squared plus
vr squared minus e r squared is
equal to this constant.
If we divide by r squared and
multiply by u, we can simplify
that and we can finally write
the equation in this form
that I've written
it at the bottom
where we have the
second derivative of u
with respect to r squared.
That looks like a
potential and --
that looks like a
kinetic energy,
excuse me, in 1 dimension.
Then we have minus
e squared times u
of r. That's the potential,
e squared over 4 pi epsilon,
not r times u of r.
That's the potential.
Then we have another term times
u of r which is h-bar squared
over 2 mu r squared
times l times l plus 1.
This whole thing then
should equal e times u of r.
And that's just a 1
dimensional problem now.
On the interval r is
equal to zero to infinity.
So rather than minus
infinity to infinity
like x, slightly different.
It's zero to infinity.
And we can write it simply
as a pseudo-kinetic
energy operator operating
on this wave function u of
r plus v effective on u of r
and v effective is the real
electrostatic potential,
which has a 1 over r dependence,
plus h-bar squared l times l
plus 1 over 2 mu r squared.
And that looks just like a
centrifuge - centrifugal term.
If I have something moving
in a circle, it has an energy
of l squared over 2 i and that's
exactly what we would expect
if we have non zero l. So
it has angular momentum.
Then we have to add the energy
for the angular momentum
before we solve it for u.
But that's the nice
simplification.
We still have to solve for u
and then we have to solve for r
in order to see what kind of
electron density we expect
to find for -- as a
function of little r,
which is the separation.
>> So that's what we
would like to know is
if we've got a hydrogen atom,
where's the electron
hanging out?
And what's the chance of finding
it here or here and so forth.
The case when l is not equal
to zero is harder for us to do.
Why? Because while we have
that kinetic energy term,
they have the potential energy
1 over r, well we still have
to figure out about that because
last time we either had no
potential energy for
the particle in a box
or 1/2 k x squared for
the harmonic oscillator,
now we've got something a
little bit more challenging,
1 over r. In particular you
might worry a little bit
that when r is zero, is that
blows up and so maybe it's going
to be a little bit difficult to
solve the differential equation.
It turns out that is
not such a big worry.
And then we've got another term
that's l times l plus 1 upon r
squared and that -- so
that adds another thing in.
So now we have 1 over r and
1 over r squared and we have
to find the function
that does that.
We're going to explore that,
how to do that step by step
in the next lecture
where we'll talk
about these radial
distribution functions.
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