PROFESSOR: So far, so good.
We've really calculated.
And we could stop here.
But there is a
nice interpretation
of this Darwin term
that gives you intuition
as to what's really happening.
See, when you look at
these terms overall--
these interactions
with this class--
you say, well, the kinetic
energy was not relativistic.
That gives rise to this term.
There is a story for
the spin orbit as well.
We think of the proton--
creates an electric field.
You, the electron,
are moving around.
As you move inside the
static-electric field,
you see a magnetic field.
The magnetic field interacts
with your dipole moment
of the electron.
That's the story here.
So what's the story
for this term?
Where does it come from?
What is the physics behind it?
That's what I want
to discuss now.
So the physics interpretation
of the Darwin term
is that the electron
behaves not as
if it would be a point
particle but as if it would be,
kind of, a little ball where
the charge is spread out.
We know that any mechanical
model of the electron, where
you think of it as
a ball, the spinning
doesn't work for the spin.
But somehow, here, we can see
that this extra correction
to the energy is
the correction that
would appear if, somehow, the
electron, instead of feeling
the potential of the
nucleus at one point,
as if it would be
a point particle,
it's as if it would be spread.
So let me make a drawing here.
Here is the proton.
And it creates a potential.
And we will put an
electron at the point r.
There's the electron.
But now, we're going to think
that this electron is really
a cloud, like this.
And the charge is
distributed over there.
And now, I'm going
to try to estimate
what happens to the
potential energy
if it's really
behaving like that.
So for that, I'm going to
make another coordinate system
sort of starting here.
I'm going to call
the vector that
goes to an arbitrary
point here vector u.
So from the origin or center
of this charge distribution,
we put the vector u.
And then I have this vector
over here, which is r plus u.
And this u points to a little
bit of charge, a little cubit
here, for example.
So this is our setup.
Now, the potential--
due to the proton--
proton-- we have a
potential V of r,
which is equal to minus the
charge of the electron times
the scalar potential, phi of r--
it's minus e times e over r,
which is our familiar minus e
squared over r.
So the potential energy--
this is the potential energy.
And so if you had
a point particle--
for a point particle--
you have that the proton creates
a potential at a distance r.
You multiply by the charge.
And you get the
potential energy.
So how do we do it a
little more generally?
I'm going to call
this V tilde of r.
The true potential energy--
potential energy-- when the
center of the electron is
at r--
electron at r.
So what is the true
potential energy
when the center of
the electron is at r?
Well, I would have
to do an integral
over the electron of the
little amount of charges
times the potential
at those points.
For every piece
of charge, I must
multiply the charge times
the potential generated
by the proton.
And that would give me the
total energy of this electron
in this potential, V of r.
So let me use that
terminology here.
I will describe the
charge density--
density-- rho of u--
the density of charge at a
position given by a u vector.
I'll write it as minus
e times rho 0 of u--
a little bit of notation.
I apologize, but it's
necessary to do things cleanly.
The charge density
of the electron--
electron-- is given by
rho of u minus e rho 0.
And then it should be
true that the integral
over the electron of d cube
u rho 0 of u is equal to 1.
Look at that integral.
Why should it be equal to 1?
It should be equal to 1, because
then the integral of the charge
density--
the integral of this over all of
space would be minus e times 1.
And that's exactly
what you expect.
So rho 0 is a unit free--
well, 1 over length cubed
is a charge-free quantity.
The charge is carried by
this constant over here.
So what happens then
with this V of r?
This V of r is the charge--
little charge q is d cubed
u times rho of u times phi
at r plus u.
Look, the charge is this
at some little element.
And then the potential there
is the potential at r plus u.
So one more step--
V of r-- if you take from
rho this extra minus e,
minus e times phi
of r is what we call
the potential energy due to r.
So take the minus
e out and append it
to the capital Phi here, so that
we have integral d cube u rho
0 of u V of r plus u.
OK, this was our goal.
It gives you the new energy--
the true potential
energy of this electron
when this center is
at r as the smearing
of the potential energy
of a point particle
smeared over the electron.
So this is a formula that
represents our intuition.
And moreover, it
has the rho 0 here
that tells you the weight that
you should apply at any point,
because this rho 0 is
proportional to the charge--
very good.
We have a formula.
But I was supposed
to explain that term.
And we have not
explained it yet.
But now, it's time
to explain it.
We're going to try to
do a computation that
helps us do that.
The idea is that we're going
to expand the potential
around the point r and treat
this as a small deviation
because, after all, we expect
this little ball that I drew
big here for the
purposes of illustration
to be rather small.
So let's write V of r plus u
as V of r in Taylor series--
plus the next term
is the derivative
of V with respect to
position evaluated
at r multiplied by the
deviation that you moved.
So sum over i dv dx i
evaluated at r times u i.
This is the component
of this vector u.
That's the first term
in the Taylor series.
And we need one more--
plus 1/2 sum over i and j,
d 2nd V vx i vx j evaluated
at r ui uj.
And then we have
to integrate here.
We substituted in
here and integrate.
So V tilde of r--
let's see what it is.
Well, I can put
this whole thing--
we don't want to write so much.
So let's try to do
it a little quick.
We're integrating over u.
And here, let's think
of the first term here.
When you plug in the first
term, this function of r
has nothing to do with u.
So it goes out and you get V of
r times the integral d cube u
rho 0 of u.
For the next term in
here, these derivatives
are evaluated at r-- have
nothing to do with u.
They go out.
So plus sum over I dv dx
i of r integral d cube u
rho 0 of u, ui.
Last term-- these
derivatives go out.
They're evaluated at r.
They have nothing
to do with you.
So plus 1/2 sum over i
and j d second V dx i dx
j evaluated at r integral
d cube u rho of u ui uj.
OK-- all these terms!
Happily we can interpret
much of what we have here.
So what is this first integral?
V of r equals--
this integral is our
normalization integral.
Over here, that's equal to 1.
So this is very nice.
That's what you
would expect that
to first approximation, the
total energy of the electron,
when it is at r--
as if it would be a
point particle at r.
Now, let's look
at the next terms.
Now, we will assume that
the distribution of charge
is spherically symmetric.
So that means that
rho of u vector
is actually a
function rho 0 of u,
where u is the length
of the u vector.
So it just depends on the
distance from the point
that you're looking.
That's the charge.
Why would it be
more complicated?
If that is the case, if this
is spherically symmetric,
you can already see that
these integrals would vanish,
because you're integrating the
spherically symmetric quantity
times a power of a coordinate.
That's something you
did in the homework
as well for this Stark effect.
You realized that integrals of
spherically symmetric functions
then powers of x, y,
and z-- well, they get
killed, unless those
are even powers.
So this is an odd power.
And that's 0.
So this term is gone.
And this integral is
interesting as well.
When you have a spherically
symmetric quantity
and you have i and j different,
the integral would be 0.
It's like having a power
of x and a power of y
in your homework.
So if that integral would be 0,
it would only be non-zero if i
is equal to j-- the first
component, for example--
1 u1 u1.
But that integral would be
the same as u2 u2 or u3 u3.
So in fact, each
of these integrals
is proportional to delta
i j, but equal to 1/3
of the integral of
d cube u rho of u--
I'll put the delta ij here--
times u squared.
u squared is-- u1 squared plus
u2 squared plus u3 squared--
each one gives
the same integral.
Therefore, the result
has a 1/3 in front.
So what do we get?
Plus 1/6-- and here,
we get a big smile.
Why?
Because delta i j,
with this thing,
is the Laplacian of
V evaluated at r.
They have the same
derivative summed.
So what do we get here?
1/6 of the Laplacian
of V evaluated
at r times the integral d
cube u rho 0 of u u squared.
Very nice-- already starting
to look like what we wanted,
a Laplacian.
Can we do a little better?
Yes, we can.
It's possible.
Let's assume the charged
particle has a size.
So let's assume the
electron is a ball of radius
the Compton wavelength,
which is h bar over mc.
So that means rho 0
of u is non-zero when
u is less than this lambda.
And it's 0 when u is
greater than this lambda.
That's a ball-- some
density up to there.
And this must integrate to 1.
So it's 1 over the volume of
that ball, 4 pi lambda cubed
over 3.
Do this integral.
This integral gives you,
actually, 3/5 of lambda
squared with that function.
So the end result is that V
tilde of r is equal to V of r.
Plus 3/5 of this thing
times that is 1/10 h
squared m squared c
squared Laplacian of V.
The new correction to the
potential, when this mole came
out to this 1/10--
that's pretty close, I must say.
There's an 1/8 there.
We assume the electron is
a ball of size, the Compton
wavelength of the electron.
It's a very rough assumption.
So even to see the number not
coming off by a factor of 100
is quite nice.
So that's the interpretation.
The known locality--
the spread out
of the electron into a little
ball of its Compton size
is quite significant.
The Compton wavelength
of the electron
is fundamental in
quantum field theory.
The Compton wavelength
of an electron
is the size of a photon whose
energy is equal to the rest
mass of the electron.
So if you have a photon
of this size equal
of wavelength equal to Compton
length of the electron,
that photon packs
enough energy to create
electron-positron pairs.
That's why, in quantum field
theory, you care about this.
And the quantum field
theory is really
the way you do relativistic
quantum mechanics.
So it's not too surprising
that, in relativistic analysis
such as that of
the Dirac equation,
we find a role for the Compton
wavelength of the electron
and a correction
associated with it.
