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M2L7g.txt
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M2L7g.txt
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#
# File: content-mit-8-421-2x-subtitles/M2L7g.txt
#
# Captions for 8.421x module
#
# This file has 190 caption lines.
#
# Do not add or delete any lines.
#
#----------------------------------------
The second term is spin orbit interaction,
and as such, it is, I would say, more interesting,
because it now really addresses the spin.
So yes, the term comes from the Dirac equation
and appears separately as an identifiable term.
If you do the Pauli approximation
to the Dirac equation, so I could just stop there and say here,
that's what it is.
But I think it's insightful if I re-derive this term for you
by using a physical picture, what
do we have to add to the Schrodinger equation to get it.
So do what we have to add -- what we have to add to the Schrodinger
equation is, of course, you have to put in
that the electron has a spin.
the Dirac equation is formulated in spin
or it's automatically formulated for spin one-half particles.
But if you start with the Schrodinger equation,
you can first write it down for spinless particle,
and now you have to add the spin by hand.
So we have to add to the Schrodinger equation,
or to the Bohr model, that we have
an electron with an intrinsic spin,
and this spin will couple to the rest of the world
because it has a magnetic moment, which
is the G factor-- we'll talk about G factors later in more
detail-- the Bohr magneton times the spin,
and if you add a it to the Schrodinger equation,
we have to assume the experimental value,
or you can say, the value which, the approximate value
of the exact value from the Dirac equation,
which is the G factor of two.
OK.
So if you put in, by hand, that the electron has a spin,
the spin is associated with a magnetic moment,
now we have a coupling because an electron which
moves in the Coulomb field of the nucleus,
sees in its moving frame a magnetic field.
And this magnetic field couples to the spin.
So the moving electron sees a motional magnetic field.
Well, you know when you have an electric field,
and you look at the-- you have an electric field in one frame,
and you have a second frame, special relativity, which
moves with a velocity v, then in the moving frame,
you observe a motional magnetic field.
And well, our moving frame is now the electron.
So that looks like coupling to the Coulomb field,
but we can immediately write it in a form
where we recognize angular momentum.
So this is the Coulomb field.
And it involves v cross r, and v cross
r is just minus L, the optical angular momentum.
So this is now our motional magnetic field.
And then the spin couples to it, and we get spin orbit coupling.
By the way, an equivalent picture--
because you will read about it or hear
about it-- is you can get the magnetic field, which couples
to the spin, by simply transforming the Coulomb
field into the frame of the moving electron.
But there is a more naive picture,
which is correct too, if you are in the frame of the electron,
then the proton is orbiting around you.
And the orbiting proton is a ring current,
which creates a magnetic field.
And this magnetic field is, or it
has to be, the same as we just got
by doing the relativistic transformation of the Coulomb
field into the frame of the moving electron.
So, therefore, if you want the spin orbit Hamiltonian,
it's now the magnetic moment of the electron
times this B-field, and just collecting
the terms, e squared, h bar squared, m squared, c squared,
1 over r cubed, the magnetic moment
is proportional to s, the spin, the motional B-field
is proportional to L, and now we have the famous spin orbit
coupling term.
OK, that would be too simple.
So we need, we have to now consider one more thing,
and this is, I told you that we can transform
the electric field into the moving frame of the electron,
but those transformations from one Lorentz frame
into the other, are simple when you go from one frame
to another, and then move with a constant velocity
relative to each other.
The electron is not moving at constant velocity,
it actually has a constant acceleration because it's
on a circular orbit.
So therefore, we have to use a relativistic transformation
between coordinate frames-- and it is in textbooks
on special relativity-- what happens
is, if you do relativistic transformations
between different moving coordinate systems,
then you find that-- you know, if you
have a velocity like this, velocity like this,
and do subsequent transformations,
you're not just going into a Lorentz frame, which
has the resultant relativistic addition of the velocities.
In addition, you're final Lorentz frame,
is rotated against the first frame.
So adding a number of velocities which are not colinear--
and this is what we have to do if the electron goes
in a circle, we have to sort of follow the electron,
follow different velocities, and do
relativistic transformations-- several relativistic Lorentz
transformation, executed in sequence,
are not a simple Lorentz transformation,
they are Lorentz transformation plus rotation.
And that means that if we go, and the result can
be summarized as follows, there is a rotational angular
velocity, which can be written as follows,
you see immediately that this term
is zero, when a, acceleration and velocity are the same.
So if you were moving frame and you accelerate the moving
frame, you don't get this term, if everything
is in one dimension.
But if the acceleration is not in one dimension--
and that's the case when the electron goes on a circle,
you get an additional contribution--
and if the acceleration is constant,
you get-- and this is on the left-hand side--
a constant angular velocity, the precession velocity.
And the letter t indicates that it's
the Thomas precession.
Now, we have learned-- and it's nice to come back
to the beginning of this course-- we have discussed
that if you have a magnetic field,
we can go into a rotating frame, and the magnetic field
disappears.
Well, now, due to relativistic kinematics,
we have an additional rotation, but using the same argument
we used at the beginning of the course
this additional rotation corresponds
to an additional magnetic field.
So therefore, this precession is identical to the effect of
an additional magnetic field.
So there is an additional magnetic field,
with the letter T, the Thomas magnetic field,
those magnetic fields are related to the precession
frequency by the gyromagnetic ratio.
And, if we use the value for the electron's gyromagnetic ratio,
and in addition, we use the fact that, the acceleration
term for this relativistic kinematics,
is nothing else than the Coulomb acceleration,
then we find that the Thomas magnetic field is
just minus one-half the motional magnetic field
we derived earlier.
So therefore, if you go back and want
to find an expression for this spin orbit coupling,
we have to use the magnetic field, which
is the sum of the two contributions
we discussed, the motional contribution
for-- if the electrons would move
in a linear way, plus the Thomas precession
due to the rotational motion due to relativistic kinematics--
and this will just reduce the final result
by a factor of one-half.
Yes?
Small one.
What letter is the gyromagnetic ratio denoted by?
Sorry for the handwriting, which we have usually used lower case
gamma, and the E means the electron,
so gamma electron is the gyromagnetic ratio
of the electron.
Then my other question are more substantively concerned,
or confused, I guess.
I wasn't sure where the h bar comes
from in the expression for angular momentum,
I guess one line up.
You go from v cross r to h bar over m L,
and I wasn't sure where the h bar came from.
Oh, It now depends-- you're correct,
angular momentum is r cross v , r cross v gives angular
momentum, and because it's r cross v and not v cross r,
the minus sign is absorbed into it.
Now, the angular momentum, it depends now,
it's purely definitional.
If you want that the angular momentum has eigenvalues
of L times L plus 1, which as dimensionless,
then we have to account for h bar explicitly.
So question is, do we want the angular momentum operator
to have this kind of integer, or L time L plus 1.
Yeah, in this integer spectrum, or do we want to measure,
or do we want to use operators, which have units of h bar.
That's where the h bar comes from.
Other questions? Ok, so this is spin orbit coupling.