M2L6a.txt

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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.
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 lengths, for energy,
for velocity, for electric field,
and all these units should be-- 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 to construct our natural units?
c.
c?
Good question.
What about C, the speed of light?
Should the speed of light be part
of our system of atomic units?

Could we not set it equal to one?
Let's not go there, because, if you set it to one,
you have make a choice, you have constrained your system,
and you 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 h bar
and you can just play the game of combinatorics
and see can do find a length which consist 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 electronvolt, or two
Rydberg, 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 a binding energy.
But then half of it is kinetic energy,
and therefore the total energy of this 1s electron
is half the Coulomb energy, and it's that.
But now-- OK, so so far, no c, the energy,
the energy levels, the wavefunction,
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 at one, sure, it wouldn't-- I mean,
if you have a relativistic equation that 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 is-- 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 lengths,
is h bar over m c, 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 units of lengths,
and the Compton radius, we have to multiply
with a dimensionless unit, which is h c over e squared,
or if I take the reverse, e squared over h plus c.
And similarly, here the dimensionless quantity
to multiply is that.
So what do we find now here is, 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 wavelength.
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 and its kinetic energy is m v squared--
and if said m v squared, I want to skip all factors of unity,
so that's not one-half, it's just m v squared--
and if I said m v squared 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 over 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 is
non-relativistic-- we have solved
the non-relativistic Schrodinger equation for it--
and consistently we find that the velocity of the electron,
is one over 137 of the speed of light.
It's actually physics trivia.
If somebody asks you, how fast is the electron
in the hydrogen atom?
Well, 1% of the speed of light.
Of course, if you had solved the nonlinearistic--
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'll realize, gosh, I've
solved the wrong equation, because
in the non-relativistic Schrodinger equation, when the solution is
that something moves at the speed of light,
I better start with a different equation.
But here we find we are consistent, an electron
in the hydrogen atom, for low nuclear charge
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 volt per centimeter.
So everything I've constructed here out of the three
fundamental constants, e, m, then h,
are typical for the 1s electron for the ground
state of hydrogen. Because 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.
If a constant has dimension, like h bar, like c,
it 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, this
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, which should ultimately
predict the value of alpha, which ultimately be predicted
by a complete theory.

So alpha, the fine-structure constant, is smaller than 1.
It's 1 over 137.
And the fact that alpha is smaller than 1,
is often phrased in these words, that since alpha
is much smaller than 1, 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, 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?
Weak as compared to strong interactions
or other interactions.
Well, let me try to justify it as follows.
We cannot go to arbitrary small distances.
You 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 similar particles prepared
with an energy uncertainty, which would spontaneously
create more particles.
So therefore, let me postulate that if energy,
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, than the concept
of single particles breaks down.
Of course, that's not the end of physics.
You need now field theory for particles, where particles are
just excitations in a 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 m c square?
Well, that means the momentum uncertainty is
on the order of m c, and with this momentum uncertainty,
I can localize particles to within h by over m c,
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 this Coulomb energy is e squared
over the radius, the Compton radius,
and this turns out to be e squared m c 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, m c squared,
and well, if I take what I had above e squared,
m c over h bar, I divide by m c squared,
I find that e squared-- that the result, the ratio
is e squared over h plus 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'll 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 iron 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.