PROFESSOR: Today, let's catch up with what we were doing before.
And last time, we were talking about hydrogen ionization.
And we went through a whole discussion
of how it would happen.
It was ionized because there would
be an electromagnetic wave within the hydrogen atom.
The initial state was the ground state of the hydrogen atom.
The final state was a plane wave.
Our main work was computing this matrix element,
and that's what we did last time.
It took us some work, because there
is a spatial integral here that was quite complicated, given
the various directions that are going on.
But that was our result. Here is the formula.
And a few things to notice here were
that there's an angle theta that we recognized
as the angle between the electric field polarization
and the momentum of the electron.
So k is the momentum of the electron,
and when k doesn't have a narrow,
it's the magnitude of the electron momentum.
As any k-- well, it's a little bit of an exaggeration. h bar k
is a momentum.
But we understand when we say that k's a momentum.
k times a, it has no units, which
is appropriate of the units of this quantity,
our units of energy.
And then another quantity here, E naught,
is the magnitude of the electric field defined by this formula,
with a 2 E naught here.
We discussed that the photon had to have an energy that it
was not supposed to be too big, so
that the wavelength of the photon
would be smaller, or much smaller
than the atom, in which case the spatial dependence
of the electromagnetic field would be relevant.
So for simplicity, with took photons
not to be too energetic.
And we also took photons to be energetic enough
that, when they would ionize the atom,
the electron that would go out would not be too affected
by the cooling potential, and we could treat it like a plane
wave to a good approximation--
a free plane wave.
Otherwise, you would have to use more complicated plane
waves associated with a hydrogen atom.
So under this ranges, these ranges
translate to this ranges for the momentum of the electron.
And if the-- as required, the energy of the photon
is significantly bigger than a Rydberg, which is the 13.6 eV.
This is like 10 Rydbergs in here.
Then the momentum of the electron
is given by this formula to a good approximation.
It's essentially the energy of the photon, then
the square root of that.
The last ingredient in our computation
is the density of states.
This was calculated also a couple of weeks ago,
so you have to keep track of some formulas here.
There's some formulas that they're a little complicated,
but we'll have to box them and just be ready to use them,
or to spend five minutes really writing them.
And this was the formula for the density
of states with some energy E. With that energy E,
these were free plane wave states.
Energy E, momentum k associated with E, and being shot
into some solid angle in space.
And this is the solid angle.
So I wrote it as a solid angle in here.
So this is basically where we stood and, at this moment,
we want to complete the discussion
and get a simple expression, and simplify it, and get what
we need to have for the ray.
So, if you remember, Fermi's golden rule
expresses a rate omega in terms of 2
pi over h bar, rho parenthesis E, times Hfi squared.
So that's Fermi's golden rule.
And this is what we want to apply.
Well, we've calculated the matrix element.
This would be Hfi prime here.
That's what we have up there.
We have rho.
We have this quantity, 2 pi over h bar is just a constant.
So we can simply plug this here, square that.
There's lots of h bars and things
that I don't think you'll benefit
if I go through them in front of your eyes.
So I'll write the answer.
But in writing the answer, remember
this rho is a differential, in some sense.
Some people might put you in a d rho
here, because there is a d omega.
So think of this, when I substitute rho,
I have the d omega, and I think of this
as the d w, a little rate.
So dw, d omega.
So I simply pass from w to I called it dw,
and substituted the rho here for d omega.
And here is what we get--
256 over pi, e E naught a naught squared over
h bar, m a naught squared over h bar squared,
k a naught cubed over 1 plus k a naught squared
to the 6th, and cosine squared theta.
OK, it's still a little complicated.
But it's mainly complicated because of constants
that have been grouped in the best possible way,
in my opinion, to make it understandable.
This is an energy.
And this is an h bar.
This is actually a Rydberg with a factor of 2.
So in a second, you can see that this thing
has units of 1 over time.
There's also an important factor here,
and that's an intuitively interesting fact,
that the emission of the electron
is preferentially in the direction
that the electric field is polarized.
There's a cosine squared theta.
There's no electrons emitted orthogonally
to the electric field.
And that's kind of intuitive.
It's almost like the electric field is shaking,
the electron, well, it kicks it out.
OK, this is a differential rate.
So the total rate w is the integral of this differential
rate over solid angle.
And for that, you need to know that the integral
over solid angle of cosine squared theta is 4 pi.
If you didn't have the cosine squared,
you would have the 4 pi.
But usually cosine squared, this is multiplied by 1/3.
You can do the calculation.
It just doesn't take any real time.
But many people remember this by thinking
of the sphere, or planet Earth.
Cosine squared theta is large near the North Pole.
It's large near the South Pole.
That doesn't amount to much.
As opposed to sine squared theta,
which is large all over the big equatorial region.
So it's a little bigger, and it turns out cosine squared,
the average over the sphere is 1/3.
And sine squared, the average over the sphere is 2/3.
Well, it's kind of not a bad thing to know,
because it saves you a minute or two from doing this integral.
And now, I'm also going to apply something that basically--
you know, a formula sometimes gives you more things
than you should really trust.
And I would say here, this 1 is not to be trusted basically,
because k a naught must be significantly bigger
than this number probably for this to be accurate.
So under most circumstances, this 1 is not worth it.
In fact, if you calculate this thing
using different approximations, people sometimes
don't get this 1.
And you may see that in books.
So let's ignore this 1.
And then this answer is 512 over 3 e E naught a naught squared,
over h bar Rydberg, 1 over k a naught to the 9th.
Pretty high power.
And the answer starts to be reasonably simple.
This 9 arises because of 12 minus 3 here.
This is still probably not ideal,
if you want to play intuition about what's going on,
and the scale of the effects.
You know, once you have a formula,
and you've worked so hard to get it--
this calculation is doing it reliably
is a couple of hours of work--
and you might as well manipulate it
and try to make it look reasonably nice.
And this is what people that do atomic physics
do with this rate.
So they write it in the following way.
Again, a short calculation to get this 256 over 3.
And they put in atomic units, Ep over E star--
I will explain what these numbers are--
squared, 1 over t star, 1 over k a naught to the 9th.
I think it's important to notice as well that you could say, OK,
so how did the photon energy, or photon frequency,
affect this result?
Well, actually, the photon omega doesn't
seem to be anywhere here.
It has disappeared.
But it is implicit in the electron momentum,
because basically the momentum of the electron
is what is obtained from using the energy of the photon.
It ionizes the electron, liberates electron, and then
gives it some kinetic energy.
And, roughly, k equals like the square root of omega.
So there is an omega dependence here.
Now, what are these other quantities?
These quantities are simple.
Ep is the peak electric field.
It's peak electric in the wave that you've sent in.
So it's 2 E naught.
E star is the atomic electric field.
And that's defined as the electric field
that the proton creates at the electron.
So it's equal to E over a 0 squared.
That's the definition.
Or, if you want, in terms of Rydbergs,
2 Rydbergs over E a naught, and it's about 5.14 times 10
to the 11 volts per meter.
So it's a nice quantity.
It's in the land you're comparing your electric field
from your laser to the typical electric field in the atom.
And that ratio is meaningful.
And then t star is the time the electron takes
to travel a distance a naught.
So it's a naught over the velocity
of the electron, which is roughly the fine structure
constant times c.
You remember the electron roughly
has a beta parameter equal to 1 over 137.
This is [INAUDIBLE] alpha equal beta,
meaning the fine structure constant in alpha
is the beta of the electron.
So this is t star.
And it's actually two Rydbergs over h bar,
and it's 2.42 times 10 to the minus 17 seconds.
That's the time the electron, which is not moving that fast,
takes to move a Bohr radius.
So this is the last form we'll take.
Atomic physics books would consider
that the best way to describe the physics of the problem.
And this is really all we're going to say about ionization.
It's kind of a precursor of field theory calculations
you will do soon--
not in 806-- in which you do reasonably
complicated calculations, matrix elements, Feynman diagrams.
And, at the end of the day, by the time you're all done,
the answer simplifies to something rather reasonable,
and not complicated.