WEBVTT

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PROFESSOR: Today, let's catch up
with what we were doing before.

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And last time, we were talking
about hydrogen ionization.

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And we went through
a whole discussion

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of how it would happen.

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It was ionized
because there would

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be an electromagnetic wave
within the hydrogen atom.

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The initial state was the ground
state of the hydrogen atom.

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The final state
was a plane wave.

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Our main work was computing
this matrix element,

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and that's what
we did last time.

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It took us some
work, because there

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is a spatial integral here that
was quite complicated, given

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the various directions
that are going on.

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But that was our result.
Here is the formula.

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And a few things
to notice here were

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that there's an angle
theta that we recognized

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as the angle between the
electric field polarization

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and the momentum
of the electron.

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So k is the momentum
of the electron,

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and when k doesn't
have a narrow,

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it's the magnitude of
the electron momentum.

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As any k-- well, it's a little
bit of an exaggeration. h bar k

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is a momentum.

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But we understand when we
say that k's a momentum.

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k times a, it has
no units, which

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is appropriate of the
units of this quantity,

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our units of energy.

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And then another
quantity here, E naught,

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is the magnitude of the electric
field defined by this formula,

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with a 2 E naught here.

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We discussed that the photon
had to have an energy that it

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was not supposed
to be too big, so

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that the wavelength
of the photon

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would be smaller,
or much smaller

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than the atom, in which
case the spatial dependence

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of the electromagnetic
field would be relevant.

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So for simplicity,
with took photons

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not to be too energetic.

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And we also took photons
to be energetic enough

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that, when they would
ionize the atom,

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the electron that would go
out would not be too affected

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by the cooling potential, and
we could treat it like a plane

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wave to a good approximation--

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a free plane wave.

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Otherwise, you would have to
use more complicated plane

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waves associated
with a hydrogen atom.

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So under this
ranges, these ranges

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translate to this ranges for
the momentum of the electron.

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And if the-- as required,
the energy of the photon

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is significantly bigger than a
Rydberg, which is the 13.6 eV.

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This is like 10
Rydbergs in here.

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Then the momentum
of the electron

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is given by this formula
to a good approximation.

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It's essentially the
energy of the photon, then

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the square root of that.

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The last ingredient
in our computation

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is the density of states.

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This was calculated also
a couple of weeks ago,

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so you have to keep track
of some formulas here.

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There's some formulas that
they're a little complicated,

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but we'll have to box them
and just be ready to use them,

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or to spend five minutes
really writing them.

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And this was the
formula for the density

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of states with some energy
E. With that energy E,

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these were free
plane wave states.

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Energy E, momentum k associated
with E, and being shot

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into some solid angle in space.

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And this is the solid angle.

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So I wrote it as a
solid angle in here.

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So this is basically where
we stood and, at this moment,

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we want to complete
the discussion

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and get a simple expression,
and simplify it, and get what

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we need to have for the ray.

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So, if you remember,
Fermi's golden rule

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expresses a rate
omega in terms of 2

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pi over h bar, rho parenthesis
E, times Hfi squared.

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So that's Fermi's golden rule.

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And this is what
we want to apply.

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Well, we've calculated
the matrix element.

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This would be Hfi prime here.

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That's what we have up there.

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We have rho.

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We have this quantity, 2 pi
over h bar is just a constant.

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So we can simply plug
this here, square that.

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There's lots of
h bars and things

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that I don't think
you'll benefit

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if I go through them
in front of your eyes.

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So I'll write the answer.

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But in writing the
answer, remember

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this rho is a differential,
in some sense.

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Some people might
put you in a d rho

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here, because
there is a d omega.

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So think of this,
when I substitute rho,

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I have the d omega,
and I think of this

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as the d w, a little rate.

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So dw, d omega.

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So I simply pass from
w to I called it dw,

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and substituted the
rho here for d omega.

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And here is what we get--

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256 over pi, e E naught
a naught squared over

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h bar, m a naught squared
over h bar squared,

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k a naught cubed over 1
plus k a naught squared

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to the 6th, and
cosine squared theta.

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OK, it's still a
little complicated.

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But it's mainly complicated
because of constants

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that have been grouped
in the best possible way,

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in my opinion, to make
it understandable.

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This is an energy.

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And this is an h bar.

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This is actually a Rydberg
with a factor of 2.

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So in a second, you
can see that this thing

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has units of 1 over time.

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There's also an
important factor here,

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and that's an intuitively
interesting fact,

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that the emission
of the electron

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is preferentially
in the direction

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that the electric
field is polarized.

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There's a cosine squared theta.

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There's no electrons
emitted orthogonally

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to the electric field.

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And that's kind of intuitive.

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It's almost like the
electric field is shaking,

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the electron, well,
it kicks it out.

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OK, this is a differential rate.

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So the total rate w is the
integral of this differential

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rate over solid angle.

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And for that, you need
to know that the integral

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over solid angle of cosine
squared theta is 4 pi.

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If you didn't have
the cosine squared,

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you would have the 4 pi.

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But usually cosine squared,
this is multiplied by 1/3.

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You can do the calculation.

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It just doesn't
take any real time.

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But many people remember
this by thinking

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of the sphere, or planet Earth.

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Cosine squared theta is
large near the North Pole.

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It's large near the South Pole.

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That doesn't amount to much.

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As opposed to sine
squared theta,

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which is large all over
the big equatorial region.

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So it's a little bigger, and
it turns out cosine squared,

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the average over
the sphere is 1/3.

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And sine squared, the average
over the sphere is 2/3.

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Well, it's kind of not
a bad thing to know,

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because it saves you a minute
or two from doing this integral.

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And now, I'm also going to
apply something that basically--

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you know, a formula sometimes
gives you more things

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than you should really trust.

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And I would say here, this 1
is not to be trusted basically,

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because k a naught must
be significantly bigger

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than this number probably
for this to be accurate.

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So under most circumstances,
this 1 is not worth it.

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In fact, if you
calculate this thing

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using different approximations,
people sometimes

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don't get this 1.

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And you may see that in books.

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So let's ignore this 1.

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And then this answer is 512 over
3 e E naught a naught squared,

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over h bar Rydberg, 1 over
k a naught to the 9th.

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Pretty high power.

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And the answer starts
to be reasonably simple.

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This 9 arises because
of 12 minus 3 here.

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This is still
probably not ideal,

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if you want to play intuition
about what's going on,

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and the scale of the effects.

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You know, once you
have a formula,

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and you've worked
so hard to get it--

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this calculation is
doing it reliably

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is a couple of hours of work--

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and you might as
well manipulate it

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and try to make it
look reasonably nice.

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And this is what people
that do atomic physics

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do with this rate.

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So they write it in
the following way.

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Again, a short calculation
to get this 256 over 3.

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And they put in atomic
units, Ep over E star--

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I will explain what
these numbers are--

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squared, 1 over t star, 1
over k a naught to the 9th.

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I think it's important to notice
as well that you could say, OK,

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so how did the photon
energy, or photon frequency,

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affect this result?

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Well, actually, the
photon omega doesn't

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seem to be anywhere here.

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It has disappeared.

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But it is implicit in
the electron momentum,

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because basically the
momentum of the electron

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is what is obtained from using
the energy of the photon.

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It ionizes the electron,
liberates electron, and then

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gives it some kinetic energy.

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And, roughly, k equals like
the square root of omega.

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So there is an omega
dependence here.

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Now, what are these
other quantities?

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These quantities are simple.

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Ep is the peak electric field.

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It's peak electric in the
wave that you've sent in.

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So it's 2 E naught.

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E star is the atomic
electric field.

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And that's defined
as the electric field

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that the proton creates
at the electron.

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So it's equal to E
over a 0 squared.

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That's the definition.

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Or, if you want, in
terms of Rydbergs,

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2 Rydbergs over E a naught,
and it's about 5.14 times 10

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to the 11 volts per meter.

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So it's a nice quantity.

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It's in the land you're
comparing your electric field

00:14:12.750 --> 00:14:19.260
from your laser to the typical
electric field in the atom.

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And that ratio is meaningful.

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And then t star is the
time the electron takes

00:14:31.230 --> 00:14:34.770
to travel a distance a naught.

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So it's a naught
over the velocity

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of the electron, which is
roughly the fine structure

00:14:41.670 --> 00:14:43.800
constant times c.

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You remember the
electron roughly

00:14:47.590 --> 00:14:52.410
has a beta parameter
equal to 1 over 137.

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This is [INAUDIBLE]
alpha equal beta,

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meaning the fine structure
constant in alpha

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is the beta of the electron.

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So this is t star.

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And it's actually two
Rydbergs over h bar,

00:15:12.850 --> 00:15:18.405
and it's 2.42 times 10
to the minus 17 seconds.

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That's the time the electron,
which is not moving that fast,

00:15:29.400 --> 00:15:33.480
takes to move a Bohr radius.

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So this is the last
form we'll take.

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Atomic physics
books would consider

00:15:39.360 --> 00:15:44.215
that the best way to describe
the physics of the problem.

00:15:47.430 --> 00:15:53.200
And this is really all we're
going to say about ionization.

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It's kind of a precursor of
field theory calculations

00:15:58.510 --> 00:16:01.470
you will do soon--

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not in 806-- in which
you do reasonably

00:16:07.660 --> 00:16:12.550
complicated calculations, matrix
elements, Feynman diagrams.

00:16:12.550 --> 00:16:16.120
And, at the end of the day,
by the time you're all done,

00:16:16.120 --> 00:16:22.220
the answer simplifies to
something rather reasonable,

00:16:22.220 --> 00:16:24.090
and not complicated.