PROFESSOR: What is a Rydberg atom?
Well, it's a-- an atom can be a Rydberg atom if the outermost
electron is in a very high principal quantum number.
That's a definition of a Rydberg atom.
The last electron is an n--
little n, very large.
Now that is very interesting because when
that happens, Rydberg atoms--
when that happens, you have a nucleus with charge ze
and you have lots of electrons.
And the last electron-- suppose the last electrode
is in an orbit further out.
You see the orbits do become further out in general.
And suppose the last electron is in some large value
of little n, so it's somewhat out.
So here is the last electron.
This electron sees a nucleus with charge z,
but it also sees z minus 1 electrons, all
the other electrons.
The last electron is outside the nucleus and the cloud
of the other electrons.
So the last electron sees a charge plus 1 z from here
and z minus 1 of the electrons.
So the last electron sees charge 1.
So in some sense, to a good approximation,
the last electron says, oh, there's a hydrogen atom here.
I'm part of a hydrogen atom.
I don't see my friends.
The other electrons are too close to the nucleus.
And I'm out there going around as if I were hydrogen.
So that's a very nice application of hydrogen atoms.
So the first question that we want to understand,
that it's, again conceptually, is what is the size.
If n is large what is the size of the atom?
Now I must say, I myself, when I look
at these things after a few years
that I don't teach quantum mechanics, I look at here
and say, OK, this is the solution.
Well, maybe the size is na0.
Unfortunately, that's completely wrong.
And we're going to try to explain what
was wrong in looking there.
You know, you see your hydrogen atom wave function.
There's nothing like really the size,
there's no such precise definition as the size.
But you could say, what's the expected value of the radius.
That's a reasonable definition of size.
And you know from this wave function,
is going to come out to a0, a0 over 2, 2a0, maybe pi over 2a0.
Something like that.
So from here you would say, well,
it's going to come out to pi and na0 or something like that.
That is not true.
So how do we see that it's not true?
We'll take a little time in a few minutes.
But how can we get the size of that atom
correctly in an intuitive way?
Again, we want to just understand
a few results about hydrogen atom
that become part of your intuition.
So the important result here is the virial theorem.
Now whenever I think of the virial theorem I say,
oh, there was a factor of 2 or a 1/2 there.
How did it go?
It takes me a few seconds to try to reconstruct that it's this.
For the hydrogen system, for any 1 over r potential,
there is this relation within the expectation
value of the kinetic energy, the potential energy.
You've seen that result probably a couple of times
already in this course.
Because it's a real important result.
It looks like OK, just a theorem about these things.
But the intuition is really important.
So how does one remember or picture that?
One way is the following.
And that's the way I like it.
I imagine the energy line here.
And here is 0.
And the one thing I remember is that, yes,
there is a kinetic energy here.
And actually the bound state energy
is exactly the same, but the opposite.
That's a way to remember that.
That's really what is going on.
That's the key thing.
The kinetic energy and the bound state energy
are just of opposite signs.
Why do we see that?
Because then we have that V, from this equation,
is minus 2T.
And if V is minus 2T, means that T plus V is T minus 2T.
And T plus V is the total energy.
So T plus V is equal to the total energy
E of the bound state.
For any stationary state, this is true.
And this thing, given this condition, is minus T.
So that's the way we think of it.
So we then have a nice result because the expectation value
of the potential would then be minus 2 times the expectation
value of T, which is itself minus the energy.
So you have a minus 2 times minus the energy.
And it's equal to the energy.
And this is correct.
The expectation value of V is the energy
and the expectation value of Eb is negative,
and the energy's negative.
So what is this?
This is the expectation value of minus e squared over r.
In fact, of-- yeah, I'll leave it like that.
So I'm back to z equal 1, because we're
talking about Rydberg atoms.
Oh, 2Eb, I'm sorry, here.
Thank you.
OK, so expectation value of the potential we have here.
And 2 times the energy--
yeah, I would have gotten this wrong.
Thanks for correcting it before it
did damage to the derivation.
We have it there, e squared over 2a0, 1 over n squared.
So what can we cancel?
Well, the 2's cancel.
The e squares cancel.
The signs cancel.
And we get expectation value of 1 over r
is equal to 1 over n squared a0, which
is suggesting very clearly that the typical radius is not na0.
It's n square a0.
So this is exact.
The virial theorem is exact.
The energy is exact.
This is exact.
That's not quite the expectation value of r.
The expectation value of r is not
the inverse of the expectation value of 1 over r.
It's somewhat related.
But there's no theorem, because it would be false,
that the expectation value of 1 over a random variable
is the expectation value of--
is 1 over the expectation value of the random variable.
It's just not true.
This nice result for 1 over r is exactly true.
And it's l independent.
On the other hand, the expectation value of r
can be calculated with a bit more effort--
a lot more effort.
And it's equal to this.
It's just the same thing and 1, with a little correction,
which is 1/2 1 minus l times l plus 1 over n squared.
So actually, the expectation value of r in the hydrogen atom
is l dependent.
Not terribly strongly l dependent, but somewhat l
dependent.
To get an idea, this is equal to n squared a0 times 3/2
for l equals 0.
When l is equal to 0, you see this whole bracket becomes 3/2.
And for the maximum l, l equals n minus 1, this is roughly 1.
This is roughly 0, to a good approximation but not exactly.
It becomes n squared a0 with corrections
that are very, very small.
It's pretty accurate.
All right.
So, first thing we've learned is that we
got a radius, expected radius, that goes back n squared a0.
So it's kind of interesting to see
what went wrong if you would have thought
with a form of the solution.
So psi nlm equal Ar to the l Wnl, a polynomial,
e to the minus r over na0 Ylm of theta and phi
or fnl of r times Ylm of the solid angle.
And what do we know about this polynomial?
It's of degree n minus l plus 1.
That's it.
And it depends on r.
So what we're looking at is what was the error in thinking
that the typical r was na0.
And you see, when you make a mistake like this--
the mistake I made of saying oh, na0 must be right--
and you find that it's wrong, it's very important
to go back and learn why did you get the wrong answer.
That's what we're doing now.
So the one question I can ask to begin
with is what is a probability density
to find the electron between some radius r and a radius r
plus dr. So, you know, this is a probability.
And it depends on theta and phi.
And it's very complicated.
How about giving me a probability
along r that they can integrate along r and visualize
how this is dependant on r?
So the probability to find the electron in this shell
must be equal to the value of the wave function squared
times the volume element.
And the volume element here is psi
squared r squared dr times--
you would say 4 pi, but it's not spherically symmetric.
So you have to integrate over solid angle.
That is the volume element.
And since I have to integrate over solid angle,
I must have psi squared here.
So that's the right equation.
If you want to make things look perfect,
put the d cubed x before the psi squared here.
And the problem is that the d cubed
x is big enough that it, in some sense, has partial integrals.
The So notation is not perfect, but somehow you
must imagine this whole volume element
that is still infinitesimal but involves some integral already.
So you have this.
And you get then r squared dr fnl squared
and you have the integral, the omega of this Y star lm Ylm.
And that integral is exactly 1.
Spherical harmonics are normalized.
So now I can cancel the dr. And I
get that the radial probability distribution, which
is a nice concept, is really r squared fnl of r squared.
Radial probability.
So our mistake-- my mistake must have
been that I didn't include all that was relevant.
The exponential is one part.
But there is the polynomial.
And the polynomial must be causing the trouble.
Indeed, that's what is happening.
Let's look at fnl from the top blackboard.
That includes r to the l times that polynomial.
And it's a polynomial of that degree.
So it begins like a0 plus up to coefficient a prime r
to the n minus l plus--
well, minus l minus 1.
And then I have the exponential.
OK.
Now I cannot do this--
I don't want to do this calculation exactly.
It's too complicated.
So let's ignore the lower part of the polynomial.
And we're thinking the radius is going to be reasonably big.
So it's a reasonable idea to keep
the power of the polynomial that is the largest.
So what is the largest?
And here you see a nice thing, actually r
to the l times this polynomial is a polynomial that
begins with r to the l and finishes
with r to the n minus 1.
So it has like equal number of terms as reaches
a value of n minus 1.
The last term in the polynomial, when you multiply it in,
this begins like r to the l and the last term
is r to the n minus 1.
So let's take this to be proportional
to r to the n minus 1 equal to minus r na0.
And here is the fight that actually changes the answer.
Because this is fnl.
So actually I can write p of r is proportional up
to a constant of normalization, and the approximation
they have then r squared times this polynomial squared.
But I'm taking the last term of the polynomial,
so we get r to the 2n e to the minus 2r over na0.
That's a probability distribution.
And with this probability distribution,
then you see what's happening is that there
is a fight between an exponential that
has a typical length where it decays to half its value, that
is related to na0 but the maximum
is delayed because it's multiplied by a function
that the higher the value of n, the slower it is to take off.
x squared or r squared takes off slower than r,
takes of slower than r to the 4th.
So here you have a thing that just grows like that,
but it takes forever to take off.
And the result of this is a function that just picks up
at some point over here.
And we want to find the maximum.
So the maximum comes from taking a derivative.
So the maximum of p of r is determined
by setting the derivative of this equal to 0.
So you get 2n over r times the same r to the 2n e
to the minus 22 over na0.
So the first term, the derivative
is 2r and r to the 2n minus 1, which is divided
by r and the same thing.
The second derivative gives you minus 2 over na0.
And that's it.
So from these two you get that n over r is equal to 1 over na0.
So r equals n squared a0, as we had predicted.
So it's the interplay of the polynomial
with the other thing that makes the radius go very high.
So what about this Rydberg atoms in nature?
Well, they've been observed in interstellar gases.
You see, there was this phenomenon of recombination
when protons captured electrons as the universe cooled off,
and formed hydrogen atoms.
And that recombination sometimes works in such a way
that the proton can capture an electron.
And it captures it in a very high quantum number.
And it keeps happening as we observe this electron.
So people have observed in astrophysics n equal 350.
And [INAUDIBLE] here the electron being captured,
just not in the usual size but almost a million times bigger.
So what happens for this thing r is equal 0.53 times
10 to the minus 10 meters.
That's a0.
And then you have 350 squared.
And that gives you 6.5 microns.
That's actually pretty big.
A blood cell is 8 microns.
A red blood cell is 8 microns.
They are not stable because eventually they spiral in.
But if you have an atom that, for example, has
an n equal to 5, it jumps to n equals to 1 in 10
to the minus 7 seconds.
These atoms are rather stable.
Instead of lasting 1/10 of a millionth of a second,
they can last a millisecond, 1/10 of a second,
sometimes even 1 second.
It takes a long time to go down that spiral.
The energy levels-- if you have energies that go like 1 over n
squared, the energy levels, the separation between them,
goes like 1 over n cubed.
So lots and lots of states there.
And it takes a long time for it to decay from one to another.
So they observe them in the lab in different ways.
They create them with lasers now.
You have the outermost electron, you kick it
to another orbit with one--
they use three lasers.
In the lab at MIT, three lasers.
One and two to kick it to an n equal 10,
and then the third one to kick it to n equals 60.
And they detect those atoms by ionization.
A normal atom, you can't ionize.
You would need millions of volts per centimeter
to ionize it with an electric field.
These atoms you can ionize very easily.
So they can see that they've been created that way.
So also in terms of sizes, the diameter of hair
is about 50 microns.
Hair is very thin.
But you see it.
So you're about a factor of 5 or 10
to being able to see that atom with your naked eye.
It's pretty impressive.
Incredible, in fact.
So a nice laboratory, those are almost semi-classical atoms.
All what Bohr was doing of calculating,
you can derive this law by assuming
that the transitions between orbits in the hydrogen atom
emit photons of the right frequency.
It's all kinds of fun things.