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M2L10zm.txt
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M2L10zm.txt
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# File: content-mit-8-421-2x-subtitles/M2L10zm.txt
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# Captions for 8.421x module
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# This file has 76 caption lines.
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This is not a coincidence, that here what we obtain
is the Bohr radius cube, which is pretty much the volume
of the hydrogen atom.
But we can now do an approximation.
It's not really relevant, but it has a historic name,
Unsold's approximation.
It's just nice to show how things work out.
We have a second order matrix element,
so we couple this state, n, with the operator, z, to a state, m.
But if we assume that all energy denominators can be taken out
of the summation, by assuming that we have
some kind of average excited energy,
then the sum of-- maybe I should've said it--
the sum nzm, mzn.
Which we sum over m, and it just cancels out.
And so, what we have is, if you take an average energy
and denominator out of the summation, what we find
is that what matters is the matrix element z square.
And we can even assume that in the energy denominator,
the excited state energies is negligible.
The hydrogen atom has a binding energy of 1 Rydberg
and the first excited state has a quarter Rydberg.
So at the 25% level, we can set that to 0.
So I'm waving all my hands, but I'm
getting a simple expression for the polarize in the ground
state.
And this goes as follows.
The ground state energy is-- we've
discussed Coulomb energy, virial's theorem, and all
of it.
We need just 1 over r in the ground state.
And for the z square matrix element,
we can simply say, for an s state,
that it is x square, y square, z square,
it is a third or r square in the current state.
So therefore, continuously waving our hands
and making approximations, we find that the polarizability
is r square expectation value divided by an r
to the minus one expectation value.
So this is some cube expectation value,
which is an atomic value.
So you see the nature of the perturbative expression
suggests that it cannot be anything else than the atomic
volume.
I sort of like that, because when
people discuss, for instance, in my group, that's
lithium or rubidium, have a bigger polarizability.
Well, the bigger atom has a bigger volume,
and the more fluffier atoms have the larger polarizability,
and that's pretty much based on that result.
Now, let me finally, do a comparison.
There is another system for which
you have done calculations of the dipole moment.
And this is in classical e and m for conducting sphere.
For conducting sphere in an electric field,
you can exactly solve the boundary conditions,
the boundary value problem, get the electric field,
and find the dipole moment.
And the exact result is that the dipole moment
is the electric field times the cube of the sphere.
So in other words, the dipole moment, or the polarizability
of this field is a neglecting factors,
which are only factors of unity, the dipole moment.
So the polarizability of a conducting sphere
is the volume of the sphere.
The dipole moment of a hydrogen atom
will use [INAUDIBLE] approximation
for all simple atoms, it's the volume of the atom.
So I find it sort of interesting that, when
it comes to dipole moments and to polarizability,
that atoms pretty much behave like metallic conducting
spheres of the same volume.
Any questions?