M2L10zk.txt

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Let's follow the standard approach,
and let's do second order perturbation theory,
and calculate the energy, calculate the dipole moment,
and see that everything is as we expect now.
So we want to do second order perturbation theory.
We know already the first order term is 0.
This was a discussion about parity.
And in second order perturbation theory,
the state, N, has an energy, which
is the unperturbed energy.
And then in second order, we have the matrix element
to all other states.
We square it.
We divide by the energy denominator.
We sum over all states, M. But I make a prime here.
Of course, we are not summing over the state.
We exclude N from the summation.

And the prefactor here is electron charge squared
times electric field squared.
OK.
Pretty much that's sorts it out in second order perturbation
theory.
So this is the energy.
And we want to relate the energy to the dipole moment.
So the next step is now is we calculate D.
And we calculate D from the first order wave function
because we already get an effect in first order.
And everything here is about leading order.
So the expectation value of the dipole operator--
so we take the expectation value of the dipole operator and we
use the 0s order, the unperturbed state,
plus the first order correction.
And we know already that the diagonal terms do not
contribute.
This is the parity selection rule.
So we get contributions from the cross term,
which is N0, the dipole operator with N1.

So let's just suppress vectorial notation.
We know everything is along the z-axis.
So we have the 0s all the way function.
Our operator is Z. And now we have
to write down the first order correction to the wave
function.
And the first order correction is
the sum over all other states.
We may connect mixture of the state, M.
And this mixture uses a matrix element.
And here we have the energy denominator.
So what we obtain is-- we have the electron charge here
from the dipole moment.
We have the electron charge due to the perturbation operator.
So it's electron squared.
we have the electric field.
And then this is due to the mixture of the wave function
with the dipole operator.
And now because we take the dipole matrix element
of the dipole operator, we get another occurrence
of the dipole operator.
So therefore, we do first order perturbation theory,
but we take the first order result
and ask what is the expectation value for the dipole moment.
And that means the dipole operator or the perturbation
operator appears twice.
And our result is as expected, quadratic in the matrix
element.
And it has it's energy denominator.
So the definition is that the dipole moment is alpha
times the electric field.
So therefore, all of that equals alpha.
And if you compare now the result for the dipole moment
with the second order perturbation
theory for the electric field and for the energy
we find-- here is a factor of two,
but there is no factor of two up there--
we find that the energy of the energy shift delta En,
it has exactly the same matrix element as the polarizability.

It is this, one half alpha x squared.

Since the perturbation operator--
I'm just writing it down here-- was dipole moment times
electron, that means that the energy shift is--
and this is what we expected now--
is one half times the expectation
value of the dipole moment times the electric field.
So now we have obtained the quantum mechanical calculation,
the result I told you that the energy is
one half-- the energy shift-- the energy levels,
is one half the dipole moment times the electric field.

Let me just redo the calculation in the way I like.
And this is-- I want to determine now the total energy
but solve the terms in a little bit different way.
So I want to know what is the energy in our result.
And what we do is we are calculating the energy using
our wave functions.

We take the total Hamiltonian and take the wave function.
This leads us to three terms.
One is the unperturbed energy.

The unperturbed energy, the energy contribution
of the first order correction.
So this is the part.
This is the part due to H naught.
And the part due to H prime is simply the dipole moment
times the electric field.
So the first part is of course simply
the energy E naught times the norm
of wave function in naught.
For the second term we use a first order perturbation theory
for N1.
This is our sum over M. Em minus E naught.
M H prime 0.
Because M1 is on either side.
This is the amplitude of the state and 1.
We have to square it.
And since we calculate what is the expectation
value of H naught, we multiply with the energy Em.
OK.
So we're done.
We have recalculate the total energy.
We get three terms.
One is the unperturbed energy.
One is the dipole energy in the electric field.
And one is the extra term, which I want to discuss.
And it is actually-- this term is the internal energy, which
would correspond to the stretching
of the spring in Hooke's law.
Now, in order to show it to you explicitly,
I want to use E naught equals 0 for the energy
because then this term is 0.
I can collect this.
And one of the squares, 1 over Em squared cancels with the Em
So it confused me for a while if I don't set E naught to 0.
The result looks different.
But what happens is, if you do perturbation theory,
there are certain issues with the normalization of the wave
function.
And the wave function in naught has to be the contribution.
If you look at the save function and perturbation
theory of a state, the 0s of the wave function
has an amplitude of 1.
And this amplitude of 1 only changes in second order.
So since I'm doing a second order calculation here,
I have to include those nonstandard terms.
But I can also bypass it by setting E naught to 0.
Then the second order determine.
The norm doesn't matter.
So in other words, if you set E naught to 0,
you make your life easier.
If you do not set E naught to 0, you
have to include some more terms in your calculation.
But the result is this one here.
It is a positive energy.
You can immediately inspect that these positive energy
is the dipole moment times the electric field over 2.
This is exactly analogues to the energy of the spring
in the gravitational problem.
So in other words, this is the energy, the internal energy,
because we mixed excited states to the ground set.
This caused energy.
And it exactly accounts for the occurrence
of the vectors of one half.
Anyway, this is just the standard theory
of the DC stock effect of the atomic polarizibility.
But I put a little bit of emphasis
on those vectors of one half, and tried
to explain in greater detail the contributions to the AC stock
effect, to the DC stock effect, which
come from the electrostatic energy,
and which come from the internal energy.
Questions?

Yes?
I just have a question.
What is allowing you to use non-degenerate perturbation
theory?
[INAUDIBLE]

Well, I'm looking-- what allows me
to do non-degenerate perturbation theory?
Well, I assume we don't have [INAUDIBLE]
If you would go to very high Rydberg states--
and actually we do that.
No.
Not today, but on Monday-- we are looking at a situation
where the splitting between states of different L
become so small that the electric field mixes them.
Then we have to degenerate perturbation theory.
And it means we get now a linear term, a linear stock
effect, and not a quadratic stock effect.
Here I would say, you're doing perturbation theory
of the ground state.
It's an S state.
It's non-degenerate.

Maybe your question is also addressing--
but we have multiple ground states.
We have hyperfine structure.
However, the electronics, the AC-- so
the stock effect, the electric field,
does not couple to the spin at all.
So therefore, all the magnetic energies,
the hyperfine energies are completely unaffected.
And also, if you apply an electric field,
all of the hyperfine states experience the same shift.
And there's no coupling between them.
So also we have multiple ground states.
We have hyperfine structure.
It's a non-degenerate problem because there
is no coupling between the different hyperfine states.
In other words, the theory or the discussion of the DC stock
shift is you have an S state.
You couple with an electric field.
And there is not degeneracy in the S state.
Other questions?

Well, then we've talked about alpha.
The only parameter which comes out of this treatment is alpha.
And now we want to discuss how big is alpha, or first,
what are the units of alpha?
Well, the units of alpha, where-- you can go back
to the second order perturbation result-- but the units of alpha
were the charge times the lengths.
This was the dipole operator.
it was squared.
And in perturbation theory, we divide it by energy
because we have an energy denominator.
Well, we can write that as Q squared over L times L cubed.
But Q squared over L is a Coulomb energy.
And therefore, when I'm interested in the units,
the units cancel.
So therefore, we find that the unit of the polarizability,
at least in [INAUDIBLE] units or atomic units, which I've chosen
here, is simply the volume.
The question is, what volume?
if you would calculate the polarizability for height
option, and simply make the assumption
that the only important matrix element goes from the S
to the P state, then we have a matrix element, which
is on the order of A in A0.
And the energy splitting between the first ground state
and first excited state is three quarter
of the Rydberg constant.

So for a height option in the 1s state,
if you only use the coupling to the 2p state,
we find that alpha is the Bohr radius cube.
And the prefactor is 2.96.
If you do the summation over all states,
the prefactor would be 4.5 because there
are higher states, especially continuum states, which
contribute to the sum.