M3L16c.txt

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Rotating wave approximation-- revisited.
Again, rotating wave approximation.
And what I want to discuss can be discussed in
the fully quantized picture, but also
in the semi-classical picture.

In the fully quantized picture-- just a reminder,
what we discussed earlier, was that when
we have the atomic raising and lowering
operator and the photonic raising and lowering operator,
we got four terms, and two of the terms
are co-rotating to a counter-rotating.
But I can get exactly the same number
of four terms in the same classical picture,
and I want you to see both.
But in the quantized picture, it's
easier, because when you see a and a dagger,
you know immediately one is absorption, one is emission.
So therefore, let me explain to you
what I want to tell you about the rotating wave
approximation, using the same classic picture,
because then you immediately know
how to apply it to the quantized picture.
So what I want to bring in the here, in addition
to what we have discussed about light-atom interaction,
we had sort of a dipole Hamiltonian,
is the fact that we have circular-polarized
light, left-handed and right-handed light.
And I want to use that and combine it
with angular momentum selection rules, which, as you remember,
we discussed after our discussion of dipole,
quadropole, and magnetic dipole transitions.
So I put now all those parts together,
and revisit the rotating wave approximation.
So what I hope for is, it tells you a little bit
how selection rules, angular momentum,
circular polarization, and semi-classical field
which rotate in one direction, how they're all connected.
I have to set up the situation by saying
that we use as a quantization axis, the direction key, which
is either the direction of the propagation of the light beam,
or, in general, it's orthogonal to the polarization
of the electric and magnetic field.
And I can talk about an electric field driving
an electric-dipole transition.
I can talk about a magnetic field driving
a magnetic transition, it doesn't really matter.
I will use B and C amplitude, but you can also immediately
think electric dipole.
And in this field is linearly polarized.
But I want to immediately decompose this field
into right-handed and left-handed field.
Or a linearly-polarized field can
regarded as a superposition over field which
circulates this way, plus one which circulates the other way.
And ultimately, the message we will see
is that if you have linearly-polarized light,
we always get counter-rotating term, we always have
a Bloch-Siegert shift and such.
But if you use rotating fields, or circularly-polarized light,
selection rules may actually lead to the result
that there is no counter-rotating term at all.
So this is, eventually, what I'm aiming for,
and this will be the final point of the discussion.
The field, which rotates in the right-handed direction,
well a rotating field is a superposition of x and y
or i and j, and one rotates has cosine omega t and one
has sine omega t.
I don't need to write down the left-handed part,
because it's just a minus sign.
Or this will actually become very handy.
I do all the discussion for the right-handed part,
but I can always obtain the expression
for the left-handed part by replacing omega by minus omega.

Which will mean that some emission
process by the right-handed part will be an absorption
process by the left-handed part, but we'll see.
You can change angular momentum by plus one
by absorbing a right-handed photo-- we'll see,
we'll get there.
So anyway, those signs will become important.
Let me now take the above the explanation
for the right-handed part, and replace cosine omega t and sine
omega t, by e to the i omega t.

So this was the i component, this was the j component,
we divide by two.
Just to avoid confusion, I want to emphasize
I've started with a real field, so I'm not
using-- as we often do in E and M-- complex field,
and the real fields are the real part.
I have not started out by adding imaginary parts to the field.
I started out with linearly-polarized field
in the x direction, cosine omega t.
And I've decomposed it into two real fields,
one is right-handed, one is left-handed.
Complex numbers only come because I
want to use complex exponential to replace cosine omega t
and sine omega t.
We are almost done with the decomposition of the field.
I just want to-- we have now four terms,
and I want to recoup them.
I minus imaginary unit j.
I plus imaginary unit j.
This is e to the i omega t, and this
is e to the minus i omega t.
So, what have we done?
Well, we've just started with linearly-polarized light,
and have rewritten the expression twice.
And now we are looking only at one of the circular components,
and in the end what we have is four terms.
Well, that's also what we had in the fully-quantized
Hamiltonian, and we now want to identify
what those four terms are.
Two will be co-rotating, two will be counter-rotating,
but it's very helpful to analyze those terms.
But the two things we have to look at now,
one is we have an e to the i omega t,
and well, it's probably a sign convention,
but trust me if you put that in the Schrodinger equation,
it means that you increase the energy of the atom
if you drive it with e to the i omega t.
You take it from a ground state to an excited state, which
differs in frequency by omega.
And therefore this means you have increased
the energy of the system, and this corresponds to absorption.
Whereas this one here means we take an atom
from an excited state to a ground state,
and this is the situation of stimulated emission.
Remember in selection rules, we take our field
and we multiply it was a dipole moment--
electric, magnetic dipole moment,
whatever-- but now we want to use also this spherical tensor
decomposition of those dipole moments.
It's a complicated word, but what
it means is those terms are dotted with the dipole moment,
and if you do it now component by component,
we retrieve selection rules.
Because this peaks out the-- let me just write it down,
the x plus y tensor component of the matrix element,
and this corresponds to delta m equals plus 1.
We change angular momentum by one unit,
and of course this term used then delta m equals minus 1.