Quadrupole

[Mat' G.]

Hello,

I do not manage to visualize the link betwen the component of the
quadrupole Q_ij and the spatial distribution of the electric quadrupole
field.

I was told to imagine the Q_ij as an ellipsoid, which I understand (the
ellipsoid "radius" in a given direction being the strength of the
quadrupole along this direction). Yet what is the link between the Q_ij
and the usual representation in Slide 12 of this file?:
http://www.cems.uvm.edu/~oughstun/LectureNotes141/Topic_09%20%28ElectrostaticMultipoles%29.pdf

In particular, I want to find out when does the gradient \nabla_k Q_ij
equal zero? When i,j =!k ?

Pleeeease, help!

[Jos Bergervoet]

On 5/12/2012 8:39 PM, Mat' G. wrote:
> Hello,
>
> I do not manage to visualize the link betwen the component of the
> quadrupole Q_ij and the spatial distribution of the electric quadrupole
> field.
>
> I was told to imagine the Q_ij as an ellipsoid, which I understand (the
> ellipsoid "radius" in a given direction being the strength of the
> quadrupole along this direction).

That isn't in general such a good advise. It would be
fine for diagonal elements, like Q_xx (with positive
charge stretched in both directions along x-axis,
balanced by a central negative charge).

But for Q_xy, the picture of a square in the xy-plan
with alternating charges on the corners is better.

> Yet what is the link between the Q_ij
> and the usual representation in Slide 12 of this file?:
> http://www.cems.uvm.edu/~oughstun/LectureNotes141/Topic_09%20%28ElectrostaticMultipoles%29.pdf

In that file they clearly are showing the field of
a square with alternating charges on the corners!

> In particular, I want to find out when does the gradient \nabla_k Q_ij
> equal zero? When i,j =!k ?

In that case you probably talk about a quadrupole
density. With quadrupoles in varying strengths
present throughout a whole region it does of course
become more complicated.

But more importantly, it is not possible for the
readers here to infer *how* your quadrupoles are
distributed in space (because you do not tell!)
And your question about \nabla_k Q_ij is in
fact a question about how they are distributed.
We can only guess!

Could it be that you use light in a nonlinear
medium? Then the light itself can create quadrupole
density (also dipole density, etc.) and in turn
some new light will be generated by those dipoles
and quadrupoles (most likely at the double or
triple frequency of the original). It is up
to you to explain what you are doing!

--
Jos

[Oliver Jennrich]

Jos Bergervoet <jos.ber...@xs4all.nl> writes:

> On 5/12/2012 8:39 PM, Mat' G. wrote:
>> Hello,
>>
>> I do not manage to visualize the link betwen the component of the
>> quadrupole Q_ij and the spatial distribution of the electric quadrupole
>> field.
>>
>> I was told to imagine the Q_ij as an ellipsoid, which I understand (the
>> ellipsoid "radius" in a given direction being the strength of the
>> quadrupole along this direction).
>
> That isn't in general such a good advise. It would be
> fine for diagonal elements, like Q_xx (with positive
> charge stretched in both directions along x-axis,
> balanced by a central negative charge).

Well, Q_ij is symmetric, hence diagonalizable. So there always exist a
coordinate system in which the axes of the ellipsoid are the coordinate
axes. So if you accept that picture for e.g. xx, then yiu have to accept
it for all orientations of the ellipsoid.

>
> But for Q_xy, the picture of a square in the xy-plan
> with alternating charges on the corners is better.

Different picture. The ellipsoid picture is used to visualise the
*strength* of the quadrupole tensor. Your picture is based on the
*sources* for the quadrupole tensor.

--
Space - The final frontier

[Mat' G.]

>> In particular, I want to find out when does the gradient \nabla_k Q_ij
>> equal zero? When i,j =!k ?
>
> In that case you probably talk about a quadrupole
> density. With quadrupoles in varying strengths
> present throughout a whole region it does of course
> become more complicated.
>
> But more importantly, it is not possible for the
> readers here to infer *how* your quadrupoles are
> distributed in space (because you do not tell!)
> And your question about \nabla_k Q_ij is in
> fact a question about how they are distributed.
> We can only guess!
>
> Could it be that you use light in a nonlinear
> medium? Then the light itself can create quadrupole
> density (also dipole density, etc.) and in turn
> some new light will be generated by those dipoles
> and quadrupoles (most likely at the double or
> triple frequency of the original). It is up
> to you to explain what you are doing!

That is it exactly: Nonlinear optics.
Fundamental oscillating at \omega=w, and SHG at 2w.
The effective nonlinear polarization term is given by (here limiting
myself to the quadrupole contribution at 2w):

\vec{P}(2w)_eff \propto \nabla \cdot \tensor{Q}
= \nabla \cdot \chi(2w,w,w)^(qee) : E(w)E(w)

or equivalently, looking at the individual components:

P_i \propto \nabla_i Q_ij
= \nabla_i \chi_ijkl E_k E_l

Because one knows the structure symmetry being investigated, *the non
vanishing Q_ij are known.*
Concretely, let say the \chi susceptibility tensor elements with ijkl=
xxxx, yyxx, zzxx,
xxyy, yyyy, zzyy,
xyxy = xyyx = yxxy = yxyx,
(+other terms which cannot be excited in our setup configuration)
are non zero.

For pure excitation along x (E_k E_l=E_x E_x), only Q_xx, Q_yy and Q_zz
are non zero and the tensor Q_ij is diagonal.
For pure excitation along y (E_k E_l=E_y E_y), only Q_xx, Q_yy and Q_zz
are non zero and the tensor Q_ij is also diagonal.
For mixed excitation along x and y (E_k E_l=E_x E_y=E_y E_x), only Q_xx,
Q_yy, Q_zz, and Q_xy=Q_yx are non zero.


The general matrix form for \nabla \cdot \tensor{Q} is:
(\nabla_x Qxx + \nabla_y Qyx + \nabla_z Qzx)
(\nabla_x Qxy + \nabla_y Qyy + \nabla_z Qzy)
(\nabla_x Qxz + \nabla_y Qyz + \nabla_z Qzz)

I have explained above that most terms vanish because of the symmetry of
the sample.
In the example where the excitation is purely along x or purely along y,
this reduces to:
(\nabla_x Qxx)
(\nabla_y Qyy)
(\nabla_z Qzz)

In the example where the excitation is mixed, this reduces to:
(\nabla_x Qxx + \nabla_y Qyx)
(\nabla_x Qxy + \nabla_y Qyy)
(\nabla_z Qzz)

but, *here comes finally my question*, is there not another limitation
that \nabla identifies with the wavevector \vec{k} giving the direction
of propagation?
So that for \vec{k}=(0,0,1), our examples would reduce to
\vec{P}_eff = \nabla \cdot \tensor{Q} = (0,0,Qzz), and since P_z cannot
be analyzed, then not SHG signal is measurable.

In other words, one can only measure a signal when Q_xz=Q_zx=!0 or
Q_yz=Q_zy=!0

Is this true or did I make a mistake somewhere?

Thank you for reading that far!