Question

Electrostatics: Permanent multipole moment interactions and atomic dipole polarization.

Answer 1

@Michele_Laino

Answer 2

we have to approximate, using the Taylor expansion, this quantity:
\[\int {d{\mathbf{r}}'} \frac{{{{\left| {\Psi \left( {{\mathbf{r}}'} \right)} \right|}^2}}}{{\left| {{\mathbf{r}} - {\mathbf{r}}'} \right|}}\]
which compares into the formula for the potential at point \(r\):
\[\phi = \sum\limits_i {\frac{{{Z_i}}}{{\left| {{\mathbf{r}} - {{\mathbf{R}}_i}} \right|}}} - \int {d{\mathbf{r}}'} \frac{{{{\left| {\Psi \left( {{\mathbf{r}}'} \right)} \right|}^2}}}{{\left| {{\mathbf{r}} - {\mathbf{r}}'} \right|}}\]

Answer 3

such potential is written in units \(e=1\), wherein \(e\) is the electronic charge

Answer 4

now, using the Taylor expansion, I approximate this quantity:
\[\frac{1}{{\left| {{\mathbf{r}} - {\mathbf{r}}'} \right|}}\]
before integrating

Answer 5

so I can write this:
\[\frac{1}{{\left| {{\mathbf{r}} - {\mathbf{r}}'} \right|}} \simeq \frac{1}{r} + \frac{{r'}}{{{r^2}}}\cos \theta + \frac{1}{{{r^3}}}\frac{{3{{\left( {\cos \theta } \right)}^2} - 1}}{2}r{'^2}\]

Answer 6

wherein \(\theta\) is the angle between the vectors \(r\) and \(r'\)

Answer 7

such expansion is up to the second order, of course we can go beyond, namely we can compute the term corresponding to the third order, and so on
Now I substitute into such integral, so I get:

Answer 8

\[\begin{gathered}
\phi = \sum\limits_i {\frac{{{Z_i}}}{{\left| {{\mathbf{r}} - {{\mathbf{R}}_i}} \right|}}} - \int {dr'} \frac{{{{\left| {\Psi \left( {{\mathbf{r}}'} \right)} \right|}^2}}}{{\left| {{\mathbf{r}} - {\mathbf{r}}'} \right|}} \simeq \hfill \\
\hfill \\
\simeq \sum\limits_i {\frac{{{Z_i}}}{{\left| {{\mathbf{r}} - {{\mathbf{R}}_i}} \right|}}} - \frac{1}{r}\int {dr'} {\left| {\Psi \left( {{\mathbf{r}}'} \right)} \right|^2} - \frac{1}{{{r^2}}}\int {dr'} {\left| {\Psi \left( {{\mathbf{r}}'} \right)} \right|^2}r'\cos \theta + \hfill \\
\hfill \\
- \frac{1}{{{r^3}}}\int {dr'} {\left| {\Psi \left( {{\mathbf{r}}'} \right)} \right|^2}\frac{{3{{\left( {\cos \theta } \right)}^2} - 1}}{2}r{'^2} \hfill \\
\hfill \\
\end{gathered} \]

Answer 9

the second term of our expansion, is like a dipole moment

Answer 10

now, it is more easy to compute the electric field, due to such potential, by means of this equation:
\[\Large {\mathbf{E}} = - \nabla \phi \]

Answer 11

maybe, we have, in the similar way, to expand this part also:
\[\Large \sum\limits_i {\frac{{{Z_i}}}{{\left| {{\mathbf{r}} - {{\mathbf{R}}_i}} \right|}}} \]

Answer 12

I was just about to ask, why wouldn't we expand the sum of nuclei?

Answer 13

you question is right! I think that better is if we expand the first contribution

Answer 14

here is the corresponding formula, up to the second term:

Answer 15

\[\begin{gathered}
\sum\limits_i {\frac{{{Z_i}}}{{\left| {{\mathbf{r}} - {{\mathbf{R}}_i}} \right|}}} \simeq \frac{1}{r}\sum\limits_i {{Z_i} + \frac{1}{{{r^2}}}} \sum\limits_i {{Z_i}{R_i}\cos \theta } + \hfill \\
\hfill \\
+ \frac{1}{{{r^3}}}\sum\limits_i {{Z_i}R_i^2\frac{{3{{\left( {\cos \theta } \right)}^2} - 1}}{2}} \hfill \\
\end{gathered} \]

Answer 16

now, you have to put together all such terms at the right side, and then to compute the electrostatic field using this formula:
\[{\mathbf{E}} = - \nabla \phi \]

Answer 17

technically speaking, what we have done is the multipole expansion of the electric potential, due to the nuclear charges and all the electrons charges

Answer 18

oops.. of course, into the second expantion, \(\theta\) is the angle between the vectors \(r\) and \(R_i\)

Answer 19

expansion*

Answer 20

Ahh I see. But this is an expansion around the electric field as I also tried to follow a little bit in a tutorial video. What confused me was when they went to matrix notation in the book I sent you and evaluate the energy of the different multipoles. Any way you could explain that?

Answer 21

Or rather multipole interactions.

Answer 22

I'm looking at your textbook

Answer 23

sorry, where is such matrix?

Answer 24

sorry, it is the polarizability tensor

Answer 25

I was specifically thinking what was happening in equation 2.27 and afterwards.

Answer 26

Cause I can see that must be the way he derive the expression:
\[E_{el}=q_0\phi -\bf \mu_0 F - \frac{ 1 }{ 2 } \Theta_0 F'-\frac{ 1 }{ 2 } \alpha F^2\]
Where it should include the first 3 terms of the multipole expansion and the fifth term.

Answer 27

here we have an interaction between other molecules, located at othr sites, and a specific molecule located at a specific site . Such molecules create an electric field \(E^fic molecule. Such induced electric dipole moment \(p^{ind}\), is given by the subsequent tensor equation:
\[\Huge {p^{ind}}_i = {\alpha _{ij}}E_j^0\]
wherein the quantities:
\[\huge {\alpha _{ij}}\]
are the so called polarizability tensor

Answer 28

The polarizability happens from an induced dipole right?

Answer 29

Ah what you wrote. Sorry my bad,

Answer 30

that's right!
oops.. I meant this:
Such molecules create an electric field \(E^0\). Such induced electric dipole moment\(p^{ind}\)

Answer 31

we can think the induced electric dipole oment as an answer from a specific molecule under the action of an external electric field. In our case such external electric field is the electric field due to the other molecules

Answer 32

moment*

Answer 33

I see.That explains why would should do external potential in eq. 2.27, but because my linear algebra sucks I must ask: Why do we do both with and without an external potential?

Answer 34

Ah charge transfer due to external potential and then solved by Coulombs law.

Answer 35

in order to check such formula, you have to perform this computation:
\[\Large {\mathbf{E}} = - \nabla \phi \]
when we want to compute the induced electric dipole moment by a specific molecule, we need to know the electric potential creates by other molecules at the site at which is our molecule

Answer 36

please keep in mind that electric fields have the property of superposition

Answer 37

Oh, I actually didn't think about that, nor knew really.

Answer 38

here is a simple example:

each of the four water molecules create an electric field, described by the formula above
and the total electric field at position of molecule A, is given by the vector sum of each vector field from the single molecules 1, 2, 3 or 4, as if each molecule 1, 2 3 or 4 is present alone without the others

Answer 39

namely, we can write:
\[{{\mathbf{E}}_A} = {{\mathbf{E}}_1} + {{\mathbf{E}}_2} + {{\mathbf{E}}_3} + {{\mathbf{E}}_4}\]
wherein each of four fields \(E_i,\;i=1,2,3,4\) is given by the computation above

Answer 40

I see. That does indeed make sense yes.

Answer 41

such field \(E_A\) is the field which produces the induced electric dipole moment of molecule A

Answer 42

Right.

Answer 43

Another I'd like to ask is: why is eq. 2.30 right? I feel I am missing the link between the energy and the electric field.

Answer 44

It is the energy needed to produce such dipole inside an external electric field. Such energy \(W\) is given by the subsequent formula:
\[\huge W = \frac{1}{2}{\mathbf{P}} \cdot {\mathbf{E}}\]
wherein \(P\) is the induced electric dipole moment, and \(E\) is the external field

Answer 45

from the theory of electricity, and in particular from dielectrics, we can write this little proof:
\[\begin{gathered}
W = \int\limits_V {dV} \int\limits_0^{{E_0}} {PdE} = \int\limits_V {dV} \int\limits_0^{{E_0}} {\chi EdE} = \int\limits_V {\frac{{\chi E_0^2}}{2}dV} \hfill \\
\hfill \\
\frac{W}{V} = \frac{{\chi E_0^2}}{2} = \frac{1}{2}PE \hfill \\
\end{gathered} \]
wherein \(W\) is the needed energy to polarize a volume \(V\)

Answer 46

oops.. we can write this \(simple\)* proof

Answer 47

e \(P=\alpha F\) and \(E_0=F\), so, we get:
\[\huge \frac{W}{V} = \frac{{\chi E_0^2}}{2} = \frac{1}{2}P{E_0} = \frac{1}{2}\alpha {F^2}\]

Answer 48

furthermore \(\chi\) is the electrical susceptibility of the medium

Answer 49

that is a simple reasoning in order to understand the factor \(1/2\)

Answer 50

nevertheless, what is important, is that it represents the needed energy to make such induced electric dipole moment

Answer 51

of course, the susceptibility \(\chi\) is proportional to the polarizability \(\alpha\)

Answer 52

I see. That explains eq. 2.30 much more!

Answer 53

Last thing I wanted to ask is:
Do you have any idea what the expansion is in the picture I send you and how to derive it? I guess it must be a continuation of the expansion we have talked about?

The very last equation in the picture with the missing symbols unfortunately. It is a lecture given by the same person who wrote this book.

Answer 54

Like:
\(q_0\phi \) is the monopole interaction?
\( -\bf \mu_0 F\) is the dipole interaction?
And so on?

Answer 55

While:
\(-\frac{ 1 }{ 2 } \alpha F^2\) is the first order polarizability?

Answer 56

I thik that we have to continue my procedure above, namely after the expansion of the integral and the summation, we have to apply this formula:
\[\Large {\mathbf{E}} = - \nabla \phi \]
Please try to perform such computation, and let me know what you get. I will try to do such computation on my own and I will tag you when I got the expected result
Furthermore, if you need to know some basic of electromagnetism, please refer to this textbook:
\[\begin{gathered}
{\text{E}}{\text{. M}}{\text{. Purcell}} \hfill \\
{\mathbf{Electricity and Magnetism}} \hfill \\
{\text{The Berkeley Physics Course}}{\text{. Vol}}{\text{.II}} \hfill \\
\end{gathered} \]

Answer 57

at first sight I think you are right, and now we can understand the reason of such terms

Answer 58

or, you can ask me for electromagnetism, and I will be happy to help you

Answer 59

Any mathematical fundament required to read the book besides calculus and linear algebra?

Answer 60

This topic is basically what I plan on doing for my master degree so I bet there will come many funny questions :)

Answer 61

only calculus, since it is a very simple, and well written textbook

Answer 62

ok! :)

Answer 63

And thank you so much for your help so far it helped quite a lot. :)