Quadrupole-quadrupole interaction?

[neo143]

Could anyone please tell me how to calculate quadrupole quadrupole interaction between two molecules having dipole moment zero? As quadrupole is a tensor quantity I get a 3*3 matrix for a molecule.

 

[kleinwolf]

I don't know much about quadrupole. Which kind of molecule are you treating, does it suffice to have 3 atoms ?

 

[neo143]

suppose the molecule is benzene?

 

[reilly]

One sure-fire way is to work with multipole expansions. Expand both field and molecular charge distribution in spherical harmonics -- the quadrapole moment is the coefficient of Y(L=2, M; theta, phi) This might be done in Jackson, and surely is done in books on nuclear physics, and on angular momentum (Edmunds, Angular momentum in QM.)
Regards,
Reilly Atkinson

 

[Meir Achuz]

The answer is a bit messy, even for symmetric quadrupoles.
I give it here for two symmetric quadrupoles Q and Q' with symmetry axes k and k' a distance r apart. k,k,r are all vectors and Q and Q' are the Qzz component of the Q tensor (with Qxx=Qyy=-Qzz/2).
The energy is
U=(3QQ'/4r^5)[35(k.r)^2(k'.r)^2-20(k'.r)(k.r)(k.k')+2(k.k')^2+(k.k')].
(The k,k',r in the square bracket are all unit vectors.)
I got this by combining Eqs. (2.118) and (2.112) in "Classical Electromagnetism" by Franklin (Addison Wesley).

 

[kleinwolf]

Did sdy know if we take ponctual charges, it is clear that 2 are building what is called a dipole...but do 3 point charges suffice to have non-vanishing quadrupole moment ?? Oh..sorry, this is not quantum atomics..

 

[Meir Achuz]

Even two point charges can have a quad moment. If you want the system to be neutral, at least three charges are needed.

 

[neo143]

Thanks Meir for the reply,
I have one more doubt. Suppose I want to calculate quadrupole moment interaction between two symmetrical molecules(with Qxx=Qyy=-Qzz/2).Rest of the components are zero. Should I consider every possible pair of quadrupole moments for two molecules?
like xx for the first molecule and xx for the second
then xx for the first and yy for the second
then xx for the first and zz for the second
....and so on...total terms will be 9.

Thanks once again
Regards

 

[azag]

The answer is a bit messy, even for symmetric quadrupoles.
I give it here for two symmetric quadrupoles Q and Q' with symmetry axes k and k' a distance r apart. k,k,r are all vectors and Q and Q' are the Qzz component of the Q tensor (with Qxx=Qyy=-Qzz/2).
The energy is
U=(3QQ'/4r^5)[35(k.r)^2(k'.r)^2-20(k'.r)(k.r)(k.k')+2(k.k')^2+(k.k')].
(The k,k',r in the square bracket are all unit vectors.)
I got this by combining Eqs. (2.118) and (2.112) in "Classical Electromagnetism" by Franklin (Addison Wesley).

I believe there is a mistake in that formula. By combining those two mentioned formulas in Franklin's book, I got
U=(3QQ'/4r^5)[35(k.r)^2(k'.r)^2-20(k'.r)(k.r)(k.k')-5(k.r)^2-5(k'.r)^2+2(k.k')^2+1]
which differs a bit. I checked it on two parallel and two crossed model linear quadrupoles. While for parallel case both formulas give same result, for crossed case my gives non-zero interaction and Meir's gives zero interaction. Calculating the energy by simple Coulomb's law one gets non-zero interaction.

 

[Vanadium 50]

Thanks for the correction, but this post is 5 years old.