1. Nov 21, 2005

### 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.

2. Nov 22, 2005

### kleinwolf

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

3. Nov 22, 2005

### neo143

suppose the molecule is benzene???

4. Nov 23, 2005

### 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

5. Nov 27, 2005

### Meir Achuz

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).

6. Nov 28, 2005

### 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..

7. Nov 28, 2005

### 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.

8. Dec 6, 2005

### neo143

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

9. Dec 6, 2010

### azag

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.

10. Dec 6, 2010