1. Jul 12, 2009

### Petar Mali

Is this $$\delta_{i,j}$$ Kronecker delta?

In my notebook I have relation:

$$Q_{i,j}=\frac{3}{2}eQ_0(x_ix_j-\frac{1}{3}\delta_{i,j})$$

When direction of external field are the direction of symmetry axis $$Q=Q_0$$.

2. Jul 12, 2009

### Meir Achuz

$$\delta_{i,j}$$ is the Kronecker delta.
The relation from your notebook is not quite right. It should be
$$Q_{i,j}=\frac{3}{2}Q_0(\delta_{i,3}\delta_{j,3}-\frac{1}{3}\delta_{i,j})$$
for a symmetric quadrupole aligned along the z (or 3) axis, having quadrupole moment Q_0.
There is a full discussion of quadrupoles in Section 2.4 of Franklin, "Classical Electromagnetism" (AW.com).

3. Jul 13, 2009

### Petar Mali

Thanks!

4. Jul 17, 2009

### Petar Mali

I found that in "Non relativistic quantum mechanics" of Landau. There is formulation:
$$Q_{i,j}=\frac{3}{2}Q_0(n_in_j-\frac{1}{3}\delta_{i,j})$$

where $$n_i,n_j$$ are components of unit vector $$\vec{n}$$.

5. Jul 17, 2009

### clem

My formula is the reduction of Landau's when the unit vector n is in the z direction.