I have following problem:
calculate the quadrupole moment of following arrangement,
where $e$ is the charge and $a$ is the distance between the charges.
Hint: The quadrupole moment is defined as:
$$Q_{ij}=\int_V\rho(\vec{x})(3x_ix_j – r^2\delta_{ij})d^3x$$
$\hskip2in$
I know that the charge density is:
$$\rho(\vec{r})=e\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)]$$
and the solution to this problem is:
$$Q_{11}=Q_{22}=-2a^2e,\quad Q_{33}=4a^2e$$
I am having difficulties solving and understanding this integral
my approach is:
$$Q_{11}=\int e\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)(3x_1x_1-r^2\delta_{11})dxdydz$$
my guess is that in this case $x_1=r=a$ since it represents the distance from the origin to the first charge, which delivers:
$$2a^2e\int\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)]dxdydz$$
what bothers me is that I am not sure how to evaluate this integral, since there is a delta function present my best guess is that we integrate from $-\infty$ to $+\infty$, and apply the sifting property of the delta function
$$2a^2e[\int_V \delta(x)\delta(y)\delta(z-a)dV-2\int_V \delta(x)\delta(y)\delta(z)dV+\int_V \delta(x)\delta(y)\delta(z+a)dV]$$
$$=2a^2e[-a-2+a]=-4a^2e$$
Which contradicts the solution.
Any help is appreciated.
Best Answer
$$Q_{ij}=\int_Ve\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)](3x_ix_j - (x^2+y^2+z^2)\delta_{ij})dxdydz$$
is zero for $i \ne j$.
Set $i=x$:
$$Q_{xx}=\int_Ve\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)](2x^2 - (y^2+z^2))dxdydz$$
integrate over $y$:
$$Q_{xx}=\int_Ve\delta(x)[\delta(z-a)-2\delta(z)+\delta(z+a)](2x^2 - (0^2+z^2))dxdz$$
integrate over $x$:
$$Q_{xx}=\int_Ve[\delta(z-a)-2\delta(z)+\delta(z+a)](0^2 - (0^2+z^2))dz$$
finally, $z$:
$$Q_{xx}=-e[a^2-2\cdot0^2+(-a)^2]=-2ea^2$$
And likewise $yy$
Of course, since you are dealing with points, you can skip the delta function and integrals and just add up moments of points, so for $Q_{zz}$ you want to sum over points with the weighting function:
$$ w(x, y, z) = q(3z^2-r^2) = q(2z^2 - x^2 -y^2) \propto qr^2Y^0_2(\theta, \phi)$$
since $x=y=0$ we are just summing:
$$ \sum_i{2ez_i^2}= 2e[a^2+(-2\cdot 0)^2+(-a)^2] = 4ea^2$$
Now the reason I included the $\propto = Y_2^0$ is because rank-2 tensors in Cartesian coordinates are obtuse. There are 9 of them, and one transforms like scalar ($\delta_{ij}$), then 3 look like a vector (the antisymmetric parts), but there are 5 that are pure natural-form (i.e. traceless) rank-2:
$Y_2^0$: that is how oblate or prolate your distributions is. (Your distribution is pure $Y_2^0$).
$Y_2^{\pm 2}$: represents a lack of cylindrical symmetry along the $z$-axis, specifically with 180 degree rotation symmetry (0 for this problem).
Finally:
$Y_2^{\pm 1}$ means you have not diagonalized your axes, and these can be removed by the correct choice of axes.