# Why isn't the electric quadrupole moment always zero?

We know that the electric quadrupole moment ($Q$) describes whether the nucleus of an atom is prolate ($Q<0$), oblate ($Q>0$) or spherical ($Q=0$). We also know that $$Q=\frac{2j-1}{2(j+1)}Q_0,$$ where \begin{align} Q_0 &=\frac{1}{e}\int_V \rho(r)(3\cos^2\theta-1)r^2\:\mathrm dV \\ &=\frac{1}{e}\int_0^{2\pi}\int_0^{\pi}\int_0^R\rho(r)(3\cos^2\theta-1)r^2r^2\sin\theta \:\mathrm dr\mathrm d\theta \mathrm d\phi \end{align} and $$Q_0=\frac{1}{e}\int_0^{2\pi}\mathrm d\phi\int_0^r r^4\mathrm dr\int_0^\pi (3\cos^2\theta\sin\theta-\sin\theta)\:\mathrm d\theta.$$ But we also have the following: $$\int_0^\pi (3\cos^2\theta\sin\theta-\sin\theta)\:\mathrm d\theta=0,$$ which means that $Q_0=0$ and thus $Q=0$, meaning that every atomic nucleus is spherical... What's the catch?

• What is $Q_0$ supposed to represent? Nov 1 '16 at 10:42
• Are you sure that last integral is zero? It doesn't look so to me. Nov 1 '16 at 10:43
• @garyp Yes, it's zero - it's the inner product of $Y_{2,0}$ with $Y_{0,0}$ over the sphere. Why the OP has dropped the angular dependence of $\rho$, on the other hand, is beyond me. Nov 1 '16 at 10:46
• Wolfram alpha agrees... $Q_0$ is the classical definition of the electric quadrupole moment, and extending it to $Q$ gives the quantum mechanical interpretation.
– ODP
Nov 1 '16 at 10:47
• @ODP Why on Earth would you assume there's no angular dependence? You are explicitly trying to describe non-spherical nuclei, right? Nov 1 '16 at 10:56

The full form of the expression for $Q_0$ is $$Q_0=\frac{1}{e}\int_0^{2\pi}\mathrm d\phi\int_0^r r^4\mathrm dr\int_0^\pi (3\cos^2(\theta)-1)\sin(\theta)\:\rho(r,\theta,\phi)\:\mathrm d\theta.$$ If you assume that $\rho(r,\theta,\phi)$ has no angular dependence, i.e. that your nucleus is spherical, then indeed the inner integral reduces to $$\int_0^\pi (3\cos^2(\theta)-1)\sin(\theta) \:\mathrm d\theta=0,$$ giving you $Q=0$. If you do not makes such an assumption, you cannot get rid of the $\theta$ dependence of $\rho$ in that inner integral, which can make it nonzero.