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?

  • $\begingroup$ What is $Q_0$ supposed to represent? $\endgroup$
    – gj255
    Commented Nov 1, 2016 at 10:42
  • $\begingroup$ Are you sure that last integral is zero? It doesn't look so to me. $\endgroup$
    – garyp
    Commented Nov 1, 2016 at 10:43
  • $\begingroup$ @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. $\endgroup$ Commented Nov 1, 2016 at 10:46
  • $\begingroup$ Wolfram alpha agrees... $Q_0$ is the classical definition of the electric quadrupole moment, and extending it to $Q$ gives the quantum mechanical interpretation. $\endgroup$
    – ODP
    Commented Nov 1, 2016 at 10:47
  • 2
    $\begingroup$ @ODP Why on Earth would you assume there's no angular dependence? You are explicitly trying to describe non-spherical nuclei, right? $\endgroup$ Commented Nov 1, 2016 at 10:56

1 Answer 1


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.

In other words, yes: spherical nuclei are spherical. Non-spherical nuclei are not.

  • $\begingroup$ Ah OK, so my mistake here is in assuming that the charge density has no angular dependence... $\endgroup$
    – ODP
    Commented Nov 1, 2016 at 10:49

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