# Why do spin-$\frac{1}{2}$ nuclei have zero electric quadrupole moment?

Why do spin-$\frac{1}{2}$ nuclei have zero electric quadrupole moment? How does this come about, and how can one tell in general whether a spin-$j$ nucleus can have a nonzero quadrupole (or higher multipole) moment?

• I'm pretty certain you can show that $\langle\frac{1}{2}|Q|\frac{1}{2}\rangle = 0$, where $Q$ is the quadrupole operator. However the details have slipped from what I laughingly refer to as my memory. This might give you enough clues to Google the answer. – John Rennie Dec 14 '14 at 10:56
• As far as I've googled, it looks not that simple. I found no credible answer. – xiaohuamao Dec 18 '14 at 8:18
• – rob Mar 18 '16 at 2:49

1. $|j_1-j_2| \leq j_\mathrm{total} \leq j_1+j_2$
2. $j_1+j_2+j_\mathrm{total}$ is an integer.
Well, the Wigner-Eckart theorem teaches us that similar rules apply involving operators. The electric quadrupole operator $Q$ can be expressed as a rank-2 spherical tensor operator. When it acts on a spin-1/2 system, the resulting spin can only be $\frac{5}{2}$ or $\frac{3}{2}$. So if you're calculating $\langle\frac{1}{2}| Q|\frac{1}{2}\rangle$, you can group it like $\langle \frac{1}{2}| \;\;\; Q |\frac{1}{2}\rangle$. In other words, this is the inner product of a spin-$\frac{1}{2}$ system with $Q|\frac{1}{2}\rangle$ (which is spin $\frac{3}{2}$ or $\frac{5}{2}$). So they're orthogonal; the inner product is zero. $\langle\frac{1}{2}| Q |\frac{1}{2}\rangle=0$.