# Does the electric quadrupole vanish if $|\psi|^2$ is spherically symmetric?

In the book Nuclear and Particle Physics by B.R. Martin it is said that the quantum mechanical analogue to the electric quadrupole $$Q \equiv \frac{1}{e}\sum\limits_i\int\psi^*q_i(3z_i^2-r^2)\psi d^3\bar{x}$$ is zero if $|\psi|^2$ is spherically symmetric. Could someone give a hint as to how I might be able to verify this?

• Your notation is inconsistent - what is $i$ indexing over, what coordinates are you integrating over, and why does $z_i$ have an index while $r$ does not? I appreciate that the author uses it, but that doesn't magically make the notation meaningful - and it doesn't exempt you from the requirement to fully define all the terms that you use. Apr 20 '18 at 19:07

Generally speaking all the multipoles of a spherically-symmetric distribution will vanish; this is because a spherical distribution is completely monopolar (i.e. it can be seen as an $\ell=0$ function) and multipolar functions are orthogonal over the sphere.
For your specific case, the vanishing integral is easy to check with simple means: if $\psi(\mathbf r)$ is spherically symmetric, then it must be invariant under the simpler symmetry $$\psi(x,y,z) = \psi(y,z,x) = \psi(z,x,y)$$ which permutes the three coordinate axes, i.e. a rotation by 120° about an axis $\arccos(1/\sqrt{3}) = 54.7$° out from your initial $z$ axis; this single discrete symmetry is all that's required to show that the quadrupoles vanish. To see that, simply observe that swapping out the $z$ coordinate in the weighing function for an $x$ or a $y$ coordinate cannot change the value of the integral (because the density doesn't change under the rotation), so therefore \begin{align} Q & = \int(3z^2-r^2)|\psi(\mathbf r)|^2 \mathrm d^3\mathbf r \\& = \int(3y^2-r^2)|\psi(\mathbf r)|^2 \mathrm d^3\mathbf r \\& = \int(3x^2-r^2)|\psi(\mathbf r)|^2 \mathrm d^3\mathbf r. \end{align} From here, the trick is to add all three versions of this equality together, giving one each of the $x^2$, $y^2$ and $z^2$ terms and three of the $r^2$ terms, \begin{align} 3Q & = \int(3(x^2+y^2+z^2)-3r^2)|\psi(\mathbf r)|^2 \mathrm d^3\mathbf r \\ & = \int(3r^2-3r^2)|\psi(\mathbf r)|^2 \mathrm d^3\mathbf r \\ & = 0 \end{align} which cancel out exactly because the anisotropic terms add up to a simple radial term whose coefficient cancels out the combined contribution of the initial isotropic terms. (And yes, this is where that $3$ comes from, of course.)