Electric quadrupole and octupole moments for nuclei I am getting slightly confused as to which nuclei cab exhibit quadrupole and octupole excitations. In This link it says closed shell nuclei cannot exhibit quadrupole oscillations because if their spherical symmetry (why not? I can imagine flexing a spherically symmetric ball in the way for an electric quadrupole!) Can they exhibit octupole excitations?
What about general even-even nuclei? My understanding is that the situation is quite a bit more complex for odd nucleon numbers, but understandable otherwise.
What about 
 A: Edited following comments. I will explain why quadrupole excitations cannot occur in spherically symmetric nuclei under electromagnetic excitations. Other types of excitations, presumably weak/strong force are not covered.
The quadrupole terms arise from the follwing tensor:
$K^{\alpha\beta}=\int d^3 r\, r^\alpha r^\beta \rho\left(\mathbf{r}\right)$
where $\rho$ is the charge density (operator) of your nucleus, $r^\alpha,\,r^\beta$ are simply the position vectors, and itegral is over the nucleus.
We should find the irreps of $K^{\alpha\beta}$ under SO(3) transforms: $3\otimes 3=1\oplus 3\oplus 5$. Spherical symmetry occurs only for the singlet irreps, of which there is one:
$K_{l=0}=\int d^3 r\, r^2 \rho\left(\mathbf{r}\right)\quad (1)$
Next, I will say that for all intents and purposes $\rho$ occupies very small point, so it can be approximated with delta functions and derivatives thereof. One can then invert the above expression to find:
$\rho = \frac{K_{l=0}}{6}\nabla^2\delta\left(\mathbf{r}\right)\quad (2)$
To test simply substitute (2) -> (1) and apply integration by parts. A suitable current density for this charge density, assuming oscillation frequency $\omega$, is ($i\omega\rho+\boldsymbol{\nabla}.\mathbf{J}=0$):
$\mathbf{J}=-\frac{i\omega K_{l=0}}{6}\boldsymbol{\nabla}\delta\left(\mathbf{r}\right)\quad (3)$
Assuming such current exists, will it radiate? The magnetic field will be:
$\mathbf{B}=\boldsymbol{\nabla}\times\int d^3 r' G(\mathbf{r}-\mathbf{r}')\mathbf{J}\left(\mathbf{r'}\right)=\int d^3 r' G(\mathbf{r}-\mathbf{r}')\boldsymbol{\nabla}'\times\mathbf{J}\left(\mathbf{r'}\right)=\mathbf{0}$
Since the curl of the current density vanishes. Above $G$ is the Greens function for finding the vector potential, and I implicitly applied integration by parts to transfer the curl from the Greens' function onto the current density.
So, we have established that current density of the spherically symmetric quadrupole-order excitation will not radiate. It also means it will not couple to free-space radiation. Thus you will not be able to excite it if indeed your nucleus obeys spherical symmetry (Hamiltonian is spherically symmetric). In fact, normally quadrupoles are defined to be traceless, so the singlet component is removed by definition.
Alternatively you could consider the interaction energy:
$\int d^3 r \left(\phi \rho - \mathbf{J}.\mathbf{A}\right)$
for scalar and vector potentials $\phi,\,\mathbf{A}$ and arrive at the same result. Throughout this text I treat charge and current densities as functions, but they could also be operators over the quantum field of charged particles.
