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


When we talk about a nucleus undergoing collective quadrupole or octupole excitations (including rotations and vibrations), we are giving a semiclassical description of a wavefunction. These excitations will arise naturally in the nuclear shell model, which is an actual microscopic model, and in order to address your questions it's instructive to think about the way in which they would arise.

In a shell model calculation, you would start by making a basis of states consisting of particle-hole excitations. You have a state with no particle-hole excitations, states with 1 particle-hole excitation, and so on. So to generate energy and angular momentum, you have the valence particles, plus $2n$ particles and holes in a state with $n$ particle-hole excitations.

Now you take this basis and, working in this basis, you diagonalize the hamiltonian. When you do this, the results can be complicated, but we can observe certain heuristics. One such heuristic is that when you impose an interaction, levels that can interact (because they have the same spin and parity) tend to "repel" each other (but this not a physical repulsion). This "level repulsion" is strongest when states are close together, and when they're closer together, the wavefunctions also mix a lot.

In a nucleus with a lot of valence particles outside a closed shell, there is a large number of such states that can mix, and a lot of these states are at relatively low excitation energies. The result tends to be that we get states that are coherent superpositions, and we can describe these as things like rotations and vibrations. These states are collective.

A good experimental measure of their collective nature is that we get electromagnetic transitions that are strong when expressed in Weisskpof units, e.g., in a rotor you get E2 transitions that are many W.u. in strength. This is because many particles are cooperating coherently in order to radiate.

In a nucleus that's near a closed shell, we don't have very many particle-hole excitations at low energies. Therefore they don't mix as much when you let them interact, and the eigenstates of the hamiltonian tend to be fairly similar to the basis states. They're just noncollective particle-hole excitations.


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.

  • $\begingroup$ This is wrong. The question of whether these excitations can exist is a separate question from their electromagnetic decay properties. $\endgroup$ – Ben Crowell May 7 '19 at 22:57
  • $\begingroup$ IMHO. Excitations that cannot be excited do not exist. Modes do, sure, but not the (observable) excitations. Can you please tell me what the right answer would be? $\endgroup$ – Cryo May 7 '19 at 23:06
  • $\begingroup$ You can excite things without exciting them electromagnetically. I've written an answer. $\endgroup$ – Ben Crowell May 7 '19 at 23:08
  • $\begingroup$ But the OP asked about " it says closed shell nuclei cannot exhibit quadrupole oscillations because if their spherical symmetry (why not?)". You seem to be saying qaudrupole oscillations can be exhibited (in principle). This is not quite answering the question. My answer was that spherically symmetric nuclei cannot exhibit quadrupole excitations under electromagnetic interactions. $\endgroup$ – Cryo May 7 '19 at 23:16

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