Dipolar charge distributions are essentially all the same: regardless of how one adds up a combination of the form $$ \sigma(\theta,\varphi) = \operatorname{Re}\left[\sum_{m=-1}^1 a_m Y_{1m}(\theta,\varphi)\right], $$ you only ever get a rotated version of the canonical $Y_{10}(\theta,\varphi) \propto \cos(\theta)$, multiplied by some global constant.

Quadrupolar charge distributions, on the other hand, are more interesting, because you can have 'double peanut' distributions of the form $$ \sigma(\theta,\varphi) = \sin(\theta)\cos(\varphi)\cos(\theta) = \frac{xz}{r^2}, $$ but you can also have the usual two-balls-and-a-ring distribution of the form $$ \sigma(\theta,\varphi) = 3\cos^2(\theta)-1 = \frac{2z^2-x^2-y^2}{r^2}, $$ and those two are not rotations of each other. With that in mind, then:

  • If you consider two charge distributions to be equivalent if they differ only by a rigid rotation or by a global constant, how many real parameters do you need to describe quadrupolar charge distributions on the unit sphere of the form $$\sigma(\theta,\varphi) = \operatorname{Re}\left[\sum_{m=-2}^2 a_m Y_{2m}(\theta,\varphi)\right]?$$ What is the dimensionality of this manifold, and what is its topology? Can this reduction be done in a systematic way? And, if so, how? What is the physical significance of those parameters, and do they have set names in the literature?

  • Similarly, how do those questions look like for octupoles?

  • What about for general multipoles?

Edit: I guess, upon further reflection, what I'm asking for is the essential topology of the orbit space of the action of $\mathrm{SO}(3)$ on its irreducible representations: is the quotient space a manifold? if so, what are its dimensions and its topology? if not, why not? I would like to see this for arbitrary $\ell$ but with a specific emphasis on both $\ell=2$ and $\ell=3$, which strike me as the first two nontrivial examples. (I don't anticipate $\ell=4$ to be much more complicated than the octupole representation, but I do think that the octupole brings in nontrivial wrinkles compared to the quadrupole.) If people can comment on what happens for half-integral $j$ that would be awesome as well.

I am aware, in particular, of the reduction of the quadrupole representation by diagonalizing its coefficient matrix, but it is not at all clear to me how one would generalize this to the octupole layer's rank-3 tensors and beyond, and I would definitely like answers to address this generalization.

On the other hand, I do want the discussion to also address the particulars of these orbit spaces: what do the different points represent, and how do those charge distributions actually look like, at least at the 'extremal' points (like $Y_{20}$ and $\operatorname{Re}(Y_{22})$). The dimension-counting argument in Logan's answer is interesting, but it indicates that there are $2\ell-2$ non-equivalent distributions at $\ell\geq 2$, and this means that several of those dimensions are not encompassed by the usual spherical harmonics: since $\operatorname{Re}(Y_{\ell ,\pm m})$ and $\operatorname{Im}(Y_{\ell ,\pm m})$ are rotation-equivalent, when taken neat the $Y_{\ell m}$ can only produce up to $\ell+1$ distinct distributions (with one knocked out at $\ell=2$ by rotational equivalence), this means that from $\ell=4$ onwards there need to be at least $\ell-3$ independent linear combinations of the $Y_{\ell m}$ that are not rotationally equivalent to any of them. What do these combinations look like? I would like explicit examples for the first nontrivial cases, as well as systematic methods to get them for arbitrary $\ell$.

Now, I realize that this whole package is a big ask, but I do think it's interesting and worth exploring. I'll probably add some fake-internet-points sweetener in a few days, but I do want more in-depth answers than the current ones.


2 Answers 2


The answer for quadrupoles is $2$. The best way to think of a quadrupole is to consider the elements $Y_{2m}$ as linear combinations of entries in a symmetric traceless matrix: $$ {\cal Q}=\left(\begin{array}{ccc} r^2-x^2&xy&xz\\ xy&r^2-y^2&yz\\ xz&yz&r^2-z^2\end{array}\right) $$ Since this matrix is symmetric, it can be brought to diagonal form with a rotation, so up to rotations there remains two eigenvalues to parametrize your quadrupole. These are usually related to $Y_{20}(\theta,\phi)$ and some linear combinations of $Y_{2\pm 2}(\theta,\phi)$ components so that, in the usual notation $$ {\cal Q}_0\sim 2z^2-x^2-y^2 \, ,\qquad {\cal Q}_2\sim x^2-y^2\, . $$ When ${\cal Q}_2=0$ the figure is axially symmetric, either oblate or prolate depending on the sign of ${\cal Q}_0$. When ${\cal Q}_2\ne 0$ the figure has no symmetry axis (basically, a non-symmetric top.)

The principal octupole moment usually discussed in the literature is proportional to: $$ Y_{30}(\theta,\phi)\sim 5\cos(\theta)^3-3\cos(\theta)\, , $$ but there ought to be at least one more ($\sim Y_{33}(\theta,\phi)-Y_{3,-3}(\theta,\phi)$?) to probe the distribution in the $\phi$ direction.

The place to look for this is the nuclear physics literature since higher multipoles reveal information about nuclear shapes, but the standard textbooks I have handy (Ring and Schuck, Krane) do not discuss the $m\ne 0$ components of the higher multipoles. Octupole moments are used probe the "pear-shapedness" of the figure, and the pear being axially symmetric it may be that the $m\ne 0$ components are too small to bother.

(See also this post for additional comments).

  • $\begingroup$ Thanks for this. I would like a more in depth discussion of the general $\ell$ case (particularly since the diagonalization procedure is nowhere near as obvious) and of the topology of the manifold, though. $\endgroup$ Apr 17, 2017 at 15:07
  • $\begingroup$ Yeah... I know what you mean. I've never seen a discussion on the topology of this (presumably you would need to coset by $SO(3)$), and it just happens to be convenient that $Q$ is a symmetric tensor. The octupole is in the symmetric part of $(\ell=1)^{\otimes 3}$ but there is also another $L=1$ tensor in there and I don't see how one can easily get rid of it. Possibly you could fit the components as the symmetric part of a $4\times 4$ matrix but it's NOT clear at all. I will ask around but if you do find something please let me know. $\endgroup$ Apr 17, 2017 at 15:19

There is an analogous situation in quantum mechanics. It is of course well known that the finite-dimensional irreducible complex projective representations of $SO(3)$ are parametrized by a non-negative half-integer $s=0,1/2,1,3/2,\ldots$ with dimension $2s+1$. However, on these representations the projective action of $SO(3)$ is only transitive if $s=0$ or $s=1/2$. Understanding the orbits of $SO(3)$ on these representations is thus a question of interest.

A simple solution was found by Majorana, commonly known as the Majorana stellar representation. A general nonzero vector in the spin $s$ representation can be written as the symmetrized tensor product of $2s$ nonzero vectors in the spin-$1/2$ representation. This is unique up to scaling the vectors by factors which multiply to $1$ and by reordering the vectors. Passing to the projective spaces, a general state is represented by a collection of $2n$ unlabeled (not necessarily distinct) points on the Riemann sphere. The action of $SO(3)$ here is just rotations of the sphere.

Thus a dense subset of the projective space $P^{2s}$ is mapped to the nondegenerate configurations of $2s$ unlabelled points on a sphere, which is a well-studied example of a configuration space. The topology of configuration spaces is well-understood by mathematicians. For example, its fundamental group is the quotient of the braid group on $2s$ strands by a single relator (see this paper for a proof). That configuration space inherits the $SO(3)$ action and the orbit space is the quotient of that action. There are also degenerate orbits when $2$ or more of the points coincide, and these prevent the orbit space from being a legitimate manifold (it is, however, an orbifold).

To relate this back to multipole distributions, actually we don't really need to do much work at all. The $Y_l^m$ spherical harmonics for $m=-l,\ldots,l$ span a copy of the $s=l$ representation of $SO(3)$. Here we are looking at a real representation rather than a complex representation. As such we must restrict the space to the appropriate real section spanned by real spherical harmonics. Additionally, in this case it is an ordinary representation, not a projective representation. This means that we get $1$ degree of freedom back from rescaling by real numbers (rescaling by complex numbers is disallowed). Finally, the quotient by the $SO(3)$ action kills 3 real degrees of freedom for sufficiently large $l$ such that the action is faithful (in this case already $l=2$ is sufficient). In particular then, the dimensionality of the orbit space is, for $l=0,1,2,3,4,\ldots$, $$d_l=1,1,2,4,6,\ldots.$$

  • $\begingroup$ This is interesting, but I feel it doesn't go enough into detail. As in the edited question, when taken neat the $Y_{\ell m}$ provide at most $\ell +1$ non-equivalent distributions, but your dimension counting to $2\ell-2$ leaves a shortfall of $\ell-3>0$ (for $\ell\geq 4$) independent combinations of the $Y_{\ell m}$ that are not rotationally equivalent to any of them. What do these look like? $\endgroup$ Apr 17, 2017 at 22:00
  • $\begingroup$ @EmilioPisanty Unfortunately I don't really know. Actually I'm not so sure about my counting now after reading your edits and rereading my answer. It seems that the part where you take the "real section" may be more complicated than I anticipated. The case of complex projective representations is quite a bit simpler but I'm not sure how to extract the data for the real representations from it. I will keep thinking about it. $\endgroup$
    – Logan M
    Apr 17, 2017 at 22:24

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