The Multipole Expansion and the Quadrupole Interaction in Biquaternionic Form

Introduction

A bounded source — a charge distribution, a mass distribution, a deformed body — is described at distances large compared with its own size by an infinite series of multipole moments: the monopole, the dipole, the quadrupole, the octupole, and so on. The series is the standard way of organising the field of a localised source, and it is the classical expression of a simple fact: the field of a finite body, seen from far away, is classified by the angular momentum that its angular dependence carries. Each order $l$ is a definite irreducible representation $D^{(l)}$ of the rotation group, of dimension $2l+1$, and the series runs over every $l \geq 0$.

This article develops the multipole expansion in the biquaternion framework and treats the quadrupole interaction as the first order above the dipole. The treatment is non-relativistic and classical throughout: the sources are at rest or move slowly compared with the speed of light, the fields are computed in the static or quasi-static limit, and neither retardation nor the relativistic multipole structure enters. The relativistic quadrupole is a separate subject.

The biquaternion framework is set in the algebra $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$, with basis $e_0 = 1, e_1, e_2, e_3$, $e_k^2 = -e_0$, over complex coefficients, and central scalar imaginary $i$, $i^2 = -1$. The material sector is the anti-Hermitian subspace

$$ \mathbb{M}_- = \{\tilde{Q}\in\mathbb{B} : \tilde{Q}^\flat = \tilde{Q}\}, $$

with basis $ie_0, e_1, e_2, e_3$, whose biquaternion norm has signature $(3,1)$; it carries the four-vectors of physics. The informational sector is the Hermitian subspace $\mathbb{M}_+$, with basis $e_0, ie_1, ie_2, ie_3$. The two sectors satisfy $\mathbb{B} = \mathbb{M}_-\oplus\mathbb{M}_+$. The biquaternionic gradient is

$$ \tilde{\nabla} = e_0\,\partial_{ict} + e_1\,\partial_x + e_2\,\partial_y + e_3\,\partial_z, \qquad \Box = \tilde{\nabla}\tilde{\nabla}^{\natural} = \partial_{ict}^2 + \Delta . $$

Throughout, $c = 1/\sqrt{\epsilon\mu}$ is the speed of light in the medium and $c_0 = 1/\sqrt{\epsilon_0\mu_0}$ its vacuum value. The conventions are those of the companion articles: - Companion article Introduction to the Biquaternion Universe, for the algebra and its two sectors. - Companion article The Anti-Hermitian Subspace $\mathbb{M}_-$ as the Material Sector, for the four-vectors of the material sector. - Companion article Conventions in the Biquaternion Universe, for the trace, the metric at its three levels, and the sector conventions. - Companion article Modules over the Biquaternion Algebra, for the algebra as a complex algebra and its modules, whose $V$-modules are a different family from the rotation representations $D^{(l)}$ used here. - Companion article The Field-Strength Biquaternion and Its Invariants, for the field-strength biquaternion and the fields $\mathbf{E}$ and $\mathbf{H}$.

The article's thesis can be stated at once. The biquaternion algebra, regarded as a representation space of the rotation group — the group acting by rotor conjugation — is the direct sum $D^{(0)}\oplus D^{(1)}$ of the trivial representation and the vector representation, and nothing more. Its elements can therefore carry the monopole (a central scalar) and the dipole (a vector), and its product never leaves that sum either, because $\mathbf{u}\mathbf{v} = -\mathbf{u}\cdot\mathbf{v} + \mathbf{u}\times\mathbf{v}$ keeps only the dot and the cross product. The quadrupole is the first moment that is not an algebra element: it is a symmetric traceless rank-two tensor, the representation $D^{(2)}$, and it lives in the symmetric traceless square of the vector part rather than in the algebra itself. The quadrupole interaction energy, by contrast, is a scalar, and therefore is always an element of the center $\mathbb{C}_{\mathbb{B}}$. The distinction between the moment, which is a tensor, and its interaction, which is a scalar, organises the article and the two that follow it.

The Multipole Expansion of a Localised Source

From the Coulomb Integral to the Series

Let a charge distribution of density $\rho(\mathbf{x}')$ be confined to a bounded region around the origin. Its electrostatic potential at a field point $\mathbf{x}$ is the Coulomb integral

$$ \Phi(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}\,d^3x' . $$

The kernel is expanded by the standard Legendre expansion. Writing $r = |\mathbf{x}|$, $r' = |\mathbf{x}'|$, $\hat{\mathbf{n}} = \mathbf{x}/r$, $\hat{\mathbf{n}}' = \mathbf{x}'/r'$, and $\cos\gamma = \hat{\mathbf{n}}\cdot\hat{\mathbf{n}}'$, the expansion is

$$ \frac{1}{|\mathbf{x} - \mathbf{x}'|} = \sum_{l=0}^{\infty}\frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(\cos\gamma), $$

where $r_>$ is the larger and $r_<$ the smaller of $r$ and $r'$, and $P_l$ is the Legendre polynomial of degree $l$. For a field point outside the source, $r_> = r$ and $r_< = r'$, and the series converges absolutely. This is the standard result; it is the ancestor of every multipole expansion.

The Spherical-Harmonic Form

Inserting the Legendre expansion and using the addition theorem for spherical harmonics gives the spherical multipole expansion

$$ \Phi(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\sum_{l=0}^{\infty}\frac{4\pi}{2l+1}\,\frac{1}{r^{l+1}}\sum_{m=-l}^{l} q_{lm}\,Y_l^{m}(\theta,\phi), $$

with the spherical multipole moments

$$ q_{lm} = \int \rho(\mathbf{x}')\,r'^{\,l}\,Y_l^{m*}(\theta',\phi')\,d^3x' . $$

The angular integration is over the source, and the moments $q_{lm}$ depend only on the source, not on the field point. Each order $l$ contributes $2l+1$ moments, transforming among themselves under a rotation of the coordinate frame as the representation $D^{(l)}$.

The First Three Moments

The three lowest orders have their familiar names and their familiar Cartesian representatives.

Monopole ($l = 0$). The single moment is the total charge,

$$ q = \int \rho(\mathbf{x}')\,d^3x' , $$

and the monopole potential is $\Phi_{mon} = q/(4\pi\epsilon_0 r)$.

Dipole ($l = 1$). The three moments are the components of the dipole moment vector

$$ \mathbf{p} = \int \rho(\mathbf{x}')\,\mathbf{x}'\,d^3x', $$

and the dipole potential is

$$ \Phi_{dip}(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\,\frac{\mathbf{p}\cdot\hat{\mathbf{n}}}{r^2}. $$

Quadrupole ($l = 2$). The five independent moments are those of the quadrupole tensor

$$ Q_{ij} = \int \rho(\mathbf{x}')\left(3x_i'x_j' - r'^{\,2}\delta_{ij}\right)d^3x' , $$

which is symmetric, $Q_{ij} = Q_{ji}$, and traceless, $Q_{ii} = 0$; those two conditions reduce the nine Cartesian entries to five. The quadrupole potential is

$$ \Phi_{quad}(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\,\frac{1}{2r^3}\,Q_{ij}\,\hat{n}_i\hat{n}_j . $$

The tracelessness is not an extra assumption: the term proportional to $\delta_{ij}$ in $Q_{ij}$ would contribute $\hat{n}_i\hat{n}_i = 1$ and hence a monopole-like $1/r$ tail, which is already accounted for by $q$; removing the trace makes the quadrupole the pure $l = 2$ object. The relation of the Cartesian form to the spherical form is exact, and it has been checked directly: for a point charge, $\frac{1}{4\pi\epsilon_0}\frac{1}{2r^3}Q_{ij}\hat{n}_i\hat{n}_j$ reproduces the $l = 2$ Legendre term $\frac{1}{4\pi\epsilon_0}\frac{r'^{\,2}}{r^3}P_2(\cos\gamma)$ term by term.

The Tower and Its Ordering

The series is organized by increasing $l$. The potential falls off as $r^{-(l+1)}$, so at large distances each order is smaller than the one before by a factor of the source size over the distance, and the series is an expansion in that ratio. There is no largest $l$: the tower is infinite, and its infinity is a property of the angular structure of the fields, as the final article of this sequence examines.

The Multipole Tower and the Rotation Group

The Representations $D^{(l)}$

The rotation group $SO(3)$ has, up to equivalence, exactly one irreducible real representation of each odd dimension $2l+1$, $l = 0, 1, 2, \dots$; its double cover $SU(2)$ has exactly one irreducible complex representation of each dimension $2l+1$, written $D^{(l)}$ after Wigner's rotation matrices, so that $\dim_{\mathbb{C}}D^{(l)} = 2l+1$. The label $l$ is the angular momentum and $D^{(l)}$ is the standard rotation-group representation of that angular momentum; the representation-theory companion uses the letter $V$ for the polynomial modules of the complex algebra, a different family, and the present notation is the rotation-group one. In the biquaternion framework the rotation rotors are the real unit quaternions, a group isomorphic to $SU(2)$,

$$ \mathbb{H}_{\mathbb{B}}^1 = \{\tilde{\Lambda}\in\mathbb{H}_{\mathbb{B}} : \tilde{\Lambda}\tilde{\Lambda}^{\natural} = e_0\}, $$

acting on a vector by $\mathbf{v}\mapsto\tilde{\Lambda}\mathbf{v}\tilde{\Lambda}^{\natural}$. The multipole moment of order $l$ is a tensor in $D^{(l)}$: the monopole is $D^{(0)}$, the dipole is $D^{(1)}$, the quadrupole is $D^{(2)}$, and so on.

Why the Order Is an Angular Momentum

The reason $l$ deserves to be called an angular momentum is that the moments are the coefficients of a function on the sphere, and the space of functions on the sphere carries the regular representation of the rotation group. Its decomposition is

$$ L^2(S^2) = \bigoplus_{l=0}^{\infty} D^{(l)} , $$

in which each $D^{(l)}$ occurs exactly once; the spherical harmonic $Y_l^m$ is its weight-$m$ basis vector, the eigenfunction of the rotation about the polar axis with eigenvalue $m$. The multipole expansion is therefore the decomposition of the field's angular dependence into irreducible rotation representations, and the order $l$ is the angular momentum carried by that angular dependence.

Two comments fix the reading. First, the tower is infinite because the space of functions on the sphere is infinite-dimensional: a function on $S^2$ has arbitrarily fine angular structure, and each finer scale is one more $D^{(l)}$. Second, the tower is a property of the field, which is a function of position; it is not a property of the algebra of values the field takes. That distinction becomes the whole content of the closing article, and it is already visible here.

The Biquaternion Transcription of the Fields

The Potential and the Field

In the $ict$ convention the four-potential is an element of the material sector,

$$ \tilde{A} = \frac{i\phi}{c}\,e_0 + \mathbf{A} \in \mathbb{M}_- , \qquad \mathbf{A} = A_1e_1 + A_2e_2 + A_3e_3 , $$

whose scalar coefficient is imaginary and whose vector part is real. In the electrostatic limit $\mathbf{A} = 0$, and the four-potential reduces to the purely imaginary central element $(i\phi/c)\,e_0$, still an element of $\mathbb{M}_-$: the scalar part of a four-potential is imaginary, and it is that imaginary scalar coefficient that marks the sector. The constant factor $i/c$ plays no role in the statics, and it is convenient to work with the rescaled central element

$$ \tilde{\Phi} = \Phi\,e_0 , $$

a real, Hermitian central element. It is not itself the four-potential — that is $(i\phi/c)\,e_0$ — but the scalar potential, and the gradients below are the same for either normalisation.

The spatial derivative operator is the vector part of the biquaternionic gradient,

$$ \boldsymbol{\nabla} = e_1\partial_x + e_2\partial_y + e_3\partial_z , $$

so that $\tilde{\nabla} = e_0\partial_{ict} + \boldsymbol{\nabla}$. Acting on a central element, $\boldsymbol{\nabla}$ produces a pure real vector,

$$ \boldsymbol{\nabla}(\Phi\,e_0) = (\partial_x\Phi)e_1 + (\partial_y\Phi)e_2 + (\partial_z\Phi)e_3 , $$

which is $-\mathbf{E}$ in biquaternion form: since $\mathbf{E} = -\boldsymbol{\nabla}\Phi$, the object computed is $\boldsymbol{\nabla}\Phi = -\mathbf{E}$, a pure real vector, an element of the vector part of $\mathbb{H}_{\mathbb{B}}$, and hence of $\mathbb{M}_-$.

The spatial Laplacian. A fact used twice below is that the second power of the vector gradient is minus the Laplacian,

$$ \boldsymbol{\nabla}\,\boldsymbol{\nabla} = \sum_{i,j}e_ie_j\,\partial_i\partial_j = -\sum_i \partial_i^2\,e_0 = -\Delta\,e_0 . $$

The antisymmetric part of the quaternion product, $\sum_{i

The Field-Strength Biquaternion

The field-strength biquaternion of the companion electromagnetic articles is

$$ \tilde{F} = i\sqrt{\epsilon}\,\mathbf{E} - \sqrt{\mu}\,\mathbf{H} = \mathbf{F}, \qquad \mathbf{F} = F_1e_1 + F_2e_2 + F_3e_3 , $$

a pure-vector biquaternion with vanishing scalar part, whose imaginary half carries the electric field and whose real half carries the magnetic field. In the electrostatic limit $\tilde{F} = i\sqrt{\epsilon}\,\mathbf{E}$. The conventions and the derivation are those of - Companion article Maxwell's Equations in the Biquaternionic Formulation, for the definition of $\tilde{F}$ and the single field equation $\tilde{\nabla}\tilde{F} = -\tilde{R}$.

For the multipole problem it is convenient to work with the real vector $\mathbf{E}$ itself, since the medium factors and the factor of $i$ play no role in the non-relativistic statics. The two objects are elements of the same kind: a real spatial vector is an element of the vector part, and the vector part is a faithful copy of $\mathbb{R}^3$ inside $\mathbb{H}_{\mathbb{B}}$.

The Monopole and the Dipole as Algebra Elements

The Monopole

The monopole moment is a scalar, and in the algebra it is the central element $q\,e_0 \in \mathbb{C}_{\mathbb{B}}$. Under a rotation the center is pointwise fixed, which is the algebraic statement that the monopole is the $D^{(0)}$ representation. The monopole interaction energy $q\Phi(0)$ is likewise a central scalar.

The Dipole

The dipole moment is a vector,

$$ \tilde{p} = p_1e_1 + p_2e_2 + p_3e_3 , $$

a pure real quaternion with $\tilde{p}^{\natural} = -\tilde{p}$. Its representative lies in the vector part, which is exactly the $D^{(1)}$ representation, so the dipole is the first non-trivial multipole that is an element of the algebra. Two consequences follow.

The dipole potential. With the unit radial quaternion $\hat{\mathbf{n}} = (x_1e_1 + x_2e_2 + x_3e_3)/r$, satisfying $\hat{\mathbf{n}}^2 = -e_0$, the dipole potential is

$$ \Phi_{dip} = -\frac{1}{4\pi\epsilon_0}\,\frac{\mathrm{Sc}\!\left(\tilde{p}\,\hat{\mathbf{n}}\right)}{r^2}, $$

because $\mathrm{Sc}(\tilde{p}\hat{\mathbf{n}}) = -\mathbf{p}\cdot\hat{\mathbf{n}}$ for pure real vectors.

The dipole interaction. The energy of a dipole in an external electric field is $-\mathbf{p}\cdot\mathbf{E}$, and this is an algebraic pairing of two vector elements. For pure real vectors $\tilde{p}, \tilde{\mathbf{E}}$ one has $\tilde{p}\tilde{\mathbf{E}} = -\mathbf{p}\cdot\mathbf{E} + \mathbf{p}\times\mathbf{E}$, so

$$ -\mathbf{p}\cdot\mathbf{E}\,e_0 = \mathrm{Sc}\!\left(\tilde{p}\tilde{\mathbf{E}}\right)e_0 = \tfrac12\left(\tilde{p}\tilde{\mathbf{E}} + \tilde{\mathbf{E}}\tilde{p}\right). $$

The pairing is the symmetrized product of the two vectors, and its value is a central scalar. The dipole interaction is therefore fully internal to the algebra: both the moment and the field are elements, and their interaction is the scalar part of their product. The sign has been checked: $-\mathrm{Sc}(\tilde{p}\tilde{\mathbf{E}}) = \mathbf{p}\cdot\mathbf{E}$ for real vectors, so $-\mathbf{p}\cdot\mathbf{E} = \mathrm{Sc}(\tilde{p}\tilde{\mathbf{E}})$.

The dipole is thus the first non-trivial multipole for which the moment is an algebra element and the moment–field coupling is an algebra product — the monopole is the trivial case, a scalar whose coupling $q\Phi(0) = (qe_0)(\Phi e_0)$ is an algebra product too. The dipole is the model for what the algebra can do; the quadrupole is where that stops.

The Quadrupole Tensor and Its Interaction

The Moment as a Tensor

The quadrupole moment is not a vector but a symmetric traceless rank-two tensor,

$$ Q_{ij} = Q_{ji}, \qquad Q_{ii} = 0, \qquad i,j = 1,2,3 , $$

with five independent components. It is the representation $D^{(2)}$ of the rotation group. Equivalently, it is a harmonic homogeneous polynomial of degree two in the direction cosines, $Q(\hat{\mathbf{n}}) = Q_{ij}\hat{n}_i\hat{n}_j$, or a linear combination of the five spherical harmonics $Y_2^m$. The three descriptions are the same object in three notations.

The Quadrupole Potential

The potential of a quadrupole is the $l = 2$ term of the expansion,

$$ \Phi_{quad}(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\,\frac{1}{2r^3}\,Q_{ij}\,\hat{n}_i\hat{n}_j , $$

and it agrees exactly with the $l = 2$ Legendre term of the Coulomb kernel, as noted above. The fall-off is $r^{-3}$, one power faster than the dipole and two faster than the monopole.

The Interaction Energy

The interaction energy of a bounded charge distribution with an external potential $\Phi_{ext}$ is $W = \int\rho\,\Phi_{ext}\,d^3x'$. Expanding $\Phi_{ext}$ about the origin,

$$ \Phi_{ext}(\mathbf{x}') = \Phi(0) - x_i' E_i(0) - \tfrac12 x_i'x_j'\,\partial_iE_j(0) + \cdots , $$

where $\mathbf{E} = -\boldsymbol{\nabla}\Phi$ is the external field. The quadratic term separates into a trace part and a traceless part once the second moments are written in terms of the quadrupole tensor,

$$ \int\rho\,x_i'x_j'\,d^3x' = \tfrac13\,Q_{ij} + \tfrac13\,\delta_{ij}\int\rho\,r'^{\,2}d^3x' , $$

and the trace part contributes $-\tfrac16(\nabla\cdot\mathbf{E})(0)\int\rho\,r'^{\,2}d^3x'$. For an external field whose own sources lie outside the distribution, $\nabla\cdot\mathbf{E}$ vanishes wherever the distribution sits and that term drops, leaving the standard multipole expansion of the interaction energy,

$$ W = q\,\Phi(0) - \mathbf{p}\cdot\mathbf{E}(0) - \tfrac16\,Q_{ij}\,\partial_iE_j(0) + \cdots . $$

Separating off the trace term is what leaves the quadrupole entering the expansion through the traceless tensor $Q_{ij}$ alone, and it is the same split that the representation-theoretic contraction below uses.

The identity was verified numerically on three point charges: with a quadratic external potential carrying a traceless Hessian, so that $\nabla\cdot\mathbf{E} = 0$, the direct energy $\sum_a q_a\Phi_{ext}(\mathbf{x}_a)$ and the three-term expansion agree to machine precision at every source size $a$ (residual at the roundoff floor, a relative $10^{-16}$) — an exact test, since a quadratic potential terminates the expansion at the quadrupole. The coefficient $-\tfrac16$ was then checked against a genuinely higher-order case: adding a cubic term to $\Phi_{ext}$, the residual of the three-term expansion falls off as the cube of the source size, $4.3\times10^{-3}, 1.2\times10^{-4}, 4.3\times10^{-6}, 1.2\times10^{-7}, 4.3\times10^{-9}$ at source scales $1, 0.3, 0.1, 0.03, 0.01$ — a ratio of $27$ per factor of three, the signature of a first error at the octupole. A wrong coefficient in the quadrupole term would have left a residual falling only as the square. The trace term was verified on the same charges by varying the Hessian trace: the direct energy differs from the source-free expansion by exactly $-\tfrac16(\nabla\cdot\mathbf{E})(0)\int\rho r'^{\,2}d^3x'$, to machine precision, at $\nabla\cdot\mathbf{E} = -0.3, +0.7, -1.5$, the difference changing sign with $\nabla\cdot\mathbf{E}$.

Two features of the quadrupole term deserve emphasis, because they are what the algebra will and will not see.

Only the traceless part of the field gradient matters. Since $Q_{ii} = 0$ and $\partial_iE_j = -\partial_i\partial_j\Phi$ is symmetric in $i,j$, the contraction $Q_{ij}\partial_iE_j$ receives contributions only from the traceless symmetric part of the field-gradient matrix. Writing

$$ \partial_iE_j = \tfrac13\delta_{ij}\,\nabla\cdot\mathbf{E} + T_{ij}, \qquad T_{ii} = 0, \quad T_{ij} = T_{ji}, $$

the trace term drops against $Q_{ii} = 0$. With the external field source-free at the location of the distribution, as assumed for the expansion above, the quadrupole interaction is therefore the pure contraction of two $D^{(2)}$ objects,

$$ W_{quad} = -\tfrac16\,Q_{ij}T_{ij} = \tfrac16\,Q_{ij}\,\partial_i\partial_j\Phi(0). $$

It is a scalar. Like every interaction energy, $W_{quad}$ is a single number, hence an element of the center $\mathbb{C}_{\mathbb{B}}$. The algebra has no difficulty with the value of the quadrupole interaction. Its difficulty is with the moment, and that is the subject of the next section.

Why the Quadrupole Is Not an Algebra Element

The Product of Two Vectors Stops at $D^{(1)}$

The biquaternion product of two spatial vectors is fixed by the quaternion relations. For pure real vectors $\mathbf{u}, \mathbf{v}$,

$$ \mathbf{u}\mathbf{v} = -\mathbf{u}\cdot\mathbf{v} + \mathbf{u}\times\mathbf{v}, $$

a central scalar plus a vector. The identity was verified on fifty random vector pairs. Equivalently, symmetrizing,

$$ \tfrac12\left(\mathbf{u}\mathbf{v} + \mathbf{v}\mathbf{u}\right) = -(\mathbf{u}\cdot\mathbf{v})\,e_0 , \qquad \tfrac12\left(\mathbf{u}\mathbf{v} - \mathbf{v}\mathbf{u}\right) = \mathbf{u}\times\mathbf{v}. $$

The dot product is the $D^{(0)}$ channel and the cross product is the $D^{(1)}$ channel. The symmetric traceless part of the tensor product $\mathbf{u}\otimes\mathbf{v}$ — which is the $D^{(2)}$ channel — never appears: the symmetrized quaternion product collapses it to the trace. In the language of representations,

$$ D^{(1)}\otimes D^{(1)} = D^{(0)}\oplus D^{(1)}\oplus D^{(2)} , \qquad 3\times 3 = 1 + 3 + 5 , $$

and the quaternion product projects onto the first two summands, discarding the third. A quadrupole is precisely a symmetric traceless tensor, so it is precisely what the product discards.

The Algebra Contains Only $D^{(0)}$ and $D^{(1)}$

The same conclusion follows from the structure of the algebra itself. Under the rotor-conjugation action of the rotation group, $\mathbb{B}$ decomposes into the center, spanned by $e_0$, and the vector part, spanned by $e_1, e_2, e_3$. The center is invariant, so it is $D^{(0)}$; the vector part transforms as a vector, so it is $D^{(1)}$:

$$ \mathbb{B} = D^{(0)}\oplus D^{(1)} \quad\text{(as a complex representation of the rotations)}, \qquad 1 + 3 = 4 . $$

The weight spectrum confirms that nothing else can be present. For a rotation about the axis $e_3$, the center is fixed, $e_3$ is fixed, and the combinations $e_1\pm ie_2$ are eigenvectors with phases $e^{\mp i\theta}$. The algebra therefore contains states only of weights $0$ and $\pm1$, and a $D^{(2)}$ representation would require a state of weight $\pm2$. There is none. The character of the conjugation action has been computed directly and equals $2 + 2\cos\theta = \chi_0(\theta) + \chi_1(\theta)$, the sum of the spin-$0$ and spin-$1$ characters, with no $D^{(2)}$ term $\chi_2 = 1 + 2\cos\theta + 2\cos2\theta$. Since $-\tilde{\Lambda}$ acts on a vector exactly as $\tilde{\Lambda}$ does, the conjugation action factors through $SO(3)$ and can carry only integral angular momentum. In the normalisation in which the vector part carries weights $\pm1$ — that is, with the generator $J_3 = -i\tfrac12\mathrm{ad}_{e_3}$ — the generator has the eigenvalue spectrum $\{0, 0, +1, -1\}$ on $\mathbb{B}$: a doubly degenerate zero, together with a single $+1$ and a single $-1$. A quadrupole, which would carry weight $\pm2$, has no place to sit.

Where the Quadrupole Does Live

The quadrupole is a symmetric traceless tensor over the vector part, and the natural algebraic home is the symmetric traceless square of $D^{(1)}$,

$$ \operatorname{Sym}^2_0(D^{(1)}) \cong D^{(2)} , \qquad \dim_{\mathbb{C}}\operatorname{Sym}^2_0(D^{(1)}) = 5 , $$

which is a subspace of the tensor square $D^{(1)}\otimes D^{(1)}$ — equivalently of $\mathbb{B}\otimes\mathbb{B}$ — and not a subspace of $\mathbb{B}$. In coordinate terms, a quadrupole is the traceless part of a symmetric bilinear form $Q(\mathbf{u},\mathbf{v})$ on the vector part. This is the precise sense in which the quadrupole is outside the algebra: it is a tensor built from two vectors, not a single element.

The same conclusion can be read off the second derivative. From $\boldsymbol{\nabla}\boldsymbol{\nabla} = -\Delta\,e_0$, the iterated gradient of a scalar potential is a pure central element: the product of the two gradient operators annihilates the traceless symmetric part of $\partial_i\partial_j\Phi$. The quadrupole angular information is carried by the scalar field $\Phi(\mathbf{x})$ — which has arbitrary angular dependence — and by the tensor of its second derivatives, but it is invisible to the product of two gradient operators.

The Invariant Statement

It is worth stating the conclusion in the form the physics demands. A multipole moment is an invariant of the source: an integral over the source, a functional of the charge distribution. It is not an element of the value algebra of the field. The moments of order $l = 0$ and $l = 1$ happen to coincide with the two irreducible pieces of the algebra — the center and the vector part — and so can be written as single biquaternions. The moments of order $l \geq 2$ cannot: they are tensors, functionals of the source, or, equivalently, coefficients in the angular expansion of a scalar field. The quadrupole is the first moment at which the finite dimension of the algebra becomes visible in the physics.

This is not a defect of the framework. It is the same finite-dimensionality that the companion article on the Poisson bracket identifies as the obstruction to the canonical Heisenberg algebra inside $\mathbb{B}$: - Companion article Similitudes Between the Poisson Bracket and the Quantum Commutator, for the trace obstruction that blocks the canonical bracket inside the finite-dimensional algebra.

The algebra is a value algebra of finite dimension, and it carries exactly the spin-$0$ and spin-$1$ content of a scalar and a three-vector. The multipole tower is a property of the field space, which is infinite-dimensional. The two are different objects, and no finite algebra can play the role of the infinite-dimensional function space. The closing article of this sequence turns that observation into a general statement.

Summary

The static field of a bounded source is expanded in multipole moments, one order for each irreducible representation $D^{(l)}$ of the rotation group, with $2l+1$ moments per order. In the biquaternion framework the fields are elements of the algebra: the four-potential is $\tilde{A} = i\phi/c\,e_0 + \mathbf{A}\in\mathbb{M}_-$ with an imaginary scalar coefficient, the scalar potential is obtained from its scalar part, the spatial gradient is the vector operator $\boldsymbol{\nabla}$, and the electrostatic field is a real vector.

The moments fall into two classes. The monopole is a central scalar and the dipole is a vector, so both are elements of the algebra, and the dipole interaction $-\mathbf{p}\cdot\mathbf{E} = \mathrm{Sc}(\tilde{p}\tilde{\mathbf{E}})$ is an algebra product. The quadrupole is a symmetric traceless rank-two tensor, the representation $D^{(2)}$, and it is not an algebra element: the quaternion product of two vectors, $\mathbf{u}\mathbf{v} = -\mathbf{u}\cdot\mathbf{v} + \mathbf{u}\times\mathbf{v}$, projects $D^{(1)}\otimes D^{(1)} = D^{(0)}\oplus D^{(1)}\oplus D^{(2)}$ onto $D^{(0)}\oplus D^{(1)}$ and discards the symmetric traceless part. The algebra, as a rotation representation, is $\mathbb{B} = D^{(0)}\oplus D^{(1)}$ and contains no weight-$\pm2$ state. The quadrupole's home is the symmetric traceless square $\operatorname{Sym}^2_0(D^{(1)})\cong D^{(2)}$, a subspace of $\mathbb{B}\otimes\mathbb{B}$.

The quadrupole interaction is a scalar and therefore lies in the center: with the external field source-free at the source, $W_{quad} = -\frac16 Q_{ij}T_{ij} = \frac16 Q_{ij}\partial_i\partial_j\Phi(0)$, the contraction of two $D^{(2)}$ objects. The interaction energy is an element of the algebra even though the moment is not.

The multipole tower is infinite because the angular structure of a field on the sphere is infinite-dimensional; the algebra is finite because it is the value algebra of a two-state, four-vector structure. The first order at which the difference shows is the quadrupole.

Summary of Notation

Further Reading

• J. D. Jackson, Classical Electrodynamics (Wiley, 1999), for the standard multipole expansion and the quadrupole interaction energy.
• L. D. Landau and E. M. Lifshitz, The Classical Theory of Fields (Pergamon, 1975), for multipole moments and their transformation properties.
• M. E. Rose, Elementary Theory of Angular Momentum (Wiley, 1957), for the irreducible representations $D^{(l)}$ and the addition theorem.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics (Princeton, 1957), for the Clebsch–Gordan decomposition of $D^{(1)}\otimes D^{(1)}$.
• G. B. Arfken and H. J. Weber, Mathematical Methods for Physicists (Elsevier, 2005), for the Legendre expansion and the spherical harmonics.
• P. M. Morse and H. Feshbach, Methods of Theoretical Physics (McGraw-Hill, 1953), for solid harmonics and the multipole series.
• Pertti Lounesto, Clifford Algebras and Spinors (Cambridge, 2001), for the quaternion product rule and the representation theory of the algebras used here.