Multipole Operators

Multipole operators organize how a localized quantum system couples to long-wavelength electromagnetic probes. For selection rules, the most important facts are not the detailed normalization constants but the transformation labels:

• rotational rank $\lambda$;
• spherical component $q=-\lambda,\ldots,\lambda$;
• parity, even or odd under spatial inversion.

An electric multipole of order $\lambda$ is denoted $E\lambda$. A magnetic multipole of order $\lambda$ is denoted $M\lambda$. The familiar examples are electric dipole $E1$, magnetic dipole $M1$, and electric quadrupole $E2$.

This page classifies multipole operators by symmetry. Atomic Selection Rules applies $E1$, $M1$, and $E2$ rules to electronic and hyperfine levels, polarization, metastability, and state mixing. Transition rates, oscillator strengths, line shapes, and detailed spectroscopy require dynamical matrix elements and belong to transition-theory and spectroscopy pages.

Rotational Rank and Parity of Multipoles

The standard symmetry labels are:

Operator	Rotational rank	Parity
$E\lambda$	$\lambda$	$(-1)^\lambda$
$M\lambda$	$\lambda$	$(-1)^{\lambda+1}$

Thus:

Multipole	Rank	Parity	Usual name
$E0$	$0$	even	electric monopole
$E1$	$1$	odd	electric dipole
$M1$	$1$	even	magnetic dipole
$E2$	$2$	even	electric quadrupole
$M2$	$2$	odd	magnetic quadrupole
$E3$	$3$	odd	electric octupole

The rank controls angular-momentum selection rules. The parity controls whether the operator connects states of the same or opposite parity.

Electric Multipoles

For point charges in a fixed origin convention, the electric multipole operator can be written schematically as

$$Q_q^{(\lambda)} = \sum_a e_a r_a^\lambda C_q^{(\lambda)}(\hat{\mathbf r}_a), \qquad q=-\lambda,\ldots,\lambda,$$

where $C_q^{(\lambda)}$ is a Racah-normalized spherical harmonic:

$$C_q^{(\lambda)}(\hat{\mathbf r}) = \sqrt{\frac{4\pi}{2\lambda+1}}\, Y_\lambda^q(\hat{\mathbf r}).$$

The exact normalization differs across atomic, molecular, nuclear, and radiation conventions. The symmetry content is stable: $Q_q^{(\lambda)}$ is an irreducible spherical tensor of rank $\lambda$.

Under parity,

$$\mathbf r\mapsto-\mathbf r, \qquad Y_\lambda^q(-\hat{\mathbf r}) = (-1)^\lambda Y_\lambda^q(\hat{\mathbf r}),$$

so

$$\Pi Q_q^{(\lambda)}\Pi^{-1} = (-1)^\lambda Q_q^{(\lambda)}.$$

That is the parity rule for $E\lambda$.

For $\lambda=0$ this static moment is simply the total charge, $Q_0^{(0)}=\sum_a e_a$. In a fixed-charge sector it is proportional to the identity and has no matrix element between distinct orthogonal states. The name E0 transition in nuclear spectroscopy refers to scalar charge-density transition operators such as $\sum_a e_ar_a^2$, not to this constant total charge. Such transitions can proceed through internal conversion or, above threshold, electron–positron pair creation; a single real photon has no electric-monopole radiation mode (Martin, 2016, Appendix D).

Electric Dipole

The electric dipole moment is

$$\mathbf d = \sum_a e_a\mathbf r_a.$$

It is a rank-$1$ polar vector and is parity odd:

$$\Pi\mathbf d\Pi^{-1} = -\mathbf d.$$

In spherical components,

$$d_0=d_z, \qquad d_{\pm1} = \mp\frac{1}{\sqrt2}(d_x\pm id_y).$$

The electric-dipole selection rules are developed in Dipole Transitions. In short, $E1$ transitions obey rank-$1$ angular rules and require opposite parity.

Electric Quadrupole

The electric quadrupole is the rank-$2$ part of two powers of position. In Cartesian form a common traceless quadrupole tensor is

$$Q_{ij} = \sum_a e_a \left( 3r_{a,i}r_{a,j} - \delta_{ij}r_a^2 \right).$$

It is symmetric and traceless:

$$Q_{ij}=Q_{ji}, \qquad \sum_i Q_{ii}=0.$$

The five independent components correspond to a rank-$2$ irreducible spherical tensor $Q_q^{(2)}$ with $q=-2,-1,0,1,2$. Since it contains two powers of position, it is parity even:

$$\Pi Q_q^{(2)}\Pi^{-1} = Q_q^{(2)}.$$

Thus electric-quadrupole matrix elements connect states of the same parity, subject to rank-$2$ angular selection rules.

Magnetic Multipoles

Magnetic multipoles involve currents, orbital angular momentum, spin, and magnetization rather than only charge density. Their detailed forms depend more strongly on the physical system and approximation.

For symmetry work, the key fact is that magnetic multipoles have parity

$$\eta_{M\lambda} = (-1)^{\lambda+1}.$$

The simplest example is the magnetic dipole $M1$. For electrons in a nonrelativistic atom, neglecting nuclear magnetism and orbital corrections, the leading operator is

$$\boldsymbol\mu = -\frac{\mu_B}{\hbar} \left( \mathbf L+g_s\mathbf S \right),$$

where $\mu_B=e\hbar/(2m_e)>0$, $g_s\simeq2$ is positive, and $\mathbf L,\mathbf S$ carry angular-momentum units. Other particles have their own charge, mass, and magnetic-moment coefficients. Both $\mathbf L$ and $\mathbf S$ are axial vectors, so $\boldsymbol\mu$ is parity even:

$$\Pi\boldsymbol\mu\Pi^{-1} = \boldsymbol\mu.$$

Therefore $M1$ transitions connect states of the same parity, subject to rank-$1$ angular rules.

Angular Selection Rules

If a multipole operator is an irreducible tensor of rank $\lambda$, the Wigner–Eckart theorem gives the angular constraints

$$|J_i-\lambda| \le J_f \le J_i+\lambda$$

and

$$M_f=M_i+q, \qquad q=-\lambda,\ldots,\lambda.$$

The triangle rule is necessary, not sufficient. The reduced matrix element may vanish for dynamical reasons or because of additional symmetries.

Some common low-rank consequences are:

Multipole	Angular rank rule	Parity rule
$E1$	$\Delta J=0,\pm1$ subject to triangle constraints	opposite parity
$M1$	$\Delta J=0,\pm1$ subject to triangle constraints	same parity
$E2$	$\Delta J=0,\pm1,\pm2$ subject to triangle constraints	same parity
$M2$	$\Delta J=0,\pm1,\pm2$ subject to triangle constraints	opposite parity

For $J_i=0$, the triangle rule requires $J_f=\lambda$. Thus E2 excludes not only $0\to0$ but also $0\to1$ and $1/2\to1/2$. The shorthand $\Delta J$ ranges in the table must always be read together with the full triangle constraint.

Orbital Examples

For central-potential scalar orbital states, parity is

$$\pi_\ell=(-1)^\ell.$$

An electric multipole $E\lambda$ has parity $(-1)^\lambda$, so it requires

$$(-1)^{\ell_f-\ell_i} = (-1)^\lambda.$$

Thus $\ell_f-\ell_i$ has the same even/odd character as $\lambda$.

For $E1$, $\lambda=1$, so $\ell_f-\ell_i$ is odd. Combining with the rank-$1$ triangle rule gives

$$\Delta\ell=\pm1.$$

For $E2$, $\lambda=2$, so $\ell_f-\ell_i$ is even. Combining with the rank-$2$ triangle rule gives

$$\Delta\ell=0,\pm2,$$

where the values are restricted by nonnegative angular momentum and possible vanishing angular coefficients.

Long-Wavelength Hierarchy

Multipoles also appear as an expansion in the size of the system relative to the wavelength of the probe. If $a$ is a characteristic size and $k$ is the radiation wavenumber, then the long-wavelength regime has

$$ka\ll1.$$

The E2 coupling typically brings an extra field gradient and is smaller than E1 by a factor of order $ka$ in amplitude for comparable charge matrix elements. Magnetic couplings also depend on currents and magnetic moments; their size cannot be inferred from $ka$ alone. E1 often dominates when it is allowed. But “usually smaller” is not a selection rule. If $E1$ is forbidden by parity, angular momentum, spin, or another symmetry, a weaker $M1$, $E2$, two-photon, or symmetry-breaking mechanism may become the leading observed process.

For quantitative size estimates, see the controlled field expansion in Dipole Approximation and the comparison of mode geometry and free-space rates in Multipole Expansion.

Origin and Approximation Caveats

Multipole operators are defined relative to an origin. In atomic physics the origin is often the nucleus or center of mass. In molecular and condensed systems one must choose the origin consistently with the symmetry being used.

The parity classification assumes the origin is at the inversion center. If the origin is shifted away from that center, different multipole orders can mix in their Cartesian expansion, and a naive parity argument can be obscured.

Gauge choices and center-of-mass separation also affect detailed operator forms. The rank and parity selection rules are robust only after the physical approximation has been stated.

Common Mistakes

• Treating “multipole order” as only a power of $r$ and forgetting rotational rank.
• Forgetting that $E\lambda$ and $M\lambda$ have different parity for the same $\lambda$.
• Calling an $E1$-forbidden line impossible rather than forbidden at electric-dipole order.
• Applying parity rules without placing the origin at the inversion center.
• Assuming a symmetry-allowed multipole matrix element must be large.
• Mixing Cartesian quadrupole tensors with irreducible rank-$2$ spherical components without removing the trace.

Exercises

1. Fill in the parity of $E1$, $M1$, $E2$, and $M2$.

Solution

Use

$$\eta_{E\lambda}=(-1)^\lambda, \qquad \eta_{M\lambda}=(-1)^{\lambda+1}.$$

Thus

$$E1:\ -1, \qquad M1:\ +1, \qquad E2:\ +1, \qquad M2:\ -1.$$

2. Which of $E1$, $M1$, $E2$, and $E0$ can connect $J_i=0$ to $J_f=0$ by angular momentum alone?

Solution

A rank-$\lambda$ operator must satisfy

$$|J_i-\lambda| \le J_f \le J_i+\lambda.$$

With $J_i=J_f=0$, this requires $\lambda=0$. Therefore $E1$, $M1$, and $E2$ are excluded by angular momentum, while a rank-$0$ operator such as $E0$ is not excluded by this angular rule. The static total-charge operator nevertheless has zero off-diagonal matrix elements in a fixed-charge sector. E0 nuclear transitions involve other scalar charge-density matrix elements and cannot emit a single real photon.

3. For scalar orbital states in a central potential, combine parity and rank for an $E2$ operator. What $\Delta\ell$ values survive?

Solution

An $E2$ operator has rank $2$ and even parity. The rank-$2$ triangle rule permits

$$\ell_f=\ell_i,\ell_i\pm1,\ell_i\pm2$$

where valid. Even parity requires $\ell_f-\ell_i$ even. The surviving values are

$$\Delta\ell=0,\pm2,$$

again subject to nonnegative angular momentum and possible vanishing coefficients in special boundary cases.

4. Why is a magnetic dipole parity even although it is a vector under rotations?

Solution

A magnetic dipole is an axial vector. In nonrelativistic models it is built from angular momentum operators such as $\mathbf L$ and $\mathbf S$. Since

$$\mathbf L=\mathbf R\times\mathbf P$$

and both $\mathbf R$ and $\mathbf P$ change sign under parity, $\mathbf L$ is unchanged. Spin is also parity even in the usual nonrelativistic representation. Thus $\boldsymbol\mu$ is even under parity even though it has rotational rank $1$.

References

• C. Cohen-Tannoudji, J. Dupont-Roc, and G. Grynberg, Atom-Photon Interactions, Wiley, 1992.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998.
• M. J. Martin, compiler, “Monopole Transition Strength,” Appendix D of Guidelines for Evaluators, pp. 70–71, Oak Ridge National Laboratory, September 2016. Evaluator guidelines hosted by the IAEA.
• M. E. Rose, Elementary Theory of Angular Momentum, Wiley, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• I. I. Sobelman, Atomic Spectra and Radiative Transitions, 2nd ed., Springer, 1992.