Tensor Operators and Selection Rules

A transition depends on both its initial and final states and the operator that connects them. An electric dipole and a magnetic dipole obey different parity rules even though both are rotational vectors. Tensor operators organize this transformation behavior, and the Wigner–Eckart theorem separates the angular dependence of matrix elements from the remaining dynamics. A permitted matrix element can still be zero or small; symmetry supplies restrictions, not a transition rate by itself.

Helpful background. Angular-momentum addition supplies the coupling coefficients, and parity distinguishes polar and axial vectors.

Enter this chapter

Question	Read next
How do operators transform under rotations?	Scalar, vector, and tensor operators
Which components form one irreducible multiplet?	Spherical tensors and commutators with angular momentum
How can many matrix elements share one calculation?	Wigner–Eckart theorem
Which zeros follow from symmetry?	Selection rules and parity selection rules
Which electromagnetic operator causes a transition?	Dipole transitions and multipoles
How do the rules enter spectra?	Atomic spectra and molecular rotations

Operator rank describes transformation behavior

A scalar obeys $[J_i,O]=0$ for all rotation generators. A vector obeys $[J_i,V_j]=i\hbar\sum_k\epsilon_{ijk}V_k$. These are statements about rotations; neither implies $[H,O]=0$ or conservation. For example, $r^2$ is a rotational scalar but is generally not constant under radial motion.

An irreducible spherical tensor of rank $k$ has components $T_q^{(k)}$, $q=-k,\ldots,k$, satisfying

$$[J_z,T_q^{(k)}]=\hbar qT_q^{(k)},$$ $$[J_\pm,T_q^{(k)}] =\hbar\sqrt{k(k+1)-q(q\pm1)}\,T_{q\pm1}^{(k)}.$$

Thus the operators themselves form an angular-momentum multiplet. Tensor rank is not matrix rank. Nor does counting Cartesian indices settle the irreducible rank: a two-index Cartesian tensor decomposes into scalar, vector, and symmetric traceless pieces of dimensions $1,3,5$.

For a vector, our spherical convention is $V_0=V_z$ and $V_{\pm1}=\mp(V_x\pm iV_y)/\sqrt2$. A polar vector such as position and an axial vector such as angular momentum both have rotational rank one; their parities differ. The operator classification and spherical-tensor page derive the finite transformation laws and conventions.

A commutator already gives one selection rule

Take initial and final states with $J_z$ eigenvalues $\hbar m$ and $\hbar m'$. Inserting them into the first commutator gives

$$\hbar(m'-m)\langle f|T_q^{(k)}|i\rangle =\hbar q\langle f|T_q^{(k)}|i\rangle.$$

Therefore $m'=m+q$ is necessary for a nonzero matrix element. This is why the spherical component is useful: it directly labels the magnetic projection transferred by that component. The name assigned to a circular polarization depends on propagation and polarization conventions; the equation $\Delta m=q$ for the operator component is unambiguous.

Wigner–Eckart separates angles from dynamics

For irreducible angular-momentum states, use the convention

$$\langle\alpha',j',m'|T_q^{(k)}|\alpha,j,m\rangle =\frac{\langle j,m;k,q|j',m'\rangle}{\sqrt{2j'+1}} \times\langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle.$$

The additional labels $\alpha,\alpha'$ include radial, configuration, parity, or multiplicity information. The reduced matrix element is independent of $m,m',q$, but need not be independent of these other labels. Check the convention before importing a reduced matrix element from a table: some definitions put the dimension factor elsewhere.

The angular coefficient requires

$$|j-k|\leq j'\leq j+k,\qquad m'=m+q,$$

with the allowed angular-momentum coupling steps and projection ranges. It can also have additional zeros. The theorem relates components of a tensor regardless of whether every term in a Hamiltonian is rotationally invariant; applying it directly to energy eigenstates with sharp $j,m$ requires those states to have the assumed labels.

For example, a rank-one tensor connecting $j=0,m=0$ to $j'=1$ gives amplitude $\langle\alpha',1\lVert T^{(1)}\rVert\alpha,0\rangle/\sqrt3$ for $m'=q$, since $\langle0,0;1,q|1,q\rangle=1$. The three magnetic components share one reduced calculation. This does not mean an arbitrary illumination drives them equally: its component amplitudes and the occupied initial states also matter. Edmonds and Zare develop this factorization and its spectroscopic applications.

Parity and rotational rank give independent restrictions

If the states have parities $\pi_i,\pi_f$ and $\Pi O\Pi^{-1}=\eta_OO$, then a nonzero matrix element requires $\pi_f\eta_O\pi_i=1$. Electric-dipole ($E1$) operators have rank one and odd parity. Hence their familiar total-angular-momentum rules are

$$\Delta J=0,\pm1,\qquad 0\not\leftrightarrow0, \qquad \pi_f=-\pi_i.$$

For spinless central-potential states, $\pi=(-1)^\ell$, so parity removes the rotationally possible $\Delta\ell=0$ channel and leaves $\Delta\ell=\pm1$. There is no universal electric-dipole restriction on $\Delta n$; radial dynamics supplies the remaining amplitude.

Magnetic-dipole ($M1$) and electric-quadrupole ($E2$) operators are parity even and have ranks one and two, respectively. A transition forbidden at electric-dipole order can therefore occur through another operator. In general, electric and magnetic multipoles of rank $\lambda$ have parities $(-1)^\lambda$ and $(-1)^{\lambda+1}$. A higher multipole’s suppression in a long-wavelength approximation is a separate size-to-wavelength estimate, not itself a symmetry zero.

Spectroscopy also needs the actual states and operator

In approximate LS coupling, the leading electric dipole is spin independent, so it preserves $S$. Spin–orbit or other spin-dependent mixing can weaken the rule when energy eigenstates cease to have sharp $S$. Spin-independent configuration mixing alone does not break total-spin conservation; it may change reduced amplitudes while preserving the spin rule. Parity-conserving spin–orbit coupling likewise does not by itself remove a parity restriction.

For an ideal polar linear rotor in a nondegenerate electronic/vibrational state, a permanent body-fixed dipole produces pure rotational $E1$ transitions with $\Delta J=\pm1$. A homonuclear diatomic lacks that permanent dipole, so its rotational energies do not automatically produce an ordinary pure rotational microwave $E1$ spectrum. Anisotropic polarizability can instead give rotational Raman shifts with $\Delta J=\pm2$. The molecular application explains the distinct operators.

When weak perturbations mix states, compare matrix elements of the perturbation with the relevant energy separation. A small coefficient can produce large mixing near degeneracy. Once the new states are known, apply the exact surviving symmetries and calculate the formerly forbidden amplitude from the admixtures. To turn that amplitude into a rate, continue to selection rules in transition rates.

Exercises

1. A rank-$2$ tensor component with $q=+1$ acts on an initial state with $j_i=1/2$ and $m_i=1/2$. Which final $j_f,m_f$ values survive the basic rotational rules?

Solution

The magnetic rule gives

$$m_f=m_i+q = \frac12+1 = \frac32.$$

The triangle rule gives

$$\left| \frac12-2 \right| \le j_f \le \frac12+2,$$

so

$$j_f=\frac32 \quad\hbox{or}\quad j_f=\frac52.$$

Both multiplets contain $m_f=3/2$. Thus the basic rotational rules permit

$$\left( j_f,m_f \right) = \left( \frac32,\frac32 \right), \qquad \left( \frac52,\frac32 \right).$$

An additional symmetry or a zero reduced matrix element could still remove either channel.

2. Two states have the same parity, with $J_i=0$ and $J_f=2$. Among $E1$, $M1$, $E2$, and $M2$, which multipole first passes both the rank and parity tests?

Solution

A transition from $J_i=0$ to $J_f=2$ requires an operator whose rank can couple $0$ to $2$. Rank $1$ cannot do so, so both $E1$ and $M1$ fail the triangle rule.

Rank $2$ passes the angular test. The states have the same parity, so the operator must be parity even. $E2$ has parity

$$(-1)^2=+1,$$

whereas $M2$ has parity

$$(-1)^{2+1}=-1.$$

Therefore $E2$ is the first of the listed multipoles that passes both tests. This does not determine the size of its reduced matrix element.

3. An ideal homonuclear diatomic molecule has rotational levels $E_J=BJ(J+1)$ but no permanent electric dipole. Explain why the energy spacing does not imply an ordinary microwave $E1$ spectrum, and name a tensor operator that can reveal rotational structure by another mechanism.

Solution

The level spacing determines possible photon energies, but an $E1$ transition also requires a nonzero electric-dipole matrix element. An ideal homonuclear diatomic molecule has no permanent body-fixed dipole, so the leading pure rotational $E1$ operator is absent even though the rotor levels exist.

The anisotropic polarizability contains a rank-$2$ irreducible tensor. It can produce rotational Raman transitions with characteristic shifted branches

$$\Delta J=\pm2.$$

This is a different operator and a different light–matter process, not a violation of the electric-dipole rule.

References

• C. Cohen-Tannoudji, J. Dupont-Roc, and G. Grynberg, Atom–Photon Interactions: Basic Processes and Applications, Wiley, 1992.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 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.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.