Selection Rule Problems

These solved problems practice selection rules as symmetry-enforced zeros of matrix elements. The aim is to identify which rule is being used, which assumptions it depends on, and what remains after the symmetry test.

Use Selection Rules for the conceptual home, Dipole Transitions for electric-dipole rules, and Wigner–Eckart Theorem for the angular-momentum theorem. Molecular rotational examples link to Rigid Rotor and Rotational Spectra.

Conventions

A transition, coupling, or mixing amplitude often contains a matrix element

$$M_{fi} = \langle f|O|i\rangle.$$

For a unitary symmetry $U$, if

$$U\lvert i\rangle=u_i\lvert i\rangle, \qquad U\lvert f\rangle=u_f\lvert f\rangle, \qquad UOU^\dagger=u_OO,$$

then a nonzero matrix element requires

$$u_f^*u_Ou_i=1.$$

For parity, with eigenvalues $\pi_i,\pi_f,\pi_O=\pm1$, this becomes

$$\pi_f\pi_O\pi_i=1.$$

For an irreducible spherical tensor component $T_q^{(k)}$, the rotational selection rules are

$$m_f=m_i+q,$$

and

$$\lvert j_i-k\rvert \le j_f \le j_i+k.$$

Electric-dipole operators are odd under parity and transform as rank-$1$ tensors. For scalar central-potential orbital states, the leading electric-dipole rules are

$$\Delta\ell=\pm1, \qquad \Delta m=q=0,\pm1.$$

Skill Map

Problem group	Skills	Preparation
Discrete symmetries	parity eigenvalues, operator parity	Parity
Electric dipole	rank-$1$ tensors, odd parity, polarization	Dipole Transitions
Multipoles	$E\lambda/M\lambda$ rank and parity	Multipole Operators
Wigner–Eckart checks	magnetic rule, triangle rule, component validity	Wigner–Eckart Theorem
Rigid rotors	$J,M$ labels, dipole spectra, permanent dipoles	Rotational Spectra
Approximate rules	forbidden versus weakly allowed	Approximate Symmetry

Problems

1. Parity Filter

Two states have parity eigenvalues $\pi_i=-1$ and $\pi_f=+1$. Decide whether a matrix element can be nonzero for an even operator and for an odd operator.

Solution

The parity condition for a nonzero matrix element is

$$\pi_f\pi_O\pi_i=1.$$

For an even operator, $\pi_O=+1$, so

$$(+1)(+1)(-1)=-1.$$

The matrix element must vanish.

For an odd operator, $\pi_O=-1$, so

$$(+1)(-1)(-1)=+1.$$

Parity does not force the matrix element to vanish. The matrix element may still be zero for another reason, such as angular momentum, a radial integral, or an additional symmetry.

2. Position and Position-Squared

In a one-dimensional parity-symmetric potential, stationary states have definite parity. If $\lvert a\rangle$ and $\lvert b\rangle$ have the same parity, which of the matrix elements

$$\langle a|X|b\rangle, \qquad \langle a|X^2|b\rangle$$

is ruled out by parity?

Solution

The position operator is odd:

$$\Pi X\Pi^{-1}=-X.$$

The operator $X^2$ is even:

$$\Pi X^2\Pi^{-1}=X^2.$$

For states of the same parity,

$$\pi_a\pi_b=+1.$$

For $X$,

$$\pi_a(-1)\pi_b=-1,$$

so

$$\langle a|X|b\rangle=0$$

by parity.

For $X^2$,

$$\pi_a(+1)\pi_b=+1.$$

Parity allows $\langle a|X^2|b\rangle$. It is not guaranteed to be nonzero; it is merely not killed by parity.

3. Hydrogenic Electric-Dipole Tests

In a spinless central-potential model, test the following electric-dipole transitions:

1. $2s\to1s$,
2. $2p,m=1\to1s,m=0$ with a $q=-1$ dipole component,
3. $2p,m=1\to1s,m=0$ with a $q=0$ dipole component.

Assume energy conservation is handled separately.

Solution

For electric-dipole transitions in this model,

$$\Delta\ell=\pm1, \qquad \Delta m=q.$$

For $2s\to1s$, both states have $\ell=0$, so

$$\Delta\ell=0.$$

This is forbidden for electric dipole transitions. Equivalently, both states have even parity and the electric dipole operator is odd.

For $2p,m=1\to1s,m=0$, the angular labels are

$$\ell_i=1, \qquad \ell_f=0.$$

Thus

$$\Delta\ell=-1,$$

which is allowed by the orbital electric-dipole rule.

The magnetic change is

$$\Delta m=m_f-m_i=0-1=-1.$$

Thus the $q=-1$ component is allowed by magnetic quantum number, while the $q=0$ component is forbidden for this pair of magnetic sublevels.

4. Polarization Chooses the Magnetic Rule

An electric-dipole operator component $d_q$ acts on an orbital state with $\ell_i=1$, $m_i=-1$. The final orbital state has $\ell_f=2$, $m_f=0$. Which value of $q$ is required, and are the angular and parity rules satisfied?

Solution

The magnetic rule is

$$m_f=m_i+q.$$

Therefore

$$q=m_f-m_i=0-(-1)=+1.$$

The angular momentum change is

$$\Delta\ell=+1,$$

which is allowed for an electric-dipole transition.

Parity also works. The initial parity is

$$(-1)^{\ell_i}=(-1)^1=-1,$$

and the final parity is

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

The parities are opposite, as required for an odd electric-dipole operator. Thus this transition is allowed by these selection rules for the $q=+1$ component.

5. Wigner–Eckart Component Test

A rank-$2$ tensor component $T_{-2}^{(2)}$ acts on a state with $j_i=3/2$ and $m_i=1/2$. List the possible $j_f$ values allowed by the triangle rule. Then apply the magnetic rule and remove any final $j_f$ values that cannot support the required $m_f$.

Solution

The triangle rule is

$$\lvert j_i-k\rvert \le j_f \le j_i+k.$$

Here $j_i=3/2$ and $k=2$, so

$$\left|\frac32-2\right| \le j_f \le \frac32+2.$$

Thus

$$\frac12 \le j_f \le \frac72.$$

The allowed half-integer values are

$$j_f=\frac12,\frac32,\frac52,\frac72.$$

The magnetic rule is

$$m_f=m_i+q.$$

With $m_i=1/2$ and $q=-2$,

$$m_f=\frac12-2=-\frac32.$$

A final multiplet with $j_f=1/2$ cannot contain $m_f=-3/2$. Therefore that value is removed for this component. The component can be nonzero only for

$$j_f=\frac32,\frac52,\frac72, \qquad m_f=-\frac32,$$

subject to any additional parity or dynamical rules.

6. Scalar Perturbation

A perturbation $V(r)$ is a rotational scalar in a spinless central-potential problem. Using selection-rule language, explain why it cannot connect states with different $\ell$ or different $m$, although it may connect different radial labels.

Solution

A rotational scalar is a rank-$0$ tensor:

$$k=0, \qquad q=0.$$

The magnetic rule gives

$$m_f=m_i+0=m_i.$$

The triangle rule gives

$$\lvert \ell_i-0\rvert \le \ell_f \le \ell_i+0,$$

so

$$\ell_f=\ell_i.$$

Therefore a radial scalar cannot change $\ell$ or $m$.

The Wigner–Eckart theorem separates angular labels from additional labels. The reduced matrix element can still depend on radial quantum numbers, so a scalar radial perturbation can connect or shift states with different radial labels when the radial integral is nonzero and the physical setting allows that mixing.

7. Pure Rotational Spectrum of a Polar Rotor

For an ideal polar linear rigid rotor, electric-dipole rotational matrix elements obey

$$\Delta J=\pm1, \qquad \Delta M=q.$$

Starting from $J=2$, $M=1$, list the allowed final states for a matrix element with $q=0$ and with $q=+1$. For absorption from this level, which $J$ branch goes upward in energy?

Solution

For $q=0$,

$$\Delta M=0,$$

so

$$M_f=1.$$

The electric-dipole rotational rule allows

$$J_f=1 \quad \text{or} \quad J_f=3.$$

Thus the allowed matrix-element targets are

$$(J_f,M_f)=(1,1),(3,1).$$

For $q=+1$,

$$M_f=M_i+1=2.$$

The $J_f=1$ option is impossible because a $J=1$ multiplet has only

$$M=-1,0,1.$$

The allowed target is therefore

$$(J_f,M_f)=(3,2).$$

The ideal rotor energy is proportional to

$$J(J+1).$$

For absorption from $J=2$, the upward branch is

$$J_f=3.$$

The $J_f=1$ matrix element belongs to downward emission or stimulated emission, not absorption from the $J=2$ level to a higher rotational energy.

8. Why Homonuclear Rotors Lack Pure Electric-Dipole Lines

An ideal homonuclear diatomic molecule has rotational levels but no permanent electric dipole moment. Why does the absence of a pure rotational electric-dipole spectrum not mean that the rotational levels are absent?

Solution

Energy levels and radiative matrix elements are different questions. The ideal rigid-rotor Hamiltonian gives rotational levels

$$E_J = B_EJ(J+1),$$

with angular wavefunctions labeled by $J,M$. Those levels exist as eigenstates of the rotational Hamiltonian.

Electric-dipole pure rotational spectroscopy requires a nonzero dipole operator that can couple those levels to radiation. A homonuclear diatomic molecule has no permanent electric dipole in the body-fixed frame, so the leading electric-dipole pure rotational matrix element is absent.

This does not remove the rotor spectrum. It says that this particular probe and approximation do not see the levels through pure electric-dipole rotational lines. Other mechanisms, such as Raman scattering, quadrupole effects, vibration-rotation coupling, collisions, or external-field mixing, belong to more detailed molecular spectroscopy.

9. Electric Quadrupole Escape Route

An electric-dipole transition between two states of the same parity is forbidden. Suppose an electric quadrupole operator is considered instead. Treat it as an even rank-$2$ tensor. What parity rule does it obey, and what orbital changes are allowed by combining parity with the rank-$2$ triangle rule?

Solution

An electric quadrupole operator has even parity:

$$\pi_O=+1.$$

The parity condition

$$\pi_f\pi_O\pi_i=1$$

therefore requires

$$\pi_f=\pi_i.$$

For scalar orbital states, this means $\ell_f$ and $\ell_i$ must have the same parity, so $\Delta\ell$ must be even.

The rank-$2$ angular triangle rule gives

$$\lvert\ell_i-2\rvert \le \ell_f \le \ell_i+2.$$

Combining the triangle rule with same parity gives the usual possibilities

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

subject to the endpoint restrictions of the triangle rule. For example, $\ell_i=0$ can connect only to $\ell_f=2$ through a rank-$2$ tensor, not to $\ell_f=0$.

This is why an electric-dipole forbidden line can sometimes appear through a weaker electric-quadrupole mechanism.

Common Mistakes

• Calling a transition forbidden without stating the operator and symmetry assumptions.
• Treating an allowed selection rule as a guarantee of a large matrix element.
• Forgetting that electric-dipole $\Delta\ell=\pm1$ uses both rotation and parity.
• Confusing the light polarization label $q$ with a universal convention for $\sigma^\pm$ names.
• Applying spinless orbital rules after total angular momentum has become the good label.
• Forgetting that $m_f=m_i+q$ must also satisfy $\lvert m_f\rvert\le j_f$.
• Mistaking absence of an electric-dipole molecular line for absence of a rotational level.

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.
• C. J. Foot, Atomic Physics, Oxford University Press, 2005.