Questions: Transition Probabilities and Selection Rules

Question 1

The electric dipole transition operator is proportional to the position vector r, which is odd under parity. What does this imply about which atomic transitions are allowed?

• A. Only transitions between states of the same parity are allowed, because the integrand must be even to be nonzero
• B. Only transitions between states of opposite parity are allowed, because the integrand must be even overall to be nonzero
• C. All transitions are equally allowed, since parity has no effect on the integral value
• D. Transitions are allowed only if both states have odd parity

Answer: (B)

For the matrix element ∫ ψ*_f r ψ_i dV to be nonzero, the integrand must not have a definite odd symmetry. Since r is odd under parity, the product ψ*_f r ψ_i must be even overall — which requires ψ*_f and ψ_i to have opposite parity (odd × odd × even = even, or even × odd × odd = even). States with the same parity give an odd integrand that integrates to zero. This is why Δl = ±1: changing l by an odd number changes the parity of the orbital wavefunction.

Question 2

The 2s state of hydrogen has a lifetime of ~100 ms, roughly 10⁸ times longer than typical allowed transitions. What best explains this?

• A. The 2s state has higher energy than 2p, making decay energetically unfavorable
• B. The 2s → 1s electric dipole transition is selection-rule forbidden, so decay must occur via much weaker mechanisms
• C. The selection rule Δn = ±1 prohibits 2s → 1s, since n changes by 1
• D. The 2s state is the ground state of hydrogen and cannot decay further

Answer: (B)

The 2s → 1s transition violates the electric dipole selection rule Δl = ±1 because both states have l = 0 (s orbitals). The matrix element ⟨1s|r|2s⟩ vanishes by parity symmetry. The transition is 'forbidden' — meaning the electric dipole amplitude is zero, not that the transition is impossible. It eventually occurs via two-photon emission (a higher-order process), which is far slower, explaining the anomalously long lifetime. 'Forbidden' means the dominant mechanism is blocked, not that physics prevents the event.

Question 3

A 'forbidden' transition in quantum mechanics is one that is prohibited by conservation of energy and therefore cannot occur.

• T. True
• F. False

Answer: False

This is the most common misconception about selection rules. 'Forbidden' does not mean energetically impossible — it means the electric dipole matrix element ⟨f|H'|i⟩ vanishes due to symmetry, so the transition cannot occur via the electric dipole mechanism. The transition can still occur through weaker mechanisms: magnetic dipole, electric quadrupole, or higher-order processes. Each is suppressed by additional powers of the fine structure constant α ≈ 1/137. Forbidden transitions are slow, not impossible.

Question 4

Selection rules for electric dipole transitions follow directly from the mathematical condition that the transition matrix element ⟨f|H'|i⟩ must be nonzero — they are not independently postulated constraints.

• T. True
• F. False

Answer: True

This is the key insight about the origin of selection rules. They are not additional axioms of quantum mechanics; they are consequences of when the integral ∫ ψ*_f H' ψ_i dV is nonzero versus zero. The dipole operator H' ∝ r has specific symmetry properties; the vanishing of the integral for certain combinations of initial and final states follows from those symmetries. The Δl = ±1 rule, for example, falls out of the parity argument — it is the mathematical content of 'the integrand must be even to be nonzero.'

Question 5

Explain in physical terms why the selection rule Δl = ±1 holds for electric dipole transitions, rather than just stating the rule.

Model answer: The electric dipole perturbation H' is proportional to r, which is an odd function under parity inversion (r → −r). For the matrix element ⟨f|r|i⟩ to be nonzero, the full integrand ψ*_f r ψ_i must be even under parity. Since r is odd, ψ*_f and ψ_i must have opposite parity. The parity of a hydrogen orbital is (−1)^l, so initial and final states need l values that differ by an odd number — the minimal such difference is 1, giving Δl = ±1.

The rule is not memorized; it is derived from the symmetry of the integral. A student who just remembers 'Δl = ±1' without knowing why will struggle when confronted with other multipole transitions or other perturbation operators. The key is that the operator's symmetry under parity determines which combinations of states give a nonvanishing integral. This same logic extends to other selection rules (Δm_l = 0, ±1 from the angular part of the integral, etc.).