A student calculates the second-order energy correction for the ground state of a quantum system and obtains a positive value. What can you immediately conclude?
AThe perturbation destabilizes the ground state, which is unusual but possible
BThe calculation contains an error — the second-order correction to the ground state is always negative or zero
CThe result is physically reasonable because perturbations can raise or lower energy at second order
DThe ground state is degenerate, requiring a different formalism
For the ground state, every term in E_n^(2) = Σ_{k≠n} |⟨k|H'|n⟩|² / (E_n^(0) − E_k^(0)) has a negative denominator: since the ground state has the lowest energy, E_n^(0) < E_k^(0) for all k ≠ n, making every denominator E_n^(0) − E_k^(0) < 0. The numerator |⟨k|H'|n⟩|² is always non-negative. Therefore every term is non-positive, and the total sum is negative (or zero if the perturbation has no off-diagonal matrix elements). A positive second-order ground-state correction is impossible — any such result indicates a computational error.
Question 2 Multiple Choice
Two neutral atoms with no permanent electric dipoles exhibit a weak attractive interaction proportional to −C/r⁶. At what order of perturbation theory does this London dispersion force first appear, and why not at lower order?
AZeroth order — it is a direct classical electrostatic effect between the electron distributions
BFirst order — the dipole-dipole interaction Hamiltonian gives a first-order energy correction
CSecond order — the first-order correction averages to zero because the atoms have no permanent dipoles, but second-order terms capture virtual excitations that create and destroy induced dipoles
DThird order — dipole-dipole coupling is too weak to appear at second order
The perturbation here is the dipole-dipole interaction between the two atoms. At first order, E_n^(1) = ⟨n|H'|n⟩ — the expectation value of the dipole-dipole interaction in the unperturbed ground state. Since the atoms have no permanent dipoles, this expectation value averages to zero over all orientations (the ground state has no preferred dipole orientation). At second order, however, the perturbation mixes in excited states with non-zero dipole matrix elements. These 'virtual excitations' create temporary induced dipoles that attract, giving a net second-order energy shift proportional to −C/r⁶. The van der Waals force is genuinely a second-order quantum effect.
Question 3 True / False
Second-order perturbation theory is primarily needed when the first-order energy correction E_n^(1) = ⟨n|H'|n⟩ is exactly zero.
TTrue
FFalse
Answer: False
Second-order perturbation theory captures corrections that are of order (H')² — a systematically smaller contribution than the first-order (H')¹ term. You may need it even when the first-order correction is nonzero, if you want a more accurate approximation. More commonly, it is *most crucial* when the first-order correction happens to be zero (as for the van der Waals force), since in that case second order gives the leading nonzero correction. But 'needing' second order does not require first order to vanish — it depends on the required precision.
Question 4 True / False
For an excited state (not the ground state), the second-order energy correction can be either positive or negative depending on the perturbation and the energy spectrum.
TTrue
FFalse
Answer: True
For an excited state with energy E_n^(0), some states |k⟩ have E_k^(0) > E_n^(0) (higher excited states), giving negative denominators and negative contributions; other states have E_k^(0) < E_n^(0) (lower states including the ground state), giving positive denominators and positive contributions. The total sign of E_n^(2) therefore depends on which contributions dominate — determined by the matrix elements ⟨k|H'|n⟩ and the energy gaps. Only for the ground state is the sign guaranteed (always negative), because no state is lower in energy.
Question 5 Short Answer
Explain the physical picture of 'virtual transitions' that underlies the second-order energy correction formula. Why do states close in energy to the perturbed state contribute more to the correction than distant states?
Think about your answer, then reveal below.
Model answer: In the second-order picture, the perturbation H' can temporarily mix state |n⟩ with other eigenstates |k⟩. Even if this mixing does not produce a permanent transition (H' is not strong enough to permanently change the state), the transient mixing shifts the energy. This temporary excursion to state |k⟩ and back is called a 'virtual transition.' The energy cost of this excursion is the gap |E_n^(0) − E_k^(0)|, which appears in the denominator of E_n^(2) = Σ_{k≠n} |⟨k|H'|n⟩|² / (E_n^(0) − E_k^(0)). States close in energy have small denominators, making their contribution large; states far away have large denominators that suppress their contribution. The formula encodes the intuition that 'cheap' virtual transitions (small energy gap to borrow) have the largest effect.
The energy-denominator structure is central to many quantum mechanical phenomena beyond perturbation theory. It appears in Fermi's golden rule, in the derivation of effective Hamiltonians, and in the renormalization of physical quantities by virtual processes in quantum field theory. The key insight is always the same: coupling to nearby states is energetically cheap and therefore large; coupling to distant states is energetically costly and therefore small.