Chapter 17 Quiz: Time-Independent Perturbation Theory — Non-Degenerate Case

Multiple Choice

Q1. The first-order energy correction $E_n^{(1)}$ in non-degenerate perturbation theory is:

(a) The expectation value of $\hat{H}_0$ in the perturbed state $|n\rangle$ (b) The expectation value of $\hat{H}'$ in the unperturbed state $|n^{(0)}\rangle$ (c) The expectation value of $\hat{H}'$ in the perturbed state $|n\rangle$ (d) The matrix element $\langle n^{(0)}|\hat{H}_0|n^{(1)}\rangle$

Q2. The first-order wavefunction correction $|n^{(1)}\rangle$ involves a sum over all states $|m^{(0)}\rangle$ with $m \neq n$. The coefficient of $|m^{(0)}\rangle$ in this expansion is proportional to:

(a) $\frac{1}{E_m^{(0)}}$ (b) $\frac{1}{E_n^{(0)} - E_m^{(0)}}$ (c) $\frac{1}{(E_n^{(0)} - E_m^{(0)})^2}$ (d) $E_n^{(0)} - E_m^{(0)}$

Q3. The second-order energy correction for the ground state of any quantum system is:

(a) Always positive (b) Always negative or zero (c) Always zero (d) Can be positive or negative depending on the perturbation

Q4. The first-order Stark effect (energy shift linear in the electric field $\mathcal{E}$) for the hydrogen ground state is:

(a) Nonzero and proportional to $\mathcal{E}$ (b) Zero, because the ground state is non-degenerate (c) Zero, because the ground state has definite parity and the perturbation is odd (d) Zero, because the selection rule $\Delta l = \pm 1$ forbids it

Q5. In the anharmonic oscillator $\hat{H} = \hat{H}_{\text{QHO}} + \lambda\hat{x}^4$, the first-order energy correction for the $n$-th state depends on $n$ as:

(a) Linearly in $n$ (b) Quadratically in $n$ (for large $n$) (c) Exponentially in $n$ (d) Is independent of $n$

Q6. Non-degenerate perturbation theory fails when:

(a) The perturbation is time-dependent (b) Two or more unperturbed states share the same energy (c) The unperturbed Hamiltonian has a continuous spectrum (d) The perturbation is a polynomial of degree higher than 2

Q7. Which of the following perturbations gives a nonzero first-order energy correction for all harmonic oscillator states $|n\rangle$?

(a) $\hat{H}' = \lambda\hat{x}$ (linear) (b) $\hat{H}' = \lambda\hat{x}^3$ (cubic) (c) $\hat{H}' = \lambda\hat{x}^4$ (quartic) (d) $\hat{H}' = \lambda\hat{p}$ (linear in momentum)

Q8. The static polarizability $\alpha$ of hydrogen in its ground state tells us:

(a) The permanent electric dipole moment of the atom (b) The energy of the atom in the absence of any field (c) How the energy shifts in response to an applied electric field, with $\Delta E = -\frac{1}{2}\alpha\mathcal{E}^2$ (d) The first-order Stark shift of the ground state

Q9. In the first-order wavefunction correction (Eq. 17.15), states with energies close to $E_n^{(0)}$ contribute:

(a) Less than distant states, because the matrix elements are smaller (b) More than distant states, because the energy denominator is smaller (c) The same as distant states, because both effects cancel (d) Not at all, because of orthogonality

Q10. The perturbation series for the anharmonic oscillator ground-state energy:

(a) Converges for all $\lambda > 0$ (b) Converges only for $|\lambda| < \lambda_c$ for some critical $\lambda_c > 0$ (c) Has zero radius of convergence but is asymptotic (d) Diverges for all $\lambda$ including the first-order term

Q11. "Level repulsion" in second-order perturbation theory refers to the phenomenon where:

(a) Two coupled states push each other further apart in energy (b) Two coupled states are attracted to the same energy (c) Energy levels cross as the perturbation increases (d) The perturbation creates new energy levels between existing ones

Q12. The second-order energy correction $E_n^{(2)}$ requires knowledge of:

(a) Only the unperturbed state $|n^{(0)}\rangle$ (b) The first-order state correction $|n^{(1)}\rangle$ (or equivalently, all matrix elements $\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle$) (c) The second-order state correction $|n^{(2)}\rangle$ (d) The exact perturbed state $|n\rangle$

True/False

Q13. True or False: If the perturbation $\hat{H}'$ has odd parity, then the first-order energy correction $E_n^{(1)}$ vanishes for all states $|n^{(0)}\rangle$ that have definite parity.

Q14. True or False: The first-order corrected energy $E_n^{(0)} + E_n^{(1)}$ is always an upper bound to the true energy $E_n$ for the ground state.

Q15. True or False: The intermediate normalization convention $\langle n^{(0)}|n\rangle = 1$ implies that the correction terms $|n^{(k)}\rangle$ (for $k \ge 1$) are orthogonal to the unperturbed state $|n^{(0)}\rangle$.

Q16. True or False: If all matrix elements $\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle$ vanish for $m \neq n$, then perturbation theory gives the exact energy $E_n = E_n^{(0)} + \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$ at first order, with no corrections at any higher order.

Short Answer

Q17. Explain in two or three sentences why the second-order correction to the ground-state energy is always negative (or zero). Your answer should reference both the sign of the numerator and denominator in the sum.

Q18. A quantum system has three non-degenerate states with energies $E_1^{(0)} = 0$, $E_2^{(0)} = 2\hbar\omega$, $E_3^{(0)} = 5\hbar\omega$, and the perturbation has matrix elements $\langle 2^{(0)}|\hat{H}'|1^{(0)}\rangle = 0.1\hbar\omega$, $\langle 3^{(0)}|\hat{H}'|1^{(0)}\rangle = 0.3\hbar\omega$, and $\langle 1^{(0)}|\hat{H}'|1^{(0)}\rangle = 0$. Compute the energy of the ground state through second order.

Q19. What is the physical meaning of the statement "perturbation theory connects the solvable to the realistic"? Give a specific example from this chapter.

Q20. Explain the difference between a convergent series and an asymptotic series. Why is it important for a physicist to know the difference, even though asymptotic series are formally divergent?

Answer Key

Q1. (b) — The first-order energy correction is $E_n^{(1)} = \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$, the expectation value of the perturbation in the unperturbed state.

Q2. (b) — The coefficient is $c_m^{(1)} = \langle m^{(0)}|\hat{H}'|n^{(0)}\rangle / (E_n^{(0)} - E_m^{(0)})$.

Q3. (b) — For the ground state, all energy denominators $E_0^{(0)} - E_m^{(0)}$ are negative, and the numerators $|\cdots|^2$ are non-negative, so every term in the sum is $\le 0$.

Q4. (c) — The hydrogen ground state has even parity ($l=0$), and the Stark perturbation $e\mathcal{E}z$ has odd parity. The matrix element of an odd operator between even-parity states vanishes.

Q5. (b) — The first-order correction is proportional to $6n^2 + 6n + 3$, which grows quadratically for large $n$.

Q6. (b) — Degenerate unperturbed levels cause the denominators in the perturbation formulas to vanish.

Q7. (c) — The quartic perturbation $\hat{x}^4$ is even-parity and gives $E_n^{(1)} = \lambda(\hbar/2m\omega)^2(6n^2 + 6n + 3) \neq 0$ for all $n$. The odd-parity perturbations (a), (b), (d) give $E_n^{(1)} = 0$ for all states with definite parity.

Q8. (c) — The polarizability characterizes the induced dipole energy: $\Delta E = -\frac{1}{2}\alpha\mathcal{E}^2$. Hydrogen has no permanent dipole moment.

Q9. (b) — The coefficient $c_m^{(1)} \propto 1/(E_n^{(0)} - E_m^{(0)})$, so states with small energy gaps contribute disproportionately.

Q10. (c) — The series has zero radius of convergence because the physics is singular at $\lambda = 0^-$ (potential becomes unbounded below), but the first few terms give excellent approximations for small $\lambda > 0$.

Q11. (a) — When two states are coupled by a perturbation, the lower state is pushed down and the upper state is pushed up (at second order), increasing their energy separation.

Q12. (b) — $E_n^{(2)} = \langle n^{(0)}|\hat{H}'|n^{(1)}\rangle = \sum_{m\neq n} |\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle|^2/(E_n^{(0)} - E_m^{(0)})$.

Q13. True — If $\hat{H}'$ is odd-parity and $|n^{(0)}\rangle$ has definite parity (even or odd), then $|\hat{H}'|n^{(0)}\rangle$ has the opposite parity and is orthogonal to $|n^{(0)}\rangle$, so $\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle = 0$.

Q14. False — Only the variational method (Chapter 19) guarantees an upper bound. First-order perturbation theory does not provide a variational bound.

Q15. True — Expanding $\langle n^{(0)}|n\rangle = 1$ gives $1 + \lambda\langle n^{(0)}|n^{(1)}\rangle + \lambda^2\langle n^{(0)}|n^{(2)}\rangle + \cdots = 1$, requiring each coefficient to vanish.

Q16. True — If $\hat{H}'$ is diagonal in the unperturbed basis, then $\hat{H} = \hat{H}_0 + \hat{H}'$ is also diagonal, and the exact energies are $E_n = E_n^{(0)} + \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$.

Q17. The second-order correction is $E_0^{(2)} = \sum_{m\neq 0} |\langle m^{(0)}|\hat{H}'|0^{(0)}\rangle|^2/(E_0^{(0)} - E_m^{(0)})$. The numerator $|\cdots|^2$ is always non-negative, and the denominator $E_0^{(0)} - E_m^{(0)}$ is always negative for the ground state (since $E_0^{(0)}$ is the lowest energy). Therefore every term is non-positive, and the sum is $\le 0$.

Q18. $E_1 = E_1^{(0)} + E_1^{(1)} + E_1^{(2)} = 0 + 0 + \left[\frac{(0.1\hbar\omega)^2}{0 - 2\hbar\omega} + \frac{(0.3\hbar\omega)^2}{0 - 5\hbar\omega}\right] = \frac{0.01}{-2}\hbar\omega + \frac{0.09}{-5}\hbar\omega = -0.005\hbar\omega - 0.018\hbar\omega = -0.023\hbar\omega$.

Q19. The statement means that perturbation theory allows us to go from idealized, exactly solvable problems (like the hydrogen atom or harmonic oscillator) to physically realistic ones (like an atom in an electric field or a molecule with anharmonic vibrations) by systematically computing corrections order by order. Example: the hydrogen atom is exactly solvable, but placing it in an electric field (the Stark effect) makes it unsolvable in closed form. Perturbation theory yields the polarizability and energy shift to any desired accuracy.

Q20. A convergent series approaches the true answer as more terms are added: $\sum_{k=0}^N a_k \lambda^k \to f(\lambda)$ as $N \to \infty$. An asymptotic series improves initially as terms are added, but eventually diverges: there is an optimal truncation order beyond which additional terms make the approximation worse. Physicists must know the difference because most perturbation series in quantum mechanics (and quantum field theory) are asymptotic, not convergent. Blindly computing higher orders can give worse results. Knowing when to stop is itself a skill.