---
title: "Perturbation Theory"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 13
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/13-perturbation-theory
---

# Chapter 13 — Perturbation Theory

Only a handful of quantum problems dissolve exactly: the box, the oscillator, hydrogen, the two-level system. Everything else — an atom in a field, a molecule’s stiffening bond, a level nudged by a neighbour — is one of those solvable skeletons wearing a small extra term. [Perturbation theory](#thm-b3-perturbation-theory-corrections) is the art of treating the extra term as what it is: a correction, computed order by order in its smallness. Its first-order rule is a one-line average; its second-order rule explains why levels *repel*, why every atom is polarisable, and why van der Waals attraction is universal; its degenerate variant handles the delicate cases where near-equal levels reorganise entirely; and its time-dependent form — Fermi’s golden rule — is the equation behind every absorption line, decay rate, and scattering cross-section measured in a laboratory. As a running reward, the chapter ends by computing the colour of the sky.

## 13.1 Small shifts: the non-degenerate theory

**Theorem 13.1 (Perturbative corrections).**

Let $\hat H = \hat H_0 + \hat V$ with $\hat H_0$ solved ($\hat H_0\ket n = E_n\ket n$, spectrum non-degenerate) and $\hat V$ small. Then, order by order in $\hat V$,

$$
E_n' = E_n + \bra n\hat V\ket n
 + \sum_{m \neq n}\frac{|\bra m\hat V\ket n|^2}{E_n - E_m}
 + \cdots
$$

and the state acquires admixtures $\ket{n'} = \ket n + \sum_{m\neq
n}\dfrac{\bra m\hat V\ket n}{E_n - E_m}\,\ket m + \cdots$ *First order*: the shift is just the average of the perturbation in the unperturbed state. *Second order*: couplings to other levels push $E_n$ *away* from them — each $m$ above pushes down, each below pushes up — so the ground state’s second-order shift is always negative. Validity: the mixing fractions $|\bra m\hat V\ket n/(E_n - E_m)|$ must be small.

**Partial proof.** Expand $E' = E + \epsilon_1 + \epsilon_2 + \cdots$ and $\ket{\psi} =
\ket n + \ket{\psi_1} + \cdots$ in powers of $\hat V$, insert into $(\hat H_0 + \hat V)\ket\psi = E'\ket\psi$, and match orders. Projecting the first-order equation on $\bra n$ gives $\epsilon_1 =
\bra n\hat V\ket n$; projecting on $\bra m$ ($m \neq n$) gives the admixture coefficients; carrying those into the second-order equation and projecting on $\bra n$ again gives $\epsilon_2$. The sign statement for the ground state: every denominator $E_0 - E_m$ is negative while the numerators are squares. ∎

**Proposition 13.2 (A perturbation with an exact answer).**

An oscillator of charge $q$ in a uniform field $\mathcal E$ has $\hat V = -q\mathcal E\hat x$. Completing the square solves it exactly: the potential is the same parabola, shifted, with all levels lowered by $q^2\mathcal E^2/2m\omega^2$. [Perturbation theory](#thm-b3-perturbation-theory-corrections) must agree — and does: first order vanishes ($\langle\hat
x\rangle_n = 0$), and the second-order sum, in which $\hat x$ couples $n$ only to $n \pm 1$, gives exactly $-q^2\mathcal E^2/2m\omega^2$ for every level ([Exercise 13.2](#exo-b3-perturbation-theory-2)). One solvable case, checked both ways, is the best calibration a method can have.

**Proof.** The two matrix elements and their denominators give

$$
\frac{|\bra{n+1}\hat x\ket{n}|^2}{-\hbar\omega}
 + \frac{|\bra{n-1}\hat x\ket n|^2}{+\hbar\omega}
 = \frac{\hbar}{2m\omega}\,\frac{-(n+1) + n}{\hbar\omega}
 = -\frac{1}{2m\omega^2} ;
$$

multiply by $q^2\mathcal E^2$. ∎

![Level repulsion: two coupled levels never cross — as the coupling (or a parameter sweeping the bare energies) varies, the exact energies ±√ 2 + v2 avoid each other. The second-order formula is the far wing of this hyperbola; the “avoided crossing” at the centre is where perturbation theory hands over to exact diagonalisation.](https://one-course.com/images/onecourse/chapters/physics-5/b3-perturbation-theory/fig-91fab2efa68f.svg)

*Level repulsion: two coupled levels never cross — as the coupling (or a parameter sweeping the bare energies) varies, the exact energies $\pm\sqrt{\Delta^2 + v^2}$ avoid each other. The second-order formula is the far wing of this hyperbola; the “avoided crossing” at the centre is where [perturbation theory](#thm-b3-perturbation-theory-corrections) hands over to exact diagonalisation.*

## 13.2 When levels are degenerate

**Theorem 13.3 (Degenerate perturbation theory).**

If $E_n$ is degenerate, the formula above divides by zero — the perturbation may reorganise the degenerate states completely, and the cure is to let it: *diagonalise $\hat V$ restricted to the degenerate subspace*. Its [eigenvalues](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) are the first-order shifts; its eigenvectors are the correct zeroth-order states, the combinations the perturbation itself selects.

**Proof.** *Admitted at this level.* ∎

**Example 13.4 (The linear Stark effect).**

Hydrogen’s $n = 2$ level holds the degenerate $2s$ and $2p$ states. In a field $\mathcal E\vect e_z$, $\hat V = e\mathcal E\hat z$ couples $2s$ to $2p_0$ with $\bra{2s}\hat z\ket{2p_0} = -3a_0$: diagonalising the $2\times2$ block gives shifts

$$
\Delta E = \pm\,3e a_0\mathcal E ,
$$

*linear* in the field — while the non-degenerate ground state shifts only quadratically. At $\mathcal E = 10^{7}\,\mathrm{V}/\mathrm{m}$: $\pm
1.6\,\mathrm{meV}$, an easily resolved splitting. The states doing the splitting are $(\ket{2s} \mp \ket{2p_0})/\sqrt2$ — hybrids with a permanent electric dipole, possible only because degeneracy left the atom free to polarise at zeroth order. (The same $s$–$p$ hybridisation, driven by neighbours instead of a field, is the carbon chemistry of every organic molecule.)

![The linear Stark effect in hydrogen’s n = 2 shell: the field hybridises the degenerate 2s and 2p_0 into permanent-dipole states shifted by ±3ea_0 E, leaving 2p_±1 untouched at first order.](https://one-course.com/images/onecourse/chapters/physics-5/b3-perturbation-theory/fig-a98697d4c24e.svg)

*The linear [Stark effect](#ex-b3-perturbation-theory-stark) in hydrogen’s $n = 2$ shell: the field hybridises the degenerate $2s$ and $2p_0$ into permanent-dipole states shifted by $\pm3ea_0\mathcal E$, leaving $2p_{\pm1}$ untouched at first order.*

## 13.3 Transitions: the golden rule

**Theorem 13.5 (Time-dependent perturbations).**

Switch on $\hat V(t) = \hat V\cos\omega t$ at $t = 0$. To first order, the probability of finding the system in $\ket f$ at time $t$, having started in $\ket i$, is

$$
\mathcal P_{i\to f}(t) = \frac{|\bra f\hat V\ket i|^2}{4\hbar^2}\;
\frac{\sin^2\!\big[(\omega_{fi} - \omega)\,t/2\big]}
{\big[(\omega_{fi} - \omega)/2\big]^2} ,
\qquad \omega_{fi} = \frac{E_f - E_i}{\hbar} :
$$

a resonance sharply peaked at $\hbar\omega = E_f - E_i$ — absorption when $E_f > E_i$, *stimulated emission* for the companion term with $E_f < E_i$. When the final states form a continuum of density $\rho(E)$, the peak’s growth becomes a constant *rate*:

$$
\Gamma_{i\to f} = \frac{2\pi}{\hbar}\,
|\bra f\hat V\ket i|^2\,\rho(E_f)
$$

— *Fermi’s golden rule*, the master formula of decay rates, absorption coefficients and cross-sections.

**Partial proof.** First-order amplitude: $c_f(t) = -\tfrac{\iu}{\hbar}\int_0^t\bra
f\hat V(t')\ket i\,\eu^{\iu\omega_{fi}t'}\dd t'$; keeping the near-resonant exponential and integrating gives the stated $\sin^2$ form. For the continuum: as a function of the detuning $x$, the resonance factor has height $t^2$ and width $\sim 2\pi/t$ — area $2\pi t$ — so summing over a smooth continuum replaces it by $2\pi t\,\rho$: probability linear in $t$, i.e. a rate. ∎

![Left: the first-order transition probability against detuning — an ever-narrower, ever-taller resonance whose area grows linearly in time. Right: into a continuum, that linear growth is a constant decay rate: the golden rule.](https://one-course.com/images/onecourse/chapters/physics-5/b3-perturbation-theory/fig-8a152fc55efc.svg)

*Left: the first-order transition probability against detuning — an ever-narrower, ever-taller resonance whose area grows linearly in time. Right: into a continuum, that linear growth is a constant decay rate: the golden rule.*

**Example 13.6 (What the golden rule runs).**

Selection rules: a transition happens only if the matrix element $\bra f\hat V\ket i$ survives — for light, $\hat V \propto \hat
z$, whose parity and angular momentum kill everything except $\Delta l = \pm1$, $\Delta m = 0, \pm1$: the rules invoked since [Chapter 10](https://one-course.com/books/physics/5/en/chapter/10-quantum-angular-momentum#ch-b3-quantum-angular-momentum), now theorems. Rates: the hydrogen $2p$ state’s spontaneous decay computes to $1.6\,\mathrm{ns}$ ([Exercise 13.8](#exo-b3-perturbation-theory-8)); ammonia’s maser transition, with its microwave $\omega^3$, takes months — the $\omega^3$ in radiated rates is why radio lines ([Example 12.7](https://one-course.com/books/physics/5/en/chapter/12-spin-and-two-level-systems#ex-b3-spin-two-level-hyperfine)) live for megayears while ultraviolet lines flash in nanoseconds. Absorption spectra, photoionisation, nuclear beta decay ([Chapter 25](https://one-course.com/books/physics/5/en/chapter/25-nuclear-physics#ch-b3-nuclear-physics)), scattering cross-sections ([Chapter 15](https://one-course.com/books/physics/5/en/chapter/15-scattering-theory#ch-b3-scattering-theory)): all are this one formula with different matrix elements and state counts.

**Method 13.7 (Choosing the right perturbative tool).**

(1) Static question, non-degenerate level: average $\hat V$ (first order); if that vanishes by symmetry, second order — expect level repulsion. (2) Degenerate level: diagonalise $\hat V$ in the degenerate block first; the symmetry-adapted combinations do the splitting. (3) Transition rates: golden rule — one matrix element, one [density of states](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-counting); check selection rules before computing anything. (4) Always test the expansion: mixing fraction $|V_{mn}/(E_n - E_m)| \ll 1$, else diagonalise exactly (two-level formula). (5) Exact solvable limits (oscillator in a field) are free calibrations: use them.

![Sunset as perturbation theory: air’s molecules scatter blue light far more readily than red (the 4 of this chapter’s golden-rule kin), so the long path at the horizon strains the blue away and mails it to someone else’s noon sky.](https://one-course.com/images/onecourse/chapters/physics-5/b3-perturbation-theory/img-e8ce01ebd30d.jpg)

*Sunset as [perturbation theory](#thm-b3-perturbation-theory-corrections): air’s molecules scatter blue light far more readily than red (the $\omega^4$ of this chapter’s golden-rule kin), so the long path at the horizon strains the blue away and mails it to someone else’s noon sky.*

## 13.4 Exercises

**Exercise 13.1 ★.**

A particle in a box $[0, a]$ feels the extra potential $V(x) =
\lambda x/a$. (a) Compute the first-order shift of every level (use $\langle x\rangle = a/2$ in any box state — justify it). (b) Why is the shift the same for all levels here? (c) What is the first-order change of the *spacings*, and why could a measurement of transition frequencies miss this perturbation entirely? (d) When does first order stop being trustworthy?

**Solution of Exercise 13.1.**

(a) Every $|\varphi_n|^2$ is symmetric about $a/2$, so $\langle
x\rangle = a/2$ and $\Delta E_n = \lambda/2$ for all $n$. (b) The symmetry, not the dynamics. (c) A common shift cancels in every difference: spectroscopy, which measures spacings, sees nothing at first order. (d) When $\lambda$ rivals the level spacings, the neglected state mixing (second order) matters.

**Exercise 13.2 ★.**

Carry out the second-order computation of [Proposition 13.2](#prop-b3-perturbation-theory-oscillator-field) in full: matrix elements of $\hat x$, the two terms, the cancellation of $n$, and the exact agreement with completing the square. Why does the *state*’s first-order correction not vanish although the energy’s does?

**Solution of Exercise 13.2.**

$\bra{n\pm1}\hat x\ket n = \sqrt{\hbar/2m\omega}\,\sqrt{n + \tfrac12
\pm \tfrac12}$; the two second-order terms give

$$
\frac{q^2\mathcal E^2\hbar}{2m\omega}\Big[\frac{n+1}{-\hbar\omega}
+ \frac{n}{\hbar\omega}\Big] = -\frac{q^2\mathcal E^2}{2m\omega^2} ,
$$

independent of $n$ — the exact answer. The state *is* corrected at first order (admixtures of $n \pm 1$ displace the wave function sideways); only the energy’s first-order piece vanishes, by parity.

**Exercise 13.3 ★.**

The two-level [Hamiltonian](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#def-b3-hamiltonian-mechanics-hamiltonian) $\begin{pmatrix}E_1 & v\\ v &
E_2\end{pmatrix}$, $E_1 < E_2$, $v$ real. (a) Find the exact [eigenvalues](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable). (b) Expand for $|v| \ll E_2 - E_1$ and identify the second-order formula. (c) Show the levels repel: the gap always exceeds $E_2 - E_1$. (d) At $E_1 = E_2$: what does [perturbation theory](#thm-b3-perturbation-theory-corrections) give, what does the exact answer give, and which theorem of the chapter reconciles them?

**Solution of Exercise 13.3.**

(a) $E_\pm = \bar E \pm \sqrt{(\Delta/2)^2 + v^2}$ with $\Delta =
E_2 - E_1$. (b) $E_- \approx E_1 - v^2/\Delta$, $E_+ \approx E_2 +
v^2/\Delta$: the second-order formula, term by term. (c) The gap $2\sqrt{(\Delta/2)^2 + v^2} \ge \Delta$, with equality only at $v =
0$. (d) The expansion diverges; the exact answer gives $\pm|v|$ — which is precisely what diagonalising $\hat V$ in the degenerate block ([Theorem 13.3](#thm-b3-perturbation-theory-degenerate)) prescribes.

**Exercise 13.4 ★.**

For [Theorem 13.1](#thm-b3-perturbation-theory-corrections): (a) show the first-order state correction is orthogonal to $\ket n$; (b) show the mixing coefficient of $\ket m$ is $V_{mn}/(E_n - E_m)$ and state the validity criterion; (c) a level $1\,\mathrm{eV}$ from its nearest neighbour is coupled to it by $0.1\,\mathrm{eV}$: estimate the admixture probability; (d) same with a $10\,\mathrm{meV}$ gap — which tool should replace the expansion?

**Solution of Exercise 13.4.**

(a) The expansion assigns $\ket n$ coefficient $1$; normalisation to first order forces the correction into the orthogonal complement. (b) From the projection on $\bra m$; small mixing requires $|V_{mn}| \ll |E_n - E_m|$. (c) $(0.1)^2 = 1\%$. (d) $(0.1/0.01)^2 = 100$: nonsense — diagonalise the two-level block exactly.

**Exercise 13.5 ★★.**

Anharmonic bonds. Add $\hat V = \beta\hat x^3 + \gamma\hat x^4$ to the oscillator. (a) Show the $x^3$ term shifts no level at first order (parity). (b) Show the $x^4$ term gives $\Delta E_n =
3\gamma(\hbar/2m\omega)^2(2n^2 + 2n + 1)$ (expand $\hat x^4$ in [ladder operators](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#def-b3-harmonic-oscillator-ladder); only the “balanced” terms survive). (c) For a real bond the effective $\gamma$ is negative (the well softens outward): show the level spacing then *shrinks* with $n$. (d) Which observed feature of molecular spectra ([Exercise 9.4](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#exo-b3-harmonic-oscillator-4)) does this reproduce?

**Solution of Exercise 13.5.**

(a) $\hat x^3$ is odd: its average in any parity eigenstate vanishes. (b) Writing $\hat x^4 \propto (\hat a + \hat
a^\dagger)^4$ and keeping the six terms with equal raisings and lowerings gives $\langle(\hat a + \hat a^\dagger)^4\rangle_n = 6n^2
+ 6n + 3$: the stated shift. (c) With $\gamma < 0$ the shift grows more negative like $n^2$: successive spacings $E_{n+1} - E_n$ decrease linearly in $n$. (d) The converging rungs of real vibrational ladders — overtones slightly less than multiples of the fundamental, and a finite dissociation limit.

**Exercise 13.6 ★★.**

In the square two-dimensional box, the degenerate pair $\ket{1,2},
\ket{2,1}$ is perturbed by $\hat V = \lambda\,xy$. (a) Explain why non-degenerate theory fails. (b) Compute the $2\times2$ matrix of $\hat V$ in the pair (the integrals factorise; $\int_0^a
x\sin(\pi x/a)\sin(2\pi x/a)\dd x = -8a^2/9\pi^2$). (c) Find the splittings and the correct zeroth-order states. (d) Compare with the symmetry analysis of [Exercise 8.9](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#exo-b3-quantum-formalism-9): which answer did symmetry give for free?

**Solution of Exercise 13.6.**

(a) The pair is degenerate: denominators vanish. (b) Diagonal elements $\lambda(a/2)^2$; off-diagonal $\lambda c^2$ with $c =
-8a^2/9\pi^2$ — the integrals factorise into the two one-dimensional pieces. (c) Shifts $\lambda[(a/2)^2 \pm
(8a^2/9\pi^2)^2/a^2\cdot a^2]$ — i.e. $\lambda a^2(\tfrac14 \pm
64/81\pi^4)$ — with eigenstates $(\ket{1,2} \pm \ket{2,1})/
\sqrt2$. (d) Symmetry had already named the eigenstates (the swap-even and swap-odd combinations); the perturbation could only choose those, and it did.

**Exercise 13.7 ★★.**

[Stark effects](#ex-b3-perturbation-theory-stark), linear and quadratic. (a) Using [Example 13.4](#ex-b3-perturbation-theory-stark), compute the $n = 2$ splitting at $\mathcal E = 2.5 \times 10^{6}\,\mathrm{V}/\mathrm{m}$. (b) Why does the ground state show *no* linear shift (two reasons: parity, and no degenerate partner)? (c) Its quadratic shift defines the polarisability, $\Delta E = -\tfrac12\alpha\mathcal E^2$: estimate $\alpha \sim e^2a_0^2/E_{\text{I}}$ and evaluate in units of $a_0^3$ (the exact answer is $4.5\,a_0^3 \times
4\pi\varepsilon_0$). (d) Which everyday property of matter — its dielectric response — did you just compute the atomic seed of?

**Solution of Exercise 13.7.**

(a) $3ea_0\mathcal E = 3 \times 5.29 \times 10^{-11} \times 2.5 \times 10^{6}
= 0.40\,\mathrm{meV}$ each way. (b) $\bra{100}\hat z\ket{100} = 0$ by parity, and no degenerate partner exists to hybridise with. (c) $\alpha \sim 2e^2a_0^2/E_{\text{I}} \to \alpha/4\pi\varepsilon_0
\sim 2a_0^3$, the right order beside the exact $4.5\,a_0^3$. (d) The dielectric constant: every capacitor’s $\varepsilon_{\text{r}}$ is atoms answering fields by exactly this second-order yielding.

**Exercise 13.8 ★★.**

Spontaneous emission (with one admitted input): an excited state decays at $A = \omega^3|d_{fi}|^2/3\pi\varepsilon_0\hbar c^3$, where $d_{fi} = e\bra f\hat z\ket i$-type matrix elements set the dipole. (a) For hydrogen $2p \to 1s$: with $|d| \approx 0.74\,
ea_0$ and $\lambda = 121.6\,\mathrm{nm}$, compute $A$ and the lifetime. (b) Scale to the sodium D line. (c) Scale to the hyperfine 21 cm transition ($|d| \to$ a magnetic moment, rate down by another $\alpha^2$-ish factor; accept $A \approx
2.9 \times 10^{-15}\,\mathrm{s}^{-1}$): check the “ten million years”. (d) From the $\omega^3$: why are ultraviolet lines fast, radio lines eternal, and why did that make the [21 cm line](https://one-course.com/books/physics/5/en/chapter/12-spin-and-two-level-systems#ex-b3-spin-two-level-hyperfine) *predictable* but almost undetectable in a laboratory?

**Solution of Exercise 13.8.**

(a) $A = \omega^3|d|^2/3\pi\varepsilon_0\hbar c^3 \approx
6 \times 10^{8}\,\mathrm{s}^{-1}$: $\tau = 1.6\,\mathrm{ns}$. (b) $\omega$ smaller by $4.8$: $\omega^3$ by $110$; with its somewhat larger dipole the D line comes out at $\sim16\,\mathrm{ns}$ — as measured. (c) $1/A = 3.4 \times 10^{14}\,\mathrm{s} \approx 11$ million years. (d) $\omega^3$ spans $10^{20}$ between ultraviolet and radio: fast UV flashes, geological radio patience — so the [21 cm line](https://one-course.com/books/physics/5/en/chapter/12-spin-and-two-level-systems#ex-b3-spin-two-level-hyperfine) was predicted (van de Hulst, 1944) from theory and sought in the sky, where $10^{66}$ atoms compensate the patience.

**Exercise 13.9 ★★.**

Selection rules as integrals. (a) Show $\bra{100}\hat z\ket{200} =
0$ by parity. (b) Show $\bra{100}\hat z\ket{21m} = 0$ unless $m =
0$ (the $\varphi$ integral). (c) State the resulting rules $\Delta
l = \pm1$, $\Delta m = 0, \pm1$ and their physical reading (the photon’s spin). (d) The $2s$ state can reach $1s$ by no dipole route: what does that predict for its lifetime, and what “forbidden” two-photon path nature actually takes?

**Solution of Exercise 13.9.**

(a) Both states even times odd $\hat z$: odd integrand. (b) The $\varphi$ integral $\int_0^{2\pi}\eu^{\iu m\varphi}\dd\varphi$ vanishes unless $m = 0$. (c) The photon carries one $\hbar$ and odd parity: $l$ must change by one, $m$ by at most one. (d) With every one-photon door closed, $2s$ lives $\sim0.12\,\mathrm{s}$ — eight orders beyond $2p$ — decaying by simultaneous emission of *two* photons, a faint continuum actually observed in planetary nebulae.

**Exercise 13.10 ★★★.**

Polarisability, honestly. The second-order shift of hydrogen’s ground state in a field is $\Delta E = -e^2\mathcal E^2\sum_{m\neq
0}|\bra m\hat z\ket 0|^2/(E_m - E_0)$. (a) Bound the sum by replacing every denominator by the smallest gap $E_2 - E_1 =
\tfrac34 E_{\text{I}}$ and using the closure relation $\sum_m|\bra
m\hat z\ket0|^2 = \langle z^2\rangle_0 = a_0^2$: obtain $\alpha
\le \tfrac{16}3\,a_0^3$ (in Gaussian-style units of $4\pi\varepsilon_0$). (b) Compare with the exact $\tfrac92 a_0^3$. (c) From $\alpha$, estimate the refractive index of hydrogen gas at atmospheric density ($n - 1 \approx N\alpha/2\varepsilon_0
\cdot e^2\dots$ — use $n - 1 = N\alpha_{\text{SI}}/2
\varepsilon_0$) and compare with the measured $1.3 \times 10^{-4}$. (d) State in one sentence the chain atom $\to$ polarisability $\to$ refraction that Part II of the weekend problem will ride.

**Solution of Exercise 13.10.**

(a) Replacing denominators by the smallest, $\tfrac34
E_{\text{I}}$, and using closure: $\alpha/4\pi\varepsilon_0 \le
2\,a_0^2\,(2E_{\text{I}})\,/\,(\tfrac34 E_{\text{I}})\cdot\tfrac12
= \tfrac{16}3\,a_0^3$. (b) The exact $4.5\,a_0^3$ sits below the bound, as it must. (c) $n - 1 = N\alpha_{\text{SI}}/2\varepsilon_0
\approx 10^{-4}$ against the measured $1.3 \times 10^{-4}$: the atom’s second-order softness *is* the gas’s refraction. (d) One matrix-element sum fixes how an atom yields to a field, and the collective yielding of $10^{25}$ atoms per cubic metre bends light: [perturbation theory](#thm-b3-perturbation-theory-corrections) $\to$ polarisability $\to$ optics.

**Exercise 13.11 ★★★.**

Avoided crossings and slow sweeps. A two-level system has bare energies $\pm\lambda t$ crossing at $t = 0$, coupled by $v$. (a) Sketch the exact levels against $t$: an avoided crossing of gap $2|v|$. (b) If the sweep is *slow*, the system follows the lower curve (adiabatic theorem, admitted): what state does it end in? (c) The Landau–Zener criterion compares the sweep time across the gap region, $\sim v/\lambda$, with the internal clock $\hbar/v$: give the condition for adiabatic following. (d) Two uses: a qubit’s state transferred by a slow chirp; an atomic collision hopping charge between nuclei — explain one of them in two sentences.

**Solution of Exercise 13.11.**

(a) Two hyperbola branches, closest approach $2|v|$ at $t = 0$. (b) In the state that follows the lower curve continuously — the initial “diabatic” state has swapped character: complete transfer. (c) Crossing time $\sim v/\lambda$ long against $\hbar/v$: adiabatic when $v^2/\hbar\lambda \gg 1$. (d) Qubit: sweep a control field slowly through the avoided crossing and the population rides the lower branch from one basis state to the other — a transfer robust against timing errors, used daily in adiabatic state preparation.

**Exercise 13.12 ★★★.**

Second order is a pessimist, and other theorems. (a) Prove that the second-order correction to the *ground* state is always negative. (b) Conclude (with [Exercise 9.10](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#exo-b3-harmonic-oscillator-10)) why van der Waals forces are always attractive between ground-state atoms. (c) Show that first-order theory *overestimates* every ground-state energy: it is a variational bound (evaluate $\bra{\psi_0}\hat
H\ket{\psi_0}$ with the unperturbed state as trial). (d) A consistency check on [Example 13.4](#ex-b3-perturbation-theory-stark): why is the *linear* [Stark effect](#ex-b3-perturbation-theory-stark) not a violation of (a)?

**Solution of Exercise 13.12.**

(a) Every term of the sum has a square over a negative denominator. (b) The leading interatomic effect of two ground-state atoms is second order in their dipole–dipole coupling: negative — always attraction, for any pair of species. (c) $\bra{\psi_0}\hat H\ket{
\psi_0} = E_0 + \bra{\psi_0}\hat V\ket{\psi_0}$ is the energy of a trial state, hence $\ge$ the true ground energy: first order can only overshoot. (d) The Stark-split $n = 2$ states are excited states; the theorem concerns the true ground state, whose linear shift vanishes — no conflict.

## 13.5 Problem: The colour of the sky

**Problem 13.1.**

Weekend problem — Rayleigh scattering from perturbed atoms

Why is the day sky blue, the sunset red, and the cloud white? The complete answer is a chain through this chapter: a light wave perturbs each air molecule; the induced dipole re-radiates (the dipole radiation of the Year 2 volume); and interference decides what survives. Data: molecular polarisability of air $\alpha/4\pi\varepsilon_0 \approx 2.2 \times 10^{-30}\,\mathrm{m}^{3}$; number density $N = 2.5 \times 10^{25}\,\mathrm{m}^{-3}$; visible range $450$–$700\,\mathrm{nm}$.

**Part I — The perturbed molecule.**

1. A static field $\mathcal E$ shifts a molecule’s ground state by $-\tfrac12\alpha\mathcal E^2$ : connect this statement to second-order [perturbation theory](#thm-b3-perturbation-theory-corrections) (which formula, and why negative?).
2. The same $\alpha$ gives the induced dipole $p =  \alpha\mathcal E$ : for sunlight’s field $\mathcal E \sim  800\,\mathrm{V}/\mathrm{m}$ , compute the induced dipole of one nitrogen molecule and compare it with $ea_0$ .
3. Why may the optical field (frequency $5 \times 10^{14}$ Hz) be treated with the *static* polarisability for molecules whose electronic transitions lie in the ultraviolet? (Compare $\hbar\omega$ with the gap; this is the far-off-resonance limit of the golden-rule denominator.)
4. An oscillating dipole $p_0\cos\omega t$ radiates the average power $P = p_0^2\omega^4/12\pi\varepsilon_0c^3$ (Year 2 volume). Combining with $p_0 = \alpha\mathcal  E_0$ : how does the scattered power depend on $\omega$ at fixed illumination?
5. Compute the ratio of scattered powers at $450\,\mathrm{nm}$ and $700\,\mathrm{nm}$ .
6. In one sentence: why is the sky blue and not violet (two ingredients: the Sun’s spectrum falling toward the violet, and the eye’s sensitivity)?

**Part II — From one molecule to the sky.**

7. The scattering cross-section works out to $\sigma =  \dfrac{8\pi}{3}\Big(\dfrac{\alpha}{4\pi\varepsilon_0}  \Big)^2\dfrac{\omega^4}{c^4}$ : verify its dimensions and evaluate it at $550\,\mathrm{nm}$ .
8. The attenuation length is $\ell = 1/N\sigma$ : evaluate it at $550\,\mathrm{nm}$ and compare with the thickness of the atmosphere ( $\sim8\,\mathrm{km}$ equivalent at sea-level density).
9. At $450\,\mathrm{nm}$ versus $700\,\mathrm{nm}$ , what fraction of a vertical sunbeam is scattered out? (Use $1 -  \eu^{-L/\ell}$ with $L = 8\,\mathrm{km}$ .)
10. At sunset the path is thirty times longer: recompute the blue’s survival and explain the colour of the low Sun — and of the light that reddens the Moon in a lunar eclipse.
11. Why is the sky’s light *polarised* at $90^\circ$ from the Sun (recall the dipole radiation pattern: no emission along the dipole’s axis)?
12. Bees and Vikings allegedly navigated by this polarisation: what does an overcast day do to it, and why?

**Part III — Why the sky is not brighter.**

13. A paradox: in a perfectly uniform medium, the wavelets from neighbouring volume elements interfere destructively sideways, and *no* light would scatter at all (only refraction survives). What actually breaks the uniformity of air — what is randomly distributed?
14. Density fluctuations of an ideal gas make the scattered intensities of independent molecules *add* : state why independence (random positions, wavelength-scale separations) kills the destructive interference.
15. From your $\ell$ at $550\,\mathrm{nm}$ : what fraction of sunlight does one clear-day atmosphere scatter — does the number match the everyday brightness of the sky against the Sun?
16. A cloud droplet ( $10\,\text{µ}\mathrm{m}$ ) holds $\sim10^{12}$ molecules within a wavelength: their wavelets add *coherently* . How does the scattered power then scale with molecule number, and why does the $\omega^4$ colour selection disappear (all wavelengths scatter strongly): what colour is the cloud?
17. Milk, fog, and white paint are white for the same reason: state the general rule — scatterers small against $\lambda$ colour the light, scatterers large against $\lambda$ whiten it.
18. Critical opalescence: near a liquid’s critical point (the phase-change chapters of the Year 1 volume, and [Chapter 21](https://one-course.com/books/physics/5/en/chapter/21-phase-transitions#ch-b3-phase-transitions) ahead), density fluctuations grow to optical scales and the fluid turns milky: connect to item 13.

**Part IV — The chain, completed.**

19. The forward-scattered wavelets do *not* cancel: they interfere with the beam and slow its phase — the refractive index. Using $n - 1 = N\alpha/2\varepsilon_0$ (with $\alpha$ in SI), evaluate $n - 1$ for air and compare with the measured $2.9 \times 10^{-4}$ .
20. One constant $\alpha$ , computed by second-order [perturbation theory](#thm-b3-perturbation-theory-corrections) , thus fixes three phenomena at once: name them (shift in a static field; the sky’s blue; the bending of light in air).
21. Ozone absorption, aerosols and multiple scattering complicate real skies: name one observable each (twilight’s colours; milky horizon; the sky’s residual brightness at the zenith after sunset).
22. Why does the Moon, airless, carry a *black* daytime sky — and what did every photograph from its surface thereby confirm about the origin of ours?
23. Mars’s daytime sky is butterscotch and its sunsets are *blue* : its thin air scatters little, and suspended micrometre dust grains do the work instead. Using the rule of item 18, explain how dust reverses the colour logic.
24. For X-rays, $\hbar\omega$ far exceeds every molecular transition: the electrons respond as if free and the scattering (Thomson) loses its $\omega^4$ . What happens to the “sky-blue” mechanism in that regime — and why is that flat response exactly what makes X-rays clean probes of electron density?
25. Summarise the named result: a $2.2 \times 10^{-30}\,\mathrm{m}^{3}$ polarisability, squared and multiplied by $\omega^4$ , gives a $60\,\mathrm{km}$ -scale scattering length in the green — long enough that the noon sky is gentle, short enough that sunsets redden and the heavens are blue, with the polarisation pattern of a forest of driven dipoles.

**Solution of Problem 13.1.**

**1.** $\hat V = e\mathcal E\hat z$ has no first-order average (parity); the second-order sum, all denominators negative for the ground state, gives $-\tfrac12\alpha\mathcal E^2$ with $\alpha$ the polarisability — negative because levels repel from above. **2.** $\alpha_{\text{SI}} = 4\pi\varepsilon_0 \times
2.2 \times 10^{-30} = 2.4 \times 10^{-40}\,\mathrm{F}\,\mathrm{m}^{2}$: $p = 2 \times 10^{-37}\,\mathrm{C}\,\mathrm{m}
\approx 2 \times 10^{-8}\,ea_0$ — a hundred-millionth of an atomic dipole, per molecule. **3.** $\hbar\omega \approx 2.3\,\mathrm{eV}$ against ultraviolet gaps $\gtrsim10\,\mathrm{eV}$: the perturbative denominators barely feel the drive, so the static $\alpha$ serves. **4.** $p_0 = \alpha\mathcal E_0$ is frequency-flat, so $P
\propto \omega^4$: Rayleigh’s law. **5.** $(700/450)^4 = 5.9$: blue light scatters six times more than red. **6.** The Sun emits less violet than blue and the eye’s violet sensitivity is poor: the scattered sky peaks, *for us*, at blue. **7.** $[\alpha/4\pi\varepsilon_0] = \mathrm{m}^{3}$ and $\omega^4/c^4 = \mathrm{m}^{-4}$: an area. At $550\,\mathrm{nm}$: $\sigma \approx 7 \times 10^{-31}\,\mathrm{m}^{2}$. **8.** $\ell = 1/N\sigma \approx 58\,\mathrm{km}$: several atmospheres thick — air is *almost* transparent. **9.** Blue ($\ell \approx 26\,\mathrm{km}$): $1 -
\eu^{-8/26} = 26\%$ scattered; red ($\ell \approx 150\,\mathrm{km}$): $5\%$. The sky is built from that difference. **10.** Over $\sim240\,\mathrm{km}$ of low-Sun path the blue survives $\eu^{-9} \sim 10^{-4}$ while red keeps $\sim20\%$: the Sun sets red — and the same red, refracted by Earth’s ring of sunsets, paints the eclipsed Moon copper. **11.** A dipole radiates nothing along its own axis: at $90^\circ$ scattering, the component of molecular oscillation along the line of sight cannot contribute, and the surviving light is polarised perpendicular to the Sun–molecule–eye plane. **12.** Clouds impose multiple scattering, which scrambles the geometry: the polarisation compass dies under overcast — which is what makes its alleged Viking use a feat. **13.** The molecules’ *positions*: air is a random gas, its density fluctuating on every scale — perfectly ordered matter (a crystal, ideally) scatters only into the refracted beam. **14.** Random, wavelength-scale-separated scatterers add with random phases: cross terms average away and intensities add — the destructive conspiracy needs order, and disorder repeals it. **15.** $1 - \eu^{-8/58} \approx 13\%$ of sunlight feeds the blue dome — consistent with a sky thousands of times dimmer than the solar disc yet bright enough to read by. **16.** Within a droplet the wavelets add *amplitudes*: power $\propto N^2$, overwhelming and nearly wavelength-blind (the droplet is far larger than $\lambda$): clouds scatter everything — white. **17.** Sub-wavelength scatterers select colour ($\omega^4$); super-wavelength scatterers reflect geometrically and whiten: milk, fog, paint, snow. **18.** Near the critical point the density fluctuations themselves grow to optical size: the fluid becomes its own cloud — critical opalescence, Rayleigh’s mechanism amplified to opacity. **19.** $n - 1 = 2.5 \times 10^{25} \times 2.4 \times 10^{-40}/(2 \times
8.85 \times 10^{-12}) \approx 3.4 \times 10^{-4}$, against the measured $2.9 \times 10^{-4}$: the forward wavelets, coherent with the beam, slow its phase. **20.** The Stark shift of a level; the blue of the sky; the refraction (and mirages, and lenses of air) of light — one $\alpha$, three phenomena. **21.** Ozone’s absorption colours the zenith twilight blue; aerosols whiten the horizon; multiply scattered light keeps the sky faintly luminous after sunset. **22.** No air, no dipoles: the Sun blazes in a black sky — proving by absence that our blue dome is scattered sunlight, not a glowing property of “space”. **23.** Martian dust grains are large against $\lambda$: they scatter red light efficiently in all directions (tinting the sky) while diffracting blue most strongly *forward* — so the sky is butterscotch and the halo around the setting Sun is blue: the grain size flips the logic of item 18. **24.** Thomson scattering is frequency-flat: no colour selection, no blue — but a response proportional simply to *electron density* is precisely what crystallography and radiography want in a probe. **25.** One second-order constant, $2.2 \times 10^{-30}\,\mathrm{m}^{3}$, squared and weighted by $\omega^4$, yields a $\sim60\,\mathrm{km}$ green-light scattering length: enough transparency for noon, enough scattering for a blue dome, polarised like a field of driven dipoles, reddening every sunset on schedule.