Part 3 · One-dimensional problems

3.5 The harmonic oscillator

Source: the analytic solution, Hermite polynomials and the ladder operators from the 2024/25 boards #12 and #13; this year’s course has not reached it yet

Objectives

Understand why anything near a stable equilibrium is an oscillator, and why its levels are equally spaced.
Recipes: solve the oscillator with a series; solve it again with ladder operators.
Picture: a ladder of equally spaced levels inside a parabola, each eigenfunction leaking past its turning points.
Theorems: $E_{n} = ℏ ω (n + \frac{1}{2})$ ; $\overset{a}{^} ∣ n ⟩ = n ∣ n - 1 ⟩$ , $\overset{a}{^}^{+} ∣ n ⟩ = n + 1 ∣ n + 1 ⟩$ .
Vocabulary: Hermite polynomials, zero-point energy, creation and annihilation operators, number operator.

The classical oscillator

A mass on a spring, sliding along $x$ . ==The force always points back to the rest position, and grows with the displacement== (Hooke’s law): $\overset{ˉ}{F} = - k x \overset{u}{^}_{x} = m \overset{a}{ˉ} ⟹ \frac{d ^{2} x}{d t ^{2}} + ω^{2} x = 0, ω^{2} = \frac{k}{m},$ with solution $x (t) = A sin (ω t + φ)$ , period $2 π / ω$ . The amplitude $A$ and phase $φ$ are fixed by how the motion starts; $ω$ depends only on $k$ and $m$ .

Why it is everywhere. Expand any force about an equilibrium $x_{0}$ : $f (x) = f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \dots$ , and $f (x_{0}) = 0$ . For small displacements the linear term wins, $f \approx - k (x - x_{0})$ with $k = - f^{'} (x_{0})$ . ==Anything slightly displaced from a stable equilibrium is a harmonic oscillator==: atoms in a bond, atoms in a crystal, a pendulum.

The potential energy is $U = - \int F d x = \frac{1}{2} k x^{2}$ , and $E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$ . The ball rolls between the turning points $\pm A$ where $\frac{1}{2} k A^{2} = E$ . Classically any $E \geq 0$ is allowed, the lowest being the ball at rest at the bottom (beyond the boards). Unlike the finite well, the parabola never levels off: every state is bound. Like the infinite well it confines forever, but its walls slope, so the region the particle can reach widens with $E$ . That is why the levels come out evenly spaced rather than growing as $n^{2}$ .

The analytic solution

With $k = m ω^{2}$ , the Hamiltonian is $\hat{H} = - \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}} + \frac{1}{2} m ω^{2} x^{2} .$ Multiply $\hat{H} u_{E} = E u_{E}$ by $- 2 m / ℏ^{2}$ , call $α = mω /ℏ$ , divide by $α^{2}$ , and change to the dimensionless position $ε = αx$ : $(\frac{d ^{2}}{d ε ^{2}} - ε^{2} + \frac{2 E}{ℏ ω}) u_{E} = 0.$ $α$ is an inverse length, so the energy appears only as $2 E /ℏ ω$ : ==the oscillator’s natural unit of energy is $ℏ ω$ == (beyond the boards).

Far from the centre, $ε^{2}$ beats any fixed energy, and $u^{''} - ε^{2} u \approx 0$ gives $u \approx e^{- ε^{2} /2}$ ; the other solution, $e^{+ ε^{2} /2}$ , cannot be normalised. (The board justifies dropping the energy term with “if $E ≪ ℏ ω$ ”; what makes it negligible is $ε^{2} ≫ 2 E /ℏ ω$ .)

Theorem 3.5.1Energies of the harmonic oscillator2024/25 board #12

$E_{n} = ℏ ω (n + \frac{1}{2}), u_{n} (ε) = N_{n} H_{n} (ε) e^{- ε^{2} /2}, n = 0, 1, 2, \dots$ with the Hermite polynomials and normalisation $H_{n} (ε) = (- 1)^{n} e^{ε^{2}} \frac{d ^{n}}{d ε ^{n}} e^{- ε^{2}}, N_{n} = (\frac{α}{π n ! 2 ^{n}})^{1/2} .$

The series must stop

The board goes straight from the series to the result; the steps are beyond the boards.

Put $u = H (ε) e^{- ε^{2} /2}$ into the equation. The $ε^{2} H$ terms cancel and, with $λ = 2 E /ℏ ω$ , $H^{''} - 2 ε H^{'} + (λ - 1) H = 0$ .
Insert $H = \sum_{j} a_{j} ε^{j}$ . Each power gives $a_{j + 2} = \frac{2 j + 1 - λ}{( j + 1 ) ( j + 2 )} a_{j} .$
If the series never stops, $a_{j + 2} / a_{j} \approx 2/ j$ for large $j$ , the ratio of the series of $e^{ε^{2}}$ . Then $u \sim e^{+ ε^{2} /2}$ blows up.
So it must stop: at some $j = n$ the numerator vanishes, $λ = 2 n + 1$ , which is $E_{n}$ . $H$ is then a polynomial of degree $n$ .

$N_{n}$ is stated on the board without proof; $N_{0}$ is derived on board 13 (Theorem 3.5.8).

==Normalisability is what quantises the energy==, as the walls do in the infinite well.

Key observation 3.5.2Equally spaced, never zero2024/25 board #12

==The levels are equally spaced, by $ℏ ω$ ==: one quantum of the classical frequency.
==The lowest is $\frac{1}{2} ℏ ω$ , not zero.== The classical lowest state, at rest at the bottom, would have position and momentum both exactly 0, which $[\overset{x}{^}, \overset{p}{^}] = i ℏ$ forbids (Theorem 2.4.3). This zero-point energy is the zero-point vibration of a chemical bond.

$n$	$E_{n}$	$u_{n}$	nodes	parity
0	$\frac{1}{2} ℏ ω$	$N_{0} e^{- ε^{2} /2}$	none	even
1	$\frac{3}{2} ℏ ω$	$N_{1} 2 ε e^{- ε^{2} /2}$	$ε = 0$	odd
2	$\frac{5}{2} ℏ ω$	$N_{2} (4 ε^{2} - 2) e^{- ε^{2} /2}$	$ε = \pm \frac{1}{2}$	even

$H_{1} = - e^{ε^{2}} \frac{d}{d ε} e^{- ε^{2}} = 2 ε$ by the chain rule; $H_{2} = 4 ε^{2} - 2$ by differentiating once more. At the exam the polynomials are given. $u_{n}$ has $n$ nodes and parity $(- 1)^{n}$ . Each spreads wider as $n$ grows, with turning points at $ε = \pm 2 n + 1$ , and its tails go past them into the forbidden region.

Loading the demo…

Figure 3.5.1 An electron in a parabola with

ℏ ω = 0.2

eV. Shoot for the levels: they come at 0.1, 0.3, 0.5 eV, evenly spaced, the first half a step above the bottom. Each

u_{n}

has

n

nodes and pokes past the parabola into the shaded forbidden region.

Example 3.5.3A parabolic quantum dotbeyond the boards

An electron with $1/ α = 1$ nm: $ω = ℏ α^{2} / m = 1.16 \times 1 0^{14}$ s⁻¹, so $ℏ ω = 0.076$ eV and $E_{0} = 0.038$ eV. Room-temperature $k_{B} T = 0.026$ eV is smaller than the spacing: the electron sits mostly in the ground state.

Ladder operators

The second solution uses only operator algebra: no differential equation, no series. The idea (beyond the boards): in dimensionless variables the energy is $\frac{ℏ ω}{2} (Q^{2} + P^{2})$ , and for numbers $Q^{2} + P^{2} = (Q - i P) (Q + i P)$ . For operators that factorisation costs an extra $\frac{1}{2}$ , because $\hat{Q}$ and $\hat{P}$ do not commute.

Rescale. With $\hat{Q} = α \overset{x}{^} = \overset{x}{^} mω /ℏ$ and $\hat{P} = \overset{p}{^} / m ℏ ω$ , both dimensionless and Hermitian, $\hat{H} = \frac{p ^ ^{2}}{2 m} + \frac{1}{2} m ω^{2} \overset{x}{^}^{2} = \frac{ℏ ω}{2} (\hat{P}^{2} + \hat{Q}^{2}), [\hat{Q}, \hat{P}] = \frac{1}{ℏ} [\overset{x}{^}, \overset{p}{^}] = i .$

Definition 3.5.4Annihilation and creation operators2024/25 board #12

$\overset{a}{^} = \frac{1}{2} (\hat{Q} + i \hat{P}), \overset{a}{^}^{+} = \frac{1}{2} (\hat{Q} - i \hat{P}) .$

$\overset{a}{^}^{+}$ is the adjoint of $\overset{a}{^}$ : the example $\hat{G} = \hat{A} + i \hat{B}$ of Definition 1.4.8. So == $\overset{a}{^}$ is not Hermitian and not an observable==; it moves between states.

Definition 3.5.5Number operator

$\hat{N} = \overset{a}{^}^{+} \overset{a}{^}, \hat{H} = ℏ ω (\hat{N} + \frac{1}{2}), \hat{N} ∣ n ⟩ = n ∣ n ⟩ .$

The Hamiltonian in ladder operators

$\overset{a}{^}^{+} \overset{a}{^} = \frac{1}{2} (\hat{Q} - i \hat{P}) (\hat{Q} + i \hat{P}) = \frac{1}{2} (\hat{Q}^{2} + \hat{P}^{2}) + \frac{i}{2} [\hat{Q}, \hat{P}] = \frac{1}{2} (\hat{Q}^{2} + \hat{P}^{2}) - \frac{1}{2} .$ The four products are $\hat{Q}^{2}$ , $i \hat{Q} \hat{P}$ , $- i \hat{P} \hat{Q}$ and $+ \hat{P}^{2}$ ; for numbers the commutator would be zero. Set beside $E_{n} = ℏ ω (n + \frac{1}{2})$ , the energy eigenstates $∣ n ⟩$ are eigenstates of $\hat{N}$ with eigenvalue $n$ .

==The $\frac{1}{2}$ is the zero-point energy, and it comes from $[\overset{x}{^}, \overset{p}{^}] \neq = 0$ .== $\hat{N}$ counts the quanta $ℏ ω$ above the ground state. It is never negative, since $n = ⟨ n ∣ \overset{a}{^}^{+} \overset{a}{^} ∣ n ⟩$ is the squared length of $\overset{a}{^} ∣ n ⟩$ (§1.4 Hermitian operators and Dirac notation, beyond the boards).

Theorem 3.5.6Commutators of the ladder operators2024/25 board #12

$[\overset{a}{^}, \overset{a}{^}^{+}] = \hat{I}, [\overset{a}{^}, \hat{H}] = ℏ ω \overset{a}{^}, [\overset{a}{^}^{+}, \hat{H}] = - ℏ ω \overset{a}{^}^{+} .$

Proof

$\overset{a}{^} \overset{a}{^}^{+}$ is the same expansion in the other order, which flips the sign of the commutator term: $\frac{1}{2} (\hat{Q}^{2} + \hat{P}^{2}) + \frac{1}{2}$ . Subtracting $\overset{a}{^}^{+} \overset{a}{^}$ leaves 1. Then $[\overset{a}{^}, \hat{H}] = ℏ ω (\overset{a}{^} \overset{a}{^}^{+} \overset{a}{^} - \overset{a}{^}^{+} \overset{a}{^} \overset{a}{^}) = ℏ ω [\overset{a}{^}, \overset{a}{^}^{+}] \overset{a}{^} = ℏ ω \overset{a}{^};$ the $\frac{1}{2}$ commutes with everything. $[\overset{a}{^}^{+}, \hat{H}]$ the same way, factoring $\overset{a}{^}^{+}$ out to the left (beyond the boards).

Lowering and raising

Theorem 3.5.7The ladder2024/25 boards #12–#13

$\hat{H} (\overset{a}{^} ∣ n ⟩) = (E_{n} - ℏ ω) \overset{a}{^} ∣ n ⟩, \hat{H} (\overset{a}{^}^{+} ∣ n ⟩) = (E_{n} + ℏ ω) \overset{a}{^}^{+} ∣ n ⟩,$ and with the phases chosen real and positive, $\overset{a}{^} ∣ n ⟩ = n ∣ n - 1 ⟩, \overset{a}{^}^{+} ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ .$

Proof

$\overset{a}{^}$ lowers. $\hat{H} \overset{a}{^} = \overset{a}{^} \hat{H} - [\overset{a}{^}, \hat{H}] = \overset{a}{^} (\hat{H} - ℏ ω)$ , so $\hat{H} \overset{a}{^} ∣ n ⟩ = (E_{n} - ℏ ω) \overset{a}{^} ∣ n ⟩$ : == $\overset{a}{^} ∣ n ⟩$ is again an energy eigenstate, one rung down.== In 1D each energy has a single state, so $\overset{a}{^} ∣ n ⟩ = c_{n} ∣ n - 1 ⟩$ , and the bra is $⟨ n ∣ \overset{a}{^}^{+} = c_{n}^{*} ⟨ n - 1∣$ . $\overset{a}{^}^{+}$ raises the same way (derived beyond the boards): $\overset{a}{^}^{+} ∣ n ⟩ = D_{n} ∣ n + 1 ⟩$ .
$D_{n}$ . $⟨ n ∣ \overset{a}{^} \overset{a}{^}^{+} ∣ n ⟩ = ∣ D_{n} ∣^{2}$ , acting left and right; and $\overset{a}{^} \overset{a}{^}^{+} = \hat{N} + \hat{I}$ gives $n + 1$ . So $D_{n} = n + 1$ . (Board 13 writes $(n + 1) ⟨ n + 1∣ n + 1 ⟩$ , copying the bracket from the line above; both brackets are 1.)
$c_{n}$ . $⟨ n ∣ \overset{a}{^}^{+} \overset{a}{^} ∣ n ⟩ = ∣ c_{n} ∣^{2} = ⟨ n ∣ \hat{N} ∣ n ⟩ = n$ .

$\overset{a}{^}$ is the lowering or annihilation operator, $\overset{a}{^}^{+}$ the raising (the board writes “rising”) or creation operator. The levels are the rungs of a ladder: $\overset{a}{^}^{+}$ climbs one, adding a quantum $ℏ ω$ ; $\overset{a}{^}$ steps down one. Read as adding or removing one photon in a mode of light, or one phonon in a crystal, they create and annihilate (beyond the boards). The spin operators $\hat{S}_{\pm}$ of §4.1 Spin do the same between spin states.

Theorem 3.5.8The ground state2024/25 board #13

$\overset{a}{^} ∣0 ⟩ = 0, ∣ n ⟩ = \frac{1}{n !} (\overset{a}{^}^{+})^{n} ∣0 ⟩, u_{0} (x) = (\frac{1}{x _{0} π})^{1/2} e^{- x^{2} /2 x_{0}^{2}},$ with $x_{0} = 1/ α = ℏ/ mω$ .

==Lowering the ground state gives the zero vector, not another state: the ladder has a bottom.== (The board crosses out $∣ - 1 ⟩$ .) It also pins the spectrum: a state with non-integer $n$ , lowered again and again, would never hit $c = 0$ and would reach negative $n$ , which cannot be. So $n = 0, 1, 2, \dots$ from the algebra alone (beyond the boards). Each step up brings $k + 1$ , which is where the $n!$ comes from.

The ground-state wavefunction

In position, with $\overset{p}{^} = - i ℏ \frac{d}{d x}$ , $\overset{a}{^} = \frac{1}{2} (\frac{x}{x _{0}} + x_{0} \frac{d}{d x})$ ; $i (- i) = 1$ and $ℏ/ m ℏ ω = x_{0}$ .
$⟨ x ∣ \overset{a}{^} ∣0 ⟩ = 0$ becomes ==a first-order equation instead of the second-order Schrödinger equation==: $x u_{0} + x_{0}^{2} \frac{d u _{0}}{d x} = 0.$
Separate: $\frac{d u _{0}}{u _{0}} = - \frac{x d x}{x _{0}^{2}}$ , so $u_{0} = A e^{- x^{2} /2 x_{0}^{2}}$ , the Gaussian of the analytic route (the integration is beyond the boards).
Normalise with $\overset{ˉ}{X} = x / x_{0}$ : $A^{2} x_{0} \int e^{- \overset{ˉ}{X}^{2}} d \overset{ˉ}{X} = A^{2} x_{0} π = 1$ .

The board sets the result beside $N_{0} = (α / π)^{1/2}$ : ==the two methods give the same ground state.==

$x_{0}$ is the turning point of a classical oscillator with the zero-point energy. The probability of finding the particle beyond it is $erfc (1) \approx 0.16$ : 16% of the time the ground state is where a classical particle of the same energy could never be (beyond the boards). The excited states follow by applying $\overset{a}{^}^{+} = \frac{1}{2} (\frac{x}{x _{0}} - x_{0} \frac{d}{d x})$ : $u_{1} = 2 \frac{x}{x _{0}} u_{0}$ , the analytic $N_{1} H_{1} e^{- ε^{2} /2}$ .

Exercises

Exercise 3.5.1Q4, sample questions

(i) Write the Hamiltonian and the time-independent Schrödinger equation. (ii) Find the energy levels and sketch the first three eigenfunctions. (iii) Derive the eigenfunctions (Hermite polynomials given). (iv) Analogies and contrasts between classical and quantum results. (v) Write $\hat{H}$ with ladder operators and discuss them. (vi) Find $[\overset{a}{^}, \overset{a}{^}^{+}]$ , $[\overset{a}{^}^{+}, \hat{H}]$ , $[\overset{a}{^}, \hat{H}]$ . (vii) Write the $n$ -th state from $∣0 ⟩$ .

Solution

(i) $\hat{H} = - \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}} + \frac{1}{2} m ω^{2} x^{2}$ , from $U = \frac{1}{2} k x^{2}$ , and $\hat{H} u_{E} = E u_{E}$ . (ii) $E_{n} = ℏ ω (n + \frac{1}{2})$ (Theorem 3.5.1); sketches from the table, each on its own level inside the parabola, with 0, 1, 2 nodes and tails past it. (iii) $u_{n} = N_{n} H_{n} (αx) e^{- α^{2} x^{2} /2}$ : dimensionless form, Gaussian at large $ε$ , series, termination. (iv) Same $ω$ ; classical $E$ any value $\geq 0$ , quantum discrete with spacing $ℏ ω$ ; classical minimum 0, quantum $\frac{1}{2} ℏ ω$ ; the quantum particle leaks past the turning points; the classical density peaks at the turning points, where the ball is slowest, the quantum ground state at the centre. At large $n$ the averaged $∣ u_{n} ∣^{2}$ approaches the classical curve. (v) $\hat{H} = ℏ ω (\hat{N} + \frac{1}{2})$ ; $\overset{a}{^}$ removes one quantum, $\overset{a}{^}^{+}$ adds one; not Hermitian, so not observables. (vi) $\hat{I}$ , $- ℏ ω \overset{a}{^}^{+}$ , $ℏ ω \overset{a}{^}$ (Theorem 3.5.6). (vii) $∣ n ⟩ = (\overset{a}{^}^{+})^{n} ∣0 ⟩ / n!$ .

Exercise 3.5.2Q5, sample questions

At $t = 0$ , $ψ = \frac{1}{50} (5 ϕ_{0} + 4 ϕ_{1} + 3 ϕ_{2})$ in the oscillator. (a) $⟨ E ⟩$ . (b) $ψ (x, t)$ and $⟨ E ⟩$ again. © $⟨ \overset{x}{^} ⟩ (t)$ .

Solution

(a) $\frac{25 + 16 + 9}{50} = 1$ ; $⟨ E ⟩ = \frac{ℏ ω}{50} (12.5 + 24 + 22.5) = 1.18 ℏ ω$ . (b) $ψ (x, t) = \frac{1}{50} (5 ϕ_{0} e^{- iω t /2} + 4 ϕ_{1} e^{- 3 iω t /2} + 3 ϕ_{2} e^{- 5 iω t /2})$ ; $⟨ E ⟩$ is unchanged. © $\overset{x}{^} = \frac{x _{0}}{2} (\overset{a}{^} + \overset{a}{^}^{+})$ links only neighbours: $⟨ 0∣ \overset{x}{^} ∣1 ⟩ = \frac{x _{0}}{2}$ , $⟨ 1∣ \overset{x}{^} ∣2 ⟩ = x_{0}$ . Each pair and its mirror add to a cosine: $⟨ \overset{x}{^} ⟩ = \frac{x _{0}}{50} (202 + 24) cos ω t \approx 1.05 x_{0} cos ω t .$ ==The average position oscillates at the classical frequency.== A single eigenstate would give $⟨ \overset{x}{^} ⟩ = 0$ : motion needs a superposition of neighbouring levels.

Exercise 3.5.3Exercise 3, sample3

Two non-interacting identical spin-½ particles in the same oscillator are in $Ψ = \frac{2}{π x _{0}^{2}} (x_{1} + x_{2}) exp (- \frac{x _{1}^{2} + x _{2}^{2}}{2 x _{0}^{2}}) χ (s_{1}, s_{2})$ . (a) The spin part. (b) The total energy. © Its value in eV for $x_{0} = 1$ nm and an electron’s mass.

Solution

(a) The spatial part is symmetric under $x_{1} \leftrightarrow x_{2}$ ; fermions need the whole state antisymmetric, so $χ$ is the singlet (§5.1 Identical particles). (b) $u_{1} (x_{1}) u_{0} (x_{2}) + u_{0} (x_{1}) u_{1} (x_{2}) = N_{0} N_{1} \frac{2}{x _{0}} (x_{1} + x_{2}) e^{- (x_{1}^{2} + x_{2}^{2}) /2 x_{0}^{2}}$ is the spatial part: one particle in $n = 0$ , one in $n = 1$ . $E = \frac{1}{2} ℏ ω + \frac{3}{2} ℏ ω = 2ℏ ω$ . (The sheet’s prefactor makes $\int ∣Ψ ∣^{2} = 2$ ; normalised it is $\frac{1}{π x _{0}^{2}}$ . The energy is unaffected.) © $ω = ℏ/ m x_{0}^{2} = 1.16 \times 1 0^{14}$ s⁻¹, $ℏ ω = 0.076$ eV, $E = 0.15$ eV.

Exercise 3.5.4beyond the boards

Show that in $∣ n ⟩$ , $⟨ \overset{x}{^} ⟩ = 0$ and $⟨ \overset{x}{^}^{2} ⟩ = x_{0}^{2} (n + \frac{1}{2})$ , and that the ground state has $Δ x Δ p = ℏ/2$ .

Solution

$\overset{x}{^} = \frac{x _{0}}{2} (\overset{a}{^} + \overset{a}{^}^{+})$ changes $n$ by $\pm 1$ , so its diagonal element is 0. In $\overset{x}{^}^{2} = \frac{x _{0}^{2}}{2} (\overset{a}{^}^{2} + \overset{a}{^} \overset{a}{^}^{+} + \overset{a}{^}^{+} \overset{a}{^} + \overset{a}{^}^{+ 2})$ only the middle two keep $n$ : $(n + 1) + n$ . The same for $\overset{p}{^} = \frac{i ℏ}{2 x _{0}} (\overset{a}{^}^{+} - \overset{a}{^})$ : $⟨ \overset{p}{^}^{2} ⟩ = \frac{ℏ ^{2}}{x _{0}^{2}} (n + \frac{1}{2})$ . So $Δ x Δ p = ℏ (n + \frac{1}{2})$ , the minimum $\frac{ℏ}{2}$ at $n = 0$ , a Gaussian (Key observation 2.5.4).

Exercise 3.5.5beyond the boards

Evaluate $\overset{a}{^}^{+} \overset{a}{^}^{2} ∣3 ⟩$ and $⟨ 2∣ \overset{a}{^}^{+} \overset{a}{^} ∣2 ⟩$ . Is $\overset{a}{^}$ Hermitian?

Solution

$\overset{a}{^}^{2} ∣3 ⟩ = 32 ∣1 ⟩$ ; then $\overset{a}{^}^{+}$ gives $62 ∣2 ⟩ = 23 ∣2 ⟩$ . $⟨ 2∣ \hat{N} ∣2 ⟩ = 2$ . No: $\overset{a}{^}^{+} \neq = \overset{a}{^}$ ; the Hermitian combinations are $\overset{a}{^} + \overset{a}{^}^{+} \propto \overset{x}{^}$ and $i (\overset{a}{^}^{+} - \overset{a}{^}) \propto \overset{p}{^}$ .

Exercise 3.5.6beyond the boards

A diatomic molecule vibrates with $ℏ ω = 0.27$ eV. Zero-point energy, and the photon that excites $0 \to 1$ ?

Solution

$E_{0} = 0.135$ eV; the photon carries $0.27$ eV, $λ = 1240/0.27 \approx 4.6$ µm, infrared. With equal spacing every $n \to n + 1$ step absorbs the same photon.