Part 3 · One-dimensional problems

3.2 The infinite well

Source: Centini’s whiteboard of 9 Oct, p. 11 (the set-up); the solution, normalisation, symmetric well, lowest states, classical limit and the superposition exercise from the 2024/25 boards #6, #7 and #20

Objectives

Understand why the energy of a confined particle is quantised, and why it can never be zero.
Recipe: solve the well, normalise its eigenfunctions, and expand a state in them.
Pictures: the lowest standing waves and their densities; a superposition sloshing back and forth.
Theorem: $E_{n} = \frac{π ^{2} ℏ ^{2}}{2 m a ^{2}} n^{2}$ , $u_{n} = 2/ a sin (nπ x / a)$ .
Vocabulary: zero-point energy, node, parity, correspondence principle.

A particle shut in a one-dimensional box with impenetrable walls. It is the simplest picture of an electron confined in a nanostructure: a thin semiconductor layer (a quantum well) or a nanocrystal (a quantum dot) (beyond the boards).

The well

$U (x) = {0 + \infty 0 \leq x \leq a x < 0 or x > a$ Inside, no force: the particle moves freely. The walls are infinitely high: no energy is enough to get past them. This year’s board marks the allowed energies as discrete lines $E_{0}, E_{1}, E_{2}$ inside the box. Classically, a ball bouncing between two perfect walls can have any energy $E \geq 0$ and is equally likely to be anywhere (beyond the boards).

Outside, $u = 0$ (Exercise 2.4.2). Inside, $U = 0$ and the equation is the free particle’s. Divide by $- ℏ^{2} /2 m$ and call $k^{2} = 2 m E / ℏ^{2}$ : $\frac{d ^{2} u}{d x ^{2}} + k^{2} u = 0.$ ==The second derivative is minus $k^{2}$ times the function==, so $u = A sin k x + B cos k x$ . Here $k$ is $2 π / λ$ , $ℏ k$ the size of the momentum, and $E = ℏ^{2} k^{2} /2 m$ is all kinetic.

The walls quantise $k$

Theorem 3.2.1The infinite well2024/25 board #20

The energies and normalised eigenfunctions of the infinite well of width $a$ are $E_{n} = \frac{π ^{2} ℏ ^{2}}{2 m a ^{2}} n^{2}, u_{n} (x) = \frac{2}{a} sin \frac{nπ x}{a}, n = 1, 2, 3, \dots$

Proof

$u$ is continuous and $0$ outside, so it vanishes at both walls. $u (0) = B = 0$ : only the sine survives.
$u (a) = A sin ka = 0$ , so $ka = nπ$ : $k_{n} = nπ / a$ , and $k_{n}^{2} = 2 m E_{n} / ℏ^{2}$ gives $E_{n}$ .
Normalise over $[0, a]$ , with $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ : $A^{2} \int_{0}^{a} sin^{2} \frac{nπ x}{a} d x = A^{2} [\frac{a}{2} - \frac{1}{2} \int_{0}^{a} cos \frac{2 nπ x}{a} d x] = A^{2} \frac{a}{2} = 1.$ The cosine runs through $n$ whole periods and integrates to zero, so $A = 2/ a$ , the same for every $n$ .

The board lists $n = 0, 1, 2, \dots$ ; but $n = 0$ gives $u = 0$ everywhere, no particle, and negative $n$ give $- u_{∣ n ∣}$ , the same state. Board 20 also writes the argument of $sin^{2}$ as $\frac{πa}{x}$ and adds the cosine term instead of subtracting it; neither matters, since it is zero.

Key observation 3.2.2The wave must fit

$k = nπ / a$ is $a = n \frac{λ}{2}$ : ==a whole number of half-wavelengths between the walls==, like a guitar string fixed at both ends (beyond the boards). The energy grows as $n^{2}$ and as $1/ a^{2}$ : a smaller box has higher, more widely spaced levels. The lowest is not $0$ but $E_{1} = π^{2} ℏ^{2} /2 m a^{2}$ , the zero-point energy.

Why $E_{1} > 0$ (beyond the boards). Confined to a width $a$ , the particle has $Δ x ≲ a$ , so by Theorem 2.5.1 $Δ p_{x} ≳ ℏ/2 a$ . Even with $⟨ p_{x} ⟩ = 0$ , $⟨ p_{x}^{2} ⟩ = (Δ p_{x})^{2} > 0$ , so $⟨ E ⟩ > 0$ . A confined particle cannot be at rest.

Example 3.2.3An electron in a 1 nm wellbeyond the boards

$ℏ^{2} /2 m = 0.0381$ eV nm², so $E_{1} = π^{2} \cdot 0.0381 = 0.376$ eV, $E_{2} = 1.50$ eV, $E_{3} = 3.38$ eV. A jump from 2 to 1 emits $1.13$ eV, $λ = 1240/1.13 = 1100$ nm, near infrared. Halve the box and every energy is four times larger: the photon moves to 275 nm. ==This is the size tuning of quantum dots.==

The 9 Oct exercise (Exercise 2.2.1) uses exactly this spectrum: $\hat{H} ∣ ϕ_{n} ⟩ = n^{2} E_{0} ∣ ϕ_{n} ⟩$ is the well with $E_{0} = π^{2} ℏ^{2} /2 m a^{2}$ .

Orthogonality (beyond the boards). For $n \neq = m$ , $\int_{0}^{a} sin \frac{nπ x}{a} sin \frac{mπ x}{a} d x = \frac{1}{2} \int_{0}^{a} [cos \frac{( n - m ) π x}{a} - cos \frac{( n + m ) π x}{a}] d x = 0,$ as it must be for a Hermitian $\hat{H}$ with different eigenvalues (Theorem 1.4.2). Together the $u_{n}$ are a basis: any state in the box is $\sum_{n} c_{n} u_{n}$ .

Remark

Compare the periodic box of Theorem 1.5.1. There the ends are joined and the eigenfunctions are travelling waves $e^{i p x /ℏ}$ with $p_{n} = 2 π ℏ n / L$ . Hard walls force standing waves that vanish at the ends instead: $k_{n} = nπ / a$ , half the spacing.

The symmetric well: even and odd states

Board 6 puts the origin in the middle: $U = 0$ for $- a \leq x \leq a$ , width $L = 2 a$ . This is also the set-up of the exam questions, with $L$ for the width.

The two solutions inside have opposite parity: $sin$ is odd, $cos$ is even. The well is symmetric, $U (- x) = U (x)$ , so if $u (x)$ solves the equation so does $u (- x)$ , and a state of one energy is either even or odd (beyond the boards).

Odd, $A sin k x$ : $sin ka = 0$ , $k = jπ / a$ , $E = \frac{π ^{2} ℏ ^{2}}{2 m a ^{2}} j^{2}$ , $j = 1, 2, \dots$
Even, $B cos k x$ : $cos ka = 0$ , $k = (2 j + 1) \frac{π}{2 a}$ , $E = \frac{π ^{2} ℏ ^{2}}{8 m a ^{2}} (2 j + 1)^{2}$ , $j = 0, 1, \dots$
Both normalise to $1/ a$ over $[- a, a]$ , the same as $2/ L$ .

With $l = 2 j + 1$ for the even states and $l = 2 j$ for the odd, $E_{l} = \frac{π ^{2} ℏ ^{2}}{8 m a ^{2}} l^{2} = \frac{π ^{2} ℏ ^{2}}{2 m L ^{2}} l^{2}, l = 1, 2, 3, \dots,$ alternating even, odd, even: in terms of the width, exactly Theorem 3.2.1. Moving the origin changes nothing physical. (Board 6 drops the square on $a$ in the boxed $E_{l}$ ; the recap above it has $a^{2}$ .)

The lowest states

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Figure 3.2.1 The three lowest levels of a 1 nm well, each

u_{n}

drawn on its energy. The ground state is half a wave with its maximum in the middle; the first excited state a full wave with a node at the centre; the second one and a half waves, with two nodes.

Ground state (even, $l = 1$ ). ==Most likely in the middle, never at the walls==: nothing like a bouncing ball, whose density is flat, $1/2 a$ .
First excited state (odd, $l = 2$ , $4 E_{1}$ ). Two humps in $∣ u ∣^{2}$ and a zero at the centre, circled on the board: found on both sides, never in the middle.
Second excited state (even, $l = 3$ , $9 E_{1}$ ). Three humps. (The board labels it “ $E_{II} = 9 E_{1}$ ”; $E_{III} = 9 E_{I}$ is meant.)

The $l$ -th state has $l$ half-wavelengths and $l - 1$ nodes: more nodes, more curvature, more kinetic energy (Definition 2.4.4).

The classical limit

A classical particle bounces, going right with $v$ and back with $- v$ . The quantum stationary state is a standing wave, and a standing wave is two travelling waves: $ψ (x, t) = \frac{1}{a} cos k x e^{- i Et /ℏ} = \frac{1}{2 a} (forward e^{i (k x - ω t)} + backward e^{- i (k x + ω t)}) .$ A momentum measurement gives $\pm ℏ k$ , 50% each, like the ball moving right or left (Exercise 2.5.4).

As $l \to \infty$ , $∣ u ∣^{2}$ oscillates ever faster between $0$ and $1/ a$ . A measurement of finite resolution sees the average, and $cos^{2}$ averages $\frac{1}{2}$ : $\overline{P_{quantum}} = \frac{1}{a} \cdot \frac{1}{2} = \frac{1}{2 a} = P_{classic} .$ ==At high $l$ the averaged quantum density becomes the classical flat one==: the correspondence principle. (The board’s oval reads “ $P_{classic} \to P_{quantum}$ ”; the limit goes the other way.) The levels also merge relatively, $(E_{l + 1} - E_{l}) / E_{l} = (2 l + 1) / l^{2} \to 0$ (beyond the boards).

A superposition in the well

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Figure 3.2.2 The state of Exercise 3.2.1:

\frac{2}{7}

\frac{3}{7}

\frac{2}{7}

on the three lowest levels. Each term turns at its own rate

E_{n} /ℏ

, so the

density

sloshes and

⟨ x ⟩

swings across the well, while the energy probabilities, the sliders, never change. Set two of them to zero for a stationary state.

Exercises

Exercise 3.2.1Board 7, Exercise 12024/25 board #7

At $t = 0$ a particle in the well on $[0, a]$ is in $ψ (x, 0) = \frac{2}{7 a} sin \frac{π x}{a} + \frac{6}{7 a} sin \frac{2 π x}{a} + \frac{2}{7 a} sin \frac{3 π x}{a} .$ (a) Energy values and probabilities. (b) $⟨ E ⟩$ . © $ψ (x, t)$ . (d) The probability of finding it in $φ = 2/ a sin (3 π x / a) e^{- i E_{3} t /ℏ}$ .

Solution

Recognise the eigenfunctions without their $2/ a$ : divide each prefactor by it. $ψ = \frac{2}{7} ϕ_{1} + \frac{3}{7} ϕ_{2} + \frac{2}{7} ϕ_{3}$ , normalised since $\frac{2 + 3 + 2}{7} = 1$ . (a) $E_{1}$ , $4 E_{1}$ , $9 E_{1}$ with $\frac{2}{7}$ , $\frac{3}{7}$ , $\frac{2}{7}$ . (b) $⟨ E ⟩ = \frac{E _{1}}{7} (2 + 12 + 18) = \frac{32}{7} E_{1} = \frac{16 π ^{2} ℏ ^{2}}{7 m a ^{2}}$ : not an eigenvalue. © Each term gets $e^{- i E_{n} t /ℏ}$ (Key observation 2.4.8). (d) $\frac{2}{7} e^{- i E_{3} t /ℏ}^{2} = \frac{2}{7}$ , at every time. (Beside the problem the board writes $E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{8 m a ^{2}}$ by a segment marked $2 a$ : the symmetric well’s. This well has width $a$ .)

Exercise 3.2.2Exercise n.2, sample3

$ψ (x, 0) = \frac{3}{30} ϕ_{0} + \frac{4}{30} ϕ_{1} + \frac{1}{6} ϕ_{4}$ , where $ϕ_{n}$ is the $n$ -th excited state of a well of width $a$ , “with energy $E_{n} = π^{2} ℏ^{2} n^{2} / (2 m a^{2})$ ”. (a) $⟨ E ⟩$ . (b) $ψ (x, t)$ and $⟨ E ⟩$ later.

Solution

“ $n$ -th excited state” makes $ϕ_{0}$ the ground state, which the sheet’s formula would give zero energy, impossible. The consistent reading: $ϕ_{n}$ has energy $(n + 1)^{2} E_{1}$ . Normalised: $\frac{9 + 16 + 5}{30} = 1$ . (a) $⟨ E ⟩ = \frac{9 + 64 + 125}{30} E_{1} = 6.6 E_{1}$ . (Taking the formula literally gives $3.2 E_{1}$ ; say which reading you use.) (b) Each $ϕ_{n}$ times $e^{- i E^{(n)} t /ℏ}$ ; $⟨ E ⟩$ is the same at every $t$ : energy is conserved.

Exercise 3.2.3Question 3 (i)–(iv), sample questions; Theory 1(ii), sample3

The well $V = 0$ for $- L /2 \leq x \leq L /2$ . (i) Draw the three lowest wavefunctions. (ii) Find the even and odd solutions, and (iii) normalise them. (iv) Where is the particle most likely in $ψ_{1}$ , and why is this unclassical?

Solution

(i) The three curves of the figure, centred: half a cosine, a full sine, one and a half cosines; all zero at $\pm L /2$ . (ii)–(iii) $k = nπ / L$ , $E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}}$ . Odd $n$ : $ψ_{n} = 2/ L cos (nπ x / L)$ , even functions. Even $n$ : $ψ_{n} = 2/ L sin (nπ x / L)$ , odd functions. Mind the naming. (iv) $∣ ψ_{1} ∣^{2} = \frac{2}{L} cos^{2} (π x / L)$ is largest at the centre and zero at the walls; classically the density is flat, $1/ L$ . And it cannot have zero energy.

Exercise 3.2.4beyond the boards

An electron is in the ground state of a 1 nm well. Probability of finding it in the middle third? Compare with the classical value.

Solution

$\frac{2}{a} \int_{a /3}^{2 a /3} sin^{2} \frac{π x}{a} d x = [\frac{x}{a} - \frac{1}{2 π} sin \frac{2 π x}{a}]_{a /3}^{2 a /3} = \frac{1}{3} + \frac{3}{2 π} = 0.609$ . Classically $\frac{1}{3}$ .

Exercise 3.2.5beyond the boards

For $u_{1}$ find $Δ x$ and $Δ p_{x}$ , and check the uncertainty principle.

Solution

$⟨ x ⟩ = a /2$ ; $(Δ x)^{2} = a^{2} (\frac{1}{12} - \frac{1}{2 π ^{2}})$ , $Δ x = 0.181 a$ . $u_{1}$ is real, so $⟨ p_{x} ⟩ = 0$ , and $⟨ p_{x}^{2} ⟩ = 2 m E_{1} = π^{2} ℏ^{2} / a^{2}$ : $Δ p_{x} = π ℏ/ a$ . Product $0.57ℏ \geq 0.5ℏ$ : close to the minimum, never below.

Exercise 3.2.6beyond the boards

$ψ = \frac{1}{2} (ϕ_{1} + ϕ_{2})$ on $[0, a]$ . With what period does $∣ ψ ∣^{2}$ repeat? Numbers for an electron in a 1 nm well.

Solution

The cross term goes as $cos ((E_{2} - E_{1}) t /ℏ)$ : $T = 2 π ℏ/3 E_{1} = h /3 E_{1}$ . With $3 E_{1} = 1.13$ eV, $T = 4.14 \times 1 0^{- 15}$ eV s $/ 1.13$ eV $= 3.7$ fs. In the figure, a third of the unit $h / E_{1}$ .

3.2 The infinite well

Objectives

The well

The walls quantise k

The symmetric well: even and odd states

The lowest states

The classical limit

A superposition in the well

Exercises

The walls quantise $k$