Part 1 · Mathematical tools

1.5 Momentum in a box, Fourier and the Dirac delta

Source: Centini’s whiteboard of 6 Oct, p. 3–5; the box from $- L$ to $L$ , its orthogonality integral, the Dirac delta and continuum normalisation from the 2024/25 board #5

Objectives

Understand why confining a particle makes its momentum discrete.
Recipe: normalise the box eigenfunctions; expand a state in them (a Fourier series).
Pictures: a function rebuilt from plane waves; a spike of area 1 getting narrower.
Theorem: $p_{n} = 2 π ℏ n / L$ ; the sifting property of $δ$ ; $⟨ u_{p} ∣ u_{p^{'}} ⟩ = δ (p - p^{'})$ .
Vocabulary: periodic boundary conditions, Fourier series and transform, Dirac delta, continuum normalisation.

The momentum eigenfunctions $C e^{i p x /ℏ}$ (Theorem 1.3.8) exist for every real $p$ , and none of them can be normalised on the whole line. Put the particle in a box and both problems go away.

A periodic box

Confine the particle to $[- L /2, L /2]$ and ask the wave to join up smoothly with itself at the ends, as on a ring of circumference $L$ : periodic boundary conditions, $u_{p} (- L /2) = u_{p} (L /2) .$

Theorem 1.5.1Momentum in a box

The momenta allowed by periodic boundary conditions on a box of length $L$ are $p_{n} = \frac{2 π ℏ}{L} n, n = 0, \pm 1, \pm 2, \dots,$ with normalised eigenfunctions $u_{n} (x) = \frac{1}{L} e^{i p_{n} x /ℏ} .$

Proof

$C e^{- i p L /2ℏ} = C e^{i p L /2ℏ}$ ; $C$ cancels.
Two exponentials are equal when their exponents differ by $2 πin$ (): $\frac{i p L}{ℏ} = 2 πin$ , so $p = 2 π ℏ n / L$ . ( $n = 0$ is included; the board drops a sign when collecting terms, harmless since $n$ runs over all integers.)
Normalise: $\int_{- L /2}^{L /2} ∣ C ∣^{2} e^{- i p x /ℏ} e^{i p x /ℏ} d x = ∣ C ∣^{2} L = 1$ , so $C = 1/ L$ . The phases cancel and only the length of the box is left.

==Confining the particle makes its momentum discrete==: only waves that fit a whole number of wavelengths into $L$ survive, $λ_{n} = h / p_{n} = L / n$ (beyond the boards). The $p_{n}$ are real, as eigenvalues of a Hermitian operator must be, and the spacing $2 π ℏ/ L$ shrinks as the box grows.

Remark

The 2024/25 box runs from $- L$ to $L$ , length $2 L$ , and gives $p = π ℏ m / L$ and $u_{p} = e^{i p x /ℏ} / 2 L$ : the same with $L$ replaced by $2 L$ . Only the length matters, not where the box starts. Each $u_{p}$ is also an energy eigenstate of the free particle, $E = p^{2} /2 m$ , so the energies are discrete too: $E_{n} = \frac{2 π ^{2} ℏ ^{2}}{m L ^{2}} n^{2}$ in this year’s notation.

Example 1.5.2An electron on a 1 nm ringbeyond the boards

$p_{1} = 2 π ℏ/ L = h / L = 6.63 \times 1 0^{- 25}$ kg m/s, and $p_{1}^{2} /2 m = (6.63 \times 1 0^{- 25})^{2} / (2 \cdot 9.11 \times 1 0^{- 31}) = 2.4 \times 1 0^{- 19}$ J $= 1.5$ eV. A visible-light-sized energy: at nanometre scales the steps between levels are large enough to measure.

The board’s side note: an electron bound to a nucleus by the Coulomb force $\overset{ˉ}{F} = - \frac{e ^{2}}{4 π ε _{0} r ^{2}} \overset{u}{^}_{r}$ is confined too, and its energies come as a ladder of discrete levels. The box is the simplest version of that.

Orthogonality and the Fourier series

Different $p_{n}$ , so $⟨ u_{n} ∣ u_{m} ⟩ = 0$ for $n \neq = m$ (Theorem 1.4.2). Directly, with $(p_{m} - p_{n}) /ℏ = 2 π (m - n) / L$ : $⟨ u_{n} ∣ u_{m} ⟩ = \frac{1}{L} \int_{- L /2}^{L /2} e^{2 πi (m - n) x / L} d x = \frac{sin π ( m - n )}{π ( m - n )} = 0,$ since the sine of a whole multiple of $π$ is zero. $e^{2 πi (m - n) x / L}$ winds $m - n$ full turns around the unit circle across the box, and the turns average out. (The 2024/25 board does this integral on $[- L, L]$ ; for $j = k$ it uses $sin x / x \to 1$ to get the normalisation.)

Recipe 1.5.3Expanding in momentum eigenstates

Any state in the box is $ψ (x) = n \sum a_{n} u_{n} = \frac{1}{L} n \sum a_{n} e^{2 πin x / L}, a_{n} = ⟨ u_{n} ∣ ψ ⟩ = \frac{1}{L} \int_{- L /2}^{L /2} ψ (x) e^{- 2 πin x / L} d x .$ The minus sign comes from conjugating $u_{n}$ . This is the Fourier series of $ψ$ , and $∣ a_{n} ∣^{2}$ is the probability of measuring $p_{n}$ (Example 2.2.7).

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Figure 1.5.1 A normalised

ψ

in the box, and its

partial sum

with

∣ n ∣ \leq N

. Below, the probabilities

∣ a_{n} ∣^{2}

of each momentum

p_{n} = 2 π ℏ n / L

; the orange bars are the ones kept. Raise

N

and the sum closes in on

ψ

, because the plane waves are a complete basis. Compare the box, with its sharp edges, to the smooth Gaussian.

Letting the box grow (6 Oct, p. 5). As $L \to \infty$ the $p_{n}$ crowd together and the sum becomes an integral, $ψ (x) = \frac{1}{2 π ℏ} \int_{- \infty}^{+ \infty} a (p) e^{i p x /ℏ} d p,$ the Fourier transform, the same superposition as a wave packet. The board writes $1/ 2 π$ , the prefactor for the variable $k = p /ℏ$ ; with $p$ it is $1/ 2 π ℏ$ , which the 2024/25 board derives (below).

The Dirac delta

On $[- L, L]$ , with $p_{x} = π ℏ m / L$ , the 2024/25 board rewrites the orthogonality integral in momenta and calls $ε = ℏ/ L$ : $⟨ u_{k} ∣ u_{j} ⟩ = 2 A^{2} ℏ π \frac{sin ( ( p _{j} - p _{k} ) / ε )}{π ( p _{j} - p _{k} )} .$ As $L \to \infty$ , $ε \to 0$ , and the fraction becomes something new.

Definition 1.5.4Dirac delta2024/25 board #5

$δ (x - x_{0}) = ε \to 0 lim \frac{sin ( ( x - x _{0} ) / ε )}{π ( x - x _{0} )} .$

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Figure 1.5.2 Drag the

peak

up:

ε

shrinks. The peak’s height is

1/ (π ε)

and its first zeros are at

\pm π ε

, so it gets taller as fast as it gets narrower; the area stays 1. (The board labels the height

π / ε

; it is

1/ (π ε)

Theorem 1.5.5Properties of the delta2024/25 board #5

$\int_{- \infty}^{+ \infty} δ (x - x_{0}) d x = 1, \int_{- \infty}^{+ \infty} δ (x - x_{0}) f (x) d x = f (x_{0}) (sifting) .$

Why the area is 1beyond the boards

Substitute $u = (x - x_{0}) / ε$ : the integral becomes $\frac{1}{π} \int \frac{s i n u}{u} d u = \frac{π}{π} = 1$ , for every $ε$ . The sifting property follows: the spike is zero except at $x_{0}$ , so only $f (x_{0})$ counts, times the area 1.

The delta is the continuous Kronecker delta: $\sum_{n} δ_{nm} b_{n} = b_{m}$ and $\int δ (x - x_{0}) f (x) d x = f (x_{0})$ do the same job (beyond the boards).

Key observation 1.5.6Continuum normalisation2024/25 board #5

With the delta, $⟨ u_{k} ∣ u_{j} ⟩ = 2 A^{2} ℏ π δ (p_{j} - p_{k})$ . Asking the coefficient of the delta to be 1 gives $A = 1/ 2 π ℏ$ : $u_{p} (x) = \frac{1}{2 π ℏ} e^{i p x /ℏ}, ⟨ u_{p} ∣ u_{p^{'}} ⟩ = δ (p - p^{'}) .$ ==Discrete normalisation $δ_{nm}$ becomes continuum normalisation $δ (p - p^{'})$ .==

The condition is on the coefficient: at $p = p^{'}$ the delta itself is infinite, not 1 (the board’s “ $= 1$ , $j = k$ ”). A plane wave over the whole line has infinite length, $\int ∣ e^{i p x /ℏ} ∣^{2} d x = \infty$ , so it is not strictly in Hilbert space; normalising to a delta is the replacement (beyond the boards).

box	$u_{p} (x)$	$p$
periodic on $(- L, L)$	$\frac{1}{2 L} e^{i p x /ℏ}$	$π ℏ m / L$ , discrete
$L \to \infty$	$\frac{1}{2 π ℏ} e^{i p x /ℏ}$	continuous

These are the $u_{p}$ a Gaussian packet is projected on to find its momentum distribution (Theorem 2.5.3).

Exercises

Exercise 1.5.1beyond the boards

In the 1 nm ring, what is $⟨ u_{1} ∣ u_{- 1} ⟩$ ? Expand $ψ = \frac{1}{L} cos (2 π x / L)$ in the $u_{n}$ . Is it normalised?

Solution

$0$ : different eigenvalues $\pm p_{1}$ . $cos θ = \frac{1}{2} (e^{i θ} + e^{- i θ})$ , so $ψ = \frac{1}{2} u_{1} + \frac{1}{2} u_{- 1}$ : $a_{\pm 1} = \frac{1}{2}$ , all others $0$ . Not normalised: $∣ a_{1} ∣^{2} + ∣ a_{- 1} ∣^{2} = \frac{1}{2}$ ; multiply $ψ$ by $2$ .

Exercise 1.5.2beyond the boards

Evaluate $\int_{- \infty}^{\infty} δ (x - 2) x^{2} d x$ and $\int_{0}^{\infty} δ (x + 1) e^{x} d x$ .

Solution

$2^{2} = 4$ by sifting. The second is $0$ : the spike is at $x = - 1$ , outside the range of integration.

Exercise 1.5.3beyond the boards

Show that the 2024/25 and 2026/27 boxes give the same physics.

Solution

2024/25: length $2 L$ , $p = π ℏ m / L$ , $A = 1/ 2 L$ . Call the length $L^{'} = 2 L$ : $p = 2 π ℏ m / L^{'}$ and $A = 1/ L^{'}$ , exactly Theorem 1.5.1.

Exercise 1.5.4beyond the boards

In the figure, the box function has bars that shrink slowly with $n$ while the Gaussian’s die out fast. Which needs more momentum states to describe, and why?

Solution

The box: its jumps are features of zero width, and building a sharp edge out of waves needs short wavelengths, large $∣ p_{n} ∣$ . Its $∣ a_{n} ∣^{2}$ fall only like $1/ n^{2}$ . A smooth function’s coefficients fall off much faster. Sharp in $x$ means wide in $p$ , the bandwidth relation again.