Part 1 · Mathematical tools

1.2 Waves and wave packets

Source: Centini’s whiteboard of 2 Oct, p. 2–4; the wave equation, forward and backward waves, the initial phase, the complex amplitude and the plane wave as a particle from the 2024/25 boards #3 and #4

Objectives

Recognise a wave as a shape that travels: anything of the form $f (x - v t)$ .
Recipe: read $k$ , $ω$ , $λ$ , $T$ , $ν$ and $v$ off a harmonic wave.
Pictures: a crest moving at $ω / k$ ; a packet built from many plane waves.
Theorem: $f (x \mp v t)$ solves the wave equation; $Δ t Δ ω \sim 1$ and $Δ k Δ y \sim 1$ .
Vocabulary: wave equation, wavenumber, angular frequency, wavevector, complex amplitude, wave packet, spherical wave.

A particle in quantum mechanics is described by a wave. Before the quantum part, the classical facts about waves that it uses.

The wave equation

Definition 1.2.1Wave equation2024/25 board #3

$\nabla^{2} ψ - \frac{1}{v ^{2}} \frac{\partial ^{2} ψ}{\partial t ^{2}} = 0, \nabla^{2} = \overset{ˉ}{\nabla} \cdot \overset{ˉ}{\nabla} = \frac{\partial ^{2}}{\partial x ^{2}} + \frac{\partial ^{2}}{\partial y ^{2}} + \frac{\partial ^{2}}{\partial z ^{2}} .$ $ψ (x, y, z, t)$ is whatever is waving, $v$ the speed of the wave, $\overset{ˉ}{\nabla} = (\partial_{x}, \partial_{y}, \partial_{z})$ is nabla and $\nabla^{2}$ the Laplacian.

$\nabla^{2} ψ$ measures how curved the profile is; $\partial^{2} ψ / \partial t^{2}$ how fast the value accelerates. Where the profile bends, the value is pulled back towards its neighbours, and the disturbance moves on (beyond the boards). In one dimension only $\partial^{2} / \partial x^{2}$ is left. (The 2024/25 board drops the $\partial^{2} / \partial t^{2}$ from the second term; a slip.)

Theorem 1.2.2Forward and backward waves2024/25 board #3

For any twice-differentiable $f$ and $g$ , $ψ (x, t) = f (x - v t) (forward), ψ (x, t) = g (x + v t) (backward)$ solve the one-dimensional wave equation.

Proofbeyond the boards

Put $s = x - v t$ . By the chain rule $\partial_{x} f = f^{'} (s)$ and $\partial_{t} f = - v f^{'} (s)$ , so $\partial_{x}^{2} f = f^{''}$ and $\partial_{t}^{2} f = v^{2} f^{''}$ , and $f^{''} - \frac{1}{v ^{2}} v^{2} f^{''} = 0$ . The same with $x + v t$ .

$f (x - v t)$ is the shape $f (x)$ slid right by $v t$ : it travels towards $+ x$ at speed $v$ without changing shape. ==The shape never enters, only the combination $x - v t$ .== The general solution is a forward plus a backward wave; a standing wave is such a sum (§3.2 The infinite well).

Harmonic plane waves

Definition 1.2.3Harmonic plane wave

$ψ (x, t) = A cos (k x - ω t + φ_{0}) = A cos (k (x - v t) + φ_{0}), k v = ω .$

$A$ is the amplitude and $φ_{0}$ the initial phase (2024/25 board #3).
$k$ is the wavenumber, in $m^{- 1}$ : how fast the phase advances with distance.
$ω$ is the angular frequency, in $s^{- 1}$ : how fast the phase advances with time.

The board writes it for light, $E = E_{0} sin (k x - ω t)$ . “Harmonic”: one sine. “Plane”: at any moment the value is the same on each whole plane $x =$ const, the wavefronts.

Recipe 1.2.4Reading a harmonic wave

Freeze time. At $t = 0$ the argument $k x$ grows by $2 π$ over the wavelength $λ = 2 π / k$ .
Sit at one point. At $x = 0$ the argument $ω t$ grows by $2 π$ in the period $T = 2 π / ω = 1/ ν$ ; $ν$ is the frequency, in Hz, and $ω = 2 π ν$ .
Speed. $k x - ω t = k (x - v t)$ requires $v = ω / k = λ ν$ : in one period the wave moves one wavelength.

Loading the demo…

Figure 1.2.1 Drag the

crest

up and down to change

A

, sideways to change

φ_{0}

. Change

k

and

ω

, then press Play: the dotted line follows one crest, which moves at

v = ω / k

. Switch the direction to see

k x + ω t

travel the other way.

A crest of $cos (k (x - v t))$ sits where $k (x - v t) = 0$ , at $x = v t$ : it moves with speed $v$ . In a medium of refractive index $n$ , light has $v = c / n$ .

Example 1.2.5Green light2024/25 board #3

$λ = 500$ nm in vacuum. The board computes $ν$ ; the rest is beyond the boards.

$k = 2 π / λ = 1.26 \times 1 0^{7} m^{- 1}$ .
$ν = c / λ = 3 \times 1 0^{8} /5 \times 1 0^{- 7} = 6 \times 1 0^{14}$ Hz; $ω = 2 π ν = 3.77 \times 1 0^{15} s^{- 1}$ .
$T = 1/ ν = 1.7 \times 1 0^{- 15}$ s. Check: $ω / k = 3.0 \times 1 0^{8}$ m/s $= c$ .

In three dimensions

Definition 1.2.6Wavevector

$ψ (\overset{r}{ˉ}, t) = A cos (\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t), \overset{ˉ}{k} \cdot \overset{r}{ˉ} = k_{x} x + k_{y} y + k_{z} z .$ The wavevector $\overset{ˉ}{k}$ points where the wave travels; its length $∣ \overset{ˉ}{k} ∣ = 2 π / λ$ is the wavenumber. $\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t$ is the phase.

The wave along $x$ is $\overset{ˉ}{k} = (k_{x}, 0, 0)$ , with $\overset{ˉ}{k} \cdot \overset{r}{ˉ} = k_{x} x$ . The wavefronts $\overset{ˉ}{k} \cdot \overset{r}{ˉ} =$ const are planes perpendicular to $\overset{ˉ}{k}$ , one wavelength apart.

Complex form

$cos θ = Real (e^{i θ})$ by Euler, so a harmonic wave is the real part of an exponential: $A cos (\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t) = Real (A e^{i (\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t)}) .$ Dropping the “Real” gives the complex wavefunction $\tilde{ψ} (\overset{r}{ˉ}, t) = ψ_{0} e^{i (\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t)}$ .

Two reasons to bother (beyond the boards):

Derivatives of an exponential are multiplications: $\partial_{x} e^{ik x} = ik e^{ik x}$ , $\partial_{t} e^{- iω t} = - iω e^{- iω t}$ .
For light the complex form is a convenience and the field is its real part. ==In quantum mechanics the wavefunction is complex==: nothing is thrown away, and probabilities come from $∣ \tilde{ψ} ∣^{2}$ .

The 2024/25 board gets there another way, by writing the cosine itself with exponentials.

Definition 1.2.7Complex amplitude2024/25 board #3

From $cos θ = \frac{1}{2} (e^{i θ} + e^{- i θ})$ (which is Proposition 1.1.4 with $c = e^{i θ}$ ), $A cos (k x - ω t + φ_{0}) = \tilde{ψ} + \tilde{ψ}^{*}, \tilde{ψ} = \tilde{A} e^{i (k x - ω t)}, \tilde{A} = \frac{1}{2} A e^{i φ_{0}} .$ $\tilde{A}$ is the complex amplitude: ==one complex number holds both the amplitude, $∣ \tilde{A} ∣ = A /2$ , and the initial phase, its angle.== The exponential carries only the travelling part. “ $+ c.c.$ ” on the board means “plus the complex conjugate”.

Loading the demo…

Figure 1.2.2 A complex amplitude

c

and the turning point

c e^{iω t}

. Press Play: its

real part

, traced on the right, is the oscillation

∣ c ∣ cos (ω t + θ)

. Drag

c

: its length sets the height of the curve, its angle where the curve starts.

Example 1.2.8Amplitude 2, initial phase

π /2

beyond the boards

$\tilde{A} = \frac{1}{2} \cdot 2 \cdot e^{iπ /2} = i$ . With $θ = k x - ω t$ : $\tilde{ψ} + \tilde{ψ}^{*} = i e^{i θ} - i e^{- i θ} = i (2 i sin θ) = - 2 sin θ$ , and indeed $2 cos (θ + π /2) = - 2 sin θ$ .

Wave packets

A plane wave fills all of space and lasts for ever. Something localised, a pulse or a beam through a hole, is a superposition of plane waves:

Definition 1.2.9Wave packet

$ψ (\overset{r}{ˉ}, t) = \int ψ_{0} (\overset{ˉ}{k}) e^{i (\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t)} d^{3} k .$ Each $\overset{ˉ}{k}$ contributes a plane wave with its own amplitude $ψ_{0} (\overset{ˉ}{k})$ . Plotted against $k$ , the amplitudes form a bump of width $Δ k$ : the spectrum of the packet.

Remark

A plane wave is everywhere (2024/25 board #4). A particle with a definite momentum $p_{x}$ has the wavefunction $e^{i p_{x} x /ℏ}$ (Theorem 1.3.8). Its modulus is $1$ at every $x$ : nothing singles out any point, and the particle could be anywhere. A packet is localised near some $x_{0}$ , at the price of mixing several $p_{x}$ .

Loading the demo…

Figure 1.2.3

N

plane waves

cos k x

with

k

spread evenly over a

band

of width

Δ k

around

k_{0}

, added and divided by

N

. Drag the edge of the band: a wider band gives a narrower packet, and the half-width times

Δ k

stays

2 π

. With few waves the packet comes back further out; more waves push the copies away.

Theorem 1.2.10Bandwidth

A pulse lasting about $Δ t$ contains frequencies spread over $Δ ω$ , and a beam confined to $Δ y$ contains wavenumbers $k_{y}$ spread over $Δ k$ , with $Δ t Δ ω = 1, Δ k Δ y = 1.$

==A short pulse needs a broad range of frequencies; a pure frequency needs an infinitely long wave.== An infinite sinusoid is the limit $Δ t \to \infty$ , with a single line in its spectrum, $Δ ω \to 0$ .

Why, and why only roughlybeyond the boards

To build a pulse, the component waves must add up inside $Δ t$ and cancel outside it. Two components whose frequencies differ by $Δ ω$ drift out of step by a full cycle in a time of about $2 π /Δ ω$ , so to cancel quickly the frequencies must spread widely. The constant depends on how the widths are defined: in the figure the half-width to the first zero gives $2 π$ , and for a Gaussian with standard deviations it is $\frac{1}{2}$ (Figure 2.5.1). “ $= 1$ ” is the order of magnitude.

The second relation is diffraction: a plane wave through a slit of width $Δ y$ comes out with wavevectors fanned over $Δ k$ ; the narrower the slit, the wider the fan. With de Broglie’s $p = ℏ k$ (Definition 2.4.1) it becomes $Δ p_{y} Δ y \sim ℏ$ : the uncertainty principle (§2.5 The uncertainty principle).

Example 1.2.11A 10 fs laser pulsebeyond the boards

$Δ t = 1 0^{- 14}$ s gives $Δ ω \approx 1 0^{14} s^{- 1}$ , $Δ ν = Δ ω /2 π \approx 1.6 \times 1 0^{13}$ Hz. Ultrashort pulses are never one colour.

Spherical waves

Definition 1.2.12Spherical wave

A point source emits $ψ (\overset{r}{ˉ}, t) = \frac{ψ _{0}}{r} e^{i (k r - ω t)}, r = ∣ \overset{r}{ˉ} ∣.$ The wavefronts are spheres around the source, and $\overset{ˉ}{k}$ points radially outwards everywhere.

Why $1/ r$ (beyond the boards): the energy goes as $∣ ψ ∣^{2} \propto 1/ r^{2}$ and spreads over a sphere of area $4 π r^{2}$ , so the same energy crosses every sphere. Far away, a small patch of a sphere is nearly flat: a distant source looks like a plane wave.

Exercises

Exercise 1.2.1beyond the boards

Show that $e^{i (\overset{ˉ}{k} \cdot \overset{r}{ˉ} - ω t)}^{2} = 1$ everywhere. What does that say about where a plane-wave particle is?

Solution

$e^{i θ} e^{- i θ} = 1$ for every real $θ$ . The probability density is the same everywhere: a particle with one exact $\overset{ˉ}{k}$ is completely delocalised. Localising it needs a packet, a spread $Δ k$ .

Exercise 1.2.2beyond the boards

Red light, $λ = 650$ nm, enters glass with $n = 1.5$ . Frequency, speed and wavelength inside?

Solution

$ν = c / λ = 4.6 \times 1 0^{14}$ Hz, unchanged by the glass: the boundary oscillates at one rate. $v = c / n = 2 \times 1 0^{8}$ m/s. $λ_{glass} = v / ν = 433$ nm $= 650/1.5$ .

Exercise 1.2.3beyond the boards

A slit of width $Δ y = 1 μ$ m. Roughly what spread in $k_{y}$ , and what angle of spread for $λ = 500$ nm?

Solution

$Δ k \approx 1/Δ y = 1 0^{6} m^{- 1}$ , and $k = 2 π / λ = 1.26 \times 1 0^{7} m^{- 1}$ . The angle is about $Δ k / k \approx 0.08$ rad, $5^{\circ}$ .

Exercise 1.2.4beyond the boards

Show that $A sin (k (x + v t))$ solves the one-dimensional wave equation. Which way does it move? And $e^{- (x - v t)^{2}}$ ?

Solution

$\partial_{x}^{2} ψ = - k^{2} ψ$ and $\partial_{t}^{2} ψ = - k^{2} v^{2} ψ$ , so $- k^{2} ψ - \frac{1}{v ^{2}} (- k^{2} v^{2} ψ) = 0$ . It is a $g (x + v t)$ and moves towards $- x$ . The Gaussian is an $f (x - v t)$ (Theorem 1.2.2): a bump of fixed shape sliding towards $+ x$ at speed $v$ .

Exercise 1.2.5beyond the boards

$ψ (x, t) = 3 cos (k x - ω t + π /3)$ . Find the complex amplitude and write $ψ = \tilde{ψ} + \tilde{ψ}^{*}$ .

Solution

$\tilde{A} = \frac{3}{2} e^{iπ /3} = \frac{3}{4} + \frac{3 3}{4} i$ , and $\tilde{ψ} = \frac{3}{2} e^{iπ /3} e^{i (k x - ω t)}$ . Then $\tilde{ψ} + \tilde{ψ}^{*} = 3 cos (k x - ω t + π /3)$ by $cos θ = \frac{1}{2} (e^{i θ} + e^{- i θ})$ .