Waves · Quantum Mechanics (Phillips)

A wave with one wavelength has no location, and a wave with a location has no one wavelength. Dispersion then decides whether the compromise between them survives being carried.

Chapter 1 established that a quantum particle has a wave character, and that Planck’s constant is the exchange rate between the wave description and the particle one. The obvious next question is what equation that wave obeys — and Phillips frames the answer with an analogy worth keeping:

The role of the Schrödinger equation in quantum mechanics is analogous to that of Newton’s Laws in classical mechanics. Both describe motion… In addition, both were postulated and then tested by experiment.

Before building it, §2.1 reviews the mathematics of waves. That review is doing more work than it looks like: almost every idea in it turns out to be load- bearing, and if you have done signal processing you already own most of it.

Sinusoidal waves

The simplest wave has one wavelength and one period:

\Psi(x,t) = A\cos(kx - \omega t)\tag{2.1}

with wave number $k = 2\pi/\lambda$ and angular frequency $\omega = 2\pi/\tau$ .

What this really says — everything is in the combination kx − ωt, which is why the shape simply moves

The whole of Eq. (2.1) is in the combination $kx - \omega t$ . Because $x$ and $t$ only ever appear that way, increasing $t$ has exactly the same effect as increasing $x$ by $(\omega/k)\,t$ — the shape is unchanged, just moved along. Phillips calls it a Mexican wave, which is the right image: nothing travels except the pattern.

Hold the phase fixed at any value and you get $x = (\omega/k)t + \text{const}$ . So every feature of the wave — every crest, every trough, every zero — moves at the same speed $\omega/k$ . That equal-speed fact is exactly what breaks down later on this page, and its breakdown is what makes quantum waves interesting.

$\sin(kx-\omega t)$ describes the same kind of wave, a quarter-cycle out of step, and the most general sinusoid with this $k$ and $\omega$ is a combination of the two:

\Psi(x,t) = A\cos(kx-\omega t) + B\sin(kx-\omega t)\tag{2.2}

Why complex exponentials

Very often in classical physics, and invariably in quantum physics, sinusoidal waves are written as complex exponentials:

\Psi(x,t) = A\,e^{i(kx-\omega t)}\tag{2.3}

Physics you need first — a complex exponential is a rotating arrow

If $e^{i\theta}$ has never quite settled in your head, this is the picture that settles it. Euler’s relation

e^{i\theta} = \cos\theta + i\sin\theta

says that $e^{i\theta}$ is the point on the unit circle at angle $\theta$ — a unit arrow pointing in the direction $\theta$ . Its shadow on the real axis is $\cos\theta$ ; its shadow on the imaginary axis is $\sin\theta$ .

So $A e^{i(kx-\omega t)}$ is an arrow of length $A$ whose angle advances steadily with time (at rate $\omega$ ) and with position (at rate $k$ ). Watch the shadow and you see a cosine wave. One rotating arrow contains both the sine and the cosine at once, which is exactly why it is more convenient than either.

Two properties make it the natural language for waves:

Multiplying is adding angles: $e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1+\theta_2)}$ . Phase shifts become multiplications.
Differentiating just multiplies: $\frac{d}{dt}e^{-i\omega t} = -i\omega\,e^{-i\omega t}$ . Derivatives never change the form of the function, only scale it — which is why substituting an exponential into a differential equation turns calculus into algebra. §2.2 does exactly this.

For a classical wave this is a convenience. For a quantum wave it will not be.

This distinction is the single most important thing on the page, and Phillips states it flatly.

The pressure in a sound wave is a real number. We may choose to write it as the real part of $Ae^{i(kx-\omega t)}$ because the algebra is easier, and at the end we take the real part and throw the imaginary part away. Nothing is lost because nothing physical was ever in there.

In quantum physics, however, the use of complex numbers is not an option.

In §2.2 the wave equation we construct will turn out to have no real solutions at all — $\cos(kx-\omega t)$ and $\sin(kx-\omega t)$ simply do not satisfy it, and you will be asked to verify that in problem 3. The imaginary part is not scaffolding to be discarded; it is half the state.

What that means — how a complex-valued function can describe a real particle — is the subject of chapter 3.

Superposing waves

Standing waves

Add two identical waves travelling in opposite directions and the trigonometry collapses:

A\cos(kx-\omega t) + A\cos(kx+\omega t) = 2A\cos kx\,\cos\omega t

Now $x$ and $t$ live in separate factors. The wave still undulates in space and still oscillates in time, but the two are no longer locked together, so nothing travels. It is a standing wave — in Phillips’ phrase, “a non-Mexican wave which merely stands and waves.” The zeros of $\cos kx$ are nailed to fixed positions.

Wave packets

A single sinusoid stretches to infinity in both directions, which is useless for describing a particle that is somewhere. To localize it, superpose a band of wave numbers — the result is a wave packet :

\Psi(x,t) = \int_{k-\Delta k}^{k+\Delta k} A\cos(k'x - \omega' t)\,dk'\tag{2.4}

At $t = 0$ that integral can be done in closed form, and the answer is a carrier times an envelope:

\Psi(x,0) = S(x)\cos kx, \qquad S(x) = 2A\,\Delta k\,\frac{\sin(\Delta k x)}{(\Delta k x)}\tag{2.5}

Drag the band width below. The three preset buttons reproduce the three panels of the book’s Fig. 2.1 exactly.

Ψ(x,t)
envelope S(x), Eq. (2.5)
a crest — moves at ω/k
envelope peak — moves at dω/dk

band Δkk/8components41

ω(k): ω = ck
length 2π/Δk: 8.4
ω/k: 1.00
dω/dk: 1.00(1.00×)

What this really says — you cannot have both

Slide the components control down to 1 and you have a single plane wave: perfectly definite wavelength, spread over all space, no location at all. Slide it back up and the components interfere — constructively near $x = 0$ , destructively elsewhere — and a localized lump appears out of nothing but addition.

Then slide the band width. A wide band ( $k/2$ ) gives a short, sharp packet. A narrow band ( $k/40$ ) gives a long, lazy one. In the limit $\Delta k \to 0$ the packet becomes infinitely long — a monochromatic wave again.

That trade — narrow band ⟺ long packet — is not a quirk of the sinc function. It is a theorem about any function and its Fourier transform, and it is the same trade-off you met as $\Delta x\,\Delta p \ge \hbar/2$ in §1.4. §2.2 will make the connection explicit and quantitative.

Engineer's bridge — §2.1 is a DSP lecture in disguise

Every object on this page has a name you already use:

Phillips calls it	You call it
a sinusoidal wave $e^{i(kx-\omega t)}$	a complex exponential / a single bin
a wave packet	a windowed carrier — a tone gated by an envelope
the envelope $S(x) = \mathrm{sinc}(\Delta k\,x)$	the transform of a rectangular window, which is exactly where sinc comes from
packet length $\approx 2\pi/\Delta k$	the time–bandwidth product, again
the dispersion relation $\omega(k)$	the system’s phase response
group velocity $d\omega/dk$	group delay
non-dispersive, $\omega = ck$	linear phase — the condition for a filter not to smear a pulse

The sinc envelope is not a coincidence: Eq. (2.4) integrates a flat amplitude over a finite band, which is precisely multiplying the spectrum by a rectangle, and the transform of a rectangle is a sinc. If you have ever wondered why truncating a signal produces ringing sidelobes, look at the small ripples either side of the packet in the widget above — same phenomenon, same mathematics.

Likewise, “dispersion spreads a packet” is the statement that a non-linear phase response smears a pulse. Engineers design filters for linear phase for exactly the reason light keeps its shape in vacuum.

Where the analogy breaks. In DSP the signal is real and the spectrum is a tool for analysing it. Here, as the caution above insists, $\Psi$ is irreducibly complex — and worse, it is not observable at all. There is no oscilloscope you can attach. What the two share is the mathematics of superposition; what they do not share is what the function is.

Dispersion: whether a packet survives

Everything now depends on one function — how $\omega$ depends on $k$ . Phillips calls $\omega(k)$ the dispersion relation , “because it determines whether the waves are dispersive or non-dispersive.”

Non-dispersive means $\omega = ck$ with $c$ constant. Then every component has the same speed $\omega/k = c$ , so the packet propagates without changing shape. Light in vacuum is the standard example, and such waves obey the classical wave equation :

\nabla^2\Psi - \frac{1}{c^2}\frac{\partial^2\Psi}{\partial t^2} = 0, \qquad \text{where}\quad \nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\tag{2.6}

and in one dimension

\frac{\partial^2\Psi}{\partial x^2} - \frac{1}{c^2}\frac{\partial^2\Psi}{\partial t^2} = 0\tag{2.7}

A dispersive wave is anything else. Then components travel at different speeds, and the packet as a whole moves at the group velocity

v_{\text{group}} = \frac{d\omega}{dk}\tag{2.8}

while an individual crest moves at the phase velocity $\omega/k$ .

step 1 of 3

Phillips derives the group velocity by asking where two sinusoids of nearby wave number reinforce each other. No calculus is needed until the last line.

1Two components are in phase — and so interfere constructively — exactly where their phases are equal.
$k_{1} x - ω_{1} t = k_{2} x - ω_{2} t$
2Solve for the position of that point of constructive interference. It is not fixed; it moves.
$x = (\frac{ω _{1} - ω _{2}}{k _{1} - k _{2}}) t$
At t = 0 the reinforcement sits at x = 0, and thereafter it travels at the bracketed speed.
3Take the two wave numbers close together. A ratio of small differences is a derivative.
$v_{group} = k_{2} \to k_{1} lim \frac{ω _{1} - ω _{2}}{k _{1} - k _{2}} = \frac{d ω}{d k}$
Equation (2.8). Two waves give infinitely many such points; many waves give one localized region — the packet — moving at this speed.

Switch the dispersion relation below and watch the two markers. The amber one rides a crest; the violet one rides the envelope peak.

Ψ(x,t)
envelope S(x), Eq. (2.5)
a crest — moves at ω/k
envelope peak — moves at dω/dk

band Δkk/8components41

ω(k): ω = ħk²/2m
length 2π/Δk: 8.4
ω/k: 3.00
dω/dk: 6.00(2.00×)

Put a number on it — Phillips' water waves

For long-wavelength water waves, gravity provides the restoring force and

\omega = \sqrt{gk}

The phase velocity is

v_{\text{phase}} = \frac{\omega}{k} = \sqrt{\frac{g}{k}}

and the group velocity is its derivative:

v_{\text{group}} = \frac{d\omega}{dk} = \frac{1}{2}\sqrt{\frac{g}{k}} = \tfrac{1}{2}\,v_{\text{phase}}

The group travels at exactly half the speed of the ripples that compose it. This is not an abstraction — it is visible in any ship’s wake or pond ripple: individual crests appear at the back of a group, sweep forward through it, and vanish at the front.

A wave of wavelength 100 m in deep water has $k = 0.063\ \mathrm{m^{-1}}$ , so $v_{\text{phase}} = 12.5\ \mathrm{m\,s^{-1}}$ and the wave group — which is what actually carries the energy across the ocean — travels at only $6.2\ \mathrm{m\,s^{-1}}$ .

Problem 1 at the end of the chapter does the short-wavelength case, where surface tension takes over and the ratio flips to $\tfrac32$ : the crests run backwards through the packet. Same water, opposite behaviour, purely because the restoring force changed.

Reproduce it in NumPy

Phase and group velocities for the four dispersion relations in the widget above, computed by numerical differentiation so nothing is taken on trust.

import numpy as np

def velocities(omega, k, h=1e-6):
    v_phase = omega(k) / k
    v_group = (omega(k + h) - omega(k - h)) / (2 * h)
    return v_phase, v_group

g, T, rho, hbar, m = 9.81, 0.072, 1000.0, 1.0, 1.0
cases = [("light  w = ck",          lambda k: k,                        6.0),
         ("deep water w=sqrt(gk)",  lambda k: np.sqrt(g * k),           0.625),
         ("ripples w=sqrt(Tk^3/r)", lambda k: np.sqrt(T * k**3 / rho),  300.0),
         ("free particle w=hk^2/2m", lambda k: hbar * k**2 / (2 * m),   6.0)]

print("dispersion         v_phase    v_group    ratio")
for name, omega, k in cases:
    vp, vg = velocities(omega, k)
    print(f"{name} {vp:9.3f} {vg:9.3f} {vg/vp:9.2f}")

k = 2 * np.pi / 100.0                       # a 100 m ocean swell
print(f"\nocean swell, wavelength 100 m:")
print(f"  v_phase = {np.sqrt(g/k):.1f} m/s   v_group = {0.5*np.sqrt(g/k):.1f} m/s")

printsdispersion v_phase v_group ratio light w = ck 1.000 1.000 1.00 deep water w=sqrt(gk) 3.962 1.981 0.50 ripples w=sqrt(Tk^3/r) 0.147 0.220 1.50 free particle w=hk^2/2m 3.000 6.000 2.00 ocean swell, wavelength 100 m: v_phase = 12.5 m/s v_group = 6.2 m/s

The phase velocity of a de Broglie wave is not the particle's speed

Select free particle in the widget and read the numbers: the group velocity comes out twice the phase velocity. That is not a rounding artefact, and it matters.

For a non-relativistic particle $\omega = \hbar k^2/2m$ , so

v_{\text{phase}} = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m} = \frac{v_{\text{particle}}}{2}

The crests of a de Broglie wave move at half the speed of the particle they describe. Nothing physical travels at that speed, and no measurement will ever return it. The particle’s velocity is the group velocity — and §2.2 turns that statement into the requirement that fixes the Schrödinger equation.

If you are wondering whether the phase velocity means anything at all: for a relativistic free particle it comes out greater than $c$ , which alarms people until they notice that a single infinite plane wave carries no information, so nothing is being signalled. Only the group velocity, the envelope of a genuine packet, is constrained by relativity — and it comes out correctly below $c$ (problem 6).

What §2.2 needs from this

Three results carry forward, and everything on the next page is built from them:

A packet of width $\Delta k$ has length $\approx 2\pi/\Delta k$ . Combined with $p = \hbar k$ this will reproduce the uncertainty principle without any new assumption.
A packet moves at $d\omega/dk$ , not $\omega/k$ . Demanding that this equal the particle’s velocity is the one physical requirement that pins down the equation.
Complex exponentials turn derivatives into multiplication. This is what makes “find the equation with this dispersion relation” a solvable problem rather than a guessing game.

§2.2 puts the three together.

Check yourself

0 / 6 answered

Try the three Fig. 2.1 preset buttons.
1.You narrow the band of wave numbers in the packet widget from $k /8$ to $k /32$ . What happens?
2.For a free particle, $ω = ℏ k^{2} /2 m$ . What is the phase velocity of its de Broglie wave?
3.Why does the envelope of the packet in Eq. (2.5) come out as a sinc function?
4.Phillips writes that in classical physics complex exponentials are a convenience but in quantum physics "the use of complex numbers is not an option". What makes the difference?
5.Deep-water waves have $v_{group} = \frac{1}{2} v_{phase}$ ; capillary ripples have $\frac{3}{2}$ . What does an observer see in each case?
6.In the standing wave $2 A cos k x cos ω t$ , what is structurally different from a travelling wave?