Photons and de Broglie Waves · Quantum Mechanics (Phillips)

One equation read in two directions. Light was a wave and turned out to carry momentum; matter was a particle and turned out to have a wavelength, and λ = h/p says both.

Classical physics had two kinds of thing in it. A particle was a discrete object with a definite position and momentum, moving by Newton’s laws. An electromagnetic wave was an extended field, present everywhere at once, changing by Maxwell’s laws. The division was clean and it worked: particles made up the world, waves lit it.

It stopped working in 1900. Explaining the spectrum of thermal radiation forced Max Planck to assume that atoms emit and absorb energy only in discrete lumps $\epsilon = h\nu$ , with

h = 6.626\times10^{-34}\ \mathrm{J\,s}

These two sections are about what that constant does. It turns out to be the exchange rate between wave language and particle language — and once you have an exchange rate, the two currencies are not really different.

Physics you need first — black-body radiation, in one box

Heat something up and it glows: a stove ring goes red, a filament goes white. A black body is the idealization — a perfect absorber, hence a perfect emitter — and the question in 1900 was what the spectrum of its radiation looks like.

Classical physics gave a confident and catastrophically wrong answer. Treat the radiation in a cavity as a set of standing-wave modes, give each mode the same average energy $kT$ (the classical equipartition theorem), and count the modes. Short wavelengths fit in a cavity in far more ways than long ones — the mode count grows without limit — so the predicted energy density diverges toward the ultraviolet. Taken literally, opening an oven door would irradiate you with infinite energy.

Planck’s fix was to assume a mode of frequency $\nu$ cannot hold any energy it likes, only integer multiples of $h\nu$ . Now a high-frequency mode is expensive: you cannot excite it at all unless $kT$ can afford one whole quantum. The ultraviolet modes are priced out of the market, the divergence disappears, and the predicted spectrum matches the measured one exactly.

You do not need the derivation for this chapter — only the shape of the argument, because you will see it again and again: a continuous classical prediction goes wrong; restricting a quantity to discrete values fixes it.

1.1 Photons

A photon is a particle-like quantum of electromagnetic radiation. It travels at $c$ , and it carries momentum and energy

p = \frac{h}{\lambda} \quad\text{and}\quad \epsilon = \frac{hc}{\lambda}\tag{1.1}

What this really says — each equation puts a particle property on one side and a wave property on the other

Look at what kind of statement this is. On the left of each equation is a particle property — momentum, energy, the sort of thing you attribute to a billiard ball. On the right is a wave property, a wavelength. There is no mechanism here, no picture of a photon “being” one or the other. There is only an exchange rate, and its value is $h$ .

That is the whole content of the chapter title. Set $h = 0$ and the two descriptions decouple completely: every wavelength maps to zero momentum, and you are back in the classical world where waves and particles are separate things. The reason the classical picture worked for two centuries is not that it was right — it is that $h$ is $10^{-34}$ .

Put a number on it — a visible photon

Take orange-red light, $\lambda = 663\ \mathrm{nm}$ :

p = \frac{6.626\times10^{-34}}{663\times10^{-9}} = 1.0\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}

\epsilon = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{663\times10^{-9}} = 3.0\times10^{-19}\ \mathrm{J} = 1.87\ \mathrm{eV}

Both are absurdly small by everyday standards — which is why nobody noticed light was granular until they looked very carefully at very faint light. The electronvolt exists precisely to make the second number readable: visible photons are a couple of eV, X-ray photons about 10 keV. Those two landmarks are worth memorising.

The Compton effect

The decisive evidence came in 1923. A. H. Compton fired X-rays at electrons and found the scattered X-rays came back with a longer wavelength — the Compton effect — which no classical wave should do. A classical wave shakes an electron at frequency $\nu$ ; the electron re-radiates at frequency $\nu$ . The colour should not change.

It changes exactly as it would if a particle had bounced off another particle.

step 1 of 9

Treat the collision as elastic between two relativistic particles — a photon of energy ε = pc and a stationary electron of rest energy mₑc². Conserve both energy and momentum. This is the book's problem 1, one move at a time.

1Conserve momentum. Nothing else has entered or left, so the vector sum is unchanged.
$p_{i} = p_{f} + P_{f} ⟹ P_{f} = p_{i} - p_{f}$
Lower case p is the photon, upper case P the electron; i and f are before and after. The electron starts at rest, so it carries no initial momentum.
2Square it, to turn vectors into the magnitudes we can measure. The cross term brings in the scattering angle.
$P_{f}^{2} = p_{i}^{2} + p_{f}^{2} - 2 p_{i} p_{f} cos θ$
This is just the cosine rule on the momentum triangle in the diagram — no physics, only geometry.
3Multiply through by c² and use ε = pc, which holds for a photon because it is massless.
$P_{f}^{2} c^{2} = ϵ_{i}^{2} + ϵ_{f}^{2} - 2 ϵ_{i} ϵ_{f} cos θ$
Everything is now in energies. Keep this line — it is one of the two we will subtract.
4Now conserve energy. The electron starts at rest, so it starts with only its rest energy.
$ϵ_{i} + m_{e} c^{2} = ϵ_{f} + E_{f} ⟹ E_{f} = ϵ_{i} - ϵ_{f} + m_{e} c^{2}$
5Square that too, so both conservation laws are expressed in the same squared quantities.
$E_{f}^{2} = ϵ_{i}^{2} + ϵ_{f}^{2} + m_{e}^{2} c^{4} - 2 ϵ_{i} ϵ_{f} + 2 m_{e} c^{2} (ϵ_{i} - ϵ_{f})$
6Subtract the momentum line from the energy line. The left-hand side is exactly the electron's energy–momentum relation, which we know equals m²c⁴.
$= m_{e}^{2} c^{4} E_{f}^{2} - P_{f}^{2} c^{2} = m_{e}^{2} c^{4} - 2 ϵ_{i} ϵ_{f} (1 - cos θ) + 2 m_{e} c^{2} (ϵ_{i} - ϵ_{f})$
This is the move that makes the whole derivation work: squaring both conservation laws lets the unknown electron momentum cancel against the unknown electron energy.
7The m²c⁴ terms cancel from both sides, leaving a relation between the two photon energies and the angle.
$m_{e} c^{2} (ϵ_{i} - ϵ_{f}) = ϵ_{i} ϵ_{f} (1 - cos θ)$
8Divide by ε_i ε_f. The energies now appear only as reciprocals, which is the form that converts cleanly to wavelength.
$m_{e} c^{2} (\frac{1}{ϵ _{f}} - \frac{1}{ϵ _{i}}) = 1 - cos θ$
9Finally use ε = hc/λ, so 1/ε = λ/hc — and the reciprocal energies become wavelengths directly.
$Δ λ = λ_{f} - λ_{i} = \frac{h}{m _{e} c} (1 - cos θ)$
Equation (1.2). Note what survived: no property of the incoming photon appears on the right. The shift depends only on the angle and on the mass of what was hit.

The result is the Compton shift:

\Delta\lambda = \frac{h}{m_ec}\,(1 - \cos\theta)\tag{1.2}

The prefactor is a constant of nature in its own right, the Compton wavelength of the electron:

\frac{h}{m_ec} = 2.43\times10^{-12}\ \mathrm{m}

Reproduce it in NumPy

The angular dependence of the Compton shift, and why it is an X-ray effect.

import numpy as np

h, c, me = 6.626e-34, 2.998e8, 9.109e-31
lam_C = h / (me * c)
print(f"Compton wavelength h/(me c) = {lam_C:.4e} m")

lam_xray = h*c / (10e3 * 1.602e-19)      # a 10 keV X-ray
for deg in (30, 90, 180):
    d = lam_C * (1 - np.cos(np.radians(deg)))
    print(f"theta = {deg:3d} deg:  dlambda = {d*1e9:.6f} nm   "
          f"({100*d/lam_xray:.2f}% of a 10 keV X-ray)")

vis = 663e-9
print(f"same shift on 663 nm visible light: "
      f"{100*lam_C*2/vis:.5f}% -- unmeasurable")

printsCompton wavelength h/(me c) = 2.4263e-12 m theta = 30 deg: dlambda = 0.000325 nm (0.26% of a 10 keV X-ray) theta = 90 deg: dlambda = 0.002426 nm (1.96% of a 10 keV X-ray) theta = 180 deg: dlambda = 0.004853 nm (3.91% of a 10 keV X-ray) same shift on 663 nm visible light: 0.00073% -- unmeasurable

But light also interferes

So light is granular. The difficulty is that the same light, sent through two slits, does what only a wave can do.

Fig. 1.2 — two slits separated by $d$ , a screen at distance $D$ . Constructive interference at $P$ when the path difference $R_1 - R_2 = xd/D$ is an integer number of wavelengths, so the fringes are evenly spaced by $\lambda D/d$ .

Nothing about that is unusual for a wave; Young used it to measure the wavelength of light in 1801. What is unusual is what you see when the light is made very faint. The pattern does not fade — it granulates. Individual photons arrive at individual points, apparently at random, and the interference pattern assembles itself out of thousands of them.

What this really says — and what it does not

Be careful about what the simulation shows. Every dot is a particle event: one photon, one place, one time. Nothing wave-like happens to any individual photon that you can see. The wave is visible only in the statistics — in where thousands of photons choose to land.

And yet the photons are not simply being sprayed through one slit or the other. If they were, the pattern would be the sum of two single-slit patterns: one smooth hump. Turn on the which-path detector and that is exactly what you get. The fringes require every photon’s possibilities to interfere, which is only possible if each photon’s alternatives — slit 1 and slit 2 — are both live until detection.

Phillips puts it exactly: when presented with two possible trajectories, the photon “seems to pass along both trajectories, arrive at a random point on the screen, and build up an interference pattern.”

Engineer's bridge — this is a two-tap filter

The arithmetic here is one you have done. Each slit contributes a complex amplitude at the screen — the same size, differing only in the phase picked up along the path:

A_{\text{total}} = A e^{i\phi_1} + A e^{i\phi_2}, \qquad I = |A_{\text{total}}|^2 = 4A^2\cos^2\!\Big(\frac{\phi_1-\phi_2}{2}\Big)

That is a two-tap FIR filter: two copies of a signal, one delayed relative to the other, summed. Its magnitude response is a raised cosine with nulls where the delay is half a period — a comb. The interference pattern is that comb, plotted against position instead of frequency.

The path difference $xd/D$ plays the role of the delay, and $2\pi(xd/D)/\lambda$ is the phase. Change $\lambda$ or $d$ in the simulation above and the comb stretches or compresses exactly as the filter’s response would.

Where the analogy breaks. In DSP you add amplitudes because the signal genuinely is the sum of two copies — there is one waveform, physically present, going down both paths. Here there is one photon, and it is not physically divided. What is being added is not two pieces of a thing but two alternatives for the same thing. The formalism is identical; the ontology is not, and that gap is what §1.4 spends its pages on.

The failure of the analogy has an observable consequence, which is the whole point of the which-path toggle: if the apparatus records which path was taken, the alternatives stop being alternatives, and you must add $|A_1|^2 + |A_2|^2$ instead of $|A_1 + A_2|^2$ . No FIR filter behaves that way.

1.2 De Broglie waves

In 1923, Louis de Broglie asked the obvious question in the other direction. If a wave carries momentum $p = h/\lambda$ , might a thing with momentum $p$ carry a wavelength $\lambda = h/p$ ?

\lambda = \frac{h}{p}\tag{1.3}

This is the de Broglie wavelength , and it is proposed for matter — electrons, atoms, cricket balls.

To connect $\lambda$ to something measurable, start from the relativistic energy–momentum relation

\epsilon^2 - p^2c^2 = m^2c^4\tag{1.4}

and eliminate $p$ in favour of $\epsilon$ :

\lambda = \frac{hc}{\sqrt{(\epsilon - mc^2)(\epsilon + mc^2)}}\tag{1.5}

That single formula covers everything, and it has two limits worth having by heart.

Ultra-relativistic ( $\epsilon \gg mc^2$ , or a massless photon): drop $mc^2$ and the square root collapses to $\epsilon$ :

\lambda = \frac{hc}{\epsilon}\tag{1.6}

which is Eq. (1.1) again — as it must be. A photon is the $m = 0$ case of the same formula.

Non-relativistic ( $\epsilon = mc^2 + E$ with $E \ll mc^2$ ): the two brackets become $E$ and $2mc^2$ , and

\lambda = \frac{h}{\sqrt{2mE}}\tag{1.7}

For an electron, substituting $m_e$ and measuring $E$ in electronvolts gives a formula worth memorising:

\lambda = \sqrt{\frac{1.5}{E}}\ \mathrm{nm} \quad (E \text{ in eV})\tag{1.8}

so 1.5 eV gives exactly 1 nm, and 15 keV gives 0.01 nm.

What this really says — why the shaded band is the whole story

The band across that plot is the spacing between atoms in a solid, a few tenths of a nanometre. It is there because a wave only shows you it is a wave when it meets something about its own size. Sound bends round a doorway because the doorway is comparable to its wavelength; light does not, and casts sharp shadows.

Read the plot as a single question: for this particle, at this energy, is $\lambda$ anywhere near the band? An electron at a few tens of eV lands right in it — so a crystal is a diffraction grating for electrons, and that is a laboratory experiment. A cricket ball is thirty orders of magnitude below the band, and no apparatus that will ever exist can bring it back.

Nothing changes about the physics as you slide along that curve. Only whether you can see it.

Put a number on it — why Davisson and Germer worked

Davisson and Germer fired 54 eV electrons at a nickel crystal. Equation (1.8) gives

\lambda = \sqrt{\frac{1.5}{54}}\ \mathrm{nm} = 0.167\ \mathrm{nm}

The rows of atoms on the nickel surface are $D = 0.215\ \mathrm{nm}$ apart. The condition for constructive interference from a row of scatterers is $D\sin\phi = n\lambda$ , so the first-order peak sits at

\sin\phi = \frac{0.167}{0.215} = 0.775 \quad\Longrightarrow\quad \phi = 51°

They measured a strong beam at 50°. That is the experiment that settled it, and it is the subject of problem 9.

The agreement is not a coincidence of scale — it is the point. Equation (1.8) says an electron becomes a 0.1 nm wave at a few tens of eV, and a few tens of eV is a perfectly ordinary voltage to put across a vacuum tube. Nature put the electron’s wave nature within reach of 1920s laboratory equipment.

Reproduce it in NumPy

The de Broglie wavelength across thirty orders of magnitude. Atomic spacings are 0.1-0.3 nm; compare each row against that.

import numpy as np

h, eV, u = 6.626e-34, 1.602e-19, 1.661e-27
me, mn    = 9.109e-31, 1.675e-27

def lam_from_E(m, E_eV):            # non-relativistic, Eq. (1.7)
    return h / np.sqrt(2 * m * E_eV * eV)

def lam_from_v(m, v):               # straight from Eq. (1.3)
    return h / (m * v)

kT = 1.381e-23 * 300 * 1.5 / eV     # thermal neutron at 300 K
rows = [("electron",     "54 eV",   lam_from_E(me, 54)),
        ("electron",     "40 keV",  lam_from_E(me, 40e3)),
        ("electron",     "7 eV",    lam_from_E(me, 7)),
        ("neutron",      "300 K",   lam_from_E(mn, kT)),
        ("C60 molecule", "200 m/s", lam_from_v(720*u, 200)),
        ("cricket ball", "40 m/s",  lam_from_v(0.16, 40))]
for name, at, lam in rows:
    print(f"{name:13s} at {at:8s}: lambda = {lam*1e9:9.3e} nm")

printselectron at 54 eV : lambda = 1.669e-01 nm electron at 40 keV : lambda = 6.133e-03 nm electron at 7 eV : lambda = 4.636e-01 nm neutron at 300 K : lambda = 1.452e-01 nm C60 molecule at 200 m/s : lambda = 2.770e-03 nm cricket ball at 40 m/s : lambda = 1.035e-25 nm

It is not just electrons

Particle⇅	Who, when⇅	λ⇅	What was seen⇅
electron	Davisson & Germer, 1927	0.167 nmⓘ	diffracted off the atoms on a nickel surfaceⓘ
electron	G. P. Thomson, 1927ⓘ	0.0061 nmⓘ	passed through a polycrystalline foil; diffracted by randomly oriented microcrystals
electron	Tonomura et al., 1989	—	a genuine two-slit pattern, built one electron at a timeⓘ
neutronⓘ	Gähler & Zeilinger, 1991	0.145 nmⓘ	two-slit interference with neutrons
atom (He)	Carnal & Mlynek, 1991	—	two-slit interference with whole atomsⓘ
C₆₀ moleculeⓘ	Arndt et al., 1999	0.0028 nmⓘ	interference fringes from a molecule of 60 carbon atoms

Engineer's bridge — momentum IS spatial frequency

Rewrite $\lambda = h/p$ using the wavenumber $k = 2\pi/\lambda$ — the spatial frequency, radians per metre, exactly the $k$ you use in a spatial Fourier transform:

\boxed{\;p = \hbar k\;}

Momentum is the Fourier conjugate of position, and $\hbar$ is the conversion factor. That one line is worth more than the rest of this page. It means:

a particle with a definite momentum is a single spatial frequency — an infinite sine wave, spread over all space;
a particle localized in space is a packet of many frequencies, i.e. many momenta;
therefore you cannot have both, and the trade-off is governed by the time–bandwidth theorem you already know. That is the uncertainty principle, arriving in §1.4 with no new physics required.

If you have ever looked at a spectrogram and understood why a click has a broad spectrum while a pure tone lasts forever, you have already understood $\Delta x\,\Delta p \ge \hbar/2$ . Chapter 3 makes the correspondence exact: $\psi(x)$ and $\phi(p)$ are literally a Fourier transform pair.

Where the analogy breaks. In signal processing the signal is the thing; the spectrum is a representation of it. Here neither is privileged. $\psi(x)$ and $\phi(p)$ contain identical information and there is no fact of the matter about which one the particle “really” is. There is no underlying waveform being analysed — the analysis is all there is.

What §§1.1–1.2 have established

Phillips ends §1.2 with the observation that ties the two sections together, and it is worth stating flatly:

If Planck’s constant were zero, all de Broglie wavelengths would be zero and particles of matter would only exhibit classical, particle-like properties.

Everything strange in this book is downstream of one small number being non-zero. And because the same $h$ appears in $p = h/\lambda$ for light and in $\lambda = h/p$ for matter, light and matter are not two kinds of thing behaving oddly in two different ways. They are one kind of thing — Phillips calls it a quantum particle — and the classical particle and the classical wave are the two limits in which you can get away with forgetting that.

What the next chapters must now supply is the machinery: a wave equation whose solutions have exactly these properties. That is chapter 2.

Check yourself

0 / 6 answered

1.The Compton shift for a 90° scattering is 0.0024 nm. What is it for a 90° scattering of a photon with ten times the wavelength?
Try the toggle in the widget above.
2.In the two-slit simulation, turning on the which-path detector destroys the fringes. Which statement describes what changed mathematically?
The neutron is about 1839 times heavier. Use $λ = h / 2 m E$ .
3.An electron and a neutron are given the same kinetic energy. Which has the longer de Broglie wavelength, and by roughly what factor?
4.Why did nobody detect the wave nature of a cricket ball?
5.The book prints the momentum of a 663 nm photon as $1 0^{- 27} J s$ . What is wrong, and how would you catch it?
6.What does $p = ℏ k$ tell a signal-processing engineer?