de Broglie Waves

In 1927 two engineers at Bell Labs were bouncing electrons off a lump of nickel and measuring how many came back at each angle. A bottle of liquid air broke, air got into their vacuum tube, the nickel oxidised, and to clean it they baked it hot enough to turn the surface into a few large crystals. When they ran the experiment again the smooth curve they had been getting was gone. In its place were sharp peaks at particular angles.

Sharp peaks at particular angles is what a wave does when it meets a regular array of scatterers. It is what X-rays do to a crystal, and what light does to a diffraction grating. But electrons are particles. They have a definite mass, you can count them one at a time, and Millikan had measured the charge on a single one. This lesson is about what has to be true for both of those things to hold at once, and about the single equation that makes it quantitative.

Running the photon relation backwards

You already know one object that is both. Light is a wave: it diffracts, it interferes, it has a wavelength you can measure with a ruler and two slits. And light is also lumpy: the photoelectric effect only makes sense if light arrives in packets of energy $E = hf$, where $h = 6.63 \times 10^{-34}$ J·s is Planck's constant and $f$ is the frequency.

A photon carries momentum as well as energy. For anything moving at $c$, relativity gives $E = pc$, so the momentum of a photon is $p = E/c = hf/c$. And since $f = c/\lambda$ for a wave travelling at $c$, that becomes

$p = \frac{h}{\lambda}$

Read this as a sentence about a wave: short wavelength means large momentum. Nothing in it mentions light specifically. It is a dictionary entry translating between a wave word, $\lambda$, and a particle word, $p$.

De Broglie's proposal in 1924 was that the dictionary works in both directions. If a thing you thought was a wave turns out to carry momentum in lumps, then a thing you know carries momentum should have a wavelength. Solve the relation the other way round and you get the whole content of this lesson:

$\lambda = \frac{h}{p}$

That $\lambda$ is called the de Broglie wavelength. The claim is not that an electron is "really" a wave or "really" a particle. It is that whenever you set up an experiment sensitive to wavelength (a grating, a crystal, a pair of slits) the electron will behave with this wavelength, the one set by its momentum.

Notice which quantity is on the bottom. It is momentum, not speed and not energy. Every mistake people make with this formula comes from reaching for the wrong one of those three.

Why a baseball has no visible wavelength

Put in a thrown baseball: $m = 0.145$ kg at $v = 40.0$ m/s. Its momentum is $p = mv = 5.80$ kg·m/s, so

$\lambda = \frac{6.63 \times 10^{-34}}{5.80} = 1.14 \times 10^{-34}\ \text{m}$

A proton is about $10^{-15}$ m across. The baseball's wavelength is nineteen orders of magnitude smaller than that. Diffraction only shows up when the wave meets an obstacle or an opening comparable to its own wavelength, and there is no aperture in the universe that small. The wave nature is not switched off for big objects. It is present and permanently unmeasurable, which in practice is the same thing.

Now the other extreme. An electron, $m_e = 9.11 \times 10^{-31}$ kg, moving at $1.00 \times 10^6$ m/s. Then $p = (9.11 \times 10^{-31})(1.00 \times 10^6) = 9.11 \times 10^{-25}$ kg·m/s and

$\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-25}} = 7.28 \times 10^{-10}\ \text{m}$

Seven ångströms. That is a few atomic diameters: precisely the scale of the gaps between atoms in a crystal. Matter waves are hidden from you not because they are weak but because $h$ is tiny, so you need a tiny momentum to get a usable wavelength, and tiny momentum means a very light particle.

Getting the wavelength from an accelerating voltage

Most of the time you are not handed a speed. You are told an electron was accelerated from rest through a potential difference $V$, or that a particle has kinetic energy $K$. Convert to momentum first.

Non-relativistically, $K = p^2/2m$, so $p = \sqrt{2mK}$ and

$\lambda = \frac{h}{\sqrt{2mK}}$

There is a version of this that is much easier to use in your head. Multiply top and bottom by $c$:

$\lambda = \frac{hc}{\sqrt{2(mc^2)K}}$

Now everything is an energy. With $hc = 1240$ eV·nm and the rest energy $m_ec^2 = 511\,000$ eV, an electron pushed through $V$ volts has $K = eV$, i.e. $K$ in electron-volts is numerically $V$. So

$\lambda = \frac{1240}{\sqrt{2(511\,000)V}}\ \text{nm} = \frac{1240}{1011\sqrt{V}}\ \text{nm} = \frac{1.226}{\sqrt{V}}\ \text{nm}$

That is the formula quoted in half the problems you will meet: $\lambda = 1.226/\sqrt{V}$ nm, with $V$ in volts. One volt gives 1.226 nm. A hundred volts gives 0.1226 nm, about one atomic spacing. A hundred thousand volts (an electron microscope) gives 3.88 pm, which is why electron microscopes see atoms and light microscopes never will.

Keep the $hc$ and $mc^2$ version in mind even for other particles. For a neutron with $K = 0.0250$ eV (a "thermal" neutron, the kind that comes out of a reactor), $m_nc^2 = 939.6$ MeV $= 9.396 \times 10^8$ eV, so

$\lambda = \frac{1240}{\sqrt{2(9.396 \times 10^8)(0.0250)}} = \frac{1240}{\sqrt{4.698 \times 10^7}} = \frac{1240}{6854} = 0.181\ \text{nm}$

Again an atomic spacing. That is not luck: it is why neutron scattering is a standard tool for finding out where atoms sit in a solid.

Where the peaks come from

Back to the nickel crystal. A crystal is a stack of parallel planes of atoms, separated by a distance $d$ of a couple of ångströms. A wave coming in at a glancing angle $\theta$ to those planes reflects a little from each plane. The beam reflecting from the second plane travels an extra distance $2d\sin\theta$ compared with the one from the first. Those two beams reinforce each other only when that extra distance is a whole number of wavelengths:

$2d\sin\theta = n\lambda$

This is Bragg's law, and it was already familiar from X-rays. The startling thing Davisson and Germer found was that electrons obey it too, with the $\lambda$ that de Broglie's formula predicts from the accelerating voltage. Nothing else in the experiment knows about $h$.

Worked example: an electron Bragg peak. Electrons accelerated through 200 V reflect from planes 0.200 nm apart. Where is the first-order peak?

First the wavelength: $\lambda = 1.226/\sqrt{200} = 1.226/14.14 = 0.0867$ nm.

Then Bragg with $n = 1$:

$\sin\theta = \frac{\lambda}{2d} = \frac{0.0867}{2(0.200)} = 0.2167$

so $\theta = \arcsin(0.2167) = 12.5^\circ$. If you had used the $n = 2$ peak you would need $\sin\theta = 0.433$, giving $25.7^\circ$; if $\lambda$ were ever bigger than $2d$, no peak exists at all, because a sine cannot exceed one.

Not every diffraction problem is Bragg. If the electrons go through two slits a distance $d$ apart onto a screen a distance $L$ away, the fringe spacing is the same one you learned for light, $\Delta y = \lambda L / d$. The physics of the interference is unchanged; only the source of $\lambda$ is new.

Worked example: an electron double slit. 50.0 keV electrons, slits 1.00 µm apart, screen 1.00 m away.

$\lambda = \frac{1.226}{\sqrt{50\,000}}\ \text{nm} = \frac{1.226}{223.6} = 5.48 \times 10^{-3}\ \text{nm} = 5.48 \times 10^{-12}\ \text{m}$

$\Delta y = \frac{\lambda L}{d} = \frac{(5.48 \times 10^{-12})(1.00)}{1.00 \times 10^{-6}} = 5.48 \times 10^{-6}\ \text{m}$

Fringes 5.5 µm apart. Small, but a microscope can resolve that, and the experiment has been done with electrons, with neutrons, and with whole C$_{60}$ molecules of 720 atoms.

Worked example: a 1.00 MeV electron. Its rest energy is 0.511 MeV, so this is firmly relativistic and the shortcut formula would be badly wrong.

Total energy: $E = K + m_ec^2 = 1.00 + 0.511 = 1.511$ MeV.

$(pc)^2 = E^2 - (m_ec^2)^2 = (1.511)^2 - (0.511)^2 = 2.283 - 0.261 = 2.022\ \text{MeV}^2$

so $pc = 1.422$ MeV $= 1.422 \times 10^6$ eV. Then

$\lambda = \frac{hc}{pc} = \frac{1240\ \text{eV·nm}}{1.422 \times 10^6\ \text{eV}} = 8.72 \times 10^{-4}\ \text{nm} = 0.872\ \text{pm}$

The non-relativistic formula would have given $1.226/\sqrt{10^6} = 1.23$ pm, about 40% too long.

Standing waves, and why energy comes in levels

A wavelength attached to a particle explains something Bohr had only been able to assert. In the Bohr model you were told that angular momentum comes in units of $\hbar$, with no reason given. De Broglie gave the reason. An electron going round a circular orbit of radius $r$ is a wave chasing its own tail. If the circumference is not a whole number of wavelengths, the wave comes back out of step with itself and cancels. So the surviving orbits are the ones with

$2\pi r = n\lambda = \frac{nh}{p}$

Rearrange: $pr = nh/2\pi$, that is, $mvr = n\hbar$. Bohr's rule, derived rather than decreed.

The same argument in a straight line gives you the single most useful toy model in quantum mechanics. Trap a particle between two hard walls a distance $L$ apart. The wave must vanish at both walls, so exactly a whole number of half-wavelengths fits in: $L = n\lambda/2$, giving $\lambda = 2L/n$ for $n = 1, 2, 3, \dots$. Feed that back through de Broglie: $p = h/\lambda = nh/2L$, and then $K = p^2/2m$ gives

$E_n = \frac{n^2h^2}{8mL^2} = \frac{n^2(hc)^2}{8(mc^2)L^2}$

The energy cannot be anything you like. It comes in levels, labelled by how many half-waves fit, and the lowest one is not zero. A confined particle always has some kinetic energy simply because confinement forces a curve into its wave.

Worked example: a proton in a nucleus. Model a proton as confined to a one-dimensional box of length $L = 5.00$ fm $= 5.00 \times 10^{-6}$ nm. Its rest energy is $m_pc^2 = 938.3$ MeV $= 9.383 \times 10^8$ eV. The ground state is $n = 1$:

$E_1 = \frac{(1240)^2}{8(9.383 \times 10^8)(5.00 \times 10^{-6})^2}$

Numerator: $1240^2 = 1.538 \times 10^6$ eV²·nm². Denominator: $8 \times 9.383 \times 10^8 \times 2.50 \times 10^{-11} = 1.877 \times 10^{-1}$ eV·nm². Dividing, $E_1 = 8.19 \times 10^6$ eV $= 8.19$ MeV.

Megaelectron-volts, from nothing but a box and a wavelength. That is the right order of magnitude for nuclear binding energies, and it is why nuclear physics is million-fold more energetic than chemistry: the box is a hundred thousand times smaller, and $E \propto 1/L^2$.

Uncertainty as a consequence

A wave with one exact wavelength has no location: a pure sine stretches forever. To build something localised within a distance $\Delta x$ you have to add up waves covering a spread of wavelengths, and through $p = h/\lambda$ that is a spread of momenta. The sharper you make the position, the wider the spread of momentum you needed. Quantitatively,

$\Delta x\,\Delta p \ge \frac{\hbar}{2}, \qquad \hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34}\ \text{J·s}$

For estimates, drop the factor of two and use $\Delta p \approx \hbar/\Delta x$. This turns "how big is it" into "how much momentum must it have", and it is often enough to get an answer within a factor of two without solving anything.

Worked example: estimating hydrogen's ground state. Say the electron is confined to within a radius $r$ of the proton. Then its momentum is at least about $p \approx \hbar / r$, so its kinetic energy is about $p^2/2m_e = \hbar^2/(2m_er^2)$, and its electrical potential energy is $-ke^2/r$. The total:

$E(r) = \frac{\hbar^2}{2m_er^2} - \frac{ke^2}{r}$

Squeeze the atom and the first term blows up; let it spread and the second term becomes feeble. Something in between is the minimum. Differentiate and set to zero:

That is exactly the Bohr radius, $0.529 \times 10^{-10}$ m. Put it back in. At that radius the kinetic term is $\tfrac{1}{2}m_ek^2e^4/\hbar^2$ and the potential term is $-m_ek^2e^4/\hbar^2$, so

$E_{\min} = -\frac{m_ek^2e^4}{2\hbar^2} = -13.6\ \text{eV}$

The measured ionisation energy of hydrogen, from a one-line estimate. The atom is the size it is because that size is the truce between confinement energy and electrical attraction. And the same argument tells you why electrons are not inside nuclei: confine one to $10^{-14}$ m and $pc \approx \hbar c/\Delta x = 197.3\ \text{eV·nm} / 10^{-5}\ \text{nm} = 19.7$ MeV, far more energy than any nucleus could hold onto.

Recognising when this is the tool

The signals are specific.

A wavelength is asked for, or given, about something with mass. Electron, neutron, proton, atom, molecule, buckyball. That is $\lambda = h/p$ and nothing else. Your only real work is getting $p$: from $mv$ if you are given speed, from $\sqrt{2mK}$ if you are given energy or a voltage, from $\sqrt{K^2 + 2Kmc^2}/c$ if the kinetic energy is comparable to $mc^2$.

Angles, fringes, slits or crystal planes with particles going in. Find $\lambda$ first, then use whichever optics formula the geometry calls for: $2d\sin\theta = n\lambda$ for reflection off planes, $d\sin\theta = n\lambda$ for a grating or a surface row, $\Delta y = \lambda L/d$ for two slits, $a\sin\theta = \lambda$ for the first minimum of a single slit of width $a$. The wave part is ordinary optics; the quantum part is one line at the start.

A particle is described as confined, trapped, in a well, or "inside" something of a stated size. Fit half-waves: $\lambda = 2L/n$ and $E_n = n^2(hc)^2 / (8mc^2L^2)$. If the question says "estimate" rather than "find", use $\Delta p \approx \hbar/\Delta x$ instead and expect agreement to a factor of a few.

The word "estimate" plus a size. That is uncertainty-principle work. Write $p \approx \hbar/r$, build an energy, and minimise if there is a competing term.

Work in $hc = 1240$ eV·nm and $\hbar c = 197.3$ eV·nm wherever you can. They let you stay in electron-volts and nanometres from start to finish and skip the powers of ten that generate most wrong answers.

Next you will meet the one place where the energies get larger still, and where particles convert into one another rather than merely diffract: radioactivity and nuclear decay.