1.5 de Broglie and wave–particle duality

The previous section ended on an open question: what justifies Bohr’s $L=n\hbar$ ?

In 1923 the 31-year-old de Broglie gave an answer in his doctoral thesis. The thesis is a few dozen pages long and its core is a single line.

An argument from symmetry

De Broglie’s starting point was an observed asymmetry.

The picture

Light had been treated as a wave for two centuries (interference and diffraction settle it). In 1905 Einstein said it was simultaneously a particle.

So what about the electron, treated as a particle for two centuries — might it also be a wave?

At the time there was essentially no experimental support for this. De Broglie later said the dual nature of light had “provoked” him: if nature likes symmetry, this should not be happening to light alone.

The mathematics

For a photon, combine Einstein’s $E=h\nu$ with the relativistic $E=pc$ :

p=\frac{E}{c}=\frac{h\nu}{c}=\frac{h}{\lambda}\tag{1.5.1}

De Broglie’s move was to turn this round and apply it to massive particles:

\boxed{\ \lambda=\frac{h}{p}\ }\tag{1.5.2}

Any particle of momentum $p$ has an associated wave of wavelength $h/p$ .

Bohr’s rule becomes a standing-wave condition

This is where the one-line formula first pays for itself.

2\pi r=n\lambda=n\frac{h}{p}=n\frac{h}{mv}\tag{1.5.3}

Rearranged:

mvr=\frac{nh}{2\pi}=n\hbar\tag{1.5.4}

Exactly Bohr’s angular-momentum quantisation condition.

So why can we not see the electron’s wave?

Because $h$ is tiny. The numbers below are the thing most worth computing yourself in this section.

Work it out: three de Broglie wavelengthsbasic~6 min

Using $\lambda=h/p$ with $h=6.626\times10^{-34}\ \text{J}\cdot\text{s}$ .

1. A 100 eV electron (a typical energy in an electron microscope)

Non-relativistically $p=\sqrt{2mE}$ ; it is easier to work with $pc=\sqrt{2mc^2E}$ :

pc=\sqrt{2\times511000\times100}\ \text{eV}=1.01\times10^4\ \text{eV}\tag{1.5.5}

\lambda=\frac{hc}{pc}=\frac{1240\ \text{eV}\cdot\text{nm}}{1.01\times10^4\ \text{eV}}=0.123\ \text{nm}=1.23\ \text{Å}\tag{1.5.6}

That is exactly the spacing between atoms in a crystal. So electrons striking a crystal diffract just as X-rays do — not a coincidence, but the reason the experiment below could work at all.

2. A helium atom at room temperature ( $m=4\times1.66\times10^{-27}$ kg, $T=300$ K)

With thermal momentum $p\approx\sqrt{3mk_BT}$ this gives $\lambda\approx0.7\ \text{Å}$ . Also atomic-scale — which is why matter-wave interference is achievable in cold-atom experiments.

3. A walking person ( $m=70$ kg, $v=1$ m/s)

\lambda=\frac{6.626\times10^{-34}}{70\times1}\approx9\times10^{-36}\ \text{m}\tag{1.5.7}

For comparison, a proton has a radius of $10^{-15}$ m — twenty orders of magnitude larger.

Conclusion: the wave nature has not disappeared for macroscopic objects, its wavelength is simply too small for anything to serve as a diffraction grating. To watch a person diffract you would need a doorway $10^{-35}$ metres wide.

Confirmation: an experiment saved by an accident

The Davisson–Germer experiment (1927)

At Bell Labs, Davisson and Germer fired an electron beam at a nickel target and measured the angular distribution of scattered electrons. At first they got an unremarkable smooth distribution.

In 1925 a liquid-air bottle burst, air rushed into the vacuum system and the nickel target was badly oxidised. To rescue the sample they heated it for a long time to reduce it — and in the process unintentionally annealed the polycrystalline nickel into a few large crystals.

On remeasuring, sharp peaks appeared in the scattering curve.

It took them two years to work out what had happened: this was electron diffraction off the crystal lattice, with peaks obeying the Bragg condition exactly, and the wavelength from $\lambda=h/p$ matching perfectly.

The same year, G. P. Thomson passed electrons through a polycrystalline film and obtained Debye rings identical to those from X-rays.

There is a rare coincidence here: J. J. Thomson won a Nobel Prize for showing the electron is a particle, and his son G. P. Thomson won one for showing the electron is a wave. Both were right.

Check: where was the Davisson–Germer peakadvanced~5 min

Experimental conditions: accelerating voltage 54 V, nickel plane spacing $d=0.091$ nm, strong peak observed at a scattering angle of $50°$ .

First the wavelength:

pc=\sqrt{2\times511000\times54}\ \text{eV}=7430\ \text{eV} \quad\Longrightarrow\quad \lambda=\frac{1240}{7430}=0.167\ \text{nm}\tag{1.5.8}

Now the diffraction condition. For grazing diffraction off rows of surface atoms the constructive condition is

D\sin\theta=n\lambda\tag{1.5.9}

where $D=0.215$ nm is the spacing of atomic rows on the nickel surface and $\theta=50°$ :

D\sin\theta=0.215\times0.766=0.165\ \text{nm}\tag{1.5.10}

against $\lambda=0.167$ nm, a difference of about 1%.

A formula written down from a pure argument about symmetry, confirmed to 1% by a laboratory accident.

"An electron is sometimes a particle and sometimes a wave" — not quite

The usual phrasing is that an electron is normally a particle and turns into a wave in a double-slit experiment.

That suggests it switches between two identities, which it does not. Closer to the truth:

An electron is always an electron. “Wave” and “particle” are two models borrowed from macroscopic experience, and each works for only part of what electrons do. An electron is neither a small ball nor a ripple on water; it is something with no macroscopic counterpart.

What actually describes it is the wavefunction — a complex-valued function that evolves according to a wave equation (its wave side), and whose modulus squared gives the probability of finding one whole electron somewhere (its particle side). The two sides never conflict, because they were talking about different things all along.

That is the language chapter 2 sets up.

Key formulas

de Broglie relation
$\lambda=\frac{h}{p}$
Holds for any particle; large p means short wavelength
Non-relativistic electron wavelength
$\lambda=\frac{hc}{\sqrt{2mc^2E}}=\frac{1.226\ \text{nm}}{\sqrt{E/\text{eV}}}$
A 100 eV electron is about 0.12 nm — exactly the atomic spacing
Standing wave → Bohr quantisation
$2\pi r=n\lambda\ \Longrightarrow\ mvr=n\hbar$
Turns an unmotivated rule into "the wave must join up with itself"
Bragg diffraction condition
$D\sin\theta=n\lambda$
How Davisson and Germer verified the electron wavelength

Self-check4 questions

What comes next

The evidence now runs both ways: light knocks out electrons one at a time, and electrons diffract like waves.

There is one experiment that puts both facts in the same picture and forces you to confront the tension between them. Feynman said it “contains the only mystery” of quantum mechanics.

de Broglie and wave–particle duality

Recommended first

After this section you should be able to

An argument from symmetry

The picture

The mathematics

Bohr’s rule becomes a standing-wave condition

So why can we not see the electron’s wave?

Confirmation: an experiment saved by an accident

Key formulas

Self-check4 questions

What comes next

An argument from symmetry

◑The picture

∑The mathematics

Bohr’s rule becomes a standing-wave condition

So why can we not see the electron’s wave?

Confirmation: an experiment saved by an accident

∎Key formulas

What comes next

The picture

The mathematics

Key formulas