Part 1 · Oscillators and waves

1.1 Why quantum mechanics

Source: Postorino, week 1 slides, pp. 1–8, and Feynman’s talk posted with them

Objectives

Picture. Say what condensed matter physics studies, and the course’s plan: quantum mechanics → atom → molecule → solid → macroscopic properties.
Vocabulary. Electronvolt, ionisation energy, Avogadro’s number, thermal energy $K_{B} T$ .
Recipe. Compare an atomic quantity with an everyday one by orders of magnitude, and decide whether classical physics, non-relativistic or relativistic quantum mechanics applies.

Condensed matter

Condensed matter physics (struttura della materia) matters for three reasons:

it is the basis of materials science, so of industry and engineering;
it is common ground for many sciences: chemistry, biology, electronics;
about a third of the world’s physicists work in it.

The plan of the course (the “conceptual scheme”):

quantum mechanics says how to treat the elementary particles: electrons, protons, neutrons;
with it, build up in degree of aggregation: atom → molecule → solid;
from the solid, derive the macroscopic properties: transport, optical, mechanical, magnetic.

The field builds theoretical models of how electrons and nuclei interact, since that interaction decides how particles aggregate, from the atom to the solid. Its elementary models show how things work; refined, they become realistic enough to design new materials.

Key observation 1.1.1When classical physics worksslides w1, p. 2

Classical physics describes and predicts the world at everyday scales:

sizes and energies large compared with atomic ones;
speeds small compared with the speed of light in vacuum.

At the atomic scale the dynamics and the interactions are very different. To understand, and above all to predict, behaviour there you need quantum mechanics (QM), or relativistic quantum mechanics when speeds approach $c$ .

The rest of this section puts numbers on “atomic scale”: sizes, masses, energies, speeds.

Sizes and masses

Particle	Mass	Charge
proton	$M_{P} = 1.67 \times 1 0^{- 27}$ kg	$e^{+} = + 1.60 \times 1 0^{- 19}$ C
neutron	$M_{N} = 1.67 \times 1 0^{- 27}$ kg	$0$
electron	$m_{e} = 9.1 \times 1 0^{- 31}$ kg	$e^{-} = - 1.60 \times 1 0^{- 19}$ C

$\frac{M _{P}}{m _{e}} \approx 1800 (1836 more precisely) .$

Key observation 1.1.2Electrons and nuclei decoupleslides w1, p. 5

Nuclei are about 2000 times heavier than electrons, so the dynamics of electrons and nuclei are decoupled. Under comparable forces the electrons accelerate about 2000 times more; as far as they are concerned, the nuclei stand still.

The same fact comes back when only the electrons follow the field of a light wave (Key observation 3.1.4).

The planetary picture of the atom is not realistic. Its relative sizes of nucleus and orbit are completely wrong:

nucleus: $R_{nucleus} \sim 0.87 fm = 0.87 \times 1 0^{- 15}$ m;
electron orbit in hydrogen (Bohr’s model, §5.4 Bohr's atom): $R_{orbit} \sim 52.9 pm = 0.529 \times 1 0^{- 10}$ m $= 0.529$ Å.

$\frac{R _{orbit}}{R _{nucleus}} \sim 6 \times 1 0^{4} .$

The nucleus is a coffee bean in the middle of a stadium. An atom is mainly empty space.

Example 1.1.3Checking the coffee beanbeyond the lectures

$0.529 \times 1 0^{- 10} /0.87 \times 1 0^{- 15} = 6.1 \times 1 0^{4}$ . A stadium is 100–200 m across and a coffee bean about 1 cm long: a ratio of $1 0^{4}$ , the right order of magnitude.

Energies

The energy of the electron in the ground state of hydrogen is $I_{1} = 13.6$ eV. This is the ionisation energy: the energy needed to pull the electron out.

Definition 1.1.4Electronvoltslides w1, p. 7

The electronvolt is the kinetic energy an electron gains crossing a potential difference of one volt: $\frac{1}{2} m v^{2} = e V, 1 eV = 1.6 \times 1 0^{- 19} C \times 1 V = 1.6 \times 1 0^{- 19} J .$

Compared with the macroscopic world. A mass of 1 g moving at 1 m/s has kinetic energy $\frac{1}{2} (1 0^{- 3} kg) (1 m/s)^{2} = 5 \times 1 0^{- 4} J \approx 3 \times 1 0^{15} eV .$

WarningSlide slip

The slide gives $0.005 J = 5 \times 1 0^{- 3}$ J. $\frac{1}{2} \cdot 0.001 \cdot 1^{2}$ is $5 \times 1 0^{- 4}$ J. The point stands either way.

So one atom’s energy is tiny on the macroscopic scale. But a macroscopic piece of matter holds an enormous number of atoms, of the order of Avogadro’s number, $N = 6.022 \times 1 0^{23}$ atoms or molecules per mole, and so a lot of energy.

Mole. The slide calls it “the mass in grams equal to the atomic number of the substance”. Strictly, it is the amount whose mass in grams equals the substance’s atomic (or molecular) mass in u; for hydrogen that is 1 g beyond the lectures. Ionising one mole of hydrogen takes $13.6 \times 1.60 \times 1 0^{- 19} J \times 6.022 \times 1 0^{23} = 1.31 \times 1 0^{6} J .$

Key observation 1.1.5Atoms are stable at room temperatureslides w1, p. 8

At room temperature the thermal energy is $K_{B} T = 1.38 \times 1 0^{- 23} J/K \times 300 K = 4.1 \times 1 0^{- 21} J = 0.025 eV .$ Thermal agitation carries about $1/500$ of the energy needed to ionise hydrogen: atoms are stable at room temperature.

How unlikely is thermal ionisation?beyond the lectures

The fraction of systems with energy $E$ above the ground state goes as the Boltzmann factor $e^{- E / K_{B} T}$ (§4.2 The Boltzmann distribution). Here $E / K_{B} T = 13.6/0.025 = 544$ , so $e^{- 544} \approx 1 0^{- 236}$ . There are about $1 0^{80}$ atoms in the observable universe: thermal ionisation of hydrogen at room temperature never happens.

Speeds

The speed of the electron going round the proton in hydrogen is $v_{1} ≪ c$ , with $c = 3 \times 1 0^{8}$ m/s. Bohr’s model gives $v_{1} = c /137 \approx 2.2 \times 1 0^{6}$ m/s (Ground-state numbers).

Recipe 1.1.6Which theory?slides w1, p. 8

Atomic sizes and energies, $v ≪ c$ : non-relativistic quantum mechanics.
Atomic scale with $v$ close to $c$ : relativistic quantum mechanics.
Everyday sizes and energies, $v ≪ c$ : classical mechanics.

This course deals essentially only with non-relativistic quantum mechanics.

Feynman’s manifesto

There’s Plenty of Room at the Bottom, a talk given at Caltech in 1959 by Richard Feynman (Nobel Prize in Physics 1965), is the theoretical manifesto of nanotechnology. It imagines manipulating matter atom by atom. But, as the slide quotes him, atoms on a small scale behave like nothing on a large scale, because they obey quantum mechanics. Going down to atoms means working with different laws, and so making things differently: not just circuits, but systems built on quantised energy levels or interacting quantised spins.

Achieved so far (the slide’s list):

STM (scanning tunnelling microscope) and AFM (atomic force microscope); local spectroscopies
nanolithography, atomic patterning, quantum dots
MBE (molecular beam epitaxy), nanofabrication, 2D materials
nanomotors, molecular machines (Nobel Prize in Chemistry 2016)
single-molecule spectroscopy, nanoparticles
magnetic memories, flash, PCM (phase-change memory)
molecular biology, DNA nanotechnology, nanomedicine

Extra reading: the talk itself

The Moodle PDF is the article version, Engineering and Science, February 1960. Its arguments:

The Britannica on a pin head (p. 1–2). A pin head 1/16 inch across, magnified 25 000 times, has the area of all the Britannica’s pages. The smallest half-tone dot, shrunk 25 000 times, is still 80 Å across: 32 atoms in a row, about 1000 in its area. It could be written with a focused ion beam and read with an electron microscope.
All the world’s books (p. 2–3). Coded at about 100 atoms per bit (a $5 \times 5 \times 5$ cube), the $1 0^{15}$ bits of all books fit in a speck 1/200 inch across. DNA already stores information at about 50 atoms per bit.
Better electron microscopes (p. 3–4). The electron wavelength is about 1/20 Å, so diffraction does not forbid seeing atoms. Seeing them would settle biology’s questions and turn chemical analysis into “just look”.
Miniaturised computers (p. 4–5). Wires 10–100 atoms across; the speed of light forces fast computers to be small.
Small machines (p. 5–7). Scaling changes the physics: strength grows relative to weight, heat escapes fast, Van der Waals forces make parts stick. Build them with chains of master–slave hands, each a quarter the size of the last.
Rearranging the atoms (p. 7–8). Place atoms one by one; at the atomic scale quantised levels and spins become design tools, and copies can be exactly identical.
Prizes (p. 8). $1000 each for shrinking a book page 25 000 times, and for a working motor in a 1/64-inch cube.

Exercises

Exercise 1.1.1beyond the lectures

Express the room-temperature thermal energy in eV, and say why atoms don’t ionise at room temperature.

Solution

$K_{B} T = 1.38 \times 1 0^{- 23} \times 300 = 4.14 \times 1 0^{- 21}$ J, and $4.14 \times 1 0^{- 21} /1.6 \times 1 0^{- 19} = 0.026$ eV. That is about $1/500$ of 13.6 eV, and the Boltzmann factor $e^{- 544}$ makes thermal ionisation impossible.

Exercise 1.1.2beyond the lectures

At what temperature would $K_{B} T$ equal hydrogen’s ionisation energy?

Solution

$T = 13.6 \times 1.6 \times 1 0^{- 19} /1.38 \times 1 0^{- 23} \approx 1.6 \times 1 0^{5}$ K, far above any solid’s melting point. (In practice hydrogen is largely ionised already near $1 0^{4}$ K, because there are so many more free states than bound ones; that needs the statistics of §4.2 The Boltzmann distribution.)

Exercise 1.1.3beyond the lectures

Why can electron and nuclear motions be treated separately?

Solution

$M_{P} / m_{e} \approx 1836$ . Under forces of the same size the electrons accelerate about 2000 times more, and respond much faster, so the nuclei look frozen to them.

Exercise 1.1.4beyond the lectures

An electron is accelerated through 100 V. What is its speed? Is it relativistic?

Solution

$\frac{1}{2} m v^{2} = 100$ eV $= 1.6 \times 1 0^{- 17}$ J, so $v = 2 \cdot 1.6 \times 1 0^{- 17} /9.1 \times 1 0^{- 31} = 5.9 \times 1 0^{6}$ m/s, about $0.02 c$ . Not relativistic: $γ - 1 \approx v^{2} /2 c^{2} = 2 \times 1 0^{- 4}$ .

Exercise 1.1.5beyond the lectures

How many atoms thick is a 10 nm film, if atoms are about 0.25 nm apart? And how many electronvolts are in 1 J?

Solution

$10/0.25 = 40$ atoms. $1/1.6 \times 1 0^{- 19} = 6.25 \times 1 0^{18}$ eV.