Theory of Metals

An energy band is a group of very closely spaced allowed energy levels that forms when a large number of atoms come together to make a solid. It forms because the outer orbitals of neighbouring atoms overlap and, by the Pauli exclusion principle, each atomic level must split into many slightly different levels.

Lithium atom

Lithium (Z = 3) has the configuration $1s^2\,2s^1$. The 1s shell is full; the 2s orbital has one electron and room for one more.

Two Li atoms

When two Li atoms come close, their 2s orbitals overlap. The two atomic 2s orbitals combine (LCAO) into two molecular orbitals:

• a bonding orbital, with lower energy than the atomic 2s level, and
• an antibonding orbital, with higher energy.

So the single 2s level splits into 2 levels. The 1s orbitals lie close to the nuclei and hardly overlap, so they split very little.

N Li atoms

For $N$ atoms (about $10^{23}$ per cm³ in a solid) the 2s level splits into N separate levels. They are so close together (spacing of order $10^{-22}$ eV) that they form an almost continuous 2s band.

 Energy
   ^         separated atoms     solid (N atoms)
   |  2p .........................######  2p band
   |                       ######### (overlap)
   |  2s ..................#########  2s band
   |                       #########  (N levels)
   |
   |  1s ..........................-   1s band
   |                                  (narrow, full)
   +-----------------------|-----------------> r
                          r0     interatomic distance

• Each level holds 2 electrons (spin up/down), so the 2s band has $2N$ states.
• N Li atoms supply only $N$ valence electrons, so the 2s band is only half filled.
• The 1s band ($2N$ states, $2N$ electrons) is full and very narrow because the 1s orbitals overlap very little.
• At the equilibrium spacing $r_0$, the 2s and 2p bands overlap, giving one wide band with many empty states just above the filled ones.

Result

Because the highest band is partly filled, electrons can easily move into empty states just above them when a small field is applied. This is why lithium (and other alkali metals) are good conductors. The band width depends on how much the orbitals overlap: outer orbitals give wide bands, inner orbitals give narrow ones.

Quantum tunneling is the passing of a particle through a potential barrier whose height $V_0$ is greater than the particle's energy $E$. In classical mechanics this is impossible, but in quantum mechanics the wave function does not become zero inside the barrier. It decays, so a small part of it comes out on the other side and there is a finite chance of finding the particle there.

Barrier and regions

 V(x)
  ^
  |            +----------+   V = V0
  |            |          |
  |  - - - - - |- - - - - |- - - - -  E < V0
  |  Region I  | Region II| Region III
  |  V = 0     | V = V0   | V = 0
--+------------+----------+-----------> x
               0          a

The time-independent Schrödinger equation is

$$\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E - V)\psi = 0$$

Region I ($x < 0$, $V = 0$)

$$\psi_1 = A e^{jkx} + B e^{-jkx}, \qquad k = \frac{\sqrt{2mE}}{\hbar}$$

$A e^{jkx}$ is the incident wave and $B e^{-jkx}$ the reflected wave. The wave function is oscillatory.

Region II ($0 \le x \le a$, $V = V_0 > E$)

$$\frac{d^2\psi_2}{dx^2} - \alpha^2\psi_2 = 0, \qquad \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$$ $$\psi_2 = C e^{\alpha x} + D e^{-\alpha x}$$

The solution is exponential, not oscillatory. For a wide barrier the decaying term dominates, so $\psi$ falls off roughly as $e^{-\alpha x}$ inside the barrier.

Region III ($x > a$, $V = 0$)

$$\psi_3 = F e^{jkx}$$

Only a transmitted wave travels to the right (there is nothing to reflect it back). It is oscillatory with the same $k$ (same energy) but a much smaller amplitude.

Boundary conditions

$\psi$ and $d\psi/dx$ must be continuous at $x = 0$ and $x = a$. These four conditions fix $B, C, D, F$ in terms of $A$.

Transmission probability

$$T = \frac{|F|^2}{|A|^2} \approx T_0\, e^{-2\alpha a}, \qquad T_0 = 16\frac{E}{V_0}\left(1 - \frac{E}{V_0}\right)$$

(valid when $\alpha a \gg 1$). $T$ falls very quickly as the barrier width $a$, the mass $m$ or $(V_0 - E)$ increases.

 psi(x)
     Region I       Region II     Region III
  /\    /\    /\ |\            |
 /  \  /  \  /  \| \           |
/    \/    \/    |  \__        |/\/\/\/\/\/\
                 |     ``--..__|  (small amp.)
                 0             a

Examples

Alpha decay of nuclei, the tunnel diode, the scanning tunneling microscope (STM), and conduction through thin oxide layers between metal contacts all rely on tunneling.

A particle in an infinite potential well of width $L$ is free inside the well ($V = 0$ for $0 < x < L$) but cannot leave it ($V = \infty$ outside). Applying the Schrödinger equation with the boundary conditions shows that only certain discrete energies are allowed.

 V = inf |                    | V = inf
         |                    |
         |      V = 0         |
         |   (particle here)  |
         +--------------------+
         0                    L    x

Solving the Schrödinger equation

Outside the well $\psi = 0$. Inside, with $V = 0$:

$$\frac{d^2\psi}{dx^2} + k^2\psi = 0, \qquad k^2 = \frac{2mE}{\hbar^2}$$

General solution:

$$\psi(x) = A\sin kx + B\cos kx$$

Applying boundary conditions

• At $x = 0$: $\psi(0) = 0 \Rightarrow B = 0$, so $\psi = A\sin kx$.
• At $x = L$: $\psi(L) = A\sin kL = 0$. Since $A \neq 0$, $\sin kL = 0$:

$$kL = n\pi \quad\Rightarrow\quad k_n = \frac{n\pi}{L}, \qquad n = 1, 2, 3, \ldots$$

($n = 0$ gives $\psi = 0$ everywhere, i.e. no particle, so it is excluded.)

Quantized energy

$$\begin{aligned} E_n &= \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m}\cdot\frac{n^2\pi^2}{L^2} \\ &= \frac{n^2 h^2}{8 m L^2} \end{aligned}$$

Since $n$ can only be a whole number, the energy can take only the values $E_1, 4E_1, 9E_1, \ldots$ where $E_1 = h^2/(8mL^2)$. The energy is quantized. The lowest energy $E_1$ is not zero (zero-point energy).

Normalized wave function

The particle must be somewhere in the well:

$$\begin{aligned} \int_0^L |\psi|^2 dx &= A^2\int_0^L \sin^2\frac{n\pi x}{L}\,dx = A^2\cdot\frac{L}{2} = 1 \\ A &= \sqrt{\frac{2}{L}} \end{aligned}$$ $$\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L}, \qquad 0 \le x \le L$$

Graphs of $\psi_n$ and $|\psi_n|^2$

        psi_n(x)                 |psi_n(x)|^2
 n=3 |  _         _  |      |  _    _    _  |  E3=9E1
     | / \       / \ |      | / \  / \  / \ |
     |/   \     /   \|      |/   \/   \/   \|
     0     \_ _/     L      0               L

 n=2 |   _           |      |   _       _   |  E2=4E1
     |  / \          |      |  / \     / \  |
     |_/   \       _/|      |_/   \___/   \_|
     0      \_____/  L      0               L

 n=1 |    .----.     |      |     .--.      |  E1
     |  /        \   |      |   /      \    |
     |/            \ |      |__/        \___|
     0               L      0               L

• $\psi_n$ has $(n - 1)$ nodes inside the well; it fits exactly $n$ half-wavelengths, $L = n\lambda/2$.
• $|\psi_n|^2$ gives the probability density. For $n = 1$ the particle is most likely at the centre; for $n = 2$ it is never found at $x = L/2$, which is quite unlike a classical particle.

The effective mass $m^*$ of an electron in a crystal is the mass it appears to have when it responds to an external force, after the effect of all the internal forces from the lattice ions and other electrons has been included in it. It lets us treat a crystal electron as if it were a free particle with mass $m^*$.

Why it is needed

Inside a crystal an electron feels the external force $F_{ext}$ (for example $-eE$ from an applied field) and also the internal periodic force $F_{int}$ from the lattice. Newton's law gives

$$F_{ext} + F_{int} = m_e a$$

$F_{int}$ is very complicated and changes from point to point. We therefore write

$$F_{ext} = m^* a$$

and put all the effect of the lattice into $m^*$.

Expression for $m^*$

The electron is a wave packet that moves with the group velocity

$$v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk}$$

Its acceleration is

$$a = \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt}$$

In time $dt$ the external force does work $dE = F_{ext}v_g\,dt$, so $\frac{dE}{dk}dk = F_{ext}\frac{1}{\hbar}\frac{dE}{dk}dt$, giving

$$F_{ext} = \hbar\frac{dk}{dt}$$

Substituting $dk/dt = F_{ext}/\hbar$:

$$a = \frac{1}{\hbar^2}\frac{d^2E}{dk^2}F_{ext} \quad\Rightarrow\quad m^* = \frac{\hbar^2}{d^2E/dk^2}$$

Meaning

• $m^*$ depends on the curvature of the E–k curve. A sharply curved band gives a small $m^*$ (light, fast electron); a flat band gives a large $m^*$.
• For a free electron $E = \hbar^2k^2/2m_e$, so $d^2E/dk^2 = \hbar^2/m_e$ and $m^* = m_e$.
• Near the bottom of a band $m^*$ is positive; near the top it is negative (this is why holes are used); at the inflection point it is infinite.

Typical values

The small $m^*$ of electrons in GaAs is why GaAs devices are faster than Si devices.

For an electron in an infinite well the allowed energies are $E_n = \dfrac{n^2h^2}{8mL^2}$.

Given: $L = 0.1\ \text{nm} = 1\times10^{-10}$ m, $m = 9.1\times10^{-31}$ kg, $h = 6.626\times10^{-34}$ J s, $\hbar = h/2\pi = 1.055\times10^{-34}$ J s.

Ground state energy ($n = 1$)

$$\begin{aligned} E_1 &= \frac{h^2}{8mL^2} = \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}\times(10^{-10})^2} \\ &= \frac{4.390\times10^{-67}}{7.28\times10^{-50}} = 6.03\times10^{-18}\ \text{J} \\ &= \frac{6.03\times10^{-18}}{1.6\times10^{-19}} = 37.7\ \text{eV} \end{aligned}$$

Radian frequency

$$\begin{aligned} \omega_1 &= \frac{E_1}{\hbar} = \frac{6.03\times10^{-18}}{1.055\times10^{-34}} \\ &= 5.72\times10^{16}\ \text{rad/s} \end{aligned}$$

Putting the electron in the third level

$$\begin{aligned} E_3 &= 3^2E_1 = 9\times37.7 = 339.2\ \text{eV} \\ \Delta E &= E_3 - E_1 = 8E_1 = 301.5\ \text{eV} = 4.82\times10^{-17}\ \text{J} \end{aligned}$$

The electron must absorb exactly this energy. It can be given by a photon of wavelength

$$\lambda = \frac{hc}{\Delta E} = \frac{6.626\times10^{-34}\times3\times10^{8}}{4.82\times10^{-17}} = 4.12\times10^{-9}\ \text{m} = 4.12\ \text{nm}$$

(a soft X-ray photon), or by a collision with another particle that transfers 301.5 eV.

Answer: $E_1 = 6.03\times10^{-18}$ J $= 37.7$ eV; $\omega = 5.72\times10^{16}$ rad/s; to reach $n = 3$ the electron must absorb $301.5$ eV, e.g. a photon of $\lambda \approx 4.12$ nm.

The effective mass $m^*$ is the mass an electron in a crystal appears to have when it is accelerated by an external force; it includes the effect of the periodic lattice potential. It is given by $m^* = \hbar^2/(d^2E/dk^2)$.

Derivation

An electron in a crystal is a wave packet. Its velocity is the group velocity:

$$v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk} \qquad (E = \hbar\omega)$$

If an external force $F$ (e.g. $-eE$) acts for time $dt$, the work done raises the electron's energy:

$$dE = F\,v_g\,dt = F\frac{1}{\hbar}\frac{dE}{dk}dt$$

Since $dE = \frac{dE}{dk}dk$, this gives

$$F = \hbar\frac{dk}{dt}$$

The acceleration is

$$\begin{aligned} a &= \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d}{dt}\left(\frac{dE}{dk}\right) = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt} \\ &= \frac{1}{\hbar^2}\frac{d^2E}{dk^2}\,F \end{aligned}$$

Comparing with $F = m^*a$:

$$m^* = \frac{\hbar^2}{\dfrac{d^2E}{dk^2}}$$

Positive and negative effective mass

In a crystal the E–k curve of a band is not a simple parabola. In the tight-binding picture a band can be written as

$$E(k) = E_0 - 2\gamma\cos(ka), \qquad \gamma > 0$$

Then

$$\frac{d^2E}{dk^2} = 2\gamma a^2\cos(ka), \qquad m^* = \frac{\hbar^2}{2\gamma a^2\cos(ka)}$$

   E
   ^      top of band: curve bends down
   |  ___                         ___
   | /   \  m* < 0          m*<0 /   \
   |      \                     /
   |       \ <- inflection:    /
   |        \   m* = infinite /
   |         \               /
   |          \__         __/ <- bottom: m* > 0
   |             '--___--'
   +---|---------|---------|-------> k
     -pi/a       0       +pi/a

• Near the bottom of the band ($k \approx 0$): $\cos ka \approx 1$, the curve is concave up, $d^2E/dk^2 > 0$, so $m^*$ is positive. Here $E \approx E_{min} + \hbar^2k^2/2m^*$, like a free electron.
• Near the top of the band ($k \approx \pm\pi/a$): $\cos ka \approx -1$, the curve is concave down, $d^2E/dk^2 < 0$, so $m^*$ is negative. An applied force makes the electron accelerate opposite to the force, because it is being Bragg-reflected by the lattice.
• At the inflection points ($k = \pm\pi/2a$): $d^2E/dk^2 = 0$, so $m^* \to \infty$; the electron does not accelerate at all.

Significance

An electron with negative mass and negative charge near the top of the valence band behaves like a particle with positive mass and positive charge. This is the idea of the hole, used to describe conduction in the valence band of semiconductors.

According to molecular orbital (MO) theory, when two atoms come close, their atomic orbitals combine (Linear Combination of Atomic Orbitals, LCAO) to form molecular orbitals that belong to the whole molecule. The H₂ molecule forms because its two electrons can occupy a bonding orbital of lower energy than the separate 1s orbitals.

Molecular orbitals of H₂

Let $\psi_{1s}(r_A)$ and $\psi_{1s}(r_B)$ be the 1s orbitals of atoms A and B. They combine in two ways:

$$\psi_\sigma = \psi_{1s}(r_A) + \psi_{1s}(r_B) \qquad \text{(bonding)}$$ $$\psi_{\sigma^*} = \psi_{1s}(r_A) - \psi_{1s}(r_B) \qquad \text{(antibonding)}$$

 bonding  psi_s              antibonding psi_s*
      .--.   .--.               .--.
     /    \_/    \             /    \
    /      ^      \     ------/------\------- 0
   A  (overlap)    B          A       \    /B
                                       '--'
 |psi_s|^2 high between A,B   |psi_s*|^2 = 0 at mid-point

• In $\psi_\sigma$ the waves add between the nuclei, so electron density builds up there. This negative charge attracts both protons and holds them together. Energy is lowered.
• In $\psi_{\sigma^*}$ the waves cancel at the mid-point (a node), there is little charge between the nuclei, and the protons repel. Energy is raised.

Energy level diagram

 Energy
   |              sigma* (empty)
   |             /  ____  \
   |  1s  __ ---          --- __  1s
   |  (H_A) ^ \            / ^ (H_B)
   |           \  _^v_  /
   |              sigma (2 electrons,
   |                     opposite spins)

The two electrons (one from each H atom) go into $\sigma$ with opposite spins (Pauli principle). The antibonding orbital stays empty.

Energy versus separation

 E
 ^  \
 |   \  sigma* (repulsive, no minimum)
 |    \___
 |        ''----.....______________
0+-------------------------------------> R
 |    \              ____...----
 |     \        _---'   sigma (bonding)
 |      \    _-'
 |       '--'  <- minimum: R0 = 0.074 nm
 |              binding energy = 4.5 eV

The bonding curve has a minimum at $R_0 = 0.074$ nm. At this separation the H₂ molecule is about 4.5 eV lower in energy than two separate H atoms, so H₂ is stable. Bond order $= \frac{1}{2}(2 - 0) = 1$, a single covalent bond.

• The number of states $S(E)$ is the total number of allowed electron states (including spin) with energy from 0 up to $E$ (usually per unit volume).
• The density of states $g(E)$ is the number of states per unit energy per unit volume at energy $E$: $g(E) = dS_v/dE$. So $g(E)\,dE$ is the number of states between $E$ and $E + dE$ per unit volume.

Electron in a three-dimensional box

Treat the free electrons of a metal as particles in a cubic box of side $L$ (infinite well in 3D). The energies are

$$E = \frac{h^2}{8mL^2}\left(n_1^2 + n_2^2 + n_3^2\right), \qquad n_1, n_2, n_3 = 1, 2, 3, \ldots$$

Each set $(n_1, n_2, n_3)$ is one orbital state, i.e. one point in "n-space" with unit volume per point. Define

$$n^2 = n_1^2 + n_2^2 + n_3^2 = \frac{8mL^2E}{h^2}$$

All states with energy $\le E$ lie inside a sphere of radius $n$. Since $n_1, n_2, n_3$ are positive, only one octant (1/8) of the sphere counts.

   n2
   ^
   | . . . .
   | . . . . .    one octant of a sphere
   | . . . . . .  of radius n; each dot
   | . . . . . .  = one orbital state
   +--------------> n1

Number of states

Each orbital state holds 2 electrons (spin), so

$$\begin{aligned} S(E) &= 2\times\frac{1}{8}\times\frac{4}{3}\pi n^3 = \frac{\pi}{3}n^3 \\ &= \frac{\pi}{3}\left(\frac{8mL^2E}{h^2}\right)^{3/2} \end{aligned}$$

Dividing by the volume $L^3$:

$$S_v(E) = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}E^{3/2}$$

Density of states

$$\begin{aligned} g(E) &= \frac{dS_v}{dE} = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}\cdot\frac{3}{2}E^{1/2} \\ &= \frac{\pi}{2}\left(\frac{8m}{h^2}\right)^{3/2}E^{1/2} = \frac{8\sqrt{2}\,\pi\,m^{3/2}}{h^3}E^{1/2} \end{aligned}$$

So $g(E) \propto \sqrt{E}$.

 g(E)
  ^                 __..--
  |           _.--''
  |       _.-'      g(E) ~ sqrt(E)
  |    .-'
  |  /
  | /
  +-------------------------> E

Use

Multiplying $g(E)$ by the Fermi–Dirac function $f(E)$ gives the number of electrons per unit energy, $n_E = g(E)f(E)$. Integrating up to $E_F$ at 0 K gives the Fermi energy $E_F = \frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{2/3}$.

Effective mass $m^*$ is the apparent mass of an electron in a crystal when it is accelerated by an external force; it includes the effect of the lattice's internal forces. It is defined as

$$m^* = \frac{\hbar^2}{d^2E/dk^2}$$

Free electron

For an electron moving freely in vacuum (or a "free" electron with $V = 0$), all its energy is kinetic:

$$E = \frac{1}{2}m_ev^2 = \frac{p^2}{2m_e}$$

By de Broglie, $p = \hbar k$, so

$$E = \frac{\hbar^2k^2}{2m_e}$$

This is a parabola in the E–k diagram. Differentiating twice:

$$\frac{dE}{dk} = \frac{\hbar^2k}{m_e}, \qquad \frac{d^2E}{dk^2} = \frac{\hbar^2}{m_e}$$

Substituting in the definition:

$$m^* = \frac{\hbar^2}{\hbar^2/m_e} = m_e$$

Hence, for a free electron the effective mass equals the ordinary mass of the electron in vacuum, $m_e = 9.1\times10^{-31}$ kg. This is expected: with no lattice there are no internal forces, so the only force is the external one and Newton's law $F = m_ea$ holds directly. In a real crystal the E–k curve is not exactly parabolic, so $m^*$ differs from $m_e$ (e.g. $1.08\,m_e$ for electrons in Si).

The Heisenberg uncertainty principle says that position and momentum cannot both be known exactly. In the form used in Kasap's textbook,

$$\Delta x\,\Delta p_x \ge \hbar$$

For an electron confined in a well of width $L$, its position is uncertain by at most the width of the well, so take $\Delta x = L$.

Given: $L = 0.1\ \text{nm} = 1\times10^{-10}$ m, $\hbar = 1.055\times10^{-34}$ J s, $m = 9.1\times10^{-31}$ kg.

Uncertainty in momentum

$$\begin{aligned} \Delta p_x &\approx \frac{\hbar}{\Delta x} = \frac{1.055\times10^{-34}}{1\times10^{-10}} \\ &= 1.055\times10^{-24}\ \text{kg m s}^{-1} \end{aligned}$$

Kinetic energy

The momentum of the electron must be at least of the order of its uncertainty, so the corresponding kinetic energy is

$$\begin{aligned} KE &\approx \frac{(\Delta p_x)^2}{2m} = \frac{(1.055\times10^{-24})^2}{2\times9.1\times10^{-31}} \\ &= 6.11\times10^{-19}\ \text{J} = \frac{6.11\times10^{-19}}{1.6\times10^{-19}} = 3.82\ \text{eV} \end{aligned}$$

Comment

• A confined electron can never be at rest: confining it to 0.1 nm (about the size of an atom) forces it to have a kinetic energy of a few eV. This is why electrons in atoms have energies in the eV range.
• This is an order-of-magnitude estimate. With the stricter form $\Delta x\,\Delta p \ge \hbar/2$, the values are half and one quarter: $\Delta p = 5.27\times10^{-25}$ kg m/s and $KE \approx 0.95$ eV. The exact ground-state energy of a 0.1 nm infinite well is $h^2/8mL^2 = 37.7$ eV.

Answer: $\Delta p \approx 1.06\times10^{-24}$ kg m s⁻¹, $KE \approx 6.11\times10^{-19}$ J $\approx 3.8$ eV.

Bonding and antibonding molecular orbitals

When two atomic orbitals $\psi_A$ and $\psi_B$ overlap, they combine (LCAO) into two molecular orbitals:

Each molecular orbital can hold at most two electrons, with opposite spins (Pauli exclusion principle).

Why H₃ is not stable

H₂: Each H atom has one 1s electron. The two 1s orbitals form $\sigma$ and $\sigma^*$. Both electrons go into the bonding orbital $\sigma$ with opposite spins, and $\sigma^*$ stays empty. The energy falls by about 4.5 eV, so H₂ is stable.

Adding a third H atom: The third electron cannot enter $\sigma$, because $\sigma$ is already full with two electrons of opposite spin (Pauli principle). It must go into the higher-energy antibonding orbital $\sigma^*$.

 Energy
   |          sigma*  _^_   <- 3rd electron forced here
   |                 /    \
   |  1s  __  ---           ---  __ 1s
   |                \      /
   |          sigma  _^v_   <- full (2 electrons)
   |
   |  H2 + H:  bonding gain is cancelled by the
   |           energy cost of the sigma* electron

• The electron in $\sigma^*$ puts its charge outside the region between the nuclei, so it causes repulsion and raises the energy.
• The energy gained from the bonding pair is largely cancelled by the energy cost of the antibonding electron, so the total energy of H₃ is not lower than that of H₂ + H.
• A system always goes to the state of lowest energy, so H₃ breaks up into a stable H₂ molecule and a free H atom.

(In a full three-orbital treatment of linear H₃, the third electron goes into a non-bonding/antibonding level above the bonding one, and the result is the same: H₃ has no energy minimum and is not a stable molecule.)

Same idea for He₂

He has two 1s electrons, so He₂ would have 2 electrons in $\sigma$ and 2 in $\sigma^*$. Bond order $= \frac{1}{2}(2 - 2) = 0$, so He₂ does not form. In general, a molecule is stable only if bonding electrons outnumber antibonding electrons enough to lower the total energy.

Quantum mechanics is the branch of physics that describes the behaviour of very small particles such as electrons, atoms and photons, using wave functions and probabilities instead of exact paths.

Importance of quantum mechanics

Classical mechanics works for large objects but fails at atomic scale. It could not explain:

• black-body radiation, the photoelectric effect and the Compton effect;
• the stability of atoms and their sharp spectral lines;
• the specific heat and electrical conductivity of metals.

Quantum mechanics explains all of these. For electrical engineering materials it is essential because it explains:

• energy bands and why some solids are conductors, others semiconductors or insulators;
• effective mass, holes and carrier concentration in semiconductors;
• tunneling, used in tunnel diodes, flash memory and the STM;
• the operation of lasers, LEDs, solar cells, transistors and superconductors.

Classical versus quantum mechanics

Examples

• Ball in a box vs electron in a well: a ball in a box can have any speed, even zero. An electron in a 0.1 nm well can only have $E_n = n^2(37.7\text{ eV})$ and can never be at rest.
• Barrier: a car that cannot climb a hill never appears on the other side, but an electron with 7 eV can pass through a 10 eV, 3 nm oxide layer between copper wires with a small probability.
• Wavelength: a 50 g golf ball at 20 m/s has $\lambda = 6.6\times10^{-34}$ m, far too small to observe, so classical mechanics is enough; an electron accelerated through 100 V has $\lambda = 0.123$ nm, comparable to atomic spacing, so its wave nature (electron diffraction) is clearly seen.

Classical mechanics is a limiting case of quantum mechanics for large masses and large quantum numbers (correspondence principle).

In a body centred cubic (BCC) unit cell there is an atom at each of the 8 corners and one atom at the centre of the cube. The atoms touch each other along the body diagonal.

        o-----------o
       /|          /|
      / |         / |      o = corner atom
     o-----------o  |      @ = body-centre atom
     |  |   @    |  |
     |  o--------|--o      atoms touch along
     | /         | /       the body diagonal
     |/          |/
     o-----------o

Let $R$ = atomic radius and $a$ = lattice constant (edge of the cube).

Lattice constant

Along the body diagonal there is one corner atom radius, one full centre atom diameter and another corner radius:

$$\begin{aligned} \text{Body diagonal} &= \sqrt{3}\,a = R + 2R + R = 4R \\ a &= \frac{4R}{\sqrt{3}} = 2.309\,R \end{aligned}$$

Face diagonal

$$\text{Face diagonal} = \sqrt{2}\,a = \sqrt{2}\times\frac{4R}{\sqrt{3}} = 4\sqrt{\frac{2}{3}}\,R = 3.266\,R$$

Body diagonal

$$\text{Body diagonal} = \sqrt{3}\,a = 4R$$

Packing density

Number of atoms per cell: $8\times\frac{1}{8} + 1 = 2$.

$$\begin{aligned} \text{APF} &= \frac{\text{volume of atoms in cell}}{\text{volume of cell}} = \frac{2\times\frac{4}{3}\pi R^3}{a^3} \\ &= \frac{\frac{8}{3}\pi R^3}{\left(\frac{4R}{\sqrt{3}}\right)^3} = \frac{\frac{8}{3}\pi R^3\cdot3\sqrt{3}}{64R^3} = \frac{\sqrt{3}\,\pi}{8} \\ &= 0.68 \end{aligned}$$

Answer: $a = 4R/\sqrt{3} = 2.309R$; face diagonal $= 3.266R$; body diagonal $= 4R$; packing density $= \sqrt{3}\pi/8 = 0.68$ (68%). Coordination number is 8. Examples: Fe ($\alpha$), Cr, Na, W.

In a face centred cubic (FCC) unit cell there is an atom at each of the 8 corners and one at the centre of each of the 6 faces. The atoms touch along the face diagonal.

         o------------o
        /|    x      /|
       / |          / |     o = corner atom
      o------------o  |     x = face-centre atom
      |  |  x      |x |
      |x |    x    |  |     atoms touch along
      |  o---------|--o     each face diagonal
      | /    x     | /
      |/           |/
      o------------o

  One face:   o-------o
              | \   / |    face diagonal
              |   x   |    = R + 2R + R = 4R
              | /   \ |
              o-------o

Let $R$ = atomic radius, $a$ = lattice constant.

Lattice constant

Along a face diagonal: corner radius + face atom diameter + corner radius.

$$\sqrt{2}\,a = 4R \quad\Rightarrow\quad a = \frac{4R}{\sqrt{2}} = 2\sqrt{2}\,R = 2.828\,R$$

Body diagonal

$$\begin{aligned} \text{Body diagonal} &= \sqrt{3}\,a = \sqrt{3}\times2\sqrt{2}\,R \\ &= 2\sqrt{6}\,R = 4.899\,R \end{aligned}$$

Number of atoms per cell

$$N = 8\times\frac{1}{8} + 6\times\frac{1}{2} = 1 + 3 = 4$$

Packing density

$$\begin{aligned} \text{APF} &= \frac{N\times\frac{4}{3}\pi R^3}{a^3} = \frac{4\times\frac{4}{3}\pi R^3}{(2\sqrt{2}R)^3} \\ &= \frac{\frac{16}{3}\pi R^3}{16\sqrt{2}\,R^3} = \frac{\pi}{3\sqrt{2}} = 0.74 \end{aligned}$$

Answer: body diagonal $= 2\sqrt{6}\,R = 4.90R$ (with $a = 2\sqrt{2}R$), packing density $= \pi/(3\sqrt{2}) = 0.74$ (74%). FCC is a close-packed structure with coordination number 12. Examples: Cu, Al, Ag, Au, Ni.

In the photoelectric effect, a photon of energy $hf = hc/\lambda$ frees an electron only if $hf \ge \phi$ (work function). Einstein's equation is $KE_{max} = hf - \phi$. The longest wavelength that can cause emission is the threshold wavelength $\lambda_0$, for which $KE = 0$.

Given: $\lambda_0 = 522$ nm, $\lambda = 250$ nm, $h = 6.626\times10^{-34}$ J s, $c = 3\times10^{8}$ m/s, $1\text{ eV} = 1.6\times10^{-19}$ J.

Work function of sodium

$$\begin{aligned} \phi &= \frac{hc}{\lambda_0} = \frac{6.626\times10^{-34}\times3\times10^{8}}{522\times10^{-9}} \\ &= 3.808\times10^{-19}\ \text{J} = \frac{3.808\times10^{-19}}{1.6\times10^{-19}} = 2.38\ \text{eV} \end{aligned}$$

Energy of the UV photon

$$\begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\times3\times10^{8}}{250\times10^{-9}} \\ &= 7.951\times10^{-19}\ \text{J} = 4.97\ \text{eV} \end{aligned}$$

Kinetic energy of photoelectrons

$$\begin{aligned} KE_{max} &= E - \phi = 4.97 - 2.38 = 2.59\ \text{eV} \\ &= 4.14\times10^{-19}\ \text{J} \end{aligned}$$

Answer: work function of sodium $\phi = 3.81\times10^{-19}$ J $= 2.38$ eV; maximum kinetic energy of photoelectrons with 250 nm UV $= 4.14\times10^{-19}$ J $= 2.59$ eV.

Quantum mechanics is the theory that describes the behaviour of very small particles (electrons, atoms, photons). In it, a particle is described by a wave function $\psi$, its energy is quantized when it is bound, and only probabilities of position and momentum can be predicted. A key idea is wave–particle duality: every moving particle has a de Broglie wavelength

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

Wavelength of the golf ball

Given: $m = 50\ \text{g} = 0.05$ kg, $v = 20$ m/s, $h = 6.626\times10^{-34}$ J s.

$$\begin{aligned} p &= mv = 0.05\times20 = 1\ \text{kg m s}^{-1} \\ \lambda &= \frac{h}{mv} = \frac{6.626\times10^{-34}}{1} = 6.626\times10^{-34}\ \text{m} \end{aligned}$$

Answer: $\lambda = 6.63\times10^{-34}$ m.

Meaning

This wavelength is more than $10^{18}$ times smaller than the size of a proton ($\sim10^{-15}$ m). No slit or crystal is small enough to show diffraction of the golf ball, so its wave nature can never be observed and classical mechanics describes it perfectly. For comparison, an electron accelerated through 100 V has $\lambda \approx 0.123$ nm, similar to the spacing between atoms, so electrons show clear diffraction and must be treated by quantum mechanics.

Copper has an FCC unit cell: 4 atoms per cell, atoms touching along the face diagonal, so $\sqrt{2}a = 4R$.

Given: $R = 0.13$ nm, $M = 63.55$ g/mol $= 63.55\times10^{-3}$ kg/mol, $N_A = 6.022\times10^{23}$ mol⁻¹, $h = 6.626\times10^{-34}$ J s, $m_e = 9.1\times10^{-31}$ kg. Copper is monovalent (one free electron per atom).

Lattice constant:

$$a = 2\sqrt{2}\,R = 2\sqrt{2}\times0.13 = 0.3677\ \text{nm} = 3.677\times10^{-10}\ \text{m}$$ $$a^3 = 4.971\times10^{-29}\ \text{m}^3$$

(i) Packing density

$$\text{APF} = \frac{4\times\frac{4}{3}\pi R^3}{a^3} = \frac{4\times\frac{4}{3}\pi R^3}{(2\sqrt{2}R)^3} = \frac{\pi}{3\sqrt{2}} = 0.74$$

(ii) Density of copper

$$\begin{aligned} \rho &= \frac{N M}{N_A a^3} = \frac{4\times63.55\times10^{-3}}{6.022\times10^{23}\times4.971\times10^{-29}} \\ &= 8.49\times10^{3}\ \text{kg m}^{-3} = 8.49\ \text{g cm}^{-3} \end{aligned}$$

(iii) Atomic concentration

$$n_{at} = \frac{N}{a^3} = \frac{4}{4.971\times10^{-29}} = 8.05\times10^{28}\ \text{m}^{-3}$$

(iv) Fermi energy

With one conduction electron per atom, electron concentration $n = n_{at} = 8.05\times10^{28}$ m⁻³. At 0 K,

$$\begin{aligned} E_F &= \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} \\ &= \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}}\left(\frac{3\times8.05\times10^{28}}{\pi}\right)^{2/3} \\ &= 6.03\times10^{-38}\times1.807\times10^{19} = 1.09\times10^{-18}\ \text{J} \\ &= \frac{1.09\times10^{-18}}{1.6\times10^{-19}} = 6.81\ \text{eV} \end{aligned}$$

Answer: (i) APF $= 0.74$; (ii) $\rho = 8.49\times10^3$ kg m⁻³ (8.49 g cm⁻³); (iii) $n_{at} = 8.05\times10^{28}$ m⁻³; (iv) $E_F = 1.09\times10^{-18}$ J $= 6.81$ eV.

(With the more accurate $R = 0.128$ nm these become 8.89 g cm⁻³, $8.43\times10^{28}$ m⁻³ and 7.0 eV, close to measured values.)

A particle of energy $E$ meeting a barrier of height $V_0 > E$ and width $a$ can tunnel through with probability

$$T \approx T_0\,e^{-2\alpha a}, \qquad \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}, \qquad T_0 = 16\frac{E}{V_0}\left(1 - \frac{E}{V_0}\right)$$

Given: $m = 0.5\ \text{g} = 5\times10^{-4}$ kg, release height 5 m, barrier height 8 m, width $a = 1.5$ m, $g = 9.81$ m/s², $\hbar = 1.055\times10^{-34}$ J s.

Energies

The ball released from rest at 5 m has total energy (its kinetic energy at ground level):

$$E = mgh_1 = 5\times10^{-4}\times9.81\times5 = 0.02453\ \text{J}$$

To climb an 8 m barrier it would need

$$V_0 = mgh_2 = 5\times10^{-4}\times9.81\times8 = 0.03924\ \text{J}$$ $$V_0 - E = 0.01472\ \text{J}$$

Classically, $E < V_0$, so the ball can never cross.

Decay constant

$$\begin{aligned} \alpha &= \frac{\sqrt{2\times5\times10^{-4}\times0.01472}}{1.055\times10^{-34}} = \frac{3.836\times10^{-3}}{1.055\times10^{-34}} \\ &= 3.64\times10^{31}\ \text{m}^{-1} \end{aligned}$$ $$2\alpha a = 2\times3.64\times10^{31}\times1.5 = 1.09\times10^{32}$$

Transmission probability

$$T_0 = 16\times\frac{0.02453}{0.03924}\left(1 - \frac{0.02453}{0.03924}\right) = 16\times0.625\times0.375 = 3.75$$ $$\begin{aligned} T &\approx 3.75\,e^{-1.09\times10^{32}} \\ &\approx 10^{-4.74\times10^{31}} \approx 0 \end{aligned}$$

Answer: $T \approx e^{-1.09\times10^{32}} \approx 10^{-4.7\times10^{31}}$, which is zero for all practical purposes.

The ball will never be seen tunneling through the barrier, because its mass is huge compared with an electron and the barrier is very wide. Tunneling matters only for very light particles and barriers of nanometre width.

The probability that a state of energy $E$ is occupied by an electron is given by the Fermi–Dirac distribution function:

$$f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}$$

Given: the state is $3kT$ above the Fermi level, so $E - E_F = 3kT$.

$$\begin{aligned} f(E) &= \frac{1}{1 + e^{3kT/kT}} = \frac{1}{1 + e^{3}} \\ &= \frac{1}{1 + 20.09} = \frac{1}{21.09} = 0.0474 \end{aligned}$$

Answer: $f(E) = 0.0474$, i.e. about 4.74 %.

The result does not depend on temperature, because the energy is given in units of $kT$. A state only $3kT$ above $E_F$ (about 0.078 eV at 300 K) is already rarely occupied; the probability that it is empty is $1 - 0.0474 = 0.9526$.

Effective mass $m^* = \hbar^2/(d^2E/dk^2)$ is the mass an electron in a crystal appears to have under an external force. Infinite effective mass means $d^2E/dk^2 = 0$ (the inflection point of the E–k curve): there the external force produces no acceleration at all, because the force from the lattice exactly balances it.

Derivation of $m^*$

The electron is a wave packet moving with group velocity ($E = \hbar\omega$):

$$v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk}$$

An external force $F$ acting for time $dt$ does work $dE = F v_g\,dt$:

$$\frac{dE}{dk}dk = F\cdot\frac{1}{\hbar}\frac{dE}{dk}dt \quad\Rightarrow\quad F = \hbar\frac{dk}{dt}$$

Acceleration:

$$\begin{aligned} a &= \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt} = \frac{1}{\hbar^2}\frac{d^2E}{dk^2}F \end{aligned}$$

Comparing with $F = m^*a$:

$$m^* = \frac{\hbar^2}{d^2E/dk^2}$$

Positive, negative and infinite $m^*$ from the E–k diagram

In a crystal the band has a cosine-like shape. Taking $E = E_0 - 2\gamma\cos ka$,

$$\frac{dE}{dk} = 2\gamma a\sin ka, \qquad \frac{d^2E}{dk^2} = 2\gamma a^2\cos ka$$

   E                  (a) E vs k
   ^  __                              __
   |    \  m*<0                m*<0  /
   |     \                          /
   |      * <- inflection,  ->     *
   |       \   m* = infinite      /
   |        \                    /
   |         `--.__ m*>0 __.--'
   +----|------------|------------|----> k
      -pi/a          0          +pi/a

   v_g ~ dE/dk        (b) velocity
   ^           max at +pi/2a
   |             .--.
   |           /      \
   +----------*--------*---------> k
   |  \      /  0      +pi/a
   |    '--'

• Lower half of band ($|k| < \pi/2a$): curve concave up, $d^2E/dk^2 > 0$, $m^* > 0$. Velocity rises with $k$; the electron accelerates in the direction of the force, like a normal particle.
• Inflection point ($|k| = \pi/2a$): $d^2E/dk^2 = 0$, $m^* = \infty$. Velocity is maximum; increasing $k$ does not increase $v_g$. The applied force gives zero acceleration.
• Upper half of band ($\pi/2a < |k| < \pi/a$): curve concave down, $d^2E/dk^2 < 0$, $m^* < 0$. Velocity decreases as $k$ increases, so the electron accelerates opposite to the applied force. Physically, the electron is being Bragg-reflected by the lattice and the lattice pushes back harder than the external force.
• At $k = \pm\pi/a$ (band edge), $v_g = 0$: a standing wave.

Significance

• Near the bottom of the conduction band electrons have positive $m^*$ and behave like free electrons.
• Near the top of the valence band electrons have negative $m^*$. An electron missing from there behaves as a hole with positive charge and positive mass, which is the basis of hole conduction in semiconductors.