Physics / Solid State Physics II Quantum Theory of Solids 100% Free Open Access
Chapter 1 • Theory & Derivations

Quantum Electronic Band Theory: Bloch Theorem, Kronig-Penney & Effective Mass

This foundational unit develops the quantum mechanical theory of electrons propagating through a periodic crystalline lattice. We establish Bloch's theorem from translational symmetry and derive the Central Equation in reciprocal space. We examine how electron Bragg reflection at Brillouin zone boundaries leads to forbidden energy band gaps in the nearly free electron model and solve the transcendental Kronig-Penney model across all barrier limits. Finally, we formulate the semiclassical equations of motion, derive the effective mass tensor, introduce the concept of holes with positive charge, and classify metals, semimetals, semiconductors, and insulators based on their electronic band structure.

§1.1Periodic Lattice Potentials, Translational Invariance & Bloch's Theorem

1. The Crystal Hamiltonian & Discrete Translational Symmetry

In an ideal, defect-free single crystal, the atomic nuclei are arranged in a regular Bravais lattice defined by real-space lattice translation vectors:

$$\vec{R} = n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3 \vec{a}_3, \quad n_1, n_2, n_3 \in \mathbb{Z}$$

The single-particle crystal potential experienced by an electron satisfies exact spatial periodicity:

$$V(\vec{r} + \vec{R}) = V(\vec{r})$$

The single-electron stationary Schrödinger equation is:

$$\hat{H} \psi(\vec{r}) = \left[ -\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}) \right] \psi(\vec{r}) = E \psi(\vec{r})$$

We define the real-space translation operator $\hat{T}_{\vec{R}}$ acting on an arbitrary wavefunction $\psi(\vec{r})$:

$$\hat{T}_{\vec{R}} \psi(\vec{r}) \equiv \psi(\vec{r} + \vec{R})$$

Because $V(\vec{r} + \vec{R}) = V(\vec{r})$ and the Laplacian $\nabla^2$ is invariant under rigid translations, $\hat{T}_{\vec{R}}$ commutes with the Hamiltonian:

$$[\hat{H}, \hat{T}_{\vec{R}}] = 0, \quad [\hat{T}_{\vec{R}}, \hat{T}_{\vec{R}'}] = 0$$

Therefore, the Hamiltonian and the translation operators share a complete set of simultaneous stationary eigenfunctions.

2. Mathematical Proof of Bloch's Theorem

Let $\psi(\vec{r})$ be an eigenstate of $\hat{T}_{\vec{R}}$ with eigenvalue $C(\vec{R})$:

$$\hat{T}_{\vec{R}} \psi(\vec{r}) = \psi(\vec{r} + \vec{R}) = C(\vec{R}) \psi(\vec{r})$$

Applying successive translations $\vec{R}_1$ and $\vec{R}_2$:

$$\hat{T}_{\vec{R}_1 + \vec{R}_2} \psi(\vec{r}) = C(\vec{R}_1 + \vec{R}_2) \psi(\vec{r}) = C(\vec{R}_1) C(\vec{R}_2) \psi(\vec{r})$$ $$\implies C(\vec{R}_1 + \vec{R}_2) = C(\vec{R}_1) C(\vec{R}_2)$$

Under periodic Born-von Kármán boundary conditions for a macroscopic crystal of dimensions $L_i = N_i a_i$ along basis directions $\vec{a}_i$, the wavefunction must remain finite and normalizeable everywhere:

$$\psi(\vec{r} + N_i \vec{a}_i) = \psi(\vec{r}) \implies [C(\vec{a}_i)]^{N_i} = 1 \implies |C(\vec{R})| = 1$$

The only function satisfying this multiplicative property and unitarity is a complex phase:

$$C(\vec{R}) = e^{i \vec{k} \cdot \vec{R}}$$

where $\vec{k}$ is a real vector in reciprocal space known as the crystal wavevector. Hence:

$$\psi_{\vec{k}}(\vec{r} + \vec{R}) = e^{i \vec{k} \cdot \vec{R}} \psi_{\vec{k}}(\vec{r})$$

Defining $u_{\vec{k}}(\vec{r}) \equiv e^{-i \vec{k} \cdot \vec{r}} \psi_{\vec{k}}(\vec{r})$, we verify its periodicity:

$$u_{\vec{k}}(\vec{r} + \vec{R}) = e^{-i \vec{k} \cdot (\vec{r} + \vec{R})} \psi_{\vec{k}}(\vec{r} + \vec{R}) = e^{-i \vec{k} \cdot \vec{r}} e^{-i \vec{k} \cdot \vec{R}} e^{i \vec{k} \cdot \vec{R}} \psi_{\vec{k}}(\vec{r}) = u_{\vec{k}}(\vec{r})$$

This establishes Bloch's Theorem: The eigenstates of a one-electron Hamiltonian in a periodic potential can be chosen in the form of a plane wave modulated by a cell-periodic amplitude:

$$\psi_{\vec{k}}(\vec{r}) = e^{i \vec{k} \cdot \vec{r}} u_{\vec{k}}(\vec{r}), \quad \text{where } u_{\vec{k}}(\vec{r} + \vec{R}) = u_{\vec{k}}(\vec{r})$$

§1.2The Central Equation: Reciprocal Lattice Fourier Expansion & Zone Boundaries

1. Fourier Expansion in Reciprocal Space

Any function possessing the full translational periodicity of the Bravais lattice, such as the potential $V(\vec{r})$ and the periodic Bloch factor $u_{\vec{k}}(\vec{r})$, can be expanded in a Fourier series over the reciprocal lattice vectors $\vec{G}$:

$$V(\vec{r}) = \sum_{\vec{G}} V_{\vec{G}} e^{i \vec{G} \cdot \vec{r}}, \quad \vec{G} \cdot \vec{R} = 2\pi n, \quad n \in \mathbb{Z}$$

Because $V(\vec{r})$ is real-valued, the Fourier components satisfy $V_{-\vec{G}} = V_{\vec{G}}^*$. By shifting the zero of energy, we can set $V_{\vec{G}=0} = 0$.

Similarly, the full Bloch wavefunction $\psi_{\vec{k}}(\vec{r})$ can be written as a plane-wave expansion:

$$\psi_{\vec{k}}(\vec{r}) = \sum_{\vec{G}} C(\vec{k} - \vec{G}) e^{i (\vec{k} - \vec{G}) \cdot \vec{r}}$$

2. Derivation of the Central Equation

Substituting the Fourier expansions into the Schrödinger equation:

$$\sum_{\vec{G}} \frac{\hbar^2}{2m} (\vec{k} - \vec{G})^2 C(\vec{k} - \vec{G}) e^{i (\vec{k} - \vec{G}) \cdot \vec{r}} + \sum_{\vec{G}'} V_{\vec{G}'} e^{i \vec{G}' \cdot \vec{r}} \sum_{\vec{G}''} C(\vec{k} - \vec{G}'') e^{i (\vec{k} - \vec{G}'') \cdot \vec{r}} = E \sum_{\vec{G}} C(\vec{k} - \vec{G}) e^{i (\vec{k} - \vec{G}) \cdot \vec{r}}$$

Setting $\vec{G} = \vec{G}'' - \vec{G}'$ in the second term, multiplying by $e^{-i (\vec{k} - \vec{G}) \cdot \vec{r}}$, and integrating over the crystal volume using the orthogonality of plane waves:

$$\left( \frac{\hbar^2}{2m} (\vec{k} - \vec{G})^2 - E \right) C(\vec{k} - \vec{G}) + \sum_{\vec{G}'} V_{\vec{G} - \vec{G}'} C(\vec{k} - \vec{G}') = 0$$

Letting $\lambda_{\vec{k}-\vec{G}} \equiv \frac{\hbar^2}{2m} (\vec{k} - \vec{G})^2$ represent the free-electron kinetic energy, this infinite set of algebraic equations is known as the Central Equation:

$$(\lambda_{\vec{k}-\vec{G}} - E) C(\vec{k} - \vec{G}) + \sum_{\vec{G}'} V_{\vec{G}'} C(\vec{k} - \vec{G} - \vec{G}') = 0$$

Nontrivial solutions require the vanishing of the infinite secular determinant, generating the discrete band energies $E_n(\vec{k})$ indexed by the band index $n$.

§1.3Electron Bragg Reflection & Band Gap Opening in the Nearly Free Electron Model

1. Semiclassical Bragg Reflection of Matter Waves

In the nearly free electron (NFE) approximation, the lattice potential $V(\vec{r})$ is regarded as a weak perturbation relative to the kinetic energy: $|V_{\vec{G}}| \ll E_F$.

Away from the Brillouin zone boundaries, the kinetic energies $\lambda_{\vec{k}}$ and $\lambda_{\vec{k}-\vec{G}}$ are widely separated, so mixing is negligible. However, when $\vec{k}$ approaches a Bragg plane in reciprocal space:

$$\lambda_{\vec{k}} \approx \lambda_{\vec{k}-\vec{G}} \implies |\vec{k}|^2 \approx |\vec{k} - \vec{G}|^2 \implies 2\vec{k} \cdot \vec{G} = |\vec{G}|^2$$

This is precisely the Von Laue condition for Bragg reflection of electron waves off the crystal planes. At this boundary, the forward-propagating plane wave $e^{i \vec{k} \cdot \vec{r}}$ and backscattered wave $e^{i (\vec{k} - \vec{G}) \cdot \vec{r}}$ interfere constructively.

2. Secular Determinant & Energy Gap Derivation

Retaining only the two strongly degenerate plane wave states $|\vec{k}\rangle$ and $|\vec{k} - \vec{G}\rangle$, the Central Equation reduces to a $2 \times 2$ secular matrix:

$$\begin{pmatrix} \lambda_{\vec{k}} - E & V_{\vec{G}} \\ V_{\vec{G}}^* & \lambda_{\vec{k}-\vec{G}} - E \end{pmatrix} \begin{pmatrix} C(\vec{k}) \\ C(\vec{k}-\vec{G}) \end{pmatrix} = 0$$

Setting the determinant to zero:

$$(\lambda_{\vec{k}} - E)(\lambda_{\vec{k}-\vec{G}} - E) - |V_{\vec{G}}|^2 = 0$$ $$E^2 - (\lambda_{\vec{k}} + \lambda_{\vec{k}-\vec{G}}) E + \lambda_{\vec{k}} \lambda_{\vec{k}-\vec{G}} - |V_{\vec{G}}|^2 = 0$$

Solving the quadratic equation:

$$E_{\pm}(\vec{k}) = \frac{\lambda_{\vec{k}} + \lambda_{\vec{k}-\vec{G}}}{2} \pm \sqrt{ \left( \frac{\lambda_{\vec{k}} - \lambda_{\vec{k}-\vec{G}}}{2} \right)^2 + |V_{\vec{G}}|^2 }$$

Exactly at the Brillouin zone boundary where $\lambda_{\vec{k}} = \lambda_{\vec{k}-\vec{G}}$:

$$E_{\pm} = \lambda_{\vec{k}} \pm |V_{\vec{G}}|$$

The energy difference defines the fundamental energy band gap $E_g$:

$$E_g = E_+ - E_- = 2 |V_{\vec{G}}|$$

The two standing wave eigenstates at the zone boundary correspond to:

$$\psi_+(\vec{r}) \sim \cos(\vec{G} \cdot \vec{r} / 2), \quad \psi_-(\vec{r}) \sim \sin(\vec{G} \cdot \vec{r} / 2)$$

The $\psi_+$ state piles electron charge directly atop the attractive ionic cores ($V < 0$), lowering its energy, whereas $\psi_-$ concentrates probability density midway between the ions, raising its energy. This electrostatic potential difference opens the forbidden energy gap.

§1.4The Kronig-Penney Model: 1D Dirac Delta & Square Well Potentials

1. Formulation of the 1D Kronig-Penney Model

To understand how discrete atomic energy levels evolve continuously into energy bands, Kronig and Penney (1931) modeled an infinite 1D crystal with an array of rectangular potential barriers of height $V_0$, width $b$, and lattice period $a$ (well width $w = a - b$):

$$V(x) = \begin{cases} 0, & 0 < x < a - b \\ V_0, & a - b < x < a \end{cases}, \quad V(x + a) = V(x)$$

In the well region ($0 < x < a - b$), where $V = 0$:

$$\psi_1(x) = A e^{i K_1 x} + B e^{-i K_1 x}, \quad K_1 = \sqrt{\frac{2mE}{\hbar^2}}$$

In the barrier region ($a - b < x < a$), for $E < V_0$:

$$\psi_2(x) = C e^{K_2 x} + D e^{-K_2 x}, \quad K_2 = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}$$

By Bloch's theorem, $\psi(x + a) = e^{i k a} \psi(x)$. Applying boundary conditions of continuity of $\psi(x)$ and $d\psi/dx$ at $x = 0$ and $x = a - b$:

2. The Dirac Delta-Barrier Limit & Transcendental Dispersion

Taking the delta-function limit where the barrier width $b \to 0$ and height $V_0 \to \infty$ such that the barrier area $b V_0$ remains finite, we define the dimensionless barrier strength parameter $P$:

$$P \equiv \lim_{b \to 0, V_0 \to \infty} \frac{m V_0 b a}{\hbar^2}$$

The determinant condition yields the famous Kronig-Penney dispersion relation:

$$P \frac{\sin(\alpha a)}{\alpha a} + \cos(\alpha a) = \cos(k a)$$

where $\alpha = \sqrt{2mE/\hbar^2}$ and $k$ is the crystal wavevector in the first Brillouin zone ($-\pi/a \le k \le \pi/a$).

  • Since $|\cos(ka)| \le 1$, energy solutions $\alpha$ are only allowed where: $$-1 \le P \frac{\sin(\alpha a)}{\alpha a} + \cos(\alpha a) \le 1$$
  • Whenever the left-hand side exceeds $+1$ or drops below $-1$, no real wavevector $k$ exists. These forbidden energy intervals constitute the forbidden energy band gaps.
  • Free Electron Limit ($P \to 0$): $\cos(\alpha a) = \cos(ka) \implies \alpha = k \implies E = \frac{\hbar^2 k^2}{2m}$ (Continuous parabolic band).
  • Tight-Binding Atomic Limit ($P \to \infty$): $\sin(\alpha a) = 0 \implies \alpha a = n\pi \implies E_n = \frac{\hbar^2 \pi^2 n^2}{2m a^2}$ (Discrete atomic bound states in an infinite square well).

§1.5Semiclassical Electron Dynamics: Wavepackets, Group Velocity & Crystal Momentum

1. Electron Wavepackets & Group Velocity

An electron in a crystal is represented by a wavepacket composed of Bloch states localized in both real space and crystal momentum space. The physical velocity of the electron is the group velocity $v_g$ of the envelope:

$$\vec{v}_g(\vec{k}) = \frac{1}{\hbar} \nabla_{\vec{k}} E(\vec{k})$$

In one dimension:

$$v_g(k) = \frac{1}{\hbar} \frac{dE}{dk}$$

Key physical consequences of group velocity:

  • At the bottom of an energy band ($k = 0$), $dE/dk = 0$, so $v_g = 0$.
  • Near the center of the band, $v_g$ reaches a maximum value.
  • At the Brillouin zone boundary ($k = \pm \pi/a$), Bragg reflection forces $dE/dk = 0$, so $v_g = 0$. The electron forms a standing wave and cannot transport net charge forward!

2. Semiclassical Equation of Motion

When an external electric field $\vec{\mathcal{E}}$ or magnetic field $\vec{B}$ is applied over macroscopic scales much larger than the lattice constant $a$, the rate of work done on the electron wavepacket is:

$$\frac{dE}{dt} = \vec{F}_{\text{ext}} \cdot \vec{v}_g = \vec{F}_{\text{ext}} \cdot \left( \frac{1}{\hbar} \nabla_{\vec{k}} E \right)$$

Using the chain rule:

$$\frac{dE}{dt} = \nabla_{\vec{k}} E \cdot \frac{d\vec{k}}{dt}$$

Comparing the two expressions gives the fundamental semiclassical equation of motion:

$$\hbar \frac{d\vec{k}}{dt} = \vec{F}_{\text{ext}} = -e \left( \vec{\mathcal{E}} + \vec{v}_g \times \vec{B} \right)$$

Here, $\hbar\vec{k}$ represents the crystal momentum. It is not the total kinematic momentum $m\vec{v}$ of the electron, because the periodic lattice potential can absorb or impart discrete momentum quanta $\hbar\vec{G}$ through Bragg diffraction.

§1.6The Effective Mass Tensor, Negative Effective Mass & Concept of Positive Holes

1. Derivation of the Effective Mass Tensor

Differentiating the group velocity $\vec{v}_g$ with respect to time gives the semiclassical acceleration:

$$a_i = \frac{d v_{g,i}}{dt} = \frac{d}{dt} \left( \frac{1}{\hbar} \frac{\partial E}{\partial k_i} \right) = \frac{1}{\hbar} \sum_j \frac{\partial^2 E}{\partial k_i \partial k_j} \frac{d k_j}{dt}$$

Substituting $\hbar \frac{dk_j}{dt} = F_j$:

$$a_i = \sum_j \left[ \frac{1}{\hbar^2} \frac{\partial^2 E}{\partial k_i \partial k_j} \right] F_j$$

Comparing this with Newton's second law in tensor form $a_i = \sum_j (m^*)^{-1}_{ij} F_j$, we define the inverse effective mass tensor:

$$\left( \frac{1}{m^*} \right)_{ij} \equiv \frac{1}{\hbar^2} \frac{\partial^2 E(\vec{k})}{\partial k_i \partial k_j}$$

In an isotropic band in one dimension:

$$m^*(k) = \frac{\hbar^2}{\frac{d^2 E}{dk^2}}$$

The effective mass encapsulates the entire dynamic interaction between the electron and the periodic periodic potential of the ionic lattice. The external force alone governs the acceleration, provided $m$ is replaced by $m^*$.

2. Negative Effective Mass & The Concept of Positive Holes

Near the top of an energy band, the band curvature is downward:

$$\frac{d^2 E}{dk^2} < 0 \implies m^* < 0$$

An electron with negative effective mass accelerates opposite to the applied force $\vec{F} = -e\vec{\mathcal{E}}$. This paradoxical behavior occurs because the electron is Bragg-scattered backwards by the lattice more strongly than the forward pull of the external field!

In a nearly filled band, instead of tracking $10^{22}\text{ cm}^{-3}$ electrons with negative effective mass, it is mathematically and physically equivalent to treat the empty states as quasiparticles called holes:

  • Charge: $q_h = -q_e = +e$ (positive elementary charge)
  • Wavevector: $\vec{k}_h = -\vec{k}_e$
  • Energy: $E_h(\vec{k}_h) = -E_e(\vec{k}_e)$
  • Effective Mass: $m_h^* = -m_e^* > 0$ (positive effective mass!)
  • Velocity: $\vec{v}_h = \vec{v}_e$

§1.7Electronic Classification of Solids: Metals, Semimetals, Insulators & Intrinsic Semiconductors

1. Band Filling & Electrical Conductivity

According to the Pauli exclusion principle, each spatial Bloch orbital $|\psi_{\vec{k}} angle$ can accommodate at most two electrons of opposite spin ($s_z = \pm 1/2$). In a 1D crystal with $N$ primitive unit cells, each Brillouin zone contains exactly $N$ allowed $\vec{k}$ states, yielding a capacity of:

$$\text{Capacity per band} = 2N \text{ electrons}$$

A completely filled band carries zero net electrical current under an applied electric field, because for every electron moving with velocity $+v_g(\vec{k})$, there exists another electron with opposite velocity $-v_g(-\vec{k})$:

$$\vec{J} = -e \sum_{\vec{k} \in \text{filled}} \vec{v}_g(\vec{k}) = -\frac{e}{\hbar} \int_{\text{BZ}} \nabla_{\vec{k}} E(\vec{k}) \frac{d^3k}{(2\pi)^3} = 0$$

2. Taxonomy of Solids

  • Metals (Good Conductors): Possess a partially filled band (e.g. monovalent alkali metals like Na, Cu with 1 valence electron per atom filling half the zone), or overlapping conduction and valence bands (divalent alkaline earth metals like Mg, Ca). The Fermi level $E_F$ cuts through an allowed band, providing continuous unoccupied states immediately adjacent in energy, enabling high electrical conductivity $\sigma \sim 10^7\text{ S/m}$ at low temperatures.
  • Insulators: Have completely filled valence bands separated from completely empty conduction bands by a large fundamental energy gap $E_g > 3.5\text{ eV}$ (e.g. diamond with $E_g = 5.47\text{ eV}$, $\text{SiO}_2$ with $E_g = 9\text{ eV}$). At room temperature ($k_B T \approx 0.026\text{ eV}$), thermal excitation across the gap is negligible ($e^{-E_g/2k_B T} \sim 10^{-46}$), yielding electrical resistivity $\rho > 10^{12}\ \Omega\cdot\text{m}$.
  • Intrinsic Semiconductors: Possess the exact same band topology as insulators, but with a modest energy band gap $E_g \lesssim 2.0\text{ eV}$ (e.g. silicon $E_g = 1.12\text{ eV}$, germanium $E_g = 0.66\text{ eV}$, gallium arsenide $E_g = 1.42\text{ eV}$). At $T = 0\text{ K}$, semiconductors are perfect insulators; at room temperature, thermal excitation promotes electrons into the conduction band while leaving mobile holes in the valence band.
  • Semimetals: Have a very small negative band gap where the top of the valence band slightly overlaps the bottom of the conduction band at different points in reciprocal space (e.g. bismuth, antimony, graphite), producing small equal numbers of electron and hole pockets with low carrier densities ($n \sim 10^{17}-10^{19}\text{ cm}^{-3}$).

Honors Examination Worked Problems & Solutions

Rigorous step-by-step mathematical proofs and solutions to university degree examination problems.

SOLVED PROBLEM 1.1

Central Equation Formulation & Band Gap of a Periodic Cosine Potential

An electron moves in a 1D lattice of constant $a$ under a weak periodic potential $V(x) = 2V_1 \cos(2\pi x / a) = V_1 (e^{i G x} + e^{-i G x})$, where $G = 2\pi/a$ is the shortest reciprocal lattice vector.\n\n(a) Write down the Central Equation for the Fourier coefficients $C(k)$ and $C(k-G)$ near the Brillouin zone boundary $k \approx G/2$.\n(b) Solve the $2\times 2$ secular determinant to determine the exact energy eigenvalues $E_\pm(k)$.\n(c) Calculate the exact magnitude of the energy band gap $E_g$ opened at $k = G/2 = \pi/a$, and state the explicit wavefunctions $\psi_+(x)$ and $\psi_-(x)$ corresponding to the band edges.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Central Equation Matrix:** The Fourier components of the potential are $V_G = V_{-G} = V_1$, and all other $V_{G'} = 0$. Near $k = \pi/a = G/2$, the two kinetic energies $\lambda_k = \frac{\hbar^2 k^2}{2m}$ and $\lambda_{k-G} = \frac{\hbar^2 (k-G)^2}{2m}$ are nearly degenerate. Retaining only these two states in the Central Equation: $$(\lambda_k - E) C(k) + V_1 C(k-G) = 0$$ $$V_1 C(k) + (\lambda_{k-G} - E) C(k-G) = 0$$ **(b) Secular Determinant & Energy Dispersion:** For non-trivial solutions: $$\det \begin{pmatrix} \lambda_k - E & V_1 \\ V_1 & \lambda_{k-G} - E \end{pmatrix} = 0$$ $$(\lambda_k - E)(\lambda_{k-G} - E) - V_1^2 = 0$$ $$E^2 - (\lambda_k + \lambda_{k-G}) E + \lambda_k \lambda_{k-G} - V_1^2 = 0$$ Solving using the quadratic formula: $$E_\pm(k) = \frac{\lambda_k + \lambda_{k-G}}{2} \pm \sqrt{ \left(\frac{\lambda_k - \lambda_{k-G}}{2}\right)^2 + V_1^2 }$$ **(c) Band Gap at Zone Boundary & Wavefunctions:** At the zone boundary $k = G/2 = \pi/a$: $$\lambda_k = \lambda_{k-G} = \frac{\hbar^2 (\pi/a)^2}{2m} \equiv E_0$$ Substituting this into the dispersion relation: $$E_\pm = E_0 \pm V_1$$ The fundamental energy band gap is: $$E_g = E_+ - E_- = (E_0 + V_1) - (E_0 - V_1) = 2 V_1$$ For the lower energy state $E_- = E_0 - V_1$, the eigenvector yields $C(k-G) = -C(k)$, producing the standing wave: $$\psi_-(x) \propto e^{i G x/2} - e^{-i G x/2} \propto \sin(Gx/2) = \sin(\pi x / a)$$ For the upper state $E_+ = E_0 + V_1$, $C(k-G) = C(k)$, giving: $$\psi_+(x) \propto e^{i G x/2} + e^{-i G x/2} \propto \cos(Gx/2) = \cos(\pi x / a)$$ The lower state piles electron probability at $x = a/2$ where $V(x) = -2V_1$, while the upper state piles probability at $x = 0$ where $V(x) = +2V_1$, accounting physically for the $2V_1$ energy separation.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.

SOLVED PROBLEM 1.2

Kronig-Penney Delta-Barrier Energy Eigenvalues & Band Widths

Consider the Kronig-Penney model with periodic delta-function potential barriers $V(x) = \frac{\hbar^2 P}{m a} \sum_n \delta(x - n a)$.\n\n(a) From the dispersion relation $P \frac{\sin \alpha a}{\alpha a} + \cos \alpha a = \cos k a$, calculate the lowest allowed energy $E_1$ (at $k = 0$) in the limit of small barrier strength $P \ll 1$.\n(b) Calculate the energy gap $E_g$ opened at the first Brillouin zone boundary ($k = \pi/a$) for small $P \ll 1$.\n(c) In the opposite tight-binding limit $P \gg 1$, derive the width of the lowest energy band $\Delta E_1 = E(\pi/a) - E(0)$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Lowest Energy Level for $P \ll 1$ at $k = 0$:** At $k = 0$, $\cos(ka) = 1$. The Kronig-Penney relation is: $$P \frac{\sin \xi}{\xi} + \cos \xi = 1, \quad \xi \equiv \alpha a$$ For small $P$ and small $\xi$, expand $\cos\xi \approx 1 - \xi^2/2$ and $\sin\xi \approx \xi$: $$P \frac{\xi}{\xi} + 1 - \frac{\xi^2}{2} = 1 \implies P - \frac{\xi^2}{2} = 0 \implies \xi^2 = 2P$$ Since $\xi = \alpha a = \sqrt{2mE/\hbar^2} a$: $$\alpha^2 a^2 = \frac{2mE}{\hbar^2} a^2 = 2P \implies E_1(k=0) = \frac{\hbar^2 P}{m a^2}$$ Notice that this matches the spatial average potential $\langle V \rangle = \frac{1}{a} \int_0^a V(x) dx = \frac{\hbar^2 P}{m a^2}$. **(b) Energy Gap at the First Zone Boundary ($k = \pi/a$):** At $k = \pi/a$, $\cos(ka) = -1$. Let $\xi = \pi + \delta$. Then $\sin(\pi + \delta) = -\sin\delta \approx -\delta$, and $\cos(\pi + \delta) = -\cos\delta \approx -(1 - \delta^2/2)$: $$-P \frac{\delta}{\pi} - \left(1 - \frac{\delta^2}{2}\right) = -1 \implies \frac{\delta^2}{2} - \frac{P}{\pi}\delta = 0$$ The two roots are $\delta_1 = 0$ and $\delta_2 = \frac{2P}{\pi}$. The two boundary energies are: $$E_- = \frac{\hbar^2 \pi^2}{2m a^2}, \quad E_+ = \frac{\hbar^2 (\pi + \delta_2)^2}{2m a^2} \approx \frac{\hbar^2 \pi^2}{2m a^2} \left(1 + \frac{2\delta_2}{\pi}\right) = \frac{\hbar^2 \pi^2}{2m a^2} + \frac{2\hbar^2 P}{m a^2}$$ The energy gap opened at the first zone boundary is: $$E_g = E_+ - E_- = \frac{2\hbar^2 P}{m a^2}$$ **(c) Lowest Band Width for $P \gg 1$:** When $P \gg 1$, the state approaches the infinite square well: $\xi_n \approx n\pi$. For the $n=1$ band, let $\xi = \pi - \epsilon$. $$P \frac{\sin(\pi - \epsilon)}{\pi} + \cos(\pi - \epsilon) = \cos(ka)$$ $$\frac{P}{\pi} \epsilon - 1 = \cos(ka) \implies \epsilon = \frac{\pi}{P} [1 + \cos(ka)]$$ At $k = 0$, $\cos(ka) = 1 \implies \epsilon(0) = \frac{2\pi}{P}$, so $\xi(0) = \pi - \frac{2\pi}{P}$. At $k = \pi/a$, $\cos(ka) = -1 \implies \epsilon(\pi/a) = 0$, so $\xi(\pi/a) = \pi$. The band width is: $$\Delta E_1 = E(\pi/a) - E(0) = \frac{\hbar^2}{2m a^2} [\pi^2 - (\pi - 2\pi/P)^2] \approx \frac{\hbar^2}{2m a^2} \left( \frac{4\pi^2}{P} \right) = \frac{2\pi^2 \hbar^2}{m a^2 P}$$ As barrier strength $P \to \infty$, the band width shrinks inversely as $1/P$, narrowing into a discrete atomic bound state.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.

SOLVED PROBLEM 1.3

Effective Mass Tensor & Cyclotron Frequency in Anisotropic Tight-Binding Crystals

An electron in an orthorhombic crystal with lattice constants $a, b, c$ has the 3D anisotropic tight-binding dispersion:\n$$E(\vec{k}) = E_0 - 2t_x \cos(k_x a) - 2t_y \cos(k_y b) - 2t_z \cos(k_z c)$$\n\n(a) Compute the three diagonal components of the effective mass tensor $m_{xx}^*, m_{yy}^*, m_{zz}^*$ near the band bottom $\vec{k} = 0$.\n(b) Determine the group velocity vector $\vec{v}_g(\vec{k})$ at an arbitrary point in the Brillouin zone.\n(c) A uniform magnetic field $\vec{B} = B_0 \hat{z}$ is applied along the $z$-axis. Derive the cyclotron resonance frequency $\omega_c$ and the cyclotron effective mass $m_c^*$ in terms of $m_{xx}^*$ and $m_{yy}^*$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Diagonal Effective Mass Components:** Expand the cosine terms near the band minimum $\vec{k} = 0$: $$\cos(k_x a) \approx 1 - \frac{k_x^2 a^2}{2}, \quad \cos(k_y b) \approx 1 - \frac{k_y^2 b^2}{2}, \quad \cos(k_z c) \approx 1 - \frac{k_z^2 c^2}{2}$$ $$E(\vec{k}) \approx (E_0 - 2t_x - 2t_y - 2t_z) + t_x a^2 k_x^2 + t_y b^2 k_y^2 + t_z c^2 k_z^2$$ The effective mass components are: $$\frac{1}{m_{xx}^*} = \frac{1}{\hbar^2} \frac{\partial^2 E}{\partial k_x^2} = \frac{2 t_x a^2}{\hbar^2} \implies m_{xx}^* = \frac{\hbar^2}{2 t_x a^2}$$ $$\frac{1}{m_{yy}^*} = \frac{1}{\hbar^2} \frac{\partial^2 E}{\partial k_y^2} = \frac{2 t_y b^2}{\hbar^2} \implies m_{yy}^* = \frac{\hbar^2}{2 t_y b^2}$$ $$\frac{1}{m_{zz}^*} = \frac{1}{\hbar^2} \frac{\partial^2 E}{\partial k_z^2} = \frac{2 t_z c^2}{\hbar^2} \implies m_{zz}^* = \frac{\hbar^2}{2 t_z c^2}$$ All off-diagonal components vanish by orthorhombic reflection symmetry: $m_{ij}^* = 0$ for $i \ne j$. **(b) Group Velocity Vector:** Using $\vec{v}_g = \frac{1}{\hbar} \nabla_{\vec{k}} E$: $$v_{gx} = \frac{1}{\hbar} \frac{\partial E}{\partial k_x} = \frac{2 t_x a}{\hbar} \sin(k_x a)$$ $$v_{gy} = \frac{1}{\hbar} \frac{\partial E}{\partial k_y} = \frac{2 t_y b}{\hbar} \sin(k_y b)$$ $$v_{gz} = \frac{1}{\hbar} \frac{\partial E}{\partial k_z} = \frac{2 t_z c}{\hbar} \sin(k_z c)$$ $$\vec{v}_g(\vec{k}) = \frac{2}{\hbar} \left[ t_x a \sin(k_x a) \hat{x} + t_y b \sin(k_y b) \hat{y} + t_z c \sin(k_z c) \hat{z} \right]$$ **(c) Cyclotron Resonance Frequency:** Under $\vec{B} = B_0 \hat{z}$, the semiclassical equation of motion is: $$\hbar \frac{d\vec{k}}{dt} = -e (\vec{v}_g \times \vec{B}) = -e (v_{gy} B_0 \hat{x} - v_{gx} B_0 \hat{y})$$ For small $k$, $v_{gx} \approx \frac{\hbar k_x}{m_{xx}^*}$ and $v_{gy} \approx \frac{\hbar k_y}{m_{yy}^*}$: $$\hbar \frac{dk_x}{dt} = -e B_0 \left(\frac{\hbar k_y}{m_{yy}^*}\right) \implies \frac{dk_x}{dt} = -\frac{e B_0}{m_{yy}^*} k_y$$ $$\hbar \frac{dk_y}{dt} = +e B_0 \left(\frac{\hbar k_x}{m_{xx}^*}\right) \implies \frac{dk_y}{dt} = +\frac{e B_0}{m_{xx}^*} k_x$$ Differentiating the first equation with respect to $t$: $$\frac{d^2 k_x}{dt^2} = -\frac{e B_0}{m_{yy}^*} \frac{dk_y}{dt} = -\frac{e^2 B_0^2}{m_{xx}^* m_{yy}^*} k_x$$ This is simple harmonic motion $\frac{d^2 k_x}{dt^2} + \omega_c^2 k_x = 0$ with cyclotron frequency: $$\omega_c = \frac{e B_0}{\sqrt{m_{xx}^* m_{yy}^*}}$$ Defining the cyclotron effective mass $m_c^* \equiv \frac{e B_0}{\omega_c}$: $$m_c^* = \sqrt{m_{xx}^* m_{yy}^*} = \frac{\hbar^2}{2 a b \sqrt{t_x t_y}}$$ This proves that the cyclotron mass in an anisotropic crystal is the geometric mean of the transverse effective mass components.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.