Let's walk through the derivation step by step for the Kronig-Penney model. The goal is to **derive the dispersion relation** for an electron in a periodic potential (the Kronig-Penney model) using the given wavefunctions and periodic boundary conditions. ### **Given:** #### In regions: - **Region I (0 < x < a):** \(\psi_I(x) = Ae^{iKx} + Be^{-iKx}\) - **Region II (a < x < a+b):** \(\psi_{II}(x) = Ce^{Qx} + De^{-Qx}\) - **Region III:** \(\psi_{III} = \psi_I \exp[ik(a+b)]\) Where: - \(K = \sqrt{2mE}/\hbar\) (free region) - \(Q = \sqrt{2m(V_0-E)}/\hbar\) (potential barrier) - The lattice constant \(d = a + b\) --- ### **1. Bloch's Theorem** By Bloch's theorem: \[ \psi(x + d) = e^{ikd}\psi(x) \] --- ### **2. Apply Boundary Conditions** At \(x = 0\) and \(x = a\), the wavefunction and its derivative must be continuous. #### At \(x = 0\): - \(\psi_I(0) = \psi_{II}(0)\) - \(\psi'_I(0) = \psi'_{II}(0)\) So, \[ A + B = C + D \] \[ iK(A - B) = Q(C - D) \] --- #### At \(x = a\): Similarly, continuity at \(x = a\): - \(\psi_I(a) = \psi_{II}(a)\) - \(\psi'_I(a) = \psi'_{II}(a)\) So, \[ Ae^{iKa} + Be^{-iKa} = Ce^{Qa} + De^{-Qa} \] \[ iK[Ae^{iKa} - Be^{-iKa}] = Q[Ce^{Qa} - De^{-Qa}] \] --- #### Periodic (Bloch) Boundary Condition at \(x = a + b\): By periodicity, \[ \psi_{II}(a + b) = \psi_I(0) e^{ikd} \] \[ \psi'_{II}(a + b) = \psi'_I(0) e^{ikd} \] --- ### **3. Express in Matrix Form** Set up the matching conditions in matrix form. Commonly, you relate the coefficients \(A, B\) at \(x=0\) to those at \(x=d\): \[ \begin{pmatrix} A' \\ B' \end{pmatrix} = M \begin{pmatrix} A \\ B \end{pmatrix} \] where \(M\) is the transfer matrix for one unit cell. But with the Bloch condition, \[ \begin{pmatrix} A' \\ B' \end{pmatrix} = e^{ikd} \begin{pmatrix} A \\ B \end{pmatrix} \] So, \[ \text{Tr}(M) = 2\cos(kd) \] --- ### **4. Find the Transfer Matrix** After applying all boundary conditions and eliminating the coefficients, the standard result is: \[ \cos(kd) = \cos(Ka)\cosh(Qb) - \frac{Q^2 - K^2}{2QK} \sin(Ka)\sinh(Qb) \] --- ### **5. Final Result** Comparing with your question: \[ \left[\frac{Q^2 - K^2}{2QK}\right] \sinh(Qb)\sin(Ka) + \cosh(Qb)\cos(Ka) = \cos[k(a+b)] \] This is **exactly** the standard Kronig-Penney dispersion relation. --- ## **Summary of Steps/Key Points** 1. **Write the wavefunctions** in each region. 2. **Apply boundary conditions** at 0 and \(a\): match wavefunctions and derivatives. 3. **Use Bloch's theorem** for periodicity: \(\psi(x+d) = e^{ikd}\psi(x)\) 4. **Eliminate coefficients** to get a relation only involving \(K, Q, k, a, b\). 5. **Arrive at the dispersion relation:** \[ \left[\frac{Q^2 - K^2}{2QK}\right]\sinh(Qb)\sin(Ka) + \cosh(Qb)\cos(Ka) = \cos[k(a+b)] \] --- ### **Reference** This derivation can be found in many quantum mechanics textbooks, e.g., Griffiths, "Introduction to Quantum Mechanics," and Ashcroft & Mermin, "Solid State Physics," under the Kronig-Penney model. If you need the explicit step-by-step algebra for eliminating the coefficients, let me know!