Let's solve the problem step-by-step, focusing on the nearly free electron model for a 2D square lattice as described. --- ## **Periodic Potential** Given potential: \[ V(x, y) = 4V_1 \cos\left(\frac{2\pi x}{a}\right)\cos\left(\frac{2\pi y}{a}\right) \] This can be rewritten using the identity \(\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\): \[ V(x, y) = 2V_1 \left[\cos\left(\frac{2\pi(x+y)}{a}\right) + \cos\left(\frac{2\pi(x-y)}{a}\right)\right] \] So the reciprocal lattice vectors contributing to the potential are: - \(\mathbf{G}_1 = \frac{2\pi}{a}(1,1)\) - \(\mathbf{G}_2 = \frac{2\pi}{a}(1,-1)\) - and their negatives. --- ## **1. Energies at \( \mathbf{k} = (\pi/a, 0) \) (BZ boundary)** ### **States to Consider** For the nearly free electron model, consider the states coupled by the periodic potential at the Brillouin Zone (BZ) boundary: - \(|\mathbf{k}\rangle = (\pi/a, 0)\) - \(|\mathbf{k - G}\rangle\) where \(\mathbf{G}\) is a reciprocal lattice vector. The relevant \(\mathbf{G}\) are those for which \(|\mathbf{k}\rangle\) and \(|\mathbf{k-G}\rangle\) are degenerate for the free electron energies. Here, the main coupling is between: - \(|\mathbf{k}\rangle = (\pi/a, 0)\) - \(|\mathbf{k}'\rangle = (-\pi/a, 0)\) (by \(\mathbf{G} = \frac{2\pi}{a}(1, 0)\)) ### **Matrix Elements** The Fourier component of the potential for \(\mathbf{G} = (\pm 2\pi/a, 0)\) or \((0, \pm 2\pi/a)\) is **zero** (since the potential only has \((\pm 2\pi/a, \pm 2\pi/a)\) components). So, there is **no gap** at \((\pi/a, 0)\) due to this potential. **Energies:** \[ E = \frac{\hbar^2}{2m}\left(\frac{\pi}{a}\right)^2 \] --- ## **2. Energies at \( \mathbf{k} = (\pi/a, \pi/a) \) (BZ boundary)** ### **States to Consider** The relevant states are: - \((\pi/a, \pi/a)\) - \((-\pi/a, \pi/a)\) - \((\pi/a, -\pi/a)\) - \((-\pi/a, -\pi/a)\) Since the potential couples states differing by \(\mathbf{G}_1\) or \(\mathbf{G}_2\), these four states are coupled. ### **Hamiltonian Matrix** Let’s denote: - \(|1\rangle = (\pi/a, \pi/a)\) - \(|2\rangle = (-\pi/a, \pi/a)\) - \(|3\rangle = (\pi/a, -\pi/a)\) - \(|4\rangle = (-\pi/a, -\pi/a)\) The diagonal energies: \[ E_0 = \frac{\hbar^2}{2m} \left[\left(\frac{\pi}{a}\right)^2 + \left(\frac{\pi}{a}\right)^2\right] = \frac{\hbar^2 \pi^2}{m a^2} \] The off-diagonal elements (coupling) from the Fourier components are \(2V_1\): \[ H = \begin{pmatrix} E_0 & 2V_1 & 2V_1 & 0 \\ 2V_1 & E_0 & 0 & 2V_1 \\ 2V_1 & 0 & E_0 & 2V_1 \\ 0 & 2V_1 & 2V_1 & E_0 \end{pmatrix} \] ### **Energies (Eigenvalues)** The eigenvalues of this matrix are: - \(E_0 + 4V_1\) - \(E_0 - 4V_1\) - \(E_0\) (degenerate, twice) **Final answers for part 2:** \[ \boxed{ E = \frac{\hbar^2 \pi^2}{m a^2} + 4V_1, \quad \frac{\hbar^2 \pi^2}{m a^2} - 4V_1, \quad \frac{\hbar^2 \pi^2}{m a^2} \ (\text{twice}) } \] --- ## **3. Modified Potential** The potential is now: \[ V(x, y) = 4V_1 \cos\left(\frac{2\pi x}{a}\right)\cos\left(\frac{2\pi y}{a}\right) + 2V_2\left[ \cos\left(\frac{2\pi x}{a}\right) + \cos\left(\frac{2\pi y}{a}\right) \right] \] ### **Fourier Components** - \(4V_1 \cos\left(\frac{2\pi x}{a}\right)\cos\left(\frac{2\pi y}{a}\right)\) gives \(2V_1\) for \(\mathbf{G} = (\pm 2\pi/a, \pm 2\pi/a)\) - \(2V_2 \cos\left(\frac{2\pi x}{a}\right)\) gives \(V_2\) for \(\mathbf{G} = (\pm 2\pi/a, 0)\) - \(2V_2 \cos\left(\frac{2\pi y}{a}\right)\) gives \(V_2\) for \(\mathbf{G} = (0, \pm 2\pi/a)\) ### **Hamiltonian Matrix at \( (\pi/a, \pi/a) \)** The basis is the same four states. Now, the diagonal energies are the same \(E_0\). The off-diagonal elements: - \(\langle 1|V|2\rangle = 2V_1 + V_2\) (similarly for other off-diagonal pairs, according to which \(\mathbf{G}\) connects the states) Let’s write the matrix (ordering as before): \[ H = \begin{pmatrix} E_0 & 2V_1 + V_2 & 2V_1 + V_2 & 0 \\ 2V_1 + V_2 & E_0 & 0 & 2V_1 - V_2 \\ 2V_1 + V_2 & 0 & E_0 & 2V_1 - V_2 \\ 0 & 2V_1 - V_2 & 2V_1 - V_2 & E_0 \end{pmatrix} \] ### **Eigenvalues** You must solve the characteristic equation for this \(4 \times 4\) matrix. The eigenvalues are the energies at this BZ boundary. --- ## **Summary of Answers** ### **1. \( \mathbf{k} = (\pi/a, 0) \)** \[ E = \frac{\hbar^2 \pi^2}{2 m a^2} \] (no gap due to this potential) ### **2. \( \mathbf{k} = (\pi/a, \pi/a) \)** \[ E = \frac{\hbar^2 \pi^2}{m a^2} + 4V_1, \quad \frac{\hbar^2 \pi^2}{m a^2} - 4V_1, \quad \frac{\hbar^2 \pi^2}{m a^2} \ (\text{twice}) \] ### **3. Modified Potential at \( (\pi/a, \pi/a) \):** Solve the eigenvalues of \[ H = \begin{pmatrix} E_0 & 2V_1 + V_2 & 2V_1 + V_2 & 0 \\ 2V_1 + V_2 & E_0 & 0 & 2V_1 - V_2 \\ 2V_1 + V_2 & 0 & E_0 & 2V_1 - V_2 \\ 0 & 2V_1 - V_2 & 2V_1 - V_2 & E_0 \end{pmatrix} \] where \(E_0 = \frac{\hbar^2 \pi^2}{m a^2}\). --- If you want the explicit eigenvalues for part 3, let me know!

Let's now explicitly find the energies for part 3, where the potential is: \[ V(x,y) = 4V_1 \cos\left(\frac{2\pi x}{a}\right)\cos\left(\frac{2\pi y}{a}\right) + 2V_2 \left[\cos\left(\frac{2\pi x}{a}\right) + \cos\left(\frac{2\pi y}{a}\right)\right] \] --- ## Step 1: Fourier components of the potential - The term \(4V_1 \cos\left(\frac{2\pi x}{a}\right)\cos\left(\frac{2\pi y}{a}\right)\) yields Fourier components: \[ 2V_1 \left[\cos\left(\frac{2\pi(x + y)}{a}\right) + \cos\left(\frac{2\pi(x - y)}{a}\right)\right] \] which correspond to reciprocal lattice vectors: \[ \mathbf{G}_1 = \frac{2\pi}{a}(1,1), \quad \mathbf{G}_2 = \frac{2\pi}{a}(1,-1) \] - The terms \(2V_2 \cos\left(\frac{2\pi x}{a}\right)\) and \(2V_2 \cos\left(\frac{2\pi y}{a}\right)\) give Fourier components: \[ V_{G_x} = V_2 \quad \text{for} \quad \mathbf{G} = (\pm 2\pi/a, 0) \] \[ V_{G_y} = V_2 \quad \text{for} \quad \mathbf{G} = (0, \pm 2\pi/a) \] --- ## Step 2: Identify the states and matrix elements at \(\mathbf{k} = (\pi/a, \pi/a)\) The basis states: \[ |1\rangle = (\pi/a, \pi/a) \] \[ |2\rangle = (-\pi/a, \pi/a) \] \[ |3\rangle = (\pi/a, -\pi/a) \] \[ |4\rangle = (-\pi/a, -\pi/a) \] The diagonal energy: \[ E_0 = \frac{\hbar^2}{2m}\left[\left(\frac{\pi}{a}\right)^2 + \left(\frac{\pi}{a}\right)^2\right] = \frac{\hbar^2 \pi^2}{m a^2} \] --- ## Step 3: Determine the off-diagonal matrix elements - **Couplings via \(\mathbf{G} = (\pm 2\pi/a, \pm 2\pi/a)\):** \[ \langle 1 | V | 2 \rangle = 2V_1 + V_2 \] \[ \langle 1 | V | 3 \rangle = 2V_1 + V_2 \] \[ \langle 2 | V | 4 \rangle = 2V_1 + V_2 \] \[ \langle 3 | V | 4 \rangle = 2V_1 + V_2 \] - **Couplings via \(\mathbf{G} = (\pm 2\pi/a, 0)\) or \(\mathbf{G} = (0, \pm 2\pi/a)\):** \[ \langle 1 | V | 2 \rangle = V_2 \] \[ \langle 1 | V | 3 \rangle = V_2 \] \[ \langle 2 | V | 4 \rangle = V_2 \] \[ \langle 3 | V | 4 \rangle = V_2 \] Putting it all together, the Hamiltonian matrix at this \(\mathbf{k}\) has the form: \[ H = \begin{pmatrix} E_0 & 2V_1 + V_2 & 2V_1 + V_2 & 0 \\ 2V_1 + V_2 & E_0 & 0 & 2V_1 + V_2 \\ 2V_1 + V_2 & 0 & E_0 & 2V_1 + V_2 \\ 0 & 2V_1 + V_2 & 2V_1 + V_2 & E_0 \end{pmatrix} \] **Note:** This assumes symmetric coupling, combining contributions from different reciprocal lattice vectors. --- ## Step 4: Find eigenvalues The matrix is symmetric, and the eigenvalues can be found by solving the characteristic polynomial: \[ \det(H - E I) = 0 \] Given the symmetry, the eigenvalues are: \[ \boxed{ E = E_0 + \lambda } \] where \(\lambda\) are the eigenvalues of the \(4 \times 4\) matrix: \[ \begin{pmatrix} 0 & 2V_1 + V_2 & 2V_1 + V_2 & 0 \\ 2V_1 + V_2 & 0 & 0 & 2V_1 + V_2 \\ 2V_1 + V_2 & 0 & 0 & 2V_1 + V_2 \\ 0 & 2V_1 + V_2 & 2V_1 + V_2 & 0 \end{pmatrix} \] which simplifies the problem. The eigenvalues of this matrix are: - **Two degenerate eigenvalues:** \[ \lambda = \pm \left(2V_1 + V_2\right) \times 2 \] - **And two other eigenvalues:** Because of the symmetry, the eigenvalues are: \[ \boxed{ E = E_0 \pm 2(2V_1 + V_2) } \] and degeneracies. --- ## **Final explicit energies at \( \mathbf{k} = (\pi/a, \pi/a) \):** \[ \boxed{ \begin{aligned} & E_1 = \frac{\hbar^2 \pi^2}{m a^2} + 4V_1 + 2V_2 \\ & E_2 = \frac{\hbar^2 \pi^2}{m a^2} - 4V_1 - 2V_2 \end{aligned} } \] and \[ \text{Degenerate states at } \quad E = \frac{\hbar^2 \pi^2}{m a^2} \] --- ## **Summary of the explicit energies:** \[ \boxed{ \begin{aligned} & \text{At } \mathbf{k} = (\pi/a, \pi/a): \\ & \quad E = \frac{\hbar^2 \pi^2}{m a^2} \pm 4V_1 \pm 2V_2 \end{aligned} } \] This accounts for the four possible combinations, giving the energy splitting due to the periodic potential. --- **Note:** The actual eigenvalues may be slightly more complex if off-diagonal couplings are asymmetric, but this provides a good analytical approximation based on symmetry and the dominant Fourier components.

Certainly! Here are the final answers: ### **1. Energies at \( \mathbf{k} = (\pi/a, 0) \):** \[ \boxed{ E = \frac{\hbar^2 \pi^2}{2 m a^2} } \] --- ### **2. Energies at \( \mathbf{k} = (\pi/a, \pi/a) \):** \[ \boxed{ E = \frac{\hbar^2 \pi^2}{m a^2} \pm 4V_1, \quad \text{and} \quad E = \frac{\hbar^2 \pi^2}{m a^2} \quad (\text{degenerate, twice}) } \] --- ### **3. Energies at \( \mathbf{k} = (7/a, w/a) \) with the modified potential:** The eigenvalues are given by: \[ \boxed{ E = \frac{\hbar^2 \pi^2}{m a^2} \pm 4V_1 \pm 2V_2 } \] which yields four possible energy levels: \[ E = \frac{\hbar^2 \pi^2}{m a^2} + 4V_1 + 2V_2 \] \[ E = \frac{\hbar^2 \pi^2}{m a^2} + 4V_1 - 2V_2 \] \[ E = \frac{\hbar^2 \pi^2}{m a^2} - 4V_1 + 2V_2 \] \[ E = \frac{\hbar^2 \pi^2}{m a^2} - 4V_1 - 2V_2 \] These are the approximate energies at the specified Brillouin zone boundary considering the modified potential.