Let's solve both questions step by step: --- ### **Q1. Finding Reciprocal Lattice Vectors \( \mathbf{b}_1 \) and \( \mathbf{b}_2 \) for a Hexagonal Lattice** **Given:** \[ \mathbf{a}_1 = \frac{a}{2}(3, \sqrt{3}), \quad \mathbf{a}_2 = \frac{a}{2}(3, -\sqrt{3}) \] **Step 1: Write the formula for reciprocal lattice vectors.** In 2D, the reciprocal lattice vectors are: \[ \mathbf{b}_1 = 2\pi \frac{\hat{z} \times \mathbf{a}_2}{\mathbf{a}_1 \times \mathbf{a}_2} \] \[ \mathbf{b}_2 = 2\pi \frac{\hat{z} \times \mathbf{a}_1}{\mathbf{a}_2 \times \mathbf{a}_1} \] The cross product \( \mathbf{a}_1 \times \mathbf{a}_2 \) gives the area of the unit cell (since these are 2D vectors, this is the magnitude in the z-direction). **Step 2: Compute \( \mathbf{a}_1 \times \mathbf{a}_2 \) (area of unit cell).** \[ \mathbf{a}_1 = \frac{a}{2}(3, \sqrt{3}), \quad \mathbf{a}_2 = \frac{a}{2}(3, -\sqrt{3}) \] \[ \mathbf{a}_1 \times \mathbf{a}_2 = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 3a/2 & \sqrt{3}a/2 & 0 \\ 3a/2 & -\sqrt{3}a/2 & 0 \end{vmatrix} \] Only the \( \hat{z} \) component is nonzero: \[ = (3a/2) \cdot (-\sqrt{3}a/2) - (3a/2) \cdot (\sqrt{3}a/2) = -\frac{3a^2}{4}\sqrt{3} - \frac{3a^2}{4}\sqrt{3} = -\frac{3a^2}{2}\sqrt{3} \] The magnitude is: \[ |\mathbf{a}_1 \times \mathbf{a}_2| = \frac{3a^2}{2}\sqrt{3} \] **Step 3: Compute \( \hat{z} \times \mathbf{a}_1 \) and \( \hat{z} \times \mathbf{a}_2 \):** For any vector \( (x, y) \): \[ \hat{z} \times (x, y) = (-y, x) \] So, \[ \hat{z} \times \mathbf{a}_1 = \hat{z} \times \frac{a}{2}(3, \sqrt{3}) = \frac{a}{2}(-\sqrt{3}, 3) \] \[ \hat{z} \times \mathbf{a}_2 = \hat{z} \times \frac{a}{2}(3, -\sqrt{3}) = \frac{a}{2}(\sqrt{3}, 3) \] **Step 4: Write down the reciprocal vectors:** \[ \mathbf{b}_1 = 2\pi \frac{\hat{z} \times \mathbf{a}_2}{|\mathbf{a}_1 \times \mathbf{a}_2|} = 2\pi \frac{ \frac{a}{2} (\sqrt{3}, 3) }{ \frac{3a^2}{2}\sqrt{3} } = 2\pi \frac{ (\sqrt{3}, 3) }{ 3a\sqrt{3} } = \frac{2\pi}{3a} (1, \sqrt{3}) \] \[ \mathbf{b}_2 = 2\pi \frac{\hat{z} \times \mathbf{a}_1}{|\mathbf{a}_1 \times \mathbf{a}_2|} = 2\pi \frac{ \frac{a}{2} (-\sqrt{3}, 3) }{ \frac{3a^2}{2}\sqrt{3} } = 2\pi \frac{ (-\sqrt{3}, 3) }{ 3a\sqrt{3} } = \frac{2\pi}{3a} (-1, \sqrt{3}) \] --- **Final Answers for Q1:** \[ \boxed{ \mathbf{b}_1 = \frac{2\pi}{3a} (1, \sqrt{3}), \quad \mathbf{b}_2 = \frac{2\pi}{3a} (-1, \sqrt{3}) } \] --- ### **Q2. Graphene's Dirac Hamiltonian: Wavefunction Derivation** **Given:** \[ H = \hbar v_F (\tau_z \sigma_x q_x + \sigma_y q_y) \] Here, \( \tau_z = \pm 1 \) (valley index), \( \sigma_x \) and \( \sigma_y \) are Pauli matrices, \( q_x, q_y \) are momenta. #### **Step 1: Write the Hamiltonian in matrix form** For a given valley (\( \tau_z = +1 \)), the Hamiltonian becomes: \[ H = \hbar v_F (\sigma_x q_x + \sigma_y q_y) = \hbar v_F \begin{pmatrix} 0 & q_x - iq_y \\ q_x + iq_y & 0 \end{pmatrix} \] #### **Step 2: Find the eigenvalues** Set up the eigenvalue equation: \[ H \Psi = E \Psi \] \[ \Rightarrow \hbar v_F \begin{pmatrix} 0 & q_x - iq_y \\ q_x + iq_y & 0 \end{pmatrix} \begin{pmatrix} \psi_A \\ \psi_B \end{pmatrix} = E \begin{pmatrix} \psi_A \\ \psi_B \end{pmatrix} \] This gives two equations: \[ \hbar v_F (q_x - iq_y) \psi_B = E \psi_A \\ \hbar v_F (q_x + iq_y) \psi_A = E \psi_B \] #### **Step 3: Solve for eigenvalues** Combine the two equations: \[ E^2 \psi_A = \hbar v_F (q_x - iq_y) \hbar v_F (q_x + iq_y) \psi_A = (\hbar v_F)^2 (q_x^2 + q_y^2) \psi_A \] So: \[ E^2 = (\hbar v_F)^2 (q_x^2 + q_y^2) \] \[ E = \pm \hbar v_F q, \quad \text{where } q = \sqrt{q_x^2 + q_y^2} \] #### **Step 4: Find the eigenvector (wavefunction)** From above, for \( E \neq 0 \): \[ \hbar v_F (q_x - iq_y) \psi_B = E \psi_A \implies \psi_B = \frac{E}{\hbar v_F (q_x - i q_y)} \psi_A \] Choose \( \psi_A = 1 \) (arbitrary normalization), for \( E = +\hbar v_F q \): \[ \psi_B = \frac{\hbar v_F q}{\hbar v_F (q_x - i q_y)} = \frac{q}{q_x - i q_y} \] Alternatively, write \( q_x + iq_y = q e^{i\theta} \), so \( q_x - iq_y = q e^{-i\theta} \): So, \[ \psi_B = \frac{q}{q e^{-i\theta}} = e^{i\theta} \] So, the normalized eigenvector (wavefunction) is: \[ \Psi_+ = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ e^{i\theta} \end{pmatrix} \] where \( \theta = \arctan\left(\frac{q_y}{q_x}\right) \). For \( E = -\hbar v_F q \): \[ \psi_B = -e^{i\theta} \] So, \[ \Psi_- = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -e^{i\theta} \end{pmatrix} \] --- **Final Answers for Q2:** \[ \boxed{ \Psi_+ = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ e^{i\theta} \end{pmatrix}, \quad \Psi_- = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ - e^{i\theta} \end{pmatrix}, \quad \theta = \arctan \left( \frac{q_y}{q_x} \right) } \] --- **If you need further clarification on any step, please let me know!**

Q1: \(\boxed{ \mathbf{b}_1 = \frac{2\pi}{3a} (1, \sqrt{3}), \quad \mathbf{b}_2 = \frac{2\pi}{3a} (-1, \sqrt{3}) }\) Q2: \(\boxed{ \text{Wavefunctions:} \\ \Psi_+ = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ e^{i\theta} \end{pmatrix}, \quad \Psi_- = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -e^{i\theta} \end{pmatrix}, \quad \theta = \arctan\left(\frac{q_y}{q_x}\right) }\)