1. **Problem Statement:** We want to derive Eq. (14.19) for the conductivity tensor $ \sigma_{\alpha \gamma}(H) $ in the presence of a magnetic field, starting from Eq. (14.17) and using the rotation operator form in Eq. (14.18). 2. **Starting Point:** Eq. (14.17) expresses the conductivity as $$\sigma_{\alpha \gamma}(H) = e^{2} V^{-1} \sum_{k} v_{k \alpha} f^{1}_{\epsilon}(k) \sum^{\infty}_{n=0} [-\tau(\epsilon_{k})]^{n+1} \omega_{c}^n \Omega^{n}_{k \gamma}.$$ 3. **Rotation Operator and Its Powers:** From Eq. (14.18), the rotation operator is $$\Omega_{k} = v_{k1} \frac{\partial}{\partial v_{k2}} - v_{k2} \frac{\partial}{\partial v_{k1}}.$$ The powers of $ \Omega_k $ act on velocity components $ v_{k \gamma} $ as given: $$\Omega^{2l+2}_k v_{k \gamma} = -(-)^l v_{k \gamma} (\delta_{\gamma 1} + \delta_{\gamma 2}),$$ $$\Omega^{2l+1}_k v_{k \gamma} = (-)^l (v_{k1} \delta_{\gamma 2} - v_{k2} \delta_{\gamma 1}),$$ valid for $ l \geq 0 $. 4. **Separating the $ n=0 $ Term and Reindexing:** Isolate the $ n=0 $ term in the sum and rewrite the sums over $ n $ as sums over $ l $ with $ n=2l+2 $ and $ n=2l+1 $ for even and odd powers respectively. 5. **Substitute the Powers of $ \Omega_k $ into Eq. (14.17):** We get \[ \sigma_{\alpha \gamma}(H) = -e^{2} V^{-1} \sum_k v_{k \alpha} \sigma_f(\epsilon_k) \tau(\epsilon_k) v_{k \gamma} - \sum_{l=0}^\infty \tau^{2l+2}(\epsilon_k) \omega_c^{2l+2} (-)^l v_{k \gamma} (\delta_{\gamma 1} + \delta_{\gamma 2}) - \sum_{l=0}^\infty \tau^{2l+1}(\epsilon_k) \omega_c^{2l+1} (-)^l (v_{k1} \delta_{\gamma 2} - v_{k2} \delta_{\gamma 1}). \] 6. **Summing the Geometric Series:** Recognize the sums over $ l $ as geometric series in $ \tau^2(\epsilon_k) \omega_c^2 $ and $ \tau(\epsilon_k) \omega_c $. Summing these series yields expressions involving $$\frac{\tau^2(\mu) \omega_c^2}{1 + \tau^2(\mu) \omega_c^2} \quad \text{and} \quad \frac{\tau(\mu) \omega_c}{1 + \tau^2(\mu) \omega_c^2}.$$ 7. **Evaluating the $ k $-Sum at Zero Temperature:** At $ T \to 0 $, use $$-f'_0(\epsilon) = \delta(\epsilon - \mu),$$ and the angular average $$\langle v_{k \alpha} v_{k \gamma} \rangle = \frac{v_F^2}{3} \delta_{\alpha \gamma}.$$ This simplifies the $ k $-sum to $$-e^{2} V^{-1} \sum_k v_{k \alpha} v_{k \gamma} \tau^{n+1}(\epsilon_k) f'_0(\epsilon_k) = \sigma_0 \tau(\mu) \delta_{\alpha \gamma}.$$ 8. **Putting All Pieces Together:** The final expression for the conductivity tensor is $$\sigma_{\alpha \gamma}(H) = \sigma_0 \left[ 1 - \frac{\tau^2(\mu) \omega_c^2}{1 + \tau^2(\mu) \omega_c^2} (\delta_{\alpha 1} + \delta_{\gamma 2}) \right] \delta_{\alpha \gamma} - \sigma_0 \frac{\tau(\mu) \omega_c}{1 + \tau^2(\mu) \omega_c^2} (\delta_{\alpha 1} \delta_{\gamma 2} - \delta_{\alpha 2} \delta_{\gamma 1}),$$ which is Eq. (14.19). This shows how the magnetic field modifies the conductivity tensor through the cyclotron frequency $ \omega_c $ and scattering time $ \tau(\mu) $.