这是一份bath巴斯大学PH20017/PH20063作业代写的成功案

We consider the case when the transport relaxation time defined as
$$
\frac{1}{\tau(E)} \equiv 2 \pi n_{\mathrm{im}} \sum_{\boldsymbol{k}^{\prime}}(1-\cos \Theta) V_{\mathrm{sc}}^{2}\left(\boldsymbol{k}^{\prime}-k\right) \delta\left(E_{\boldsymbol{k}^{\prime}}-E_{k}\right)
$$
depends only on energy $E$. Here $\Theta$ is the angle between $v^{\prime}=\partial E_{\boldsymbol{h}^{\prime}} / \partial \boldsymbol{k}^{\prime}$ and $v$. Then keeping the terms linear in the electric field $E$, the Boltzmann equation for the function $F$ becomes
$$
F \cdot v=-e \tau(E)\left[E \cdot v-F \cdot\left(v \times B \cdot \nabla_{k}\right) v\right]
$$
If the magnetic field is sufficiently weak ( $\omega \tau \ll 1 ; \omega \propto B$ is the Larmour frequency), one can keep only the terms linear in $B$ with the following result for the non-equilibrium part of the distribution function:
$$
F \cdot v=-e \tau(E)\left[E \cdot v+e \tau(E) E \cdot\left(v \times B \cdot \nabla_{k}\right) v\right]
$$
Using the current density
$$
\boldsymbol{j}=-2 e \sum_{k} v f(\boldsymbol{r}, \boldsymbol{k}, t)=2 e \sum_{k} v \frac{\partial n_{k}}{\partial E} v \cdot \boldsymbol{F}
$$
one obtains the longitudinal conductivity $\sigma_{x x}$,
$$
\sigma_{x x}=-2 e^{2} \sum_{k} \frac{\partial n_{k}}{\partial E} \tau(E) v_{x}^{2}
$$

PPH20017/PH20063 COURSE NOTES ：

$$
A(\boldsymbol{r})=-\frac{\nabla \Phi}{e^{}} $$ Then the magnetic flux becomes $$ \Phi_{\mathrm{B}}=\oint_{C} \mathrm{~d} \boldsymbol{l} \cdot \boldsymbol{A}(\boldsymbol{r})=\frac{\delta \Phi}{e^{}}
$$
Here $\delta \Phi$ is a change of the phase in the round trip along the contour. The wavefunction is single-valued if $\delta \Phi=2 \pi p$ where $p=0,1,2, \ldots .$ Hence, the flux is quantized ( $\Phi_{\mathrm{B}}=p \Phi_{0}$ ) and the flux quantum (in ordinary units)
$$
\Phi_{0}=\frac{\pi \hbar c}{e}=2.07 \times 10^{-7} \mathrm{G} \mathrm{cm}^{2}
$$